Solution Manual For Introduction to Renewable Energy Conversions 2nd Edition By Sergio Capareda

Solutions Manual for Introduction to Renewable Energy Conversions 2nd Edition

Sergio C. Capareda CRC Press, Taylor and Francis Group Boca Raton, Florida, USA

September 2025

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Chapter 1 Introduction to Renewable Energy 1.7 Solved Problems Problem 1.7.1 Carbon Dioxide Required to Make Carbohydrates Problem 1.7.1 Determine the amount of carbohydrate or glucose produced (in tonnes) for every tonne of carbon dioxide used by way of Equation 1.2.

Solution: a. Equation 1.2 may be converted into units of weight using the molecular weights of the compounds involved. The molecular weights of carbon is 12 g/mole and that for hydrogen is 1 g/mol and oxygen is 16 g/mol. b. 6𝐶𝐶𝑂𝑂2 + 6𝐻𝐻2 𝑂𝑂 + 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 → 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 + 6𝑂𝑂2

c. Thus, the amount of glucose (C6H12O6) produced per tonne of CO2 consumed is calculated as follows: 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 1 × [(6 × 12) + (12 × 1) + (6 × 16)] 180 𝑔𝑔 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 �= =� = 0.75 6𝐶𝐶𝑂𝑂2 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝐶𝐶𝑂𝑂2 364 𝑔𝑔 6 × �12 + 2 × (16)�

d. Theoretically, it would require about one-and-a-half tonnes of CO2 to generate a tonne of carbohydrate (glucose) sugar.

Problem 1.7.2. Kinetic Energy of a Mass of Wind Problem 1.7.2 Determine the kinetic energy available from a 100 kg of air moving at a speed of 12 m/s. Report the answer in units of Joules and in Btu. Solution a. Using Equation 1.1, the given values are simply plugged in the equation as follows: b. 𝐾𝐾𝐾𝐾 =

1 2

× 100 𝑘𝑘𝑘𝑘 × �

12 𝑚𝑚 2 𝑠𝑠

𝑁𝑁×𝑠𝑠 2

� × 𝑘𝑘𝑘𝑘×𝑚𝑚 = 7,200 𝑁𝑁 − 𝑚𝑚 = 7.2 𝑘𝑘𝑘𝑘 2

c. Thus, the equivalent unit in English system is shown below. 𝐾𝐾𝐾𝐾 = 7,200 𝐽𝐽 ×

𝐵𝐵𝐵𝐵𝐵𝐵 = 6.82 𝐵𝐵𝐵𝐵𝐵𝐵 1055 𝐽𝐽

d. Note the difference in the order of magnitude of Joules and Btu. One should always remember the conversion of Joules to Btu using the relationship 1055 Joules/Btu.

Problem 1.7.3 Carbon Dioxide Production During Ethanol Fermentation Problem 1.7.3 Determine the amount of carbon dioxide produced (kgs) for every kg of glucose (𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 ) consumed in its conversion into ethanol. Use Equation 1.4 shown below in your calculations. Convert these units into English units of lbs CO2 per lb glucose. 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 + 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 → 2𝐶𝐶2 𝐻𝐻5 𝑂𝑂𝑂𝑂 + 2𝐶𝐶𝑂𝑂2 + ℎ𝑒𝑒𝑒𝑒𝑒𝑒

Solution:

a. The amount of carbon dioxide produced per weight of glucose is given as follows: 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 2𝐶𝐶𝑂𝑂2 2 × [(1 × 12) + (2 × 16)] 88 𝑔𝑔 �= = =� [(6 × 12) + (12 × 1) + (6 × 16)] 𝑘𝑘𝑘𝑘 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 180 𝑔𝑔 = 0.49

𝑘𝑘𝑘𝑘 𝐶𝐶𝑂𝑂2 𝑘𝑘𝑘𝑘 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6

b. The amount of CO2 produced per lb of glucose consumed should also be equal to 0.49 kg/kg or lb/lb.

Problem 1.7.4. Theoretical and Actual Power from Water Stream Problem 1.7.4 Determine the theoretical and actual power that can be derived from a water stream with a volumetric flow rate of 55 L/s and with a dynamic head of 2.0 m. Assume an overall conversion efficiency of 45%.

Solution:

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a. Equation 1.5 will be used and the density of water value at 1 kg/L and other conversion factors to arrive at the correct units of power in Watts. This will require the use of gravitational constant 9.8 m/s2 as well as the Newton’s Law relationship as shown below. 𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊 ) =

55 𝐿𝐿 1 𝑘𝑘𝑘𝑘 2 𝑚𝑚 𝑁𝑁𝑠𝑠 2 9.8 𝑚𝑚 𝑊𝑊𝑊𝑊 × × × × 2 × = 1,078 𝑊𝑊 𝑠𝑠 𝐿𝐿 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑠𝑠 𝑁𝑁𝑁𝑁

b. The actual power that could be derived will be less due to conversion efficiencies as shown below. 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊 ) = 𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 × 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 = 1,078 𝑊𝑊 × 0.45 = 485 𝑊𝑊

Problem 1.7.5 Theoretical Thermal Conversion Efficiency of Rankine Cycle Problem 1.7.5 Determine the theoretical thermal conversion efficiency of a Rankine Cycle with the following data: (a) Wout = 650 kJ/kg, (b) Win = 3.5, and (c) the heat input from the steam, Qin = 2,500 kJ/kg. The mass flow rate of the steam is around 22,000 kg/hr.

Solution: a. It will be intuitive to convert all values in energy units rather than energy per unit of weight, even though the energy per unit of weight may also be used for efficiency calculations. Thus, the work output may be converted into kW as follows: 𝑊𝑊𝑜𝑜𝑜𝑜𝑜𝑜 = 650

𝑘𝑘𝑘𝑘 22,000 𝑘𝑘𝑘𝑘 1 ℎ𝑟𝑟 𝑘𝑘𝑘𝑘 × × × = 3,972 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 3600 𝑠𝑠 𝑘𝑘𝑘𝑘/𝑠𝑠

b. Likewise, Win and Qin will be converted the same way as follows: 𝑊𝑊𝑖𝑖𝑖𝑖 = 3.5

𝑘𝑘𝑘𝑘 22,000 𝑘𝑘𝑘𝑘 1 ℎ𝑟𝑟 𝑘𝑘𝑘𝑘 × × × = 21.39 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 3600 𝑠𝑠 𝑘𝑘𝑘𝑘/𝑠𝑠

𝑄𝑄𝑖𝑖𝑖𝑖 = 2,500

𝑘𝑘𝑘𝑘 22,000 𝑘𝑘𝑘𝑘 1 ℎ𝑟𝑟 𝑘𝑘𝑘𝑘 × × × = 15,278 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 3600 𝑠𝑠 𝑘𝑘𝑘𝑘/𝑠𝑠 4

c. Thus, the efficiency may be calculated as follows: 𝜂𝜂𝑡𝑡ℎ =

𝑊𝑊𝑜𝑜𝑜𝑜𝑜𝑜 − 𝑊𝑊𝑖𝑖𝑖𝑖 3,972 𝑘𝑘𝑘𝑘 − 21.39 𝑘𝑘𝑘𝑘 × 100% = × 100% = 25.86% 𝑄𝑄𝑖𝑖𝑖𝑖 15,278 𝑘𝑘𝑘𝑘

Problem 1.7.6. Fuel Cell Efficiencies Problem 1.7.6 Calculate the actual electrical energy efficiency of converting hydrogen and oxygen into electrical energy via fuel cell technology using the following data: (a) the electricity produced was 180 kJ/mol and (b) the heating value of hydrogen was 285.8 kJ/mol. Compare this efficiency with the practical electricity production from hydrogen fuel cell having an electrical output of only 130 kJ/mol.

Solution: a. Electrical energy efficiency (EEE%) is simply the ratio of output over energy input as follows: 𝐸𝐸𝐸𝐸𝐸𝐸 (%) =

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 180 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚 × 100% = × 100% = 63% 𝐻𝐻𝐻𝐻𝐻𝐻 𝑜𝑜𝑜𝑜 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 285.8 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚

b. Thus, the efficiency is still higher than when hydrogen is combusted in an internal combustion engine with overall combustion efficiencies of around 30% to 35%. c. The practical efficiency is calculated as follows: 130 kJ / mol Practical EEE (% ) = × 100% = 45.5% 285.8 kJ / mol 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐸𝐸𝐸𝐸𝐸𝐸 (%) =

130 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚 × 100% = 45.5% 285.8 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚

d. Practical efficiency is likewise higher than internal combustion engines.

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Problem 1.7.7. Tidal Power Calculations Problem 1.7.7 Residents of a household living near the shore would like to build a reservoir that could generate tidal power for their needs. The yearly average tide is about 1.5 meters. The owner built a reservoir that can contain 8,000 cubic meters of sea water and they plan to drop about 35 liters/sec of this water when the tide is lowest with an average head of 1 meter for about 12 hours. Estimate the amount of actual power generated during that period if the overall conversion efficiency is 60%. Assume the density of seawater to be about 1.03 kg/L. How much energy was produced during this period?

Solution: a. Equation 1.5 will be used and the density of sea water value at 1.03 kg/L and other conversion factors to arrive at the correct units of power in Watts. This will require the use of gravitational constant 9.8 m/s2 as well as the Newton’s Law relationship as shown below. 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊 ) =

35 𝐿𝐿 1.03 𝑘𝑘𝑘𝑘 1 𝑚𝑚 𝑁𝑁𝑠𝑠 2 9.8 𝑚𝑚 𝑊𝑊𝑊𝑊 × × × × 2 × × 0.60 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑠𝑠 𝐿𝐿 𝑠𝑠 𝑁𝑁𝑁𝑁

= 212 𝑊𝑊

b. The energy produced is calculated as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑘𝑘𝑘𝑘ℎ) = 212 𝑊𝑊 × 12 ℎ𝑟𝑟𝑟𝑟 ×

1 𝑘𝑘𝑘𝑘 = 2.54 𝑘𝑘𝑘𝑘ℎ 1,000 𝑊𝑊

Problem 1.7.8 Solar Water Heater Conversion Efficiency Problem 1.7.8 Determine the efficiency of converting solar energy received on a solar water heater if the average incoming solar radiation for an hour was measured to be 900 W/m2. The

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water was heated from 25oC to 75oC for this period with an average mass flow rate of 4 kg/s on a 1m2 collector area.

Solution: a. From the basic heat energy equation 𝑄𝑄 = 𝑚𝑚𝐶𝐶𝑝𝑝∆𝑡𝑡 , we have the following calculations with given constants (Cp=4.190 MJ/kgoC). 𝑄𝑄 = 𝑚𝑚𝐶𝐶𝑝𝑝∆𝑡𝑡 = 4

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 × 4,190 𝑜𝑜 × (75𝑜𝑜 𝐶𝐶 − 25𝑜𝑜 𝐶𝐶 ) = 838 𝑘𝑘𝑘𝑘 𝑠𝑠 𝑘𝑘𝑔𝑔 𝐶𝐶

b. The input energy may be calculated as follows and converted into the same unit as the output energy (in Watts): 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑘𝑘𝑘𝑘) =

900 𝑊𝑊 𝐽𝐽 3600 𝑠𝑠 𝑘𝑘𝑘𝑘 × 1 𝑚𝑚2 × × × 1 ℎ𝑟𝑟 × = 3,240 𝑘𝑘𝑘𝑘 2 𝑚𝑚 𝑊𝑊𝑊𝑊 1 ℎ𝑟𝑟 1,000 𝐽𝐽

c. Thus, the conversion efficiency is calculated as follows: 𝜂𝜂 (%) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 838 𝑘𝑘𝑘𝑘 × 100% = × 100% = 25.9% 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 3,240 𝑘𝑘𝑘𝑘

d. Only 25.9% of the incoming solar radiation was effective used to bring the water temperature from 25oC to 75oC at the given mass flow rate.

Problem 1.7.9. OTEC Energy Conversion Problem 1.7.9 Ammonia was used as a working fluid in an OTEC system. Actual energy conversion efficiency, which is defined as the net power (wnet) divided by the heat supplied to the system (Qh). The net power was found to be 35,000 kW while the heat supplied to the ocean water was found to be 1,300,000 kW. Solution: a. Since the units are already the same, the thermodynamic efficiency is calculated simply as follows: 7

𝜂𝜂𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 (%) =

𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 35,000 𝑘𝑘𝑘𝑘 × 100% = × 100% = 2.69% 𝑄𝑄ℎ 1,300,000 𝑘𝑘𝑘𝑘

b. Hence, actual power may only be in a single digit range for OTECs.

Problem 1.7.10. Solar PV Conversion Efficiency Problem 1.7.10 Determine the efficiency of converting solar energy received on a solar PV panel if the incoming solar radiation was measured to be 1,100 W/m2 and the electrical output of the solar panel having an area of 1 m2 was measured to be 40 volts and 3.5 Amps.

Solution: a. From basic electrical power equation, Volt x Ampere = Watts and thus the electrical power output could be easily calculated as follows: Power (Watts) = 40 Volts x 3.5 Amperes = 140 Watts b. The input energy may be calculated as follows and converted into the same unit as the output energy (in Watts): 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑊𝑊 ) =

1,100 𝑊𝑊 × 1 𝑚𝑚2 = 1,100 𝑊𝑊 𝑚𝑚2

c. Thus, the conversion efficiency is calculated as follows: 𝜂𝜂 (%) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 140 𝑊𝑊 × 100% = × 100% = 12.7% 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 1,100 𝑊𝑊

d. Only 12.7% of the incoming solar radiation is converted into electrical power.

1.7.11 Wind Energy Kinetic Energy P1.11 Determine the kinetic energy (in units of Btu) available from 110 lbs of moving air at a speed of 32.8 ft/s. a. 5,000 8

b. 2,500 c. 2.36 d. 5.25 Solutions: a. The governing equation is as follows: 1 𝐾𝐾𝐾𝐾 (𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽) = 𝑚𝑚𝑣𝑣 2 2 1 32.8 𝑓𝑓𝑓𝑓 2 1 𝐵𝐵𝐵𝐵𝐵𝐵 𝑠𝑠 2 𝐾𝐾𝐾𝐾 (𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽) = × 110 𝑙𝑙𝑙𝑙𝑙𝑙 × � � × × = 2.36 𝐵𝐵𝐵𝐵𝐵𝐵 2 𝑠𝑠 778 𝑓𝑓𝑓𝑓 𝑙𝑙𝑙𝑙 32.2 𝑓𝑓𝑓𝑓 1.7.12 Power from Small Hydro Power Plant P1.12 Determine the actual amount of power (Watts) that could be derived from a very small hydro power plant with a head of 2 m and a flow of 70 L/s. Assume an overall efficiency of 65%. a. 686 b. 411.6 c. 808.5 d. 891.8 Solution: a. The governing equation is shown and with values substituted. 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑄𝑄𝑄𝑄𝑄𝑄 𝑃𝑃𝑎𝑎 = 𝑄𝑄𝑄𝑄𝑄𝑄 =

70 𝐿𝐿 𝑠𝑠

× 2 𝑚𝑚 ×

1 𝑘𝑘𝑘𝑘 𝐿𝐿

𝑁𝑁𝑠𝑠 2

× 𝑘𝑘𝑘𝑘 𝑚𝑚 ×

9.8 𝑚𝑚

9

𝑠𝑠 2

×

𝐽𝐽

𝑁𝑁 𝑚𝑚

×

𝑊𝑊 𝑠𝑠 𝐽𝐽

× 0.65 = 891.8 𝑊𝑊

1.7.13 Photosynthesis and Use of Carbon Dioxide P1.13 Photosynthesis is the conversion of carbon dioxide and water to produce the biomass for energy utilization. Assuming the biomass produced is a simple carbohydrate sugar with a chemical formula of C6H12O6, determine the amount of carbon dioxide needed (kg) to generate a kg of this carbohydrate sugar. a. 1.5 b. 5.0 c. 7.5 d. 10.0 Solution: a. This problem needs a simple balanced equation for biomass production as shown and take the ratio of input CO2 and output carbohydrate.

6𝐶𝐶𝐶𝐶2 + 6𝐻𝐻2 𝑂𝑂 + 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 → 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 + 6𝑂𝑂2

b. The CO2 needed is calculated as follows:

6 × [(12 × 1) + (16 × 2)] 264 6𝐶𝐶𝐶𝐶2 𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 = = = 1.47 ~1.5 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 [(12 × 6) + (1 × 12) + (16 × 6)] 180 𝑙𝑙𝑙𝑙 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 1.7.14 Photosynthesis and Oxygen Generated P1.14 Photosynthesis is the conversion of carbon dioxide and water to produce the biomass used for energy utilization. Assuming the biomass produced is a simple carbohydrate sugar with a chemical formula of C6H12O6, determine the amount of oxygen produced (kg) for every kg of carbohydrate sugar produced during this biomass production process. a. 1.5 10

b. 1.07 c. 5.0 d. 7.5 Solution: a. This problem needs a simple balanced equation for biomass production as shown and take the ratio of out O2 and output carbohydrate.

6𝐶𝐶𝐶𝐶2 + 6𝐻𝐻2 𝑂𝑂 + 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 → 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 + 6𝑂𝑂2

b. The CO2 needed is calculated as follows:

6 × [(16 × 2)] 192 6𝑂𝑂2 𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 = = = 1.07 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 [(12 × 6) + (1 × 12) + (16 × 6)] 180 𝑙𝑙𝑙𝑙 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 1.7.15 Photosynthesis and Amount of Water Needed P1.15 Photosynthesis is the conversion of carbon dioxide and water to produce the biomass used for energy utilization. Assuming the biomass produced is a simple carbohydrate sugar with a chemical formula of C6H12O6, determine the amount of water needed (kg) to generate a kg of this carbohydrate sugar. a. 1.5 b. 5.0 c. 0.60 d. 1.07 Solution: a. This problem needs a simple balanced equation for biomass production as shown and take the ratio of input H2O and output carbohydrate.

6𝐶𝐶𝐶𝐶2 + 6𝐻𝐻2 𝑂𝑂 + 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 → 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 + 6𝑂𝑂2 11

b. The CO2 needed is calculated as follows: 6𝐻𝐻2 𝑂𝑂 6 × [(1 × 2) + (16 × 1)] 108 𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 = = = 0.60 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 [(12 × 6) + (1 × 12) + (16 × 6)] 180 𝑙𝑙𝑙𝑙 𝐶𝐶6 𝐻𝐻12 𝑂𝑂6 ASSESS YOUR KNOWLEDGE 1. Which of the following renewables is not a consequence of solar energy? a. Biomass b. Salinity gradient c. Tidal energy d. Wave energy 2. What causes wave energy? a. Geothermal b. OTEC c. Wind energy d. Hydro power 3. Solar radiation received at the surface of the earth comes in two forms the direct radiation and what is the other form? a. Reflected b. Transmitted c. Diffuse d. Absorbed 4. On cloudy days the amount of solar radiation received on the earth ranged from what values in Watts/m2? a. 1100-1400 12

b. 600-900 c. 900-1000 d. 100-200 5. What do you call the quantity of solar energy received at normal incidence and outside of the earth’s atmosphere at a mean distance between the sun and the earth? a. Solar constant b. Bowen ratio c. Albedo d. Reflectance 6. Renewable energy is defined as energy resources that are replenished at the same rate they are used. Which of the listed renewable energy is not a consequence of solar energy? a. Solar b. Salinity gradient c. Wind energy d. Hydropower 7. What is the range of conversion efficiency for hydro power? a. 60-80 b. 40-50 c. 80-95 d. 10-30 8. Determine the theoretical amount of power (Watts) derived from a moving air mass having a density of 1.2 kg/m3 and a cross-sectional area of 10 m2 and moving at a speed of 12 m/s? a. 600

13

b. 10,368 c. 10.3 d. 6,000 Solutions: a. The governing equation is below and substituting values we have the following: 1 𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 𝜌𝜌𝜌𝜌𝑣𝑣 3 2

𝐽𝐽 1 1 1.2 𝑘𝑘𝑘𝑘 12 𝑚𝑚 3 𝑁𝑁𝑠𝑠 2 𝑊𝑊 𝑠𝑠 3 2 𝑃𝑃𝑡𝑡 = 𝜌𝜌𝜌𝜌𝑣𝑣 = × × 10 𝑚𝑚 × � � × × × = 10,368 𝑊𝑊 𝑚𝑚3 𝑠𝑠 𝑘𝑘𝑘𝑘 𝑚𝑚 𝑁𝑁 𝑚𝑚 2 2 𝐽𝐽

9. What is the current recommended value of solar constant in units of W/m2? a. 1353 b. 750 c. 3.6 d. 1367

10. What is the average solar radiation received in the state of Texas in units of kWh/m2/day? a. 10.5 b. 1367 c. 5.11 d. 750 11. Hydro power is a function of what two parameters? a. Volume and flow rate b. Volumetric flow rate and dynamic head. c. Elevation and density of water d. Dynamic head and pressure drop 14

12. The old value of solar constant was 1353 W/m2. Convert the value of the old solar constant into English units of Btu/hr-ft2. a. 429 b. 924 c. 866 d. 858 Solution: a. 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 =

1353 𝑊𝑊 𝑚𝑚 2

×

𝐽𝐽

𝑊𝑊−𝑠𝑠

×

𝐵𝐵𝐵𝐵𝐵𝐵

1055 𝐽𝐽

�

1 𝑚𝑚

3.28 𝑓𝑓𝑓𝑓

2

� ×

3600 𝑠𝑠 ℎ𝑟𝑟

= 429

𝐵𝐵𝐵𝐵𝐵𝐵

ℎ𝑟𝑟−𝑓𝑓𝑓𝑓 2

13. Determine the theoretical amount of power (Watts) that could be derived from a very small hydro power plant with a head of 1.5 m and a flow of 55 L/s. a. 686 b. 411.6 c. 808.5 d. 285.1 Solution: a. The governing equation is shown and with values substituted. 𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑄𝑄𝑄𝑄 𝑃𝑃𝑡𝑡 = 𝑄𝑄𝑄𝑄 =

55 𝐿𝐿 𝑠𝑠

× 1.5 𝑚𝑚 ×

1 𝑘𝑘𝑘𝑘 𝐿𝐿

𝑁𝑁𝑠𝑠 2

× 𝑘𝑘𝑘𝑘 𝑚𝑚 ×

9.8 𝑚𝑚 𝑠𝑠 2

14. Convert 1,367 Watts into units of horsepower. a. 1.83 b. 0.545 c. 1.02 d. 3.66 15

𝐽𝐽

𝑊𝑊 𝑠𝑠

× 𝑁𝑁 𝑚𝑚 × 𝐽𝐽 = 808.5 𝑊𝑊

Solution: a. This is a simple conversion as follows: 𝐻𝐻𝐻𝐻 = 1367 𝑊𝑊 ×

1 ℎ𝑝𝑝 = 1.83 ℎ𝑝𝑝 746 𝑊𝑊

15. Determine the actual amount of power (Watts) that could be derived from a very small hydro power plant with a head of 1.5 m and a flow of 55 L/s. Assume an overall efficiency of 60%. a. 686 b. 411.6 c. 808.5 d. 485.1 Solution: a. The governing equation is shown and with values substituted. 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑄𝑄𝑄𝑄𝑄𝑄 𝑃𝑃𝑎𝑎 = 𝑄𝑄𝑄𝑄𝑄𝑄 =

55 𝐿𝐿 𝑠𝑠

× 1.5 𝑚𝑚 ×

1 𝑘𝑘𝑘𝑘 𝐿𝐿

𝑁𝑁𝑠𝑠 2

× 𝑘𝑘𝑘𝑘 𝑚𝑚 ×

9.8 𝑚𝑚 𝑠𝑠 2

×

𝐽𝐽

𝑁𝑁 𝑚𝑚

×

𝑊𝑊 𝑠𝑠 𝐽𝐽

× 0.60 = 485.1 𝑊𝑊

16. How much energy (MJ) is captured by a biomass plant from an acre of land in a year. The biomass variety propagated was high tonnage sorghum biomass with a yield of 15 tons/acre/year? Assume the heating value of biomass is 7,540 Btu/lb [17.5 MJ/kg]. a. 238,641 b. 163,630 c. 159,094 d. 425,288 Solution: a. This is simply a conversion as shown.

16

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) = 1 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ×

15 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 2,000 𝑙𝑙𝑙𝑙𝑙𝑙 7,540 𝐵𝐵𝐵𝐵𝐵𝐵 1055 𝐽𝐽 𝑀𝑀𝑀𝑀 × × × × 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑙𝑙 𝐵𝐵𝐵𝐵𝐵𝐵 1 × 106 𝐽𝐽

= 238,641, 𝑀𝑀𝑀𝑀

17. Determine the conversion efficiency (in%) of biomass production from a square meter of an area with the following data: Solar Constant = 1,367 W/m2; actual solar radiation received on earth = 20% of solar constant throughout the year; biomass energy captured = 175,000 MJ/acre/yr; assume sunshine is received by biomass for an average of 10 hrs a day and use 365 days in a year of good sunshine. a. 0.37% b. 1.2% c. 5.1% d. 3.71% Solution: a. This is also a simple unit and conversion problem to match units in W/m2 as shown.

175,000 𝑀𝑀𝑀𝑀 1 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 3.28 𝑓𝑓𝑓𝑓 2 1 𝑦𝑦𝑦𝑦 1 𝑑𝑑𝑑𝑑𝑑𝑑 1 ℎ𝑟𝑟 1 × 106 𝐽𝐽 × × � � × × × × 43,560 𝑓𝑓𝑓𝑓 2 𝑚𝑚 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 10 ℎ𝑟𝑟𝑟𝑟 3600 𝑠𝑠 𝑀𝑀𝑀𝑀 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑦𝑦 ×

𝑊𝑊 𝑠𝑠 𝑊𝑊 = 3.29 2 𝐽𝐽 𝑚𝑚

b. The efficiency equation is used as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (%) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 3.29 𝑊𝑊/𝑚𝑚2 × 100% = × 100% = 1.2% 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 1367 × 0.2 𝑊𝑊/𝑚𝑚2

18. Determine the kinetic energy (Joules) available from 50 kg of moving air at a speed of 10 m/s. a. 5000 b. 2500 17

c. 1250 d. 255 Solutions: a. The governing equation is a s follows: 1 𝐾𝐾𝐾𝐾 (𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽) = 𝑚𝑚𝑣𝑣 2 2

1 10 𝑚𝑚 2 𝑁𝑁𝑠𝑠 2 𝐾𝐾𝐾𝐾 (𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽) = × 50 𝑘𝑘𝑘𝑘 × � � × = 2,500 𝐽𝐽 2 𝑠𝑠 𝑘𝑘𝑘𝑘 𝑚𝑚

19. Determine the actual power (in Watts) derived from a moving air mass having a density of 1.2 kg/m3 and a cross-sectional area of 10 m2 and moving at a speed of 12 m/s? Assume an overall conversion efficiency of 45%. a. 600 b. 10.4 c. 10,368 d. 4,665.6 Solutions: a. The governing equations are as follows including substitutions.

1 𝑃𝑃𝑡𝑡 = 𝜌𝜌𝜌𝜌𝑣𝑣 3 2

𝐽𝐽 1 1.2 𝑘𝑘𝑘𝑘 12 𝑚𝑚 3 𝑁𝑁𝑠𝑠 2 𝑊𝑊 𝑠𝑠 2 𝑃𝑃𝑡𝑡 = × × 10 𝑚𝑚 × � � × × × = 10,368 𝑊𝑊 2 𝑚𝑚3 𝑠𝑠 𝑘𝑘𝑘𝑘 𝑚𝑚 𝑁𝑁 𝑚𝑚 𝐽𝐽

b. The actual power is calculated as follows:

1 𝑃𝑃𝑎𝑎 = 𝜌𝜌𝜌𝜌𝑣𝑣 3 𝐸𝐸 = 10,368 × 0.45 = 4,665.6 𝑊𝑊 2

18

20. A 10 m2 solar water heater receives solar energy with an hourly average of 650 W/m2. It was used to heat 25 kg of water from 25oC to 55oC for that one-hour period. Estimate the percent conversion efficiency of this solar collector. Assume specific heat of water to be 4.190 kJ/kgoC and there are 3.6 kJ/Wh. a. 45.2% b. 25.5% c. 13.4% d. 34.9% Solution: a. The output energy is calculated as follows: 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑘𝑘𝑘𝑘) = 𝑄𝑄 = 𝑚𝑚𝐶𝐶𝑝𝑝 𝛥𝛥𝑡𝑡

𝑄𝑄 = 25 𝑘𝑘𝑘𝑘 × 4.190

𝑘𝑘𝑘𝑘 × (55 − 25)𝑜𝑜 𝐶𝐶 = 3,142.5 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘𝑜𝑜 𝐶𝐶

b. The energy input is calculated as follows: 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑘𝑘𝑘𝑘) = 650

𝑊𝑊 3.6 𝑘𝑘𝑘𝑘 2 × 10 𝑚𝑚 × 1 ℎ𝑟𝑟 × = 23,400 𝑘𝑘𝑘𝑘 𝑚𝑚2 𝑊𝑊 ℎ𝑟𝑟

c. Hence, the efficiency is calculated as follows: 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 =

3,142.5 𝑘𝑘𝑘𝑘 × 100% = 13.4% 23,400 𝑘𝑘𝑘𝑘

19

Chapter 2 Solar Energy 2.13 Solved Problems Problem 2.13.1. Extraterrestrial (ET) Solar Radiation Problem 2.13.1 Determine the extraterrestrial solar radiation received on the earth on June 15.

Plot the variation of extraterrestrial solar radiation received with time of year on a normal plane.

Use Gsc = 1367 W/m2.

Solution: a. The equation for ET radiation is given as follows: 𝐺𝐺𝑜𝑜𝑜𝑜 = 𝐺𝐺𝑠𝑠𝑠𝑠 × �1 + 0.033 × 𝑐𝑐𝑐𝑐𝑐𝑐

360 × 𝑛𝑛 � 365

b. n = 166 and substituting the values for n results to the following 𝐺𝐺𝑜𝑜𝑜𝑜 = 1,367 × �1 + 0.033 × 𝑐𝑐𝑐𝑐𝑐𝑐 c.

360 × 166 𝑊𝑊 � = 1,324 2 365 𝑚𝑚

If the value of n varies from 1 to 365 and the resulting values plotted, the graph will be the same as that in Example Problem 2.1.

Problem 2.13.2 Solar Time Problem 2.13.2 Determine the solar time corresponding to 8:30 am Central Time on June 15th 2014 in College Station, Texas. The longitude of College Station, Texas is 96.32 West and this town’s time was based on 90o meridian.

Solution: a. For June 15th, 2014, n = 116 b. B is then calculated using Equation 2.4 as follows: 20

𝐵𝐵 = (116 − 1) ×

360 = 113. 4𝑜𝑜 365

c. The equation of time E is then calculated using Equation 2.3 as shown. 𝐸𝐸 = 229.2 × (0.000075 + 0.00186 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 − 0.032077 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − 0.014615 cos 2𝐵𝐵 − 0.04089 sin 2𝐵𝐵)

𝐸𝐸 = 229.2 × (0.000075 + 0.00186 cos (113.4) − 0.032077 𝑠𝑠𝑠𝑠𝑠𝑠 (113.4) 𝐸𝐸 = 2.22

− 0.014615 cos 2(113.4) − 0.04089 sin(2 × 113.4)

d. The longitude correction is calculated as follows: Longitude correction = 4 x (90-96.32) = -25.28 minutes e. The solar time is calculated as follows.

Solar Time = 8.50 + (-25.28/60) + (2.22/60) Solar Time = 8.50 -0.42 + 0.037 = 8.116 f. Note that the longitude correction and the equation of time, E, is converted into units of hours to be consistent with the calculation of solar time. The fraction may be brought back into units of minutes by multiplying this by 60 minutes to an hour (0.116 x 60 = 7) g. Therefore, the solar time is 8.12 or 8:07 am.

Problem 2.13.3. Solar Declination Angle Problem 2.13.3 Calculate the solar declination angle on June 15, 2014.

Solution: 1. Using the declination Equation 2.3, and with a value of n = 116, the resulting answer is shown here. 𝛿𝛿 = 23.45 × 𝑠𝑠𝑠𝑠𝑠𝑠 �360 ×

284+𝑛𝑛 365

�, in degrees 21

𝛿𝛿 = 23.45 × 𝑠𝑠𝑠𝑠𝑠𝑠 �360 ×

284 + 116 � = 13. 2𝑜𝑜 365

2. The solar declination angle is 13.29 degrees.

Problem 2.13.4. Angle of Incidence Problem 2.13.4 Calculate the angle of incidence of beam radiation on a solar collector surface located in College Station, Texas at 8:30 am (solar time) on June 15, 2014, if the surface is tilted at an angle of 30o and pointed 25o west of south. The latitude of College Station, Texas is 30.61oN. Solution: a. First, calculate the value of n = 116 b. The other values are as follows: ω = -52.5o; γ = 25o; β = 30o and φ = 30.61o. c. The solar declination angle is then calculated as follows: 𝛿𝛿 = 23.45 × 𝑠𝑠𝑠𝑠𝑠𝑠 �360 ×

284 + 116 � = 13.29𝑜𝑜 365

d. Hence, with all the parameters known, Equation 2.6 may be used to estimate the incidence angle as follows: 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝜙𝜙 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 − 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = [sin(13.29) × sin(30.61) × 𝑐𝑐𝑐𝑐𝑐𝑐(30)]

− [𝑠𝑠𝑠𝑠𝑠𝑠(13.29) × 𝑐𝑐𝑐𝑐𝑐𝑐(30.61) × 𝑠𝑠𝑠𝑠𝑠𝑠(30) × 𝑐𝑐𝑐𝑐𝑐𝑐(25)]

+ [𝑐𝑐𝑐𝑐𝑐𝑐[13.29] × 𝑐𝑐𝑐𝑐𝑐𝑐(30.61) × 𝑐𝑐𝑐𝑐𝑐𝑐(30) × 𝑐𝑐𝑐𝑐𝑐𝑐(−52.5)]

+ [𝑐𝑐𝑐𝑐𝑐𝑐(13.29) × 𝑠𝑠𝑠𝑠𝑠𝑠(30.61) × 𝑠𝑠𝑠𝑠𝑠𝑠(30) × 𝑐𝑐𝑐𝑐𝑐𝑐(25) × 𝑐𝑐𝑐𝑐𝑐𝑐(−52.5)] + [𝑐𝑐𝑐𝑐𝑐𝑐(13.29) × 𝑠𝑠𝑠𝑠𝑠𝑠(30) × 𝑠𝑠𝑠𝑠𝑠𝑠(25) × 𝑠𝑠𝑠𝑠𝑠𝑠(−52.5)] 22

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = [0.230 × 0.0509 × 0.861] − [0.230 × 0.861 × 0.50 × 0.906] + [0.230 × 0.861 × 0.866 × 0.609]

+ [0.230 × 0.509 × 0.50 × 0.906 × 0.609]

e.

+ [0.230 × 0.50 × 0.423 × (−0.793)] = 0.110

Cos θ = 0.11 and θ = 83.68o

f. If the magnitude of solar energy received on the surface is known, only about 11% of this is absorbed by the collector at this time of the day.

Problem 2.13.5. Hour Angle, Time of Sunrise, and Number of Daylight Hours Problem 2.13.5 Calculate the hour angle, time of sunrise and number of daylight hours on June 15, 2014 in College Station, Texas. The latitude in College Station, Texas is 30.61o.

Solution: a. The solar declination angle is first calculated as follows: 𝛿𝛿 = 23.45 × 𝑠𝑠𝑠𝑠𝑠𝑠 �360 ×

284 + 116 � = 13.29𝑜𝑜 365

b. The hour angle is first determined using Equation 2.10. 𝑐𝑐𝑐𝑐𝑐𝑐𝜔𝜔𝑠𝑠 = −tan (𝛿𝛿) × tan (𝜙𝜙)

𝜔𝜔𝑠𝑠 = 𝑐𝑐𝑐𝑐𝑠𝑠 −1 (−tan (30.61) × tan(−13.29)) = 98.03𝑜𝑜

c. This hour angle is then recalculated to correspond to the solar time by assuming that at sunrise hour angle of 90o, the sunrise would be 6:00 am (or the sunset at 6:00 pm, solar time). Since the angle is more than 90o, one would already speculate that the real sunrise time should be before 6:00 am. d. The fraction of angle between 90o and 98.03o (90-98.03 = -8.03o) is simply converted by ratio and proportion to the time value to give the number of minutes shown below.

23

𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 = −8.03𝑜𝑜 ×

ℎ𝑟𝑟 60 𝑚𝑚𝑚𝑚𝑚𝑚 × = −32.12 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 15𝑜𝑜 1 ℎ𝑟𝑟

e. Therefore, the sunrise should be (6.00 - 32.12/60=5.465 min) 5:27.88 a.m. (solar time). f. The number of possible daylight hours is calculated from Equation 2.11 and the results are shown below. 𝑁𝑁 =

2 𝑐𝑐𝑐𝑐𝑠𝑠 −1 (−𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡) 15 2

𝑁𝑁 = 15 𝑐𝑐𝑐𝑐𝑠𝑠 −1 �− tan(13.29) × 𝑡𝑡𝑡𝑡𝑡𝑡(30.61)� = 13.07 ℎ𝑟𝑟𝑟𝑟

g. For this day (summer in College Station, Texas), one would expect to have more than 13 hours of sunshine (unless blocked by clouds at some point during the day).

Problem 3.13.6. Theoretical Daily Solar Radiation, Ho Problem 3.13.6 Determine the Ho, the day’s solar radiation on a horizontal surface in the absence of atmosphere in College Station, Texas on June 15, 2014. The latitude of College Station, Texas is 30.61o. Use the results from previous problems if possible.

Solution: a. The following have already been given or calculated, or they will have to be calculated for this problem. Solar constant = 1,367 W/m2; n = 116; φ = 30.61o and δ = 13.29o and ωs = 98.03o. Substituting these values in Equation 2.12 lead to the following results for Ho. 𝐻𝐻𝑜𝑜 =

24 × 3600 × 𝐺𝐺𝑠𝑠𝑠𝑠 360𝑛𝑛 × �1 + 0.033 × 𝑐𝑐𝑐𝑐𝑐𝑐 � 365 𝜋𝜋 × �𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝜔𝜔𝑠𝑠 +

𝜋𝜋𝜔𝜔𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠� 180

24

𝐻𝐻𝑜𝑜 =

24 × 3600 × 1367 360 × 116 × �1 + 0.033 × 𝑐𝑐𝑐𝑐𝑐𝑐 � 3.1416 365

× �cos (30.61) × cos (−13.29) × sin (92.03) +

3.1416 × 98.03 sin (30.61) × sin (−13.29)� 180

𝐻𝐻𝑜𝑜 = 37,595,111 × (0.9864)

× ( (0.861) × (0.973) × (0.9994) + (1.711) × (0.509) × (−0.2299))

𝐻𝐻𝑜𝑜 = 38,473,349

𝐽𝐽 𝑀𝑀𝑀𝑀 = 38.47 2 2 𝑚𝑚 𝑚𝑚

b. As such, in the absence of atmosphere, one would expect to receive a solar radiation of approximately 38.5 MJ/m2, a high number in the summer compared to winter values.

Problem 2.13.7. Theoretical Hourly Solar Radiation Problem 2.13.7 Determine the Io, the day’s solar radiation on a horizontal surface in the absence of atmosphere in College Station, Texas on June 15, 2014, between the hours of 8 and 9 am (solar time). The latitude of College Station, Texas is 30.61o. Use the results from previous example problems.

Solution: a. The following have already been given or calculated: Solar constant = 1,367 W/m2; n = 116; φ = 30.61o and δ = 13.29o and ω1 = -60o and ω1 = -45o. Substituting these values in Equation 2.13 leads to the following results for Io. 𝐼𝐼𝑜𝑜 =

12 × 3600𝐺𝐺𝑠𝑠𝑠𝑠 360𝑛𝑛 × �1 + 0.033𝑐𝑐𝑐𝑐𝑐𝑐 � 𝜋𝜋 365

× �𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐(𝑠𝑠𝑠𝑠𝑠𝑠𝜔𝜔2 − 𝑠𝑠𝑠𝑠𝑠𝑠𝜔𝜔1 ) + 25

𝜋𝜋(𝜔𝜔2 − 𝜔𝜔1 ) 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠� 180

𝐼𝐼𝑜𝑜 =

12 × 3600 × 1367 360 × 116 × �1 + 0.033 × 𝑐𝑐𝑐𝑐𝑐𝑐 � 3.1416 365

× �cos (30.61) × cos (−13.29) × (sin (−45) − sin (−60)) +

𝜋𝜋 × (−45 − (−60)) × sin (30.61) × sin (−13.29)� 180

b. The individual terms have values as follows: 𝐼𝐼𝑜𝑜 = 18,797,555 × (0.986)

× ((0.861) × (0.9739) × (−0.707 + 0.866) + (0.2618) × (0.5092) × (0.230)

𝐼𝐼𝑜𝑜 = 18,797,555 × (0.986) × (0.861) × (0.973) × (0.159) + (0.0306)) 𝐼𝐼𝑜𝑜 = 2,986,536

𝐽𝐽 𝑀𝑀𝑀𝑀 = 3.0 𝑚𝑚2 𝑚𝑚2

c. Thus, the hourly solar radiation received in this time frame will be around 3 MJ/m2 in the absence of atmosphere.

Problem 2.13.8. Clearness Index to Estimate Beam and Diffuse Radiation Problem 2.13.8 Determine the clearness index, Kt, and the fraction of solar energy that is beam and diffuse for the month of June 15, 2014, in College Station, Texas if the solar radiation received was 15 MJ/m2.

Solution: a. The clearness index, Kt, is simply calculated using Equation 2.15 by first determining the extraterrestrial solar radiation in the absence of atmosphere as illustrated in Problem 2.7.

The value of Ho = 38.47 MJ/m2.

26

𝑀𝑀𝑀𝑀 15 2 𝐻𝐻 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ) 𝑚𝑚 𝐾𝐾𝑇𝑇 = = = 0.390 𝑀𝑀𝑀𝑀 𝐻𝐻𝑜𝑜 (𝑚𝑚𝑚𝑚𝑚𝑚 ) 38.47 2 𝑚𝑚

b. The second version of Equation 2.14 will be used as follows: 𝐻𝐻𝑑𝑑 = 1.188 − 2.272𝐾𝐾𝑡𝑡 + 9.473𝐾𝐾𝑡𝑡2 − 21.865𝐾𝐾𝑡𝑡3 + 14.648𝐾𝐾𝑡𝑡4 𝐻𝐻

𝐻𝐻𝑑𝑑 = 1.188 − 2.272 × (0.39) + 9.473 × (0.39)2 − 21.865 × (0.39)3 𝐻𝐻 + 14.648 × (0.39)4

𝐻𝐻𝑑𝑑 = 1.188 − 0.886 + 1.441 − 1.297 + 0.339 = 0.785 𝐻𝐻

c. About 78.5% of the solar radiation received was diffuse radiation or 11.8 MJ/m2 d. The amount of beam or direct solar radiation is the difference of about 3.2 MJ/m2. e. Note that while the extraterrestrial solar radiation is quite high, the clearness index was quite low during this month and thus, the diffuse radiation is higher than the beam radiation.

Problem 2.13.9 Sizing Solar PV panels Problem 2.13.9 A typical barn lights up its entrance from 6 pm to 6 am, the feeding stall for two hours (6pm-8pm), and the feed water pump for two hours. The entrance lamp wattage is 10 watts, and the feeding stall lamp is 10 watts, whereas the pump consumes 2 amperes at 12 volts. (a) What is the energy requirement of the barn? (b) Compute the size of the battery and (c) the PV panel needed for the system in watt-peak (Wp) assuming CGF of 3 and overall system efficiency of 85%. (d) What size of wire is recommended for the PV-battery and battery, assuming 15 meters in length?

27

Solution: a. The total energy calculation may be easily made by creating a tabulation of loads and their wattage and hours of use to estimate energy in watt-hour as shown in the table below.

Load

Wattage

Hours

Watt-Hour

Entrance Lamp 1

10

12 hrs

120

Feeding Stall Lamp 2

10

2 hr

20

Pump

2 x 12 V =24

2 hrs

48

Total

44 watts

188 watt-hour

Therefore, the energy requirement for this barn each day is 188 watt-hours. Battery capacity is estimated using Equation 2.17 and Equation 2.18 as shown. 𝐵𝐵𝐶𝐶𝑤𝑤ℎ = 𝐷𝐷𝐷𝐷𝐷𝐷 × 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹

𝐵𝐵𝐶𝐶𝑤𝑤ℎ = 𝐷𝐷𝐷𝐷𝐷𝐷 × 10 = 188 𝑊𝑊ℎ × 10 = 1880 𝑊𝑊ℎ

𝐵𝐵𝐶𝐶𝐴𝐴ℎ ==

𝐵𝐵𝐶𝐶𝑤𝑤ℎ 𝑉𝑉

=

𝐷𝐷𝐷𝐷𝐷𝐷×𝐷𝐷𝐷𝐷𝐷𝐷 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉

=

1880 𝑊𝑊ℎ 12 𝑉𝑉

= 156.6 𝐴𝐴ℎ

The battery capacity that is selected should have values higher than this. These battery capacities (100-200 AH) are typically used for recreational vehicles (RVs). b. The pV panel Wp calculation is shown using Equation 2.19 and the assumptions given. 𝑊𝑊𝑝𝑝 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = 𝑊𝑊𝑝𝑝 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) =

𝐷𝐷𝐷𝐷𝐷𝐷 𝐶𝐶𝐶𝐶𝐶𝐶 × 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸

188 𝑊𝑊ℎ = 73.73 𝑊𝑊 3 ℎ × 0.85

28

c. The PV panel to be selected must also be higher and of nominal size, perhaps 80 watts.

Finally, the wire size may be calculated from the PV to the battery and from the battery to the load as shown below and using Equation 2.20. 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆(𝑚𝑚𝑚𝑚) = 0.04 ×

𝑊𝑊𝐿𝐿 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) 𝐿𝐿(𝑚𝑚) 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉

𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆(𝑚𝑚𝑚𝑚) = 0.04 ×

80 𝑊𝑊 × 15(𝑚𝑚) = 4 𝑚𝑚𝑚𝑚 [𝐴𝐴𝐴𝐴𝐴𝐴#6] 12 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉

𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆(𝑚𝑚𝑚𝑚) = 0.04 ×

𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆(𝑚𝑚𝑚𝑚) = 0.04 ×

73.7 𝑊𝑊 × 15(𝑚𝑚) = 3.69 𝑚𝑚𝑚𝑚 [𝐴𝐴𝐴𝐴𝐴𝐴#6] 12 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉

44 𝑊𝑊 × 15(𝑚𝑚) = 2.2 𝑚𝑚𝑚𝑚 [𝐴𝐴𝐴𝐴𝐴𝐴#11] 12 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉

One should be able to find corresponding wire sizes easily from electrical wiring stores as well as from electricity textbooks.

Problem 2.13.10 Economics of Solar Energy Problem 2.13.10 A homeowner would like to net meter his residential electrical power load using solar energy. The average solar radiation in his area is around 4 kWh/m2/day. He uses about 12,000 kWh of electricity per year. He pays for electricity at $0.12/kWh. (a) What is the minimum area of solar PV system this household would need if the overall conversion efficiency were 15%? (b) At a price of $1.5/Wp, what is the initial investment and payback period required to recover the initial capital cost only and based on the payment for electricity alone? Assume a climate generation factor (CGF) of 3 and efficiency of 85% for the calculation of peak wattage.

Solution:

29

a. The first step is to calculate the daily load requirement for this household. This can simply be calculated using the yearly requirement divided by the number of days in a year, assuming 365 days/yr for a non-leap year as shown.

12,000 𝑘𝑘𝑘𝑘ℎ 1 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 32.9 𝑘𝑘𝑘𝑘ℎ × = 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑

𝐷𝐷𝐷𝐷𝐷𝐷 (𝑘𝑘𝑘𝑘) =

b. Then, using the efficiency equation, we find the amount of solar energy needed to satisfy this DLR as follows: 𝜂𝜂 =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼

𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 �

× 100%

0.15 =

32.9 𝑘𝑘𝑘𝑘ℎ/𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼

× 100%

32.9 𝑘𝑘𝑘𝑘ℎ/𝑑𝑑𝑑𝑑𝑑𝑑 𝑘𝑘𝑘𝑘ℎ 𝑘𝑘𝑘𝑘ℎ �= × 100% = 219 𝑑𝑑𝑑𝑑𝑑𝑑 0.15 𝑑𝑑𝑑𝑑𝑑𝑑

c. The minimum area needed to satisfy this output energy is calculated as follows: 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 (𝑚𝑚2 ) =

219 𝑘𝑘𝑘𝑘ℎ 𝑚𝑚2 × 𝑑𝑑𝑑𝑑𝑑𝑑 × = 54.8 𝑚𝑚2 𝑑𝑑𝑑𝑑𝑑𝑑 4 𝑘𝑘𝑘𝑘ℎ

d. The watt-peak calculation is next found as follows, assuming CGF of 3 and efficiency of 85%. 𝑊𝑊𝑝𝑝 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = 𝑊𝑊𝑝𝑝 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) =

𝐷𝐷𝐷𝐷𝐷𝐷 𝐶𝐶𝐶𝐶𝐶𝐶 × 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸

32,900 𝑊𝑊ℎ = 12,902 𝑊𝑊 3 ℎ × 0.85

e. At $1.5/Wp, the initial investment is about $19,353, assuming 12,902 Wp. If he uses the grid power at $0.12/kWh, he will be paying around $1,1440/yr for electricity.

Consequently, recovering the initial capital investment alone and paying for net-metered electricity would require 13.4 years. 𝑝𝑝𝑝𝑝 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 ($) = 12,902 𝑊𝑊 ×

$1.5 = $19,353 𝑊𝑊𝑝𝑝 30

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 ($) = 12,000 𝑃𝑃𝑃𝑃𝑃𝑃 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌) = 2.13.11

𝑘𝑘𝑘𝑘ℎ $0.12 $1,440 × = 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑘𝑘𝑘𝑘ℎ 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 ($)

𝑁𝑁𝑁𝑁𝑁𝑁 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 (

$ ) 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

=

$19,353

$1440/𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

= 13.44 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

Solar Refrigerator Refrigerant

P2.11 This is an efficient and popular refrigerant due to its superior thermodynamic properties and low cost. It is environmentally benign, having zero GWP and zero ODP (ozone depletion potential). It is hazardous when released in large quantities due to its toxicity. It is a 3-10% more efficient refrigerant than CFCs (chloro-fluoro-carbons), so an ammonia-based system requires less electricity, resulting in lower operating costs. Name this refrigerant. a. Ammonia-water b. R134a c. R-12 d. Freon 2.13.12

Solar Water Heaters

P2.12 What is the main difference between a natural convection solar water heater versus a forced circulation water heater when it comes to the location of the storage tank? a. More efficient b. Less efficient c. The water tank is above the solar collector d. The tank is below the solar collector

31

2.13.13

Angle of Incidence

P2.13 Calculate the angle of incidence of beam radiation (in degrees) on a surface located in College Station, Texas at 11:30 am (solar time) on October 4 (leap year) if the surface is tilted 30o from the horizontal and oriented due south. The latitude of College Station is 30.61oN. a. 9.9 b. 12.5 c. 16.5 d. 17.4 Solution: a. n = 274 + 4 = 278 (leap year); ω = -7.5o; γ = 0o; β = 30o; Ф = 30.61oN b. The solar declination will be c. Given: n = 274 + 4 = 278 (leap year); ω = -7.5o; γ = 0o; β = 30o; Ф=30.61oN d. The declination is calculated as follows: 𝛿𝛿 = 23.45 × 𝑠𝑠𝑠𝑠𝑠𝑠 �360 ×

284 + 278 � = −5. 8𝑜𝑜 365

e. The incidence angle is then calculated as follows: 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝜙𝜙 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 − 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = sin(−5.8) sin(30.61) cos(30)) − sin(−5.8) cos (30.61) sin (30) cos (0) + cos (−5.8) cos (30.61) cos (30) cos (−7.5)

+ cos (−5.8) sin (30.61) sin (30) cos (0) cos (−7.5) + cos (−5.8) sin (30) sin (0) sin(−7.5)

32

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = (−0.1011)(0.5092)(0.866) − (−0.1011)(0.8607)(0.50)(1) + (0.9949)(0.8607)(0.866)(0.9914)

+ (0.9949)(0.5092)(0.50)(1)(0.9914) + (0.9949)(0.50)(0)(−0.1305)

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = −0.0446 + 0.0435 + 0.7352 + 0.2511 − 0 = 0.9852 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 0.9852 2.13.14

𝜃𝜃 = 𝐶𝐶𝐶𝐶𝐶𝐶 −1 (0.9852) = 9.87𝑜𝑜

Solar Radiation Received

P2.14 Calculate the percentage (in %) of solar radiation received on a pV surface that is in College Station, Texas at 11:30 am (solar time) on October 4 (leap year) if the surface is tilted 30o from the horizontal and oriented due South. The latitude of College Station is 30.61oN. a. 10 b. 98.5 c. 85.9 d. 59.8 Solution: a. The angle of incidence is calculated as follows: 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝜙𝜙 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 − 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = sin(−5.8) sin(30.61) cos(30)) − sin(−5.8) cos (30.61) sin (30) cos (0) + cos (−5.8) cos (30.61) cos (30) cos (−7.5)

+ cos (−5.8) sin (30.61) sin (30) cos (0) cos (−7.5) + cos (−5.8) sin (30) sin (0) sin(−7.5) 33

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = (−0.1011)(0.5092)(0.866) − (−0.1011)(0.8607)(0.50)(1) + (0.9949)(0.8607)(0.866)(0.9914)

+ (0.9949)(0.5092)(0.50)(1)(0.9914) + (0.9949)(0.50)(0)(−0.1305)

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = −0.0446 + 0.0435 + 0.7352 + 0.2511 − 0 = 0.9852 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 0.9852

b. This angle of incidence is also the percentage of solar energy received by the collector or 98.52% 2.13.15

Theoretical Daily Solar Radiation

P2.15 What is theoretical daily solar radiation outside of the earth’s atmosphere, Ho (in units of MJ/m2), the days solar radiation on a horizontal surface in the absence of interference in College Station, Texas on October 4 (leap-year). The latitude for College Station, Texas is 30.61o. a. 24.21 b. 34.5 c. 29.3 d. 20.2 Solution: a. n = 274 + 4 = 278 (a leap year); b. Gsc = 1,367 W/m2 c. 𝜙𝜙 = 30.61

d. Declination is then calculated as follows: 𝛿𝛿 = 23.45 × 𝑠𝑠𝑠𝑠𝑠𝑠 �360 ×

284 + 278 � = −5. 8𝑜𝑜 365 34

e. The sunset hour angle will be calculated as follows: 𝑐𝑐𝑐𝑐𝑐𝑐𝜔𝜔𝑠𝑠 = −𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

𝜔𝜔𝑠𝑠 = 𝑐𝑐𝑐𝑐𝑠𝑠 −1 (−𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)

𝜔𝜔𝑠𝑠 = 𝑐𝑐𝑐𝑐𝑠𝑠 −1 (−tan (30.61) × tan (−5.8)) = 𝑐𝑐𝑐𝑐𝑠𝑠 −1 (0.060096) = 86. 6𝑜𝑜

f. The remaining steps will just be plugging in of values calculated above as shown. 𝐻𝐻𝑜𝑜 =

360 × 278 24 × 3600 × 1367 × �1 + 0.033 × 𝑐𝑐𝑐𝑐𝑐𝑐 � 3.1416 365 × �cos (30.61) × cos (−5.8) × sin (86.6) +

3.1416 × 86.6 sin (30.61) × sin (−5.8)� 180

𝐻𝐻𝑜𝑜 = 37,595,111 × (1.002412)

× ( (0.86065) × (0.994881) × (0.99824)

+ (1.51146) × (0.509192) × (−0.1010563))

𝐻𝐻𝑜𝑜 = 37,595,111 × (1.002412) × (0.854737 − 0.077775) 𝐻𝐻𝑜𝑜 = 29,280,427

𝑀𝑀𝑀𝑀 𝐽𝐽 = 29.3 2 2 𝑚𝑚 𝑚𝑚

g. Thus, Ho is equal to 29.3 MJ/m2.

ASSESS YOUR KNOWLEDGE 1. In solar radiation calculation, this term refers to the angle between the vertical and the line to the sun, that is, the angle of incidence of beam radiation on a horizontal surface.

The symbol assigned is theta zed (θz). a. Hour angle b. zenith angle

35

c. angle of incidence d. solar altitude angle 2. In solar radiation calculation, this term refers to the angle between the horizontal and the line to the sun, that is, the complement of the zenith angle. The symbol assigned is alpha sub s (αs).

a. Hour angle b. angle of incidence c. solar altitude angle d. zenith angle 3. If the sunrise hour angle is -85 degrees, what time is sunrise (assume solar time)?

a. 6:20 am b. 5:40 am c. 7:40 am d. 7:20 am Solution: a. The sun moves at 15 degrees per hour and if 90 degrees is 6:00 am, then (90-85) degrees is 5 degrees is equivalent to 5/15 = 0.333 x 60 = 20 minutes. b. Hence, sunrise is 6:20 am.

4. This parameter is the time based on the apparent angular motion of the sun across the sky with solar noon the time the sun crosses the meridian of the observer. It does not coincide with local time. Name this nomenclature.

a. Standard time b. Declination c. solar time d. hour angle 36

5. Determine the solar time corresponding to 11:30 am Central Time on October 4 in College Station, Texas. The longitude of College Station is 96.32W. a. 12:07 am b. 11:21 am c. 11:16 am d. 12:12 pm

Solution: a. First, n is calculated to be equal to 273 + 4 = 277 b. Parameter B is first estimated as follows: 𝐵𝐵 = (𝑛𝑛 − 1) ×

360 360 = (277 − 1) × = 272. 22𝑜𝑜 365 365

c. E is then calculated as shown

𝐸𝐸 = 229.2 × (0.000075 + 0.00186 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 − 0.032077 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − 0.014615 cos 2𝐵𝐵 − 0.04089 sin 2𝐵𝐵)

𝐸𝐸 = 229.2 × (0.000075 + 0.00186 cos 272.22𝑜𝑜 − 0.032077 sin 272.22𝑜𝑜

− 0.014615 cos(2 × 272.22𝑜𝑜 ) − 0.04089 sin(2 × 272.22𝑜𝑜 )

𝐸𝐸 = 229.2 × (0.000075 + 0.00007205 + 0.03205 + 0.014571 + 0.004086) = 11.66 𝑚𝑚𝑚𝑚𝑚𝑚

d. The longitude correction will be = 4 x (90 - 96.32) = -25.28 min e. Then the solar time will be as follows:

Solar time = 11.5 hrs + (-25.28/60) + (11.66/60) = 11.27 [0.27/60 =16] f.

Thus, the solar time will be about 11:16 am

6. This nomenclature is considered the most important formula in solar energy course.

a. solar altitude angle b. Hour angle 37

c. zenith angle d. angle of incidence 7. This is one of the 8 thermodynamic pathways for solar energy conversion and began by the conversion of solar energy into thermal heat where chemicals are used to absorb this heat and release this heat for future use. These chemicals are simply termed absorbents whose phase may or may not change during absorption of solar energy. The classic example is the use of potassium oxide to absorb sunlight and later decomposed to generate the absorbed energy for other uses.

Name this thermodynamic pathway.

a. STEC + electrolysis b. solar thermochemical conversion c. solar thermo-mechanical conversion d. solar chemical conversion 8. Determine the solar time corresponding to 10:30 am Central Time on October 7, 2021 in College Station, Texas. The longitude of College Station is 96.32W.

a. 10:17 am b. 9:21 am c. 11.17.4 am d. 12:17 pm Solution: a. First, n is calculated to be equal to 273 + 7 = 280 b. Parameter B is first estimated as follows: 𝐵𝐵 = (𝑛𝑛 − 1) ×

360 360 = (280 − 1) × = 275. 2𝑜𝑜 365 365

c. E is then calculated as shown

38

𝐸𝐸 = 229.2 × (0.000075 + 0.00186 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 − 0.032077 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − 0.014615 cos 2𝐵𝐵 − 0.04089 sin 2𝐵𝐵)

𝐸𝐸 = 229.2 × (0.000075 + 0.00186 cos 275. 2𝑜𝑜 − 0.032077 sin 275. 2𝑜𝑜

− 0.014615 cos(2 × 275. 2𝑜𝑜 ) − 0.04089 sin(2 × 275. 2𝑜𝑜 )

𝐸𝐸 = 229.2 × (0.000075 + 0.0001686 + 0.031945 + 0.014375 + 0.007381) = 12.36 𝑚𝑚𝑚𝑚𝑚𝑚

d. The longitude correction will be = 4 x (90 - 96.32) = -25.28 min e. Then the solar time will be as follows: Solar time = 10.5 hrs + (-25.28/60) + (12.36/60) = 10.29 [0.29 x 60 = 17] f. Thus, the solar time will be about 10:17 am.

9. Determine the extra-terrestrial solar radiation to be received by the earth on October 4, 2021, in units of W/m2? a. 1370 b. 1389 c. 1367 d. 1353

Solution: a. The first step is to determine the value of n = 273 + 4 = 277 b. Then the Gon equation will be used as follows: 𝐺𝐺𝑜𝑜𝑜𝑜 = 𝐺𝐺𝑠𝑠𝑠𝑠 × �1 + 0.033 × 𝑐𝑐𝑐𝑐𝑐𝑐

360 × 𝑛𝑛 � 365

𝐺𝐺𝑜𝑜𝑜𝑜 = 1367 × �1 + 0.033 × 𝑐𝑐𝑐𝑐𝑐𝑐

c. Hence, Gon = 1370 W/m2.

360 × 277 𝑊𝑊 � = 1370 2 365 𝑚𝑚

39

10. Calculate the declination angle on September 29, 2021? a. -15.1o b. 3.82o c. 15.1o d. -3.42o

Solution: a. The first step is to find the value of n which is 243 + 29 = 272 b. Then the equation for declination is used as follows: 𝛿𝛿 = 23.45 × 𝑠𝑠𝑠𝑠𝑠𝑠 �360 ×

284 + 𝑛𝑛 284 + 272 � = 23.45 × 𝑠𝑠𝑠𝑠𝑠𝑠 �360 × � = −3.42𝑜𝑜 365 365

c. Hence, the solar declination-angle on this day is -3.42o.

11. In solar radiation calculation, this term refers to the angular displacement from south of the projection of beam radiation on the horizontal plane. The symbol assigned is gamma sub s (γs).

a. angle of incidence b. zenith angle c. solar altitude angle d. Solar surface azimuth angle 12. If the sunset hour angle is 85 degrees, what time is sunset (assume solar time)?

a. 6:20 pm b. 5:40 pm c. 5:20 pm d. 6:40 pm Solution:

40

a. The sun moves at 15 degrees per hour and if 90 degrees is 6:00 pm, then (90-85) degrees is 5 degrees and is equivalent to 5/15 = 0.333 x 60 = 20 minutes. b. Hence, sunrise is 6:00 pm- 20 min = 5:40 pm.

13. Determine the extra-terrestrial solar radiation to be received on September 29, 2021, in units of W/m2? a. 1389 b. 1366 c. 1353 d. 1386

Solution: a. The first step is to determine the value of n. For September 29, 2021 it is 243 + 29 = 272. b. Then the Gon equation will be used as follows: 𝐺𝐺𝑜𝑜𝑜𝑜 = 𝐺𝐺𝑠𝑠𝑠𝑠 × �1 + 0.033 × 𝑐𝑐𝑐𝑐𝑐𝑐

360 × 𝑛𝑛 � 365

𝐺𝐺𝑜𝑜𝑜𝑜 = 1367 × �1 + 0.033 × 𝑐𝑐𝑐𝑐𝑐𝑐

c. Hence, Gon = 1366 W/m2.

360 × 272 𝑊𝑊 � = 1366 2 365 𝑚𝑚

14. This is one of the 8 thermodynamic pathways for solar energy conversion and began by the conversion of solar energy into thermal heat followed by the conversion of heat into kinetic energy and electrical energy with the final step the conversion of a compound such as water into high energy rich chemical energy such as hydrogen gas. Name this pathway.

a. STEC + electrolysis b. solar thermal conversion c. solar electric conversion 41

d. solar chemical conversion 15. If the sunrise hour angle is -100 degrees, what time is sunrise (assume solar time)?

a. 6:20 am b. 5:40 am c. 5:20 am d. 7:20 am Solution: a. The sun moves at 15 degrees per hour and if -105 degrees is 5:00 am, then (105-100) degrees is 5 degrees is equivalent to 5/15 = 0.333 x 60 = 20 minutes b. Hence, sunrise is 5:20 am.

16. If the sunset hour angle is 100 degrees, what time is sunset (assume solar time)?

a. 6:20 pm b. 5:40 pm c. 5:20 pm d. 6:40 pm Solution: a. The sun moves at 15 degrees per hour and if 90 degrees is 6:00 pm, then (90-100) degrees is 10 degrees is equivalent to 10/15 = 0.667 x 60 = 40 minutes b. Hence, sunset is 6:40 pm.

17. What is theoretical daily solar radiation outside of the earth’s atmosphere, Ho (in units of MJ/m2), the days solar radiation on a horizontal surface in the absence of interference in College Station, Texas on October 4, 2016? The latitude for College Station, Texas is 30.61o.

a. 24.21 b. 34.5 42

c. 29.3 d. 20.2 Solutions: a. n = 274 + 4 = 278 (a leap year) b. Gsc = 1,367 W/m2 c. 𝜙𝜙 = 30.61

d. Declination is then calculated as follows: 𝛿𝛿 = 23.45 × 𝑠𝑠𝑠𝑠𝑠𝑠 �360 ×

284 + 278 � = −5. 8𝑜𝑜 365

e. The sunset hour angle will be calculated as follows: 𝑐𝑐𝑐𝑐𝑐𝑐𝜔𝜔𝑠𝑠 = −𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

𝜔𝜔𝑠𝑠 = 𝑐𝑐𝑐𝑐𝑠𝑠 −1 (−𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)

𝜔𝜔𝑠𝑠 = 𝑐𝑐𝑐𝑐𝑠𝑠 −1 (−tan (30.61) × tan (−5.8)) = 𝑐𝑐𝑐𝑐𝑠𝑠 −1 (0.060096) = 86. 6𝑜𝑜

f. The remaining steps will just be plugged in of the values calculated above as shown. 𝐻𝐻𝑜𝑜 =

24 × 3600 × 1367 360 × 278 × �1 + 0.033 × 𝑐𝑐𝑐𝑐𝑐𝑐 � 3.1416 365 × �cos (30.61) × cos (−5.8) × sin (86.6) +

3.1416 × 86.6 sin (30.61) × sin (−5.8)� 180

𝐻𝐻𝑜𝑜 = 37,595,111 × (1.002412)

× ( (0.86065) × (0.994881) × (0.99824)

+ (1.51146) × (0.509192) × (−0.1010563))

𝐻𝐻𝑜𝑜 = 37,595,111 × (1.002412) × (0.854737 − 0.077775) 𝐻𝐻𝑜𝑜 = 29,280,427

𝑀𝑀𝑀𝑀 𝐽𝐽 = 29.3 𝑚𝑚2 𝑚𝑚2

43

g. Thus, Ho is equal to 29.3 MJ/m2.

18. Determine the estimated number of daylight hours on September 29, 2016, in College Station, Texas. Assume latitude is 30.632 degrees.

a. 11.7 b. 10.94 c. 12.5 d. 13 19. What is the main difference between a natural convection solar water heater versus forced circulation water heater when it comes to location of storage tank?

a. More efficient b. tank is above the solar collector c. Less efficient d. Tank is below the solar collector 20. Calculate the percentage (in %) of solar radiation received on a pV surface that is in College Station, Texas at 11:30 am (solar time) on October 4, 2016, if the surface is tilted 30o from the horizontal and oriented due south. The latitude of College Station is 30.61oN.

a. 10 b. 98.5 c. 85.9 d. 59.8 Solution: a. n = 274 + 4 = 278 (leap year); ω = -7.5o; γ = 0o; β = 30o; Ф = 30.61oN b. The declination is calculated as follows: 𝛿𝛿 = 23.45 × 𝑠𝑠𝑠𝑠𝑠𝑠 �360 ×

284 + 278 � = −5. 8𝑜𝑜 365 44

c. The incidence angle is then calculated as follows: 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝜙𝜙 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 − 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = sin(−5.8) sin(30.61) cos(30)) − sin(−5.8) cos (30.61) sin (30) cos (0) + cos (−5.8) cos (30.61) cos (30) cos (−7.5)

+ cos (−5.8) sin (30.61) sin (30) cos (0) cos (−7.5) + cos (−5.8) sin (30) sin (0) sin(−7.5)

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = (−0.1011)(0.5092)(0.866) − (−0.1011)(0.8607)(0.50)(1) + (0.9949)(0.8607)(0.866)(0.9914)

+ (0.9949)(0.5092)(0.50)(1)(0.9914) + (0.9949)(0.50)(0)(−0.1305)

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = −0.0446 + 0.0435 + 0.7352 + 0.2511 − 0 = 0.9852 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 0.9852

𝜃𝜃 = 𝐶𝐶𝐶𝐶𝐶𝐶 −1 (0.9852) = 9.87𝑜𝑜

d. About 98.5% of the solar radiation is absorbed by the collector.

45

Chapter 3 Wind Energy 3.10 Solved Problems Problem 3.10.1 Kinetic Energy from Wind Problem 3.10.1 How much kinetic energy is contained in a 10 cubic meter of air moving at a speed of 10 m/s? Assume the air density to be 1.22 kg/m3.

Solution: a. The mass of air is first calculated as shown: 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 (𝑘𝑘𝑘𝑘) = 10 𝑚𝑚3 ×

1.22 𝑘𝑘𝑘𝑘 = 12.2 𝑘𝑘𝑘𝑘 𝑚𝑚3

b. The kinetic energy is then calculated using the equation discussed in the previous section.

1 𝐾𝐾𝐾𝐾 (𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽) = 𝑚𝑚𝑣𝑣 2 2 𝐾𝐾𝐾𝐾 =

10 𝑚𝑚 2 𝑁𝑁𝑠𝑠 2 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 1 × 12.2 𝑘𝑘𝑘𝑘 × � � × × = 610 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 2 𝑠𝑠 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑁𝑁𝑁𝑁

c. Note that all the units should cancel out to leave a unit of Joules.

Problem 3.10.2 Power from the Wind Problem 3.10.2 Determine the power that could be derived from a 4 m diameter windmill on a location with an average wind speed of 5 m/s. Assume the air density to be 1.22 kg/m3, a power coefficient of 0.45, and an efficiency of conversion to power of 60%. Express your units in Watts or kW. Solution: a. The windmill area is first calculated as shown: 𝜋𝜋𝑑𝑑 2 𝜋𝜋 × (4 𝑚𝑚)2 𝐴𝐴 (𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴) = = = 12.57 𝑚𝑚2 4 4

46

b. Power is then calculated using the equation discussed in previous sections. The theoretical power is first calculated followed by the actual power using the power coefficient as well as conversion efficiency. 𝑃𝑃𝑡𝑡 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) =

1 × 𝜌𝜌 × 𝐴𝐴 × 𝑣𝑣 3 2

𝑃𝑃𝑎𝑎 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) =

1 × 𝜌𝜌 × 𝐴𝐴 × 𝑣𝑣 3 × 𝐶𝐶𝑝𝑝 × 𝐸𝐸 2

1 1.22 𝑘𝑘𝑘𝑘 5 𝑚𝑚 3 𝑁𝑁𝑠𝑠 2 𝐽𝐽𝐽𝐽 𝑊𝑊𝑊𝑊 2 𝑃𝑃𝑡𝑡 (𝑊𝑊 ) = × × 12.57 𝑚𝑚 × � � × × × = 958.5 𝑊𝑊 3 2 𝑚𝑚 2 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑁𝑁𝑁𝑁 𝐽𝐽

𝑃𝑃𝑎𝑎 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = 958.8 W × 0.45 × 0.60 = 258.8 𝑊𝑊

c. Note that all the units should cancel out to leave a unit of Watts. If the overall efficiency is given, then there is no need to use the power coefficient.

Problem 3.10.3. Power Differential as Wind Speed is Doubled Problem 3.10.3 Compare the power difference in Watts for a location with an average wind speed of 5 m/s doubled to 10 m/s. Assume that the other parameters are the same, such as the air density of 1.22 kg.m3, area of 100 m2 and overall efficiency of 50 % (that includes the power coefficient and conversion efficiency for mechanical and electrical power).

Solution: a. Power is easily calculated using the equation discussed in previous section. 𝑃𝑃𝑎𝑎 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) =

1 × 𝜌𝜌 × 𝐴𝐴 × 𝑣𝑣 3 × 𝐸𝐸𝑜𝑜 2

1 𝑘𝑘𝑘𝑘 5 𝑚𝑚 3 𝑁𝑁𝑠𝑠 2 𝐽𝐽 𝑊𝑊𝑊𝑊 2 𝑃𝑃𝑎𝑎 (𝑊𝑊 ) = × 1.22 3 × 100 𝑚𝑚 × � � × 0.50 × × × 2 𝑚𝑚 𝑠𝑠 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑁𝑁𝑁𝑁 𝐽𝐽 = 3,812.5 𝑊𝑊

47

1 𝑘𝑘𝑘𝑘 10 𝑚𝑚 3 𝑁𝑁𝑠𝑠 2 𝐽𝐽 𝑊𝑊𝑊𝑊 2 𝑃𝑃𝑎𝑎 (𝑊𝑊 ) = × 1.22 3 × 100 𝑚𝑚 × � � × 0.50 × × × 2 𝑚𝑚 𝑠𝑠 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑁𝑁𝑁𝑁 𝐽𝐽 = 30,500 𝑊𝑊

∆𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = 30,500W − 3812 W = 26,687 W 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 =

30,500 𝑊𝑊 =8 3812.5 𝑊𝑊

b. Note that the wind speed is doubled but the power ratio is 8 times as much or there are 8 times more power when the speed is doubled.

Problem 3.10.4. Actual Power from Windmill Problem 3.10.4 Determine the actual power generated from a windmill with an effective diameter of 6 meters, air density of 1.225 kg/m3, average wind speed of 12 m/s and overall conversion efficiency of 30%.

Solution: a. Equation 3.7 is used as follows: 𝜋𝜋𝑑𝑑 2 𝜋𝜋 × (6 𝑚𝑚)2 𝐴𝐴 (𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴) = = = 28.27 𝑚𝑚2 4 4

1 𝑘𝑘𝑘𝑘 12 𝑚𝑚 3 𝑁𝑁𝑠𝑠 2 𝐽𝐽 𝑊𝑊𝑊𝑊 2 𝑃𝑃𝑎𝑎 (𝑊𝑊 ) = × 1.225 3 × 28.27 𝑚𝑚 × � � × 0.30 × × × 2 𝑚𝑚 𝑠𝑠 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑁𝑁𝑁𝑁 𝐽𝐽 = 8,976.3 𝑊𝑊

b. As such, the actual power is significantly lower due to mechanical and electrical conversion losses.

48

Problem 3.10.5 Rayleigh Distribution Estimate Problem 3.10.5 Estimate the value of c’ and the corrected c, representing the constants in Equations 3.8, 3.9, and 3.10, for the following wind speed data: N = 52; the sum of squares of all data points was 860.16.

2 4.5 3.5 0 1.8 3.0

7.1 2.6 2.0 0 4.8 3.5

0.6 1.0 3.9 2.0 5.5 5.1

3.5 4.5 7.0 0.1 4.4 7.0

3.2 5.0 0.8 8.0 2.8 1.4

2.0 3.2 4.0 5.6 0.7 1.6

3.0 1.6 5.1 1.8 4.8 1.6

0.9 8.1 10.0 2.8 4.0

1.8 2.1 0.5 7.0 0.6

Solution: a. The c’ is calculated using equation 3.12 as follows: 𝑐𝑐 ′ = �

𝑁𝑁 1 1 � 𝑥𝑥𝑖𝑖2 ≈ � × 860.16 = 2.8759 2𝑁𝑁 𝑖𝑖=1 2 × 52

b. The correction is calculated using Equation 3.13 𝑐𝑐 = 𝑐𝑐 ′ ×

4𝑁𝑁 𝑁𝑁! (𝑁𝑁 − 1)! √𝑁𝑁 452 52! (52 − 1)! √52 = 2.8759 × = 2.8804 (2𝑁𝑁)! √𝜋𝜋 (2 × 52)! √3.1416

c. The difference between the estimated value of c’ and the corrected value is quite insignificant.

Problem 3.10.6 Estimating Average Wind Speed from Rayleigh Distribution Problem 3.10.6 Estimate the value of average velocity for the Example problem 3.5 using the estimated formula given in Equation 3.9.

Solution: a. The average wind speed as calculated from Equation 3.9 is as follows:

49

𝑣𝑣̅ = 𝑐𝑐 ×

√3.1416 √𝜋𝜋 =2.8804× = 2.55 2 2

b. Note that the average value of wind speed is slightly less than the value of c. If one were to take the average value of all the wind speed data points in the above example, the average, using simple summation, is equal to 3.33. If we use this average value by simple summation, we would be overestimating the power and energy from the wind.

Problem 3.10.7 Average Wind Velocity for a Given Site and Hours of Occurrence Problem 3.10.7 The hourly wind speed data (in m/s) for one year in each location was analyzed to establish the Weibull shape factor k = 2.04 and scale factor c = 5.7. (a) What is the average velocity for this site? (b) Estimate the number of hours per year that the wind speed is 5 m/s? (c) Estimate the number of hours per year that the wind speed is less than 5 m/s.

Solution: a. The average velocity for this site is calculated using the simple formula given in Equation 3.21 as follows: 𝑣𝑣̅ = 𝑐𝑐

𝜋𝜋 3.1416 𝑚𝑚 = 5× = 7.85 2 2 𝑠𝑠

a. Equation 3.14 is used to estimate the number of hours per year that the wind speed has a value of 5 m/s. The MS Excel spreadsheet software may also be used with the following syntax — “Weibull(5.7,2.04,5,False)*8760”. This will give a value of 1109 hrs. The use of calculator is shown below with slight difference. 𝑘𝑘 𝑣𝑣 (𝑘𝑘−1) 𝑣𝑣 𝑘𝑘 𝑓𝑓(𝑣𝑣) = � � � � 𝑒𝑒𝑒𝑒𝑒𝑒 �− � � � 𝑐𝑐 𝑐𝑐 𝑐𝑐 𝑓𝑓(5) = �

2.04 5 (2.04−1) 5 2.04 �− �� � 𝑒𝑒𝑒𝑒𝑒𝑒 � � � = 0.145 5.7 5.7 5.7 50

𝑓𝑓(5) = (0.358) × (0.873) × (0.465) = 0.145 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = 𝑓𝑓(𝑣𝑣) × 8,760 ℎ𝑟𝑟𝑟𝑟 = 1,270 ℎ𝑟𝑟𝑟𝑟

𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = (0.145) × 8,760 ℎ𝑟𝑟𝑟𝑟 = 1,270 ℎ𝑟𝑟𝑟𝑟

Note that the MS Excel results may be quite different from calculator value as shown above. MS Excel returns a value of 1109.1 hours, around 12.7% difference.

b. The cumulative density function Equation 3.15 will be used to estimate the hours that wind speed is less than 6 m/s as follows: 𝑣𝑣 𝑘𝑘

𝐹𝐹(𝑣𝑣) = 1 − 𝑒𝑒𝑒𝑒𝑒𝑒 �− � � � 𝑐𝑐

5

𝐹𝐹(𝑣𝑣) = 1 − 𝑒𝑒𝑒𝑒𝑒𝑒 �− � � 5.7

2.04

� = 1 − 0.4651 = 0.5349

𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = F(𝑣𝑣) × 8,760 ℎ𝑟𝑟𝑟𝑟 = 4,685 ℎ𝑟𝑟

𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = (0.5349) × 8,760 ℎ𝑟𝑟𝑟𝑟 = 4,685 ℎ𝑟𝑟𝑟𝑟

If one were to use MS Excel spreadsheet software, the answer will be using the syntax “Weibull(5,2.04,5.7,True)*8760” equal to 4,685 hrs. Answer is very close to calculator value.

Problem 3.10.8. Estimate Weibull Parameters k and C from Linear Regression Data Problem 3.10.8 In linear regression calculations, the following results were provided: intercept = -3.3172; X variable = 2.1588. Calculate the parameters k and c from these given data. Also estimate the average value of wind speed for this location.

Solution: a. Following Equation 3.17, we know that the parameter k is simply equal to the slope of the linear regression analysis or the x variable. Hence k = 2.1588.

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b. To calculate for the constant c, Equation 3.18 will be used as follows:

c.

𝑐𝑐 = 𝑒𝑒𝑒𝑒𝑒𝑒 �

𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 −3.3172 � = 𝑒𝑒𝑒𝑒𝑒𝑒 � � = 4.6487 −𝑘𝑘 −2.1588

The average value of wind speed for this location may be found using the approximate

relationship given in Equation 3.21 as shown below. 𝑣𝑣̅ = 𝑐𝑐

𝜋𝜋 3.1416 𝑚𝑚 = 4.6487 × = 7.30 2 2 𝑠𝑠

Problem 3.10.9 Wind Speed at Different Elevation Problem 3.10.9 The wind speed at a reference height of 10 meters was 8 m/s. Predict the expected wind speed at 40 meters, assuming the windmill is installed in an area with a few trees.

Solution: a. From Table 3.1, the roughness coefficient was found to be 100 mm. Hence from Equation 3.23, we calculate the value of alpha as follows: 𝛼𝛼 = 0.096 × 𝑙𝑙𝑙𝑙𝑙𝑙10 𝑧𝑧𝑜𝑜 + 0.016 × (𝑙𝑙𝑙𝑙𝑙𝑙10 𝑧𝑧𝑜𝑜 )2 + 0.24

2

𝛼𝛼 = 0.096 × 𝑙𝑙𝑙𝑙𝑙𝑙10 (100⁄100) + 0.016 × �𝑙𝑙𝑙𝑙𝑙𝑙10 (100⁄100)� + 0.24 𝛼𝛼 = 0.24

b. Thus, using the power law Equation 3.22, the velocity at a 40-meter height was found to be as follows: 𝑉𝑉(𝑧𝑧)

𝑧𝑧 𝛼𝛼

=� � 𝑉𝑉(𝑧𝑧 ) 𝑟𝑟

𝑧𝑧𝑟𝑟

𝑧𝑧 𝛼𝛼 40 0.24 𝑚𝑚 𝑉𝑉 (𝑧𝑧) = 𝑉𝑉(𝑧𝑧𝑟𝑟 ) × � � = 8 × � � = 11.2 𝑧𝑧𝑟𝑟 10 𝑠𝑠

52

Problem 3.10.10. Payback period for Wind Machine Problem 3.10.10 Estimate the simple payback period for a small wind turbine with the following data. The average annual electricity consumption was 12,000 kWh; the average price per kWh was 10 cents, and the household invested on a 15-kW system that costs $91,500. Determine the following: a. How much does each household pay for electricity in a year b. What is the payback period using these assumptions? Comment on this value assuming a lack of government subsidies.

Solution: a. The amount being paid by each household is simply the product of its consumption and how much they pay per kWh; simply calculated below: Yearly payment for electricity = $0.10/kWh x 12,00 kWh = $1,200/yr b. If the payback period is simply based on the payment for electricity, then the calculation will be the ratio of the capital cost and the amount of electricity the household pays each year as follows: Payback period = $91,500/$1,200/yr =76.25 years c. The payback period is quite unattainable without a subsidy.

3.12.11

Wind Energy Parameter

P3.11 This wind parameter compares the windmill’s actual production over a given period with the amount of power the plant would have produced if the windmill had run at full capacity for the same amount of time. a. reliability

53

b. availability c. efficiency d. capacity factor

3.12.12

Wind Energy Parameter

P3.12 This wind energy term is a measurement of the reliability of a wind turbine. It refers to the percentage of time that a plant is ready to generate (that is, not out of service for maintenance or repairs) electrical power. a. availability b. accuracy c. capacity factor d. efficiency

3.12.13

Wind Speed at Different Height

P3.13 The wind speed from a nearby airport MET station at a standard height of 10 meters was reported at 12 m/s. What is the wind speed (in m/s) at 30 meters? Assume the terrain has lawn grass. Use the power law as the governing equation for this problem. a. 13.5 b. 36 c. 15.6 d. 20 Solution:

54

a. From table, the surface roughness for lawn grass was found to be 8 mm or 8/1000 mm/m = 0.008 m. b. Calculate the roughness coefficient first 2

8 8 𝛼𝛼 = 0.096 × 𝑙𝑙𝑙𝑙𝑙𝑙10 � � + 0.016 × �𝑙𝑙𝑙𝑙𝑙𝑙10 � �� + 0.24 = 0.109 1000 1000

c. Using the power law equation, the new wind speed is simply calculated as follows: 𝑧𝑧 𝛼𝛼 𝑚𝑚 30 0.109 𝑚𝑚 𝑉𝑉 (𝑧𝑧) = 𝑉𝑉(𝑧𝑧𝑟𝑟 ) � � = 12 × � � = 13.5 𝑧𝑧𝑟𝑟 𝑠𝑠 10 𝑠𝑠 3.12.14

Weibull Distribution

P3.14 The Hourly wind speed data (m/s) for one year was analyzed to establish the Weibull shape factor k and scale factor c as follows: k = 2 and c = 4.5. What is the average velocity for this site in m/s? Use the approximate equation for the calculations. a. 2 b. 7.07 c. 3.99 d. 6.5 Solution: a. The average velocity for this site is given by the parameter c = 4.5 m/s. 𝑣𝑣̅ = 𝑐𝑐 3.12.15

𝜋𝜋 3.1416 = 4.5 × = 7.07 2 2 Weibull Distribution

P3.15 The hourly wind speed data for one year was analyzed to establish the Weibull shape factor k and scale factor c as follows: k = 2 and c = 4.5. Estimate the number of hours per year 55

that the wind speed is around the average value. Please use the probability density function (pdf) formula for the calculations. a. 573 b. 1795 c. 518 d. 7178 Solution: a. The average wind speed for this site is given by the parameter c = 4.5 m/s. 𝑣𝑣̅ = 𝑐𝑐

𝜋𝜋 3.1416 = 4.5 × = 7.07 2 2

b. The number of hours per year is estimated using the probability density function f(v) as shown below. 𝑣𝑣 𝑘𝑘 𝑘𝑘 𝑣𝑣 (𝑘𝑘−1) 𝑓𝑓 (𝑣𝑣) = � � � � 𝑒𝑒𝑒𝑒𝑒𝑒 �− � � � 𝑐𝑐 𝑐𝑐 𝑐𝑐

7.07 (2−1) 2 7.07 2 𝑓𝑓 (𝑣𝑣) = � � � � 𝑒𝑒𝑒𝑒𝑒𝑒 �− � � � = 0.059158 4.5 4.5 4.5

c. Hence the number of hours that the velocity is around the average velocity value of 7.07 m/s is as follows: 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (𝑣𝑣 = 7.07, ℎ𝑟𝑟𝑟𝑟) = 8760 × 0.059158 = 518 ℎ𝑟𝑟𝑟𝑟 Assess Your Knowledge 1. This International Standard published by IEC sets the design requirements for wind turbines to make sure that wind turbines are appropriately engineered against damage from hazards within the planned lifetime. Identify this IEC standards number.

a. IEC 41600 56

b. ASTM 6140 c. IEC 4160 d. IEC 61400 2. This technical term determines which turbine is suitable for the normal wind conditions of a particular site. They are mainly defined by the average annual wind speed (measured at the turbine's hub height), the speed of extreme gusts that could occur over 50 years, and how much turbulence is there at the wind site. Identify this wind energy related term.

a. wind standard b. wind regime c. wind map d. wind classes 3. Which organization sets international standards for the wind speeds each wind class must withstand? This organization has 4 wind classes I-IV (high wind to very low wind).

a. The International Wind Energy Standards Organization b. The International Electrotechnical Commission. c. The International Energy Administration (IEA) d. The American Wind Energy Association (AWEA) 4. What is the value of Betz Coefficient or the maximum extractable energy from the wind using momentum theory? a. 0.593 b. 0.45 c. 0.75 d. 0.335 5. How many wind classes are set by the International Electrotechnical Commission?

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a. 1 b. 2 c. 4 d. 6 6. What was the primary use of large windpumps developed by the Dutch along the Rhine River Delta?

a. grinding grains b. cutting wood c. draining lakes d. small wind electric generators 7. Which of the following are the early recorded uses of windmills?

a. grinding grain b. generating electricity c. jousting d. cooling system for houses in summer 8. What do you call the wind speed at which the turbine starts to generate power?

a. cut-in b. cut-out c. rated d. optimal wind speed 9. As early as 5000 BC, people used to do what along the Nile River in Egypt?

a. irrigation and agriculture b. propel boats c. build pyramids 58

d. grind grain 10. What do you call the wind speed at which the turbine is shut down to keep loads and generator from being damaged?

a. cut-in b. cut-out c. rated d. optimal wind speed 11. Actual Wind Power. Determine the actual power (in units of kW) from a mass of wind passing through a cross-sectional area of 100 m2 and with a speed of 12 m/s. Assume the density of wind is 1.225 kg/m3 and overall conversion efficiency of 45%.

a. 3970 b. 6.1 c. 3.97 d. 47.6 Solution a. The wind power equation will be used as follows: 1

b. 𝑃𝑃𝑡𝑡 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = × 𝜌𝜌 × 𝐴𝐴 × 𝑣𝑣 3 2 1

c. 𝑃𝑃𝑡𝑡 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = × 2

1.225 𝑘𝑘𝑘𝑘 𝑚𝑚 3

𝑚𝑚 3

𝑁𝑁×𝑠𝑠 2

× 100 𝑚𝑚2 × �12 � × 𝑘𝑘𝑘𝑘×𝑚𝑚 ×

d. 𝑃𝑃𝑡𝑡 (𝑘𝑘𝑘𝑘) = 105,840 Watts ×

𝑠𝑠

kW

1,000 𝑊𝑊

= 105.8 𝑘𝑘𝑘𝑘

𝑊𝑊×𝑠𝑠

𝑁𝑁×𝑚𝑚

= 105,840 𝑊𝑊

e. The actual power is calculated below:

f. 𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) = 𝑃𝑃𝑡𝑡 (𝑘𝑘𝑘𝑘) × 𝐸𝐸𝑜𝑜 = 105.8 kW × 0.45 = 47.6 𝑘𝑘𝑘𝑘

12. Kinetic Energy. Determine the kinetic energy available in 150 kg mass of wind moving at a speed of 12 m/s. Express the units in Joules.

59

a. 10,800 b. 5,000 c. 15,600 d. 2,500 Solution: a. The kinetic energy of motion is given by the following equation 1

𝐾𝐾𝐾𝐾 = 𝑚𝑚𝑣𝑣 2 2

b. Thus, the KE from this mass of wind and given speed is as follows: 1 𝑚𝑚 2 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝐾𝐾𝐾𝐾 = {𝑘𝑘𝑘𝑘} � � = × 𝑚𝑚 = 𝑁𝑁 × 𝑚𝑚 = 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 𝑠𝑠 𝑠𝑠 2 2

1 𝑚𝑚 2 𝑁𝑁 × 𝑠𝑠 2 = 10,800 𝑁𝑁 × 𝑚𝑚 = 10,800 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 𝐾𝐾𝐾𝐾 = {150 𝑘𝑘𝑘𝑘} �12 � × 𝑠𝑠 𝑘𝑘𝑘𝑘 × 𝑚𝑚 2

13. Rayleigh Distribution. The average wind speed in an area was found to be 8 m/s. Determine the number of hours in a year (365 days, 24 hrs per day or 8760 hrs) that the wind speed is less than 8 m/s. Assume the wind velocity distribution follows the Rayleigh Probability Distribution.

a. 2,847 b. 5,913 c. 7,787 d. 973 Solution: a. The parameter c is first calculated such that the Rayleigh CDF may be calculated as follows: 𝑣𝑣̅ = 𝑐𝑐 ×

√𝜋𝜋 2 60

c=

2 × 𝑣𝑣̅ √𝜋𝜋

=

2× 8

√3.1416

= 9.03

b. Then, parameter c is substituted at the CDF as follows: 𝐹𝐹(𝑣𝑣) = 1 − 𝑒𝑒𝑒𝑒𝑒𝑒 �

−𝑣𝑣 2 −(8)2 ( ) � = 𝐹𝐹 𝑣𝑣 = 1 − 𝑒𝑒𝑒𝑒𝑒𝑒 � � = 0.325 2𝑐𝑐 2 2(9.03)2

c. The time that a certain speed will happen is calculated as follows: 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 �𝑣𝑣 < 8 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 �𝑣𝑣 > 8

𝑚𝑚 � = 8760 ℎ𝑟𝑟𝑟𝑟 × 0.325 = 2,847 ℎ𝑟𝑟𝑟𝑟 𝑠𝑠

𝑚𝑚 � = 8760 ℎ𝑟𝑟𝑟𝑟 − 2,847 ℎ𝑟𝑟𝑟𝑟 = 5,913 ℎ𝑟𝑟𝑟𝑟 𝑠𝑠

d. Hence, in a year of 8760 hrs, the number of hours that the wind speed is less than 8 m/s is 2,847 hrs. Thus, the number of hours that the wind speed is greater than 8 m/s must be the difference or 5913 hrs.

14. Theoretical Wind Power. Determine the available theoretical power (in units of kW) from a mass of wind passing through a cross-sectional area of 100 m2 and with a speed of 12 m/s.

Assume the density of wind is 1.225 kg/m3.

a. 105,840 b. 8.82 c. 105.8 d. 8,820 Solution a. The wind power equation will be used as follows: 𝑃𝑃𝑡𝑡 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) =

1 × 𝜌𝜌 × 𝐴𝐴 × 𝑣𝑣 3 2

1 1.225 𝑘𝑘𝑘𝑘 𝑚𝑚 3 𝑁𝑁 × 𝑠𝑠 2 𝑊𝑊 × 𝑠𝑠 2 𝑃𝑃𝑡𝑡 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = × × 100 𝑚𝑚 × �12 � × × = 105,840 𝑊𝑊 2 𝑚𝑚3 𝑠𝑠 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑁𝑁 × 𝑚𝑚 61

𝑃𝑃𝑡𝑡 (𝑘𝑘𝑘𝑘) = 105,840 Watts ×

kW = 105.8 𝑘𝑘𝑘𝑘 1,000 𝑊𝑊

15. Weibull Distribution. Hourly wind speed data for one year was analyzed to establish the Weibull shape factor k and scale factor c as follows: k = 2.5 and c = 4.5. What is the average wind speed for this site? Use the simpler approximate formula to calculate the average wind speed.

a. 2 b. 4 c. 7.07 d. 8 Solution: a. The average wind speed for this site is given by the parameter c = 4.5 m/s. 𝑣𝑣̅ = 𝑐𝑐

𝜋𝜋 3.1416 𝑚𝑚 = 4.5 × = 7.07 2 2 𝑠𝑠

16. Kinetic Energy. Determine the kinetic energy available in 150 kg mass of wind moving at a speed of 12 m/s? Express the units in Btu.

a. 10,800 b. 10.2 c. 15,600 d. 2.5 Solution: a. The kinetic energy of motion is given by the following equation 1

𝐾𝐾𝐾𝐾 = 𝑚𝑚𝑣𝑣 2 2

b. Thus, the KE from this mass of wind and given speed is as follows:

62

1 𝑚𝑚 2 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝐾𝐾𝐾𝐾 = {𝑘𝑘𝑘𝑘} � � = × 𝑚𝑚 = 𝑁𝑁 × 𝑚𝑚 = 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 2 𝑠𝑠 𝑠𝑠 2

1 𝑚𝑚 2 𝑁𝑁 × 𝑠𝑠 2 𝐾𝐾𝐾𝐾 = {150 𝑘𝑘𝑘𝑘} �12 � × = 10,800 𝑁𝑁 × 𝑚𝑚 = 10,800 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 2 𝑠𝑠 𝑘𝑘𝑘𝑘 × 𝑚𝑚

c. In English units, this is also equivalent to the following Btu calculations. 𝐾𝐾𝐾𝐾 = 10,800 J ×

𝐵𝐵𝐵𝐵𝐵𝐵

1055 𝐽𝐽

= 10.2 𝐵𝐵𝐵𝐵𝐵𝐵

17. Wind Power Production Terminology. This wind parameter compares the windmill’s actual production over a given period with the amount of power the plant would have produced if the windmill had run at full capacity for the same amount of time.

a. reliability b. availability c. efficiency d. capacity factor 18. Wind Energy Production Terminology This wind energy term is a measurement of the reliability of a wind turbine. It refers to the percentage of time that a plant is ready to generate (that is, not out of service for maintenance or repairs) electrical power. a. availability b. accuracy c. capacity factor d. efficiency 19. Weibull Distribution. Hourly wind speed data for one year (365 days, 24 hrs per day or 8760 hrs) was analyzed to establish the Weibull shape factor k and scale factor c as follows: k = 2.5 and c = 4.5. Estimate the number of hours per year that the wind speed is around 8 m/s.

a. 3,501 b. 8,630 63

c. 1,073 d. 170.6 Solution: a. The number of hours per year is estimated using the probability density function f(v) as shown below. 𝑘𝑘 𝑣𝑣 (𝑘𝑘−1) 𝑣𝑣 𝑘𝑘 𝑓𝑓(𝑣𝑣) = � � � � 𝑒𝑒𝑒𝑒𝑒𝑒 �− � � � 𝑐𝑐 𝑐𝑐 𝑐𝑐 𝑓𝑓(𝑣𝑣) = �

8 (2.5−1) 2.5 8 2.5 �� � 𝑒𝑒𝑒𝑒𝑒𝑒 �− � � � = 0.017473 4.5 4.5 4.5

b. Hence, the approximate hours that the wind speed is around 8 m/s is calculated as follows: 𝑚𝑚

c. 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 �𝑣𝑣 = 8 � = 8760 ℎ𝑟𝑟𝑟𝑟 × 0.017473 = 170.6 ℎ𝑟𝑟𝑟𝑟 𝑠𝑠

d. The syntax when using MS Excel is as follows:

𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊(8, 2.5, 4.5, 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹) × 8760 = 170.6 ℎ𝑟𝑟𝑟𝑟

20. Weibull Distribution. Hourly wind speed data for one year (365 days, 24 hrs per day or 8760 hrs) was analyzed to establish the Weibull shape factor k and scale factor c as follows: k = 2.5 and c = 4.5. Estimate the number of hours per year that the wind speed is above 8 m/s.

a. 8,630 b. 130 c. 170.6 d. 985 Solution: a. The cumulative density function may be used to estimate the hours that the wind speed is above a certain value as shown.

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𝑣𝑣 𝑘𝑘

𝐹𝐹(𝑣𝑣) = 1 − 𝑒𝑒𝑒𝑒𝑒𝑒 �− � � � 𝑐𝑐

𝑚𝑚

8

2.5

𝐹𝐹 �𝑣𝑣 < 8 � = 1 − 𝑒𝑒𝑒𝑒𝑒𝑒 �− � � � = 1 − 0.014787 = 0.985213 𝑠𝑠

4.5

b. Hence, the probability that the wind speed is less than 8 m/s is around 98.5%. Thus, the total hours is calculated as follows: 𝑚𝑚

c. 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 �𝑣𝑣 < 8 � = 8,760 ℎ𝑟𝑟𝑟𝑟 × 0.985213 = 8,630 ℎ𝑟𝑟𝑟𝑟 𝑠𝑠

d. The syntax in MS Excel is as follows:

𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊(8, 2.5, 4.5, 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇) × 8760 = 8,630 ℎ𝑟𝑟𝑟𝑟

e. The hours when the velocity is greater than 8 m/s is the difference as follows: 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = 8,760 − 8,630 = 130 ℎ𝑟𝑟𝑟𝑟

65

Chapter 4 Biomass Energy 4.8 Solved Problems Problem 4.8.1 Area Required to Build a Power Plant Problem 4.8.1 Determine the area needed to supply the yearly fuel requirement of a 3MW power plant using wood as feedstock. The power plant should operate continuously for at least 365 days in a year. The heating value of wood was found to be 22 MJ/kg and the conversion efficiency via steam generation was assumed to be 50%. The yield of the wood plantation is around 4 tonnes/hectare/year.

Solution: a.

The amount of biomass required may be calculated using the efficiency equation as follows:

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸(%)

=

𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼(𝑀𝑀𝑀𝑀)

=

50%

=

𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 × 100% 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 × 100% 3 𝑀𝑀𝑀𝑀 1 𝑀𝑀𝑀𝑀 = 6 𝑀𝑀𝑀𝑀 0.50

b. About 6MW of input feedstock is needed to operate this power plant. c. The next step is to determine the daily and yearly requirement of this biomass, assuming the heating value as well as the length of operation of the power plant in a year. This is shown below.

66

𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑀𝑀𝑀𝑀 𝑘𝑘𝑘𝑘 3,600 𝑠𝑠 24 ℎ𝑟𝑟𝑟𝑟 1 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 � = 6 𝑀𝑀𝑀𝑀 × × × × × = 23.56 𝑑𝑑𝑑𝑑𝑑𝑑 1 𝑀𝑀𝑀𝑀 − 𝑠𝑠 22 𝑀𝑀𝑀𝑀 ℎ𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑 1,000 𝑘𝑘𝑘𝑘 𝑑𝑑𝑑𝑑𝑑𝑑 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 1,000 𝑘𝑘𝑘𝑘 2.2 𝑙𝑙𝑙𝑙𝑙𝑙 𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 � � = 23.56 × × × = 25.92 𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 1 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 1 𝑘𝑘𝑘𝑘 2,000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑𝑑𝑑 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 � � = 23.56 × = 8,600 𝑦𝑦𝑦𝑦 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 � � = 25.92 × = 9,461 𝑦𝑦𝑦𝑦 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 �

d. The next step is to estimate the required area given the productivity of the wood fuel that will be used for the power plant as shown below. 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 � 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 �

ℎ𝑎𝑎 8,600 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 ℎ𝑎𝑎 − 𝑦𝑦𝑦𝑦 �= × = 2,150 ℎ𝑎𝑎 𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦 4 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 2,150 ℎ𝑎𝑎 2.47 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 �= × = 5,311 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦 1 ℎ𝑎𝑎

e. Hence, more than 2,150 hectares [5,311 acres] is required to satisfy the yearly requirement of this power plant fueled by wood.

Problem 4.8.2. Electrical Power from MSW Problem 4.8.2 An analysis of segregated municipal solid waste (MSW) by Balcones Resources of Dallas, Texas showed that the energy content of the segregated and combustible biomass from the garbage was about 20 MJ/kg. In a city like Dallas with a population of 1.3 million people, estimate the amount of garbage produced in tonnes per day if the per capita production is 2.5 kg/person/day. If this biomass is converted into electrical power with an overall conversion efficiency of 15%, what would be the output power in MW, assuming you operate 24 hrs/day.

Solution: a.

The amount of garbage generated per day in tonnes is calculated as shown.

67

𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 2.5 𝑘𝑘𝑘𝑘 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 � = 1.3 × 106 ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 × × = 3,250 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑 ℎ𝑑𝑑 − 𝑑𝑑𝑑𝑑𝑑𝑑 1,000 𝑘𝑘𝑘𝑘 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 1.1 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑀𝑀𝑀𝑀𝑀𝑀 � � = 3,250 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 × = 3,575 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑

𝑀𝑀𝑀𝑀𝑀𝑀 � b.

In each day, the total energy contained in this biomass is calculated as follows:

𝑀𝑀𝑀𝑀 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 1,000 𝑘𝑘𝑘𝑘 20 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 � = 3,250 × × = 65 × 106 𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 1 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑘𝑘𝑘𝑘 𝑑𝑑𝑑𝑑𝑑𝑑 6 6 10 𝐵𝐵𝐵𝐵𝐵𝐵 𝑀𝑀𝑀𝑀 1 × 10 𝐽𝐽 𝐵𝐵𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵𝐵𝐵 𝑀𝑀𝑀𝑀𝑀𝑀 � � = 65 × 106 × × = 6.2 × 1010 𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 1 𝑀𝑀𝑀𝑀 1,055 𝐽𝐽 𝑑𝑑𝑑𝑑𝑑𝑑 𝑀𝑀𝑀𝑀𝑀𝑀 �

c.

The electrical power output is calculated using the conversion efficiency provided and the continuous operation in a day as follows:

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃(𝑀𝑀𝑀𝑀) =

d.

65 × 106 𝑀𝑀𝑀𝑀 1 𝑑𝑑𝑑𝑑𝑑𝑑 1 ℎ𝑟𝑟 𝑀𝑀𝑀𝑀 − 𝑠𝑠 × × × × 0.15 = 113 𝑀𝑀𝑀𝑀 𝑑𝑑𝑑𝑑𝑑𝑑 24 ℎ𝑟𝑟𝑟𝑟 3,600 𝑠𝑠 𝑀𝑀𝑀𝑀

More than 110 MW power generation system may be installed in the city of Dallas using MSW alone.

Problem 4.8.3 Feedstock Requirement for a 3 MGY Biodiesel Plant Problem 4.8.3 Make an estimate of the vegetable oil requirement (in kg or lbs per year) for a three-million-gallon-per-year (11.355 million liter per year) biodiesel facility using soybean oil.

Use the governing Equation 4.1 for your calculations. Assume that the density of biodiesel is 7.3 lbs/gal [877 kg/m3].

Solution: a. Since the governing equation is in units of weight and the biodiesel plant capacity in units of volume, the density units are important. One should begin by converting the volume requirement into weight units as follows: 68

𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 �

𝑘𝑘𝑘𝑘 11.355 × 106 𝐿𝐿 877 𝑘𝑘𝑘𝑘 1 𝑚𝑚3 �= × × 𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦 𝑚𝑚3 1,000 𝐿𝐿 𝑘𝑘𝑘𝑘 𝑙𝑙𝑙𝑙𝑙𝑙 = 9.958 × 106 �21.908 × 106 � 𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦

b. The above calculation determines that one would need equal amounts of soybean oil in a year as well.

𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 �

𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑘𝑘𝑘𝑘 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 � = 9.958 × 106 × = 9,958 tonnes/yr �10,954 � 𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦 1,000𝑘𝑘𝑘𝑘 𝑦𝑦𝑦𝑦

c. One would need at least 9,958 metric tonnes of soybean oil each year for this biodiesel plant [10,954 tons/yr].

Problem 4.8.4 Sugar Needed to Produce Ethanol Problem 4.8.4 Determine the amount of ethanol produced (liters) per kg of sugar used using the governing Equation 4.2. Use both English and Metric units. Assume the density of ethanol to be 0.8 kg/L. Solution: a. The amount of sugar needed (kg) for every L of ethanol produced is shown below: 𝐿𝐿 𝑒𝑒𝑒𝑒ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐿𝐿 2 𝑚𝑚𝑚𝑚𝑚𝑚 × [(12 × 2) + (1 × 6) + (16 × 1)]1 92 𝑘𝑘𝑘𝑘 L � �= = × 𝑘𝑘𝑘𝑘 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑘𝑘𝑘𝑘 1 𝑚𝑚𝑚𝑚𝑚𝑚 × [(12 × 6) + (1 × 2) + (16 × 6)] 180 𝑘𝑘𝑘𝑘 0.8 𝑘𝑘𝑘𝑘 𝐿𝐿 = 0.64 𝑘𝑘𝑘𝑘 b.

In English units, the calculations are as follows: 𝐿𝐿 𝑒𝑒𝑒𝑒ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐿𝐿 𝐿𝐿 𝑔𝑔𝑔𝑔𝑔𝑔 1𝑘𝑘𝑘𝑘 𝑔𝑔𝑔𝑔𝑔𝑔 � � = 0.64 × × = 0.077 𝑘𝑘𝑘𝑘 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 3.785𝐿𝐿 2.2 𝑙𝑙𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙 69

Problem 4.8.5. Biogas Digester Sizing Problem 4.8.5 Determine the size of the digester for an experiment that has the following results: retention time is 30 days; the type of manure is dairy manure without co-digestion. The dairy farm population is 500 heads and approximately 25 kg [55 lbs] of fresh manure is generated each day. Use equal amounts of water mixed with fresh manure in the digester. Assume the density of slurry to be around 1.2 kg/L. Solution: a. The amount of slurry generated and to be input or fed to the digester is shown below. 𝑘𝑘𝑘𝑘 25 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 � = 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 � � × 500 ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 × 2 = 25,000 𝑑𝑑𝑑𝑑𝑑𝑑 ℎ𝑒𝑒𝑒𝑒𝑒𝑒 − 𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙𝑙𝑙 𝑘𝑘𝑘𝑘 2.2 𝑙𝑙𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙𝑙𝑙 𝑆𝑆𝑆𝑆𝑆𝑆r𝑟𝑟𝑟𝑟 � � = 25,000 × = 55,000 𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 1 𝑘𝑘𝑘𝑘 𝑑𝑑𝑑𝑑𝑑𝑑

𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 �

b. The digester size is calculated as follows:

𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 (𝑚𝑚3 )

𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 (𝑓𝑓𝑓𝑓 3 )

= 25,000 =

𝐿𝐿 1 𝑚𝑚3 𝑘𝑘𝑘𝑘 × 30 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 × × = 625 𝑚𝑚3 𝑑𝑑𝑑𝑑𝑑𝑑 1.2 𝑘𝑘𝑘𝑘 1,000 𝐿𝐿 3.28 𝑓𝑓𝑓𝑓 3 3 625 𝑚𝑚 × � � = 22,055 𝑓𝑓𝑓𝑓 3 1 𝑚𝑚

c. Free-board must be added to the digester size to ensure that the reactor will not be overfilled during times when manure production is high.

Problem 4.8.6 Residence Time for Biomass Conversion in Fluidized Bed Reactors Problem 4.8.6 Determine the residence time for the biomass being thermally converted in a fluidized bed pyrolyzer. The reactor size is 15.24 cm [6 inches] and 60.96 cm [24 inches tall] with a volumetric flow rate of 5.56 L/s [11.77 cubic feet per minute, cfm].

70

Solution: a. Using Equation 4.4, the units must be the same and the reactor volume in cubic meter is calculated as follows:

𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 (𝑚𝑚3 ) =

𝜋𝜋𝑑𝑑 2 3.1416 × (15.24 𝑐𝑐𝑐𝑐)2 = × 60.96 𝑐𝑐𝑐𝑐 = 11,120 𝑐𝑐𝑐𝑐3 4 4

b. Then, the flow rate is also converted into the same unit as follows:

𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 �

𝑐𝑐𝑐𝑐3 𝐿𝐿 𝑚𝑚3 100 𝑐𝑐𝑐𝑐 3 𝑐𝑐𝑐𝑐3 � = 5.56 × ×� � = 5,560 𝑠𝑠 1,000 𝐿𝐿 1 𝑚𝑚 𝑠𝑠 𝑠𝑠

c. The residence time in seconds is calculated as follows:

𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠) =

11,120 𝑐𝑐𝑐𝑐3 = 2 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐3 5,560 𝑠𝑠

Problem 4.8.7 Chemical Formula for Biomass.

Problem 4.8.7 Determine the approximate chemical formula for corn stover. About 100 g of bone-dry corn stover was placed in the ultimate analyzer and the analysis showed that the carbon, hydrogen, oxygen, nitrogen, sulfur, and ash content was found to be as follows: C=40g; H=5.9g; O=46.2g; N=1.3g, S=0.2g; and Ash=6.3g.

Solution: a. The chemical formula is established by calculating the number of moles of each element based on the gravimetric elemental analysis as follows: Carbon = 40g/12g/mol = 3.33 mol Hydrogen = 5.9g/1g/mol = 5.9 mol Oxygen = 46.2g/16g/mol = 2.89 mol 71

Nitrogen = 1.3g/14g/mol = 0.093 mol Sulfur = 0.2g/32g/mol = 0.006 mol b. The approximate chemical formula of the biomass is as follows: 𝐶𝐶3.33 𝐻𝐻5.9 𝑂𝑂2.89 𝑁𝑁0.093 𝑆𝑆0.006

c. From the chemical formula, the balanced (or stoichiometric combustion) equation may be calculated.

Problem 4.8.8 Air-to-Fuel Ratio (AFR) Calculations Problem 4.8.8 Determine the amount of air needed per unit weight of corn stover during complete combustion of this biomass. Use the approximate chemical formula of the biomass as 𝐶𝐶3.33 𝐻𝐻5.9 𝑂𝑂2.89 𝑁𝑁0.1 𝑆𝑆0.01

.Solution:

a. The stoichiometric combustion equation is shown below. Note that air is assumed to be composed primarily of 79% nitrogen and 21% oxygen and hence, the ratio is 3.76 as shown in the coefficient for nitrogen. 𝐶𝐶3.33 𝐻𝐻5.9 𝑂𝑂2.89 𝑁𝑁0.1 𝑆𝑆0.01 + 3.325(𝑂𝑂2 + 3.76𝑁𝑁2 ) → 3.3𝐶𝐶𝐶𝐶2 + 2.95𝐻𝐻2 𝑂𝑂 + 12.6𝑁𝑁2 + 0.01𝑆𝑆

b. The air-to-fuel (AFR) ratio is calculated as follows:

𝐴𝐴 3.325(32 + 105.28) 456.5 𝑘𝑘𝑘𝑘 = = 𝐹𝐹 [(3.33 × 12) × (5.9 × 1) × (2.89 × 16) × (0.1 × 14) × (0.01 × 32)] 93.62 𝑘𝑘𝑘𝑘 = 4.88

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘

c. It would take around 4.88 kg of air per kg of rice hull burned in the combustion mode.

72

Problem 4.8.9 Eutectic Point of Biomass Problem 4.8.9 Determine the base-to-acid ratio, the agglomeration index, and the alkali index for the biomass whose ash profile is given in the table. The heating value of this biomass is around 15.4 MJ/kg [6635 Btu/lb]. The percentage of ash in the biomass is 12.5%. Comment on the values obtained.

Basic Compounds

Values

Acidic Compounds

Values

Fe2O3

1.0

SiO2

73.2

CaO

5.0

TiO2

0

MgO

1.5

Al2O3

5.1

K2O

8.4

Na2O

0.4

Totals

16.3

Totals

78.3

Solution: a. The base-to-acid ratio is calculated using Equation 4.7 as follows:

𝑅𝑅𝑏𝑏/𝑎𝑎 = 𝑅𝑅𝑏𝑏/𝑎𝑎 =

%(𝐹𝐹𝐹𝐹2 𝑂𝑂3 + 𝐶𝐶𝐶𝐶𝐶𝐶 + 𝑀𝑀𝑀𝑀𝑀𝑀 + 𝐾𝐾2 𝑂𝑂 + 𝑁𝑁𝑁𝑁2 𝑂𝑂) %(𝑆𝑆𝑆𝑆𝑂𝑂2 + 𝑇𝑇𝑇𝑇𝑂𝑂2 + 𝐴𝐴𝐴𝐴2 𝑂𝑂3 )

%(1.0 + 5.0 + 1.5 + 8.4 + 0.4) 16.3 = 0.208 %(73.2 + 0 + 5.1) 78.3

The values are less than the unity and hence, we project slagging and fouling events to be minimal.

b. The biomass agglomeration index (BAI) is shown below using Equation 4.8 𝐵𝐵𝐵𝐵𝐵𝐵 =

%(𝐹𝐹𝐹𝐹2 𝑂𝑂3 ) %(𝐾𝐾2 𝑂𝑂 + 𝑁𝑁𝑁𝑁2 𝑂𝑂)

73

𝐵𝐵𝐵𝐵𝐵𝐵 =

%(1.0) = 0.114 %(8.4 + 0.4)

Bed agglomeration may happen on account of the low value of agglomeration index. c. The alkali index is given by Equation 4.9 and calculations are shown below. Note that the % K2O and Na2O are 8.4% and 0.4% respectively, and for every kg of ash material around 8.8% is the alkali. For every kg of biomass, 12.5% is ash, hence, the amount of this alkali per kg is the product of 12.5% and 8.8% which is 0.011 kg (0.125 x 0.088).

Thus, the heating value of GJ is simply the value of heating in MJ divided by 1,000 as shown. 𝐴𝐴𝐴𝐴 =

𝑘𝑘𝑘𝑘(𝐾𝐾2 𝑂𝑂) 0.011 𝑘𝑘𝑘𝑘 = = 0.714 𝐺𝐺𝐺𝐺 15.4/1000 𝐺𝐺𝐺𝐺

The value of AI is higher than 0.34 therefore this biomass has high slagging and fouling risk. In the calculations above, two out of the three parameters show the biomass has potential for slagging and fouling and conclusively, one must be careful when converting this biomass by thermal means. The occurrence of agglomeration, slagging, and fouling is likely, especially at high temperatures of operation.

Problem 4.9.10 Area Needed for Wood Power Problem 4.8.10 Determine the amount of area needed to produce 25 MW of electrical power from the combustion of high tonnage sorghum via gasification. Assume the heating value of high tonnage sorghum to be 16 MJ.kg and the productivity of land to be about 30 tonnes per hectare per year (1,000 kg is one tonne). Further, assume the overall conversion efficiency of about 15% using the steam cycle. The plant will be operated continuously throughout the year (365 days in a year).

74

Solution: a. The input power is first calculated using the given conversion efficiency as follows: 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 (𝑀𝑀𝑀𝑀) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 (𝑀𝑀𝑀𝑀) 25 𝑀𝑀𝑀𝑀 = = 166.7 𝑀𝑀𝑀𝑀 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑) 0.15

b. Then, input biomass in tonnes per hr is calculated. 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 � 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

37.5

𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 ℎ𝑟𝑟

� = 166.7 𝑀𝑀𝑀𝑀 ×

𝑀𝑀𝑀𝑀

1 𝑀𝑀𝑀𝑀−𝑠𝑠

×

𝑘𝑘𝑘𝑘

16 𝑀𝑀𝑀𝑀

×

3600 𝑠𝑠 ℎ𝑟𝑟

= 37,500

𝑘𝑘𝑘𝑘 ℎ𝑟𝑟

=

ℎ𝑟𝑟

c. Finally, the acreage required is calculated. 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 (ℎ𝑎𝑎) =

37.5 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 24 ℎ𝑟𝑟𝑟𝑟 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ℎ𝑎𝑎 − 𝑦𝑦𝑦𝑦 × × × = 10,950 ℎ𝑎𝑎 ℎ𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦 30 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

d. Over 10,000 hectares may be needed for this project, or a little over 400 hectares per MW of power output.

4.8.11 Carbon Sequestration (Metric Units) P4.11 A 11.355-million liter per year (3 MGY) bioethanol plant plans to sequester all the CO2 produced by the facility and inject into underground soil. It was found that the CCS technology was appropriate for the soil strata of the facility. Assume actual production rate from sugar to bioethanol of around 80 kg CO2 per 180 kg of pure sugar. This process will also generate 90 kg of bioethanol. The density of bioethanol was 0.750 kg/L. Determine the total amount of CO2 sequestered in units of metric tonnes (MT) per year for this facility and the potential maximum federal revenue. a) 7,570 b) 2,963 c) 9,565 75

d) 1,470 Solution: a. Based on the theoretical data given, the amount of CO2 produced per liter of bioethanol is calculated as follows: 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = = 𝑘𝑘𝑘𝑘 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 0.8889

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

b. The amount of CO2 produced in English units are calculated as follows: 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = ×

0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 2.2 𝑙𝑙𝑙𝑙𝑙𝑙 0.785 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 × × 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1 𝑘𝑘𝑘𝑘 𝐿𝐿

3.785 𝐿𝐿 𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 = 5.8 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

c. Next, the total amount of bioethanol produced by weight for this facility each year is calculated as follows: 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 � =

𝑀𝑀𝑀𝑀 � 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

11.355 × 106 𝐿𝐿 0.750 𝑘𝑘𝑘𝑘 𝑀𝑀𝑀𝑀 × × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝐿𝐿 1,000 𝑘𝑘𝑘𝑘

= 8,516

𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

d. Thus, the total amount of CO2 produced each year on a weight basis is calculated as follows:

76

𝐶𝐶𝑂𝑂2 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 0.8889 𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 � = 8,516 × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

= 7,570

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

4.8.12 Carbon Sequestration (English Units) P4.12 A 11.355-million liter per year (3 MGY) bioethanol plant plans to sequester all the CO2 produced by the facility and inject into underground soil. It was found that the CCS technology was appropriate for the soil strata of the facility. Assume actual production rate from sugar to bioethanol of around 80 kg CO2 per 180 kg of pure sugar. This process will also generate 90 kg of bioethanol. Assume the density of bioethanol to be 0.750 kg/L. The sequestered CO2 in units of metric tonnes (MT) per year for this facility is a potential maximum federal revenue. Determine the amount of CO2 produced (in lbs) per gallon of bioethanol produced by this facility. a) 7.5 b) 5.8 c) 9.5 d) 8.0 Solution: e. Based on the theoretical data given, the amount of CO2 produced per liter of bioethanol is calculated as follows: 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = = 𝑘𝑘𝑘𝑘 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 0.8889

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 77

f. The amount of CO2 produced in English units are calculated as follows:

𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = ×

0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 2.2 𝑙𝑙𝑙𝑙𝑙𝑙 0.785 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 × × 1 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐿𝐿

𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 3.785 𝐿𝐿 = 5.8 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

4.8.13 Government Federal Credits P4.13 A 11.355-million liter per year (3 MGY) bioethanol plant plans to sequester all the CO2 produced by the facility and inject into underground soil. It was found that the CCS technology was appropriate for the soil strata of the facility. Assume actual production rate from sugar to bioethanol of around 80 kg CO2 per 180 kg of pure sugar. This process will also generate 90 kg of bioethanol. Assume the density of bioethanol to be 0.750 kg/L. Determine the potential maximum federal credit revenue ($) for this process assuming revenue potential of $50/MT on a yearly basis. a) 757,000 b) 296,300 c) 378,500 d) 147,000 Solution: a. Based on the theoretical data given, the amount of CO2 produced per liter of bioethanol is calculated as follows: 78

𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = = 𝑘𝑘𝑘𝑘 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 0.8889

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

b. The amount of CO2 produced in English units are calculated as follows: 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = ×

0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 2.2 𝑙𝑙𝑙𝑙𝑙𝑙 0.785 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 × × 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1 𝑘𝑘𝑘𝑘 𝐿𝐿

3.785 𝐿𝐿 𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 = 5.8 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

c. Next, the total amount of bioethanol produced by weight for this facility each year is calculated as follows: 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 � =

𝑀𝑀𝑀𝑀 � 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

𝑀𝑀𝑀𝑀 11.355 × 106 𝐿𝐿 0.750 𝑘𝑘𝑘𝑘 × × 𝐿𝐿 1,000 𝑘𝑘𝑘𝑘 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

= 8,516

𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

d. Thus, the total amount of CO2 produced each year on a weight basis is calculated as follows: 𝐶𝐶𝑂𝑂2 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 0.8889 𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 � = 8,516 × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

= 7,570

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

e. The maximum potential federal revenue for this process assuming the highest reported revenue potential of $50/MT each year is calculated as follows: 79

𝐶𝐶𝐶𝐶𝐶𝐶 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 �

$ 𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 $50 � = 7,570 × = $378,500/𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑀𝑀𝑀𝑀

4.8.14 Bioethanol Production by Weight (Metric Units) P4.14 A 11.355-million liter per year (3 MGY) bioethanol plant plans to sequester all the CO2 produced by the facility and inject it into underground soil. It was found that the CCS technology was appropriate for the soil strata of the facility. Assume actual production rate from sugar to bioethanol of around 80 kg CO2 per 180 kg of pure sugar. This process will also generate 90 kg of bioethanol. Assume the density of bioethanol to be 0.750 kg/L. Determine the total amount of bioethanol produced each year by this facility in units of metric tonnes (MT) a) 7,570 b) 2,963 c) 1,475 d) 8,516 Solution: a. Based on the theoretical data given, the amount of CO2 produced per liter of bioethanol is calculated as follows: 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = = 𝑘𝑘𝑘𝑘 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 0.8889

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

b. The amount of CO2 produced in English units are calculated as follows:

80

𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = ×

0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 2.2 𝑙𝑙𝑙𝑙𝑙𝑙 0.785 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 × × 1 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐿𝐿

𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 3.785 𝐿𝐿 = 5.8 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

c. Next, the total amount of bioethanol produced by weight for this facility each year is calculated as follows: 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 � =

𝑀𝑀𝑀𝑀 � 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

𝑀𝑀𝑀𝑀 11.355 × 106 𝐿𝐿 0.750 𝑘𝑘𝑘𝑘 × × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝐿𝐿 1,000 𝑘𝑘𝑘𝑘

= 8,516

𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

4.8.15 Carbon Footprint of Dairy Facility P4.15 Determine the overall carbon footprint (units of MT CO2eq/yr) for a 400-head dairy facility in Central Texas. The estimated emissions for each dairy animal in terms of equivalent CO2e emissions per head from ground level sources were reported as 27.5 kg CO2eq hd-1 d-1 resulting from collective methane, CO2 and N2O emissions. In addition, the dairy facility utilizes energy to operate the milking, housing, manure management, feeding, and water for a total annual energy usage of 750 kWh hd-1 yr-1. Natural gas usage was reported at 5 kWh hd-1 d-1. Finally, diesel fuel usage was estimated at 155 L hd-1 yr-1. The energy content of diesel was roughly 36.9 MJ/L. a) 4,435 b) 1,652 81

c) 2,010 d) 5,108 Solution: a. The annual gaseous emission is calculated as follows: 𝐺𝐺𝐺𝐺𝐺𝐺 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 27.5 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2𝑒𝑒 365 𝑑𝑑 𝑀𝑀𝑀𝑀 �= × 400 ℎ𝑑𝑑𝑑𝑑 × × 𝑦𝑦𝑦𝑦 ℎ𝑑𝑑 − 𝑑𝑑 𝑦𝑦𝑦𝑦 1000𝑘𝑘𝑘𝑘

= 4,015

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 𝑦𝑦𝑦𝑦

b. The equivalent CO2e emissions from electrical power usage using Table 14.6 is calculated as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 � = 750

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 � 𝑦𝑦𝑦𝑦

𝑘𝑘𝑘𝑘ℎ 0.424 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2𝑒𝑒 𝑀𝑀𝑀𝑀 × 400 ℎ𝑑𝑑𝑑𝑑 × × ℎ𝑑𝑑 − 𝑦𝑦𝑦𝑦 1000𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘ℎ

= 127.2

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 𝑦𝑦𝑦𝑦

c. The equivalent CO2e emissions from the use of natural gas is calculated as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝐺𝐺𝐺𝐺𝐺𝐺 𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 � =5

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 � 𝑦𝑦𝑦𝑦

𝑘𝑘𝑘𝑘ℎ 365 𝑑𝑑 0.181 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2𝑒𝑒 𝑀𝑀𝑀𝑀 × × 400 ℎ𝑑𝑑𝑑𝑑 × × ℎ𝑑𝑑 − 𝑑𝑑 𝑦𝑦𝑦𝑦 1000𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘ℎ

= 132

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 𝑦𝑦𝑦𝑦

d. Diesel usage equivalent emissions are estimated as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 � = 155 ×

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 � 𝑦𝑦𝑦𝑦

36.9 𝑀𝑀𝑀𝑀 1 𝑘𝑘𝑘𝑘ℎ 𝐿𝐿 × 400 ℎ𝑑𝑑𝑑𝑑 × × 𝐿𝐿 3.6 𝑀𝑀𝑀𝑀 ℎ𝑑𝑑 − 𝑦𝑦𝑦𝑦

0.253 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2𝑒𝑒 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 × = 161 𝑦𝑦𝑦𝑦 1000𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘ℎ 82

e. The overall carbon footprint for this facility is the sum of all equivalent emissions as follows: 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 � = 4435

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 � = (4,015 + 127.2 + 132 + 161) 𝑦𝑦𝑦𝑦

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 𝑦𝑦𝑦𝑦

f. More than 90% of these emissions are from ground level greenhouse gas emissions (i.e., 4015/4435 = 0.9053).

g. Note that there may be other activities at the dairy farm that may contribute to additional emissions not accounted for in this example. The operator must develop a more comprehensive inventory of all possible greenhouse gas emissions.

Assess Your Knowledge 1. Determine the equivalent CO2 emissions from a conversion process in which gas emissions are as follows: 10 tonnes of CO2 and 3 tonnes of CH4 and 0.05 tonne of N2O.

a) 88.50 b) 67.42 c) 9.02 d) 27.48 Solution: a. The equation below will be used to calculate CO2 equivalents. 𝐶𝐶𝐶𝐶2 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡) = (𝐶𝐶𝐶𝐶2 × 1 + 𝐶𝐶𝐶𝐶4 × 21 + 𝑁𝑁2 𝑂𝑂 × 310) 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

b. The percentage contribution of each compound in the overall GHG potential in CO2 equivalent is calculated as follows:

𝐶𝐶𝐶𝐶2𝑒𝑒 (𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡) = (10 × 1 + 3 × 21 + 0.05 × 310) = [10 + 63 + 15.5] = 88.5 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝐶𝐶𝐶𝐶2𝑒𝑒 83

2. A 400-head hybrid dairy in Central Texas uses 750 kWh yr-1 hd-1 of electrical power. The owner plans to install an anaerobic digester (AD) to begin addressing reduction in power for a circular economy for this facility. Assume the biogas production to be around 1.27 m3 h-1 d-1 of biogas with a heating value of 22.4 MJm-3. Determine the recovered energy (kWh) per year (365 days).

Assume an overall 35% conversion efficiency from biogas to electrical power in an internal combustion engine.

a) 1,153,692 b) 750,652 c) 657,023 d) 361,123 Solution: a. The daily theoretical (Pt) and actual (Pa) power production is calculated as follows: 𝑀𝑀𝑀𝑀 1.27 𝑚𝑚3 22.4 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 𝑃𝑃𝑡𝑡 � � = 400 ℎ𝑑𝑑𝑑𝑑 × × = 11,379.2 ℎ𝑑𝑑 − 𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚3 𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑

𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) = 11379.2

𝑀𝑀𝑀𝑀

𝑑𝑑𝑑𝑑𝑑𝑑

×

1 𝑑𝑑𝑑𝑑𝑑𝑑

24 ℎ𝑟𝑟𝑟𝑟

×

1 ℎ𝑟𝑟

3600 𝑠𝑠

×

1000 𝑘𝑘𝑘𝑘 𝑀𝑀𝑀𝑀

× 0.35 = 131.7 𝑘𝑘𝑘𝑘

b. Thus, the total renewable energy produced by this AD facility over a year is calculated as follows: 𝑘𝑘𝑘𝑘 8760 ℎ𝑟𝑟𝑟𝑟 𝑘𝑘𝑘𝑘ℎ 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 � � = 131.7 𝑘𝑘𝑘𝑘 × = 1,153,692 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

3. A 400-head hybrid dairy facility in Central Texas plans to install an advanced mechanical solids separator after the morning and afternoon manure flush from its free stall. The manure will be dried (10% moisture content, MC) using part of the syngas (20% of power produced) and the remaining gaseous product will be used to generate electrical power.

Estimate the potential actual power (kW) to be generated using an overall conversion efficiency of 15%. About 20% of the input dry manure will be converted into biochar as potential fertilizer. The recoverable dry manure with 10% MC was found to be 2.3 84

kg/head/day. Its heating value was found to be 11.6 MJ/kg. The electrical power consumption for this facility is about 750 kWh yr-1 hd-1. a) 18.5 b) 33.40 c) 84.0 d) 44.8 Solution: a. The potential power is calculated as follows: 𝑃𝑃𝑡𝑡 (𝑘𝑘𝑘𝑘) = 400 ℎ𝑑𝑑𝑑𝑑 ×

2.3 𝑘𝑘𝑘𝑘 11.6 𝑀𝑀𝑀𝑀 1 𝑑𝑑𝑑𝑑𝑑𝑑 1 ℎ𝑟𝑟 𝑀𝑀𝑀𝑀 1000 𝑘𝑘𝑘𝑘 × × × × × ℎ𝑑𝑑 − 𝑑𝑑𝑑𝑑𝑑𝑑 𝑘𝑘𝑘𝑘 24 ℎ𝑟𝑟𝑟𝑟 3600 𝑠𝑠 𝑀𝑀𝑀𝑀/𝑠𝑠 1𝑀𝑀𝑀𝑀

= 123.5 𝑘𝑘𝑘𝑘

𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) = 123.5 𝑘𝑘𝑘𝑘 × 0.15 = 18.5 𝑘𝑘𝑘𝑘

4. A 400-head hybrid dairy in Central Texas uses 750 kWh yr-1 hd-1 of electrical power. The owner plans to install an anaerobic digester (AD) to begin addressing reduction in power for a circular economy for this facility. Assume the biogas production to be around 1.27 m3 h-1 d-1 of biogas with a heating value of 22.4 MJm-3. Determine the theoretical power (MJ/day) available from this facility based on the energy content of the manure.

a. 57,125.5 b. 65,751.1 c. 11,379.2 d. 14,224.0

Solution: a. The daily theoretical (Pt) and actual (Pa) power production is calculated as follows:

85

𝑀𝑀𝑀𝑀 1.27 𝑚𝑚3 22.4 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 𝑃𝑃𝑡𝑡 � � = 400 ℎ𝑑𝑑𝑑𝑑 × × = 11,379.2 𝑑𝑑𝑑𝑑𝑑𝑑 ℎ𝑑𝑑 − 𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚3 𝑑𝑑𝑑𝑑𝑑𝑑

5. Determine the % contribution of methane from the overall equivalent CO2 emissions from a conversion process in which gas emissions are as follows: 10 tonnes of CO2 and 3 tonne of CH4 and 0.05 tonne of N2O.

a) 14.56% b) 71.2% c) 9.02% d) 27.48% Solutions: a. The percentage contribution of each compound in the overall GHG potential in CO2 equivalent is calculated as follows: 𝐶𝐶𝐶𝐶2𝑒𝑒 (𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡) = (10 × 1 + 3 × 21 + 0.05 × 310) = [10 + 63 + 15.5] = 88.5 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝐶𝐶𝐶𝐶2𝑒𝑒 𝐶𝐶𝐶𝐶2 (%) =

10 × 100% = 11.3% 88.5

𝑁𝑁2 𝑂𝑂(%) =

15.5 × 100% = 17.5% 88.5

𝐶𝐶𝐶𝐶4 (%) =

63 × 100% = 71.2% 88.5

6. A 340-million liter per year (90 MGY) bioethanol plant in the US plans to sequester 50% of all the CO2 produced by the facility and inject this into the underground soil. It was found that the CCS technology was appropriate for the soil strata of the facility. Assume the actual production rate from sugar to bioethanol of around 80 kg CO2 per 180 kg of pure sugar. This process will also generate 90 kg of bioethanol. Assume the density of bioethanol to be 0.789 kg/L. Determine the total amount of CO2 to be sequestered into 86

units of English tons per year for this facility and as the potential maximum federal revenue in the future. a) 131,151 b) 238,456 c) 268,260 d) 119,228 Solution: a. Based on the theoretical data given, the amount of CO2 produced per liter of bioethanol is calculated as follows: 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = = 𝑘𝑘𝑘𝑘 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 0.8889

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

b. The amount of CO2 (lbs) produced per gallon of bioethanol in English units are calculated as follows: 𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = ×

0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 2.2 𝑙𝑙𝑙𝑙𝑙𝑙 0.789 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 × × 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1 𝑘𝑘𝑘𝑘 𝐿𝐿

3.785 𝐿𝐿 𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 = 5.84 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

c. The total amount of bioethanol produced by weight for this facility each year is calculated as follows:

87

𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑀𝑀𝑀𝑀 � 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

340 × 106 𝐿𝐿 0.789 𝑘𝑘𝑘𝑘 𝑀𝑀𝑀𝑀 = × × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝐿𝐿 1,000 𝑘𝑘𝑘𝑘 = 268,260

𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

d. The total amount of CO2 produced each year on a weight basis is calculated as follows: 𝐶𝐶𝑂𝑂2 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑀𝑀𝑀𝑀 � 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

= 268,260 = 238,456 𝐶𝐶𝑂𝑂2 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 0.8889 𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 × 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 � 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

= 238,456 = 262,302

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 1.10 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑇𝑇𝑇𝑇𝑇𝑇 × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 1 𝑀𝑀𝑀𝑀 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

e. The total amount of CO2 to be captured or sequestered each year (assuming 50% capture) on a weight basis is calculated as follows: 𝐶𝐶𝑂𝑂2 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 �

𝑀𝑀𝑀𝑀 𝐶𝐶𝑂𝑂2 238,456 𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 �= × 50% 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

= 119,228

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

88

𝐶𝐶𝑂𝑂2 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 �

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 � 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

= 119,228

= 131,151

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 1.10 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑇𝑇𝑇𝑇𝑇𝑇 × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 1 𝑀𝑀𝑀𝑀 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

7. A 340-million liter per year (90 MGY) bioethanol plant in the US plans to sequester 50% of all the CO2 produced by the facility and inject into the underground soil. It was found that the CCS technology was appropriate for the soil strata of the facility. Assume the actual production rate from sugar to bioethanol of around 80 kg CO2 per 180 kg of pure sugar. This process will also generate 90 kg of bioethanol. Assume the density of bioethanol to be 0.789 kg/L. Determine the total amount of bioethanol produced each year by this facility in units of metric tonnes (MT) a) 340,000 b) 168,260 c) 119,228 d) 268,260 Solution: a. Based on the theoretical data given, the amount of CO2 produced per liter of bioethanol is calculated as follows: 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = = 𝑘𝑘𝑘𝑘 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 0.8889

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

b. The amount of CO2 produced in English units are calculated as follows:

89

𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = ×

0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 2.2 𝑙𝑙𝑙𝑙𝑙𝑙 0.789 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 × × 1 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐿𝐿

𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 3.785 𝐿𝐿 = 5.84 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

c. Next, the total amount of bioethanol produced by weight for this facility each year is calculated as follows: 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 � =

𝑀𝑀𝑀𝑀 � 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

𝑀𝑀𝑀𝑀 340 × 106 𝐿𝐿 0.789 𝑘𝑘𝑘𝑘 × × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝐿𝐿 1,000 𝑘𝑘𝑘𝑘

= 268,260

𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

8. A 340-million liter per year (3 MGY) bioethanol plant in the US plans to sequester 50% of all the CO2 produced by the facility and inject this into the underground soil. It was found that the CCS technology was appropriate for the soil strata of the facility. Assume the actual production rate from sugar to bioethanol of around 80 kg CO2 per 180 kg of pure sugar. This process will also generate 90 kg of bioethanol. Assume the density of bioethanol to be 0.789 kg/L. The sequestered CO2 in units of metric tonnes (MT) per year for this facility is a potential maximum federal revenue. Determine the amount of CO2 produced (in lbs) per gallon of bioethanol produced by this facility. a) 7.51 b) 5.84 c) 9.55 d) 8.05 90

Solution: a. Based on the theoretical data given, the amount of CO2 produced per liter of bioethanol is calculated as follows: 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = = 𝑘𝑘𝑘𝑘 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 0.8889

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

b. The amount of CO2 produced in English units are calculated as follows: 𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = ×

0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 2.2 𝑙𝑙𝑙𝑙𝑙𝑙 0.789 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 × × 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1 𝑘𝑘𝑘𝑘 𝐿𝐿

3.785 𝐿𝐿 𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 = 5.84 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

9. Determine the overall carbon footprint (units of MT CO2eq/yr) for a 400-head dairy facility in Central Texas. The estimated emissions for each dairy animal in terms of equivalent CO2e emissions per head from ground level sources was reported as 27.5 kg CO2eq hd-1 d-1 resulting from collective methane, CO2 and N2O emissions. In addition, the dairy facility utilizes energy to operate the milking, housing, manure management, feeding and water for a total annual energy usage of 750 kWh hd-1 yr-1. Natural gas usage was reported at 5 kWh hd-1 d-1. Finally, diesel fuel usage was estimated at 155 L hd-1 yr-1.

The energy content of diesel was roughly 36.9 MJ/L. a) 4,435 b) 1,652 c) 2,010 d) 5,108 91

Solution: a. The annual gaseous emission is calculated as follows: 𝐺𝐺𝐺𝐺𝐺𝐺 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 � =

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 � 𝑦𝑦𝑦𝑦

27.5 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2𝑒𝑒 365 𝑑𝑑 𝑀𝑀𝑀𝑀 × 400 ℎ𝑑𝑑𝑑𝑑 × × ℎ𝑑𝑑 − 𝑑𝑑 𝑦𝑦𝑦𝑦 1000𝑘𝑘𝑘𝑘

= 4,015

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 𝑦𝑦𝑦𝑦

b. The equivalent CO2e emissions from electrical power usage using Table 14.6 is calculated as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 � = 750

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 � 𝑦𝑦𝑦𝑦

𝑘𝑘𝑘𝑘ℎ 0.424 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2𝑒𝑒 𝑀𝑀𝑀𝑀 × 400 ℎ𝑑𝑑𝑑𝑑 × × ℎ𝑑𝑑 − 𝑦𝑦𝑦𝑦 𝑘𝑘𝑘𝑘ℎ 1000𝑘𝑘𝑘𝑘

= 127.2

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 𝑦𝑦𝑦𝑦

c. The equivalent CO2e emissions from the use of natural gas is calculated as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝐺𝐺𝐺𝐺𝐺𝐺 𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 � =5 ×

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 � 𝑦𝑦𝑦𝑦

𝑘𝑘𝑘𝑘ℎ 365 𝑑𝑑 0.181 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2𝑒𝑒 × × 400 ℎ𝑑𝑑𝑑𝑑 × ℎ𝑑𝑑 − 𝑑𝑑 𝑦𝑦𝑦𝑦 𝑘𝑘𝑘𝑘ℎ

𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 = 132 1000𝑘𝑘𝑘𝑘 𝑦𝑦𝑦𝑦

d. Diesel usage equivalent emissions are estimated as follows:

92

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 � = 155 ×

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 � 𝑦𝑦𝑦𝑦

𝐿𝐿 36.9 𝑀𝑀𝑀𝑀 1 𝑘𝑘𝑘𝑘ℎ × 400 ℎ𝑑𝑑𝑑𝑑 × × 𝐿𝐿 3.6 𝑀𝑀𝑀𝑀 ℎ𝑑𝑑 − 𝑦𝑦𝑦𝑦

0.253 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2𝑒𝑒 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 × = 161 𝑘𝑘𝑘𝑘ℎ 1000𝑘𝑘𝑘𝑘 𝑦𝑦𝑦𝑦

e. The overall carbon footprint for this facility is the sum of all equivalent emissions as follows: 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 � 𝑦𝑦𝑦𝑦

= (4,015 + 127.2 + 132 + 161) = 4435

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2𝑒𝑒 𝑦𝑦𝑦𝑦

f. More than 90% of these emissions are from ground level greenhouse gas emissions (i.e., 4015/4435 = 0.9053). g. Note that there may be other activities at the dairy farm that may contribute to additional emissions not accounted for in this example. The operator must develop a more comprehensive inventory of all possible greenhouse gas emissions.

10. A 340-million liter per year (90 MGY) bioethanol plant in the US plans to sequester 50% of all the CO2 produced by the facility and inject this into the underground soil. It was found that the CCS technology was appropriate for the soil strata of the facility. Assume the actual production rate from sugar to bioethanol of around 80 kg CO2 per 180 kg of pure sugar. This process will also generate 90 kg of bioethanol. Assume the density of bioethanol to be 0.789 kg/L. Determine the potential maximum federal credit revenue ($) for this process assuming revenue potential of $50/MT on a yearly basis. a. $8, 296,300 93

b. $3,278,500 c. $5,961,400 d. 1,757,000 Solution: a. Based on the theoretical data given, the amount of CO2 produced per liter of bioethanol is calculated as follows: 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = = 𝑘𝑘𝑘𝑘 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 0.8889

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

b. The amount of CO2 (lbs) produced per gallon of bioethanol in English units are calculated as follows: 𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 80 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 90 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = ×

0.8889 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶2 2.2 𝑙𝑙𝑙𝑙𝑙𝑙 0.789 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 × × 𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1 𝑘𝑘𝑘𝑘 𝐿𝐿

3.785 𝐿𝐿 𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶2 = 5.84 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

c. The total amount of bioethanol produced by weight for this facility each year is calculated as follows: 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑀𝑀𝑀𝑀 � 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

340 × 106 𝐿𝐿 0.789 𝑘𝑘𝑘𝑘 𝑀𝑀𝑀𝑀 = × × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝐿𝐿 1,000 𝑘𝑘𝑘𝑘 = 268,260

𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

d. The total amount of CO2 produced each year on a weight basis is calculated 94

as follows: 𝐶𝐶𝑂𝑂2 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑀𝑀𝑀𝑀 � 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

= 268,260 = 238,456 𝐶𝐶𝑂𝑂2 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 0.8889 𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑀𝑀𝑀𝑀 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 � 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

= 238,456 = 262,302

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 1.10 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑇𝑇𝑇𝑇𝑇𝑇 × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 1 𝑀𝑀𝑀𝑀 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

e. The total amount of CO2 to be captured or sequestered each year (assuming 50% capture) on a weight basis is calculated as follows: 𝐶𝐶𝑂𝑂2 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 �

𝑀𝑀𝑀𝑀 𝐶𝐶𝑂𝑂2 238,456 𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 �= × 50% 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

= 119,228 𝐶𝐶𝑂𝑂2 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 �

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 � 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

= 119,228

= 131,151

𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 1.10 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑇𝑇𝑇𝑇𝑇𝑇 × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 1 𝑀𝑀𝑀𝑀 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

f. Hence, the potential federal credit revenue is calculated as follows: $ 𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶2 $50 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 � � = 119,228 × = $5,961,400 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑀𝑀𝑀𝑀

g. Close to $6M dollars of federal credit revenue is possible for this undertaking.

95

11. Determine the % contribution of CO2 from the overall equivalent CO2 emissions from a conversion process in which gas emissions are as follows: 10 tonnes of CO2 and 3 tonnes of CH4 and 0.05 tonne of N2O.

a) 41.56% b) 31.3% c) 11.3% d) 27.48% Solutions: a. The percentage contribution of each compound in the overall GHG potential in CO2 equivalent is calculated as follows: 𝐶𝐶𝐶𝐶2𝑒𝑒 (𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡) = (10 × 1 + 3 × 21 + 0.05 × 310) = [10 + 63 + 15.5] = 88.5 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝐶𝐶𝐶𝐶2𝑒𝑒 𝐶𝐶𝐶𝐶2 (%) =

10 × 100% = 11.3% 88.5

𝑁𝑁2 𝑂𝑂(%) =

15.5 × 100% = 17.5% 88.5

𝐶𝐶𝐶𝐶4 (%) =

63 × 100% = 71.2% 88.5

12. A 400-head hybrid dairy in Central Texas uses 750 kWh yr-1 hd-1 of electrical power. The owner plans to install an anaerobic digester (AD) to begin addressing reduction in power for a circular economy for this facility. Assume the biogas production to be around 1.27 m3 h-1 d-1 of biogas with a heating value of 22.4 MJm-3. Determine the percent reduction in purchased power for this facility assuming an overall 35% conversion efficiency from biogas to electrical power in an internal combustion engine and for 365 days in a year.

a) 57.6% 96

b) 100% c) 65.7 d) 36.1 Solution: a. The daily theoretical (Pt) and actual (Pa) power production is calculated as follows: 𝑃𝑃𝑡𝑡 �

𝑀𝑀𝑀𝑀 1.27 𝑚𝑚3 22.4 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 � = 400 ℎ𝑑𝑑𝑑𝑑 × × = 11,379.2 3 𝑑𝑑𝑑𝑑𝑑𝑑 ℎ𝑑𝑑 − 𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚 𝑑𝑑𝑑𝑑𝑑𝑑

𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) = 11379.2

𝑀𝑀𝑀𝑀

𝑑𝑑𝑑𝑑𝑑𝑑

×

1 𝑑𝑑𝑑𝑑𝑑𝑑

24 ℎ𝑟𝑟𝑟𝑟

×

1 ℎ𝑟𝑟

3600 𝑠𝑠

×

1000 𝑘𝑘𝑘𝑘 𝑀𝑀𝑀𝑀

× 0.35 = 131.7 𝑘𝑘𝑘𝑘

b. Thus, the total renewable energy produced by this AD facility over a year is calculated as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝑘𝑘𝑘𝑘 8760 ℎ𝑟𝑟𝑟𝑟 𝑘𝑘𝑘𝑘ℎ � = 131.7 𝑘𝑘𝑘𝑘 × = 1,153,692 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

c. The total fossil-based energy utilized by this facility is calculated as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝑘𝑘𝑘𝑘ℎ

𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

� = 400 ℎ𝑑𝑑𝑑𝑑 ×

750 𝑘𝑘𝑘𝑘ℎ

ℎ𝑑𝑑−𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

= 300,000

𝑘𝑘𝑘𝑘ℎ

𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

c. There is excess power indicating 100% reduction in electrical power needs and excess power may be sold to the grid for additional revenue.

13. A 400-head hybrid dairy facility in Central Texas plans to install an advanced mechanical solids separator after the morning and afternoon manure flush from its free stall. The manure will be dried (10% moisture content, MC) using part of the syngas (20% of power produced) and the remaining gaseous product (80%) will be used to generate electrical power. One may estimate the potential power to be generated using an overall conversion efficiency of 15%. About 20% of the input dry manure will be converted into biochar as potential fertilizer. The recoverable dry manure with 10% MC was found to be 97

2.3 kg/head/day. Its heating value was found to be 11.6 MJ/kg. The electrical power consumption for this facility is about 750 kWh yr-1 hd-1. Determine the yearly electrical energy (kWh/year) generated by this facility via gasification assuming the power plant is operated 365 days in a year and 24 hrs/day. a) 148,255 b) 129,823 c) 84,560 d) 44,898 Solution: a. The amount of dry manure generated each day is calculated as follows:

b.

Manure �

𝑀𝑀𝑀𝑀 2.3 𝑘𝑘𝑘𝑘 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 1 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 � = 400 hds × × × = 335.8 𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 ℎ𝑑𝑑 − 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 1,000 𝑘𝑘𝑘𝑘 𝑦𝑦𝑦𝑦

The potential power produced is calculated as follows assuming only 80% of the energy of dry manure 𝑃𝑃𝑡𝑡 �

𝑀𝑀𝑀𝑀 2.3 𝑘𝑘𝑘𝑘 11.6 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 � = 400 hds × × = 10,672 𝑑𝑑𝑑𝑑𝑑𝑑 ℎ𝑑𝑑 − 𝑑𝑑𝑑𝑑𝑑𝑑 𝑘𝑘𝑘𝑘 𝑑𝑑𝑑𝑑𝑑𝑑

𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) = 10,672

𝑀𝑀𝑀𝑀 1 𝑑𝑑𝑑𝑑𝑑𝑑 1 ℎ𝑟𝑟 1000 𝑘𝑘𝑘𝑘 × × × × 0.15 × 0.80 𝑑𝑑𝑑𝑑𝑑𝑑 24 ℎ𝑟𝑟𝑟𝑟 3600 𝑠𝑠 𝑀𝑀𝑀𝑀

= 14.82 𝑘𝑘𝑘𝑘

c. Hence, the amount of energy produced by the gasification system in a year is calculated as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝑘𝑘𝑘𝑘 8760 ℎ𝑟𝑟𝑟𝑟 𝑘𝑘𝑘𝑘ℎ � = 14.82 𝑘𝑘𝑘𝑘 × = 129,823 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

14. A 400-head hybrid dairy in Central Texas uses 750 kWh yr-1 hd-1 of electrical power. The owner plans to install an anaerobic digester (AD) to begin addressing reduction in power 98

for a circular economy for this facility. Assume the biogas production to be around 1.27 m3 h-1 d-1 of biogas with a heating value of 22.4 MJm-3. Determine the actual power (kW) generated for this facility, assuming an overall conversion efficiency from biogas to electrical power in an internal combustion engine of 35%.

a) 57.6 b) 65.7 c) 36.1 d) 131.7 Solution: a. The daily theoretical (Pt) and actual (Pa) power production is calculated as follows: 𝑃𝑃𝑡𝑡 �

𝑀𝑀𝑀𝑀 1.27 𝑚𝑚3 22.4 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 � = 400 ℎ𝑑𝑑𝑑𝑑 × × = 11,379.2 3 ℎ𝑑𝑑 − 𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚 𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑

𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) = 11379.2

𝑀𝑀𝑀𝑀

𝑑𝑑𝑑𝑑𝑑𝑑

×

1 𝑑𝑑𝑑𝑑𝑑𝑑

24 ℎ𝑟𝑟𝑟𝑟

×

1 ℎ𝑟𝑟

3600 𝑠𝑠

×

1000 𝑘𝑘𝑘𝑘 𝑀𝑀𝑀𝑀

× 0.35 = 131.7 𝑘𝑘𝑘𝑘

15. Determine the % contribution of N2O from the overall equivalent CO2 emissions from a conversion process in which gas emissions are as follows: 10 tonnes of CO2 and 3 tonne of CH4 and 0.05 tonne of N2O.

a. 14.56% b. 11.3% c. 9.02% d. 17.5% Solutions: b. The percentage contribution of each compound in the overall GHG potential in CO2 equivalent is calculated as follows: 99

𝐶𝐶𝐶𝐶2𝑒𝑒 (𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡) = (10 × 1 + 3 × 21 + 0.05 × 310) = [10 + 63 + 15.5] = 88.5 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝐶𝐶𝐶𝐶2𝑒𝑒 𝐶𝐶𝐶𝐶2 (%) =

10 × 100% = 11.3% 88.5

𝑁𝑁2 𝑂𝑂(%) =

15.5 × 100% = 17.5% 88.5

𝐶𝐶𝐶𝐶4 (%) =

63 × 100% = 71.2% 88.5

16. One lb of rice straw with a heating value of 6,550 Btu/lb was pyrolyzed at a temperature of 932oF. About 0.35 lb of char was produced with an energy content of 8,000 Btu/lb. Determine the percentage of energy retained in the char. a. 42.75% b. 36.4% c. 73.6% d. 25.9%

Solution: a. The energy in the original biomass was 6,550 Btu b. The energy that remained in the char was the product of its final weight 0.35 lbs by its heating value of 8,000 Btu/lb to get 2,800 Btu (0.35 lb x 8,000 Btu/lb = 2,800 Btu). c. Thus, the percentage of energy that remained in the char is the ratio of the energy in the char to the original energy which will be about 42.75% (2,800/6550 x100% = 42.75%) 17. Determine the %ratio of biodiesel production versus diesel consumption in the US in the year 2023. Assume the annual diesel consumption in the US in 2023 is around 3 million barrels per day and the annual biodiesel production in 2023 is around 1.6 billion gallons.

100

a. 3.5% b. 6.2% c. 5.3% d. 4.0% Solution: a. The biodiesel data is first converted into millions of barrels using the conversion a barrel is 42 gallons. 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 (2023) = 1.6 × 109 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ×

1 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 = 38.1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 42 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

b. The diesel value is converted into the same units as follows: 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 (2023) = 3 × 106

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 × = 1,095 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

c. Thus, the ratio of biodiesel production to diesel consumption is calculated as follows: % 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 �

𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 38.1 �= × 100% = 3.47%~3.5% 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 1,095

d. The amount of biodiesel produced is still a small fraction of the total diesel consumption in the US 18.

Convert the biooil yield in gallons per ton of biomass if during the slow pyrolysis experiment, the density of the resulting biooil was found to be 0.95 g/mL and the biooil yield was 35% by weight?

a. 60 b. 88.5 c. 35.5 d. 120 Solution: 101

a. The first step is to convert the density units in English system as follows: 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 �

0.95 𝑔𝑔 1,000 𝑚𝑚𝑚𝑚 𝑘𝑘𝑘𝑘 3.785 𝐿𝐿 2.2 𝑙𝑙𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙𝑙𝑙 �= × × × × = 7.91 𝑔𝑔𝑔𝑔𝑔𝑔 𝐿𝐿 1,000 𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔 𝑘𝑘𝑘𝑘 𝑔𝑔𝑔𝑔𝑔𝑔 𝑚𝑚𝑚𝑚

b. Then, the 35% yield by weight of biooil will be translated into 700 lbs of biooil per

ton of biomass to yield the following output per ton of biomass (2000 lbs/ton x 0.35 = 700 lbs). 𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 �

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 700 𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 �= × = 88.5 𝑡𝑡𝑡𝑡𝑡𝑡 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑡𝑡𝑡𝑡𝑡𝑡 7.91 𝑙𝑙𝑙𝑙𝑙𝑙 𝑡𝑡𝑡𝑡𝑡𝑡

19. Determine the stoichiometric air-to-fuel ratio for the complete combustion of municipal solid waste (MSW) (in units of lb/lb). Assume the chemical formula for MSW is C4H6O2N0.08S0.009

a. 5.17 b. 7.07 c. 4.5 d. 4.94 Solution: a. The chemical formula for the rice straw has already been calculated as shown here: C4H6O2N0.08S0.009 b. A balanced combustion equation is made as follows: 𝐶𝐶4 𝐻𝐻6 𝑂𝑂2 𝑁𝑁0.08 𝑆𝑆0.009 + 4.5(𝑂𝑂2 + 3.76𝑁𝑁2 ) → 4𝐶𝐶𝐶𝐶2 + 3𝐻𝐻2 𝑂𝑂 + 16.92𝑁𝑁2 + 0.009𝑆𝑆 c. Then, the fuel-to-air ratio is simply calculated as follows:

𝐴𝐴 4.5 × [(16 × 2) + 3.76 × (14 × 2)] 617.76 = = 𝐹𝐹 [(12 × 4) + (1 × 6) + (16 × 2) + (14 × 0.08) + (32 × 0.009)] 87.4 = 7.07

𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙

102

20. The ultimate analysis for corn cob is given as follows: C=46.2%; H=7.6%; O=42.3%; N=1.2%; and S=0.3% and the ash content is 2.4%. Determine the chemical formula for this biomass material.

a. C3.85H7.6O2.64N0.086S0.009 b. C46.2H7.6O1.2N1.2S0.3 c. C1H7.6O2.64N0.086S0.009 d. C3.7H5.7O4.1N1S0.2 Solution a. Assuming 100 g sample, the mole fractions are as follows: C = 46.2/12 = 3.85

C = 3.85 /3.85 = 1

H = 7.6/1 = 7.6

H = 7.6 /3.85 = 1.97

O = 42.3/16 = 2.64

O = 2.64 /3.85 = 0.69

N = 1.2/14 = 0.086

N = 0.086 /3.85 = 0.022

S = 0.3/32 = 0.009

S = 0.009 /3.85 = 0.0024

b. Thus, the “chemical formula” for corncob is shown in suggested formats: C3.85H7.6O2.64N0.086S0.009 or CH1.97O0.69N0.022S0.0024

103

Chapter 5 Hydro Power 5.7 Solved Problems Problem 5.10.1 Theoretical Power from Water Problem 5.10.1 Determine the theoretical power in watts for water flowing at 50 kg/s and dropped from a dynamic head of 10 meters.

Solution: a. The use of Equation 5.1 is straightforward, making use of the acceleration due to gravity constant of 9.8 m/s2 as well as Newton’s Law for the units of force as a function of mass and acceleration. 𝑃𝑃𝑡𝑡 (𝑘𝑘𝑘𝑘) =

𝑄𝑄×𝐻𝐻 102

𝑃𝑃𝑡𝑡 (𝑊𝑊 ) =

𝑄𝑄×𝐻𝐻

0.102

50 kg 10 𝑚𝑚 𝑁𝑁𝑠𝑠 2 9.8 𝑚𝑚 𝑊𝑊𝑊𝑊 𝑃𝑃𝑡𝑡 (𝑊𝑊 ) = × × × 2 × = 4,900 𝑊𝑊 𝑠𝑠 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑠𝑠 𝑁𝑁𝑁𝑁 𝑃𝑃𝑡𝑡 (𝑊𝑊 ) =

50 kg 10 𝑚𝑚 × = 4,900 𝑊𝑊 𝑠𝑠 0.102

b. Hence, this amount of water has a potential energy of around 4,900 watts.

Problem 5.10.2 Actual Efficiencies of Micro Hydro Units Problem 5.10.2 A small commercial micro-hydro power unit had the following data: the volumetric flow rate was 70 liters/sec and the dynamic head was 1.5 meters. This unit was reported to produce 500 watts of continuous power output at those conditions. Determine the actual efficiency of this unit. Also calculate the yearly output of this small device in units of kWh/year, assuming 24 hrs per day and 365 days in a year.

Solution: 104

a. The volumetric flow rate is converted into units of kg/s. 𝑘𝑘𝑘𝑘

𝑄𝑄 � � = 𝑠𝑠

70 𝐿𝐿 𝑠𝑠𝑠𝑠𝑠𝑠

×

1 𝑘𝑘𝑘𝑘 𝐿𝐿

= 70

𝑘𝑘𝑘𝑘 𝑠𝑠

b. Then, the theoretical power is calculated. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊 )𝑡𝑡 = 𝑄𝑄 × 𝐻𝐻 =

70 𝑘𝑘𝑘𝑘 𝑁𝑁𝑠𝑠 2 9.8 𝑚𝑚 𝑊𝑊 − 𝑠𝑠 × 1.5 m × × 2 × = 1,029 𝑊𝑊 𝑠𝑠𝑠𝑠𝑠𝑠 𝑘𝑘𝑘𝑘 𝑚𝑚 𝑠𝑠 𝑁𝑁 − 𝑚𝑚

c. The efficiency is calculated using the ratio of actual and theoretical power as shown. 𝜂𝜂 (%) = 𝜂𝜂 (%) =

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 × 100% 𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 500 𝑊𝑊 × 100% = 48.6% 1029 𝑊𝑊

d. Hence, the overall conversion efficiency of this unit is below 60%. e. The yearly output may be estimated as follows:

𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 �

𝑘𝑘𝑘𝑘ℎ 24 ℎ𝑟𝑟𝑟𝑟 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 1 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘ℎ � = 500 𝑊𝑊 × × × = 4,380 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 1,000 𝑊𝑊 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

Problem 5.10.3 Hydro Power Plant Calculations.

Problem 5.10.3 Estimate the power output from one of the turbines in of the Bhumibol hydro power plant with the following data: dynamic head = 150 meters, volumetric flow rate in one turbine = 97,750 kg/s. Assume an overall conversion efficiency of 80%.

Solution: a. Since the units for Q and H are those defined in Equation 5.2, the calculation of theoretical power is straightforward:

105

𝑃𝑃𝑡𝑡 (𝑘𝑘𝑘𝑘) = 𝑃𝑃𝑡𝑡 (𝑘𝑘𝑘𝑘) = 𝑃𝑃𝑡𝑡 (𝑘𝑘𝑘𝑘) =

𝑄𝑄 × 𝐻𝐻 102

97,750 𝑘𝑘𝑘𝑘 𝑁𝑁𝑠𝑠 2 9.8 𝑚𝑚 𝑊𝑊𝑊𝑊 𝑘𝑘𝑘𝑘 × 150 𝑚𝑚 × × 2 × × = 143,692.5 𝑘𝑘𝑘𝑘 𝑠𝑠𝑠𝑠𝑠𝑠 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑠𝑠 𝑁𝑁𝑁𝑁 1,000 𝑊𝑊 97,750 × 150 = 143,750 𝑘𝑘𝑘𝑘 [143.75 𝑀𝑀𝑀𝑀] 102

Note the answers using the estimated constant and the actual units and conversions.

b. Then, using the overall conversion efficiency, the actual power is calculated. 𝑃𝑃𝑎𝑎 (𝑀𝑀𝑀𝑀) = 143.75 MW × 0.80 = 115 MW

c. This power (115 MW) is the output of the Pelton turbine in the Bhumibol Dam Hydro Power plant.

Problem 5.10.4 Pump Specific Speed Problem 5.10.4 Given a pump rotational speed of 1800 rps, flow rate of 1200 m3/s, and head of 20 m, calculate the specific speed.

Solution: a. Equation 5.8 is used as follows: 𝑁𝑁𝑠𝑠 =

𝜔𝜔𝑞𝑞 0.5 (ℎ)0.75

1800 × 12000.5 𝑁𝑁𝑠𝑠 = = 1,746 (9.8 × 12)0.75

b. This is the specific speed for a typical centrifugal pump.

106

Problem 5.10.5 Volumetric Efficiency of Hydraulic Rams Problem 5.10.5 Calculate the volumetric efficiency of a hydraulic ram with a volume delivered to a higher elevation at 500 liters per minute. The total volume taken and diverted from a water source is 2 cubic meters per minute [2,000 liters per minute].

Solution: a. Equation 5.10 is used as shown below. 𝐸𝐸𝑣𝑣 = 𝐸𝐸𝑣𝑣 =

𝑞𝑞 × 100% (𝑄𝑄 + 𝑞𝑞 )

500 × 100% = 20% (2,000 + 500)

b. The volumetric efficiency is about 20%.

Problem 5.10.6 Energy Efficiency of Hydraulic Rams Problem 5.10.6 Calculate the water use efficiency for a hydraulic ram with a volume delivered to a higher elevation of 500 liters per minute, with a vertical lift of 25 meters and a horizontal fall of 8 meters. The total volume taken and diverted from a water source is 2 cubic meters per minute [2,000 liters per minute].

Solution: a. Equation 5.11 may be used as shown below. 𝐸𝐸𝐷𝐷𝐷𝐷 = 𝐸𝐸𝑣𝑣 = 𝐸𝐸𝑤𝑤 =

𝜌𝜌𝜌𝜌ℎ 𝑞𝑞ℎ × 100% = × 100% (𝑄𝑄 + 𝑞𝑞 )𝐻𝐻 𝜌𝜌(𝑄𝑄 + 𝑞𝑞 )𝐻𝐻

𝑞𝑞 × 100% (𝑄𝑄 + 𝑞𝑞 )

𝑞𝑞ℎ 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 × 100% = × 100% 𝑄𝑄𝑄𝑄 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 107

𝐸𝐸𝑤𝑤 =

500 × 25 × 100% = 78.13% 2,000 × 8

b. The water use efficiency is close to 80%.

Problem 5.10.7. Specifying Drive Pipe Size and Lengths Problem 5.10.7 Use the empirical Equation 5.12 to calculate the drive pipe length, L, if the supply head is 4 meters and the drive pipe diameter is 0.07522 meter [~3 inches]. The number of pulsations per minute, N, is about 60.

Solution: a. Equation 5.12 is used, and the length of the drive pipe is calculated as follows: 𝐿𝐿 =

𝐿𝐿 =

900𝐻𝐻 𝑁𝑁 2 𝐷𝐷

900 × 4

602 × 0.07522

= 13.3 𝑚𝑚 [43.6 𝑓𝑓𝑓𝑓]

b. Note how long the drive pipe is for this application—about 13.3 meters or over 40 feet— which is typical of drive pipe lengths.

Problem 5.10.8 Specifying Drive Pipe Size Using Rife Ram Problem 5.10.8 Calculate the drive pipe size (diameter in inches) for a hydraulic ram which would take water from a source with a maximum flow of around 360 L/min (95.1 gpm). This water will be brought 100 feet above the source—this is the vertical lift [30.48 m]. A vertical fall or supply head of 20 ft [6.10 m] is available. Use the Rife ram data for this problem as shown (Performance Chart). Estimate the amount of water that may be delivered (in gpm) as well as the pipe length assuming N of 60 beats per second.

Solution: 108

a. Referring to the performance chart for Rife Hydraulic Rams, about 12% of the water may be delivered or about 11.412 gpm of water (95.1 x 0.12 = 11.412 gpm) b. Calculate Q in L/s [(360 L/min)/(60 sec/min)] = 6 L/s and supply head of 20 ft [6.10 meters] c. The diameter of the drive pipe will be approximately 4” from the table, for a supply head of 6.10 meters (between 5.01 to 10 meters). d. The length of the drive pipe can be found as follows: 𝐿𝐿 = 𝐿𝐿 =

900𝐻𝐻 𝑁𝑁 2 𝐷𝐷

900 × 6.10 = 14.95 𝑚𝑚 [49 𝑓𝑓𝑓𝑓] 602 × (0.102)

Problem 5.10.9 Pumped Storage Power Production.

Problem 5.10.9 Determine the power generated from the Kalayaan Pumped Storage facility if the dynamic head is 290 meters and the volumetric flow rate is 120 cubic meters per second.

Assume an overall efficiency of 80%.

Solution: a. The actual power generated is calculated using Equation 5.3. 𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) =

𝑘𝑘𝑘𝑘 �×𝐻𝐻(𝑚𝑚)×𝐸𝐸𝑜𝑜 𝑠𝑠

𝑄𝑄�

𝑃𝑃𝑎𝑎 (𝑊𝑊 ) = 𝑄𝑄𝑄𝑄𝑄𝑄𝑜𝑜

= ×

102

𝑃𝑃𝑡𝑡 (𝑊𝑊 ) =

𝑘𝑘𝑘𝑘 �×𝐻𝐻(𝑚𝑚) 𝑠𝑠

𝑄𝑄�

0.102

120 𝑚𝑚3 1000 𝑘𝑘𝑘𝑘 𝑁𝑁𝑠𝑠 2 9.8 𝑚𝑚 𝑊𝑊𝑊𝑊 × × 290 𝑚𝑚 × × 2 × × 0.80 3 𝑠𝑠 𝑚𝑚 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑠𝑠 𝑁𝑁𝑁𝑁

𝑘𝑘𝑘𝑘 = 272,832 𝑘𝑘𝑘𝑘 [273 𝑀𝑀𝑀𝑀] 1,000 𝑊𝑊 109

𝑃𝑃𝑡𝑡 (𝑘𝑘𝑘𝑘) =

120,000 × 290 × 0.80 = 272,941 𝑘𝑘𝑘𝑘 [273 𝑀𝑀𝑀𝑀] 102

b. The Kalayaan Pumped Storage Power Plant generated 620 MW through two Francis Turbines. This is only about half of the installed capacity.

Problem 5.10.10 Pumped Storage Power Production Water Use. Problem 5.10.10 Determine the total maximum time (in hrs and days) in which the water from Caliraya Lake would be empty if power is generated by the Kalayaan Pumped Storage facility continuously. The volumetric flow rate is 120 cubic meters per second, and Lake Caliraya has a storage volume of 80x106 m3 during the rainy season.

Solution: a. The time (in hours) for Lake Caliraya to empty is simply the ratio of the storage volume and the amount of water used for the power plant, as shown below. 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) =

𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 (𝑚𝑚3 ) 1 ℎ𝑟𝑟 × 3 𝑚𝑚 3600 𝑠𝑠 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 � 𝑠𝑠 �

1 ℎ𝑟𝑟 80 × 106 (𝑚𝑚3 ) × = 185 ℎ𝑟𝑟𝑟𝑟 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = 3 𝑚𝑚 3600 𝑠𝑠 120 � 𝑠𝑠 �

b. Note that this storage volume would be empty after about 185 hours or over 7.7 days.

This is only half the capacity of the power plant.

5.10.11

Specific Speeds and Pump Type

P5.11 An application has a volumetric flow rate of 20 m3/s [317,040 gpm] and a dynamic head of 500 m [1640 feet]. The pump rotational speed is 700 rpm. You are to calculate the specific speed and identify the type of pump to be used. For this question, what is the specific speed for this pump?

110

a. 1746 b. 1467 c. 1529 d. 1190 Solutions: a. The governing equation is used as follows: 𝑛𝑛 × �𝑄𝑄 700 × √317,040 𝑁𝑁𝑠𝑠 = = 1529 3 = 3 4 4 (𝐻𝐻) (1640)

5.10.12

Pump Type and Specific Speeds

P5.12 An application has a volumetric flow rate of 20 m3/s [317,040 gpm] and a dynamic head of 500 m [1640 feet]. The pump rotational speed is 700 rpm. You are to calculate the specific speed and identify the type of pump to be used. For this question, what is the type of pump to be used for this application? a. Mixed impeller single suction pumps b. Propeller type pump c. Axial pumps d. Centrifugal pumps Solutions: a. The governing equation is used as follows: 𝑁𝑁𝑠𝑠 =

𝑛𝑛 × �𝑄𝑄 3 (𝐻𝐻)4

=

700 × √317,040 3

(1640)4

= 1529

b. This is the specific speed for a typical centrifugal pump.

111

5.10.13

Specific Speeds of Pumps

P5.13 Given a pump rotational speed of 1365 rpm, flow rate of 1200 gpm, and head of 20 feet.

Calculate the specific speed. a. 5000 b. 1467 c. 1190 d. 1746 Solutions: a. Equation 5.8 is used as follows: 𝑁𝑁𝑠𝑠 = 5.10.14

𝑛𝑛 × �𝑄𝑄 3 (𝑔𝑔𝑔𝑔)4

=

1365 × √1200

Pump Type

3 (20)4

= 5,000

P5.14 Given a pump rotational speed of 1365 rpm, flow rate of 1200 gpm and head of 20 feet.

Identify what type of pump to be used for this application. a. Centrifugal pumps b. Mixed impeller single suction pumps c. Propeller type pump d. Axial pumps Solutions: a. Equation 5.8 is used as follows: 𝑁𝑁𝑠𝑠 =

𝑛𝑛 × �𝑄𝑄 3 (𝑔𝑔𝑔𝑔)4

=

1365 × √1200 3

(20)4

= 5,000

b. This is the specific speed for a typical mixed flow impellers single suction pump.

112

5.10.15

Specific Speeds of Pumps

P5.15 Given a pump rotational speed of 4500 rpm, flow rate of 5000 gpm and head of 60 feet.

Calculate the specific speed. a. 5000 b. 1467 c. 14,760 d. 17460 Solutions: a. Equation 5.8 is used as follows: 𝑁𝑁𝑠𝑠 =

𝑛𝑛 × �𝑄𝑄 3 (𝑔𝑔𝑔𝑔)4

=

4500 × √5000 3 (60)4

= 14,760

b. This is the specific speed typical for propeller type or axial flow pumps.

Assess Your Knowledge 1. What is the current highest state-of-the art overall conversion efficiency (in %) for large-scale hydro power plants?

a. 50 b. 75 c. 90 d. 95 2. Select an example of an impulse turbine.

a. Francis b. Propeller c. Pelton 113

d. Kaplan 3. This type of water turbines changes the velocity of a water jet. The jet impinges on the turbine's curved blades which reverse the flow. The resulting change in momentum causes a force on the turbine blades. Since the turbine is spinning, the force acts through a distance (work) and the diverted water flow are left with diminished energy. Name this type of water turbine.

a. Impulse turbines b. reaction turbines c. impellers d. centrifugal pumps 4. What are the range of overall conversion efficiencies (in %) for large-scale hydro power plants?

a. 50-60 b. 80-90 c. 40-50 d. 60-80 5. This hydropower plant component is a sluice or gate or intake structure that controls water flow, or an enclosed pipe that delivers water to hydro turbines and sewerage systems. The term is inherited from the earlier technology of mill ponds and watermills. Name this term.

a. Dam b. Penstock c. delivery pipe d. control gate 6. This type of water turbine uses Newton’s 3rd law (reciprocal reactions or conservation of momentum) and describes the transfer of energy for this type of turbines. Most water turbines in

114

use are of this type. They are used in either low or medium-heart applications. Name this turbine type.

a. Impulse turbines b. Reaction turbines c. Impellers d. Centrifugal pumps 7. Select an example of an impulse turbine.

a. Francis b. Propeller c. Kaplan d. Turgo wheels 8. Which type of turbine considering flow types will one select if the specific speeds ranged from 2000-8000?

a. Centrifugal b. Radial c. Mixed flow d. Axial flow 9. Select an example of a reaction turbine.

a. Francis b. Pelton c. Turgo wheels d. Crossflow

115

10. In one section of a river in Turtle Island, the stream flow was estimated to be 555 gpm. The total dynamic head was measured to be 120 ft. Determine the theoretical power (in kW) that could be generated from this water stream.

a. 12.6 b. 7.92 c. 12,566 d. 7920 Solution: a. The theoretical power is calculated from the hydro power equation as follows: 𝑃𝑃𝑡𝑡 (𝑊𝑊 ) =

(555 𝑔𝑔𝑔𝑔𝑔𝑔) × (120 𝑓𝑓𝑓𝑓) 𝑄𝑄 × 𝐻𝐻 == = 12,566 𝑊𝑊 [12.6 𝑘𝑘𝑘𝑘] 5.3 5.3

𝑃𝑃𝑡𝑡 (𝑊𝑊 ) =

ℎ𝑝𝑝 𝑚𝑚𝑚𝑚𝑚𝑚 746 𝑊𝑊 555 gallons 8.34 𝑙𝑙𝑙𝑙𝑙𝑙 × × 120 𝑓𝑓𝑓𝑓 × × = 12,556 𝑊𝑊 33,000 𝑓𝑓𝑓𝑓 𝑙𝑙𝑙𝑙 min 𝑔𝑔𝑔𝑔𝑔𝑔 ℎ𝑝𝑝

b. To verify the validity of this equation we can use the original equation as follows:

11. Which type of turbine considering flow types will one select if the specific speeds ranged from 500-4000?

a. Centrifugal b. Radial c. Mixed-flow d. axial flow 12. The water stream in a river has a velocity of 15 ft/min. The approximate cross section where the velocity was measured can be approximated by a trapezoid with 25 ft on top and 15 ft as the width at the bottom with a height of 5 ft. The dynamic head was measured to be around 20 ft.

Determine the theoretical power (in hp) that could be generated from this water stream.

a. 35.5 116

b. 56.7 c. 71 d. 113.45 Solution: a. The mass flow rate is first calculated using the product of density of water, crosssectional area and velocity as follows: 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 (𝑓𝑓𝑓𝑓 2 ) = 𝑄𝑄 �

(𝐵𝐵1 + 𝐵𝐵2 ) (15 + 25) 𝑓𝑓𝑓𝑓 × 𝐻𝐻 = × 5 𝑓𝑓𝑓𝑓 = 100 𝑓𝑓𝑓𝑓 2 2 2

𝑙𝑙𝑙𝑙𝑙𝑙 62.4 𝑙𝑙𝑙𝑙𝑙𝑙 15 𝑓𝑓𝑓𝑓 𝑙𝑙𝑙𝑙𝑙𝑙 � = 𝜌𝜌𝜌𝜌𝜌𝜌 = × 100 𝑓𝑓𝑓𝑓 2 × = 93,600 3 𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚

b. Thus, the theoretical power is then calculated as follows: 𝑙𝑙𝑙𝑙𝑙𝑙 93,600 𝑚𝑚𝑚𝑚𝑚𝑚 × 20 𝑓𝑓𝑓𝑓 𝑄𝑄 × 𝐻𝐻 𝑃𝑃𝑡𝑡 (ℎ𝑝𝑝) = = = 56.7 ℎ𝑝𝑝 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙𝑙𝑙 33,000 33,000 ℎ𝑝𝑝 − 𝑚𝑚𝑚𝑚𝑚𝑚

13. A small commercial micro-hydro power unit had the following data: the volumetric flow rate was 70 liters/sec and the dynamic head was 1.5 meters. This unit was reported to produce 500 Watts of continuous power output at those conditions. Calculate the yearly output of this small device in units of kWh/year assuming 24 hrs per day and 365 days in a year.

a. 8,340 b. 4,380 c. 1,029 d. 2,190 Solution a. The volumetric flow rate is converted into units of kg/s 𝑄𝑄 �

𝑘𝑘𝑘𝑘 70 𝐿𝐿 1 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 �= × = 70 𝑠𝑠 𝑠𝑠 𝐿𝐿 𝑠𝑠

117

b. Then the theoretical power is calculated 𝑃𝑃𝑡𝑡 (𝑘𝑘𝑘𝑘) = Q × H =

70 𝑘𝑘𝑘𝑘 𝑁𝑁 × 𝑠𝑠 2 9.8 𝑚𝑚 𝑊𝑊 × 𝑠𝑠 × 1.5 𝑚𝑚 × × 2 × = 1,029 𝑊𝑊 𝑠𝑠 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑠𝑠 𝑁𝑁 × 𝑚𝑚

c. The efficiency is calculated using the ratio of actual and theoretical power as shown. 𝜂𝜂 (%) =

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 500 𝑊𝑊 × 100% = × 100% = 48.6% 1,029 𝑊𝑊 𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃

d. The yearly output may then be estimated as follows: 𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 �

𝑘𝑘𝑘𝑘ℎ 24 ℎ𝑟𝑟𝑟𝑟 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 1 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘ℎ � = 500 𝑊𝑊 × × × = 4,380 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦 𝑑𝑑𝑑𝑑𝑑𝑑 1,000 𝑊𝑊 𝑦𝑦𝑦𝑦

14. Determine the theoretical power in watts for water flowing at 50 kg/s and dropped from a dynamic head of 10 meters.

a. 9800 b. 2450 c. 7350 d. 4900 Solution: a. The use of Equation 5.1 is straight forward but making use of the acceleration due to gravity constant of 9.8 m/s2 as well as the Newton’s Law for the units of force as a function of mass and acceleration. 𝑃𝑃𝑡𝑡 (𝑊𝑊 ) =

50 𝑘𝑘𝑘𝑘 𝑁𝑁 × 𝑠𝑠 2 9.8 𝑚𝑚 𝑊𝑊 × 𝑠𝑠 × 10 𝑚𝑚 × × 2 × = 4,900 𝑊𝑊 𝑠𝑠 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑠𝑠 𝑁𝑁 × 𝑚𝑚

15. Given a pump rotational speed of 1800 rpm, flow rate of 1200 gpm and head of 20 ft. Calculate the specific speed.

a. 1746 b. 1467

118

c. 1190 d. 6593 Solution: a. The governing equation is used as follows: 𝑁𝑁𝑠𝑠 =

𝑛𝑛 × �𝑄𝑄 3 (𝐻𝐻)4

=

1800 × √1200 3

(20)4

= 6,593

b. This is the specific speed for typical propellers and axial fans.

16. The power generation facility in King Cove, Alaska has a water source that generates an average stream flow of 5 m3 per second. The dynamic head of the facility where a hydro power plant is to be installed was estimated to be 95 meters. Assume an overall conversion efficiency of 80%, estimate the actual power that could be produced from this facility in kW.

a. 840 b. 1375 c. 3725 d. 2156 Solution: a. The volumetric flow rate is first calculated as follows: 𝑄𝑄 �

𝑘𝑘𝑘𝑘 5 𝑚𝑚3 1,000 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 �= × = 5,000 3 𝑠𝑠 𝑠𝑠 𝑚𝑚 𝑠𝑠

b. Thus, the actual power may be easily calculated as follows: 𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) =

𝑄𝑄 × 𝐻𝐻 × 𝐸𝐸𝑜𝑜 5000 × 95 × 0.80 = = 3,725.5 𝑘𝑘𝑘𝑘 102 102

c. To verify the validity of the equation, the following calculation may also be used. 𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) = 5,000

𝑘𝑘𝑘𝑘 𝑁𝑁𝑠𝑠 2 9.8 𝑚𝑚 𝑊𝑊𝑊𝑊 𝑘𝑘𝑘𝑘 × 95 𝑚𝑚 × × 2 × × × 0.80 = 3,724 𝑘𝑘𝑘𝑘 𝑠𝑠 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑠𝑠 𝑁𝑁𝑁𝑁 1,000 𝑊𝑊 119

17. A small commercial micro-hydro power unit had the following data: the volumetric flow rate was 70 liters/sec and the dynamic head was 1.5 meters. This unit was reported to produce 500 Watts of continuous power output at those conditions. Determine the actual efficiency of this unit.

a. 48.6 b. 72.9 c. 60.75 d. 80 Solution a. The volumetric flow rate is converted into units of kg/s 𝑄𝑄 �

𝑘𝑘𝑘𝑘 70 𝐿𝐿 1 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 �= × = 70 𝑠𝑠 𝑠𝑠 𝐿𝐿 𝑠𝑠

b. Then the theoretical power is calculated 𝑃𝑃𝑡𝑡 (𝑘𝑘𝑘𝑘) = Q × H =

𝑁𝑁 × 𝑠𝑠 2 9.8 𝑚𝑚 𝑊𝑊 × 𝑠𝑠 70 𝑘𝑘𝑘𝑘 × 1.5 𝑚𝑚 × × 2 × = 1,029 𝑊𝑊 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑠𝑠 𝑁𝑁 × 𝑚𝑚 𝑠𝑠

c. The efficiency is calculated using the ratio of actual and theoretical power as shown. 𝜂𝜂 (%) =

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 500 𝑊𝑊 × 100% = × 100% = 48.6% 𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 1,029 𝑊𝑊

18. Determine the power (MW) generated from the Kalayaan Pumped Storage facility is the dynamic head is 290 meters and the volumetric flow rate is 270 cubic meters per second. Assume overall efficiency of 80%.

a. 273 b. 614 c. 819 d. 409.5 120

Solution: a. The actual power generated is calculated using Equation 5.3 𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) =

𝑄𝑄 × 𝐻𝐻 × 𝐸𝐸𝑜𝑜 102

𝑚𝑚3 1000 𝑘𝑘𝑘𝑘 270 𝑠𝑠 × × 290 𝑚𝑚 × 0.80 𝑚𝑚3 𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) = = 614,118 𝑘𝑘𝑘𝑘 [614 𝑀𝑀𝑀𝑀] 102

b. The Kalayaan Pumped Storage Power Plant generated 620 MW through two Francis Turbines. This value is very close to the installed capacity.

19. Estimate the power output (MW) from one of the turbines in of the Bhumibol hydro power plant with the following data: dynamic head = 150 meters, volumetric flow rate in one turbine was measured as 97,750 kg/s. Assume an overall conversion efficiency is 80%.

a. 57.5 b. 230 c. 115 d. 345 Solution: a. Since the units for Q and H are those defined in Equation 5.2, then the calculation of theoretical power is straight forward as follows. 𝑃𝑃𝑡𝑡 (𝑘𝑘𝑘𝑘) =

𝑄𝑄 × 𝐻𝐻 97,500 × 150 = = 143,750 𝑘𝑘𝑘𝑘 [143.75 𝑀𝑀𝑀𝑀] 102 102

b. Then using the overall conversion efficiency, the actual power is calculated. 𝑃𝑃𝑎𝑎 (𝑀𝑀𝑀𝑀) = 143.75 𝑀𝑀𝑀𝑀 × 0.80 = 115 𝑀𝑀𝑀𝑀

c. This power is the output of the Pelton turbine in the Bhumibol Dam Hydro Power plant (115 MW).

121

20. Determine the total maximum time (in hrs) that the water from the Caliraya Lake will be emptied when power is generated from the Kalayaan Pumped Storage facility continuously. The volumetric flow rate is 270 cubic meters per second and the Lake Caliraya has a storage volume of 80 x 106 m3 during rainy season.

a. 370 b. 82 c. 277.5 d. 562.5 Solution: a. The time (in hours) for Lake Caliraya to be emptied is simply the ratio of the storage volume and the amount of water used for the power plant as shown below. 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) =

𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 (𝑚𝑚3 ) 1 ℎ𝑟𝑟 × 3 𝑚𝑚 3600 𝑠𝑠 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 � 𝑠𝑠 �

80 × 106 𝑚𝑚3 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = = 82 ℎ𝑟𝑟𝑟𝑟 [3.4 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑] 270 𝑚𝑚3 3600 𝑠𝑠 × 𝑠𝑠 ℎ𝑟𝑟

b. Note that at this storage volume, the volume will be empty after about 82 hours or over 3.4 days. This is the full capacity of the power plant.

122

Chapter 6 Geothermal Energy 6.11 Solved Problems Problem 6.11.1 Well Selection Problem 6.11.1 Three wells were dug, and the temperature profile is shown. Select which well is most likely to be a potential geothermal site.

Well A (oC)

Well B(oC)

Well C(oC)

1 km

50

85

30

2 km

100

125

120

3 km

135

205

125

4 km

190

265

130

5 km

220

310

135

Depth (km)

Solution. a. The easiest way to select a certain well is by plotting its temperature profile. This is shown below. Well B has a good temperature profile which increases with depth and briefly may be a good choice. Well C does not have a high temperature, however, for low temperature applications, it may be a better choice.

123

Well Temperature Profile 350

Well Temperature (C)

300

-6

250 200 150 100 50 -5

-4

-3

-2

-1

0

0

Well depth (km) Well A (C)

Well B (C)

Well C(C)

b. Of course, upon inspection, Well B is superior without plotting. But Well C and A could be further inspected if well drilling is expensive, and low temperature is a target. In this case, those two wells may be ideal after all.

Problem 6.11.2. The Ideal Rankine Cycle Problem 6.11.2 Given the following data for the ideal Rankine Cycle. a. The boiler pressure, p2 = 2.5 Mpa = 2.5x106 Pa b. The boiler temperature, T3 = 280oC, and c. The condenser temperature, T1 = 60oC Determine the first law efficiency (η1).

Solution: a. Determine the properties first from tables and charts. One is shown in Figure 6.7 b. The saturated water properties table (Table A-7M at right) gives the following: p1 =19,940 Pa v1 = 1.0172x10-3 m3/kg h1=251.13 kJ/kg 124

c. From equation (1) ℎ2 = ℎ1 + 𝑣𝑣1 (𝑝𝑝2 − 𝑝𝑝2 )

3 kJ −3 m (2,500 − 19.94) kN2 = 253.65 kJ h 2 = 251.13 + 1.0172 × 10 kg kg kg m

d. From the table of compressed water properties (Table A-10M see Table 6.2) we have h = 253.315 kJ/kg at 60oC (by interpolating between 40oC and 80oC data) and 2.5 Mpa.

125

Table 6.1 Thermodynamic properties of water-saturation temperature table

126

Table 6.2 Properties of water-compressed liquid table.

e. From the superheated vapor table (Table A-9M see right) we have the following: h3 = 2,958.85 kJ/kg (interpolating between 2.0 and 3.0 MPa) at 280oC. s3 = 6.5645 kJ/kg From the previous table (A-7M) we also have s1 = 0.8226 kJ/kg s4’ =7.9096 kJ/kg The data for superheated vapor is shown in Table 6.3

127

Table 6.3 Superheated vapor table.

128

f. Using Equation 6.2 with T1 = 333.5oK (i.e., 273.15oK + 60oC) we obtain h4 = h1 + T1(s3 – s1) ℎ4 = 251.13

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 + 333.15𝑜𝑜 𝐾𝐾 × (6.5645 − 0.8226) 𝑜𝑜 = 2,164 𝑘𝑘𝑘𝑘 𝑘𝑘𝑔𝑔 𝐾𝐾 𝑘𝑘𝑘𝑘

g. The quality of steam is then estimated: 𝑥𝑥 = 𝑥𝑥 =

(𝑠𝑠3 − 𝑠𝑠1 ) (𝑠𝑠4′ − 𝑠𝑠1 )

(6.5645 − 0.8226) = 0.810 (7.9096 − 0.8226)

h. The energy flows are calculated as follows: Work input from pump 𝑤𝑤12 = (ℎ2 − ℎ1 ) = (253.65 − 251.13) Heat input from boiler

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 = 2.52 𝑜𝑜 𝑘𝑘𝑔𝑔 𝐾𝐾 𝑘𝑘𝑘𝑘

𝑞𝑞23 = (ℎ3 − ℎ2 ) = (2,958.85 − 253.65) Work output to turbine

𝑤𝑤34 = (ℎ3 − ℎ4 ) = (2,958.85 − 2,164) The heat output to the condenser

𝑞𝑞41 = (ℎ4 − ℎ1 ) = (2,164 − 251.13)

i. Thus, the first law efficiency will be: 𝜂𝜂1 =

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 = 2,705.2 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 = 794.85 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 = 1,912.87 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘

(𝑤𝑤34 − 𝑤𝑤12 ) × 100% (𝑞𝑞23 ) 129

𝜂𝜂1 =

(794.85 − 2.52) × 100% = 29.3% (2,705.2)

j. The schematic for points and their values are shown in Figure 6.8.

The summary of state points in the ideal Rankine Cycle is shown in the table below.

Summary of state points in the ideal Rankine Cycle example.

State Point

Temperature Temperature o

Pressure

v

h

S

C

o

K

kPa

10-3m3/kg

kJ/kg

kJ/kgoK

1

60

333.15

19.94

1.0172

251.13

0.8226

2

60

333.15

2,500

1.0174

253.65

?

2’

280

553.15

?

?

?

?

3’

280

553.15

?

?

?

?

3

280

553.15

2,500

?

2,958.85

6.5645

4

60

333.15

?

?

2,164

?

4’

60

333.15

19.94

19,523

?

7.9096

130

Problem 6.11.3 Efficiency of Ideal Geothermal Cycle Problem 6.11.3 Assume that an ideal power plant with 100% isentropic efficiency has a source with a temperature of 120oC and a steam rate of 150 kg/s of flow. The cooling water is assumed to enter the power plant at 10oC and leave the plant at 20oC. Assume further that the power plant can cool down the geothermal fluid to 75oC. The resulting configuration is schematically shown in the figure. Determine the efficiency of this system.

Solution: a. The efficiency of the ideal geothermal power plant is simply the ratio of the work output to the energy input as follows: 𝜂𝜂𝑡𝑡 =

𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 6,250 𝑘𝑘𝑘𝑘 × 100% = × 100% = 19.5% 𝑄𝑄𝑖𝑖𝑖𝑖 32,051 𝑘𝑘𝑘𝑘

b. The ideal efficiency is rather low.

131

Problem 6.11.4. Changes in Efficiency and Power Output.

Problem 6.11.4 Assume the same configuration as Problem 6.3, with a 100% isentropic efficiency plant which has a source with a temperature of 120oC and a steam rate of 150 kg/s of flow. The cooling water is assumed to enter the power plant at 10oC and leave the plant at 20oC. However, assume that the power plant can cool the geothermal fluid down to 45oC instead of 75oC. The resulting configuration and new values for energy output is schematically shown in the figure below. Determine the new efficiency of this system.

Solution: a. The new efficiency of the ideal geothermal power plant is, as usual, the ratio of the work output to the energy input as follows: 𝜂𝜂𝑡𝑡 =

𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 8,500 𝑘𝑘𝑘𝑘 × 100% = × 100% = 16.9% 𝑄𝑄𝑖𝑖𝑖𝑖 50,200 𝑘𝑘𝑘𝑘

132

b. The ideal efficiency went down further to 16.9% even though the output power went up to 8.5 MW. Problem 6.11.5. COP of Ideal Refrigeration Cycle Problem 6.11.5 A refrigerator is to be maintained at 5oC by removing heat at a rate of 350 kJ/min. If the required power to the refrigerator is 1.9 kW, determine the (a) COP, and (b) the rate of discharge of energy to the environment (i.e., the kitchen that houses the refrigerator).

Solution: a. The COP is given by Equation 6.9. 𝐶𝐶𝐶𝐶𝐶𝐶 =

𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑄𝑄𝐿𝐿 350,000 𝐽𝐽 𝑊𝑊𝑊𝑊 1 𝑚𝑚𝑚𝑚𝑚𝑚 = = × × =3 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑊𝑊 1,900 𝑊𝑊 × 𝑚𝑚𝑚𝑚𝑚𝑚 𝐽𝐽 60 𝑠𝑠

b. The rate of discharge to the environment will be the sum of the work done and the refrigeration effect as follows: 𝑘𝑘𝑘𝑘 60 𝑚𝑚𝑚𝑚𝑚𝑚 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑄𝑄𝐿𝐿 + 𝑊𝑊 = 𝑄𝑄𝐻𝐻 = 350 + 1.9 𝑘𝑘𝑘𝑘 × = 464 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚

133

Figure 6.9 The TS diagram for the vapor refrigeration cycle (VRC).

Figure 6.10 Vapor refrigeration cycles.

134

Problem 6.11.6. Ideal Vapor Refrigeration System Problem 6.11.6 An ideal-vapor refrigeration cycle uses Freon 12 as its working fluid. The saturation temperature in the condenser is 120oF, and the evaporator temperature is 30oF. Calculate the COP and the horsepower (hp) input necessary to produce the cooling effect of 5 tons. Draw the diagram for this process.

Solution: a. A ton refrigeration effect is defined as the heat rate to melt a ton of ice in a 24 hr period, also equal to 12,000 Btu/hr or 3.516 kW. The diagram is shown in Figure 6.11. b. The next step is to consult the properties table for the properties of the working fluid. In this example, Freon 12. c. The next step is to consult the properties table for the properties of the working fluid. In this example, Freon 12.

135

Figure 6.11 Ideal Refrigeration Cycle

136

d. The properties table gives the following data: h1 = 80.419 Btu/lbm (saturated vapor at 30oF) s1 = 0.16648 Btu/lbmoR h3 = h4=36.013 Btu/lbm p3 = p2=172.35psia e. One should consult the superheat tables/charts to obtain the value of h2 = 91.5 Btu/lbm thus, 𝐶𝐶𝐶𝐶𝐶𝐶 = 𝐶𝐶𝐶𝐶𝐶𝐶 =

𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑄𝑄𝐿𝐿 (ℎ1 − ℎ4 ) = = 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑊𝑊 (ℎ2 − ℎ1 ) (ℎ1 − ℎ4 ) 80.4 − 36.0 = =4 (ℎ2 − ℎ1 ) 91.5 − 80.4

f. Properties in the superheat region require inter/extrapolation if taken from the tables, since temperature or pressure values are not given exactly.

For this example, the pressure was found to be 172.35 psia, so the value of h2 must be interpolated between 150 psia and 175 psia. Further, the value of s1=s2=0.16648 and must also be interpolated to receive h2 = 91.5 Btu/lb.

Work input 𝑊𝑊 = ℎ2 − ℎ1 = 91.45 − 80.419 = 11.1 Total rate of flow

𝐵𝐵𝐵𝐵𝐵𝐵 𝑙𝑙𝑙𝑙

𝑄𝑄𝐿𝐿 = 𝑚𝑚̇ × (ℎ1 − ℎ4 ) 𝑚𝑚̇ =

𝑄𝑄𝐿𝐿 5 × 12,000 𝑙𝑙𝑙𝑙𝑚𝑚 = = 1,315 (ℎ1 − ℎ4 ) (80.4 − 36.0) ℎ𝑟𝑟

Total work required

𝑊𝑊𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑚𝑚̇ × (ℎ2 − ℎ1 )

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𝑊𝑊𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 1,315

𝑙𝑙𝑏𝑏𝑚𝑚 𝐵𝐵𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵𝐵𝐵 × (91.5 − 80.419) = 14,970 ℎ𝑟𝑟 𝑙𝑙𝑙𝑙𝑚𝑚 ℎ𝑟𝑟

𝑊𝑊𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 14,970

𝐵𝐵𝐵𝐵𝐵𝐵 1 ℎ𝑝𝑝 × = 5.9 ℎ𝑝𝑝 ℎ𝑟𝑟 2,544 𝐵𝐵𝐵𝐵𝐵𝐵 ℎ𝑟𝑟

Table 6.5 Table of properties for Freon 12

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Table 6.6 Table of properties for superheated Freon 12

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Problem 6.11.7. Power Consumed in Heat Pump Problem 6.11.7 A heat pump is used to meet the heating requirements of a house and maintain its temperature at 25oC [77oF]. On a day when the outdoor temperature is 10oC [50oF], the house is estimated to lose heat at a rate of 50,000 kJ/hr. If the heat pump has a COP of 3, determine (a) the power consumed by the heat pump and (b) the rate at which heat is extracted from the cold outdoor air.

Solution: a. The power consumed by the heat pump is simply the work input as follows: 𝑘𝑘𝑘𝑘 50,000 ℎ𝑟𝑟 𝑄𝑄𝐻𝐻 𝑘𝑘𝑘𝑘 [4.6 𝑘𝑘𝑘𝑘 ] 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛,𝑖𝑖𝑖𝑖 = = = 16,667 𝐶𝐶𝐶𝐶𝐶𝐶 3 ℎ𝑟𝑟

b. The rate at which energy is extracted from the house is the same energy rate to keep the temperature at 25oC [77oF] and is equal to 50,000 kJ/hr. The difference will be the heat extracted from the cold outdoor air as follows: 𝑄𝑄𝐿𝐿 = 𝑄𝑄𝐻𝐻 − 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛,𝑖𝑖𝑖𝑖 = 50,000

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 [9.26 𝑘𝑘𝑘𝑘] − 16,667 = 33,333 ℎ𝑟𝑟 ℎ𝑟𝑟 ℎ𝑟𝑟

c. Note that if all the heating comes from an electrical resistance heater, a total of 50,000 KJ/hr or 13.9 kW would be needed. As such, heat pumps are popularly used to provide heating effects. However, the initial capital cost of a heat pump can be quite high.

Problem 6.11.8. Cost Comparison.

Problem 6.11.8 A heat pump using Freon 12 as its refrigerant was used to meet the heating requirements of a facility and maintain its temperature at 120oF. The outdoor temperature was 40oF. The compressor efficiency was 80%. The heating requirements are 100,000 Btu/hr. The electricity cost was 0.10/kWh. Compare the cost of operating this heat pump with that of a fuel

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oil heater which would cost $25 per 106 Btu. Assume a 70% efficient oil heater. Find the COP as well.

Solution: a. The following parameters are taken from the table: h1 = 81.436 Btu/lbm (saturated vapor at 40oF, see earlier Table) s1 = 0.16586 Btu/lbmoR = s2 h3=h4=36.013 Btu/lbm (saturated liquid at 120oF) p3=p2=172.35 psia (saturation pressure at 120oF) b. Using s1 = s2 = 0.16586 and p = 172.5 psia and by interpolation h2 = 90.64 Btu/lbm c. To estimate h2’, the compressor efficiency equation is written as follows: 𝑥𝑥 =

(ℎ2 − ℎ1 ) (ℎ2′ − ℎ1 )

0.85 =

(90.64 − 81.436) (ℎ2′ − 81.436)

ℎ2′ = 92.264

𝐵𝐵𝐵𝐵𝐵𝐵 𝑙𝑙𝑏𝑏𝑚𝑚

d. The flow rate is then calculated. 𝑄𝑄𝐻𝐻 = 𝑚𝑚̇ × (ℎ2′ − ℎ3 ) = 100,000

𝐵𝐵𝐵𝐵𝐵𝐵 ℎ𝑟𝑟

𝐵𝐵𝐵𝐵𝐵𝐵 1000,000 ℎ𝑟𝑟 𝑄𝑄𝐻𝐻 𝑙𝑙𝑙𝑙𝑚𝑚 𝑚𝑚̇ = = = 1,778 (ℎ2′ − ℎ3 ) (92.264 − 36.013) ℎ𝑟𝑟

e. The work input will be as follows:

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𝑊𝑊 = 𝑚𝑚̇ × (ℎ2′ − ℎ1 ) = 1,778 = 19,252

𝑙𝑙𝑙𝑙𝑚𝑚 𝐵𝐵𝐵𝐵𝐵𝐵 × (92.264 − 81.436) ℎ𝑟𝑟 𝑙𝑙𝑏𝑏𝑚𝑚

𝐵𝐵𝐵𝐵𝐵𝐵 [5.64 𝑘𝑘𝑘𝑘] ℎ𝑟𝑟

f. The cost for the heat pump will be as follows: 𝐶𝐶𝐻𝐻𝐻𝐻 = �

$0.10 � × 5.64 𝑘𝑘𝑘𝑘 = $0.564/ℎ𝑟𝑟 𝑘𝑘𝑘𝑘ℎ

g. The unit heating cost (per million Btu) will be as follows: 𝐶𝐶𝐻𝐻𝐻𝐻 $0.564/ℎ𝑟𝑟 $5.64 = = 6 6 6 10 𝐵𝐵𝐵𝐵𝐵𝐵 0.10 × 10 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ𝑟𝑟 10 𝐵𝐵𝐵𝐵𝐵𝐵

h. For the oil heater, the following calculations should be made: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (%) = 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 × 100% 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼

100,000 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ𝑟𝑟 𝐵𝐵𝐵𝐵𝐵𝐵 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 × 100% = × 100% = 142,857 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (%) 70% ℎ𝑟𝑟

i. At $25/106Btu, the cost will be: 𝐶𝐶𝐹𝐹𝐹𝐹 = �

𝐵𝐵𝐵𝐵𝐵𝐵 $25 � × 0.142857 × 106 = $3.57/ℎ𝑟𝑟 6 ℎ𝑟𝑟 10 𝐵𝐵𝐵𝐵𝐵𝐵

$3.57/ℎ𝑟𝑟 $35.70 𝐶𝐶𝐹𝐹𝐹𝐹 = = 106 𝐵𝐵𝐵𝐵𝐵𝐵 0.10 × 106 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ𝑟𝑟 106 𝐵𝐵𝐵𝐵𝐵𝐵

j. The heat pump cost is 6x lower than the oil heater cost. k. The COP may be calculated as well. 𝐶𝐶𝐶𝐶𝐶𝐶𝐻𝐻 =

𝑄𝑄𝐻𝐻 100,000 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ𝑟𝑟 = = 5.194 𝑊𝑊 19,252 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ𝑟𝑟

Property values are taken from Table 6.7

The following parameters are taken from the table: a. h1 = 81.436 Btu/lbm 142

b. s1=0.16586 Btu/lbmoR c. h3 = 36.013 Btu/lbm d. h4 = 36.013 Btu/lbm e. p2 = 172.35 psia f. p3 = 172.35 psia Table 6.7 Properties of Freon 12 – Saturation temperature table.

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Problem 6.11.9 Number of Household Served by Geothermal Facility Problem 6..11.9 The average annual consumption of electricity in a developing country was reported at 3,235 kWh/yr. If the annual generation of power for a large geothermal power plant was reported at 2.4M MWh, how many households can this facility serve?

Solution: a. The number of household populations served by this geothermal facility will simply be the ratio of annual production divided by the yearly household consumption. However, the units must be the same. Hence, the production units must be converted into units of kWh/year as follows: 𝑘𝑘𝑘𝑘ℎ 2.4 × 106 𝑀𝑀𝑀𝑀ℎ 1,000 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘ℎ 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 � �= × = 2.4 × 106 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 1 𝑀𝑀𝑀𝑀 𝑦𝑦𝑦𝑦

b. The number of households served will be as follows: # 𝑜𝑜𝑜𝑜 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 = 2.4 × 109

𝑘𝑘𝑘𝑘ℎ 𝐻𝐻𝐻𝐻 × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 × = 741,931 𝐻𝐻𝐻𝐻 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 3,235 𝑘𝑘𝑘𝑘ℎ

Problem 6.11.10 ROI of Geothermal Heating and Cooling

Problem 6.11.10 Determine the return to investment for a household which invested $26,000 into geothermal heating and cooling. Their average monthly geothermal usage was 416 kWh/month and they paid an average of $0.138/kWh over the years. In how many years can they recover their initial investment for heating and cooling? If they have a rebate of $2,400, what is the ROI after the rebate?

Solution: a. The monthly savings on electricity is calculated as follows: Monthly Savings = $0.138/kWh x 416 kWh = $57.41 b. The yearly savings will be as follows: 144

Yearly Savings = $51.41/month x 12 months = $688.90/year c. Their return to investment without the rebate is calculated as follows: ROI = $26,000/$688.90/yr = 37.74 years. d. After applying the ROI, the rebate is calculated as follows: ROI = $23,600/$688.90/yr = 34.26 years. e. They have saved about 3 years after the rebate.

6.12.11

Operational Cost of a Heat Pump

P6.11 A heat pump using Freon 12 as refrigerant was used to meet the heating requirements of a house to maintain its temperature at 77oF (25oC). The outdoor temperature was around 50oF (10oC). The house heating requirements (85,000 Btu/hr) require a work input of 18,000 Btu/hr.

The electricity cost was 0.10/kWh. Determine the cost of operating this heat pump per unit of time (in $/hr). a. $0.70 b. $0.15 c. $0.53 d. $0.053 Solution: a. First, the work input is simply converted into units of kW as follows: 𝑊𝑊𝑖𝑖𝑖𝑖 = 18,000

𝐵𝐵𝐵𝐵𝐵𝐵 𝑘𝑘𝑘𝑘ℎ × = 5.28 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 3412 𝐵𝐵𝐵𝐵𝐵𝐵

b. Then the cost per hour is calculated as follows: 𝐶𝐶𝐻𝐻𝐻𝐻 =

$0.10 $0.53 × 5.28 𝑘𝑘𝑘𝑘 = 𝑘𝑘𝑘𝑘ℎ 𝑘𝑘𝑘𝑘ℎ

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6.12.12

Heat Pump Cost per Unit of Energy

P6.12 A heat pump using Freon 12 as refrigerant was used to meet the heating requirements of a house to maintain its temperature at 77oF (25oC). The outdoor temperature was around 50oF (10oC). The house heating requirements (85,000 Btu/hr) require a work input of 18,000 Btu/hr.

The electricity cost was 0.10/kWh. Determine the cost per unit of energy usage (in $/ MMBtu). a. $6.53 b. $12.42 c. $5.30 d. $6.21 Solution: a. First, the work input is simply converted into units of kW as follows: 𝑊𝑊𝑖𝑖𝑖𝑖 = 18,000

𝐵𝐵𝐵𝐵𝐵𝐵 𝑘𝑘𝑘𝑘ℎ × = 5.28 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 3412 𝐵𝐵𝐵𝐵𝐵𝐵

b. Then the cost per hour is calculated as follows: 𝐶𝐶𝐻𝐻𝐻𝐻 =

$0.10 $0.53 × 5.28 𝑘𝑘𝑘𝑘 = 𝑘𝑘𝑘𝑘ℎ ℎ𝑟𝑟

c. The cost of the heat pump per million Btu (MMBtu) will be as follows: 𝐶𝐶𝐻𝐻𝐻𝐻 $0.53 ℎ𝑟𝑟 $6.21 = × = 6 6 10 𝐵𝐵𝐵𝐵𝐵𝐵 ℎ𝑟𝑟 85,000 𝐵𝐵𝐵𝐵𝐵𝐵 10 𝐵𝐵𝐵𝐵𝐵𝐵

6.12.13

Number of Households Serving a Geothermal Facility

P6.13 The average annual consumption of electricity in a developing country was reported at 3,235 kWh/yr. If the annual generation of power for a large geothermal power plant was reported at 2.4M MWh, how many households will this facility serve? a. 741,886 b. 370,966 146

c. 1,112,896 d. 241,025 Solution: a. The number of household population served by this geothermal facility will simply be the ratio of annual production divided by the yearly household consumption. However, the units must be the same. Hence the production units must be converted into units of kWh/year as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑘𝑘𝑘𝑘ℎ 2.4 × 106 𝑀𝑀𝑀𝑀ℎ 1,000 𝑘𝑘𝑘𝑘 2.4 × 109 𝑘𝑘𝑘𝑘ℎ �= × = 𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦 1 𝑀𝑀𝑀𝑀 𝑦𝑦𝑦𝑦

b. Hence the number of households served will be as follows: 𝑁𝑁𝑁𝑁. 𝑜𝑜𝑜𝑜 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 =

6.12.14

2.4 × 109 𝑘𝑘𝑘𝑘ℎ 𝐻𝐻𝐻𝐻 − 𝑦𝑦𝑦𝑦 × = 741,886 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑦𝑦𝑦𝑦 3,235 𝑘𝑘𝑘𝑘ℎ

Return on Investment (ROI) of Geothermal Heating and Cooling

P6.14 Determine the return to investment for a household that invested $26,000 in geothermal heating and cooling. Their average monthly geothermal usage was 416 kWh/month, and they pay an average of $0.138/kWh over the years. How many years will they recover their initial investment for heating and cooling alone? a. 20 b. 38 c. 10 d. 15 Solution: a. The monthly savings on electricity is calculated as follows: Monthly Savings = $0.138/kWh x 416 kWh = $57.41 147

b. The yearly savings will be as follows: Yearly Savings = $51.41/month x 12 months = $688.90/year c. Hence their return to investment without rebate is calculated as follows: ROI = $26,000/$688.90/yr = 37.74 years. d. The ROI after rebate is calculated as follows: ROI = $23,600/$688.90/yr = 34.26 years. e. They have saved about 3 years after the rebate.

6.12.15

Return on Investment (ROI) of Geothermal Heating and Cooling

P6.15 Determine the return to investment for a household that invested $26,000 in geothermal heating and cooling. Their average monthly geothermal usage was 416 kWh/month, and they pay an average of $0.138/kWh over the years. How many years will they recover their initial investment for heating and cooling alone if they have a rebate of $2,400? a. 23 b. 15 c. 34 d. 8 Solution: a. The monthly savings in electricity is calculated as follows: Monthly Savings = $0.138/kWh x 416 kWh = $57.41 b. The yearly savings will be as follows: Yearly Savings = $51.41/month x 12 months = $688.90/year c. Hence their return to investment without rebate is calculated as follows: ROI = $26,000/$688.90/yr = 37.74 years.

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d. The ROI after rebate is calculated as follows: ROI = $23,600/$688.90/yr = 34.26 years. e. They have saved about 3 years after the rebate.

Assess Your Knowledge 1. Geothermal energy finds its way to the earth’s surface in three ways. Name one possible way to locate potential geothermal energy project.

a. Himalayas b. Geysers c. Forests d. Deep Ocean 2. Which of the following geothermal resources is not found in Texas?

a. liquid dominated b. vapor dominated c. hot dry rock d. geo-pressure 3. Geothermal energy finds its way to the earth’s surface in three ways, name one possible way to locate potential geothermal energy projects.

a. Himalayas b. Deep Ocean c. Forests d. Fumaroles 4. Where are the largest geysers found in the world?

a. Iceland’s Strokkur Geyser 149

b. New Zealand’s Lady Knox c. California Geysers d. Russia’s Valley of Geysers 5. Which region in the world is considered one of the most active geothermal areas?

a. Ring of fire b. Eurasian plate c. Antarctic Plate d. African Plate 6. Determine the enthalpy in units of (kJ/kg) at point 2, (i.e., h2) in the ideal Rankine cycle with the following conditions: boiler pressure, p2 = 3.0 Mpa = 3.0x106 Pa; boiler temperature, T3 = 250oC; condenser temperature, T1 = 45oC. Please use the tables from lecture notes or from a thermodynamics textbook.

a. 191.46 b. 211.85 c. 188.45 d. 198.04 Solution: a. Determine the properties from tables b. Saturated water properties (Table A-7M from lecture notes) give the following: p1 =9,593 Pa v1 = 1.0099x10-3 m3/kg h1 = 188.44 kJ/kg c. From equation (1) ℎ2 = ℎ1 + 𝑣𝑣1 (𝑝𝑝2 − 𝑝𝑝1 )

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𝑚𝑚3 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 −3 (3,000 − 9.593) 2 = 191.46 ℎ2 = 188.44 + 1.0099 × 10 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑚𝑚 𝑘𝑘𝑘𝑘

7. A refrigerator is to be maintained at 4oC by removing heat from it at a rate of 375 kJ/min. If the required power to the refrigerator is 1.5 kW, determine the coefficient of performance (COP).

a. 6.25 b. 2.985 c. 4.2 d. 3.125 Solution: a. The COP is given by 𝐶𝐶𝐶𝐶𝐶𝐶 =

𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑄𝑄𝐿𝐿 375,000 𝐽𝐽 𝑊𝑊 − 𝑠𝑠 1 𝑚𝑚𝑚𝑚𝑚𝑚 = = × × = 4.2 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑊𝑊 1,500 𝑊𝑊 − 𝑚𝑚𝑚𝑚𝑚𝑚 𝐽𝐽 60 𝑠𝑠

b. The rate of discharge to environment will be the sum of the work done and refrigeration effect 𝑄𝑄𝐿𝐿 + 𝑊𝑊 = 𝑄𝑄𝐻𝐻 = 375

𝑘𝑘𝑘𝑘 60 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚 𝑘𝑘𝑘𝑘 + 1.5 𝑘𝑘𝑘𝑘 × = 465 𝑚𝑚𝑚𝑚𝑚𝑚 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑚𝑚

8. An ideal-vapor refrigeration cycle uses Freon 12 as the working fluid. The condenser temperature was 160oF, and the evaporator temperature was 60oF. Determine the value of s1, in units of (Btu/lbmoR), the entropy at the saturated vapor conditions after the condenser and before the expansion valve.

a. 0.11861 b. 0.16479 c. 0.04618 d. 0.16530 Solution: 151

a. From the table for Freon 12, the following data are found: h1 = 83.409 Btu/lbm (the enthalpy at the saturated vapor conditions after the condenser and before the expansion valve) b. The entropy at point 1 is as follows: s1 = 0.16479 Btu/lbmoR (the entropy at the saturated vapor conditions after the condenser and before the expansion valve) c. The value of enthalpy at point 3 is as follows: h3 = 46.633 Btu/lbm (the enthalpy after the evaporator and before the compressor) d. The value of p2 = p3 is as follows: p2 = 279.82 psiap3 = 279.82 psia 9. A heat pump is used to meet the heating requirements of a house and maintain its temperature at 23.9oC [75oF]. On the day when the outdoor temperature is 10oC [50oF], the house is estimated to lose heat at a rate of 45,000 kJ/hr. If the heat pump has a COP of 3, determine the rate at which heat is extracted from the cold outdoor air (in units of kW).

a. 6.54 b. 8.33 c. 4.17 d. 7.50 Solution: a. The power consumed by the heat pump is simply the work input as follows: 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛, 𝑖𝑖𝑖𝑖 =

𝑄𝑄𝐻𝐻 45,000 𝑘𝑘𝑘𝑘/ℎ𝑟𝑟 𝑘𝑘𝑘𝑘 [4.167 𝑘𝑘𝑘𝑘 ] = = 15,000 𝐶𝐶𝐶𝐶𝐶𝐶 3 ℎ𝑟𝑟

152

b. The rate at which energy is extracted from the house is the same energy rate to keep the temperature at 23.9oC [75oF] and thus is equal to 45,000 kJ/hr. Thus, the difference will be the heat extracted from the cold outdoor air as follows: 𝑄𝑄𝐿𝐿 = 𝑄𝑄𝐻𝐻 − 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛, 𝑖𝑖𝑖𝑖 = 45,000

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 − 15,000 = 30,000 [8.33 𝑘𝑘𝑘𝑘 ] ℎ𝑟𝑟 ℎ𝑟𝑟 ℎ𝑟𝑟

c. Note that if all the heating comes from an electrical resistance heater, a total of 45,000 kJ/hr or 12.5 kW will be needed. Hence, the popularity of heat pumps provides the heating effect. The initial capital cost of heat pumps is just quite high.

10. An ideal-vapor refrigeration cycle uses Freon 12 as the working fluid. The condenser temperature was 160oF, and the evaporator temperature was 60oF. Determine the value of p3 and p4, the pressures at Points 3 and 4 of the cycle in units of psia.

a. 172.35 b. 249.31 c. 313.000 d. 279.82 Solution: a. From the table for Freon 12, the following data are found: h1 = 83.409 Btu/lbm (the enthalpy at the saturated vapor conditions after the condenser and before the expansion valve) b. The entropy at point 1 is as follows: s1 = 0.16479 Btu/lbmoR (the entropy at the saturated vapor conditions after the condenser and before the expansion valve) c. The value of enthalpy at point 3 is as follows: h3 = 46.633 Btu/lbm (the enthalpy after the evaporator and before the compressor) 153

d. The value of p2 = p3 is as follows: p2 = 279.82 psiap3 = 279.82 psia 11. Name the ideal thermodynamic cycle that best describes power generation from geothermal energy.

a. Brayton Cycle b. Carnot Cycle c. Stirling cycle d. Rankine Cycle 12. Name the two common ingredients of hydrothermal features.

a. water and heat b. sulfur and water c. minerals and sulfur d. magma and minerals 13. Name the most basic geothermal power cycle.

a. non-condensing b. straight condensing c. single flash d. binary fluid 14. Which country in the world had the highest geothermal output?

a. Indonesia b. Philippines c. US d. Hawaii 15. This is a device that provides heat energy to space by extracting energy from a cold environment.

154

a. HRSG b. heat pump c. CHP d. space heater 16. Determine the enthalpy in units of (kJ/kg) at point 2, (i.e., h2) in the ideal Rankine cycle with the following conditions: boiler pressure, p2 = 5 Mpa = 5x106 Pa; boiler temperature, T3 = 250oC; condenser temperature, T1 = 55oC. Please use the tables from lecture notes or from a thermodynamics textbook.

a. 211.85 b. 235.3 c. 188.45 d. 1085.4 Solution: a. Determine the properties from tables b. Saturated water properties (Table A-7M from lecture notes) gives the following: p1 =15,760 Pa v1 = 1.0146x10-3 m3/kg h1 = 230.21 kJ/kg c. From equation (1) ℎ2 = ℎ1 + 𝑣𝑣1 (𝑝𝑝2 − 𝑝𝑝1 )

𝑘𝑘𝑘𝑘 𝑚𝑚3 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 −3 (5,000 − 15.76) 2 = 235.3 ℎ2 = 230.21 + 1.0146 × 10 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑚𝑚 𝑘𝑘𝑘𝑘

155

17. A refrigerator is to be maintained at 4oC by removing heat from it at a rate of 375 kJ/min. If the required power to the refrigerator is 1.5 kW, determine the rate of discharge of energy to the environment (i.e., the kitchen that houses the refrigerator) in units of kJ/min.

a. 245 b. 377 c. 625 d. 465 Solution: a. The COP is given by 𝐶𝐶𝐶𝐶𝐶𝐶 =

375,000 𝐽𝐽 𝑊𝑊 − 𝑠𝑠 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑄𝑄𝐿𝐿 = = × × = 4.2 𝑊𝑊 1,500 𝑊𝑊 − 𝑚𝑚𝑚𝑚𝑚𝑚 60 𝑠𝑠 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝐽𝐽

b. The rate of discharge to environment will be the sum of the work done and refrigeration effect 𝑄𝑄𝐿𝐿 + 𝑊𝑊 = 𝑄𝑄𝐻𝐻 = 375

𝑘𝑘𝑘𝑘 60 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚 𝑘𝑘𝑘𝑘 + 1.5 𝑘𝑘𝑘𝑘 × = 465 𝑚𝑚𝑚𝑚𝑚𝑚 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑚𝑚

18. An ideal-vapor refrigeration cycle uses Freon 12 as the working fluid. The condenser temperature was 160oF, and the evaporator temperature was 60oF. Determine the value of h1 (Btu/lbm), the enthalpy at the saturated vapor conditions after the condenser and before the expansion valve.

a. 83.409 b. 61.643 c. 21.766 d. 80.419 Solution: a. From the table for Freon 12, the following data are found:

156

h1 = 83.409 Btu/lbm (the enthalpy at the saturated vapor conditions after the condenser and before the expansion valve) 19. An ideal-vapor refrigeration cycle uses Freon 12 as the working fluid. The condenser temperature was 160oF, and the evaporator temperature was 60oF. Determine the value of h3 and h4 (Btu/lbm), the enthalpy after the evaporator and before and after the compressor.

a. 44.373 b. 91.006 c. 46.633 d. 43.850 Solution: a. One simply looks at Table A-15 properties of Freon 12 to find the enthalpy value of 46.633 across the temperature of 160F. 20. A heat pump using Freon 12 as refrigerant was used to meet the heating requirements of a house to maintain its temperature at 77oF (25oC). The outdoor temperature was 50oF (10oC). The house heating requirements (85,000 Btu/hr) require a work input of 18,000 Btu/hr. The electricity cost was 0.10/kWh. Determine the value of the work input in units of kW.

a. 10.56 b. 2.64 c. 17.06 d. 5.28 Solution: a. The work input is simply converted into units of kW as follows: 𝑊𝑊𝑖𝑖𝑖𝑖 = 18,000

𝐵𝐵𝐵𝐵𝐵𝐵 𝑘𝑘𝑘𝑘ℎ × = 5.28 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 3412 𝐵𝐵𝐵𝐵𝐵𝐵 157

Chapter 7 Salinity Gradient 7.13 Solved Problems Problem 7.13.1 Sensible Heat from Solar Pond P7.1 Half a hectare [1.235 acre] of pond with a depth of 6.1 meters [20 ft] is heated by the sun.

The average temperature of the pond has risen from 25oC to 30oC [77oF to 86oF] on a hot day.

Determine the amount of solar energy absorbed by this pond using the sensible heat Equation 7.1. Assume the density of sea water to be 1,027 kg/m3 [8.56 lbs/gal] and its specific heat equal to 3.93 kJ/kgoC [0.94 Btu/lboF].

Solution: a. First, the mass of the body of water heated is calculated. 𝑚𝑚(𝑘𝑘𝑘𝑘) = 0.5 ℎ𝑎𝑎 ×

100 𝑚𝑚 × 100 𝑚𝑚 1,027 𝑘𝑘𝑘𝑘 × 6.1 𝑚𝑚 × = 31,323,500 𝑘𝑘𝑘𝑘 1 ℎ𝑎𝑎 𝑚𝑚3

b. The heat energy equation is used.

𝑄𝑄 = 𝑚𝑚𝐶𝐶𝑝𝑝 ∆𝑡𝑡 = 31,323,500 𝑘𝑘𝑘𝑘 × 3.93

𝑘𝑘𝑘𝑘 × (30 − 25) 𝑜𝑜𝐶𝐶 = 615,503,775 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑜𝑜𝐶𝐶

c. This is the energy absorbed by this vast area, more than 615 million kJ. Problem 7.13.2 Theoretical Carnot Cycle Efficiency

P7.2 Calculate the theoretical Carnot cycle efficiency of a solar pond which has a high temperature reservoir of 90oC and a low reservoir temperature of 25oC. If the net output or work done in the system is 200 kW, at this theoretical efficiency, how much heat is put out of the system?

Solution: a. Equation 7.2 is simply used as follows:

158

𝜂𝜂𝑐𝑐 (%) = �1 −

(273.15 + 25)𝑜𝑜 𝐾𝐾 𝑇𝑇𝐶𝐶 � × 100% = �1 − � × 100% = 17.9% (273.15 + 90)𝑜𝑜 𝐾𝐾 𝑇𝑇𝐻𝐻

b. Using Equation 7.2, the heat output of the system is calculated as follows: 𝜂𝜂𝑐𝑐 (%) = �

𝑊𝑊 � × 100% 𝑄𝑄𝐻𝐻

𝑄𝑄𝐻𝐻 (𝑘𝑘𝑘𝑘) = �

200 𝑘𝑘𝑘𝑘 200 𝑘𝑘𝑘𝑘 � × 100% = � � × 100% = 1,117.3 𝑘𝑘𝑘𝑘 𝜂𝜂𝑐𝑐 (%) 17.9

𝑄𝑄𝐻𝐻 = 1,117.3 𝑘𝑘𝑘𝑘 ×

1,000 𝑊𝑊 ℎ𝑝𝑝 × = 1,498 ℎ𝑝𝑝 1 𝑘𝑘𝑘𝑘 746 𝑊𝑊

c. About over a MW of power is transferred to the reservoir.

Problem 7.13.3 Osmotic Pressure Calculations P7.3 Determine the osmotic pressure in units of kg/cm2 for the following data. The temperature is 650oK, the molar concentration of salt ions is equal to 1.128 mol/L, and the universal gas constant is 0.82 L-bar/deg-mol.

Solution: a. Equation 7.1 is used directly to get the following answers. 𝜋𝜋 = 1.128 × 0.082 × 650 = 60 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 = 60

b. The osmotic pressure is around 60 bars.

𝑘𝑘𝑘𝑘 𝑐𝑐𝑐𝑐2

Problem 7.13.4 Work Done Against Pressure P7.4 Calculate the work done against pressure using the following data: pressure equal to 60 kg/cm2; one liter of water pushed against an area of 2 cm2 over a distance of 20 meters.

Solution: a. The work or energy per liter of saltwater is as follows: 𝑊𝑊 = 60

𝑘𝑘𝑘𝑘 𝑁𝑁𝑠𝑠 2 𝑚𝑚 2 × 2𝑐𝑐𝑐𝑐 × 20 𝑚𝑚 = 2,400 𝑘𝑘𝑘𝑘 − 𝑚𝑚 × × 9.8 2 = 23,520 𝐽𝐽 2 𝑐𝑐𝑐𝑐 𝑘𝑘𝑘𝑘 − 𝑚𝑚 𝑠𝑠 159

b. This is the minimum amount of energy required to desalinate one liter of seawater.

Problem 7.13.5 Energy Required to Boil Seawater P7.5 Determine the amount of energy required to boil a kg of salt water from 30oC to 100oC. Assume the specific heat of water to be 3.93 kJ/kgoC. Solution a. The heat energy equation is used. 𝑄𝑄 = 𝑚𝑚𝐶𝐶𝑝𝑝 ∆𝑡𝑡 = 1 𝑘𝑘𝑘𝑘 × 3.93

𝑘𝑘𝑘𝑘 𝑜𝑜 𝑜𝑜 × (100 − 30) 𝐶𝐶 = 275 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝐶𝐶

b. This is the energy required to heat up a kg of salt water.

Problem 7.13.6 Size of PRO Unit to Generate Given Power

P7.6 The pressure retarded osmosis (PRO) salinity gradient prototype has a net power density of 15 W/m2. Determine the size (in m2) of the polyamide membrane needed to power a household with a monthly requirement of 1,000 kWh. Solution: a. This is a simple energy and power calculation as follows: 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝑚𝑚2 ) =

𝑚𝑚2 1 𝑚𝑚𝑚𝑚 1 𝑑𝑑𝑑𝑑𝑑𝑑 1,000 𝑊𝑊 1,000 𝑘𝑘𝑘𝑘ℎ × × × × = 92.6 𝑚𝑚2 𝑚𝑚𝑚𝑚 15 𝑊𝑊 30 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 24 ℎ𝑟𝑟𝑟𝑟 𝑘𝑘𝑘𝑘

b. An area of about 93 m2 is needed for this application.

Problem 7.13.7 Amount of Membrane to Use to Generate Power for a Household P7.7 If the average household in the US uses around 12,000 kWh of electricity per year, how much membrane is needed to power 1 household if the yearly production of this membrane is 30 MWh/yr per m2?

Solution:

160

a. This is a simple ratio of output to input as follows: 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝑚𝑚2 ) =

12,000 𝑘𝑘𝑘𝑘ℎ 𝑚𝑚2 𝑦𝑦𝑦𝑦 × = 0.40 𝑚𝑚2 𝑦𝑦𝑦𝑦 30,000 𝑘𝑘𝑘𝑘ℎ

b. Thus, a nanotube with an area of 0.40 m2 is needed. The value seems too good to be true.

Problem 7.13.8 RED Salinity Gradient System P7.8 The San Lorenzo River in California has an average annual discharge of 480 m3/s. The average salinity at this location is 0.6 mol/L. The nearby fresh water salinity is 0.003 mol/L; the salinity ratio is reported at 1:200. The average temperature is 25oC. Use the Nernst Equation (Equation 7.4) to calculate the voltage generated. Calculate the theoretical power which may be generated from this facility if the amount of energy produced is 0.74 MJ/m3 of flow instead of the erroneously reported 1.4 MJ/m3. If the overall efficiency is 1%, determine the estimated power output from this system.

Solution: a. The voltage is calculated from Equation 7.4: 𝑉𝑉 (𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉) =

𝐾𝐾 8.314 𝑚𝑚𝑚𝑚𝑙𝑙 𝑜𝑜 𝐾𝐾 × 298.15𝑜𝑜 𝐾𝐾 96,485 1 × 𝑚𝑚𝑚𝑚𝑚𝑚

× 𝑙𝑙𝑙𝑙

[0.60 𝑚𝑚𝑚𝑚𝑚𝑚/𝐿𝐿] = 0.077 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 [0.003 𝑚𝑚𝑚𝑚𝑚𝑚/𝐿𝐿]

b. The power is estimated using the following relationship:

c.

𝑃𝑃𝑡𝑡 (𝑀𝑀𝑀𝑀) =

480 𝑚𝑚3 0.74 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 × = 355 = 355 𝑀𝑀𝑀𝑀 3 𝑠𝑠 𝑚𝑚 𝑠𝑠

The actual power will be as follows, using 1% conversion efficiency:

𝑃𝑃𝑎𝑎 (𝑀𝑀𝑀𝑀) = 355 𝑀𝑀𝑀𝑀 × 0.01 = 3.55 𝑀𝑀𝑀𝑀

Problem 7.13.9. Cost of RED Power Plants

161

P7.9 Determine the total capital cost of a 500 MW RED salinity gradient power plant if the reported unit cost is $7.84/50 kW. Hack (2011) reported that the practical cost of salinity gradient plant is around $3M/MW. Determine the capital cost if this is the new estimate.

Solution: a. The unit cost in MW is simply calculated: 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶($) $7.84 𝑀𝑀 1,000 𝑘𝑘𝑘𝑘 = $156.8 𝑀𝑀 = × = 𝑀𝑀𝑀𝑀 50 𝑘𝑘𝑘𝑘 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀

b. The cost for a 500 MW system is: 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶($) = 500 𝑀𝑀𝑀𝑀 ×

$156.8 𝑀𝑀 = $78,400 𝑀𝑀 𝑀𝑀𝑀𝑀

c. This is quite high and a more practical value for a 500 MW plant is $1,500M instead of $78.4B. Problem 7.13.10 Simple Payback Period for Salinity Gradient Power Plant P7.10 Make a rough estimate of the payback period in years for a 125 MW salinity gradient facility that costs $375M. The main revenue will come from selling electricity at $0.08/kWh. Assume the payback period is simply based on the ratio of the initial capital cost to the initial capital investment. Assume also that operation will be for 365 days in a year and 24 hours in a day.

Solution: a. The number of kWh of electrical power produced in a year is calculated as follows: 𝑘𝑘𝑘𝑘ℎ 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 24 ℎ𝑟𝑟𝑟𝑟 1,000 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘ℎ = 125 𝑀𝑀𝑀𝑀 × × × = 1,095,000,000 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑑𝑑𝑑𝑑𝑑𝑑 1 𝑀𝑀𝑀𝑀 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

b. The revenue from the sale of this electricity is calculated from the product of the energy output and the rate as shown below.

162

𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 �

$ 1,095,000,000 𝑘𝑘𝑘𝑘ℎ $0.08 �= × = $87,600,000 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑘𝑘𝑘𝑘ℎ

c. The simple payback period is the ratio of the capital cost and revenue as shown below. 𝑃𝑃𝑃𝑃𝑃𝑃 (𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦) = $375,000,000 ×

𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 = 4.28 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 $87,600,000

d. The payback is less than 5 years. Of course, this could be more if one considers the operating cost per year in the calculations.

7.13.11

Sizing Salinity Gradient Membranes

P7.11 If the average household in the US uses around 12,000 kWh of electricity per year, how much membrane is needed (in units of square meters) to power 1 household if the yearly energy throughput of this membrane is 300 MWh/yr per m2? a. 3.6 b. 0.36 c. 0.4 d. 25 Solution: a. This is a simple ratio of output to input as follows: b. 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝑚𝑚2 ) =

12,000 𝑊𝑊ℎ 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

×

𝑚𝑚 2 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

30,000 k𝑊𝑊ℎ

= 0.4 𝑚𝑚2

c. Thus, a nanotube with an area of 0.4 m2 is needed. The value seems too good to be true.

7.13.12

Output Voltage of a Salinity Gradient Power Plant

P7.12 The San Lorenzo River in California has an average annual discharge of 480 m3/s. The average salinity at this location was 0.6 mol/L. The nearby freshwater salinity was 0.003 mol/L, hence the salinity ratio was reported at 1:200. The average temperature was 25oC. Use the Nernst Equation for the calculation of voltage generated. You are also asked to calculate the theoretical 163

power that may be generated from this facility if the amount of energy produced was 0.74 MJ/m3 of flow instead of the erroneously reported 1.4 MJ/m3. Finally, you are to calculate the actual power (MW) assuming an overall efficiency of 2%. Use R = 8.314 J/mole oK and F = 96,485 Coulombs/mole. Note that J/C is a volt. For this question, what is the output voltage in units of volts? a. 3.55 b. 355 c. 7.1 d. 0.14 Solutions: a. The voltage is calculated from Equation 7.4 as follows: V (𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉) = 7.13.13

𝐽𝐽 8.314 𝑚𝑚𝑚𝑚𝑚𝑚 − 𝐾𝐾 × 298.15𝑜𝑜 𝐾𝐾 1 × 96,485 𝐶𝐶/𝑚𝑚𝑚𝑚𝑚𝑚

𝑚𝑚𝑚𝑚𝑚𝑚 � 𝐿𝐿 𝑙𝑙𝑙𝑙 = 0.136 𝑉𝑉 ≈ 0.14 𝑉𝑉 𝑚𝑚𝑚𝑚𝑚𝑚 �0.003 � 𝐿𝐿 �0.6

Theoretical Power Output of a Salinity Gradient Power Plant

P7.13 The San Lorenzo River in California has an average annual discharge of 480 m3/s. The average salinity at this location was 0.6 mol/L. The nearby freshwater salinity was 0.003 mol/L, hence the salinity ratio was reported at 1:200. The average temperature was 25oC. Use the Nernst Equation (shown in lecture notes) for the calculation of voltage generated. You are also asked to calculate also the theoretical power that may be generated from this facility if the amount of energy produced was 0.74 MJ/m3 of flow instead of the erroneously reported 1.4 MJ/m3. To calculate actual power, assume an overall efficiency of 2% and report the units in MW. Use R = 8.314 J/mole oK and F = 96,485 Coulombs/mole. Note that J/C is a volt. For this question, what is the theoretical power output in units of MW? 164

a. 355 b. 3.55 c. 7.1 d. 0.14 Solutions: a. The voltage is calculated from Equation 7.4 as follows: V (𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉) =

𝐽𝐽 8.314 𝑚𝑚𝑚𝑚𝑚𝑚 − 𝐾𝐾 × 298.15𝑜𝑜 𝐾𝐾 1 × 96,485 𝐶𝐶/𝑚𝑚𝑚𝑚𝑚𝑚

𝑚𝑚𝑚𝑚𝑚𝑚 � 𝐿𝐿 𝑙𝑙𝑙𝑙 = 0.136 𝑉𝑉 𝑚𝑚𝑚𝑚𝑚𝑚 � �0.003 𝐿𝐿 �0.6

b. The power is estimated using the following relationship 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 𝑚𝑚3 𝑃𝑃𝑡𝑡 (𝑀𝑀𝑀𝑀) = 480 × 0.74 3 × = 355 𝑀𝑀𝑀𝑀 𝑠𝑠 𝑚𝑚 𝑀𝑀𝑀𝑀/𝑠𝑠

7.13.14

Actual Power Output of a Salinity Gradient Power Plant

P7.14 The San Lorenzo River in California has an average annual discharge of 480 m3/s. The average salinity at this location was 0.6 mol/L. The nearby freshwater salinity was 0.003 mol/L, hence the salinity ratio was reported at 1:200. The average temperature was 25oC. Use the Nernst Equation (shown in lecture notes) for the calculation of voltage generated. You are also asked to calculate also the theoretical power that may be generated from this facility if the amount of energy produced was 0.74 MJ/m3 of flow instead of the erroneously reported 1.4 MJ/m3. To calculate actual power, assume an overall efficiency of 2% and report the units in MW. Use R = 8.314 J/mole oK and F = 96,485 Coulombs/mole. Note that J/C is a volt. For this question, what is the actual power output in units of MW? a. 355 b. 7.1 c. 3.5 165

d. 0.14 Solutions: a. The voltage is calculated from Equation 7.4 as follows: V (𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉) =

𝐽𝐽 8.314 𝑚𝑚𝑚𝑚𝑚𝑚 − 𝐾𝐾 × 298.15𝑜𝑜 𝐾𝐾 1 × 96,485 𝐶𝐶/𝑚𝑚𝑚𝑚𝑚𝑚

𝑚𝑚𝑚𝑚𝑚𝑚 � 𝐿𝐿 𝑙𝑙𝑙𝑙 = 0.136 𝑉𝑉 𝑚𝑚𝑚𝑚𝑚𝑚 � �0.003 𝐿𝐿 �0.6

b. The power is estimated using the following relationship 𝑃𝑃𝑡𝑡 (𝑀𝑀𝑀𝑀) = 480

𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 𝑚𝑚3 × 0.74 3 × = 355 𝑀𝑀𝑀𝑀 𝑠𝑠 𝑚𝑚 𝑀𝑀𝑀𝑀/𝑠𝑠

c. The actual power will be as follows using 2% conversion efficiency 𝑃𝑃𝑎𝑎 (𝑀𝑀𝑀𝑀) = 355 𝑀𝑀𝑀𝑀 × 0.02 = 7.1 𝑀𝑀𝑀𝑀 7.13.15

Simple Payback Period for a Salinity Gradient Power Plant

P7.15 A publication reported that the practical cost for salinity gradient is around $3M/MW. Determine the capital cost of a 200-kW system if this is the new estimate in US dollars. a. 31.36 b. 31,360 c. 600,000 d. 6M Solution: a. Hence, the cost for a 200 MW system is as follows: 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 ($) = 200 𝑘𝑘𝑘𝑘 ×

𝑀𝑀𝑀𝑀 $3,000,000 × = $600,000 1,000 𝑘𝑘𝑘𝑘 𝑀𝑀𝑀𝑀

Assess Your Knowledge 166

1. Name the main principle behind the use of salinity gradient for power generation?

a. Osmotic pressure difference between salt water and fresh water b. Difference in temperature for salt water and fresh water c. Difference in physical properties of membranes to allow water and salt water to pass through d. Difference in concentration of salt water and fresh water 2. This type of salinity gradient technology uses the transport of salt ions through membranes. The salinity gradient difference is the driving force in transporting ions that results in an electric potential which is then converted into electricity. Name this salinity gradient technology.

a. RED b. PRO c. Solar Pond d. Vant Hoff’s Technology 3. What do you call the conditions when low-salinity water floats on top of high-salinity water?

a. Salt gradient b. saline pond c. halocline d. Kelvin-cline 4. True or False. Salinity gradient is suitable only for large-scale plant applications.

a. False b. True c. It depends d. No idea 5. True or False. Salinity gradient technology is periodic, like wind and wave power.

167

a. True b. No idea c. It depends d. False 6. The total energy cost for salinity gradient technology is very sensitive to its efficiency and to the price of what input?

a. Machinery cost b. membrane cost c. conveyance cost d. operating cost 7. The specific cost of a storage tank for a salinity gradient system is a strong function of what parameter?

a. membrane cost b. storage capacity c. salinity d. material cost 8. The pressure retarded osmosis (PRO) salinity gradient prototype has a net power density of 15 W/m2. Determine the size (in m2) of the polyamide membrane to power a household with a monthly requirement of 1,200 kWh?

a. 111 b. 278 c. 69.5 d. 139 Solution: 168

a. This is a simple energy and power calculation as follows: 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝑚𝑚2 ) =

1,200 𝑘𝑘𝑘𝑘ℎ 𝑚𝑚2 1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚ℎ 1 𝑑𝑑𝑑𝑑𝑑𝑑 1,000 𝑊𝑊 × × × × = 111 𝑚𝑚2 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚ℎ 𝑘𝑘𝑘𝑘 15 𝑊𝑊 30 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 24 ℎ𝑟𝑟𝑟𝑟

9. In salinity gradient energy calculations for osmotic pressure, what famous formula is used?

a. Van Horn’s Equation b. Desalination c. Van’t Hoff Equation d. Boyle’s Law 10. What is the most serious problem for salinity gradient systems when it comes to use of membranes?

a. Slagging b. Fouling c. Biofouling d. Emissions 11. Determine the energy required (in kJ) to boil a liter of water from an original temperature of 15oC. How much more energy is required (in kJ) to boil water if the energy required to desalinate water is 2,780 J/L.

a. 335.2 b. 1390 c. 356.2 d. 2,780 Solution: a. The heat required to boil water from 15oC to 100oC is shown below. 𝑄𝑄 = 𝑚𝑚𝐶𝐶𝑝𝑝 ∆𝑡𝑡 = 1 𝑘𝑘𝑘𝑘 × 4.190

𝑘𝑘𝑘𝑘 × (100 − 15)𝑜𝑜𝑜𝑜 = 356.2 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘𝑜𝑜 𝐶𝐶 169

12. This type of salinity gradient technology uses a membrane to separate concentrated salt solution (like sea water) from fresh water. The freshwater flows through a semi-permeable membrane towards the seawater, which increases the pressure within the seawater chamber. A turbine is then spun as the pressure is compensated, and electricity is generated. Name this salinity gradient technology

a. Van’t Hoff’s Technology b. RED c. Solar Pond d. PRO 13. The baseline energy required to desalinate water is 2,780 J/L. Calculate the difference in this energy compared with that of boiling water from an original temperature of 15oC?

a. 335.2 b. 2.78 c. 1,390 d. 353.4 Solution: a. The heat required to boil water from 15oC to 100oC is shown below. 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑄𝑄𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 � � = 𝑚𝑚̇𝐶𝐶𝑝𝑝 ∆𝑡𝑡 = 1 × 4.190 𝑜𝑜 × (100 − 15)𝑜𝑜𝑜𝑜 = 356.2 𝐿𝐿 𝐿𝐿 𝑘𝑘𝑘𝑘 𝐶𝐶 𝐿𝐿

b. The units must be the same hence the desalination energy must also be converted into units of kJ/L as follows: 𝑘𝑘𝑘𝑘 𝐽𝐽 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑄𝑄𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 � � = 2,780 × = 2.78 𝐿𝐿 𝐿𝐿 1000 𝐽𝐽 𝐿𝐿

c. More energy is needed to boil water than to desalinate the same volume. Hence, the difference in energy is calculated as follows: 170

𝑄𝑄𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 𝑄𝑄𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 − 𝑄𝑄𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 356.2

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 − 2.78 = 353.4 𝐿𝐿 𝐿𝐿 𝐿𝐿

d. Thus, the difference is only 0.78% of the energy required to boil water.

14. The reported capital cost for a pressure retarded osmosis (PRO) salinity gradient membrane was $2.5M/MW. Determine the capital investment for a facility that requires a monthly energy of 12,000 kWh. The performance of the PRO unit was 15 W/m2.

a. $50,000 b. $41,667 c. $100,000 d. $83,333 Solution: a. The first step is to calculate the power in equivalent MW for this facility. 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀ℎ) =

12,000 𝑘𝑘𝑘𝑘ℎ 1 𝑚𝑚𝑚𝑚 1 𝑑𝑑𝑑𝑑𝑑𝑑 1 M𝑊𝑊 $2.5𝑀𝑀 × × × × = $41,667 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚ℎ 30 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 24 ℎ𝑟𝑟𝑟𝑟 1,000 𝑘𝑘𝑘𝑘 𝑀𝑀𝑀𝑀

b. Thus, the initial capital cost seems too good to be true.

15. Hack (2011) reported that the practical cost for salinity gradient is around $3M/MW. Determine the capital cost of a 500-kW system if this is the new estimate in US dollars.

a. 31.36 b. 31,360 c. 1.5M d. 6M Solution: a. Hence, the cost for a 500 MW system is as follows: 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 ($) = 500 𝑘𝑘𝑘𝑘 ×

𝑀𝑀𝑀𝑀 $3,000,000 × = $1,500,000 1,000 𝑘𝑘𝑘𝑘 𝑀𝑀𝑀𝑀 171

16. If the average household in the US uses around 12,000 kWh of electricity per year, how much membrane is needed (in units of square meters) to power 1 household if the yearly energy throughput of this membrane is 300 MWh/yr per m2?

a. 3.6 b. 0.36 c. 0.4 d. 25 Solution: a. This is a simple ratio of output to input as follows: b. 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝑚𝑚2 ) =

12,000 𝑊𝑊ℎ 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

×

𝑚𝑚 2 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

30,000 k𝑊𝑊ℎ

= 0.4 𝑚𝑚2

c. Thus, a nanotube with an area of 0.4 m2 is needed. The value seems too good to be true.

17. Calculate the work done (in kJ) against pressure using the following data: pressure is equal to 35 bars or 35 kg/cm2, one liter of water is pushed against an area of 1 cm2 over 10 meters.

a. 3.43 b. 3,430 c. 2,700 d. 35 Solution: a. The work or energy is as follows (per liter of saltwater): 𝑊𝑊 =

35 𝑘𝑘𝑘𝑘 𝑁𝑁𝑠𝑠 2 9.8 𝑚𝑚 2 × 1 𝑐𝑐𝑐𝑐 × 10 𝑚𝑚 = 350 𝑘𝑘𝑘𝑘 − 𝑚𝑚 × × 2 = 3,430 𝑁𝑁𝑁𝑁 = 3.43 𝑘𝑘𝑘𝑘 2 𝑐𝑐𝑐𝑐 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑠𝑠

b. This is the minimum energy required to desalinate one liter of seawater.

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18. Determine the osmotic pressure in units of kg/cm2 or bars for the following data: temperature is 303oK, molar concentration of salt ions is equal to 1.128 mol/L and the universal gas constant is 0.082 L-bar/deg-mol. Use the equation given in lecture notes for your calculations.

a. 55.6 b. 222.4 c. 16.7 d. 28 Solution a. Equation 7.3 is used directly to get the following answers. 𝜋𝜋 = 𝑐𝑐𝑐𝑐𝑐𝑐

𝜋𝜋 = 1.128

𝑚𝑚𝑚𝑚𝑚𝑚 𝐿𝐿 − 𝑏𝑏𝑏𝑏𝑏𝑏 𝑘𝑘𝑘𝑘 × 0.082 × 303𝑜𝑜 𝐾𝐾 = 28 𝑏𝑏𝑏𝑏𝑏𝑏 = 28 2 𝐿𝐿 𝐾𝐾 − 𝑚𝑚𝑚𝑚𝑚𝑚 𝑐𝑐𝑐𝑐

b. The osmotic pressure is around 28 bars

19. The pressure retarded osmosis (PRO) salinity gradient prototype has a net power density of 1.5 W/m2. Determine the size (in m2) of the polyamide membrane to power a household with a monthly requirement of 1,200 kWh.

a. 89 b. 1389 c. 111 d. 139 Solution: a. This is a simple energy and power calculation as follows: 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝑚𝑚

2)

1,500 𝑘𝑘𝑘𝑘ℎ 𝑚𝑚2 1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚ℎ 1 𝑑𝑑𝑑𝑑𝑑𝑑 1,000 𝑊𝑊 = × × × × = 1,389 𝑚𝑚2 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚ℎ 1.5 𝑊𝑊 30 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 24 ℎ𝑟𝑟𝑟𝑟 𝑘𝑘𝑘𝑘

b. About 1389 m2 of area is needed for this application.

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20. Calculate the theoretical Carnot cycle efficiency (%) of a solar pond that has a high temperature reservoir of 50oC and a low reservoir temperature of 20oC. If the net output or work done in the system is 150 kW at this theoretical efficiency, how much heat was put out of the system?

a. 8.5 b. 9.3 c. 40 d. 60 Solution: a. Equation 7.2 is simply used as follows: 𝜂𝜂𝐶𝐶 = �1 − 𝜂𝜂𝐶𝐶 = �1 −

𝑇𝑇𝐶𝐶 � × 100% 𝑇𝑇𝐻𝐻

273.15 + 20 � × 100% = 9.3% 273.15 + 50

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Chapter 8 Fuel Cells 8.12 Solved Problems Problem 8.12.1 Conversion Efficiency of Direct Methane Fuel Cell P8.1 Determine the conversion efficiency if one uses methane (CH4) as fuel for a fuel cell system. The heating value of methane was reported to be 37.7 MJ/m3. The electricity produced was 150 kWh. The test was done in 100 hours, and 28.65 cubic meters of methane was consumed. Comment on the conversion efficiency compared to when methane is used in an engine. How much average power was produced during this test?

Solution: a. Conversion efficiency is the ratio of the output energy over the input energy. The energy input is calculated first: 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) =

37.7 𝑀𝑀𝑀𝑀 × 28.65 𝑚𝑚3 = 1,080 𝑀𝑀𝑀𝑀 𝑚𝑚3

b. The output energy is converted into the same unit: 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) = 150 𝑘𝑘𝑘𝑘ℎ ×

3.6 𝑀𝑀𝑀𝑀 = 540 𝑀𝑀𝑀𝑀 𝑘𝑘𝑘𝑘ℎ

c. Efficiency is then calculated using the classical conversion efficiency equation. 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (%) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 540 𝑀𝑀𝑀𝑀 × 100% = × 100% = 50% 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 1,080 𝑀𝑀𝑀𝑀

d. The efficiency is about 50%, much higher than when methane is burned in an engine and electrical power is generated.

175

e. Power is simply the ratio of the energy produced over the period for testing. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) =

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 150 𝑘𝑘𝑘𝑘ℎ = = 1.5 𝑘𝑘𝑘𝑘 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 100 ℎ𝑟𝑟𝑟𝑟

f. The power is relatively low at 1,500 Watts.

Problem 8.12.2 Maximum Conversion Efficiency for Direct Methane Fuel Cell P8.2 Determine the maximum conversion efficiency for the direct methane fuel cell if the maximum electricity produced is 8.17 kWh for every cubic meter of methane consumed. The energy content of methane is 37.7 MJ/m3.

Solution: a. The output power must be converted first into the proper units of MJ/m3: 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝑀𝑀𝑀𝑀 8.17 𝑘𝑘𝑘𝑘ℎ 3.6 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 �= × = 29.4 3 3 3 𝑚𝑚 𝑚𝑚 𝑘𝑘𝑘𝑘ℎ 𝑚𝑚

b. The maximum conversion efficiency is simply the ratio of the maximum electricity generated versus the heating value of methane: 𝑀𝑀𝑀𝑀 3 𝑚𝑚 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (%) = × 100% = 77.98% 𝑀𝑀𝑀𝑀 37.7 3 𝑚𝑚 29.4

c. Hence, the maximum efficiency is about 78%, much higher than the Carnot Cycle efficiency, if methane is burned in an engine.

Problem 8.12.3 Heat Energy Losses in Direct Methane Fuel Cell.

P8.3 Determine the percentage of energy loss for the direct methane fuel cell if the heat produced in a fuel cell is 18.85 MJ for every cubic meter of methane consumed. The energy content of methane is approximately 50 MJ/kg and its density was reported at 0.754 kg/m3.

Solution: a. The energy units must be made uniform: 176

𝑀𝑀𝑀𝑀𝑀𝑀ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝑀𝑀𝑀𝑀 50 𝑀𝑀𝑀𝑀 0.754 𝑘𝑘𝑘𝑘 𝑀𝑀𝑀𝑀 �= × = 37.7 3 3 3 𝑚𝑚 𝑘𝑘𝑘𝑘 𝑚𝑚 𝑚𝑚

b. The percentage of energy losses is simply the ratio of heat loss to the heating value of hydrogen: % 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 (%) =

18.85 𝑀𝑀𝑀𝑀/𝑚𝑚3 × 100% = 50% 37.7 𝑀𝑀𝑀𝑀/𝑚𝑚3

Problem 8.12.4. Hydrogen Needed (in kg) to Produce a Liter of Water P8.4 Determine the theoretical amount of hydrogen needed (in kg) to produce a liter of water from the ideal reaction between hydrogen and oxygen (Equation 8.13 or Equation 8.16).

Solution: a. The equation is already balanced and the amount of hydrogen needed for every liter of water produced is simply calculated: [(2 × 1)] 2𝐻𝐻2 1 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑘𝑘𝑘𝑘 𝐻𝐻2 = = = 0.11 2𝐻𝐻2 𝑂𝑂 [(2 × 1) + (1 × 16)] 9 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂

b. About 0.11 kg of hydrogen is needed to produce a liter of water.

Problem 8.12.5 Potassium Carbonate Produced in an Alkaline Fuel Cell P8.5 Determine the amount of potassium carbonate produced for every kg of potassium hydroxide used in the alkaline fuel cell following Equation 8.9. The molecular weight of potassium is 39.0983 g/mol.

Solution: a. The amount of bicarbonate produced per weight of potassium hydroxide is simply calculated: 𝐾𝐾2 𝐶𝐶𝑂𝑂3

(39.0983)×(2)+(12)×(1)+(16)×(3)

=� 2𝐾𝐾𝐾𝐾𝐾𝐾

2×(39.0983+16+1)

�=

138.1966 112.1966

177

= 1.232

𝑘𝑘𝑘𝑘 𝐾𝐾2𝐶𝐶𝑂𝑂3 𝑘𝑘𝑘𝑘 𝐾𝐾𝐾𝐾𝐾𝐾

b. More than 1.232 kg of potassium carbonate is produced per weight of potassium hydroxide used.

Problem 8.12.6 Ideal Water and Carbon Dioxide Produced for Direct Methane Fuel Cell P8.6 Determine the amount of water (kg or liters) and carbon dioxide produced (kg or cubic meter) for every cubic meter of methane used in an ideal direct methane fuel cell. Use a density of 0.754 kg/m3. Also use the ideal combustion equation of methane in air for the calculation. Air has 79% oxygen and 21% nitrogen.

Solution: a. The ideal combustion equation of methane is given below 𝐶𝐶𝐻𝐻4 + 2 × (𝑂𝑂2 + 3.76𝑁𝑁2 ) → 𝐶𝐶𝑂𝑂2 + 2𝐻𝐻2 𝑂𝑂

b. The amount of water is simply calculated:

c.

(2) × (1 × 2 + 16) 2𝐻𝐻2 𝑂𝑂 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 0.754 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶4 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 � = 2.25 =� × = 1.7 (12) + (4) 𝐶𝐶𝐻𝐻4 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 𝑚𝑚3 𝑚𝑚3 𝐶𝐶𝐶𝐶4

About 1.7 liters of water is produced per cubic meter of methane used in an ideal direct

methane fuel cell using a water density of 1 kg/L. d. Likewise, for carbon dioxide, the calculations are as follows: (12) + (2 × 16) 𝐶𝐶𝐶𝐶2 𝑘𝑘𝑘𝑘 𝐶𝐶𝑂𝑂2 0.754 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶4 𝑘𝑘𝑘𝑘 𝐶𝐶𝑂𝑂2 � = 2.75 =� × = 2.1 (12) + (4) 𝐶𝐶𝐻𝐻4 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 𝑚𝑚3 𝑚𝑚3 𝐶𝐶𝐶𝐶4

e. If the density of carbon dioxide at standard condition (zero degrees centigrade) is 1.977 kg/m3, the amount of CO2 produced is calculated as follows: 𝐶𝐶𝐶𝐶2 𝑘𝑘𝑘𝑘 𝐶𝐶𝑂𝑂2 𝑚𝑚3 𝐶𝐶𝐶𝐶2 𝑚𝑚3 𝐶𝐶𝑂𝑂2 = 2.1 × = 1.1 𝐶𝐶𝐻𝐻4 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 1.977 𝑘𝑘𝑘𝑘 𝑚𝑚3 𝐶𝐶𝐶𝐶4

Problem 8.12.7 Zinc Needed for Every Tonne Zinc Oxide Produced in Zinc Air Fuel Cell

178

P8.7 Determine the amount of zinc needed (in kg) for every tonne of pure zinc oxide produced in zinc-air fuel cell. Use the ideal Equation 8.17 for the calculation. The molecular weight of zinc is 65.38 g/mol and that of oxygen is 16 g/mol.

Solution: a. The amount of zinc needed is simply calculated using Equation 8.10: (65.38) 𝑍𝑍𝑍𝑍 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑍𝑍𝑍𝑍𝑍𝑍 1,000 𝑘𝑘𝑘𝑘 � = 0.8034 =� × = 803.4 (65.38 + 16) 𝑍𝑍𝑍𝑍𝑍𝑍 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑘𝑘𝑘𝑘 𝑍𝑍𝑍𝑍

b. About 803 g of pure zinc is needed for every tonne of zinc oxide produced in an ideal zinc-air fuel cell system.

Problem 8.12.8 Practical Conversion Efficiency for Direct Methanol Fuel Cell P8.8 Determine the conversion efficiency when methanol (CH3OH) was used as fuel for a fuel cell system. The electricity produced was measured to be 16.2 kWh. The energy content of methanol was reported to be 19.93 MJ/kg. During the test, some 10 liters of methanol were used.

The density of methanol at standard conditions was 0.792 kg/L. Compare this efficiency with that of internal combustion engines.

Solution: a. The conversion efficiency equation, as shown in earlier chapters, is given below. 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (%) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 × 100% 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼

b. The input and output must have the same units. The input is calculated first. 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) =

19.93 𝑀𝑀𝑀𝑀 0.792 𝑘𝑘𝑘𝑘 × × 10 𝐿𝐿 = 157.85 𝑀𝑀𝑀𝑀 𝑘𝑘𝑘𝑘 𝐿𝐿

c. Then, the output energy is calculated.

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) = 16.2 𝑘𝑘𝑘𝑘ℎ ×

3.6 𝑀𝑀𝑀𝑀 = 58.32 𝑀𝑀𝑀𝑀 𝑘𝑘𝑘𝑘ℎ

179

d. Finally, the efficiency is calculated: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (%) =

58.32 𝑀𝑀𝑀𝑀 × 100% = 36.95% 157.85 𝑀𝑀𝑀𝑀

e. Unfortunately, the conversion efficiency is quite low and is like that of internal combustion engines.

Problem 8.12.9 Efficiency of Spider Lion Fuel Cell P8.9 The inaugural flight of the small Spider Lion vehicle lasted for one hour and 43 minutes using a half tank of hydrogen. The total weight of the hydrogen gas used was 8.7 g. The fuel cell used has an output power of 100 watts or 100 J/s. Determine the overall efficiency of this system, assuming the energy content of pure hydrogen to be 142 MJ/kg. Compare the efficiency to an engine running on hydrogen gas.

Solution: a. The energy in units of MJ is calculated: 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑀𝑀𝑀𝑀) = 8.7 𝑘𝑘𝑘𝑘 ×

1 𝑘𝑘𝑘𝑘 142 𝑀𝑀𝑀𝑀 × = 1.24 𝑀𝑀𝑀𝑀100% 1,000 𝑔𝑔 𝑘𝑘𝑘𝑘

b. The output power converted in the same units will be calculated: 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑀𝑀𝑀𝑀) =

100 𝐽𝐽 𝑀𝑀𝑀𝑀 3,600 𝑠𝑠 × × × 1.7167 ℎ𝑟𝑟𝑟𝑟 = 0.618 𝑀𝑀𝑀𝑀 𝑠𝑠 1,000,000 𝐽𝐽 ℎ𝑟𝑟

c. The efficiency is calculated from the conversion efficiency equation as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (%) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 0.618 𝑀𝑀𝑀𝑀 × 100% = × 100% = 49.8% 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 1.24 𝑀𝑀𝑀𝑀

d. The efficiency of this test unit is around 56% greater than engine efficiency.

Problem 8.12.10 Efficiency of Commercial Fuel Cell.

P8.10 A Ballard fuel cell has the following specifications: a. Average Power Produced During Test = 5 kW 180

b. Number of Hours of Test = 80 hrs c. Methanol consumption = 225 liters d. Fuel used = 62% methanol and 38% water Determine the overall efficiency of this system assuming the methanol mixture has a density of 0.792 kg/L with the energy content of the methanol mixture at 14.3 MJ/kg.

Solution: a. The energy in units of J/s is calculated: 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑀𝑀𝑀𝑀) = 225 𝐿𝐿 ×

14.3 𝑀𝑀𝑀𝑀 0.792 𝑘𝑘𝑘𝑘 × = 2,548.3 𝑀𝑀𝑀𝑀 𝑘𝑘𝑘𝑘 𝐿𝐿

b. The energy output is likewise converted into the same units. 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑀𝑀𝑀𝑀) = 5 𝑘𝑘𝑘𝑘 × 80 ℎ𝑟𝑟𝑟𝑟 ×

3.6 𝑀𝑀𝑀𝑀 = 1,440 𝑀𝑀𝑀𝑀 𝑘𝑘𝑘𝑘ℎ

c. The efficiency is calculated from the conversion efficiency equation as follows:

8.10.11

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (%) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 1,440 𝑀𝑀𝑀𝑀 × 100% = × 100% = 56.5% 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 2,548.3 𝑀𝑀𝑀𝑀

Energy Losses in Fuel Cell Systems

P8.11 Determine the percentage of energy loss (ratio of heat produced to the heating value of methane, in %) for the direct methane fuel cell if the heat produced in a fuel cell is 18.85 MJ for every cubic meter of methane consumed. The energy content of methane is approximately 50 MJ/kg and its density was reported at 0.657 kg/m3. a. 83.5 b. 78.8 c. 57.4 d. 65.5 Solution: 181

a. The energy units must be made uniform as follows: 𝑀𝑀𝑀𝑀𝑀𝑀ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝑀𝑀𝑀𝑀 50 𝑀𝑀𝑀𝑀 0.657 𝑘𝑘𝑘𝑘 32.85 𝑀𝑀𝑀𝑀 �= × = 3 𝑚𝑚 𝑘𝑘𝑘𝑘 𝑚𝑚3 𝑚𝑚3

b. The percentage energy loss is simply the ratio of the heat loss to the heating value of methane as follows: 𝑀𝑀𝑀𝑀 3 𝑚𝑚 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 (%) = × 100% = 57.4% 𝑀𝑀𝑀𝑀 32.85 3 𝑚𝑚 18.85

8.10.12

Hydrogen Requirement in a Hydrogen Fuel Cell

P8.12 Determine the theoretical amount of hydrogen needed (in kg) to produce a liter of water from the ideal reaction between hydrogen and oxygen in an acid electrolyte proton exchange membrane (PEM) fuel cell. a. 1.1 b. 0.909 c. 9.1 d. 0.11 Solution: a. The equation is already balanced and hence the amount of hydrogen needed for every liter of water produced is simply calculated as follows: �2 × (2)� 2𝐻𝐻2 4 𝑘𝑘𝑘𝑘 𝐻𝐻2 1 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 𝑘𝑘𝑘𝑘 𝐻𝐻2 = = × = 0.11 2𝐻𝐻2 𝑂𝑂 [2 × (2 + 16)] 36 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 1 𝐿𝐿 𝐻𝐻2 𝑂𝑂 𝐿𝐿 𝐻𝐻2 𝑂𝑂

b. About 0.11 kg of hydrogen is needed to produce a liter of water.

182

8.10.13

Potassium Carbonate Requirement in a Fuel Cell

P8.13 Determine the amount of potassium carbonate (K2CO3) produced in kg for every kg of potassium hydroxide (KOH) used in the alkaline fuel cell following the ideal equation presented in lecture notes. The molecular weight of potassium is 39.1 g/mol. a. 1.2 b. 0.812 c. 0.406 d. 2.464 Solution: a. The amount of bicarbonate produced per weight of potassium hydroxide is simply calculated as follows: 𝐾𝐾2 𝐶𝐶𝑂𝑂3 (39.1 × 2) + (12 × 1) + (16 × 3) 138.2 𝑘𝑘𝑘𝑘 𝐾𝐾2 𝐶𝐶𝑂𝑂3 = = = 1.2 2𝐾𝐾𝐾𝐾𝐾𝐾 𝑘𝑘𝑘𝑘 𝐾𝐾𝐾𝐾𝐾𝐾 2 × [(39.1 + 16 + 1)] 112.2

b. More than 1.2 kg of potassium carbonate is produced per weight of potassium hydroxide used.

8.10.14

Water Production in a Fuel Cell

P8.14 Determine the amount of water produced (in kg water) for every cubic meter of methane used in an ideal direct methane fuel cell in air. Use methane density of 0.754 kg/m3. Also use the ideal combustion equation of methane in air for the calculation. Air has 79% oxygen and 21% nitrogen. a. 3.4 b. 1.7 c. 0.85 d. 5.1 183

Solution: a. The ideal combustion equation of methane is given below 𝐶𝐶𝐻𝐻4 + 2(𝑂𝑂2 + 3.76 𝑁𝑁2 ) → 𝐶𝐶𝑂𝑂2 + 2𝐻𝐻2 𝑂𝑂 + 7.52 𝑁𝑁2

b. The amount of water is simply calculated as follows: 2𝐻𝐻2 𝑂𝑂 2 × �(1 × 2) + (1 × 16)� 36 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 2.25 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 = = = [(12 × 1) + (1 × 4)] 𝐶𝐶𝐻𝐻4 16 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4

c. About 1.7 liters of water is produced per cubic meter of methane used in an ideal direct methane fuel cell using a water density of water is 1 kg/L. 2𝐻𝐻2 𝑂𝑂 2.25 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 0.754 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 = × = 1.7 3 3 𝐶𝐶𝐻𝐻4 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 𝑚𝑚 𝐶𝐶𝐻𝐻4 𝑚𝑚 𝐶𝐶𝐻𝐻4

8.10.15

Water Production in a Methane Fuel Cell

P8.15 Determine the amount of water produced (in kg) for every kg of methane used in an ideal direct methane fuel cell in air. Use methane density of 0.754 kg/m3. Also use the ideal combustion equation of methane in air for the calculation. Air has 79% oxygen and 21% nitrogen. a. 1.7 b. 3.4 c. 2.25 d. 5.1 Solution: a. The ideal combustion equation of methane is given below 𝐶𝐶𝐻𝐻4 + 2(𝑂𝑂2 + 3.76 𝑁𝑁2 ) → 𝐶𝐶𝑂𝑂2 + 2𝐻𝐻2 𝑂𝑂 + 7.52 𝑁𝑁2

b. The amount of water is simply calculated as follows: 2𝐻𝐻2 𝑂𝑂 2 × �(1 × 2) + (1 × 16)� 36 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 2.25 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 = = = [(12 × 1) + (1 × 4)] 𝐶𝐶𝐻𝐻4 16 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 184

c. About 1.7 liters of water is produced per cubic meter of methane used in an ideal direct methane fuel cell using a water density of water is 1 kg/L. 2𝐻𝐻2 𝑂𝑂 2.25 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 0.754 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 = × = 1.7 3 3 𝐶𝐶𝐻𝐻4 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 𝑚𝑚 𝐶𝐶𝐻𝐻4 𝑚𝑚 𝐶𝐶𝐻𝐻4

Assess Your Knowledge 1. What is the current average capital cost of a commercial fuel cell unit ($/kW)?

a. 400 b. 1,000 c. 10,000 d. $4,500 2. What is the practical efficiency (%) for fuel cell systems?

a. 35 b. 45 c. 83 d. 60 3. A very high efficiency research-grade fuel cell has an output of 5 kW. This unit has a net hydrogen consumption of around 15 gallons per hour. The cost of this hydrogen was estimated to be around $0.085/gallon. Determine the cost per kWh for this unit ($/kWh). Hint. You may simply use the energy efficiency equation (output/input).

a. 3.92 b. 0.05 c. 0.34

185

d. 0.255 Solution a. This may be solved by the conversion equation as follows: 𝜂𝜂 (%) = 5 𝑘𝑘𝑘𝑘ℎ

15 𝑔𝑔𝑔𝑔𝑔𝑔

×

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼

𝑔𝑔𝑔𝑔𝑔𝑔

$0.085

=

× 100% 3.92 𝑘𝑘𝑘𝑘ℎ $

b. The cost per kWh is simply the inverse of this relationship as follows and is equal to $0.255/kWh. This is still quite higher than most utilities cost per kWh. 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 ($) 𝑘𝑘𝑘𝑘ℎ

=

$

3.92 𝑘𝑘𝑘𝑘ℎ

=

$0.255 𝑘𝑘𝑘𝑘ℎ

4. What is the primary disadvantage of fuel cell technology?

a. Emissions b. Cost c. Noise d. Scalability 5. What is the range of efficiency (in %) of electricity generation using heat engines?

a. 60-83 b. 35-45 c. 10-30 d. 80-90 6. What is the theoretical maximum conversion efficiency (%) for a hydrogen oxygen fuel cell?

a. 83 b. 54 c. 37 d. 92 186

7. The US Department of Energy formed the Solid-State Energy Conversion Alliance (SECA) whose goal is to produce a solid-state fuel cell module with a target cost of how much $/kW?

a. 4500 b. 1,000 c. 400 d. 10,000 8. This is considered one main issue with fuel cell technology.

a. high currents b. low current and power. c. high power d. high cost of electrodes 9. Determine the maximum conversion efficiency (in %) for the direct methane fuel cell if the maximum electricity produced is 8.17 kWh for every cubic meter of methane consumed. The energy content of methane is 37.7 MJ/m3.

a. 83 b. 78 c. 50 d. 65 Solution: a. The output power must be converted first into its proper units of MJ/m3 as follows: 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝑀𝑀𝑀𝑀 8.17 𝑘𝑘𝑘𝑘ℎ 3.6 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 �= × = 29.4 3 3 3 𝑚𝑚 𝑚𝑚 𝑘𝑘𝑘𝑘ℎ 𝑚𝑚

b. The maximum conversion efficiency is simply the ratio of the maximum electricity generated versus the heating value of methane as follows:

187

𝑀𝑀𝑀𝑀 3 𝑚𝑚 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (%) = × 100% = 78% 𝑀𝑀𝑀𝑀 37.7 3 𝑚𝑚 29.4

c. Hence, the theoretical efficiency of a hydrogen oxygen fuel cell is about 78% much higher than the Carnot cycle efficiency, if hydrogen is burned in an engine.

10. Determine the amount of potassium carbonate (K2CO3) produced in kg for every kg of potassium hydroxide (KOH) used in the alkaline fuel cell following the ideal equation presented in lecture notes.

The molecular weight of potassium is 39.1 g/mol.

a. 1.2 b. 0.812 c. 0.406 d. 2.464 Solution: a. The amount of bicarbonate produced per weight of potassium hydroxide is simply calculated as follows: 𝐾𝐾2 𝐶𝐶𝑂𝑂3 (39.1 × 2) + (12 × 1) + (16 × 3) 138.2 𝑘𝑘𝑘𝑘 𝐾𝐾2 𝐶𝐶𝑂𝑂3 = = = 1.2 2𝐾𝐾𝐾𝐾𝐾𝐾 𝑘𝑘𝑘𝑘 𝐾𝐾𝐾𝐾𝐾𝐾 2 × [(39.1 + 16 + 1)] 112.2

b. More than 1.2 kg of potassium carbonate is produced per weight of potassium hydroxide used.

11. Determine the theoretical amount of hydrogen needed (in kg) to produce a liter of water from the ideal reaction between hydrogen and oxygen in an acid electrolyte proton exchange membrane (PEM) fuel cell.

a. 1.1 b. 0.909 c. 9.1 188

d. 0.11 Solution: a. The equation is already balanced and hence the amount of hydrogen needed for every liter of water produced is simply calculated as follows: �2 × (2)� 2𝐻𝐻2 4 𝑘𝑘𝑘𝑘 𝐻𝐻2 1 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 𝑘𝑘𝑘𝑘 𝐻𝐻2 = = × = 0.11 2𝐻𝐻2 𝑂𝑂 [2 × (2 + 16)] 36 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 1 𝐿𝐿 𝐻𝐻2 𝑂𝑂 𝐿𝐿 𝐻𝐻2 𝑂𝑂

b. About 0.11 kg of hydrogen is needed to produce a liter of water.

12. It is an electrochemical energy conversion device. It converts the chemical energy of the fuel (e.g. hydrogen) directly into electrical energy. Fuel and oxidizing agent (oxygen) are continuously and separately supplied to the two electrodes of the cell, at which they undergo a reaction. Name this device.

a. battery b. capacitor c. fuel cell d. electrolyzer 13. Estimate the electrical efficiency (in percent) of converting hydrogen into electrical power via fuel cell. The electricity produced in an hour per kg of hydrogen used was 20 Watt-hr while the high heating value of hydrogen was found to be 143 kJ/kg.

a. 50.4 b. 94 c. 65.58 d. 83 Solution: a. The electrical energy efficiency is given by the following equation:

189

𝐸𝐸𝐸𝐸𝐸𝐸 (%) =

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 × 100% 𝐻𝐻𝐻𝐻𝐻𝐻 𝑜𝑜𝑜𝑜 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈

b. Thus, the same unit must be used. The energy produced per kg of hydrogen used is as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑘𝑘𝑘𝑘 20 𝑊𝑊ℎ 3,600 𝑘𝑘𝑘𝑘 1 𝑘𝑘𝑘𝑘ℎ 𝑘𝑘𝑘𝑘 �= × × = 72 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑘𝑘𝑘𝑘ℎ 𝑘𝑘𝑘𝑘 1,000 𝑊𝑊ℎ

c. Thus, the EEE(%) is estimated to be about 45% as shown below.

𝐸𝐸𝐸𝐸𝐸𝐸 (%) =

𝑘𝑘𝑘𝑘 72 𝑘𝑘𝑘𝑘

𝑘𝑘𝑘𝑘 143 𝑘𝑘𝑘𝑘

× 100% = 50.4%

14. This is considered one main issue with fuel cell technology.

a. slow reaction rates b. excessive currents c. excessive high power d. high cost of electrodes 15. Determine the theoretical maximum amount of water (in gallons) produced per gallon of hydrogen used in an acid electrolyte proton exchange membrane (PEM) fuel cell using the ideal equation presented in lecture notes. Assume the density of water to be 8.34 lbs/gallon and for hydrogen, there are 1.693 gallons of hydrogen per lb of hydrogen gas.

a. 1.1 b. 0.637 c. 2.2 d. 1.275 Solution: a. The ideal equation is shown below.

190

2𝐻𝐻2 + 𝑂𝑂2 → 2𝐻𝐻2 𝑂𝑂

b. Hence the balanced equation gives rise to the following balance by weight.

2𝐻𝐻2 𝑂𝑂 [2 × (2 + 16)] 36 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 = = =9 2𝐻𝐻2 4 𝑘𝑘𝑘𝑘 𝐻𝐻2 �2 × (2)�

c. Using the densities of each material we have the following calculations. About xx gallons of water is condensed for every gallon of hydrogen utilized.

9

𝑙𝑙𝑙𝑙 𝐻𝐻2 𝑂𝑂 1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐻𝐻2 𝑂𝑂 1 𝑙𝑙𝑙𝑙 𝐻𝐻2 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐻𝐻2 𝑂𝑂 × × = 0.637 8.34 𝑙𝑙𝑙𝑙𝑙𝑙 𝐻𝐻2 𝑂𝑂 1.693 𝑔𝑔𝑔𝑔𝑔𝑔 𝐻𝐻2 𝑙𝑙𝑙𝑙 𝐻𝐻2 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐻𝐻2

9

𝑙𝑙𝑙𝑙 𝐻𝐻2 𝑂𝑂 1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐻𝐻2 𝑂𝑂 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐻𝐻2 𝑂𝑂 × = 1.1 𝑙𝑙𝑙𝑙 𝐻𝐻2 𝑙𝑙𝑙𝑙 𝐻𝐻2 8.34 𝑙𝑙𝑙𝑙𝑙𝑙 𝐻𝐻2 𝑂𝑂

d. The amount of water produced per lb of hydrogen used is as follows:

16. The Nexa TM 1200 Fuel Cell system from Heliocentris has an electrical power output of 1.2 kW. The hydrogen consumption was found to be 18.5 liters per minute [4.9 gallons/min]. If the cost of hydrogen was $0.50/gallon, determine the power production cost for this unit in units of $/kWh.

a. 0.10 b. 122.5 c. 61.25 d. 50 Solution: a. Power production cost is simply the cost of hydrogen divided by the power output as follows: 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 ($) 4.9 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 $0.50 60 𝑚𝑚𝑚𝑚𝑚𝑚 1 $122.5 = × × × = 𝑘𝑘𝑘𝑘ℎ 𝑚𝑚𝑚𝑚𝑚𝑚 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ℎ𝑟𝑟 1.2 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘ℎ b. The cost per kWh is simply too high to be economical.

191

17. Determine the amount of water produced (in kg) for every kg of methane used in an ideal direct methane fuel cell in air. Use methane density of 0.754 kg/m3. Also use the ideal combustion equation of methane in air for the calculation. Air has 79% oxygen and 21% nitrogen.

a. 1.7 b. 3.4 c. 2.25 d. 5.1 Solution: a. The ideal combustion equation of methane is given below 𝐶𝐶𝐻𝐻4 + 2(𝑂𝑂2 + 3.76 𝑁𝑁2 ) → 𝐶𝐶𝑂𝑂2 + 2𝐻𝐻2 𝑂𝑂 + 7.52 𝑁𝑁2

b. The amount of water is simply calculated as follows: 2𝐻𝐻2 𝑂𝑂 2 × �(1 × 2) + (1 × 16)� 36 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 2.25 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 = = = [(12 × 1) + (1 × 4)] 𝐶𝐶𝐻𝐻4 16 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4

c. About 1.7 liters of water is produced per cubic meter of methane used in an ideal direct methane fuel cell using a water density of water is 1 kg/L. 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 2𝐻𝐻2 𝑂𝑂 2.25 𝑘𝑘𝑘𝑘 𝐻𝐻2 𝑂𝑂 0.754 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 = × = 1.7 3 3 𝐶𝐶𝐻𝐻4 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 𝑚𝑚 𝐶𝐶𝐻𝐻4 𝑚𝑚 𝐶𝐶𝐻𝐻4

18. Determine the conversion efficiency (in %) if one will use methane (CH4) as fuel for fuel cell system.

The heating value of methane was reported at 37.7 MJ/m3. The electricity produced was 150 kWh. The test was done in 100 hours and 28.65 cubic meters of methane was consumed.

a. 25 b. 83 c. 75 d. 50 Solution: 192

a. The conversion efficiency is the ratio of the output energy over the input energy.

Hence, the input energy is calculated first as follows: 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) =

37.7 𝑀𝑀𝑀𝑀 × 28.65 𝑚𝑚3 = 1,080 𝑀𝑀𝑀𝑀 𝑚𝑚3

b. The output energy is converted into the same unit as follows: 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) = 150 𝑘𝑘𝑘𝑘ℎ ×

3.6 𝑀𝑀𝑀𝑀 = 540 𝑀𝑀𝑀𝑀 𝑘𝑘𝑘𝑘ℎ

c. The efficiency is then calculated using the classical conversion efficiency equation as shown. 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (%) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 540 𝑀𝑀𝑀𝑀 × 100% = × 100% = 50% 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 1,080 𝑀𝑀𝑀𝑀

d. Hence, the efficiency so about 50%, much higher than when methane is burned in an engine and electrical power is generated.

19. What fuel cell type appears to be the most promising technology for small electric power plants over 1 kW?

a. Solid oxide fuel cell (SOFC). b. Hydrogen fuel cell (HFC) c. Biological fuel cell (BFC) d. Proton exchange membrane fuel cell (PEMFC) 20. The Nexa TM 1200 Fuel Cell system from Heliocentris has an electrical power output of 1.2 kW. The hydrogen consumption was found to be 18.5 liters per minute [4.9 gallons/min]. If the cost of hydrogen was $0.50/gallon, determine the capital cost per kW ($/kW) if this unit was sold for $10,000 a unit.

a. 122.5 b. 16,666

193

c. 8,333 d. 12,250 Solution: a. The capital cost/kW is simply $10,000/1.2kW = $8,333/kW.

194

Chapter 9 Tidal Energy 9.9 Solved Problems Problem 9.9.1 Variation of Tide Level with Time Using Sine Curve P9.1 The variations in height of the sea water tide may be described by a sine curve. The equation for height is simply as follows: Tide level = Maximum tide x sin Φ, where Φ is an angle between 0 to 90o with sine considered. Plot and make a table of the height as a function of time for a tide that has a maximum height of 10 meters. Assume that the total time for this cycle to occur is 12 hours and 25 minutes (or 12.4167 hrs).

Solution: a. The table below shows variations in height at various sine angles at increments of +15o angle for the range stated. If one were to divide the whole cycle into various segments, then the time these tide levels may be achieved are also given in the table.

195

Time

Height (m)

Time

Height

Time

(m)

Height

Time

(Meters)

Height (Meters)

6:00

0.0000

9:37

4.8296

13:14

-2.5000

16:51

-3.5355

6:31

1.2941

10:08

4.3301

13:45

-3.5355

17:22

-2.5000

7:02

2.5000

10:39

3.5355

14:16

-4.3301

17:53

-1.2941

7:33

3.5355

11:10

2.5000

14:47

-4.8296

18:24

0.0000

8:04

4.3301

11:41

1.2941

15:18

-5.0000

8:35

4.8296

12:12

0.0000

15:49

-4.8296

9:06

5.0000

12:43

-1.2941

16:20

-4.3301

b. Hence, the data may be plotted to reflect time variation in height levels of a tide as a function of time.

196

Tide Level Height with Time Chart 6.0000

2.0000 0.0000 -2.0000

6:00 6:31 7:02 7:33 8:04 8:35 9:06 9:37 10:08 10:39 11:10 11:41 12:12 12:43 13:14 13:45 14:16 14:47 15:18 15:49 16:20 16:51 17:22 17:53 18:24

Tide Height (m)

4.0000

-4.0000 -6.0000

Time of Day (hr:min)

Problem 9.9.2. Reservoir Volume Calculation.

P9.2 Determine the volume of water contained in a 1-hectare plot filled with water with a depth of 30 meters [98.4 ft]? Report your answers in cubic meters and in gallons.

Solution: a. This is a simple volumetric calculation: 𝑉𝑉 (𝑚𝑚3 ) = 𝐿𝐿 × 𝑊𝑊 × 𝐻𝐻 = 1 ℎ𝑎𝑎 ×

10,000 𝑚𝑚2 × 30 𝑚𝑚 = 300,000 𝑚𝑚3 ℎ𝑎𝑎

b. Hence, the total volume of this container is 300,000 cubic meters. In the English System the volume will be as follows: 𝑉𝑉 (𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔) = 300,000 𝑚𝑚3 ×

1,000 𝐿𝐿 1 𝑔𝑔𝑔𝑔𝑔𝑔 × = 79,260,238 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1 𝑚𝑚3 3.785 𝐿𝐿

Problem 9.9.3. Time to Release Water from Reservoir.

P9.3 If the volume of water in a 1,000-hectare plot with a depth of 30 meters [98.4 ft] is released at a rate of 800 million liters per minute [211.4 million gallons/min], how long does it take (in hrs) for this reservoir to run out of water?

197

Solution: a. The volume of the reservoir is first calculated. 𝑉𝑉 (𝑚𝑚3 ) = 𝐿𝐿 × 𝑊𝑊 × 𝐻𝐻 = 1,000 ℎ𝑎𝑎 ×

10,000 𝑚𝑚2 × 30 𝑚𝑚 = 300 × 106 𝑚𝑚3 ℎ𝑎𝑎

b. This is a simple volume over volumetric flow rate calculation: 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇(ℎ𝑟𝑟𝑟𝑟) = 300 × 106 𝑚𝑚3 ×

1,000 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚 ℎ𝑟𝑟 × × = 6.25 ℎ𝑟𝑟𝑟𝑟 1 𝑚𝑚3 800 × 106 𝐿𝐿 60 𝑚𝑚𝑚𝑚𝑚𝑚

Problem 9.9.4. Power from Tidal Reservoir

P9.4 Determine the power generated for a water storage facility having an area of 1000 hectares with a depth of 30 meters [98.4 ft] if water is released at a rate of 800 million liters per minute [211 million gpm] at a height of 30 meters [65.6 ft]. Use an overall conversion efficiency of 75%. Report your answer in kW. Solution: a. The volumetric flow rate is first converted into the proper unit: 𝑘𝑘𝑘𝑘 800 × 106 𝐿𝐿 1 𝑘𝑘𝑘𝑘 1 𝑚𝑚𝑚𝑚𝑚𝑚 13.33 × 106 𝑘𝑘𝑘𝑘 𝑄𝑄 � � = × × = 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 1 𝐿𝐿 60 𝑠𝑠 𝑠𝑠

b. The solution makes use of the hydro power equation previously presented: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) =

𝑄𝑄𝑄𝑄𝑄𝑄 102

13.33 × 106 𝑘𝑘𝑘𝑘 𝑁𝑁𝑠𝑠 2 9.8 𝑚𝑚 𝐽𝐽 𝑊𝑊𝑊𝑊 = × 30 𝑚𝑚 × 0.75 × × 2 × × 𝑠𝑠 𝑘𝑘𝑘𝑘 − 𝑚𝑚 𝑠𝑠 𝑁𝑁𝑁𝑁 𝐽𝐽 ×

𝑘𝑘𝑘𝑘 = 2.94 × 106 𝑘𝑘𝑘𝑘 1,000 𝑊𝑊

𝑄𝑄𝑄𝑄𝑄𝑄 13.33 × 106 𝑘𝑘𝑘𝑘 1 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) = = × 30 𝑚𝑚 × 0.75 × = 2.94 × 106 𝑘𝑘𝑘𝑘 102 𝑠𝑠 102

198

Problem 9.9.5. Energy from Tidal Reservoir P9.5 Determine the energy produced (in kWh) by emptying 1,000 hectares of land filled with water having a depth of 30 meters [65.6 ft] at a rate of 800 million L/min [211.4 million gpm].

Compare this value to the average annual household energy requirement of 12,000 kWh/yr in the US. Use an overall conversion efficiency of 75%.

Solution: a. The volumetric flow rate is converted into the proper units: 𝑄𝑄 �

𝑘𝑘𝑘𝑘 800 × 106 𝐿𝐿 1 𝑘𝑘𝑘𝑘 1 𝑚𝑚𝑚𝑚𝑚𝑚 13.33 × 106 𝑘𝑘𝑘𝑘 �= × × = 𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠 𝑠𝑠 1 𝐿𝐿 60 𝑠𝑠

b. The power is calculated next: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) =

𝑄𝑄𝑄𝑄𝑄𝑄 13.33 × 106 𝑘𝑘𝑘𝑘 1 = × 30 𝑚𝑚 × 0.75 × = 2.94 × 106 𝑘𝑘𝑘𝑘 102 102 𝑠𝑠

c. The time is calculated from the total reservoir volume: 𝑉𝑉 (𝑚𝑚3 ) = 𝐿𝐿 × 𝑊𝑊 × 𝐻𝐻 = 1,000 ℎ𝑎𝑎 × 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇(ℎ𝑟𝑟𝑟𝑟) = 300 × 106 𝑚𝑚3 ×

10,000 𝑚𝑚2 × 30 𝑚𝑚 = 300 × 106 𝑚𝑚3 ℎ𝑎𝑎

1,000 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚 ℎ𝑟𝑟 × × = 6.25 ℎ𝑟𝑟𝑟𝑟 1 𝑚𝑚3 800 × 106 𝐿𝐿 60 𝑚𝑚𝑚𝑚𝑚𝑚

d. Energy is simply the product of power and time, calculated as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑘𝑘𝑘𝑘ℎ) = 2.94 × 106 𝑘𝑘𝑘𝑘 × 6.25 ℎ𝑟𝑟𝑟𝑟 = 18,375,000 𝑘𝑘𝑘𝑘ℎ # 𝐻𝐻𝐻𝐻 =

18,375,000 𝑘𝑘𝑘𝑘ℎ = 1,531 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑘𝑘𝑘𝑘ℎ 12,000 𝐻𝐻𝐻𝐻

e. This energy is higher than the yearly requirements of a household. The problem is that this power is only available for a few hours and reservoir water must be replenished after that period. However, around 1,531 households will benefit from this renewable power.

199

Problem 9.9.6 Matching Household Energy Requirements P9.6 At what rate would you release water (in kg/s) for a reservoir measuring 1,000 hectares [2,470 acres] in area and 30 meters [98.4 feet] in depth if you simply want to generate 300 kW [402 hp] of electrical power? Assume an overall conversion efficiency of 75%. Convert the units to gpm.

Solution: a. The rate of water release is simply calculated as follows: 𝑄𝑄 �

𝑘𝑘𝑘𝑘 102 × 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) 102 × 300 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 �= = = 1,360 𝑠𝑠 𝐻𝐻𝐻𝐻 30 𝑚𝑚 × 0.75 𝑠𝑠

𝑄𝑄(𝑔𝑔𝑔𝑔𝑔𝑔) = 1,360

𝑘𝑘𝑘𝑘 1 𝐿𝐿 𝑔𝑔𝑔𝑔𝑔𝑔 60 𝑠𝑠 × × × = 21,559 𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠 1 𝑘𝑘𝑘𝑘 3.785 𝐿𝐿 1 𝑚𝑚𝑚𝑚𝑚𝑚

Problem 9.9.7 Water Level Decline with Time for a Given Basin

P9.7 Compare the level of water from two basins, one with a cross-sectional area of 50 m x 50 m and 5-meter height to another basin with a cross section of 5 m x 5 m and 5 m height. The rate of release is 325 L/min. Estimate the water level after 5 hours of release in both systems.

Solution. a. The volume of each basin is calculated. 𝑉𝑉𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 5 𝑚𝑚 × 5 𝑚𝑚 × 5 𝑚𝑚 = 125 𝑚𝑚3 ×

1000 𝐿𝐿 = 125,000 𝐿𝐿 1 𝑚𝑚3

𝑉𝑉𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 50 𝑚𝑚 × 50 𝑚𝑚 × 5 𝑚𝑚 = 12,500 𝑚𝑚3 ×

1000 𝐿𝐿 = 12,500,000 𝐿𝐿 1 𝑚𝑚3

b. After 5 hours, the volume released will be calculated as follows: 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 5 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜) =

325 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚

× 5 ℎ𝑟𝑟𝑟𝑟 ×

60 𝑚𝑚𝑚𝑚𝑚𝑚 1 ℎ𝑟𝑟

= 97,500 𝐿𝐿

c. Hence, the level of the small basin will be calculated as follows: 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝐿𝐿) = 125,000 𝑙𝑙 − 97,500 𝑙𝑙 = 27,500 𝐿𝐿 200

𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑚𝑚3 ) = 27,500 𝐿𝐿 ×

𝑚𝑚3 = 27.5 𝑚𝑚3 1,000 𝐿𝐿

𝑉𝑉 (𝑚𝑚3 ) 27.5 𝑚𝑚3 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑚𝑚) = = = 1.1 𝑚𝑚 𝐿𝐿(𝑚𝑚) × 𝑊𝑊 (𝑚𝑚) 5 𝑚𝑚 × 5 𝑚𝑚

125,000 L – 97,500 L = 27,500 Liters = 27.5 m3/(5m x 5m) = 1.1 meter The small basin has decreased by slightly less than 4 meters. d. The level decrease in the large basin is calculated the same way. 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝐿𝐿) = 12,500,000 𝑙𝑙 − 97,500 𝑙𝑙 = 12,402,500 𝐿𝐿 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑚𝑚

3)

𝑚𝑚3 = 12,402.5 𝑚𝑚3 = 12,402,500 𝐿𝐿 × 1,000 𝐿𝐿

12,402.5 𝑚𝑚3 𝑉𝑉 (𝑚𝑚3 ) = = 4.96 𝑚𝑚 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑚𝑚) = 𝐿𝐿(𝑚𝑚) × 𝑊𝑊 (𝑚𝑚) 50 𝑚𝑚 × 50 𝑚𝑚

The large basin only decreased by a fraction of a meter, which is minimal.

Problem 9.9.8 Power Generated from Small Basin’ P9.8 Determine the power generated when water is released at a rate of 5 million L/min [1.321 million gpm] with 1 meter [3.28 ft] of head. Assume an overall efficiency of 75%.

Solution: a. The rate of water release is converted into units of kg/s: 𝑄𝑄 �

𝑘𝑘𝑘𝑘 5 × 106 𝐿𝐿 1 𝑚𝑚𝑚𝑚𝑚𝑚 1 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 �= × × = 83,333 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 60 𝑠𝑠 1 𝐿𝐿 𝑠𝑠

b. Use the classical hydro power equation, converting units properly: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) =

𝑄𝑄𝑄𝑄𝑄𝑄 83,333 𝑘𝑘𝑘𝑘 1 = × 1 𝑚𝑚 × 0.75 × = 612.75 𝑘𝑘𝑘𝑘 102 𝑠𝑠 102

Problem 9.9.9 Power and Energy from Double Basin System

P9.9 A double basin system has two reservoirs, Basin A and Basin B, with similar dimensions of about 50 m x 50 m x 12 meters [164 x 164 x 39.36 ft] in depth (half a hectare). Basin A is the 201

higher-level basin and Basin B is the lower-level basin. The minimum dynamic head of the power generation is10 meters [32.8 ft]. (a) Determine the amount of water (kg/s) to be released such that 2 meters [6.56 ft] of head remains in Basin A as it releases its water to Basin B. In this case, Basin A will use up about 10 meters [32.8 ft] of its dynamic head in 24 hours. (b) Determine the minimum power (kW) generated based on the appropriate release rate from Basin A. Assume an overall conversion efficiency of 75%. (c) Determine the energy produced (kWh) if this system operates 24 hours per day and 365 days in a year. Report this in kWh/yr. (d) Determine the number of households that may be served by this facility if one household uses 12,000 kWh/yr.

Solution: a. If Basin A releases water to Basin B in 6 hours with 2 meters of depth remaining, the volumetric rate of release will be calculated as follows: 𝑉𝑉 � 𝑉𝑉 �

𝑚𝑚3 𝐿𝐿 (𝑚𝑚) × 𝑊𝑊(𝑚𝑚) × 𝐻𝐻(𝑚𝑚) 50 𝑚𝑚 × 50 𝑚𝑚 × 10 𝑚𝑚 25,000 𝑚𝑚3 𝑚𝑚3 �= = = = 4,167 ℎ𝑟𝑟 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 (ℎ𝑟𝑟𝑟𝑟) 6 ℎ𝑟𝑟𝑟𝑟 6 ℎ𝑟𝑟𝑟𝑟 ℎ𝑟𝑟

𝑘𝑘𝑘𝑘 𝑚𝑚3 1,000 𝑘𝑘𝑘𝑘 1 ℎ𝑟𝑟 𝑘𝑘𝑘𝑘 � = 4,167 × × = 1,157.5 3 𝑠𝑠 ℎ𝑟𝑟 1 𝑚𝑚 3600 𝑠𝑠 𝑠𝑠

b. Hence, the power is calculated as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) =

𝑄𝑄𝑄𝑄𝑄𝑄 1,157.5 𝑘𝑘𝑘𝑘 10 𝑚𝑚 = × × 0.75 ×= 85 𝑘𝑘𝑘𝑘 102 𝑠𝑠 102

c. Energy is simply the product of power and time: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑘𝑘𝑘𝑘ℎ) = 85 𝑘𝑘𝑘𝑘 ×

24 ℎ𝑟𝑟𝑟𝑟 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑘𝑘𝑘𝑘ℎ × = 744,600 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

d. This is less than a million kWh per year. If a household consumes 12,000 kWhrs/yr, then the number of households served by this facility is given below.

202

# 𝐻𝐻𝐻𝐻 =

744,600 𝑘𝑘𝑘𝑘ℎ = 62 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑘𝑘𝑘𝑘ℎ 12,000 𝐻𝐻𝐻𝐻

e. A little over 60 households may be served by this facility.

Problem 9.9.10 Cost to Recover Initial Investment.

P9.10 Determine the return to investment for a tidal barrage project in Korea (Sihwa Lake) with a rated power capacity of 254 MW (considered power input). Assume the yearly rated power output is 65% of the rated output [i.e., 203 MW] and the power is to be sold for $0.08/kWh. The plant operates 360 days a year 24 hours per day. Base the ROI on just the initial capital cost of $298 million and the sale of electricity.

Solution: a. The gross sales of electricity is calculated by first calculating energy output in a year: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝑘𝑘𝑘𝑘ℎ 360 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 24 ℎ𝑟𝑟𝑟𝑟 1,000 𝑘𝑘𝑘𝑘 � = 254 𝑀𝑀𝑀𝑀 × 0.60 × × × 𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦 𝑑𝑑𝑑𝑑𝑑𝑑 𝑀𝑀𝑀𝑀 = 1,316.7 × 106

𝑘𝑘𝑘𝑘ℎ 𝑦𝑦𝑦𝑦

b. The gross receipt after a year will be as follows: 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 ($) = 1,316.7 × 106

𝑘𝑘𝑘𝑘ℎ $0.08 × = $105.34 × 106 /𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦 𝑘𝑘𝑘𝑘ℎ

c. The RO is the ratio of the initial capital cost and the yearly gross revenue: 𝑅𝑅𝑅𝑅𝑅𝑅 (𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌) =

$298 × 106 = 2.82 𝑦𝑦𝑦𝑦𝑦𝑦 $105.34 × 106 /𝑦𝑦𝑦𝑦

d. Hence, the ROI could be less than 5 years based on simply the gross receipts.

203

Problem 9.9.11 Sizing Water Reservoirs P9.11 The average annual household energy requirement is about 10,000 kWh/yr in the US. What is the minimum size (in meters for one side) of a reservoir required (assume cubic with L x W x H in meters) if the water will continually produce power throughout the year (8760 hrs).

Assume a uniform dynamic head (H) of 15 meters and overall conversion efficiency of 80%. a. 22 b. 42 c. 67.4 d. 84 Solutions: a. Energy is simply the product of power and time and calculated as follows: 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑜𝑜𝑜𝑜 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) =

1 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 10,000 𝑘𝑘𝑘𝑘ℎ × = 1.14 𝑘𝑘𝑘𝑘 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 8760 ℎ𝑟𝑟𝑟𝑟

b. Hence, the calculation will make use of the hydro power equation to estimate Q as shown. 𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄𝑄𝑄 𝑄𝑄 � 𝑠𝑠 � × 15 𝑚𝑚 × 0.80 1.14 kW = = 102 102 𝑘𝑘𝑘𝑘

𝑄𝑄 � � = 𝑠𝑠

102 ×1.14 𝑘𝑘𝑘𝑘 15 𝑚𝑚 ×0.80

= 9.7

𝑘𝑘𝑘𝑘 𝑠𝑠

c. The volumetric flow in cubic meters is as follows: 𝑄𝑄 �

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑚𝑚3 3600 𝑠𝑠 24 ℎ𝑟𝑟𝑟𝑟 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚3 � = 9.7 × × × × = 305,899 1,000 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦 𝑠𝑠 𝑠𝑠 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

d. Hence, the volume of reservoir will be 305,899 m3 and assuming equal dimensions on each side is as follows: 3

𝐿𝐿 = 𝑊𝑊 = 𝐻𝐻 = �305,899 𝑚𝑚3 = 67.4 𝑚𝑚

204

9.9.12 Rate of Water Release in Tidal Systems P9.12 At what rate would you release water (in kg/s) for a reservoir measuring 5 acres in area and 65.6 ft [20 meters] in depth if you want to generate 500 kW of electrical power? Assume the average dynamic head is 15 m and a conversion efficiency of 80%. a. 40,423 b. 2,550 c. 101,057 d. 4,250 Solutions: a. The solution may be easily calculated using the power equation as shown below. 𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄𝑄𝑄 𝑄𝑄 � 𝑠𝑠 � × 15 𝑚𝑚 × 0.80 500 kW = = 102 102 102 × 500 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 = 4250 𝑄𝑄 � � = 𝑠𝑠 𝑠𝑠 15 𝑚𝑚 × 0.80

9.9.13 Rate of Water Release in a Tidal Facility

P9.13 At what rate would you release water (in gpm) for a reservoir measuring 5 acres in area and 65.6 ft [20 meters] in depth if you want to generate 500 kW of electrical power? Assume the average dynamic head is 15 m and a conversion efficiency of 80%. a. 67,371 b. 2,550 c. 101,057 d. 40,423 Solutions: a. The solution may be easily calculated using the power equation as shown below.

205

𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄𝑄𝑄 𝑄𝑄 � 𝑠𝑠 � × 15 𝑚𝑚 × 0.80 500 kW = = 102 102 𝑄𝑄 �

𝑘𝑘𝑘𝑘 102 × 500 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 �= = 4250 15 𝑚𝑚 × 0.80 𝑠𝑠 𝑠𝑠

𝑄𝑄 �

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑚𝑚3 3600 𝑠𝑠 24 ℎ𝑟𝑟𝑟𝑟 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚3 � = 4250 × × × × = 134 × 106 𝑠𝑠 𝑠𝑠 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 1,000 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦

b. The flow rate in m3/yr is calculated as follows:

c. The flow rate in gpm is calculated as follows: 𝑄𝑄(𝑔𝑔𝑔𝑔𝑔𝑔) = 4250

𝑘𝑘𝑘𝑘 1 𝐿𝐿 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 60 𝑠𝑠 × × × = 67,371 𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠 1 𝑘𝑘𝑘𝑘 3.785 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚

9.9.14 Water Release Rate in a Tidal Facility P9.14 At what rate would you release water (in million cubic meters/year) for a reservoir measuring 5 acres in area and 65.6 ft [20 meters] in depth if you want to generate 500 kW of electrical power? Assume average dynamic head is 15 m and a conversion efficiency of 80%. a. 67,371 b. 134 c. 255 d. 40,423 Solutions: a. The solution may be easily calculated using the power equation as shown below. 𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄𝑄𝑄 𝑄𝑄 � 𝑠𝑠 � × 15 𝑚𝑚 × 0.80 500 kW = = 102 102 𝑄𝑄 �

𝑘𝑘𝑘𝑘 102 × 500 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 �= = 4250 𝑠𝑠 15 𝑚𝑚 × 0.80 𝑠𝑠

206

b. The flow rate in m3/yr is calculated as follows: 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑚𝑚3 3600 𝑠𝑠 24 ℎ𝑟𝑟𝑟𝑟 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚3 6 𝑄𝑄 � � = 4250 × × × × = 134 × 10 𝑠𝑠 𝑠𝑠 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 1,000 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 𝑦𝑦𝑦𝑦

c. The flow rate in gpm is calculated as follows: 𝑄𝑄(𝑔𝑔𝑔𝑔𝑔𝑔) = 4250

𝑘𝑘𝑘𝑘 1 𝐿𝐿 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 60 𝑠𝑠 × × × = 67,371 𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠 1 𝑘𝑘𝑘𝑘 3.785 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚

9.9.15 Water Release Rate in a Tidal Facility

P9.15 At what rate would you release water (in gpm) for a reservoir measuring 1 acre in area and 32.8 ft [20 meters] in depth if you want to generate 2 kW of electrical power? Assume an average dynamic head of 15 meters and a conversion efficiency of 80% a. 40,423 b. 2,550 c. 269.5 d. 6,375 Solutions: a. The solution may be easily calculated using the power equation as shown below. 𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄𝑄𝑄 𝑄𝑄 � 𝑠𝑠 � × 15 𝑚𝑚 × 0.80 2 kW = = 102 102 𝑄𝑄 �

𝑘𝑘𝑘𝑘 102 × 2 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 �= = 17 𝑠𝑠 15 𝑚𝑚 × 0.80 𝑠𝑠

𝑄𝑄 �

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑚𝑚3 3600 𝑠𝑠 24 ℎ𝑟𝑟𝑟𝑟 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚3 � = 17 × × × × = 0.536 × 106 1,000 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦 𝑠𝑠 𝑠𝑠 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

b. The flow rate in m3/yr is calculated as follows:

c. The flow rate in gpm is calculated as follows: 𝑄𝑄(𝑔𝑔𝑔𝑔𝑔𝑔) = 17

𝑘𝑘𝑘𝑘 1 𝐿𝐿 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 60 𝑠𝑠 × × × = 269.5 𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠 1 𝑘𝑘𝑘𝑘 3.785 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚 207

Assess Your Knowledge 1. This is a type of tide that happens when the earth-moon and earth-sun lines are at right angles, the gravitational forces are subtractive, and the tides are subdued, and tides are usually lower than normal.

a. spring tides b. ebb tides c. neap tides d. slack tide 2. What is the duration of semi-diurnal tides in hours?

a. 12.417 b. 12 c. 24 d. 11.5 3. Name the only tidal power generation scheme in which electrical power is produced continuously.

a. Double basin b. Single basin ebb c. single basin tide d. single basin two-way 4. This is the energy powered by the natural rise and fall of ocean tides and currents. The variation in tides results from attractive forces between the earth, the sun and the moon. Name this renewable energy resource.

a. hydraulic ram b. pumped storage c. tidal energy 208

d. lunar power 5. This is a type of tide that happens during the full moon and new moon (lines are along the center of the earth), the sun and moon are linear, and tides are usually higher than normal.

a. neap tides b. spring tides c. ebb tides d. slack tide 6. Determine the volume of water contained in a 5-acre plot with a depth of 65.6 ft [20 meters]. Report your answers in millions of gallons.

a. 107,000 b. 530000 c. 53.5 d. 107 Solutions: a. This is a simple volumetric calculation as follows: 𝑉𝑉 (𝑚𝑚

3)

1 ℎ𝑎𝑎 10,000 𝑚𝑚2 × × 20 𝑚𝑚 = 404,858 𝑚𝑚3 = 𝐿𝐿 × 𝑊𝑊 × 𝐻𝐻 = 5 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 × 2.47 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1 ℎ𝑎𝑎

b. Simply convert this volume in English units as follows: 𝑉𝑉 (106 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ) = 404,858 𝑚𝑚3 ×

264.172 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 107 × 106 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1 𝑚𝑚3

𝑉𝑉 (106 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ) = 404,858 𝑚𝑚3 ×

1,000 𝐿𝐿 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 × = 106,963,804.5 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1 𝑚𝑚3 3.785 𝐿𝐿

c. The long way conversion is as follows:

𝑉𝑉 (106 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ) = 107 × 106 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 [107 million gallons] 209

7. Determine the power generated for a water storage facility having an area of 5 acres with a depth of 65.6 ft [20 meters] if water is released at a rate of 250 gpm at an average dynamic head of 49.2 ft [15 meters]. Use an overall conversion efficiency of 75%. Report your answer in kW.

a. 0.51 b. 0.85 c. 510 d. 1.74 Solutions: a. The volume of water in the reservoir is calculated as follows: 𝑉𝑉 (𝑚𝑚3 ) = 𝐿𝐿 × 𝑊𝑊 × 𝐻𝐻 = 5 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ×

1 ℎ𝑎𝑎 10,000 𝑚𝑚2 × × 10 𝑚𝑚 = 404,858 𝑚𝑚3 2.47 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1 ℎ𝑎𝑎

b. The long way conversion is as follows: 𝑉𝑉 (106 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ) = 404,858 𝑚𝑚3 ×

1,000 𝐿𝐿 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 × = 106,963,804 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1 𝑚𝑚3 3.785 𝐿𝐿

c. If this volume of water is released at 250 gpm, the time it takes to do this process is calculated as follows: 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = 106,963,804 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ×

𝑚𝑚𝑚𝑚𝑚𝑚 1 ℎ𝑟𝑟 × = 7131 ℎ𝑟𝑟𝑟𝑟 250 𝑔𝑔𝑔𝑔𝑔𝑔 60 𝑚𝑚𝑚𝑚𝑚𝑚

d. Hence, it would take over 7131 hours or 297 days to generate the power continuously. e. The mass flow rate of water is first calculated as follows: 𝑄𝑄 �

𝑘𝑘𝑘𝑘 250 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 3.785 𝐿𝐿 1 𝑘𝑘𝑘𝑘 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑘𝑘𝑘𝑘 �= × × × = 15.77 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 1 𝑔𝑔𝑔𝑔𝑔𝑔 1 𝐿𝐿 60 𝑠𝑠 𝑠𝑠

f. The output power is simply calculated making use of the hydro power equations as follows:

210

𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) =

𝑘𝑘𝑘𝑘 𝑄𝑄 � 𝑠𝑠 � × 𝐻𝐻 (𝑚𝑚) × 𝐸𝐸𝑜𝑜 (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ) 102

𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄𝑄𝑄 15.77 𝑠𝑠 × 15 𝑚𝑚 × 0.75 𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) = = = 1.74 𝑘𝑘𝑘𝑘 102 102

g. This is such as low power output but will last for 7,131 hrs or 297 days.

8. If the volume of water in a 5-acre plot with a depth of 65.6 ft [20 meters] is released at a rate of 250 gallons per minute, how long (in hrs) will this containment ran out of water?

a. 3,241 b. 1,621 c. 7,131 d. 8,103 Solutions: a. The volume is calculated as follows: 𝑉𝑉 (𝑚𝑚

3)

1 ℎ𝑎𝑎 10,000 𝑚𝑚2 × × 20 𝑚𝑚 = 404,858 𝑚𝑚3 = 𝐿𝐿 × 𝑊𝑊 × 𝐻𝐻 = 5 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 × 2.47 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1 ℎ𝑎𝑎

a. The long way conversion is as follows: 𝑉𝑉 (106 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ) = 404,858 𝑚𝑚3 ×

1,000 𝐿𝐿 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 × = 106,963,804.5 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1 𝑚𝑚3 3.785 𝐿𝐿

b. This is also a simple volume over volumetric flowrate calculations as follows: 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = 106,963,804.5 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ×

𝑚𝑚𝑚𝑚𝑚𝑚 1 ℎ𝑟𝑟 × = 7131 ℎ𝑟𝑟𝑟𝑟 250 𝑔𝑔𝑔𝑔𝑔𝑔 60 𝑚𝑚𝑚𝑚𝑚𝑚

c. Hence, it would take over 7131 hours or 297 days to empty this reservoir continuously.

9. Determine the volume of water contained in a 1-acre plot with a depth of 32.8 ft [10 meters]. Report your answers in cubic meters.

a. 40,486 211

b. 10,696,388 c. 40.5 d. 5,348,194 Solutions: a. This is a simple volumetric calculation as follows: 𝑉𝑉 (𝑚𝑚3 ) = 𝐿𝐿 × 𝑊𝑊 × 𝐻𝐻 = 1 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ×

1 ℎ𝑎𝑎 10,000 𝑚𝑚2 × × 10 𝑚𝑚 = 40,486 𝑚𝑚3 1 ℎ𝑎𝑎 2.47 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

10. Determine the volume of water contained in a 5-acre plot with a depth of 65.6 ft [20 meters]. Report your answers in cubic meters.

a. 202430 b. 10,696,388 c. 404,858 d. 5,348,194 Solutions: a. This is a simple volumetric calculation as follows: 𝑉𝑉 (𝑚𝑚3 ) = 𝐿𝐿 × 𝑊𝑊 × 𝐻𝐻 = 5 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ×

1 ℎ𝑎𝑎 10,000 𝑚𝑚2 × × 20 𝑚𝑚 = 404,858 𝑚𝑚3 2.47 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1 ℎ𝑎𝑎

11. The depth variation in tides as a function of time may be approximated by what trigonometric equation?

a. Logarithmic b. Polynomial c. Exponential d. sine curve

212

12. Where is the best place to see the highest tides in the world?

a. Indian Ocean b. Atlantic Coast c. Pacific Coast d. Bay of Fundy 13. This is a power generation scheme whereby during high tide, one generates power and the basin is filled.

a. Single basin ebb cycle b. Single basin tide cycle. c. Single basin two-way generation d. Double basin system 14. Determine the energy produced (kWh) by emptying a water storage facility having an area of 5 acres with a depth of 32.8 ft [20 meters]. The water was released at a rate of 250 gpm at an average dynamic head of 49.2 ft [15 meters]. Use an overall conversion efficiency of 75%.

a. 12,408 b. 4,131.2 c. 5,100 d. 6,202 Solutions: a. The volume of water in the reservoir is calculated as follows: 𝑉𝑉 (𝑚𝑚3 ) = 𝐿𝐿 × 𝑊𝑊 × 𝐻𝐻 = 5 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ×

1 ℎ𝑎𝑎 10,000 𝑚𝑚2 × × 10 𝑚𝑚 = 404,858 𝑚𝑚3 2.47 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1 ℎ𝑎𝑎

b. The long way conversion is as follows: 𝑉𝑉 (106 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ) = 404,858 𝑚𝑚3 ×

1,000 𝐿𝐿 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 × = 106,963,804 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1 𝑚𝑚3 3.785 𝐿𝐿 213

c. If this volume of water is released at 250 gpm, the time it takes to do this process is calculated as follows: 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = 106,963,804 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ×

𝑚𝑚𝑚𝑚𝑚𝑚 1 ℎ𝑟𝑟 × = 7131 ℎ𝑟𝑟𝑟𝑟 250 𝑔𝑔𝑔𝑔𝑔𝑔 60 𝑚𝑚𝑚𝑚𝑚𝑚

d. Hence, it would take over 7131 hours or 297 days to generate the power continuously. e. The mass flow rate of water is first calculated as follows: 𝑄𝑄 �

𝑘𝑘𝑘𝑘 250 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 3.785 𝐿𝐿 1 𝑘𝑘𝑘𝑘 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑘𝑘𝑘𝑘 �= × × × = 15.77 𝑠𝑠 1 𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 1 𝐿𝐿 60 𝑠𝑠

f. The output power is simply calculated making use of the hydro power equations as follows:

𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) =

𝑘𝑘𝑘𝑘 𝑄𝑄 � 𝑠𝑠 � × 𝐻𝐻 (𝑚𝑚) × 𝐸𝐸𝑜𝑜 (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ) 102

𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄𝑄𝑄 15.77 𝑠𝑠 × 15 𝑚𝑚 × 0.75 𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) = = = 1.74 𝑘𝑘𝑘𝑘 102 102

g. This is such as low power output but will last for 7,131 hrs or 297 days. h. Hence, the energy is the product of power and time as shown. 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑘𝑘𝑘𝑘ℎ) = 1.74 𝑘𝑘𝑘𝑘 × 7131 ℎ𝑟𝑟𝑟𝑟 = 12,408 𝑘𝑘𝑘𝑘ℎ

i. The percentage of energy provided by this tidal power is calculated as follows: % 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (%) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 12408 × 100% = × 100% = 99.3% 12500 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶

15. At what rate would you release water (in gpm) for a reservoir measuring 5 acres in area and 65.6 ft [20 meters] in depth if you want to generate 500 kW of electrical power? Assume average dynamic head is 15 m and a conversion efficiency of 80%.

a. 67,371 b. 2,550 214

c. 101,057 d. 40,423 Solutions: a. The solution may be easily calculated using the power equation as shown below. 𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄𝑄𝑄 𝑄𝑄 � 𝑠𝑠 � × 15 𝑚𝑚 × 0.80 500 kW = = 102 102 𝑄𝑄 �

𝑘𝑘𝑘𝑘 102 × 500 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 �= = 4250 15 𝑚𝑚 × 0.80 𝑠𝑠 𝑠𝑠

𝑄𝑄 �

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑚𝑚3 3600 𝑠𝑠 24 ℎ𝑟𝑟𝑟𝑟 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚3 � = 4250 × × × × = 134 × 106 𝑠𝑠 𝑠𝑠 1,000 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦

b. The flow rate in m3/yr is calculated as follows:

c. The flow rate in gpm is calculated as follows: 𝑄𝑄(𝑔𝑔𝑔𝑔𝑔𝑔) = 4250

𝑘𝑘𝑘𝑘 1 𝐿𝐿 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 60 𝑠𝑠 × × × = 67,371 𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠 1 𝑘𝑘𝑘𝑘 3.785 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚

16. At what rate would you release water (in gpm) for a reservoir measuring 1 acre in area and 32.8 ft [20 meters] in depth if you want to generate 2 kW of electrical power? Assume an average dynamic head of 15 meters and a conversion efficiency of 80%

a. 40,423 b. 2,550 c. 269.5 d. 6,375 Solution: a. The solution may be easily calculated using the power equation as shown below. 𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄𝑄𝑄 𝑄𝑄 � 𝑠𝑠 � × 15 𝑚𝑚 × 0.80 2 kW = = 102 102 215

𝑄𝑄 �

𝑘𝑘𝑘𝑘 𝑠𝑠

�=

102 × 2 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 = 17 15 𝑚𝑚 × 0.80 𝑠𝑠

b. The flow rate in m3/yr is calculated as follows: 𝑄𝑄 �

𝑘𝑘𝑘𝑘 𝑠𝑠

� = 17

𝑘𝑘𝑘𝑘 𝑚𝑚3 3600 𝑠𝑠 24 ℎ𝑟𝑟𝑟𝑟 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚3 × × × × = 0.536 × 106 𝑠𝑠 1,000 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦

c. The flow rate in gpm is calculated as follows: 𝑄𝑄(𝑔𝑔𝑔𝑔𝑔𝑔) = 17

𝑘𝑘𝑘𝑘 1 𝐿𝐿 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 60 𝑠𝑠 × × × = 269.5 𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠 1 𝑘𝑘𝑘𝑘 3.785 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚

17. At what rate would you release water (in million cubic meters/year) for a reservoir measuring 5 acres in area and 65.6 ft [20 meters] in depth if you want to generate 500 kW of electrical power? Assume average dynamic head is 15 m and a conversion efficiency of 80%.

a. 67,371 b. 134 c. 255 d. 40,423 Solutions: a. The solution may be easily calculated using the power equation as shown below. 𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄𝑄𝑄 𝑄𝑄 � 𝑠𝑠 � × 15 𝑚𝑚 × 0.80 500 kW = = 102 102

𝑄𝑄 �

𝑘𝑘𝑘𝑘 𝑠𝑠

�=

102 × 500 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 = 4250 15 𝑚𝑚 × 0.80 𝑠𝑠

b. The flow rate in m3/yr is calculated as follows: 𝑄𝑄 �

𝑘𝑘𝑘𝑘 𝑠𝑠

� = 4250

𝑘𝑘𝑘𝑘 𝑚𝑚3 3600 𝑠𝑠 24 ℎ𝑟𝑟𝑟𝑟 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚3 × × × × = 134 × 106 𝑠𝑠 1,000 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦

18. The average annual household energy requirement is about 10,000 kWh/yr in the US. What is the minimum size (in meters for one side) of a reservoir required (assume cubic with L x W x H in 216

meters) if the water will continually produce power throughout the year (8760 hrs). Assume a uniform dynamic head (H) of 15 meters and overall conversion efficiency of 80%.

a. 22 b. 42 c. 67.4 d. 84 Solution: a. Energy is simply the product of power and time and calculated as follows: 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑜𝑜𝑜𝑜 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) =

1 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 10,000 𝑘𝑘𝑘𝑘ℎ × = 1.14 𝑘𝑘𝑘𝑘 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 8760 ℎ𝑟𝑟𝑟𝑟

b. Hence, the calculation will make use of the hydro power equation to estimate Q as shown. 𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄𝑄𝑄 𝑄𝑄 � 𝑠𝑠 � × 15 𝑚𝑚 × 0.80 1.14 kW = = 102 102 𝑘𝑘𝑘𝑘

𝑄𝑄 � 𝑠𝑠 � =

102 ×1.14 𝑘𝑘𝑘𝑘 15 𝑚𝑚 ×0.80

= 9.7

𝑘𝑘𝑘𝑘 𝑠𝑠

c. The volumetric flow in cubic meters is as follows: 𝑘𝑘𝑘𝑘 𝑚𝑚3 3600 𝑠𝑠 24 ℎ𝑟𝑟𝑟𝑟 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚3 𝑄𝑄 � � = 9.7 × × × × = 305,899 𝑠𝑠 𝑠𝑠 1,000 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦 𝑘𝑘𝑘𝑘

d. Hence, the volume of reservoir will be 305,899 m3 and assuming equal dimensions on each side is as follows: 3

𝐿𝐿 = 𝑊𝑊 = 𝐻𝐻 = �305,899 𝑚𝑚3 = 67.4 𝑚𝑚

217

19. At what rate would you release water (in kg/s) for a reservoir measuring 5 acres in area and 65.6 ft [20 meters] in depth if you want to generate 500 kW of electrical power? Assume average dynamic head is 15 m and a conversion efficiency of 80%.

a. 40,423 b. 2,550 c. 101,057 d. 4,250 Solutions: a. The solution may be easily calculated using the power equation as shown below. 𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄𝑄𝑄 𝑄𝑄 � 𝑠𝑠 � × 15 𝑚𝑚 × 0.80 500 kW = = 102 102

𝑄𝑄 �

𝑘𝑘𝑘𝑘 𝑠𝑠

�=

102 × 500 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 = 4250 15 𝑚𝑚 × 0.80 𝑠𝑠

20. Determine the percentage of energy (%) being satisfied by a tidal power plant for a reservoir having an area of 5 acres with a depth of 32.8 ft [20 meters]. The water was released at a rate of 250 gpm at an average dynamic head of 49.2 ft [15 meters]. Use an overall conversion efficiency of 75%. The annual energy consumption of the owner is 12,500 kWh.

a. 75 b. 99.3 c. 85 d. 60.5 Solution: a. The volume of water in the reservoir is calculated as follows: 218

𝑉𝑉 (𝑚𝑚3 ) = 𝐿𝐿 × 𝑊𝑊 × 𝐻𝐻 = 5 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ×

1 ℎ𝑎𝑎 10,000 𝑚𝑚2 × × 10 𝑚𝑚 = 404,858 𝑚𝑚3 2.47 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1 ℎ𝑎𝑎

b. The long way conversion is as follows: 𝑉𝑉 (106 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ) = 404,858 𝑚𝑚3 ×

1,000 𝐿𝐿 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 × = 106,963,804 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1 𝑚𝑚3 3.785 𝐿𝐿

c. If this volume of water is released at 250 gpm, the time it takes to do this process is calculated as follows: 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = 106,963,804 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ×

𝑚𝑚𝑚𝑚𝑚𝑚 1 ℎ𝑟𝑟 × = 7131 ℎ𝑟𝑟𝑟𝑟 250 𝑔𝑔𝑔𝑔𝑔𝑔 60 𝑚𝑚𝑚𝑚𝑚𝑚

d. Hence, it would take over 7131 hours or 297 days to generate the power continuously. e. The mass flow rate of water is first calculated as follows: 𝑄𝑄 �

𝑘𝑘𝑘𝑘 250 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 3.785 𝐿𝐿 1 𝑘𝑘𝑘𝑘 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑘𝑘𝑘𝑘 �= × × × = 15.77 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 1 𝑔𝑔𝑔𝑔𝑔𝑔 1 𝐿𝐿 60 𝑠𝑠 𝑠𝑠

f. The output power is simply calculated making use of the hydro power equations as follows: 𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘 ) =

𝑘𝑘𝑘𝑘 𝑄𝑄 � 𝑠𝑠 � × 𝐻𝐻 (𝑚𝑚 ) × 𝐸𝐸𝑜𝑜 (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑) 102

𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄𝑄𝑄 15.77 𝑠𝑠 × 15 𝑚𝑚 × 0.75 𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘 ) = = = 1.74 𝑘𝑘𝑘𝑘 102 102

g. This is such as low power output but will last for 7,131 hrs or 297 days. h. Hence, the energy is the product of power and time as shown. 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑘𝑘𝑘𝑘ℎ) = 1.74 𝑘𝑘𝑘𝑘 × 7131 ℎ𝑟𝑟𝑟𝑟 = 12,408 𝑘𝑘𝑘𝑘ℎ

i. The percentage of energy provided by this tidal power is calculated as follows: % 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (%) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 12408 × 100% = × 100% = 99.3% 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 12500 219

Chapter 10 Wave Energy 10.13 Solved Problems Problem 10.13.1 Determine the Constant for Wave Power Equation P13.1 Determine the estimated value of constant such that g and ρ do not appear in the equation (i.e. value of ρg2/64π or units of kW/m2s) if the density for saltwater is 1,027 kg/m3 instead of 1,000 kg/m3 (density of pure water).

Solution: a. The value of constant is approximately 0.5 as shown below. 𝜌𝜌𝑔𝑔2 1,027 𝑘𝑘𝑘𝑘 9.8 𝑚𝑚 2 1 𝑁𝑁 × 𝑠𝑠 2 𝐽𝐽 𝑊𝑊 × 𝑠𝑠 𝑘𝑘𝑘𝑘 = × � � × × × × 3 2 64𝜋𝜋 𝑚𝑚 𝑠𝑠 64 × 3.1416 𝑘𝑘𝑘𝑘 × 𝑚𝑚 𝑁𝑁 × 𝑚𝑚 𝐽𝐽 1000𝑊𝑊 = 0.490

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 ≅ 0.5 3 3 𝑚𝑚 𝑠𝑠 𝑚𝑚 𝑠𝑠

b. The value of constant did not change so much, hence 0.50 is commonly used.

Problem 10.13.2 Basic Power from Wave Problem 10.12.2 Determine the power from the wave in units of kW/m if the wave height is 4 meters and the wave period is 8 seconds. Use Equation 10.1 for calculations.

Solution: a. The value of constant is approximately 0.5 as shown below and the power is given as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 � = �0.50 3 � (4 m)2 × 8 s = 64 𝑚𝑚 𝑚𝑚 𝑠𝑠 𝑚𝑚

b. The power is about half a kW per square meter per second.

220

Problem 10.13.3 Wave Power in Storms P10.3 In a major storm, the wave height could be 10 meters for a period of 10 seconds. How much will the total available power be if the wave is 50 meters long.

Solution: a. Use Equation 10.1 and simply substitute the values: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃(𝑘𝑘𝑘𝑘) = �0.50

𝑘𝑘𝑘𝑘 � (10 m)2 × 10 s × 50 m = 25,000 kW [25 MW] 𝑚𝑚3 𝑠𝑠

b. Several tens of MW are available from this wave.

Problem 10.13.4 Total Power from Wave

P10.4 Given a wave height of 5 meters and a wave period of 7 seconds, estimate the power per meter of crest head. Determine the power for every km of wave.

Solution: a. Use the following formula: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 � = �0.50 3 � (5 m)2 × 7 s = 87.5 𝑚𝑚 𝑚𝑚 𝑠𝑠 𝑚𝑚

b. There is about 87.5 kW of power potential per meter of wave crest. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃(𝑘𝑘𝑘𝑘) = 87.5

𝑘𝑘𝑘𝑘 × 1,000 𝑚𝑚 = 87,500 𝑘𝑘𝑘𝑘 [87.5 𝑀𝑀𝑀𝑀] 𝑚𝑚

c. The total power per km of wave is 87.5 MW.

Problem 10.13.5. Hydraulic Power Developed from Buoys P10.5 Determine the pressure generated if the buoy load is 8,909.1 N [2,000 lbs] and the wave height is 6 meters [19.68 ft] with a wave period of 8 seconds. Also estimate the volumetric flow rate generated assuming the diameter of the small piston cylinder is 10 cm [3.94 inch]. Please refer to Figure 10.3 for this arrangement. Finally, estimate the theoretical power generated for this action.

221

Solution: a. Using Equation 10.2, the pressure generated is as follows for both the English and metric systems of units: 𝑝𝑝 =

𝐹𝐹 2,000 𝑙𝑙𝑙𝑙𝑙𝑙 = = 164.04 𝑝𝑝𝑝𝑝𝑝𝑝 𝐴𝐴 𝜋𝜋(3.94 𝑖𝑖𝑖𝑖)2 4

𝐹𝐹𝑏𝑏 (𝑁𝑁) 8,909.1 𝑁𝑁 𝑁𝑁 100 𝑐𝑐𝑐𝑐 2 𝑝𝑝 = = = 113.4339 ×� � = 1,134.3 𝑘𝑘𝑘𝑘𝑘𝑘 𝐴𝐴𝑝𝑝 (𝑐𝑐𝑐𝑐2 ) 𝜋𝜋(10 𝑐𝑐𝑐𝑐)2 𝑐𝑐𝑐𝑐2 𝑚𝑚 4

b. The volumetric flow rate is calculated using Equations 10.3 and 10.4. 𝐿𝐿 𝐻𝐻 (𝑀𝑀) 1,000 𝐿𝐿 𝑄𝑄𝑇𝑇 � � = × 𝐴𝐴𝑝𝑝 (𝑚𝑚2 ) × 𝑠𝑠 𝑇𝑇 (𝑠𝑠) 𝑚𝑚3

𝐿𝐿 6 𝑚𝑚 𝜋𝜋 10 𝑐𝑐𝑐𝑐 2 1,000 𝐿𝐿 𝐿𝐿 𝑄𝑄𝑇𝑇 � � = × ×� � × = 5.89 3 𝑠𝑠 8 𝑠𝑠 4 100 𝑐𝑐𝑐𝑐/𝑚𝑚 𝑚𝑚 𝑠𝑠

𝑔𝑔𝑔𝑔𝑔𝑔 𝐻𝐻 (𝑓𝑓𝑓𝑓) 𝑓𝑓𝑓𝑓 2 60 𝑠𝑠 7.48 𝑔𝑔𝑔𝑔𝑔𝑔 2) ( �= × 𝐴𝐴𝑝𝑝 𝑖𝑖𝑖𝑖 × � � × × 𝑄𝑄𝑇𝑇 � 𝑚𝑚𝑚𝑚𝑚𝑚 𝑇𝑇 (𝑠𝑠) 12 𝑖𝑖𝑖𝑖 𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓 3

𝑄𝑄𝑇𝑇 �

𝑔𝑔𝑔𝑔𝑔𝑔 19.68 𝑓𝑓𝑓𝑓 𝜋𝜋 1 𝑓𝑓𝑓𝑓 2 60 𝑠𝑠 7.48 𝑔𝑔𝑔𝑔𝑔𝑔 �= × × (3.94 𝑖𝑖𝑖𝑖)2 × � � × × = 93.5 𝑔𝑔𝑔𝑔𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚 8 𝑠𝑠 4 12 𝑖𝑖𝑖𝑖 𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓 3

c. The power is calculated using Equations 10.5 and 10.6. 𝐿𝐿 𝑁𝑁 𝑄𝑄𝑖𝑖𝑖𝑖 � 𝑠𝑠 � × 𝑝𝑝 � 2 � 𝑚𝑚 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) = 1,000,000

5.89 𝐿𝐿 1,134,339 𝑁𝑁 𝑚𝑚3 𝐽𝐽 𝑊𝑊𝑊𝑊 𝑘𝑘𝑘𝑘 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) = × × × × × = 6.68 𝑘𝑘𝑘𝑘 𝑠𝑠 𝑚𝑚2 1,000 𝐿𝐿 𝑁𝑁𝑁𝑁 𝐽𝐽 1,000 𝑊𝑊

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) =

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) =

𝑄𝑄𝑖𝑖𝑖𝑖 (𝑔𝑔𝑔𝑔𝑔𝑔) × 𝑝𝑝(𝑝𝑝𝑝𝑝𝑝𝑝 ) 1714

93.5 𝑔𝑔𝑔𝑔𝑔𝑔 × 164.04 𝑝𝑝𝑝𝑝𝑝𝑝 = 8.95 ℎ𝑝𝑝 1714 222

Problem 10.13.6 Hydraulics Power P10.6 Determine the output power or power stored (in hp) for this system assuming the stroke is 3 feet at a rate of one cycle per 5 seconds, matching the wave cycle. The buoy load force is around 500 lbs. Use the dimensions in Figure 10.3 for calculations. Assume the handle’s total length is 12 feet and the distance from the handle pivot is 2 feet (the short side distance to the anchor). The pump has a 3-inch diameter piston and the load cylinder has a 6-inch diameter piston. The load will be raised 12 inches. Also determine the number of cycles required to lift this load at this height. Each complete cycle consists of two pump strokes (intake and power).

Solution: a. The buoy load F is first calculated: 𝐹𝐹𝑟𝑟𝑟𝑟𝑟𝑟 =

500 𝑙𝑙𝑙𝑙𝑙𝑙 × 12 𝑓𝑓𝑓𝑓 = 3,000 𝑙𝑙𝑙𝑙𝑙𝑙 2 𝑓𝑓𝑓𝑓

b. The pressure created from this action will be: 𝐴𝐴 = 𝑝𝑝 =

𝜋𝜋 × (3 𝑖𝑖𝑖𝑖)2 = 7.0686 𝑖𝑖𝑖𝑖2 4

𝐹𝐹

𝐴𝐴

=

3,000 𝑙𝑙𝑙𝑙𝑙𝑙

7.0686 𝑖𝑖𝑖𝑖 2

= 424 psi

c. The load on the piston cylinder is then calculated: 𝐹𝐹𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 𝑝𝑝𝑝𝑝 = 424

𝑙𝑙𝑙𝑙𝑙𝑙 𝜋𝜋 × (6 𝑖𝑖𝑖𝑖)2 = 12,000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑖𝑖𝑖𝑖2 4

d. The force load balance for each stroke of movement of the wave cycle is given by the following relationship: (𝐴𝐴 × 𝑆𝑆)𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = (𝐴𝐴 × 𝑆𝑆)𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ��

𝜋𝜋(3 𝑖𝑖𝑖𝑖)2 𝜋𝜋(6 𝑖𝑖𝑖𝑖)2 � × (3 𝑓𝑓𝑓𝑓)� × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = �� � × (1 𝑓𝑓𝑓𝑓)� 4 4 223

[(6 𝑖𝑖𝑖𝑖)2 × (1 𝑓𝑓𝑓𝑓)] 36 #𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = = = 1.33 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 [(3 𝑖𝑖𝑖𝑖)2 × (3 𝑓𝑓𝑓𝑓)] 27

e. Hence, the power is simply the product of the distance to raise the load 1 foot and the force associated with this load: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) =

𝐹𝐹𝐹𝐹 12,000 𝑙𝑙𝑙𝑙𝑙𝑙 × 1 𝑓𝑓𝑓𝑓 60 𝑠𝑠 ℎ𝑝𝑝 − 𝑚𝑚𝑚𝑚𝑚𝑚 = × × = 3.28 ℎ𝑝𝑝 𝑡𝑡 1.33 × 5 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 33,000 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙

Problem 10.13.7 Hydraulic Jack Power (Metric)

P10.7 Determine the output power or power stored (in kW) for this system assuming the stroke is 6 meters [19.68 ft] at a rate of one cycle per 8 seconds, matching the wave cycle in the previous example. The buoy load force is around 8,909.1 N [2000 lbs]. Use Figure 10.3 for calculations.

Assume the handle’s total length is 4 m [13.12 ft] and the distance from the handle pivot is 1 m (the short side distance to the anchor or 3.28 ft). The pump has a 10 cm [3.94 in]-diameter piston and the load cylinder has a 20 cm [7.87 in]-diameter piston. The load will be raised 15 m [49.2 ft]. Determine the number of cycles required to lift this load at this height. Each complete cycle consists of two pump strokes (intake and power).

Solution: a. The buoy load F is first calculated as follows: 𝐹𝐹𝑟𝑟𝑟𝑟𝑟𝑟 =

8,909.1 𝑁𝑁 × 4 𝑚𝑚 = 35,636.4 𝑁𝑁 1 𝑚𝑚

b. The pressure created out of this action will be as follows: 𝐴𝐴 = 𝑝𝑝 =

𝜋𝜋 × (10 𝑐𝑐𝑐𝑐)2 = 78.54 𝑐𝑐𝑐𝑐2 4

𝐹𝐹 35,636.4 𝑁𝑁 𝑁𝑁 𝑁𝑁 = = 453.7357 = 4,537,356 2 = 4,537.4 𝑘𝑘𝑘𝑘𝑘𝑘 2 2 𝐴𝐴 78.54 𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐 𝑚𝑚

c. The load on the other piston cylinder is calculated as follows:

224

𝐹𝐹𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 𝑝𝑝𝑝𝑝 = 453.7356

𝑁𝑁 𝜋𝜋 × (20 𝑐𝑐𝑐𝑐)2 = 142,545.6 𝑁𝑁 [142.5 𝑘𝑘𝑘𝑘] 2 𝑐𝑐𝑐𝑐 4

d. The force load balance for each stroke of movement of the wave cycle is given by the following relationship: (𝐴𝐴 × 𝑆𝑆)𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = (𝐴𝐴 × 𝑆𝑆)𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ��

𝜋𝜋(10 𝑐𝑐𝑐𝑐)2 𝜋𝜋(20 𝑐𝑐𝑐𝑐)2 � × (6 𝑚𝑚)� × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = �� � × (15 𝑚𝑚)� 4 4

[(20 𝑐𝑐𝑐𝑐)2 × (15 𝑚𝑚)] 6,000 #𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = = = 10 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 [(10 𝑐𝑐𝑐𝑐)2 × (6 𝑚𝑚)] 600

e. Hence, the power is simply the product of the distance to raise the load 12 inches, and the force associated with this load: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) =

𝐹𝐹𝐹𝐹 142,545.6 𝑁𝑁 × 15 𝑚𝑚 = = 26,727.3 𝑊𝑊 [26.73 𝑘𝑘𝑘𝑘] 𝑡𝑡 10 × 8 𝑠𝑠

Problem 10.13.8 Hydraulic Jack Power (English System)

P10.8 Determine the output power or power stored (in hp) for this system assuming the stroke is 6 meters [19.68 ft] at a rate of one cycle per 8 seconds, matching the wave cycle in the previous example. The buoy load force is around 8,909.1 N [2000 lbs]. Use Figure 10.3 for calculations.

Assume the handle’s total length is 4 m [13.12 ft] and the distance from the handle pivot is 1 m (the short side distance to the anchor or 3.28 ft). The pump has a 10 cm [3.94 in]-diameter piston and the load cylinder has a 20 cm [7.87 in]-diameter piston. The load will be raised 15 m [49.2 ft]. Determine the number of cycles required to lift this load at this height. Each complete cycle consists of two pump strokes (intake and power).

Solution: a. The buoy load F is first calculated: 𝐹𝐹𝑟𝑟𝑟𝑟𝑟𝑟 =

2,000 𝑙𝑙𝑙𝑙𝑙𝑙 × 9.84 𝑓𝑓𝑓𝑓 = 19,676.72 𝑙𝑙𝑙𝑙𝑙𝑙 3.28 𝑓𝑓𝑓𝑓 225

b. The pressure created out of this action will be: 𝐴𝐴 = 𝑝𝑝 =

𝜋𝜋 × (3.94 𝑖𝑖𝑖𝑖)2 = 3.05 𝑖𝑖𝑖𝑖2 4

𝐹𝐹 19,676.72 𝑙𝑙𝑙𝑙𝑙𝑙 = = 98.4 𝑝𝑝𝑝𝑝𝑝𝑝 3.05 𝑖𝑖𝑖𝑖2 𝐴𝐴

c. The load on the piston cylinder is then calculated: 𝐹𝐹𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 𝑝𝑝𝑝𝑝 = 98.4

𝑙𝑙𝑙𝑙𝑙𝑙 𝜋𝜋 × (4.72 𝑖𝑖𝑖𝑖)2 = 1,721.7 𝑙𝑙𝑙𝑙𝑙𝑙 𝑖𝑖𝑖𝑖2 4

d. The force load balance for each stroke of movement of the wave cycle is given by the following relationship: (𝐴𝐴 × 𝑆𝑆)𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = (𝐴𝐴 × 𝑆𝑆)𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ��

𝜋𝜋(3.94 𝑖𝑖𝑖𝑖)2 𝜋𝜋(7.87 𝑖𝑖𝑖𝑖)2 � × (19.68 𝑓𝑓𝑓𝑓)� × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = �� � × (49.2 𝑓𝑓𝑓𝑓)� 4 4

[(7.87 𝑖𝑖𝑖𝑖)2 × (49.2 𝑓𝑓𝑓𝑓)] 3047.3 #𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = = = 9.975 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 [(3.94 𝑖𝑖𝑖𝑖)2 × (19.68 𝑓𝑓𝑓𝑓)] 305.5

e. Hence, the power is simply the product of the distance to raise the load 12 inches, and the force associated with this load: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) =

𝐹𝐹𝐹𝐹 1,721.7 𝑙𝑙𝑙𝑙𝑙𝑙 × 49.2 𝑓𝑓𝑓𝑓 60 𝑠𝑠 ℎ𝑝𝑝 − 𝑚𝑚𝑚𝑚𝑚𝑚 = × × = 1.93 ℎ𝑝𝑝 𝑡𝑡 9.975 × 8 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 33,000 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙

Problem 10.13.9 Power Generated in a Piston per Stroke

P10.9 Determine the power generated in each piston for every complete stroke if the piston diameter is 1 meter and the stroke is 2 meters. The wave motion completes its cycle every 6 seconds (that is, assuming six seconds per stroke). Assume that the pressure was created by a 500-liter volume of water creating about 4900 N of load onto the piston rod, driving the piston in and out due to the wave movement. The pressure created by this force is approximately 6,239 N/m2.

226

Solution: a. This is a simple power equation given in Equation 10.2. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) =

6,239 𝑁𝑁 𝜋𝜋 2 𝑚𝑚 𝑊𝑊𝑊𝑊 × × (1 𝑚𝑚)2 × × = 1,633.4 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 2 𝑚𝑚 4 6 𝑠𝑠 𝑁𝑁𝑁𝑁

b. This is the theoretical power generated by each piston of the given dimensions.

Problem 10.13.10 Significant Wave Height (Hs) P10.10 Problem 10.12. 10 The following data were obtained from wave height measurement (in meters) 7, 9, 9, 9, 8, 6, representing the highest 1/3 of the bigger waves. Estimate the significant wave height Hs. Compare this value with the simple mean wave height.

Solution: a. The simple mean is calculated as follows:

b.

𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻ℎ𝑡𝑡 (𝑚𝑚) =

(7 + 9 + 9 + 9 + 8 + 6) = 8 𝑚𝑚 6

The sum of squares (SS) is then calculated as follows:

𝑆𝑆𝑆𝑆 = (7 − 8)2 +(9 − 8)2 + (9 − 8)2 + (9 − 8)2 + (8 − 8)2 + (6 − 8)2 = 8

c. The degrees of freedom (DF) is given as follows: 𝐷𝐷𝐷𝐷 = (𝑛𝑛 − 1) = 6 − 1 = 5

d. The standard deviation (SD) of the sample is given by the following equation and is calculated as follows: 𝑆𝑆𝑆𝑆 8 𝑆𝑆𝑆𝑆 = � =� = 1.26 (𝑛𝑛 − 1) (6 − 1)

e. Hence, the variance, which is simply the square of the standard deviation (SD) will be calculated as follows: (𝑆𝑆𝑆𝑆)2 = (1.26)2 = 1.6 227

f. Thus, the significant wave height is calculated following Equation 10.11 and as shown below. 𝐻𝐻𝑠𝑠 = 4 × √1.6 = 5 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

g. The value of significant wave height is less than the simple means of the data.

Problem 10.13.11 Capital Cost of Wave Converters P10.11 The capital cost for a wave converter is around $1,000/kW. In an area where the average wave height is 6 meters for a period of 7 seconds, how much is the capital cost for every km of wave distance?

Solution: a. The power is first calculated: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 � = �0.5 3 � × (6 𝑚𝑚)2 × (7 𝑠𝑠) = 126 𝑚𝑚 𝑚𝑚 𝑠𝑠 𝑚𝑚

b. The power per km is then calculated: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) = 126

𝑘𝑘𝑘𝑘 1,000 𝑚𝑚 × = 126,000 𝑘𝑘𝑘𝑘 𝑚𝑚 𝑘𝑘𝑘𝑘

c. Hence, the total cost will be: 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 ($) = 126,000 𝑘𝑘𝑘𝑘 ×

$1,000 = $126.6 𝑀𝑀 𝑘𝑘𝑘𝑘

d. Over $126.6 million is needed for this project.

10.13.12

Power Output of Simulated Wave Converters

P10.12 Determine the output power or power stored (in units of kW) for the system shown in the figure below, assuming the stroke is 3 feet at a rate of one cycle per 5 seconds, matching the wave cycle. The buoy load force is around 500 lbs. Assume the handle total length is 12 feet and the distance from the handle pivot is 2 feet (the short side distance to anchor). The pump has 3-

228

inch diameter piston, and the load cylinder has a 6-inch diameter piston. The load will be raised 12 inches. Each complete cycle consists of two pump strokes (intake and power).

` a. 91.8 b. 30.6 c. 2.45 d. 122.4 Solutions: a. The buoy load F is first calculated as follows: 𝐹𝐹𝑟𝑟𝑟𝑟𝑟𝑟 =

𝐹𝐹𝐴𝐴 × 𝑙𝑙 500 𝑙𝑙𝑙𝑙𝑙𝑙 × 12 𝑓𝑓𝑓𝑓 = = 3000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥 2 𝑓𝑓𝑓𝑓

b. The pressure created out of this action will be as follows: 𝐹𝐹 (𝑙𝑙𝑙𝑙𝑙𝑙)

3,000 𝑙𝑙𝑙𝑙𝑙𝑙

𝑝𝑝(𝑝𝑝𝑝𝑝𝑝𝑝 ) = 𝐴𝐴 𝑏𝑏(𝑖𝑖𝑖𝑖2 ) = 𝜋𝜋 (3 𝑖𝑖𝑖𝑖)2 = 424.4 𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝

4

c. The load on the piston cylinder is then calculated as follows: 𝐹𝐹𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 𝑝𝑝𝑝𝑝 =

424.4 𝑙𝑙𝑙𝑙𝑙𝑙 𝜋𝜋 × (6 𝑖𝑖𝑖𝑖)2 = 12,000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑖𝑖𝑖𝑖2 4 229

d. The force load balance for each stroke of movement of the wave cycle is given by the following relationship (please see example in lecture notes) (𝐴𝐴 × 𝑆𝑆)𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = (𝐴𝐴 × 𝑆𝑆)𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝜋𝜋(3 𝑖𝑖𝑖𝑖)2 𝜋𝜋(6 𝑖𝑖𝑖𝑖)2 �� � × (36 𝑖𝑖𝑖𝑖)� × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = �� � × (12 𝑖𝑖𝑖𝑖)� 4 4 𝜋𝜋(6 𝑖𝑖𝑖𝑖)2 � × (12 𝑖𝑖𝑖𝑖)� 339.3 4 #𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = = = 1.33 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝜋𝜋(3 𝑖𝑖𝑖𝑖)2 254.5 �� � × (36 𝑖𝑖𝑖𝑖)� 4 ��

e. Hence, power is simply the product of distance to raise the load 1 foot, and the force associated with this load as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 =

𝐹𝐹𝐹𝐹 𝑡𝑡

𝐹𝐹𝐹𝐹 12,000 𝑙𝑙𝑙𝑙𝑙𝑙 × 1 𝑓𝑓𝑓𝑓 60 𝑠𝑠 ℎ𝑝𝑝 − 𝑚𝑚𝑚𝑚𝑚𝑚 = × × = 3.28 ℎ𝑝𝑝 𝑡𝑡 1.33 × 5 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 33,000 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 3.28 ℎ𝑝𝑝 ×

10.13.13

0.746 𝑘𝑘𝑘𝑘 = 2.45 𝑘𝑘𝑘𝑘 ℎ𝑝𝑝

Hydraulic System Equivalents

P10.13 Determine the force on the rod acting on the hydraulic cylinder (in units of lbs) for the system shown in the figure below, assuming the stroke is 3 feet at a rate of one cycle per 5 seconds, matching the wave cycle. The buoy load force is around 500 lbs. Assume the handle total length is 12 feet and the distance from the handle pivot is 2 feet (the short side distance to anchor). The pump has a 3-inch diameter piston, and the load cylinder has a 6-inch diameter piston. The load will be raised 12 inches. Each complete cycle consists of two pump strokes (intake and power).

230

a. 1000 b. 2000 c. 6000 d. 3000 Solutions: a. The buoy load F is first calculated as follows: 𝐹𝐹𝑟𝑟𝑟𝑟𝑟𝑟 =

10.13.14

𝐹𝐹𝐴𝐴 × 𝑙𝑙 500 𝑙𝑙𝑙𝑙𝑙𝑙 × 12 𝑓𝑓𝑓𝑓 = = 3000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥 2 𝑓𝑓𝑓𝑓

Use of Hydraulic Systems Equivalents in Wave Power

P10.14 Determine the pressure created by a wave because of the piston rod (left piston) acting on the hydraulic cylinder (in units of psi) for the system shown in the figure below, assuming the stroke is 3 feet at a rate of one cycle per 5 seconds, matching the wave cycle. The buoy load force is around 500 lbs. Assume the handle total length is 12 feet and the distance from the 231

handle pivot is 2 feet (the short side distance to anchor). The pump has a 3-inch diameter piston and the load cylinder has a 6-inch diameter piston. The load will be raised 12 inches. Each complete cycle consists of two pump strokes (intake and power).

a. 424.4 b. 212.2 c. 848.8 d. 3000 Solutions: a. The buoy load F is first calculated as follows: 𝐹𝐹𝑟𝑟𝑟𝑟𝑟𝑟 =

𝐹𝐹𝐴𝐴 × 𝑙𝑙 500 𝑙𝑙𝑙𝑙𝑙𝑙 × 12 𝑓𝑓𝑓𝑓 = = 3000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥 2 𝑓𝑓𝑓𝑓

b. The pressure created out of this action will be as follows: 232

𝐹𝐹 (𝑙𝑙𝑙𝑙𝑙𝑙)

3,000 𝑙𝑙𝑙𝑙𝑙𝑙

𝑝𝑝(𝑝𝑝𝑝𝑝𝑝𝑝 ) = 𝐴𝐴 𝑏𝑏(𝑖𝑖𝑖𝑖2 ) = 𝜋𝜋 (3 𝑖𝑖𝑖𝑖)2 = 424.4 𝑝𝑝𝑝𝑝𝑝𝑝

10.13.15

𝑝𝑝

4

Hydraulic System Application in Wave Power Converters

P10.15 Determine the force created by a wave because of the pressure created at the right piston acting on the hydraulic cylinder (in units of lbs) for the system shown in the figure below, assuming the stroke is 3 feet at a rate of one cycle per 5 seconds, matching the wave cycle. The buoy load force is around 500 lbs. Assume the handle total length is 12 feet and the distance from the handle pivot is 2 feet (the short side distance to anchor). The pump has 3-inch diameter piston, and the load cylinder has a 6-inch diameter piston. The load will be raised 12 inches.

Each complete cycle consists of two pump strokes (intake and power).

a. 3000 b. 12000 c. 6000

233

d. 9000 Solutions: a. The buoy load F is first calculated as follows: 𝐹𝐹𝑟𝑟𝑟𝑟𝑟𝑟 =

𝐹𝐹𝐴𝐴 × 𝑙𝑙 500 𝑙𝑙𝑙𝑙𝑙𝑙 × 12 𝑓𝑓𝑓𝑓 = = 3000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥 2 𝑓𝑓𝑓𝑓

b. The pressure created out of this action will be as follows: 𝐹𝐹 (𝑙𝑙𝑙𝑙𝑙𝑙)

3,000 𝑙𝑙𝑙𝑙𝑙𝑙

𝑝𝑝(𝑝𝑝𝑝𝑝𝑝𝑝 ) = 𝐴𝐴 𝑏𝑏(𝑖𝑖𝑖𝑖2 ) = 𝜋𝜋 (3 𝑖𝑖𝑖𝑖)2 = 424.4 𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝

4

c. The load on the piston cylinder is then calculated as follows: 𝐹𝐹𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 𝑝𝑝𝑝𝑝 =

10.13.16

424.4 𝑙𝑙𝑙𝑙𝑙𝑙 𝜋𝜋 × (6 𝑖𝑖𝑖𝑖)2 = 12,000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑖𝑖𝑖𝑖2 4

Use of Hydraulic Systems in Wave Converters

P10.16 Determine the number of cycles for the system shown in the figure below, assuming the stroke is 3 feet at a rate of one cycle per 5 seconds, matching the wave cycle. The buoy load force is around 500 lbs. Assume the handle total length is 12 feet and the distance from the handle pivot is 2 feet (the short side distance to anchor). The pump has 3-inch diameter piston, and the load cylinder has a 6-inch diameter piston. The load will be raised 12 inches. Each complete cycle consists of two pump strokes (intake and power).

234

a. 3.5 b. 2.5 c. 1.33 d. 2.66 Solutions: a. The bouy load F is first calculated as follows: 𝐹𝐹𝑟𝑟𝑟𝑟𝑟𝑟 =

𝐹𝐹𝐴𝐴 × 𝑙𝑙 500 𝑙𝑙𝑙𝑙𝑙𝑙 × 12 𝑓𝑓𝑓𝑓 = = 3000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥 2 𝑓𝑓𝑓𝑓

b. The pressure created out of this action will be as follows: 𝐹𝐹 (𝑙𝑙𝑙𝑙𝑙𝑙)

3,000 𝑙𝑙𝑙𝑙𝑙𝑙

𝑝𝑝(𝑝𝑝𝑝𝑝𝑝𝑝 ) = 𝐴𝐴 𝑏𝑏(𝑖𝑖𝑖𝑖2 ) = 𝜋𝜋 (3 𝑖𝑖𝑖𝑖)2 = 424.4 𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝

4

c. The load on the piston cylinder is then calculated as follows: 𝐹𝐹𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 𝑝𝑝𝑝𝑝 =

424.4 𝑙𝑙𝑙𝑙𝑙𝑙 𝜋𝜋 × (6 𝑖𝑖𝑖𝑖)2 = 12,000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑖𝑖𝑖𝑖2 4 235

d. The force load balance for each stroke of movement of the wave cycle is given by the following relationship (please see example in lecture notes) (𝐴𝐴 × 𝑆𝑆)𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = (𝐴𝐴 × 𝑆𝑆)𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝜋𝜋(3 𝑖𝑖𝑖𝑖)2 𝜋𝜋(6 𝑖𝑖𝑖𝑖)2 �� � × (36 𝑖𝑖𝑖𝑖)� × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = �� � × (12 𝑖𝑖𝑖𝑖)� 4 4 𝜋𝜋(6 𝑖𝑖𝑖𝑖)2 � × (12 𝑖𝑖𝑖𝑖)� 339.3 4 #𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = = = 1.33 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝜋𝜋(3 𝑖𝑖𝑖𝑖)2 254.5 �� � × (36 𝑖𝑖𝑖𝑖)� 4 ��

10.13.17

Stored Energy in Hydraulic Systems-Based Wave Converters

P10.17 Determine the output power or power stored (in units of HP) for the system shown in the figure below, assuming the stroke is 3 feet at a rate of one cycle per 5 seconds, matching the wave cycle. The buoy load force is around 500 lbs. Assume the handle total length is 12 feet and the distance from the handle pivot is 2 feet (the short side distance to anchor). The pump has 3inch diameter piston and the load cylinder has a 6-inch diameter piston. The load will be raised 12 inches. Each complete cycle consists of two pump strokes (intake and power).

236

a. 91.8 b. 30.6 c. 61.2 d. 3.28 Solutions: a. The bouy load F is first calculated as follows: 𝐹𝐹𝑟𝑟𝑟𝑟𝑟𝑟 =

𝐹𝐹𝐴𝐴 × 𝑙𝑙 500 𝑙𝑙𝑙𝑙𝑙𝑙 × 12 𝑓𝑓𝑓𝑓 = = 3000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥 2 𝑓𝑓𝑓𝑓

b. The pressure created out of this action will be as follows: 𝐹𝐹 (𝑙𝑙𝑙𝑙𝑙𝑙)

3,000 𝑙𝑙𝑙𝑙𝑙𝑙

𝑝𝑝(𝑝𝑝𝑝𝑝𝑝𝑝 ) = 𝐴𝐴 𝑏𝑏(𝑖𝑖𝑖𝑖2 ) = 𝜋𝜋 (3 𝑖𝑖𝑖𝑖)2 = 424.4 𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝

4

c. The load on the piston cylinder is then calculated as follows: 𝐹𝐹𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 𝑝𝑝𝑝𝑝 =

424.4 𝑙𝑙𝑙𝑙𝑙𝑙 𝜋𝜋 × (6 𝑖𝑖𝑖𝑖)2 = 12,000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑖𝑖𝑖𝑖2 4

237

d. The force load balance for each stroke of movement of the wave cycle is given by the following relationship (please see example in lecture notes) (𝐴𝐴 × 𝑆𝑆)𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = (𝐴𝐴 × 𝑆𝑆)𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝜋𝜋(3 𝑖𝑖𝑖𝑖)2 𝜋𝜋(6 𝑖𝑖𝑖𝑖)2 �� � × (36 𝑖𝑖𝑖𝑖)� × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = �� � × (12 𝑖𝑖𝑖𝑖)� 4 4 𝜋𝜋(6 𝑖𝑖𝑖𝑖)2 � × (12 𝑖𝑖𝑖𝑖)� 339.3 4 #𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = = = 1.33 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝜋𝜋(3 𝑖𝑖𝑖𝑖)2 254.5 �� � × (36 𝑖𝑖𝑖𝑖)� 4 ��

e. Hence, power is simply the product of distance to raise the load 1 foot, and the force associated with this load as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 =

𝐹𝐹𝐹𝐹 𝑡𝑡

𝐹𝐹𝐹𝐹 12,000 𝑙𝑙𝑙𝑙𝑙𝑙 × 1 𝑓𝑓𝑓𝑓 60 𝑠𝑠 ℎ𝑝𝑝 − 𝑚𝑚𝑚𝑚𝑚𝑚 = × × = 3.28 ℎ𝑝𝑝 𝑡𝑡 1.33 × 5 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 33,000 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙

Assess Your Knowledge 1. The ballpark figure for the initial capital cost of wave energy converted is in what range in units of $/kW?

a. $500 b. $1,000 c. $5,000 d. $10,000 2. Which among the renewables should work well with wave energy converters?

a. Solar collectors b. Wind energy extractors.

238

c. Hydro power d. Biomass Energy 3. Hydraulic Power Calculations. Determine the power (in units of HP) generated from a wave energy converter that has the following data: rotational speed = 400 rpm; volumetric displacement = 6 in3/rev; pressure developed = 100 psi.

a. 1.2 b. 0.61 c. 1.8 d. 2.4 Solution: a. First calculate the volume of flow

𝑄𝑄𝑇𝑇 (𝑔𝑔𝑔𝑔𝑔𝑔) = 𝑄𝑄𝑇𝑇 (𝑔𝑔𝑔𝑔𝑔𝑔) =

𝑖𝑖𝑖𝑖3 𝑟𝑟𝑟𝑟𝑟𝑟 𝑉𝑉𝐷𝐷 �𝑟𝑟𝑟𝑟𝑟𝑟� × 𝑁𝑁 �𝑚𝑚𝑚𝑚𝑚𝑚� 231

𝑖𝑖𝑖𝑖3 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

𝑖𝑖𝑖𝑖3 𝑟𝑟𝑟𝑟𝑟𝑟 6 �𝑟𝑟𝑟𝑟𝑟𝑟� × 4 �𝑚𝑚𝑚𝑚𝑚𝑚� 𝑖𝑖𝑖𝑖3 231 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

= 10.4 𝑔𝑔𝑔𝑔𝑔𝑔

b. Then calculate the power 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃(ℎ𝑝𝑝) = 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃(ℎ𝑝𝑝) =

𝑄𝑄 (𝑔𝑔𝑔𝑔𝑔𝑔) × 𝑝𝑝(𝑝𝑝𝑝𝑝𝑝𝑝 ) 1714

10.4 (𝑔𝑔𝑔𝑔𝑔𝑔) × 100 (𝑝𝑝𝑝𝑝𝑝𝑝) = 0.61 ℎ𝑝𝑝 1714

4. Which type of wave energy converter concept can be placed only on the surface of the wave?

a. Oscillating wave surge converter b. Submerged pressure differential c. Oscillating water column 239

d. Attenuator 5. What do you call the machine that is used to capture energy from waves?

a. Hydro power b. Hydraulic power c. Hydraulic ram d. Wave energy converters or WECs. 6. Throughout the year, what season is wave energy potential the highest?

a. Winter b. Summer c. Fall d. Spring 7. What is the estimated worldwide resource of wave energy in units of TW?

a. 50 b. 5 c. 2 d. 100 8. Given a wave height of 5 meters and a wave period of 7 seconds, estimate the power for every mile (1 mile = 1.609 km or 1,609 meters) of crest head in units of MW.

a. 140.8 b. 87.5 c. 54.4 d. 140,788 Solution: a. Use Equation 10.1 formula and simply substitute the values as follows: 240

𝑘𝑘𝑘𝑘 𝜌𝜌𝑔𝑔2 2 𝑘𝑘𝑘𝑘 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 � �= 𝐻𝐻 𝑇𝑇 ≈ �0.50 3 � 𝐻𝐻2 𝑇𝑇 𝑚𝑚 64𝜋𝜋 𝑚𝑚 𝑠𝑠 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑘𝑘𝑘𝑘 𝑚𝑚

� = �0.50

𝑘𝑘𝑘𝑘

𝑚𝑚 3𝑠𝑠

� × 52 × 7 𝑠𝑠 = 87.5

b. The power per mile is as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃(𝑘𝑘𝑘𝑘) = 87.5

𝑘𝑘𝑘𝑘 𝑚𝑚

𝑘𝑘𝑘𝑘 × 1,609 𝑚𝑚 = 140,788 𝑘𝑘𝑘𝑘 = 140.8 𝑀𝑀𝑀𝑀 𝑚𝑚

9. Determine the significant wave height (Hs) if the variance of water level is 8.64 meters.

a. 11.8 b. 8.64 c. 2.93 d. 5.88 Solution: a. Simply use the formula for Hs as follows: 𝐻𝐻𝑠𝑠 = 4 × √8.64 = 11.8 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 [38.6 𝑓𝑓𝑓𝑓]

10. In major storms, the wave height could be 10 meters for a period of 10 seconds. How much will the total power (in MW) be available if a wave is 50 meters long?

a. 75 b. 50 c. 12.5 d. 25 Solution: a. Use Equation 10.1 formula and simply substitute the values as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑘𝑘𝑘𝑘 𝜌𝜌𝑔𝑔2 2 𝑘𝑘𝑘𝑘 �= 𝐻𝐻 𝑇𝑇 ≈ �0.50 3 � 𝐻𝐻2 𝑇𝑇 𝑚𝑚 64𝜋𝜋 𝑚𝑚 𝑠𝑠

𝑘𝑘𝑘𝑘

𝑘𝑘𝑘𝑘

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 � 𝑚𝑚 � = �0.50 𝑚𝑚3𝑠𝑠� × 102 × 10 𝑠𝑠 × 50 𝑚𝑚 = 25,000 𝑘𝑘𝑘𝑘 = 25 𝑀𝑀𝑀𝑀 241

11. Wave energy is ultimately derived from what renewable energy resource?

a. Wind b. Geothermal c. Solar d. Hydro 12. Determine the total power (in MW) generated for a wave height of 5 meters and a period of 5 seconds and capture width (or wave distance) of a kilometer.

a. 12,500 b. 2.50 c. 62.5 d. 31.3 Solution: a. The same formula is used as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃(𝑘𝑘𝑘𝑘 ) = �0.50

𝑘𝑘𝑘𝑘 � × 52 × 5 × 1,000 𝑚𝑚 = 62,500 𝑘𝑘𝑘𝑘 = 62.5 𝑀𝑀𝑀𝑀 𝑚𝑚 3 𝑠𝑠

a. Hence the power is about 62.5 MW

13. Determine the power (in kW) generated by a wave with a height of 2.5 meters and a time period of 5 seconds. The length of the wave is around 30 meters.

a. 234 b. 469 c. 469,000 d. 234,000 Solution: a. The simple formula is used as follows: 242

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃(𝑘𝑘𝑘𝑘 ) = �0.50

𝑘𝑘𝑘𝑘 � × 2.52 × 5 × 30 𝑚𝑚 = 469 𝑘𝑘𝑘𝑘 𝑚𝑚 3 𝑠𝑠

14. The Pelamis Waver Energy Converter is what type of WECS?

a. Attenuator b. Oscillating wave surge converter c. Submerged pressure differential d. Oscillating water column 15. Name the three different wind waves developed over time.

a. Capillary, gravity, and swells b. Calm, viscous, and turbulent c. Low, medium, high d. Wave shear, wave friction, and wave swell 16. The following sample data were measured for wave heights: 5, 6, 7, 8, 9, 8, 7, 6, 5 meters. Determine the value of the significant wave height Hs (standard formula, in meters) and not for regular waves.

a. 8.64 b. 4.67 c. 6.32 d. 5.6 Solution: a. The mean is calculated as follows: 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻ℎ𝑡𝑡 (𝑚𝑚) =

(5 + 6 + 7 + 8 + 9 + 8 + 7 + 6 + 5 ) 𝑚𝑚 61 = = 4.78 𝑚𝑚 9 9

b. The sum of squares is calculated as follows:

𝑆𝑆𝑆𝑆 = (5 − 4.78)2 + (6 − 4.78)2 + (7 − 4.78)2 + (8 − 4.78)2 + (9 − 4.78)2

+ (8 − 4.78)2 + (7 − 4.78)2 + (6 − 4.78)2 + (5 − 4.78)2 = 15.6 243

c. The degrees of freedom for sample is (𝑛𝑛 − 1) = 9 − 1 = 8.

d. The standard deviation (of the sample) is calculated as follows: 𝑆𝑆𝑆𝑆 15.6 𝑆𝑆𝑆𝑆 = � =� = 1.4 (𝑛𝑛 − 1) (9 − 1)

e. The variance will be as follows (just the square of SD)

𝑆𝑆𝑆𝑆2 = 1.42 = 1.96

f. Hence the significant wave height will be calculated as follows: 𝐻𝐻𝑠𝑠 = 4 × √1.96 = 5.6 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 [18.4 𝑓𝑓𝑓𝑓]

17. Determine the power (in kW/m) generated for a wave height of 2.5 meters and a period of 5 seconds.

a. 15.6 b. 54 c. 27 d. 81 Solution: a. The simple formula is used as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑘𝑘𝑘𝑘 𝜌𝜌𝑔𝑔2 2 𝑘𝑘𝑘𝑘 �= 𝐻𝐻 𝑇𝑇 ≈ �0.50 3 � 𝐻𝐻2 𝑇𝑇 𝑚𝑚 64𝜋𝜋 𝑚𝑚 𝑠𝑠

𝑘𝑘𝑘𝑘

𝑘𝑘𝑘𝑘

𝑘𝑘𝑘𝑘

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 � 𝑚𝑚 � = �0.50 𝑚𝑚3𝑠𝑠� × 2.52 × 5 = 15.6 𝑚𝑚

b. Hence the power is about 15.6 kW/m.

18. Determine the pressure generated for each stroke (in units of psi) if the buoy load to the pump cylinder is 8,909.1 N [2,000 lbs] and a wave height of 6 meters [19.68 ft], which is also considered the stroke of the piston. The wave period is 8 seconds. Assume the diameter of the small piston cylinder is 10 cm [3.94 inch]. Please refer to the figure shown below for this arrangement. a. 41 244

b. 164 c. 82 d. 328

Solutions: a. Using the Pascal Equation (p=F/A), the pressure generated is as follows for English system of units: 𝑝𝑝(𝑝𝑝𝑝𝑝𝑝𝑝 ) =

𝐹𝐹𝑏𝑏 (𝑙𝑙𝑙𝑙𝑙𝑙) 2,000 𝑙𝑙𝑙𝑙𝑙𝑙 = 𝜋𝜋 = 164 𝑝𝑝𝑝𝑝𝑝𝑝 2 𝐴𝐴𝑝𝑝 (𝑖𝑖𝑖𝑖 ) (3.94 𝑖𝑖𝑖𝑖)2 4

19. Determine the force created by a wave because of the pressure created at the right piston acting on the hydraulic cylinder (in units of lbs) for the system shown in the figure below, assuming the stroke is 3 feet at a rate of one cycle per 5 seconds, matching the wave cycle. The buoy load force is around 500 lbs. Assume the handle total length is 12 feet and the distance from the handle pivot is 2 feet (the short side distance to anchor). The pump has a 3-inch diameter piston, and the load cylinder has a 6inch diameter piston. The load will be raised 12 inches. Each complete cycle consists of two pump strokes (intake and power).

a. 3000 b. 12000 c. 6000 d. 9000 Solution: a. The buoy load F is first calculated: 𝐹𝐹𝑟𝑟𝑟𝑟𝑟𝑟 =

500 𝑙𝑙𝑙𝑙𝑙𝑙 × 12 𝑓𝑓𝑓𝑓 = 3,000 𝑙𝑙𝑙𝑙𝑙𝑙 2 𝑓𝑓𝑓𝑓

b. The pressure created from this action will be:

245

𝐴𝐴 = 𝑝𝑝 =

𝜋𝜋 × (3 𝑖𝑖𝑖𝑖)2 = 7.0686 𝑖𝑖𝑖𝑖2 4

𝐹𝐹

𝐴𝐴

=

3,000 𝑙𝑙𝑙𝑙𝑙𝑙

7.0686 𝑖𝑖𝑖𝑖 2

= 424 psi

c. The load on the piston cylinder is then calculated: 𝐹𝐹𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 𝑝𝑝𝑝𝑝 = 424

𝑙𝑙𝑙𝑙𝑙𝑙 𝜋𝜋 × (6 𝑖𝑖𝑖𝑖)2 = 12,000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑖𝑖𝑖𝑖2 4

20. Determine the output power or power stored (in units of kW) for the system shown in the figure below, assuming the stroke is 3 feet at a rate of one cycle per 5 seconds, matching the wave cycle.

The buoy load force is around 500 lbs. Assume the handle total length is 12 feet and the distance from the handle pivot is 2 feet (the short side distance to anchor). The pump has a 3-inch diameter piston and the load cylinder has a 6-inch diameter piston. The load will be raised 12 inches. Each complete cycle consists of two pump strokes (intake and power).

a. 91.8 b. 30.6 c. 61.2 d. 122.4 Solutions to Problems 15-20: a. The bouy load F is first calculated as follows: 𝐹𝐹𝑟𝑟𝑟𝑟𝑟𝑟 =

𝐹𝐹𝐴𝐴 × 𝑙𝑙 500 𝑙𝑙𝑙𝑙𝑙𝑙 × 12 𝑓𝑓𝑓𝑓 = = 3000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥 2 𝑓𝑓𝑓𝑓

b. The pressure created out of this action will be as follows: 𝐹𝐹 (𝑙𝑙𝑙𝑙𝑙𝑙)

3,000 𝑙𝑙𝑙𝑙𝑙𝑙

𝑝𝑝(𝑝𝑝𝑝𝑝𝑝𝑝 ) = 𝐴𝐴 𝑏𝑏(𝑖𝑖𝑖𝑖2 ) = 𝜋𝜋 (3 𝑖𝑖𝑖𝑖)2 = 424.4 𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝

4

c. The load on the piston cylinder is then calculated as follows: 𝐹𝐹𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 𝑝𝑝𝑝𝑝 =

424.4 𝑙𝑙𝑙𝑙𝑙𝑙 𝜋𝜋 × (6 𝑖𝑖𝑖𝑖)2 = 12,000 𝑙𝑙𝑙𝑙𝑙𝑙 4 𝑖𝑖𝑖𝑖2 246

d. The force load balance for each stroke of movement of the wave cycle is given by the following relationship (please see example in lecture notes) (𝐴𝐴 × 𝑆𝑆)𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = (𝐴𝐴 × 𝑆𝑆)𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝜋𝜋(3 𝑖𝑖𝑖𝑖)2 𝜋𝜋(6 𝑖𝑖𝑖𝑖)2 �� � × (36 𝑖𝑖𝑖𝑖)� × (#𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = �� � × (12 𝑖𝑖𝑖𝑖)� 4 4 𝜋𝜋(6 𝑖𝑖𝑖𝑖)2 � × (12 𝑖𝑖𝑖𝑖)� 339.3 4 #𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = = = 1.33 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝜋𝜋(3 𝑖𝑖𝑖𝑖)2 254.5 �� � × (36 𝑖𝑖𝑖𝑖)� 4 ��

e. Hence, the power is simply the product of distance to raise the load 1 foot and the force associated with this load as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 =

𝐹𝐹𝐹𝐹 𝑡𝑡

𝐹𝐹𝐹𝐹 12,000 𝑙𝑙𝑙𝑙𝑙𝑙 × 12 𝑖𝑖𝑖𝑖 60 𝑠𝑠 ℎ𝑝𝑝 − 𝑚𝑚𝑚𝑚𝑚𝑚 = × × = 82 ℎ𝑝𝑝 𝑡𝑡 1.33/5 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 33,000 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 82 ℎ𝑝𝑝 ×

0.746 𝑘𝑘𝑘𝑘 = 61.2 𝑘𝑘𝑘𝑘 ℎ𝑝𝑝

Figure used for Questions 18-20.

247

Chapter 11 Ocean Thermal Energy Systems 11.11 Solved Problems Problem 11.11.1. Heat Capacity of the Ocean P11.1 Determine the amount of energy absorbed by a hectare of seawater surface water with a depth of 5 meters. Assume the original temperature of 5oC and the final temperature of 25oC. Assume the specific heat of sea water to be 3.93 kJ/kgoC. If the overall conversion efficiency is 3%, estimate the actual power. If the cycle happens in 24 hours, how much power is produced?

Solution: a. The energy absorbed is simply calculated using the sensible heat energy equation: 𝑄𝑄(𝑀𝑀𝑀𝑀) = 𝑚𝑚𝐶𝐶𝑝𝑝 ∆𝑡𝑡

𝑄𝑄(𝑀𝑀𝑀𝑀) = 10,000 𝑚𝑚2 × 5 𝑚𝑚 ×

1,000 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑀𝑀𝑀𝑀 𝑜𝑜 𝑜𝑜 ) ( × 3.93 × 25 𝐶𝐶 − 5 𝐶𝐶 × 𝑚𝑚3 𝑘𝑘𝑔𝑔𝑜𝑜 𝐶𝐶 1,000 𝑘𝑘𝑘𝑘

= 3,930,000 𝑀𝑀𝑀𝑀

Thus, 3.93 million MJ of energy is absorbed by this body of water. b. If the overall conversion efficiency is 3%, the actual energy produced is given as follows: 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) = 𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 × 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) = 3,930,000 𝑀𝑀𝑀𝑀 × 0.03 = 117,900 𝑀𝑀𝑀𝑀

c. The power is calculated from actual energy per unit of time: 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑀𝑀𝑀𝑀) =

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) 117,900 𝑀𝑀𝑀𝑀 1 ℎ𝑟𝑟 𝑀𝑀𝑀𝑀 − 𝑠𝑠𝑠𝑠𝑠𝑠 = × × 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (𝑠𝑠𝑠𝑠𝑠𝑠 ) 24 ℎ𝑟𝑟𝑟𝑟 3600 𝑠𝑠𝑠𝑠𝑠𝑠 𝑀𝑀𝑀𝑀

= 1.4 𝑀𝑀𝑀𝑀

248

Problem 11.11.2 Ideal Carnot Cycle Efficiency P11.2 Determine the ideal Carnot Cycle efficiency for an ocean with a warm temperature of 25oC and a cold temperature of 5oC. Solution: a. Convert the temperature units into absolute scale as follows: Th = 25 + 273.15 = 298.15oK and Tc = 5 + 273.15 = 278.15oK. b. Using Equation 11.2 directly, we have the following: 𝜂𝜂𝐶𝐶 (%) = 𝜂𝜂𝐶𝐶 (%) =

𝑇𝑇ℎ − 𝑇𝑇𝑐𝑐 × 100% 𝑇𝑇ℎ

298.15 − 278.15 × 100% = 6.71% 298.15

c. The ideal Carnot Cycle efficiency is only 6.71% Problem 11.11.3 Volume of Water Needed for a 100 kW of Power.

P11.3 Determine the volume of ocean water needed to generate a 100 kW of electrical power assuming a 15oC temperature difference, and power produced continuously for 24 hrs a day and 365 days in a year. Assume an overall conversion efficiency of 5%. Use the simple sensible heat equation for the calculation. Compare this volume of water to the capacity of an Olympic-size swimming pool that measures 50 meters in length, 25 meters in width, with a depth of 3 meters.

How many Olympic-sized swimming pools are needed to generate a 100 kW of electrical power in a day? Assume the density of sea water to be 1,027 kg/m3.

Solution: a. The efficiency equation is first used to calculate the required energy input using the overall conversion efficiency. Hence, if 100 kW of output power is needed, a 5% conversion efficiency shows that 2,000 kW of input power, as shown below.

249

𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 100 𝑘𝑘𝑘𝑘 = = 2,000 𝑘𝑘𝑘𝑘 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑) 0.05

b. This energy must come from the thermal energy from the ocean, assuming the specific heat of ocean water to be 3.93 kJ/kgoC and a 15oC difference. 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 � =

𝑘𝑘𝑘𝑘 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �= 𝑠𝑠 𝐶𝐶𝑝𝑝 × ∆𝑡𝑡 (𝐶𝐶 )

2,000 𝑘𝑘𝑘𝑘 1,000 𝑊𝑊 𝐽𝐽 1 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 × × × = 33.9 𝑘𝑘𝑘𝑘 1 𝑘𝑘𝑘𝑘 𝑊𝑊𝑊𝑊 1,000 𝐽𝐽 𝑠𝑠 3.39 𝑘𝑘𝑔𝑔𝑜𝑜 𝐶𝐶 × 15𝑜𝑜 𝐶𝐶

c. The volume and weight of seawater that is contained in an Olympic-size swimming pool is calculated below. 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 (𝑚𝑚3 ) = 50 𝑚𝑚 × 25 𝑚𝑚 × 3 𝑚𝑚 = 3,750 𝑚𝑚3 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊ℎ𝑡𝑡 (𝑘𝑘𝑘𝑘) = 3,750 𝑚𝑚3 ×

1,027 𝑘𝑘𝑘𝑘 = 3,851,250 𝑘𝑘𝑘𝑘 𝑚𝑚3

d. If one were to produce 100 kW of electrical power using 33.9 kg/s of water, one Olympic-size swimming pool will be emptied in 31 hrs. In terms of how many swimming pools are needed to generate this power in a day, it would take just the volume of one.

The calculation is shown below. 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = 3,851,250 𝑘𝑘𝑘𝑘 ×

Problem 11.11.4 Pumping Power.

𝑠𝑠 ℎ𝑟𝑟 × = 31.56 ℎ𝑟𝑟𝑟𝑟 33.9 𝑘𝑘𝑘𝑘 3,600 𝑠𝑠

P11.4 Determine the amount of theoretical power and actual power used to move 50 kg/s of water through 100 meters of head. Assume an overall pumping efficiency of 75%. Use the basic hydro power equation from Chapter 5 for calculations.

Solution:

250

a. If the water is flowing down, the power for this event is calculated as follows, which we call the theoretical power: 𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄 50 𝑠𝑠 × 100 𝑚𝑚 𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) = = = 49 𝑘𝑘𝑘𝑘 102 102

b. When moving this water uphill, the required power should be more and we call this the actual power and is calculated as follows: 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) =

𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘 ) 49 𝑘𝑘𝑘𝑘 = = 65𝑘𝑘𝑘𝑘 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ) 0.75

c. Note that it will take more power to bring the water from the lower elevation to the higher elevation. The efficiency equation (in reverse) can be used to find a higher actual power than the lower theoretical power to accomplish the work.

Problem 11.11.5 Base Load Power P11.5 Determine the minimum size of a power plant that would satisfy the requirement of a community with a population of 12 million. The average per capita of a household (hh) was reported to be 670 kWh/year.

Solution: a. Power is simply the ratio of energy and time. The calculation is a simple conversion from the data given, converting this into power, assuming 365 days in a year and 24 hrs in a day. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑀𝑀𝑀𝑀) =

670 𝑘𝑘𝑘𝑘ℎ 1 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 1 𝑑𝑑𝑑𝑑𝑑𝑑 1 𝑀𝑀𝑀𝑀 × × × 12 × 106 ℎℎ × ℎℎ − 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 24 ℎ 1,000 𝑘𝑘𝑘𝑘

= 918 𝑀𝑀𝑀𝑀

251

b. Usually, power plants are equipped with additional power in case the requirement peaks a little bit more than the base load. For this problem, a 1,000 MW base load plant may be needed.

Problem 11.11.6 OTEC Closed Cycle P11.6 The following data are provided for an OTEC system operating at temperature between (T1) 4oC and (T3) 28oC using Refrigerant 717 as working fluid. The properties of the refrigerant at these temperatures are found on its thermodynamic table. Calculate the overall cycle efficiency as well as the Carnot efficiency. The values for the properties are summarized below: a. h1 = 199.7 kJ/kg b. h2 = 313.4 kJ/kg c. h3 = 1,467.8 kJ/kg d. h4 = 1,385 kJ/kg e. v = 0.001173 m3/kg f. P2 = 1,099 kPa (kN/m2) g. P1 = 497.5 kPa (kN/m2) h. Heat input to the evaporator, Qe = (h3 - h2) = (1,467.8 - 313.4) = 1,154.4 kJ/kg i. Heat absorbed by the condenser = Qc = (h4 – h1) = (1,385 - 199.7) = 1,185.3 kJ/kg j. Pump work = v(dP) = v (P2 – P1) = 0.001173 (1,099 – 497.5) = 0.7056 kJ/kg k. Turbine work = (h3 – h4) = (1,467.8 – 1,385) = 82.8 kJ/kg Solution: a. From the data given in the figure, the overall cycle efficiency is calculated as follows: 𝑊𝑊𝑡𝑡 = (ℎ3 − ℎ4 ) = (1,467.8 − 1,385)

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 = 82.8 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 252

𝑊𝑊𝑝𝑝 = 𝜈𝜈 (𝑃𝑃2 − 𝑃𝑃1 ) = 0.001173 × (1,099 − 497.5) 𝑄𝑄𝑒𝑒 = (ℎ3 − ℎ2 ) = (1,467.8 − 313.4) �𝑊𝑊𝑡𝑡 −𝑊𝑊𝑝𝑝 � 𝑄𝑄𝑒𝑒

× 100% =

(82.8−0.7056) 1,154.4

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 = 0.7056 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 = 1,154.4 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘

× 100% = 7.11%

b. The Carnot Efficiency is calculated below and this amounts to about 8%. 𝜂𝜂𝐶𝐶 (%) = 𝜂𝜂𝐶𝐶 (%) =

𝑇𝑇ℎ − 𝑇𝑇𝑐𝑐 × 100% 𝑇𝑇ℎ

(28 + 273.15) − (4 + 273.15) 𝑇𝑇ℎ − 𝑇𝑇𝑐𝑐 × 100% = × 100% = 7.97% 𝑇𝑇ℎ (28 + 273.15)

Problem 11.11.7 Actual Cycle Examples

P11.7 Actual Rankine cycles have numerous losses from each component part of the system. In this problem, the efficiency calculations will be made for every 1,000 kg/s of working fluid flow.

The low temperature region will be 7.5oC and the high temperature region will be assumed at 25oC. Recalculate the OTEC system efficiency as well as the Carnot Effiency for these conditions. The values for working fluid properties are listed below. Use the diagram in Figure 11.6 and Figure 11.7 for the points. a. h1 = 263 kJ/kg b. h3 = 1,463.7 kJ/kg c. v = 0.001592 m3/kg d. P2 = 958.3 kPa (kN/m2) e. P1 = 565.6 kPa (kN/m2) f. S3 = 5.049 kJ/kg g. S2,f = 0.799 kJ/kg h. Sfg = 4.984 kJ/kg 253

i. hfg = 1,353.9 kJ/kg Solutions: The actual cycle calculations are as follows: 1. At Point 3, the temperature of the working fluid is reduced to 25oC due to heat exchange losses from the condensed ammonia to the warm surface. It cannot be the same temperature as the surface water temperature. As such, the properties at this point will be: – h3 = 1,463.7 kJ/kg – s3 = 5.049 kJ/kg – P2 = P3 = 958.3 kPa, and 2. Values for enthalpy, entropy, and specific volume of the ammonia can be obtained from the saturated ammonia tables. From saturated ammonia tables we have: – h1 = 263 kJ/kg – s2,f = 0.799 kJ/kg – P1 = P4 = 565.6 kPa, and – v = 0.001592 m3/kg 3. To solve the enthalpy at Point 2, we need to use the equation for quality (x2) as follows: 𝑠𝑠3 = 𝑠𝑠2,𝑓𝑓 + 𝑥𝑥2 × �𝑠𝑠𝑓𝑓𝑓𝑓 �

5.049 = 0.799 + 𝑥𝑥2 × (4.984) 𝑥𝑥2 = 0.8527 or 85.27%

4. Hence, h4 is calculated as follows: ℎ4 = ℎ1,𝑓𝑓 + 𝑥𝑥2 × �ℎ𝑓𝑓𝑓𝑓 �

ℎ4 = 263 + 0.8527 × (1,353.9) = 1,417.44

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘

5. Next, the power output of the turbine is calculated: 254

𝑊𝑊𝑡𝑡 = 𝑚𝑚̇ × (ℎ3 − ℎ4 )

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘𝑘𝑘 × (1,463.7 − 1,417.44) × = 46,260 𝑘𝑘𝑘𝑘 𝑠𝑠 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘

𝑊𝑊𝑡𝑡 = 1,000

6. Next, the pump power input and total pump power input will be calculated along with enthalpy at Point 4: 𝑊𝑊𝑝𝑝 = 𝜈𝜈 × (𝑃𝑃2 − 𝑃𝑃1 )

𝑊𝑊𝑝𝑝 = 0.001592 × (958.3 − 565.6) = 0.6252 𝑘𝑘𝑘𝑘

𝑤𝑤𝑝𝑝 = (ℎ1 − ℎ2 )

0.6252 = 263 − ℎ2 ℎ2 = 262.4

𝑘𝑘𝑘𝑘

𝑘𝑘𝑘𝑘

7. Hence, h2 is equal to 262.4 kJ/kg and the total pump input is then calculated: 𝑊𝑊𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝−𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑚𝑚̇ × �𝑊𝑊𝑝𝑝 � 𝑊𝑊𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝−𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 1,000

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 × (0.5253) = 625.2 𝑘𝑘𝑘𝑘 𝑠𝑠 𝑘𝑘𝑘𝑘

8. The heat supplied and specific heat to the heat exchanger may now be calculated as follows: 𝑞𝑞ℎ = (ℎ3 − ℎ2 ) = 1463.7 − 2624 = 1,201.3

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘

𝑄𝑄ℎ = 𝑚𝑚̇ × (𝑞𝑞ℎ ) = 1,000 × 1,201.3 = 1,201,300 𝑘𝑘𝑘𝑘

9. The actual cycle efficiency may now be calculated: 𝜂𝜂𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 (%) =

𝑊𝑊𝑡𝑡 − 𝑊𝑊𝑝𝑝 (46,260 − 625.2) 𝑤𝑤𝑛𝑛𝑛𝑛𝑛𝑛 × 100% = = × 100% = 3.8% 𝑄𝑄ℎ 𝑄𝑄ℎ 1,201,300

10. The Carnot Efficiency is calculated as follows: 𝜂𝜂𝐶𝐶 (%) =

𝑇𝑇ℎ − 𝑇𝑇𝑐𝑐 × 100% 𝑇𝑇ℎ 255

𝜂𝜂𝐶𝐶 (%) =

(22.9 + 273.15) − (10 + 273.15) 𝑇𝑇ℎ − 𝑇𝑇𝑐𝑐 × 100% = × 100% = 4.35% (22.9 + 273.15) 𝑇𝑇ℎ

Note that the actual efficiency is quite low within 3% to 4%. To achieve significant output power, vast volumes of ocean water must be used. This will then also require larger pipe diameters for conveying the required working fluid and needs more pumping power to circulate the liquid.

Problem 11.11.8 Heat of Evaporation.

P11.8 Calculate the heat required to evaporate 100 kg of ethylene glycol and 100 kg of Refrigerant R-11. The heat of evaporation for ethylene glycol and R-11 are 800 and 232 kJ/kg, respectively.

Solution: a. Using Equation 11.8 directly, the values are as follows for water: 𝑞𝑞𝑤𝑤 (𝑘𝑘𝑘𝑘) = 𝑚𝑚 × ℎ𝑒𝑒 = 100 𝑘𝑘𝑘𝑘 × 800

𝑘𝑘𝑘𝑘 = 80,000 𝑘𝑘𝑘𝑘 [80 𝑀𝑀𝑀𝑀] 𝑘𝑘𝑘𝑘

b. The heat of vaporization for ammonia will be: 𝑞𝑞𝑎𝑎 (𝑘𝑘𝑘𝑘) = 𝑚𝑚 × ℎ𝑒𝑒 = 100 𝑘𝑘𝑘𝑘 × 232

𝑘𝑘𝑘𝑘 = 23,200 𝑘𝑘𝑘𝑘 [23.2 𝑀𝑀𝑀𝑀] 𝑘𝑘𝑘𝑘

c. It will take less energy to evaporate R-11 than it would to evaporate ethylene glycol.

Problem 11.11.9 Number of Households Served by OTEC P11.9 Determine the number of households that will be served by a 1MW OTEC power plant.

Assume that each household will use around 12,000 kWh of energy per year.

Solution: a. This is a straightforward energy and power calculation since power is energy per unit of time.

256

# 𝐻𝐻𝐻𝐻 = 1 𝑀𝑀𝑀𝑀 ×

𝐻𝐻𝐻𝐻 − 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 24 ℎ𝑟𝑟𝑟𝑟 1,000 𝑘𝑘𝑘𝑘 × × × = 730 𝐻𝐻𝐻𝐻 12,000 𝑘𝑘𝑘𝑘ℎ 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑑𝑑𝑑𝑑𝑑𝑑 1 𝑀𝑀𝑀𝑀

b. This power plant will serve around 730 households.

Problem 11.12.10 Initial Capital Cost of OTEC

P11.10 Determine the initial capital cost for a 100 kW OTEC power plant using empirical Equation 11.11. Compare this cost with the DOE’s estimate of $3,400/kW. Solution a. The direct use of empirical equation yields the following results. 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 �

$ 1 𝑀𝑀𝑀𝑀 � = 373 × 𝑙𝑙𝑙𝑙 �100 𝑘𝑘𝑘𝑘 × � + 21,687 = $20,828/𝑘𝑘𝑘𝑘 1,000 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘

b. For this facility, the initial capital cost is around $2.1M. 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 ($) =

$20,828 × 100 𝑘𝑘𝑘𝑘 = $2.1 𝑀𝑀 𝑘𝑘𝑘𝑘

c. Using the DOE’s estimate the total cost is simply equal to $100kW x $3,400/kW = $340,000, which is significantly lower than the cost found using the empirical equation.

11.11.11

Sizing OTEC Systems

P11.11 Determine the minimum size (in MW) of an OTEC power plant that would satisfy the requirement of a community with a population of 12 million. The average per capita of a household was reported to be 670 kWh/year. a. 670 MW b. 80 c. 9,180 d. 918 Solutions:

257

a. Power is simply the ratio of energy and time. Hence, the calculation is a simple conversion of the data given and converts this into power assuming 365 days in a year and 24 hrs in a day as shown. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑀𝑀𝑀𝑀) =

670 𝑘𝑘𝑘𝑘ℎ 1 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑀𝑀𝑀𝑀 × × × 12 × 106 𝐻𝐻𝐻𝐻 = 918 𝑀𝑀𝑀𝑀 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 8760 ℎ𝑟𝑟𝑟𝑟 1,000 𝑘𝑘𝑘𝑘

b. Usually power plants are sized with additional power in case the requirement peaks a little bit more than the base load. For this problem a 1,000 MW base load plant may be needed.

11.11.12

Number of Households Served by OTEC Systems

P11.12 Determine the number of households that will be served by a 1 MW OTEC power plant.

Assume that each household will use around 12,000 kWh of energy per year. a. 730 b. 730,000 c. 8330 d. 8,330,000 Solution: a. This is a straightforward energy and power calculation since power is energy per unit of time. # 𝐻𝐻𝐻𝐻 = 1 𝑀𝑀𝑀𝑀 ×

11.11.13

1 𝐻𝐻𝐻𝐻 − 𝑦𝑦𝑦𝑦 8760 ℎ𝑟𝑟𝑟𝑟 1,000 𝑘𝑘𝑘𝑘 × × = 730 𝐻𝐻𝐻𝐻 12,000 𝑘𝑘𝑘𝑘ℎ 𝑦𝑦𝑦𝑦 1 𝑀𝑀𝑀𝑀

Capital Cost of OTEC Systems

P11.13 Determine the initial capital cost per kW (in $/kW) for a 100 kW OTEC power plant using the published empirical equation shown in lecture notes. a. 3,400 258

b. 20,828 c. 2,083 d. 21,687 Solutions: a. The direct use of empirical equations yields the following results 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 �

11.11.14

$ $20,828 � = 373 × 𝑙𝑙𝑙𝑙(0.1 𝑀𝑀𝑀𝑀 ) + 21,687 = 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘

Estimating the Initial Capital Cost of OTEC Systems

P11.14 Determine the initial capital cost (in million $) for a 100 kW OTEC power plant using Equation 11.10. a. 1.1 b. 21 c. 2.1 d. 42 Solution a. The direct use of empirical equation yields the following results 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 �

$ $20,828 � = 373 × 𝑙𝑙𝑙𝑙(0.1 𝑀𝑀𝑀𝑀 ) + 21,687 = 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘

b. Hence, for this facility, the initial capital cost is around $2.1M as shown below. 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 ($) =

11.11.15

$20,828 × 100 𝑘𝑘𝑘𝑘 = $2.1 𝑀𝑀 𝑘𝑘𝑘𝑘

Estimating the Density of Seawater at Given Depth

P11.15 Estimate the density of seawater at a depth of 9,000 meters or 9 km. Report your answer in kg/m3. Please use the governing equation below.

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𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 � a. 1067.4

𝑘𝑘𝑘𝑘 � = 4.3 × 10−3 × 0 𝑚𝑚 + 1028.7 𝑚𝑚3

b. 1032.6 c. 1029.1 d. 1415.7 Solution: a. The density of seawater at half a km is calculated as follows: 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 �

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 � = 4.3 × 10−3 × 9,000 𝑚𝑚 + 1028.7 = 1067.4 3 3 𝑚𝑚 𝑚𝑚

Assess Your Knowledge 1. What regions on earth would be the ideal location of power generation systems using OTEC?

a. Between the Tropic of Capricorn and Tropic of Cancer b. Above the Tropic of Cancer c. Below the Tropic of Capricorn d. Between Arctic Center and Antarctic Center 2. What ideal cycle best represents OTEC Systems?

a. Rankine Cycle b. Carnot Cycle c. Otto Cycle d. Stirling Cycle 3. Ideally, OTEC works best if the temperature difference between the colder ocean water and warm surface water is at least how many degrees of difference in units of Centigrade?

a. 20 260

b. 15 c. 10 d. 5 4. What type of OTEC system uses only the tropical ocean’s warm water to make electricity? This ocean water is placed in a low-pressure vessel where it will boil. This will then drive low-pressure turbines that drive generators to generate electrical power.

a. Closed Cycle b. Binary Cycle c. Open Cycle d. Hybrid Systems 5. Which US Federal Facility is tasked to develop and diversify resources and facilities for energy- and ocean-related research, education, and commercial activities?

a. National Renewable Energy Laboratory (NREL) b. Argonne National Laboratory (ANL) c. Hawaii Ocean Science and Technology Park (HOST) d. Ocean Center in Daytona Beach Florida (OCDB) 6. Determine the Carnot efficiency (%) for the following given data: Th = 25oC; Tc = 3oC; Wt = 81.5 kJ/kg; Wp = 0.76123 kJ/kg; Qe = 1,347.9 kJ/kg.

a. 7.38 b. 3.69 c. 14.7 d. 5.99 Solution: a. Carnot Efficiency is calculated as follows: 261

𝜂𝜂𝐶𝐶 = 𝜂𝜂𝐶𝐶 =

(𝑇𝑇ℎ − 𝑇𝑇𝑐𝑐 ) × 100% 𝑇𝑇ℎ

(25 + 273.15) − (3 + 273.15) 298.15 − 276.15 × 100% = × 100% = 7.38% 25 + 273.15 298.15

7. Determine the Carnot efficiency (in %) for the following given data: Th = 22oC: Tc = 5oC; Wt = 7,813 kW; Wp = 68.58 kW; and Qh = 172,100 kW.

a. 4.5 b. 9.0 c. 2.25 d. 5.76 Solution: a. Carnot Efficiency is calculated as follows: 𝜂𝜂𝐶𝐶 (%) = 𝜂𝜂 (%) =

(𝑇𝑇ℎ − 𝑇𝑇𝑐𝑐 ) × 100% 𝑇𝑇ℎ

(22 + 273.15) − (5 + 273.15) 295.15 − 278.15 × 100% = = 5.76% 22 + 273.15 295.15

8. Determine the mass flow rate of ocean water needed (in kg/s) to generate a 100 kW of electrical power assuming a 15oC temperature difference and power is produced continuously for 24 hrs a day and 365 days in a year. Assume an overall conversion efficiency of 5% from sensible heat to electrical power. Use the simple sensible heat equation for the calculation. Assume the density of sea water to be 1,027 kg/m3.

a. 93.3 b. 17 c. 67.8 d. 33.9 Solution: 262

a. The efficiency equation is first used to calculate the energy required input using the overall conversion efficiency. Hence, if 100 kW of output power is needed, a 5% conversion efficiency shows that 2,000 kW of input power as shown below. 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑘𝑘𝑘𝑘) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑘𝑘𝑘𝑘) 100 𝑘𝑘𝑘𝑘 = = 2,000 𝑘𝑘𝑘𝑘 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑) 0.05

b. This energy must come from the thermal energy from the ocean assuming the specific heat of ocean water to be 3.93 kJ/kgoC and a 15oC difference. 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 = 𝑚𝑚̇ � 𝑚𝑚̇ �

𝑘𝑘𝑘𝑘 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) �= 𝑠𝑠 𝐶𝐶𝑝𝑝 ∆𝑡𝑡

𝑘𝑘𝑘𝑘𝑜𝑜 𝐶𝐶 1 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 � = 2,000 𝑘𝑘𝑘𝑘 × × 𝑜𝑜 × = 33.9 𝑠𝑠 3.93 𝑘𝑘𝑘𝑘 15 𝐶𝐶 1 𝑘𝑘𝑘𝑘 − 𝑠𝑠 𝑠𝑠

9. Determine the actual amount of power (in kW) used to move 50 kg/s of water through 100 meters of head from the bottom of ocean to the top surface. Assume an overall pumping efficiency of 75%. Use the basic hydro power equation from the Hydro Power Lecture for calculations. This will be the parasitic load required for pumping seawater alone against the given dynamic head.

a. 49 b. 65.3 c. 32.5 d. 135 Solution: a. The problem is straightforward using the pumping power equation shown. 𝑘𝑘𝑘𝑘 𝑄𝑄𝑄𝑄 50 𝑠𝑠 × 100 𝑚𝑚 𝑃𝑃𝑡𝑡 (𝑘𝑘𝑘𝑘) = = = 49 𝑘𝑘𝑘𝑘 102 102

b. The actual power required to lift this water against the head should be higher considering efficiency or losses and is calculated as follows: 263

𝑃𝑃𝑎𝑎 (𝑘𝑘𝑘𝑘) =

49 𝑘𝑘𝑘𝑘 = 65.3 𝑘𝑘𝑘𝑘 0.75

10. A hectare [2.47 acres] of seawater surface water with a depth of 5 meters [16.4 ft] was used for OTEC power generation. Assume the original temperature of 5oC (Bottom) and the final temperature is 25oC (top). Assume the specific heat of sea water to be 3.93 kJ/kgoC. If the overall conversion efficiency from sensible heat to power is 3%, estimate the actual power produced (in MW). The mass flow rate of the OTEC system was 1,000 kg/s and seawater density of 1025 kg/m3.

a. 2358 b. 2.34 c. 3,930 d. 3.93 Solution: a. The energy absorb is simply calculated using the sensible heat energy equation as shown. 𝑄𝑄(𝑘𝑘𝑘𝑘) = 𝑚𝑚̇𝐶𝐶𝑝𝑝 𝛥𝛥𝑡𝑡

𝑘𝑘𝑘𝑘

3.93 𝑘𝑘𝑘𝑘

𝑄𝑄𝑡𝑡 (𝑘𝑘𝑘𝑘 ) = 1,000 𝑠𝑠 × 𝑘𝑘𝑘𝑘𝑜𝑜 × (25𝑜𝑜 𝐶𝐶 − 5𝑜𝑜 𝐶𝐶 ) = 78600 𝑘𝑘𝑘𝑘 𝐶𝐶

𝑄𝑄𝑎𝑎 (𝑘𝑘𝑘𝑘 ) = 78600 𝑘𝑘𝑘𝑘 × 0.03 = 2358 𝑘𝑘𝑘𝑘 [2.34 𝑀𝑀𝑀𝑀]

11. True or False. Seawater becomes less dense with increasing ocean depth.

a. True b. No idea c. It depends d. False 12. What type of OTEC system uses fluids with low boiling points to generate electricity?

a. Open Cycle b. Closed Cycle

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c. Binary Cycle d. Hybrid Systems 13. Determine the minimum size (in MW) of an OTEC power plant that would satisfy the requirement of a community with a population of 12 million. The average per capita of a household was reported to be 490 kWh/year.

a. 671 b. 80 c. 9,180 d. 918 Solution: a. Power is simply the ratio of energy and time. The calculation is a simple conversion from the data given, converting this into power, assuming 365 days in a year and 24 hrs in a day. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑀𝑀𝑀𝑀) =

490 𝑘𝑘𝑘𝑘ℎ 1 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 1 𝑑𝑑𝑑𝑑𝑑𝑑 1 𝑀𝑀𝑀𝑀 × × × 12 × 106 ℎℎ × 1,000 𝑘𝑘𝑘𝑘 ℎℎ − 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 24 ℎ

= 671 𝑀𝑀𝑀𝑀

b. Usually, power plants are equipped with additional power in case the requirement peaks a little bit more than the base load. For this problem, a 1,000 MW base load plant may be needed.

14. This renewable energy is an energy conversion process that can produce electricity by using the temperature difference between deep cold ocean water and warm tropical surface waters. They pump large quantities of deep cold seawater and surface seawater to run a power cycle and produce electricity.

a. tidal energy

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b. wave energy c. OTEC d. salinity gradient 15. Determine the amount of energy absorbed (in million MJ) by a hectare [2.47 acres] of seawater surface water with a depth of 5 meters [16.4 ft]. Assume the original temperature of 5oC and the final temperature is 25oC. Assume the specific heat of sea water to be 3.93 kJ/kgoC and the seawater density of 1025 kg/m3.

a. 4 b. 2 c. 6 d. 8 Solution: a. The energy absorb is simply calculated using the sensible heat energy equation as shown. 𝑄𝑄(𝑘𝑘𝑘𝑘) = 𝑚𝑚𝐶𝐶𝑝𝑝 𝛥𝛥𝑡𝑡 𝑄𝑄(𝑘𝑘𝑘𝑘 ) = 1 ℎ𝑎𝑎 ×

10,000 𝑚𝑚 2 1 ℎ𝑎𝑎

× 5 𝑚𝑚 × 𝑀𝑀𝑀𝑀

1,025 𝑘𝑘𝑘𝑘 𝑚𝑚 3

3.93 𝑘𝑘𝑘𝑘

× 𝑘𝑘𝑘𝑘𝑜𝑜 𝐶𝐶 × (25𝑜𝑜 𝐶𝐶 − 5𝑜𝑜 𝐶𝐶 ) = 4,028,250,000 𝑘𝑘𝑘𝑘

𝑄𝑄(𝑀𝑀𝑀𝑀 ) = 4,028,250,000 𝑘𝑘𝑘𝑘 × 1,000 𝑘𝑘𝑘𝑘 = 4,028,250 𝑀𝑀𝑀𝑀

𝑄𝑄(106 𝑀𝑀𝑀𝑀 ) = 4 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑀𝑀

16. Determine the volume of water needed (in millions of cubic meter) to generate 100 kW of electrical power assuming a 15oC temperature difference and power is produced continuously for 24 hrs a day and 365 days in a year. Assume an overall conversion efficiency of 5% from sensible heat to electrical power. Use the simple sensible heat equation for the calculation. Assume the density of sea water to be 1,027 kg/m3.

a. 1 b. 1.7 266

c. 2.5 d. 3.3 Solution: a. The efficiency equation is first used to calculate the input energy required using the overall conversion efficiency. Hence, if 100 kW of output power is needed, a 5% conversion efficiency shows that 2,000 kW of input power as shown below. 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑘𝑘𝑘𝑘) =

𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑘𝑘𝑘𝑘) 100 𝑘𝑘𝑘𝑘 = = 2,000 𝑘𝑘𝑘𝑘 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑) 0.05

b. This energy must come from the thermal energy from the ocean assuming the specific heat of ocean water to be 3.93 kJ/kgoC and a 15oC difference. 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 = 𝑚𝑚̇ � 𝑚𝑚̇ �

𝑘𝑘𝑘𝑘 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑘𝑘𝑘𝑘) �= 𝑠𝑠 𝐶𝐶𝑝𝑝 ∆𝑡𝑡

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘𝑜𝑜 𝐶𝐶 1 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 � = 2,000 𝑘𝑘𝑘𝑘 × × 𝑜𝑜 × = 33.9 𝑠𝑠 3.93 𝑘𝑘𝑘𝑘 15 𝐶𝐶 1 𝑘𝑘𝑘𝑘 − 𝑠𝑠 𝑠𝑠

c. The volume of water needed for this project year-round is calculated as follows: 𝑄𝑄 �

106 𝑚𝑚3 𝑘𝑘𝑘𝑘 𝑚𝑚3 3600 𝑠𝑠 24 ℎ𝑟𝑟𝑟𝑟 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚3 � = 33.9 × × × × × 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 1027 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑 1 × 106 𝑚𝑚3 𝑠𝑠 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 = 1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

17. What is the most common low boiling point material is used in some OTEC systems?

a. Ethylene Glycol b. Ammonia c. R12 d. R134a 18. Ideally, OTEC works best if the temperature difference between the colder ocean water and warm surface water is at least how many degrees of difference in units of Fahrenheit?

267

a. 20 b. 36 c. 25 d. 10 19. Determine the actual cycle efficiency (in %) for the following given data: Th = 22oC; Tc = 5oC; Wt = 7,813 kW; Wp = 68.58 kW; and Qh = 172,100 kW.

a. 2.25 b. 9.0 c. 4.5 d. 5.76 Solution: a. Cycle efficiency is calculated as follows: 𝜂𝜂 (%) = 𝜂𝜂 (%) =

�𝑊𝑊𝑡𝑡 − 𝑊𝑊𝑝𝑝 � × 100% 𝑄𝑄𝑒𝑒

(7813 − 68.58) × 100% = 4.5% 172,100

20. Determine the number of households that will be served by a 1 MW OTEC power plant. Assume that each household will use around 10,500 kWh of energy per year.

a. 730 b. 730,000 c. 834 d. 8,330,000 Solution: a. This is a straightforward energy and power calculation since power is energy per unit of time.

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# 𝐻𝐻𝐻𝐻 = 1 𝑀𝑀𝑀𝑀 ×

𝐻𝐻𝐻𝐻 − 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 24 ℎ𝑟𝑟𝑟𝑟 1,000 𝑘𝑘𝑘𝑘 × × × = 834 𝐻𝐻𝐻𝐻 10,500 𝑘𝑘𝑘𝑘ℎ 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑑𝑑𝑑𝑑𝑑𝑑 1 𝑀𝑀𝑀𝑀

b. This power plant will serve around 834 households.

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Chapter 11 Human and Animal Power and Piezoelectrics 12.6 Solved Problems Problem 12.6.1 Power from Animals P12.1 Determine the amount of power (in units of ft-lbs/min) generated from an ox carrying a load of 120 lbs and walking at a rate of 2.5 feet/s. Convert this unit into horsepower and Watts.

Solution: a. Power is the product of force and speed: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 2.5 𝑓𝑓𝑓𝑓 60 𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 � = 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 × 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 = 120 𝑙𝑙𝑙𝑙𝑙𝑙 × × = 18,000 𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚

b. The conversion of 18,000 ft-lb/min into horsepower is shown below. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) = 18,000

𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 ℎ𝑝𝑝 − 𝑚𝑚𝑚𝑚𝑚𝑚 × = 0.545 ℎ𝑝𝑝 𝑚𝑚𝑚𝑚𝑚𝑚 33,000 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙

c. The conversion of 0.545 hp into Watts is shown below. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = 0.545 ℎ𝑝𝑝 ×

Problem 12.6.2 Power from Humans

746 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 406.9 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 1 ℎ𝑝𝑝

P12.2 Determine the amount of power generated by a human being carrying a load of 25 lbs and walking at a rate of 2 feet/sec.

Solution: a. Power is the product of force and speed: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) = 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 × 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) = 25 𝑙𝑙𝑙𝑙𝑙𝑙 ×

2 𝑓𝑓𝑓𝑓 ℎ𝑝𝑝 − 𝑚𝑚𝑚𝑚𝑚𝑚 60 𝑠𝑠𝑠𝑠𝑠𝑠 × × = 0.091 ℎ𝑝𝑝 𝑠𝑠𝑠𝑠𝑠𝑠 33,000 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚 270

b. The value 0.091 hp is equivalent to 67.8 Watts as shown below. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = 0.091 ℎ𝑝𝑝 ×

746 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 67.8 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 1 ℎ𝑝𝑝

Problem 12.6.3 Various Units of Power

P12.3 Calculate the amount of power that can be generated by a 600-kg dairy cow pulling a load of 60 kg and traveling at an average speed of 0.7 m/s. Provide your answers in kg-m/s, Watts, and horsepower.

Solution: a. Equation 11.1 may be used but with different units and a different constant: 𝑚𝑚 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 (𝑘𝑘𝑘𝑘) × 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 � 𝑠𝑠 � 𝑘𝑘𝑘𝑘 − 𝑚𝑚 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 � �= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 �

𝑘𝑘𝑘𝑘 − 𝑚𝑚 60 𝑘𝑘𝑘𝑘 × 7 𝑚𝑚 𝑘𝑘𝑘𝑘 − 𝑚𝑚 �= = 42 𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠

b. The conversion of the above units into horsepower may be done as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) =

42 𝑘𝑘𝑘𝑘 − 𝑚𝑚 𝑁𝑁 − 𝑠𝑠𝑠𝑠𝑠𝑠 2 9.8 𝑚𝑚 𝑊𝑊𝑊𝑊 × × × = 411.6 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑠𝑠𝑠𝑠𝑠𝑠 𝑘𝑘𝑘𝑘 − 𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑁𝑁𝑁𝑁

c. The conversion to horsepower is the reverse of the previous example: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) = 411.6 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 ×

ℎ𝑝𝑝 = 0.552 ℎ𝑝𝑝 746 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊

Problem 12.6.4 Power from Groups of Animals

P12.4 If a small horse can provide a draft of 80 kg, how much can be provided by 5 small horses together? Use the harnessing factor for work animals in Table 11.4 for your calculations. Report your units in kg and in Newtons. If the animals’ speed is 2 m/s, determine the power generated by the group.

271

Solution: a. The solution is straightforward, as shown below. 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 (𝑘𝑘𝑘𝑘) =

𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 (𝑘𝑘𝑘𝑘) =

𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 × # ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 × 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 80 𝑘𝑘𝑘𝑘 3.5 × 5 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 × = 280 𝑘𝑘𝑘𝑘 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 5.0

b. Note that the harnessing factor value in the table means the number out of the total population. That is, 3.8 out of 6, 3.5 out of 5, and so on. c. The power generated from this group is calculated: 2 𝑚𝑚 𝑁𝑁 − 𝑠𝑠𝑠𝑠𝑐𝑐 2 9.8 𝑚𝑚 𝑊𝑊𝑊𝑊 × × × = 5,488 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑘𝑘𝑘𝑘 − 𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑁𝑁𝑁𝑁

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = 280 𝑘𝑘𝑘𝑘 ×

Problem 12.6.5 Energy Output of a Cow in the Form of Milk

P12.5 Determine the energy in MJ produced by a cow from Wisconsin in the form of milk.

Assume the energy content of milk is 150 kcal per 250 g and 72,000 lbs was produced in a year.

If the same cow produces an average of 7 lbs of dry manure each day or 2,555 lbs in a year with a heating value of 6,000 Btu/lb, how much energy in MJ is likewise produced? How much total energy is produced by this cow in MJ? Solution: a. The milk energy is simply a conversion problem: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) = 72,000 𝑙𝑙𝑙𝑙𝑙𝑙 × = 82,158.5 𝑀𝑀𝑀𝑀

150 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 4,184 𝐽𝐽 1 𝑀𝑀𝑀𝑀 1 𝑘𝑘𝑘𝑘 × × × 6 0.250 𝑘𝑘𝑘𝑘 1 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 1 × 10 𝐽𝐽 2.2 𝑙𝑙𝑙𝑙𝑙𝑙

b. The manure energy is also estimated the same way: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) = 2,555 𝑙𝑙𝑙𝑙𝑙𝑙 ×

6,000 𝐵𝐵𝐵𝐵𝐵𝐵 1,055 𝐽𝐽 1 𝑀𝑀𝑀𝑀 × × = 16,173.15 𝑀𝑀𝑀𝑀 𝑙𝑙𝑙𝑙 𝐵𝐵𝐵𝐵𝐵𝐵 1 × 106 𝐽𝐽

c. The total energy produced by this cow is roughly 98,332 MJ. 272

Problem 12.6.6 Power of Humans Over Longer Periods of Time P12.6 A person is asked to do work for 8 hours. Estimate the power that can be generated by this person in units of horsepower (hp) and Watts. Compare this value with the results of Example 11.6.

Solution: a. Equation 11.3 is simply used: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) = 0.35 − 0.092 × log �8 ℎ𝑟𝑟𝑟𝑟 ×

b. The equivalent power in Watts is calculated: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = 0.103 ℎ𝑝𝑝 ×

60 𝑚𝑚𝑚𝑚𝑚𝑚 � = 0.103 ℎ𝑝𝑝 ℎ𝑟𝑟

746 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 77 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 1 ℎ𝑝𝑝

c. When working for a full day, the average power delivered by a human being will diminish.

Problem 12.6.7 Power from Arms and Legs of Humans P12.7 Compare a person’s power output if he works for 10 hours a day using his arms or his legs (that is, use Equations 11.3 and 11.4). Comment on the power output in the results of calculations. Provide results in hp and Watts.

Solution: a. Equation 11.3 is used: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) = 0.35 − 0.092 × log �10 ℎ𝑟𝑟𝑟𝑟 ×

b. The power in Watts is calculated:

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = 0.0944 ℎ𝑝𝑝 ×

60 𝑚𝑚𝑚𝑚𝑚𝑚 � = 0.0944 ℎ𝑝𝑝 ℎ𝑟𝑟

746 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 70.4 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 1 ℎ𝑝𝑝 273

c. Equation 11.4 is also used: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) = 0.53 − 0.13 × log �10 ℎ𝑟𝑟𝑟𝑟 ×

d. The equivalent power in Watts is calculated: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = 0.1688 ℎ𝑝𝑝 ×

60 𝑚𝑚𝑚𝑚𝑚𝑚 � = 0.1688 ℎ𝑝𝑝 ℎ𝑟𝑟

746 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 126 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 1 ℎ𝑝𝑝

e. There is almost twice as much power when the legs are used rather than arms.

Problem 12.6.8 Basic Piezoelectric Power from Numerous Repeated Cycles P12.8 Determine the energy produced for numerous cycles by a piezoelectric element. The measured voltage across the capacitor after a series of cycles was found to be 10 Volts and the capacitor has a rating of 220 microfarad, 35V capacitor. Provide your answer in Joule.

Solution: a. Equation 11.5 is simply used: 1 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽) = 𝐶𝐶𝑉𝑉 2 2 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽) =

1 220 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 × × (10 𝑉𝑉 )2 = 0.011 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 2 1 × 106

b. There is quite a small amount of energy stored in this capacitor after a series of cycles.

Problem 12.6.9 Piezoelectric Power from Single Tap P12.9 Determine the energy produced for each single tap on a piezoelectric element. The initial voltage was found to be 2.0 Volts and after each tap, the final voltage was found to be 3.5 Volts.

The capacitor has a rating or 220 microfarad and 35V. Provide your answer in Joule.

Solution: a. Equation 11.6 is simply used: 1 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽) = 𝐶𝐶𝑉𝑉 2 2

274

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 1 1 220 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 � = 𝐶𝐶𝑉𝑉 2 = × × [(3.5 𝑉𝑉 )2 − (2 𝑉𝑉 )2 ] = 0.00091 6 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 2 2 1 × 10 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

b. This is a small amount of energy stored from each deformation cycle of a piezoelectric device, a very small fraction of a Joule.

Problem 12.6.10 Charging Cell Phone by Piezoelectric Power P12.10 Determine how long it would take to charge a cell phone battery with a fully charged

load requirement of 15,000 Joules. The voltage across the capacitor was improved from 2 Volts to 6 Volts with a single cycle. Assuming this is done continuously in a day, in how many days can one achieve full charge? The capacitor has a rating of 220 microfarad and 35V. One can generate 3 cycles a second on this device.

Solution: a. Equation 11.6 is simply used: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 1 1 220 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 � = 𝐶𝐶𝑉𝑉 2 = × × [(6 𝑉𝑉 )2 − (2 𝑉𝑉 )2 ] = 0.00352 6 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 2 2 1 × 10

b. To generate 18,000 joules of energy, the number of cycles is calculated: 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 15,000 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 ×

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 4,261,363 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 0.00352 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽

c. The number of days required to get the device fully charged is calculated: 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 4,261,363 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ×

𝑠𝑠𝑠𝑠𝑠𝑠 1 ℎ𝑟𝑟 1 𝑑𝑑𝑑𝑑𝑑𝑑 × × = 9.9 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 5 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 3600 𝑠𝑠𝑠𝑠𝑠𝑠 24 ℎ𝑟𝑟𝑟𝑟

d. With this improvement, it will take about 10 days to fully charge this cell phone.

12.6.11

Charging Time for Cell Phone Battery with Piezoelectric Power

P12.11 Determine how long it would take to charge a cell phone battery with a fully charged load requirement of 20,000 joules (18.96 Btu). The voltage across the capacitor was improved from 5 275

volts to 15 volts with a single cycle. Assuming that this was done continuously in a day, in how many days will one achieve a full charge? The capacitor has a rating of 320 microfarads and 30 volts. One can generate 5 cycles in a second for this device. a. 1.15 b. 1.45 c. 2.15 d. 2.75 Solutions: a. Equation 12.6 is used as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 1 1 320 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 � = 𝐶𝐶𝑉𝑉 2 = × × [(15)2 − (5)2 ] = 0.032 6 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 2 2 1 × 10

b. To generate 20,000 joules of energy, the number of cycles is calculated as follows: 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 20,000 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 ×

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 625,000 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 0.032 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽

c. Hence, the number of days required to get this fully charged is calculated as follows: 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 625,000 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ×

12.6.12

ℎ𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑𝑑𝑑𝑑𝑑 × × = 1.45 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 5 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 3600 𝑠𝑠𝑠𝑠𝑠𝑠 24 ℎ𝑟𝑟𝑟𝑟

Power of Work Animals

P12.12 African bush elephants are the world’s largest land animals, and the strongest mammals, able to carry up to 9,000 kg [19,800 lbs] of load. If this mammal can walk at a speed of 5 mph, determine the horsepower generated in units of watts. a. 746 b. 1741 c. 2150 d. 2.33 276

Solutions: a. 𝐻𝐻𝐻𝐻 (ℎ𝑝𝑝) = 3.5 2.33 ℎ𝑝𝑝

𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

b. 𝐻𝐻𝐻𝐻 = 2.33 ℎ𝑝𝑝 ×

12.6.13

ℎ𝑟𝑟

× 250 𝑙𝑙𝑙𝑙𝑙𝑙 ×

746 𝑊𝑊 ℎ𝑝𝑝

5,280 𝑓𝑓𝑓𝑓 1 𝑚𝑚𝑚𝑚

×

1 ℎ𝑟𝑟

60 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

= 77,000

𝑓𝑓𝑓𝑓−𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚

×

ℎ𝑝𝑝−𝑚𝑚𝑚𝑚𝑚𝑚

33,000 𝑓𝑓𝑓𝑓−𝑙𝑙𝑙𝑙

=

= 1740.7 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤

Charge in a Capacitor

P12.13 Calculate the amount of charge stored in the capacitor (units of coulombs) if the value of the capacitance is 100uF and the voltage is 12 V. a. 0.0012 b. 0.000235 c. 0.12 d. 0.235 Solution: a. The charge is simply calculated as follows: 𝑄𝑄 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = 𝐶𝐶 × 𝑉𝑉 = 100 × 10−6 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 × 12 𝑉𝑉 = 0.0012 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

b. The amount of energy stored is simply calculated as follows (using Equation 12.6): 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗) =

12.6.14

1 2 𝐶𝐶𝑉𝑉 = 0.5 × 100 × 10−6 × 122 = 0.0072 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 2

Energy Stored in a Capacitor

P12.14 Calculate the amount of energy stored in the capacitor shown in the figure (units of Joules) if the value of the capacitance is 100uF and the voltage is 12 V. a. 0.0005875 b. 0.0072 c. 0.72

277

d. 0.59 Solution: a. The charge is simply calculated as follows: 𝑄𝑄 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = 𝐶𝐶 × 𝑉𝑉 = 100 × 10−6 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 × 12 𝑉𝑉 = 0.0012 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

b. The amount of energy stored is simply calculated as follows (using Equation 12.6): 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗) =

12.6.15

1 2 𝐶𝐶𝑉𝑉 = 0.5 × 100 × 10−6 × 122 = 0.0072 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 2

Longevity of Batteries

P12.15 Determine how long will a battery last (in hours) if one has a 2000 mAh battery and discharges at a constant current of 500 mA? a. 1 b. 2 c. 3 d. 4 Solutions: a. The formula to use is a simple one shown below: 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 (𝑚𝑚𝑚𝑚ℎ) = 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 (𝑚𝑚𝑚𝑚) × 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) =

2000 𝑚𝑚𝑚𝑚ℎ 500 𝑚𝑚𝑚𝑚

= 4 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜

Assess Your Knowledge 1. A large animal is working at a peak sustained speed of 3.5 mph and carrying a load of 250 lbs.

Determine the horsepower generated. Units should be in horsepower.

a. 1.15 b. 1.53 278

c. 2.15 d. 2.33 Solutions: a. 𝐻𝐻𝐻𝐻 (ℎ𝑝𝑝) = 3.5 2.33 ℎ𝑝𝑝

𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ℎ𝑟𝑟

× 250 𝑙𝑙𝑙𝑙𝑙𝑙 ×

5,280 𝑓𝑓𝑓𝑓 1 𝑚𝑚𝑚𝑚

×

1 ℎ𝑟𝑟

60 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

= 77,000

𝑓𝑓𝑓𝑓−𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚

×

ℎ𝑝𝑝−𝑚𝑚𝑚𝑚𝑚𝑚

33,000 𝑓𝑓𝑓𝑓−𝑙𝑙𝑙𝑙

=

2. Determine the amount of power g(Watts) generated by a human being carrying a load of 50 lbs and walking at a rate of 2 ft/s.

a. 0.114 b. 0.182 c. 1.2 d. 135 Solutions: a. Power is the product of force and speed 𝐻𝐻𝐻𝐻 = 50 𝑙𝑙𝑙𝑙𝑙𝑙 ×

2 𝑓𝑓𝑓𝑓 ℎ𝑝𝑝 − 𝑚𝑚𝑚𝑚𝑚𝑚 60 𝑠𝑠 × × = 0.182 ℎ𝑝𝑝 𝑠𝑠 33,000𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚

b. The equivalent units in metric is as follows: 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 0.182 ℎ𝑝𝑝 × c. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) =

746 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 135 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 1 ℎ𝑝𝑝 𝑓𝑓𝑓𝑓 𝑠𝑠

𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 (𝑙𝑙𝑙𝑙𝑙𝑙)×𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 � � 550

3. A person is asked to do work using his legs for 6 hours. Estimate the power that can be generated by this person in units of hp.

a. 0.107 b. 0.155 c. 0.197 279

d. 0.255 Solution a. Using the above equation, we have the following: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) = 0.53 − 0.13 × 𝑙𝑙𝑙𝑙𝑙𝑙(360) = 0.197 ℎ𝑝𝑝 [147.5 𝑊𝑊 ]

b. This power is significantly higher than earlier hp calculations for manual human power.

4. Calculate the charging time (hrs) of a lithium-ion battery having 3000mAh, 24W charging rate, 12V voltage and 90% charging efficiency.

a. 1.67 b. 2.38 c. 3.5 d. 4.5 Solutions: a. First you need to convert the charging current (24W) into amps as follows: 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) =

24 𝑊𝑊 = 2 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 12 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉

b. Similarly, convert the battery capacity from mAh to Ah as follows: 𝐴𝐴ℎ (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 − ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜) = 3000 𝑚𝑚𝑚𝑚ℎ ×

𝐴𝐴ℎ = 3 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 − ℎ𝑟𝑟𝑟𝑟 1000 𝑚𝑚𝑚𝑚

c. Then, the charging time is calculated as follows: 𝐶𝐶ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = 3 𝐴𝐴ℎ ×

1 = 1.67 ℎ𝑟𝑟𝑟𝑟 2 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 × 0.90

5. A large animal is working at a peak sustained speed of 3.5 mph and carrying a load of 250 lbs.

Determine the horsepower generated in units of watts.

a. 746 b. 1741 280

c. 2150 d. 2.33 Solutions: a. 𝐻𝐻𝐻𝐻 (ℎ𝑝𝑝) = 3.5 2.33 ℎ𝑝𝑝

𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

b. 𝐻𝐻𝐻𝐻 = 2.33 ℎ𝑝𝑝 ×

ℎ𝑟𝑟

× 250 𝑙𝑙𝑙𝑙𝑙𝑙 ×

746 𝑊𝑊 ℎ𝑝𝑝

5,280 𝑓𝑓𝑓𝑓 1 𝑚𝑚𝑚𝑚

×

1 ℎ𝑟𝑟

60 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

= 77,000

𝑓𝑓𝑓𝑓−𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚

×

ℎ𝑝𝑝−𝑚𝑚𝑚𝑚𝑚𝑚

33,000 𝑓𝑓𝑓𝑓−𝑙𝑙𝑙𝑙

=

= 1741 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤

6. Calculate the amount of energy stored in the capacitor shown in the figure (units of Joules) if the value of the capacitance is 100uF and the voltage is 12 V.

a. 0.0005875 b. 0.0072 c. 0.72 d. 0.59 Solution a. Apply the governing equation as follows: 1

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝐽𝐽) = × 2

100 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 1×106

× (12 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣)2 = 0.0072 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽

7. Determine how long a battery will last (in hours) if one has a 2000 mAh battery and discharges at a constant current of 500 mA?

a. 1 b. 2 c. 3 d. 4 Solutions: a. 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) =

2,000 𝑚𝑚𝑚𝑚ℎ 500 𝑚𝑚𝑚𝑚

= 4 ℎ𝑟𝑟𝑟𝑟

281

8. The new iPhone has a 30-Watt charger. How long will it take for this iPhone to fully charge in hours if the ratings are as follows: voltage is 9V, 1400 mAh energy output unit.

a. 0.5 b. 1.0 c. 1.5 d. 3 Solutions: a. First you need to convert the charging current (30W) into amps as follows: 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) =

30 𝑊𝑊 = 3.3 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 9 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉

b. Similarly, convert the battery capacity from mAh to Ah as follows: 𝐴𝐴ℎ (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 − ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜) = 1400 𝑚𝑚𝑚𝑚ℎ ×

𝐴𝐴ℎ = 1.4 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 − ℎ𝑟𝑟𝑟𝑟 1000 𝑚𝑚𝑚𝑚

c. Then, the charging time is calculated as follows: 𝐶𝐶ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (ℎ𝑟𝑟𝑟𝑟) = 1.4 𝐴𝐴ℎ ×

1 = 0.5 ℎ𝑟𝑟𝑟𝑟 3.3 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 × 0.85

9. The measured voltage across a strained piezoelectric material per cycle varies between 10 to 20 volts with a capacitor rating is 320 micro-farads. Determine the energy produced in Joules per cycle.

a. 0.00006 b. 0.000056 c. 0.0480 d. 0.00213 Solution a. Apply the governing equation as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝐽𝐽 1 320 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐽𝐽 2 2] [ �= × × 20 − 10 = 0.048 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 2 1 × 106 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 282

b. The output energy is indeed small per cycle 10. Determine how long it would take to charge a cell phone battery with a fully-charged load requirement of 20,000 joules (18.96 Btu). The voltage across the capacitor was improved from 5 volts to 15 volts with a single cycle. Assuming that this was done continuously in a day, in how many days will one achieve full charge? The capacitor has a rating of 320 microfarad and 30 volts. One can generate 5 cycles in a second for this device.

a. 1.1 b. 1.5 c. 2.1 d. 2.7 Solutions: a. Apply the governing equation as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝐽𝐽 1 320 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐽𝐽 �= × × [152 − 52 ] = 0.032 6 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 2 1 × 10 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

b. To generate 20,000 Joules, the number of cycles is calculated as follows: 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 20,000 𝐽𝐽 ×

𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 625,000 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 0.032 𝐽𝐽

c. Hence, the time required to charge this device is calculated as follows: 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 625,000 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ×

𝑠𝑠𝑠𝑠𝑠𝑠 1 𝑑𝑑𝑑𝑑𝑑𝑑 × = 1.5 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 5 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 86,400 𝑠𝑠

11. Determine the amount of power (hp) generated by a human being carrying a load of 50 lbs and walking at a rate of 2 ft/s.

a. 0.114 b. 0.182 c. 1.2 283

d. 135 Solutions: a. Power is the product of force and speed 𝐻𝐻𝐻𝐻 = 50 𝑙𝑙𝑙𝑙𝑙𝑙 ×

2 𝑓𝑓𝑓𝑓 ℎ𝑝𝑝 − 𝑚𝑚𝑚𝑚𝑚𝑚 60 𝑠𝑠 × × = 0.182 ℎ𝑝𝑝 𝑠𝑠 33,000𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚

b. The equivalent unit in metric is as follows: 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 0.182 ℎ𝑝𝑝 × c. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) =

746 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 135 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 1 ℎ𝑝𝑝 𝑓𝑓𝑓𝑓 𝑠𝑠

𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 (𝑙𝑙𝑙𝑙𝑙𝑙)×𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 � � 550

12. If an ox can provide a draft of 40 kg [88 lbs], estimate the draft that can be provided by 4 oxen together. Use the revised harnessing factors for worked animals. What is the horsepower generated if the speed is 2 mph? Report the units in hp.

a. 237 b. 4.2 c. 521 d. 1.45 Solution: a. One may use the table for this problem as follows: 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷(𝑘𝑘𝑘𝑘) = 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷(𝑘𝑘𝑘𝑘) =

𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 × # 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 × 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑂𝑂𝑂𝑂

40 𝑘𝑘𝑘𝑘 3.09 ×4 × = 123.6 𝑘𝑘𝑘𝑘 [271.9 𝑙𝑙𝑙𝑙𝑙𝑙] 𝑂𝑂𝑂𝑂 4.0

b. The power generated is calculated as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) = 271.9 𝑙𝑙𝑙𝑙𝑙𝑙 ×

2 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 5,280 𝑓𝑓𝑓𝑓 1ℎ𝑟𝑟 ℎ𝑝𝑝 − 𝑚𝑚𝑚𝑚𝑚𝑚 × × × = 1.45 ℎ𝑝𝑝 ℎ𝑟𝑟 1 𝑚𝑚𝑚𝑚 60 𝑚𝑚𝑚𝑚𝑚𝑚 33,000 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 284

c. The power in metric units is as follows: 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 1.45 ℎ𝑝𝑝 ×

746 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 1081.7 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 1 ℎ𝑝𝑝

13. If an ox can provide a draft of 40 kg [88 lbs], how much draft (in lbs) can be provided by 4 oxen together? Use the revised harnessing factors for worked animals.

a. 237 b. 123.6 c. 521 d. 132 Solution: a. One may use the table for this problem as follows: 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷(𝑘𝑘𝑘𝑘) = 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷(𝑘𝑘𝑘𝑘) =

𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 × # 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 × 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑂𝑂𝑂𝑂

3.09 40 𝑘𝑘𝑘𝑘 ×4 × = 123.6 𝑘𝑘𝑘𝑘 [271.9 𝑙𝑙𝑙𝑙𝑙𝑙] 𝑂𝑂𝑂𝑂 4.0

14. If an ox can provide a draft of 40 kg [88 lbs], estimate the draft that can be provided by 4 oxen together. Use the revised harnessing factors for worked animals. What is the horsepower generated if the speed is 2 mph. Report the units in Watts.

a. 2.37 b. 1081.7 c. 5210. d. 1.45 Solution: a. One may use the table for this problem as follows: 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷(𝑘𝑘𝑘𝑘) =

𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 × # 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 × 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑂𝑂𝑂𝑂 285

𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷(𝑘𝑘𝑘𝑘) =

40 𝑘𝑘𝑘𝑘 3.09 ×4 × = 123.6 𝑘𝑘𝑘𝑘 [271.9 𝑙𝑙𝑙𝑙𝑙𝑙] 𝑂𝑂𝑂𝑂 4.0

b. The power generated is calculated as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) = 271.9 𝑙𝑙𝑙𝑙𝑙𝑙 ×

2 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 5,280 𝑓𝑓𝑓𝑓 1ℎ𝑟𝑟 ℎ𝑝𝑝 − 𝑚𝑚𝑚𝑚𝑚𝑚 × × × = 1.45 ℎ𝑝𝑝 ℎ𝑟𝑟 1 𝑚𝑚𝑚𝑚 60 𝑚𝑚𝑚𝑚𝑚𝑚 33,000 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙

c. The power in metric unit sis as follows: 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 1.45 ℎ𝑝𝑝 ×

746 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 1081.7 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 1 ℎ𝑝𝑝

15. A person is asked to do work using his legs for 6 hours. Estimate the power that can be generated by this person in units of watts.

a. 0.20 b. 147.5 c. 200.5 d. 0.10 Solution a. Using the above equation, we have the following: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) = 0.53 − 0.13 × 𝑙𝑙𝑙𝑙𝑙𝑙(360) = 0.197 ℎ𝑝𝑝 [147.5 𝑊𝑊 ]

b. This power is significantly higher than earlier hp calculations for manual human power.

16. The measured voltage across a strained piezoelectric material was 15 volts with a capacitor rating is 320 micro-farads. Determine the energy produced in Joules.

a. 0.011 b. 0.022 c. 0.123 d. 0.036 Solution 286

a. Apply the governing equation as follows: 1

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝐽𝐽) = × 2

320 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 1×106

× (15 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣)2 = 0.036 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽

17. Determine the charging time (in hrs) for a cell phone using a USB port. The USB 2.0 output rating is 2.5 watts per hour. One has a 21-watt charger.

a. 1.5 b. 3.4 c. 5.5 d. 8.4 Solutions: a. The calculation is simply using some dimensional analysis as follows: 𝐶𝐶ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 (ℎ𝑟𝑟𝑟𝑟) = 21 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 ×

ℎ𝑟𝑟 = 8.4 ℎ𝑟𝑟𝑟𝑟 2.5 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊

b. It will take too long to charge using a USB 2.0 output that is quite low.

18. This component does the same thing as a diode, but it lights up when the current is sufficiently high.

Name this component.

a. Diode b. Capacitor c. Multimeter d. Light emitting diode 19. African bush elephants are the world’s largest land animals (and the strongest mammals), able to carry up to 9,000 kg [19,800 lbs] of load. If this mammal can walk at a speed of 5 mph, determine the horsepower generated in units of watts.

a. 74.6 b. 39.4 287

c. 215 d. 52.8 Solution: a. One may use the table for this problem as follows: 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷(𝑘𝑘𝑘𝑘) = 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷(𝑘𝑘𝑘𝑘) =

𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 × # 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 × 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑂𝑂𝑂𝑂

40 𝑘𝑘𝑘𝑘 3.09 ×4 × = 123.6 𝑘𝑘𝑘𝑘 [271.9 𝑙𝑙𝑙𝑙𝑙𝑙] 4.0 𝑂𝑂𝑂𝑂

b. The power generated is calculated as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (ℎ𝑝𝑝) = 19,800 𝑙𝑙𝑙𝑙𝑙𝑙 × = 52.8 ℎ𝑝𝑝

5 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 5,280 𝑓𝑓𝑓𝑓 1ℎ𝑟𝑟 ℎ𝑝𝑝 − 𝑚𝑚𝑚𝑚𝑚𝑚 × × × ℎ𝑟𝑟 1 𝑚𝑚𝑚𝑚 60 𝑚𝑚𝑚𝑚𝑚𝑚 33,000 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙

c. The power in metric units is as follows: 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 52.8 ℎ𝑝𝑝 ×

746 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 39,388.8 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 [39.4 𝑘𝑘𝑘𝑘] 1 ℎ𝑝𝑝

20. Calculate the amount of charges stored in the capacitor if the value of the capacitance is 100uF and the voltage is 12 V. Determine the amount of energy stored in units of joules.

a. 0.0072 b. 0.000235 c. 0.12 d. 0.235 Solution a. Apply the governing equation as follows: 1

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝐽𝐽) = × 2

100 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 1×106

× (12 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣)2 = 0.0072 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 288

Chapter 13 Cold Fusion and Gravitational Energy 13.12 Solved Problems Problem 13.12.1 Energy Balance in Electrolysis Setup P13.1 Determine the difference between the energy provided to the electrolysis setup as shown in Figure 13.1 from the energy generated by the medium using Equations 13.1 to Equation 13.3.

The following experimental data were measured.

Voltage = 5.201 volts Amperage = 0.256 Amps Duration of Experiment = 30 minutes Rise in temperature = 0.42oC Amount of medium = 1.5 kg Specific heat of medium = 4.2 kJ/kgoC Solution: a. From the input voltage and amperage data the power supplied to the setup using Equation 13.2 is as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊 ) = 𝐸𝐸 × 𝐼𝐼 = 5.201 𝑉𝑉 × 0.256 𝐴𝐴 = 1.33 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊

b. The energy used during the experiment is calculated using Equation 11.3 as shown. 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 (𝑘𝑘𝑘𝑘) = 3.6

𝑘𝑘𝑘𝑘 × 1.33 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 × 0.5 ℎ𝑟𝑟𝑟𝑟 = 2.394 𝑘𝑘𝑘𝑘 𝑊𝑊ℎ

c. Energy absorbed is calculated using Equation 13.1 as shown 𝑄𝑄 (𝑘𝑘𝑘𝑘) = 𝑚𝑚𝐶𝐶𝑝𝑝 ∆𝑡𝑡 = 1.5 𝑘𝑘𝑘𝑘 × 4.2

𝑘𝑘𝑘𝑘 × 0.42𝑜𝑜 𝐶𝐶 = 2.646 𝑘𝑘𝑘𝑘 𝑘𝑘𝑔𝑔𝑜𝑜 𝐶𝐶 289

d. The excess heat is the difference between the heat absorbed by the medium and the heat input to the setup as follows: Energy Gain = Energy Generated - Energy Supplied Energy gain = 2.646 kJ - 2.394 kJ = 0.252 kJ e. Clearly there is an excess in energy absorbed by the medium. This represents about 9.5% excess energy. The value is debatable, considering other factors beyond the control of the scientist or those that were not measured.

Problem 13.12.2 Heat Capacity of Calorimeters P13.2 A 200-Watt heater was used to heat a bomb calorimeter. The temperature rise was reported to be 48oK in 4 minutes. (a) Calculate the heat capacity of the calorimeter in units of J/K, (b) determine also the heat capacity in units of J/K-mol if water is used as medium assuming specific heat of water to be 4.184 kJ/kgoK and (c) determine the number of moles of water used in this experiment.

Solution: a. The energy utilized is first calculated as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽) = 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) × 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 (𝑚𝑚𝑚𝑚𝑚𝑚) × 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽) = 200 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 × 4 𝑚𝑚𝑚𝑚𝑚𝑚 ×

𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 60 𝑠𝑠𝑠𝑠𝑠𝑠 × 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 − 𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚

60 𝑠𝑠𝑠𝑠𝑠𝑠 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 × = 48,000 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 𝑚𝑚𝑚𝑚𝑚𝑚 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 − 𝑠𝑠𝑠𝑠𝑠𝑠

b. The heat capacity is simply the energy used divided by the rise in temperature: 𝐽𝐽

𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 � � = 𝐾𝐾

48,000 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 48𝐾𝐾

= 1,000

𝐽𝐽

𝐾𝐾

c. If water is used as a medium to absorb heat, the following calculations are also correct.

Note that the specific heat of water is 4.184 kJ/kgoK and one mole of water weighs 18 g, so that the heat capacity of water is also equivalent to 75.312 J/K mol as shown below.

290

𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 �

𝐽𝐽 4.184 𝑘𝑘𝑘𝑘 18 𝑔𝑔 1 𝑘𝑘𝑘𝑘 1,000𝐽𝐽 �= × × × = 75.312 𝑜𝑜 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 𝑘𝑘𝑔𝑔 𝐾𝐾 𝑚𝑚𝑚𝑚𝑚𝑚 1,000 𝑔𝑔 𝑘𝑘𝑘𝑘

𝐽𝐽

𝑜𝑜 𝐾𝐾 𝑚𝑚𝑚𝑚𝑚𝑚

d. One can determine the number of moles of water used in the calorimeter as shown below. # 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐻𝐻2 𝑂𝑂 =

𝑜𝑜 1,000 𝐽𝐽 𝐾𝐾 × = 13.3 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻2 𝑂𝑂 𝑜𝑜 𝐾𝐾 75.312 𝐽𝐽

Problem 13.12.3 Heat Released from Combustion of Chemicals P13.3 Benzoic acid is a common material used in the calibration of a calorimeter. The reported heating value was 26.419 kJ/g. A calorimeter with a heat capacity of 1,000 J/K was used to measure the heat of combustion from a gram of benzoic acid with a chemical formula of C7H6O2.

The temperature increase was found to be 26.4oK. Compare the heat released in units of Joules when a gram of benzoic acid is combusted in a calorimeter and the amount of heat released by a gram of this chemical in units of kJ/mol.

Solution: a. The heat released in Joules is calculated:

b.

c.

𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 (𝑘𝑘𝑘𝑘) = 26.4𝑜𝑜 𝐾𝐾 ×

1,000 𝐽𝐽 = 26,400 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 𝑜𝑜 𝐾𝐾

The amount of heat released by a gram of benzoic acid is calculated:

𝑘𝑘𝑘𝑘 26,400 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 � � = × = 26.4 𝑔𝑔 1𝑔𝑔 1,000 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 𝑔𝑔

The molecular weight of sugar is found to be (12 x 7 + 1 x 6 + 16 x 2 = 122 g/mol) 122 g/mol. The heat released by a gram of sugar in units of kJ/mol is calculated as follows: 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 �

𝑘𝑘𝑘𝑘 26.4 𝑘𝑘𝑘𝑘 122 𝑔𝑔 𝑘𝑘𝑘𝑘 �= × = 26.4 𝑚𝑚𝑚𝑚𝑚𝑚 𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚

d. Clearly the reported heating value of benzoic acid is the same as that found using the calorimeter test data.

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Problem 13.12.4 Energy Balance in N-Machine or N-Generator P13.4. The drive motor for the N-machine has the following data: Input motor amperage

=

43.5 amperes

Input motor voltage

=

115 volts

Output drive motor current at no load

=

15 amperes

Voltage output of N generator when loaded

=

1.05 VDC

Current output of the N-generator

=

4,500 amperes

Voltage output of the N-machine at no load

=

1.5 VDC

Current output at no load

=

14,500 Amps

a. Calculate the input energy in watts and in hp. b. Calculate the power output of the N-generator when loaded using the measured voltage and current. Compare this with the measured input power. c. Calculate the power output of the N-generator at no load using the measured voltage and current. Compare this with the measured input power. d. Calculate the input energy in watts and in hp. e. Calculate the power output of the N-generator when loaded using the measured voltage and current. Compare this with the measured input power. f. Calculate the power output of the N-generator at no load using the measured voltage and current. Compare this with the measured input power.

Solution: a. Calculate the input energy in watts and in hp.

The energy input is simply as shown below. 𝑃𝑃𝑖𝑖𝑖𝑖 = 43.5 𝐴𝐴 × 115 𝑉𝑉 = 5,002.5 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 [6.7 ℎ𝑝𝑝] 292

b. Calculate the power output of the N-generator when loaded using the measured voltage and current. Compare this with the measured input power.

When loaded, the power output is as follows: 𝑃𝑃𝑖𝑖𝑖𝑖 = 4,500 𝐴𝐴 × 1.05 𝑉𝑉 = 4,725 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 [6.33 ℎ𝑝𝑝]

Note that the power output is nearly as the input value. c. Calculate the power output of the N-generator at no load using the measured voltage and current. Compare this with the measured input power.

When not loaded, the power output is as follows: 𝑃𝑃𝑖𝑖𝑖𝑖 = 14,500 𝐴𝐴 × 1.5 𝑉𝑉 = 21,750 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 [29.2 ℎ𝑝𝑝]

Note that the power output exceeds power input, a violation of our basic energy conservation law.

d. Calculate the input energy in watts and in hp.

The energy input is simply as shown below. 𝑃𝑃𝑖𝑖𝑖𝑖 = 43.5 𝐴𝐴 × 115 𝑉𝑉 = 5,002.5 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 [6.7 ℎ𝑝𝑝]

e. Calculate the power output of the N-generator when loaded using the measured voltage and current. Compare this with the measured input power.

When loaded, the power output is as follows: 𝑃𝑃𝑖𝑖𝑖𝑖 = 4,500 𝐴𝐴 × 1.05 𝑉𝑉 = 4,725 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 [6.33 ℎ𝑝𝑝] Note that the power output is nearly as the input value. f. Calculate the power output of the N-generator at no load using the measured voltage and current. Compare this with the measured input power.

When not loaded, the power output is as follows:

293

𝑃𝑃𝑖𝑖𝑖𝑖 = 14,500 𝐴𝐴 × 1.5 𝑉𝑉 = 21,750 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 [29.2 ℎ𝑝𝑝]

Note that the power output exceeds power input, a violation of our basic energy conservation law.

Problem 13.12.5 Determining Magnetic Fluxes, Voltages, and Current in Conducting Coils P13.5 Determine the voltage and the current induced in a group of conducting coil with an area of two square meters and a constant magnetic field of five (5) Teslas perpendicular to the surface of the ring. Assume that there is only one turn or loop of the wire. Over the next 4 seconds the magnetic field increased to 10 Teslas. The conducting coil has a resistance of 2 ohms. Determine (a) the change on flux over this period, (b) the voltage and (c) current induced.

Solution: a. The initial and final magnetic flux is calculated as the product of the magnetic field and the area as shown. Φ𝑖𝑖 = 5 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 × 4 𝑚𝑚2 = 20 𝑇𝑇𝑚𝑚2

Φ𝑓𝑓 = 10 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 × 4 𝑚𝑚2 = 40 𝑇𝑇𝑚𝑚2

b. We calculate the voltage induced using the Faraday-Len’s Law. 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 (𝑉𝑉 ) = −𝑁𝑁 ×

(40 − 20)𝑇𝑇𝑚𝑚2 ∆Φ 𝑇𝑇𝑚𝑚2 = 11 × = −5 = −5𝑉𝑉 ∆𝑡𝑡 4 𝑠𝑠 𝑠𝑠

c. Since the voltage and resistance are known, we calculate the current as shown. 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 (𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴) =

5 𝑉𝑉 = 2.5 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 2 𝑂𝑂ℎ𝑚𝑚𝑚𝑚

Problem 13.12.6 Estimating Gravitational Forces at Various Elevation P13.6 Determine the gravitational force between the earth and a person at sea level. Assume the mass of earth is 5.98 x 10 24 kg and that of the person at 100 kg. The distance from the mean sea level and earth center is assumed to be 6.38 x 106 meters. Compare this force when the person is 294

at the top of Mt. Denali in the US which is 6,192 meters from mean sea level. Will the gravitational force increase or decrease with distance?

Solution: a. The solution is simply to substitute values: 𝐹𝐹 =

(6.674 × 10−11 𝑁𝑁𝑚𝑚2 /𝑘𝑘𝑘𝑘) × (5.98 × 1024 𝑘𝑘𝑘𝑘) × (100𝑘𝑘𝑘𝑘) (6.38 × 106 )2 𝑚𝑚

= 980.5 𝑁𝑁

b. If the distance between the masses are increased, the new gravitational force is shown below. 𝐹𝐹 =

(6.674 × 10−11 𝑁𝑁𝑚𝑚2 /𝑘𝑘𝑘𝑘) × (5.98 × 1024 𝑘𝑘𝑘𝑘) × (100𝑘𝑘𝑘𝑘) (6.386192 × 106 )2 𝑚𝑚

= 978.6 𝑁𝑁

c. The gravitational force decreases with increased distance between masses.

Problem 13.12.7 Calculating Acceleration Due to Gravity at Various Elevation P13.7 Determine the value of the acceleration due to gravity for a 75 kg person who is of a distance equal to twice the center of the earth’s distance (or 12.76 x 106 m) or virtually in outer space of the earth.

Solution a. The solution is simply to use Equation 13.5 and relate that with Equation 13.6 to calculate the value of g as shown below. 𝑚𝑚𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 × 𝑔𝑔 = 𝑔𝑔 =

𝐺𝐺 × 𝑚𝑚𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒ℎ × 𝑚𝑚𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 (𝑑𝑑 )2

𝐺𝐺 × 𝑚𝑚𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒ℎ × 𝑚𝑚𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝐺𝐺 × 𝑚𝑚𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒ℎ 6.674 × 10−11 × 5.98 × 1024 = = (𝑑𝑑 )2 × 𝑚𝑚𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 (𝑑𝑑 )2 (12.76 × 106 )2 = 2.45

𝑚𝑚 𝑠𝑠 2

295

b. This is precisely the reason why a person on the moon would weigh less due to lower acceleration due to gravity.

Problem 13.12.8 Calculating Voltages, Current and Magnetic Fluxes in Coils P13.8 Determine the voltage and the current induced in a group of conducting coil having an area of 25 square meters with a constant magnetic field of five (9) Teslas perpendicular to the surface of the ring. Assume that there are 200 turns or loops of the wire. Over the next five minutes, the magnetic field increases to 16 Teslas. The conducting coil has a resistance of 12 ohms. Determine (a) the change on flux over this period, (b) the voltage and (c) current induced.

Solution: a. The initial and final magnetic flux is calculated as the product of the magnetic field and the area: Φ𝑖𝑖 = 9 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 × 25 𝑚𝑚2 = 225 𝑇𝑇 − 𝑚𝑚2

Φ𝑖𝑖 = 16 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 × 25 𝑚𝑚2 = 400 𝑇𝑇 − 𝑚𝑚2

b. We calculate the voltage induced using the Faraday-Len’s Law. 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 (𝑉𝑉 ) = −𝑁𝑁 ×

(400 − 225) 𝑇𝑇𝑚𝑚2 ΔΦ = −200 × = −116.7 𝑉𝑉 300 Δ𝑡𝑡

c. Since the voltage and resistance are known, we calculate the current as shown. 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 (𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴) = −

116.7 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 = 9.725 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 12 𝑂𝑂ℎ𝑚𝑚𝑚𝑚

Problem 13.12.9 Calculating Input and Output Power in an Electric Motor P13.9. Determine the input power and the output power as a percentage or nameplate rated power of an old 3-phase 40 HP motor that has a reported or nameplate full-load efficiency of 90%. The average line voltages measured for each phase was 475 volts, the average amperage on each line was measured at 40 amps and an average power factor of 0.75.

296

Solution: a. The input power in Watts is calculated using the equation in Table 13.1 as follows: 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 (𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = 475 × 40 × 0.75 × 1.73 = 24,652.5 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊

b. The output power may be calculated using the power at rated load and the corresponding power factor as follows: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑎𝑎𝑎𝑎 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = 40𝐻𝐻𝐻𝐻 ×

746 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 1 × = 33,155.6 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 1𝐻𝐻𝐻𝐻 0.9

c. The output power as a % of nameplate rated power is calculated as follows: % 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 =

24,652.5 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊

33,155.6 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊

× 100% = 74.35%

Problem 13.12.10 Improving Power Factor of Resistive Motors P13.10 An electrical motor application has 150 kVA load operating at 70% lagging power factor.

To improve the efficiency of the system, a 77 kVA shunt capacitor was added to act as a reactive element thereby improving the circuit and only taking 30 kVA of power from the source.

Determine the new overall power factor for this set up. Use Figure 13.2 for the calculation.

Solution. a. If one were to draw the power triangle for this application as depicted in Figure 13.2, these values for kVA and kW are found. The original power factor is calculated below.

105 𝑘𝑘𝑘𝑘

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 (1) =

150 𝑘𝑘𝑘𝑘𝑘𝑘

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 (2) =

105 𝑘𝑘𝑘𝑘 = 0.96 109.2 𝑘𝑘𝑘𝑘𝑘𝑘

= 0.70

b. The new power factor is calculated below using the smaller triangle from the figure:

c. The new power factor with the additional shunt capacitor has improved to 96%.

297

13.12.11

Conceptualization of Radiant Energy

P13.11 Who was responsible for the development of the energy-generating device called Radiant Energy? a. Dr. T. Henry Moray b. Professor Shinichi Seike c. Bruce De Palma d. Paramahamsa Tewari 13.12.12

Mass Production Issues of Gravitational Field Energy Devices

P13.12 Professor Shinichi, Seike, Director of Gravity Research Laboratory in Japan, developed a gravitational field energy related device that has not been manufactured or commercialized.

What do you think was the primary reason for this inability to mass-produce the unit? a. No specific material was found to be adequate b. His device works only at very high electrical voltage with corresponding ultra-high alternating current frequencies c. There are no ceramic particles available based on his specifications d. The direct current magnetic field requirement is too low 13.12.13

Inventor of N-Machine or Homopolar Device

P13.13 Who invented the device called N-Machine or Homopolar generator? a. Professor Shinichi Seike b. Dr. T. Henry Moray c. Bruce De Palma d. Paramahamsa Tewari

298

13.12.14

Technical Evaluation of the N-Machine

P13.14 In 1978, a large N-machine was fabricated and tested in Santa Barbara, California following a request from the US Department of Energy. Who performed this test? a. Professor Shinichi Seike b. Bruce De Palma c. Paramahamsa Tewari d. Dr. Robert Kincheloe 13.12.15

Vacuum Spaces Energy Concept

P13.15 This is a general term applied to theories that attempt to explain the concept of tapping into the abundant power available directly from the vacuum of space itself. Name this term. a. Zero-point energy b. Tachyon field energy c. Gyroscope energy d. Negative polarity energy

Assess Your Knowledge 1. An isoperibol bomb calorimeter operates at a constant bath temperature of 25oC. The water-cooling water storage capacity is 5 liters. About 1.651 grams of standard benzoic acid was used as a calibration material with a heat of combustion of 6318 cal/g. This material produced a temperature rise of 3.047oC. There is a correction constant factor (due to heat induced by the fuse and the acid formation) established with a value of 0.070574. What is the energy equivalent of this calorimeter in units of cal/oC?

a. 2416 b. 4184 299

c. 1208 d. 2092 Solutions: a. The energy equivalent in units of cal/oC is calculated as follows using dimensional analysis: 𝑐𝑐𝑐𝑐𝑐𝑐 1.651 𝑔𝑔 × 6318 𝑔𝑔 𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐 𝑊𝑊 � � = × 0.70574 = 2416 𝑜𝑜 𝐶𝐶 3.047 𝐶𝐶 𝐶𝐶

2. Determine the potential energy of a 3 kg mass at a height of 5 meters and acted upon by the Earth’s gravity. Report your units in Joules.

a. 147.15 b. 471.5 c. 541.7 d. 1.47 Solutions: a. The potential energy is simply the product of mass, acceleration and gravitational field energy and calculated as follows: 𝑃𝑃𝑃𝑃 (𝐽𝐽) = 3 𝑘𝑘𝑘𝑘 × 5 𝑚𝑚 × 9.81

𝑚𝑚 𝑘𝑘𝑘𝑘 − 𝑚𝑚2 = 147.15 𝑠𝑠 2 𝑠𝑠 2

b. By newtons law, the following is true:

𝑚𝑚 𝑘𝑘𝑘𝑘 − 𝑚𝑚 𝐹𝐹(𝑁𝑁) = 𝑚𝑚 (𝑘𝑘𝑘𝑘) × 𝑎𝑎 � 2 � = 𝑁𝑁 = 𝑠𝑠 𝑠𝑠 2

c. The unit Joule is simply the product of force in Newtons and distance in meters. Hence, the following is also correct in the required units. 𝑃𝑃𝑃𝑃 (𝐽𝐽) = 147.15

𝑘𝑘𝑘𝑘 − 𝑚𝑚2 𝑁𝑁𝑠𝑠 2 × = 147.15 𝑁𝑁𝑁𝑁 = 147.15 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 𝑠𝑠 2 𝑘𝑘𝑘𝑘 − 𝑚𝑚 300

3. Determine the energy absorbed by the isoperibol bomb calorimeter that operates at a constant bath temperature of 25oC and when 1.651 g of standard and pure benzoic acid was used as a calibration material during combustion. Report the units in kJ. This material has a heat of combustion of 26.4345 kJ/g and produced a temperature rise of 3.047oC. Ignore the heat absorbed by the fuse and the effect of acid formation. Simply use the sensible heat equation. Heat is absorbed by 2-L deionized-distilled water.

a. 15.3 b. 24.5 c. 39.6 d. 40.2 Solution: a. Use the sensible heat equation as follows: 𝑄𝑄 = 𝑚𝑚𝐶𝐶𝑝𝑝 ∆𝑡𝑡 = 2 𝑘𝑘𝑘𝑘 × 4184

𝐽𝐽 × (3.047𝑜𝑜 𝐶𝐶 ) = 24,497.3 𝐽𝐽 [24.5 𝑘𝑘𝑘𝑘] 𝑘𝑘𝑔𝑔𝑜𝑜 𝐶𝐶

4. The acceleration due to gravity of the moon is approximately 1.625 m/s2. Determine the potential energy of a 3-kg mass of material on the moon, about 5 meters above the mean moon surface. Report your answer in Joules.

a. 147.15 b. 457.2 c. 512.9 d. 24.4 Solution: a. The potential energy is simply the product of mass, acceleration and gravitational field energy and calculated as follows:

301

𝑃𝑃𝑃𝑃 (𝐽𝐽) = 3 𝑘𝑘𝑘𝑘 × 5 𝑚𝑚 × 1.625

𝑚𝑚 𝑘𝑘𝑘𝑘 − 𝑚𝑚2 = 24.375 ≈ 24.4 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽 𝑠𝑠 2 𝑠𝑠 2

5. This is a hypothetical particle that always travels faster than light. Physicists believe that faster-thanlight particles cannot exist because they are inconsistent with the known laws of physics. If such particles did exist, they could be used to send signals faster than light. Name this particle.

a. Matter b. Tachyon c. Bradyons d. Luxons 6. What do you call an elementary subatomic particle with no net electric charge?

a. Muon neutrino b. Bradyons c. Luxons d. Tachyon 7. Determine the gravitational force in units of Newtons exerted by a 1.88 m [6 foot, 2 inches] linebacker, weighing 109 kg [240 lbs].

a. 735 b. 829 c. 920 d. 1069 Solutions: a. Simply use the Newtons law as follows: 𝐹𝐹 = 𝑚𝑚𝑚𝑚 = 109 𝑘𝑘𝑘𝑘 × 9.81

𝑚𝑚 = 1069.3 𝑁𝑁 𝑠𝑠 2 302

8. Determine the voltage (volts), current (amps) and power (watts) induced in a group of 4 conducting coils (i.e., 4 turns or 4 loops of the wire), with an area of 5 square meters and a constant magnetic field of 6 Teslas perpendicular to the surface of the ring. Over the next 10 seconds, the magnetic field increased to 15 Teslas. The conducting coils have a resistance of 3 ohms. For this question, what is the change in flux in units of Teslas-m2?

a. 20 b. 30 c. 45 d. 75 Solutions: a. The initial and final magnetic fluxes are first calculated as follows: Φ𝑖𝑖 = 6 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 × 5 𝑚𝑚2 = 30 𝑇𝑇𝑚𝑚2

Φ𝑓𝑓 = 15 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 × 5 𝑚𝑚2 = 75 𝑇𝑇𝑚𝑚2 ∆Φ = 75 − 30 = 45 𝑇𝑇𝑚𝑚2

9. Determine the voltage (volts), current (amps) and power (watts) induced in a group of 4 conducting coils (i.e., 4 turns or 4 loops of the wire), with an area of 5 square meters and a constant magnetic field of 6 Teslas perpendicular to the surface of the ring. Over the next 10 seconds, the magnetic field increased to 15 Teslas. The conducting coils have a resistance of 3 ohms. For this question, what is the voltage induced in units of volts? Please ignore the direction or sign of electrical flow.

a. 18 b. 36 c. 45 d. 75 Solutions: 303

a. The initial and final magnetic fluxes are first calculated as follows: Φ𝑖𝑖 = 6 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 × 5 𝑚𝑚2 = 30 𝑇𝑇𝑚𝑚2

Φ𝑓𝑓 = 15 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 × 5 𝑚𝑚2 = 75 𝑇𝑇𝑚𝑚2 ∆Φ = 75 − 30 = 45 𝑇𝑇𝑚𝑚2

b. The voltage induced is then calculated as follows: 𝑉𝑉 (𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣) = −𝑁𝑁 ×

∆Φ 45 𝑇𝑇𝑚𝑚2 = −4 × = −18 𝑉𝑉 Δ𝑡𝑡 10 𝑠𝑠

10. Determine the voltage (volts), current (amps) and power (watts) induced in a group of 4 conducting coils (i.e., 4 turns or 4 loops of the wire), with an area of 5 square meters and a constant magnetic field of 6 Teslas perpendicular to the surface of the ring. Over the next 10 seconds, the magnetic field increased to 15 Teslas. The conducting coils have a resistance of 3 ohms. For this question, what are the current induced units of amps?

a. 2 b. 4 c. 6 d. 8 Solutions: a. The initial and final magnetic fluxes are first calculated as follows: Φ𝑖𝑖 = 6 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 × 5 𝑚𝑚2 = 30 𝑇𝑇𝑚𝑚2

Φ𝑓𝑓 = 15 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 × 5 𝑚𝑚2 = 75 𝑇𝑇𝑚𝑚2 ∆Φ = 75 − 30 = 45 𝑇𝑇𝑚𝑚2

b. The voltage induced is then calculated as follows: 𝑉𝑉 (𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣) = −𝑁𝑁 ×

∆Φ 45 𝑇𝑇𝑚𝑚2 = −4 × = −18 𝑉𝑉 Δ𝑡𝑡 10 𝑠𝑠

c. The current is calculated as follows:

304

𝐼𝐼 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) =

𝑉𝑉 (𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣) 18 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 =− = 6 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑅𝑅 (𝑜𝑜ℎ𝑚𝑚𝑚𝑚) 3 𝑜𝑜ℎ𝑚𝑚𝑚𝑚

11. An isoperibol bomb calorimeter operates at a constant bath temperature of 25oC. The water-cooling water storage capacity is 5 liters. About 1.651 grams of standard benzoic acid was used as a calibration material with a heat of combustion of 26.4345 kJ/g. This material produced a temperature rise of 3.047oC. There is a constant correction factor (due to heat induced by the fuse and the acid formation) established with a value of 0.070574. What is the energy equivalent of this calorimeter in units of kJ/oC?

a. 2416 b. 4.184 c. 8.960 d. 20.92 Solutions: a. The energy equivalent in units of kJ/oC is calculated as follows using dimensional analysis: 𝑘𝑘𝑘𝑘 1.651 𝑔𝑔 × 26.4345 𝑔𝑔 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑊𝑊 � � = × 0.70574 = 8.96 3.047𝑜𝑜 𝐶𝐶 𝐶𝐶 𝐶𝐶

12. Who coined the word Tachyon?

a. Galileo b. Isaac Newton c. Tesla d. Gerald Feinberg 13. Determine the difference between the energy provided to the electrolysis setup similar to that shown in Figure 13.1 from the energy generated by the medium using Equations 11.1to 11.3. The measured data are as follows: 305

Voltage = 10.402 V Amperage = 0.128 amps Duration of experiment = 1 hr Rise in temperature = 0.50oC Amount of medium = 3 kg [6.6 lbs] Specific heat of medium = 4.2 kJ/kgoC [1.00315 Btu/lboF] a. 2.646 b. 5.292 c. 1.51 d. 6.50 Solutions: a. Power is calculated first as follows: 𝑃𝑃(𝑊𝑊) = 𝐸𝐸𝐸𝐸 = 10.402 𝑉𝑉 × 0.128 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 = 1.33 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊

b. The energy used is calculated as follows:

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑘𝑘𝑘𝑘) = 3.6 × 1.33 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 × 1 ℎ𝑟𝑟 = 4.79 𝑘𝑘𝑘𝑘

c. The energy absorbed is then estimated: 𝑄𝑄(𝑘𝑘𝑘𝑘) = 𝑚𝑚𝐶𝐶𝑝𝑝 ∆𝑡𝑡 = 3 𝑘𝑘𝑘𝑘 × 4.2

𝑘𝑘𝑘𝑘 × 0. 5𝑜𝑜 𝐶𝐶 = 6.3 𝑘𝑘𝑘𝑘 𝑘𝑘𝑔𝑔𝑜𝑜 𝐶𝐶

d. The difference is as follows and is positive. 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 = 6.3 − 4.79 = 1.51 𝑘𝑘𝑘𝑘

14. A small 10-mm diameter permanent magnet produces a field of 100 mT. The field drops away rapidly with distance and is negligible more than 1 mm from the surface. If the magnet moves at a speed of 1 m/s through a 100-turn coil of length 1 mm and a diameter larger than the magnet, what is the EMF induced in units of volts? Please ignore the sign or the direction of the current.

306

a. 0.15 b. 0.25 c. 0.55 d. 0.78 Solutions: a. We can use Faraday’s law of induction to find the induced EMF. This requires us to know the change in flux through the coil and how quickly the change happens. We can start by separately considering the cases when the magnet is outside and inside the coil.

Since we are told the field decays rapidly, we can assume the flux is zero when the magnet is outside the coil. Because the coil is tight and fit around the magnet, we can assume the field is always orthogonal to the coil and the flux is as follows: Φ = 𝐵𝐵𝐵𝐵

b. Because the magnet is known to be moving at 1000 mm/s, we know that it will be inside the 1 mm long coil for just 1/1000s (1 ms). So, applying the Faraday’s law, we have as follows: 𝜀𝜀 = −𝑁𝑁

(100 × 10−3 𝑇𝑇) × 𝜋𝜋 × (5 × 10−3 𝑚𝑚2 ) dΦ = 10 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 × = −0.78 𝑉𝑉 𝑑𝑑𝑑𝑑 1 × 10−3 𝑠𝑠

15. Calculate the magnitude of the induced EMF when a magnet is thrust into the coil, given the following information: the single loop coil has a radius of 6.00 cm and the average value of flux (B cos theta, given since the bar magnetic field is complex) increases from 0.0500 T to 0.250 T in 0.100 s. Ignore sign or direction of current. Report on the units of EMF in millivolts.

a. 22.6 b. 22,600 c. 11.3 307

d. 5,500 Solutions: a. We are given N = 1 and delta T = 0.100 s, but we must determine the change in flux delta theta before we can find EMF. Since the area of the loop is fixed, we see that: ∆∅ = ∆(𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵) = 𝐴𝐴∆(𝐵𝐵 cos 𝜃𝜃 )

b. Now ∆(𝐵𝐵 cos 𝜃𝜃 ) = 0.200𝑇𝑇, since it was given that ∆(𝐵𝐵 cos 𝜃𝜃) changes from 0.0500 to 0.250 T.

c. The area of the loop is calculated as follows: 𝐴𝐴 = 𝜋𝜋𝑟𝑟 2 = (3.1416) × (0.060 𝑚𝑚)2 = 1.13 × 10−2 𝑚𝑚2

d. Hence, the change in flux is calculated as follows: ∆∅ = ∆(𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵) = 1.13 × 10−2 𝑚𝑚2 × (0.200𝑇𝑇)

e. Thus, the EMF is calculated as follows: 𝜀𝜀 = −𝑁𝑁

(1.13 × 10−2 𝑚𝑚2 ) × (0.200𝑇𝑇) dΦ = 1 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 × = 0.0226 𝑉𝑉 = 22.6. 𝑚𝑚𝑚𝑚 𝑑𝑑𝑑𝑑 0.100 𝑠𝑠

16. Determine the input power and the output power as a percentage or nameplate-rated power of a new 3-phase 20 hp [14.92 kW] motor that has a reported or nameplate full-load efficiency of 95%. The average measured line voltage for each phase was 460 volts, and the average amperage on each line was measured at 24.1 amps with an average power factor (PF) of 0.80. For this question, what is the input power in units of watts?

a. 12,500 b. 23800 c. 32400 d. 15,343 Solutions: 308

a. The power is simply calculated as follows: 𝑃𝑃(𝑊𝑊) = 𝐸𝐸 × 𝐼𝐼 × 𝜂𝜂 × 𝑃𝑃𝑃𝑃 = 460 𝑉𝑉 × 24.1 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 × 0.80 × 1.73 = 15,343 𝑊𝑊

17. Determine the input power and the output power as a percentage or nameplate-rated power of a new 3-phase 20 hp [14.92 kW] motor that has a reported or nameplate full-load efficiency of 95%. The average measured line voltage for each phase was 460 volts, and the average amperage on each line was measured at 24.1 amps with an average power factor (PF) of 0.80. For this question, what is the output power in units of watts?

a. 12,500 b. 15,705 c. 32,400 d. 15,343 Solutions: a. The input power is simply calculated as follows: 𝑃𝑃𝑖𝑖 (𝑊𝑊) = 𝐸𝐸 × 𝐼𝐼 × 𝜂𝜂 × 𝑃𝑃𝑃𝑃 = 460 𝑉𝑉 × 24.1 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 × 0.80 × 1.73 = 15,343 𝑊𝑊

b. The output power at rated load is calculated as follows: 𝑃𝑃𝑜𝑜 (𝑊𝑊0 = 20 ℎ𝑝𝑝 ×

746 𝑊𝑊 ℎ𝑝𝑝

×

1

0.95

= 15,705 W

18. Determine the voltage (volts), current (amps) and power (watts) induced in a group of 4 conducting coils (i.e., 4 turns or 4 loops of the wire), with an area of 5 square meters and a constant magnetic field of 6 Teslas perpendicular to the surface of the ring. Over the next 10 seconds, the magnetic field increased to 15 Teslas. The conducting coils have a resistance of 3 ohms. For this question, what is the power induced in units of watts? Likewise, please ignore the direction of the current.

a. 50 b. 65 c. 85 309

d. 108 Solutions: a. The initial and final magnetic fluxes are first calculated as follows: Φ𝑖𝑖 = 6 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 × 5 𝑚𝑚2 = 30 𝑇𝑇𝑚𝑚2

Φ𝑓𝑓 = 15 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 × 5 𝑚𝑚2 = 75 𝑇𝑇𝑚𝑚2 ∆Φ = 75 − 30 = 45 𝑇𝑇𝑚𝑚2

b. The voltage induced is then calculated as follows: 45 𝑇𝑇𝑚𝑚2 ∆Φ 𝑉𝑉 (𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣) = −𝑁𝑁 × = −4 × = −18 𝑉𝑉 Δ𝑡𝑡 10 𝑠𝑠

c. The current is calculated as follows: 𝐼𝐼 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) =

𝑉𝑉 (𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣) 18 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 =− = 6 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑅𝑅 (𝑜𝑜ℎ𝑚𝑚𝑚𝑚) 3 𝑜𝑜ℎ𝑚𝑚𝑚𝑚

d. Power is simply the product of voltage and amperage as follows: 𝑃𝑃(𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊) = 𝑉𝑉 × 𝐼𝐼 = 18 𝑉𝑉 × 6 𝑎𝑎𝑎𝑎𝑎𝑎 = 108 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊

19. Given a bar magnet, how can you induce an electric current in a wire loop?

a. An electric current is induced if a bar magnet is placed near the wire loop. b. An electric current is induced if a wire loop is wound around the bar magnet. c. An electric current is induced if a bar magnet is moved through the wire loop. d. An electric current is induced if a bar magnet is placed in contact with the wire loop.

20. Determine the input power and the output power as a percentage or nameplate-rated power of a new 3-phase 20 hp [14.92 kW] motor that has a reported or nameplate full-load efficiency of 95%. The average measured line voltage for each phase was 460 volts, and the average amperage on each line was measured at 24.1 amps with an average power factor (PF) of 0.80. For this question, what is the output power as a percentage of nameplate-rated power?

a. 80% 310

b. 95% c. 97.7% d. 100% Solutions: a. The input power is simply calculated as follows: 𝑃𝑃𝑖𝑖 (𝑊𝑊) = 𝐸𝐸 × 𝐼𝐼 × 𝜂𝜂 × 𝑃𝑃𝑃𝑃 = 460 𝑉𝑉 × 24.1 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 × 0.80 × 1.73 = 15,343 𝑊𝑊

b. The output power at rated load is calculated as follows: 𝑃𝑃𝑜𝑜 (𝑊𝑊0 = 20 ℎ𝑝𝑝 ×

746 𝑊𝑊 ℎ𝑝𝑝

×

1

0.95

= 15,705 W

c. The output power as a percentage of nameplate-rated power is calculated as follows: % 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 =

15,343 × 100% = 97.7% 15,705

311

Chapter 14 Environmental Issues and Economics of Renewables 14.10 Solved Problems Problem 14.10.1 Area Required for Solar PV Systems P14.1. Determine the reduction in solar collector area for a 1 MW solar PV system if the overall conversion efficiency is increased from 11% to 30%. Assume the average solar radiation in the area is around 5 kWh/m2/day. The average daily solar radiation period is around 12 hrs.

Solution. a. The original area required for a 1 MW solar PV farm is calculated first. 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) =

1𝑀𝑀𝑀𝑀 𝑚𝑚2 − 𝑑𝑑𝑑𝑑𝑑𝑑 12ℎ𝑟𝑟𝑟𝑟 1,000𝑘𝑘𝑘𝑘 ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 × × × × = 2.2 ℎ𝑎𝑎 0.11 5𝑘𝑘𝑘𝑘ℎ 𝑑𝑑𝑑𝑑𝑑𝑑 1𝑀𝑀𝑀𝑀 10,000𝑚𝑚2

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) =

1𝑀𝑀𝑀𝑀 𝑚𝑚2 − 𝑑𝑑𝑑𝑑𝑑𝑑 12ℎ𝑟𝑟𝑟𝑟 1,000𝑘𝑘𝑘𝑘 ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 × × × × = 0.8 ℎ𝑎𝑎 0.30 𝑑𝑑𝑑𝑑𝑑𝑑 1𝑀𝑀𝑀𝑀 10,000𝑚𝑚2 5𝑘𝑘𝑘𝑘ℎ

b. Then with a new conversion efficiency, the required area will be calculated:

c. Hence, the difference in area with improvement in efficiency will be about 1.76 hectares.

The difference in area equates to savings in land cost when one installs a new efficient PV system.

Problem 14.10.2 Algal Oil production and Yield Calculations P14.2 Determine the potential oil production in liters per metric tonne [gallons per ton] for an algae species that has oil content by weight of 30%. Assume an oil density of about 1.2 kg/L [10 lbs/gal]. How much increase in yield will be achieved (L/tonne or Gal/ton) if there is a 20% increase in potential oil yield by weight of algae. Determine also the percentage increase in yield.

Solution: a. A 30% by weight oil in algae means that for every 100 g, there are 30 g of oil. Hence, the oil yield per tonne of material is calculated below using the density of oil.

312

𝑂𝑂𝑂𝑂𝑂𝑂 𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 �

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 0.3 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑜𝑜𝑜𝑜𝑜𝑜 𝐿𝐿 1,000 𝑘𝑘𝑘𝑘 𝐿𝐿 �= × × = 250 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1.2𝑘𝑘𝑘𝑘 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

𝑂𝑂𝑂𝑂𝑂𝑂 𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 �

𝑔𝑔𝑔𝑔𝑔𝑔 0.3 𝑡𝑡𝑡𝑡𝑡𝑡 𝑜𝑜𝑜𝑜𝑜𝑜 𝑔𝑔𝑔𝑔𝑔𝑔 2,000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑔𝑔𝑔𝑔𝑔𝑔 �= × × = 60 𝑡𝑡𝑡𝑡𝑡𝑡 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 10 𝑙𝑙𝑙𝑙𝑙𝑙 𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡

b. A 20% increase in oil yield will provide the following new calculations on yield. 𝑂𝑂𝑂𝑂𝑂𝑂 𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 �

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 0.5 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑜𝑜𝑜𝑜𝑜𝑜 𝐿𝐿 1,000 𝑘𝑘𝑘𝑘 𝐿𝐿 �= × × = 416.7 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1.2𝑘𝑘𝑘𝑘 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

𝑂𝑂𝑂𝑂𝑂𝑂 𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 �

𝑔𝑔𝑔𝑔𝑔𝑔 0.5 𝑡𝑡𝑡𝑡𝑡𝑡 𝑜𝑜𝑜𝑜𝑜𝑜 𝑔𝑔𝑔𝑔𝑔𝑔 2,000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑔𝑔𝑔𝑔𝑔𝑔 �= × × = 100 𝑡𝑡𝑡𝑡𝑡𝑡 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 10 𝑙𝑙𝑙𝑙𝑙𝑙 𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡

c. There is a 40% increase in yield per tonne.

Problem 14.10.3 Size and Cost of PV Systems for Large Commercial Applications P14.3 Determine the size and cost of a PV system for a large commercial application in the US. Assume a monthly average electrical consumption of around 7,650 kWh/month. Assume overall solar PV efficiency of 85% and a climate generation factor of 3. Make an estimate of this cost based on discussions of the above data of around $3.89/Watt.

Solution: a. The average daily load requirement (watts-hours) is first calculated, following the example shown in Chapter 1. 𝐷𝐷𝐷𝐷𝐷𝐷 (𝑊𝑊ℎ𝑟𝑟) =

7,650 𝑘𝑘𝑘𝑘ℎ 1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚ℎ 1,000 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 × × = 255,000 𝑊𝑊 − ℎ𝑟𝑟𝑟𝑟 1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚ℎ 30 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑘𝑘𝑘𝑘

b. The size of solar PV system is then calculated using the conversion efficiency as shown below. 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 �𝑊𝑊𝑝𝑝 � =

𝐷𝐷𝐷𝐷𝐷𝐷 255,000 = = 100,000 𝑊𝑊 𝑜𝑜𝑜𝑜 100 𝑘𝑘𝑘𝑘 𝐶𝐶𝐶𝐶𝐶𝐶 × 𝜀𝜀𝑠𝑠 3 × 0.85

c. The average cost for this household solar PV system will be around $61,000 as shown in the calculation below.

313

𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑃𝑃𝑃𝑃 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 ($) = 100,000 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 ×

$3.89 = $389,000 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊

Problem 14.10.4 Balance of System Cost as Percentage of PV Cost

P14.4. Determine the percentage of cost of a solar PV module compared with other associated costs, including electrical and structural balance of system (BOS) costs. Use Table 14.1 and the GTM 2015 data for the calculation.

Solution. a. The estimation of module cost compared with other costs can be done simply by adding all associated costs excluding the module. The total installed price of the module is equal to $0.75/Watt. The percentage of module cost compared with total cost is calculated as follows: % 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 (%) =

$0.75 × 100% = 21.7% $3.46

b. The solar PV module is just 21.7% of the total installed cost.

Problem 14.10.5 SO2 Daily Emissions Rate for Coal Power Plants P14.5 Calculate the daily emissions rate (ER) of particulates and SO2 from a new 1,000 MW coal-fired power plant which must meet the performance standard of 0.60 lb SO2 per million Btu heat input at approximately 87% scrubber efficiency. Assume that the coal is of semi-anthracite type with a heating value of 17,700 Btu/lb, containing 3.8% sulfur. The plant is 42.5% efficient.

Solution: a. First, one should calculate the heat input at a rate of 40% efficiency. 𝐸𝐸𝑖𝑖𝑖𝑖 =

1,000 𝑀𝑀𝑀𝑀 1,000 𝑘𝑘𝑘𝑘 24 ℎ𝑟𝑟𝑟𝑟 3,412 𝐵𝐵𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵𝐵𝐵 × × × = 1.93 × 1011 0.425 1 𝑀𝑀𝑀𝑀 𝑑𝑑𝑑𝑑𝑑𝑑 𝑘𝑘𝑘𝑘ℎ 𝑑𝑑𝑑𝑑𝑑𝑑

b. The amount of emitted particulates is then estimated:

314

𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 0.038 𝑙𝑙𝑙𝑙𝑙𝑙 𝑆𝑆𝑆𝑆2 1.93 × 1011 𝐵𝐵𝐵𝐵𝐵𝐵 1 𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑆𝑆𝑆𝑆2 � �= × × = 3.7 𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 1 × 106 𝐵𝐵𝐵𝐵𝐵𝐵 2,000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑𝑑𝑑

c. Using the dotted line in Figure 14. 4, the emission rate is about 0.6 lb SO2 per million Btu heat input. This is at about 87% scrubber efficiency. Hence, the emitted SO2 in tons per day will be calculated as follows: 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 0.60 𝑙𝑙𝑙𝑙 𝑆𝑆𝑆𝑆2 1.93 × 1011 𝐵𝐵𝐵𝐵𝐵𝐵 1 𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑆𝑆𝑆𝑆2 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 � �= × × = 57.9 𝑑𝑑𝑑𝑑𝑑𝑑 106 𝐵𝐵𝐵𝐵𝐵𝐵 𝑑𝑑𝑑𝑑𝑑𝑑 2,000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑𝑑𝑑

Problem 14.10.6 SO2 Daily Emissions Rate for Biomass Power Plants

P14.6 Calculate the daily emissions rate (ER) of particulates and SO2 from a new 1,000 MW MSW-fired power plant that must meet the performance standard of 0.60 lb SO2 per million Btu heat input at approximately 70% scrubber efficiency. Assume that the MSW has a heating value of 9,000 Btu/lb [20.9 MJ/kg] and contains 0.5% sulfur. The plant is 42.5% efficient.

Solution: a. First, one should calculate the heat input at a rate of 40% efficiency. 𝐸𝐸𝑖𝑖𝑖𝑖 =

1,000 𝑀𝑀𝑀𝑀 1,000 𝑘𝑘𝑘𝑘 24 ℎ𝑟𝑟𝑟𝑟 3,412 𝐵𝐵𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵𝐵𝐵 × × × = 1.93 × 1011 0.425 1 𝑀𝑀𝑀𝑀 𝑑𝑑𝑑𝑑𝑑𝑑 𝑘𝑘𝑘𝑘ℎ 𝑑𝑑𝑑𝑑𝑑𝑑

b. The amount of emitted particulates is then estimated as follows: 𝑆𝑆𝑆𝑆2 �

𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 0.005 𝑙𝑙𝑙𝑙𝑙𝑙 𝑆𝑆𝑆𝑆2 1.93 × 1011 𝐵𝐵𝐵𝐵𝐵𝐵 1 𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 �= × × = 0.4825 6 𝑑𝑑𝑑𝑑𝑑𝑑 1 × 10 𝐵𝐵𝐵𝐵𝐵𝐵 2,000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑

c. Using the dotted line from Figure 14.4, the emission rate is about 0.35 lb SO2 per million Btu heat input. This is at about 70% scrubber efficiency. Hence, the emitted SO2 in tons per day will be calculated as follows: 𝑆𝑆𝑆𝑆2 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 �

𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 0.35 𝑙𝑙𝑙𝑙 𝑆𝑆𝑆𝑆2 1.93 × 1011 𝐵𝐵𝐵𝐵𝐵𝐵 1 𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 �= × × = 33.8 6 𝑑𝑑𝑑𝑑𝑑𝑑 10 𝐵𝐵𝐵𝐵𝐵𝐵 𝑑𝑑𝑑𝑑𝑑𝑑 2,000 𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑𝑑𝑑

315

d. The SO2 emissions are significantly lower—about 55%—of the emissions of a coal power plant.

Problem 14.10.7 Ozone and SO2 Concentration Units from NAAQS Standards P14.7. a. Determine the 8-hour ozone standard (NAAQS) in units of µg/m3? b. Determine also the concentration of the 1-hr NAAQS standard for sulfur dioxide in units of µg/m3. c. If the temperature of SO2 emissions for the 1-hr standard is raised to 150oC, what is the concentration of SO2 in units of ppm and µg/m3?

Solution: a. First, the molecular weight of ozone is calculated. The value is the sum of its elemental weights (O3 = 16 + 16 + 16 = 48) of 48 g/mol. b. Then, the concentration of the 8-hr ozone standard of 0.070 ppm is calculated using Equation 14.3: 𝐶𝐶𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐶𝐶𝜇𝜇𝜇𝜇/𝑚𝑚3 =

1,000 𝑚𝑚𝑚𝑚 × 0.070 𝑝𝑝𝑝𝑝𝑝𝑝 × 48 𝜇𝜇𝜇𝜇 = 137.4 3 24.45 𝑚𝑚

c. Similarly, to convert units of ppm to ug/m3, one will need the MW of sulfur dioxide [32 + 32 =64] which is 64 and to use Equation 14.3 above. Note than 1 ppb = 1,000 ppm, hence 75 ppb = 0.075 ppm 𝐶𝐶𝜇𝜇𝜇𝜇/𝑚𝑚3 =

1,000 × 0.075 𝑝𝑝𝑝𝑝𝑝𝑝 × 64 𝜇𝜇𝜇𝜇 = 196.3 3 24.45 𝑚𝑚

d. The concentration values will change at different temperatures. The denominator constants will be adjusted accordingly with the temperature of oK = 273+150 = 423oK: 𝑉𝑉 0.08206 × 423𝑜𝑜 𝐾𝐾 = = 34.71 𝑛𝑛 1 𝑎𝑎𝑎𝑎𝑎𝑎

e. The new mass concentration is then recalculated. The ppm concentration will remain at 0.075 ppm, even at 150oC. 316

𝐶𝐶𝜇𝜇𝜇𝜇/𝑚𝑚3 =

1,000 × 0.075 𝑝𝑝𝑝𝑝𝑝𝑝 × 64 𝜇𝜇𝜇𝜇 = 138.3 3 34.71 𝑚𝑚

Problem 14.10.8 Net Energy Ratio (NER) for Biofuels

P14.8 Determine the net energy ratio (NER) for making bioethanol from corn via the wet milling process. Data shows that the energy used to produce ethanol via the wet mill process is as follows: a. Corn production = 5.171 MJ/L b. Corn transport = 0.586 MJ/L c. Ethanol conversion = 14.591 d. Ethanol distribution = 0.414 The heating value of bioethanol produced was reported to be 21.28 MJ/L. Solution: a. The first step is to calculate the total energy required to produce the fuel per liter. 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑡𝑡𝑡𝑡 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 = 5.171 + 0.586 + 14.591 + 0.414 = 20.762 𝑀𝑀𝑀𝑀

b. The NER is calculated based from Equation 14.2 as follows: 𝑁𝑁𝑁𝑁𝑁𝑁 =

21.28𝑀𝑀𝑀𝑀 = 1.02 20.762𝑀𝑀𝑀𝑀

c. The NER of making bioethanol via the wet milling process was estimated to be 1.02, a positive number, and we can conclude that there is more energy from the fuel produced than all the energy associated with making the fuel, barely qualifying this as a sustainable process.

317

Problem 14.10.9 Net Energy Balance (NEB) for Biofuels P14.9 Determine the net energy balance (NEB) for making bioethanol from corn via wet milling with co-products allocation or recovery. Data shows that the energy used to produce the ethanol via the wet mill process (with recovery of co-products) is as follows: a. Corn production = 4.188 MJ/L b. Corn transport = 0.474 MJ/L c. Ethanol conversion = 11.819 d. Ethanol distribution = 0.414 The heating value of bioethanol produced was reported to be 21.28 MJ/L. Solution: a. The first step is to calculate the total energy required to produce fuel per liter. 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑡𝑡𝑡𝑡 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 = 4.188 + 0.474 + 11.819 + 0.414 = 16.895 𝑀𝑀𝑀𝑀

b. The NEB is calculated based on Equation 14.3:

𝑁𝑁𝑁𝑁𝑁𝑁 = 21.28 − 16.895 = 4.385

c. The NEB of making bioethanol via the wet milling process (with co-products recovery) was estimated to be 4.385, a positive number, though slightly lower than that of the dry mill process. Still, we can deem the process as sustainable as there is more energy from the fuel produced than all the energy associated with making the fuel.

Problem 14.10.10 Return on Investment for the Production Cost of Solar PV Systems P14.10 Determine the number of years it will take for a multi-crystalline solar cell to recover the energy used for its production. The data shows that it will take 3,700 MJ of energy to produce a square meter of multi-crystalline PV cell. The overall conversion efficiency from solar energy to

318

electrical power was reported to be 14%. The average solar energy per square meter on the site per day is around 5 kWh. Solution: a. The first step is to determine the solar energy output per day per square meter using the insolation given: 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 (𝑀𝑀𝑀𝑀) = 5 𝑘𝑘𝑘𝑘ℎ ×

3.6𝑀𝑀𝑀𝑀 × 0.14 = 2.52 𝑀𝑀𝑀𝑀 𝑘𝑘𝑘𝑘ℎ

b. The number of years required to recover the energy used to produce the solar cell is simply calculated: 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑜𝑜𝑜𝑜 𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 = 3,700𝑀𝑀𝑀𝑀 ×

𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 × = 4 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 2.52𝑀𝑀𝑀𝑀 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

c. It will take 4 years to recover the energy used to produce this solar PV cell.

14.10.11

Hydroelectric Plants Powering US Power Needs

P14.11 Calculate the number of 1,300 MW hydroelectric plant that would satisfy 100% of the US end-use power of 1.8 TW in 2030. Assume that the hydroelectric plant is available 40% of the time in a year. a) 884 b) 88,400 c) 3,458 d) 34,580 Solutions: a. About 100% of the US end use demand is calculated as follows: 𝑈𝑈𝑈𝑈 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 1.8 𝑇𝑇𝑇𝑇 ×

1 × 109 𝑘𝑘𝑘𝑘 × 8,760 = 1.577 × 1013 𝑘𝑘𝑘𝑘ℎ 1 𝑇𝑇𝑇𝑇

b. The output of each 1,300-MW hydroelectric plant is calculated as follows: 319

𝐸𝐸𝑂𝑂 (𝑘𝑘𝑘𝑘ℎ) = 1,300 𝑀𝑀𝑀𝑀 × �

1,000 𝑘𝑘𝑘𝑘 � × 8,760 ℎ𝑟𝑟𝑟𝑟 × 0.40 = 4.56 × 109 𝑘𝑘𝑘𝑘ℎ 1 𝑀𝑀𝑀𝑀

e. Hence, the number of 1,300 MW hydroelectric plant is calculated as follows:

14.10.12

# 𝑜𝑜𝑜𝑜 1,300 𝑀𝑀𝑀𝑀 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 =

1.577 × 1013 = 3,458 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 4.56 × 109

Tidal Energy Powering US Electricity Requirements

P14.12 A tidal turbine is rated at 1000 MW. Calculate the number of this tidal turbine required to satisfy 1% of the US total end-use power of 1.8 TW in 2030. This unit is only available 0.237% of the time within the year. a) 6200 b) 490,000 c) 590,000 d) 7,596 Solutions: a. About 1% of the US end use demand is calculated as follows: 1 × 109 𝑘𝑘𝑘𝑘 𝑈𝑈𝑈𝑈 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 1.8 𝑇𝑇𝑇𝑇 × × 8,760 × 0.01 = 1.577 × 1011 𝑘𝑘𝑘𝑘ℎ 1 𝑇𝑇𝑇𝑇

b. The output of each 1-MW tidal plant is calculated as follows: 𝐸𝐸𝑂𝑂 (𝑘𝑘𝑘𝑘ℎ) = 1000 𝑀𝑀𝑀𝑀 × �

1,000 𝑘𝑘𝑘𝑘 � × 8,760 ℎ𝑟𝑟𝑟𝑟 × 0.00237 = 2.076 × 107 𝑘𝑘𝑘𝑘ℎ 1 𝑀𝑀𝑀𝑀

e. Hence, the number of 1000 MW tidal plant is calculated as follows:

1.577 × 1011 # 𝑜𝑜𝑜𝑜 1000 𝑀𝑀𝑀𝑀 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = = 7,596 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 2.076 × 107

320

14.10.13

Capacity Factor for Offshore Wind Turbines

P14.13 Calculate the capacity factor for a GE Haliade-X 13 MW offshore wind turbine with a 220-m rotor diameter operating at an annual wind speed (Vm) of 12 m/s. a) 0.775 b) 0.425 c) 0.294 d) 0.360 Solutions: a. The capacity factor for 12 m/s average windspeed is as follows: 𝑚𝑚

13000 𝑘𝑘𝑘𝑘

𝐶𝐶𝐶𝐶 = �0.087 × 12 � − � (220

14.10.14

𝑠𝑠

𝑚𝑚)2

� = 0.7754

Inland Wind Turbine Capacity Factor

P14.14 Calculate the capacity factor for a GE Haliade-X 13 MW wind turbine with a 220-m rotor diameter operating at an annual wind speed (Vm) of 8.5 m/s. a) 0.747 b) 0.471 c) 0.294 d) 0.360 Solutions: a. The capacity factor for the higher average wind speed is as follows: 𝐶𝐶𝐶𝐶 = �0.087 × 8.5

𝑚𝑚 13000 𝑘𝑘𝑘𝑘 �−� � = 0.471 (220 𝑚𝑚)2 𝑠𝑠

321

14.10.15

Wind Turbines Powering US Grid

P14.15 Calculate the number of wind turbines needed to satisfy 100% of the US end-use power of 1.8 TW in 2030 from the wind turbine alone. The average wind speed is 12 m/s. The wind turbine size is 13 MW with 220 m diameter. Assume the system efficiency due to array, transmission, and distribution losses to be about 85%. a) 590,000 b) 1,180,000 c) 210,000 d) 110,000 Solutions: a. The capacity factor for the average windspeed is as follows: 𝐶𝐶𝐶𝐶 = �0.087 × 12

𝑚𝑚 13,000 𝑘𝑘𝑘𝑘 �−� � = 0.7754 (220 𝑚𝑚)2 𝑠𝑠

b. Using the wind energy equation, we have the following calculations: 𝐸𝐸𝑡𝑡 = 13,000 × 0.7754 × 8,760 × 0.85 = 7.51 × 107 𝑘𝑘𝑘𝑘ℎ

c. The US end use demand in units of kWh is calculated as follows: 𝑈𝑈𝑈𝑈 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 1.8 𝑇𝑇𝑇𝑇 ×

1 × 109 𝑘𝑘𝑘𝑘 × 8,760 = 1.577 × 1013 𝑘𝑘𝑘𝑘ℎ 1 𝑇𝑇𝑇𝑇

d. Hence, the number of turbines is calculated as follows: # 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 =

(1.577 × 1013 ) = 210,000 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 7.51 × 107

𝐶𝐶𝐶𝐶 = �0.087 × 10.5

𝑚𝑚 13,000 𝑘𝑘𝑘𝑘 �−� � = 0.645 (220 𝑚𝑚)2 𝑠𝑠

e. The capacity factor for the average windspeed is as follows:

f. Using the wind energy equation, we have the following calculations: 322

𝐸𝐸𝑡𝑡 = 13,000 × 0.645 × 8,760 × 0.90 = 6.611 × 107 𝑘𝑘𝑘𝑘ℎ

g. The US end use demand in units of kWh is calculated as follows:

1 × 109 𝑘𝑘𝑘𝑘 𝑈𝑈𝑈𝑈 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 1.8 𝑇𝑇𝑇𝑇 × × 8,760 = 1.577 × 1013 𝑘𝑘𝑘𝑘ℎ 1 𝑇𝑇𝑇𝑇

h. Hence, the number of turbines is calculated as follows: # 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 =

(1.577 × 1013 ) = 238,542 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 6.611 × 107

i. In reality, and considering the capacity factor, this would be a better conservative estimate. The problem, however, did not specify the capacity factor.

Assess Your Knowledge 1. Fill in the blanks. One of the goals of circular economy is to keep materials, products, and services __________ for as long as possible.

a) At low cost b) At highest value c) In circulation d) Wasted 2. In circular economy modelling and analysis, what is meant by LIBRA?

a) Well balanced economy b) Make things orderly and aesthetically pleasing c) Lithium-Ion Battery Resources Analysis d) Learning and Industry-Based Reuse Adoption 3. In circular economy modeling and analysis, what is meant by MFI?

a) Made for Industries b) Money Flow Index 323

c) Materials and Fuels for Industries d) Materials Flow Through Industry Tool 4. Fill in the blanks. The goals of a circular economy are both ________ and _______.

a) Disruptive and economic b) Restorative and regenerative c) Rounded and aesthetic d) Social and economic 5. In circular economy modeling and analysis, what is the meaning of CELAVI?

a) Such is life b) Circular Economy Life Cycle Assessment and Visualization Framework c) Something is good, something is bad, and you have no control of it d) Simple, smart, and essential product 6. This is a circular economy modeling and analysis tool that models how interactions between a system’s actors might help maximize circularity.

a) Circular Economy Life Cycle Assessment and Visualization Framework b) Circular Economy Agent-based Model c) Photovoltaics in the Circular Economy d) Materials Flow Through Industry Tool 7. This is one of the goals of a circular economy.

a) Keep materials, products, and services in circulation for as long as possible b) It is a linear approach from beginning to end c) It is disruptive and iterative d) Aims for improved and rounded economy

324

8. What is the meaning of the acronym BEIOM?

a) Bio-Based Circular Carbon Economy Environmentally Extended Input/Output Model b) Biomass Energy for Industrial Operation and Management c) Bio-Economy in Operations Management d) Biomass-Based Economy for Industry and Orderly Management 9. In circular economy modeling and analysis, what is meant by PVICE?

a) Photovoltaics in the Circular Economy b) Photovoltaics for Internal Combustion Engines c) Product Visualization for Industry and Clean Engines d) Public Vehicles for Integrated Combustion and Exhaust 10. Calculate the capacity factor for a GE 2-MW wind turbine with a 116-m rotor diameter operating at an annual wind speed (Vm) of 10.3 m/s.

a) 0.747 b) 0.425 c) 0.294 d) 0.360 Solutions: a. The capacity factor for lower average windspeed is as follows: 𝐶𝐶𝐶𝐶 = �0.087 × 10.3

𝑚𝑚 2000 𝑘𝑘𝑘𝑘 �−� � = 0.747 (116 𝑚𝑚)2 𝑠𝑠

b. The capacity factor for the higher average wind speed is as follows: 𝑚𝑚

2000 𝑘𝑘𝑘𝑘

𝐶𝐶𝐶𝐶 = �0.087 × 8.5 � − �(116 𝑠𝑠

𝑚𝑚)2

� = 0.591

c. During the year, the turbine will produce about 74.7% to 59.1% of its maximum possible energy or an average of 66.8% 325

11. Calculate the number of wind turbines needed to satisfy 100% of the US end-use power of 1.8 TW in 2030 from the wind turbine alone. The average wind speed is 10.3 m/s. The wind turbine size is 3 MW with 130 m diameter. Assume the system efficiency due to array, transmission, and distribution losses to be about 85%.

a) 590,000 b) 1,180,000 c) 982,430 d) 110,000 Solutions: a. The capacity factor for the average windspeed is as follows: 𝐶𝐶𝐶𝐶 = �0.087 × 10.3

𝑚𝑚 3000 𝑘𝑘𝑘𝑘 �−� � = 0.7186 (130 𝑚𝑚)2 𝑠𝑠

b. Using the wind energy equation, we have the following calculations: 𝐸𝐸𝑡𝑡 = 3,000 × 0.7186 × 8,760 × 0.85 = 1.605 × 107 𝑘𝑘𝑘𝑘ℎ

c. The US end use demand in units of kWh is calculated as follows: 𝑈𝑈𝑈𝑈 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 1.8 𝑇𝑇𝑇𝑇 ×

1 × 109 𝑘𝑘𝑘𝑘 × 8,760 = 1.577 × 1013 𝑘𝑘𝑘𝑘ℎ 1 𝑇𝑇𝑇𝑇

d. Hence, the number of turbines is calculated as follows:

(1.577 × 1013 ) # 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = = 982,430 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 1.605 × 107

12. Calculate the number of wind turbines needed to satisfy 100% of the US end-use power of 1.8 TW in 2030 from the wind turbine alone. The average wind speed is 9.5 m/s. The wind turbine size is 5 MW with 130 m diameter. Assume the system efficiency due to array, transmission, and distribution losses to be about 90%.

a) 590,000

326

b) 1,180,000 c) 982,430 d) 753,464 Solutions: a. The capacity factor for the average windspeed is as follows: 𝐶𝐶𝐶𝐶 = �0.087 × 10.3

𝑚𝑚 3000 𝑘𝑘𝑘𝑘 �−� � = 0.7186 (130 𝑚𝑚)2 𝑠𝑠

b. Using the wind energy equation, we have the following calculations: 𝐸𝐸𝑡𝑡 = 3,000 × 0.7186 × 8,760 × 0.85 = 1.605 × 107 𝑘𝑘𝑘𝑘ℎ

c. The US end use demand in units of kWh is calculated as follows:

1 × 109 𝑘𝑘𝑘𝑘 𝑈𝑈𝑈𝑈 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 1.8 𝑇𝑇𝑇𝑇 × × 8,760 = 1.577 × 1013 𝑘𝑘𝑘𝑘ℎ 1 𝑇𝑇𝑇𝑇

d. Hence, the number of turbines is calculated as follows: # 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 =

(1.577 × 1013 ) = 982,430 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 1.605 × 107

e. The capacity factor for the average windspeed is as follows: 𝐶𝐶𝐶𝐶 = �0.087 × 9.5

𝑚𝑚 5000 𝑘𝑘𝑘𝑘 �−� � = 0.531 (130 𝑚𝑚)2 𝑠𝑠

f. Using the wind energy equation, we have the following calculations: 𝐸𝐸𝑡𝑡 = 5,000 × 0.531 × 8,760 × 0.90 = 2.093 × 107 𝑘𝑘𝑘𝑘ℎ

g. The US end use demand in units of kWh is calculated as follows:

1 × 109 𝑘𝑘𝑘𝑘 𝑈𝑈𝑈𝑈 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 1.8 𝑇𝑇𝑇𝑇 × × 8,760 = 1.577 × 1013 𝑘𝑘𝑘𝑘ℎ 1 𝑇𝑇𝑇𝑇

h. Hence, the number of turbines is calculated as follows: # 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 =

(1.577 × 1013 ) = 753,464 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 2.093 × 107 327

13. Calculate the capacity factor for a GE 2-MW wind turbine with a 116-m rotor diameter operating at an annual wind speed (Vm) of 8.5 m/s.

a) 0.747 b) 0.591 c) 0.294 d) 0.360 Solutions: a. The capacity factor for lower average windspeed is as follows: 𝐶𝐶𝐶𝐶 = �0.087 × 10.3

𝑚𝑚 2000 𝑘𝑘𝑘𝑘 �−� � = 0.747 (116 𝑚𝑚)2 𝑠𝑠

b. The capacity factor for the higher average wind speed is as follows: 𝑚𝑚

2000 𝑘𝑘𝑘𝑘

𝐶𝐶𝐶𝐶 = �0.087 × 8.5 � − �(116 𝑠𝑠

𝑚𝑚)2

� = 0.591

c. During the year, the turbine will produce about 74.7% to 59.1% of its maximum possible energy or an average of 66.8% 14. Calculate the number of 100-MW geothermal plant that would satisfy 100% of the US end-use power of 1.8 TW in 2030. Assume that the geothermal plant is available 86% of the time in a year.

a) 2,094 b) 20,940 c) 884 d) 88,400 Solutions: a. The output of each 100-MW geothermal plant is calculated as follows: 𝐸𝐸𝑂𝑂 (𝑘𝑘𝑘𝑘ℎ) = 100 𝑀𝑀𝑀𝑀 × �

1,000 𝑘𝑘𝑘𝑘 � × 8,760 ℎ𝑟𝑟𝑟𝑟 × 0.86 = 7.53 × 108 𝑘𝑘𝑘𝑘ℎ 1 𝑀𝑀𝑀𝑀 328

b. About 100% of the US end use demand is calculated as follows: 1 × 109 𝑘𝑘𝑘𝑘 𝑈𝑈𝑈𝑈 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 1.8 𝑇𝑇𝑇𝑇 × × 8,760 = 1.577 × 1013 𝑘𝑘𝑘𝑘ℎ 1 𝑇𝑇𝑇𝑇

c. Hence, the number of 100-MW geothermal plant is calculated as follows:

15.

# 𝑜𝑜𝑜𝑜 100 − 𝑀𝑀𝑀𝑀 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 =

1.577 × 1013 = 20,940 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 7.53 × 108

A tidal turbine is rated at 1225 MW. Calculate the number of this tidal turbine required to satisfy

1% of the US total end-use power of 1.8 TW in 2030. This unit is only available 0.237% of the time within the year.

a) 7,596 b) 490,000 c) 590,000 d) 6,200 Solutions: a. About 1% of the US end use demand is calculated as follows: 𝑈𝑈𝑈𝑈 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 1.8 𝑇𝑇𝑇𝑇 ×

1 × 109 𝑘𝑘𝑘𝑘 × 8,760 × 0.01 = 1.577 × 1011 𝑘𝑘𝑘𝑘ℎ 1 𝑇𝑇𝑇𝑇

b. The output of each 1-MW tidal plant is calculated as follows: 𝐸𝐸𝑂𝑂 (𝑘𝑘𝑘𝑘ℎ) = 1225 𝑀𝑀𝑀𝑀 × �

1,000 𝑘𝑘𝑘𝑘 � × 8,760 ℎ𝑟𝑟𝑟𝑟 × 0.00237 = 2.54 × 107 𝑘𝑘𝑘𝑘ℎ 1 𝑀𝑀𝑀𝑀

e. Hence, the number of 1000 MW tidal plant is calculated as follows: # 𝑜𝑜𝑜𝑜 1000 𝑀𝑀𝑀𝑀 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 =

1.577 × 1011 = 6,200 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 2.54 × 107

16. Calculate the capacity factor for a GE Haliade-X 13 MW offshore wind turbine with a 220-m rotor diameter operating at an annual wind speed (Vm) of 8 m/s.

a) 0.427 329

b) 0.775 c) 0.294 d) 0.360 Solutions: a. The capacity factor for 18 m/s average windspeed is as follows: 𝐶𝐶𝐶𝐶 = �0.087 × 8

𝑚𝑚 13000 𝑘𝑘𝑘𝑘 �−� � = 0.427 (220 𝑚𝑚)2 𝑠𝑠

17. Calculate the number of 510 MW hydroelectric plants that would satisfy 100% of the US end-use power of 1.8 TW in 2030. Assume that the hydroelectric plant is available 40% of the time in a year.

a) 884 b) 88,400 c) 8,825 d) 34,580 Solutions: a. About 100% of the US end use demand is calculated as follows: 𝑈𝑈𝑈𝑈 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 1.8 𝑇𝑇𝑇𝑇 ×

1 × 109 𝑘𝑘𝑘𝑘 × 8,760 = 1.577 × 1013 𝑘𝑘𝑘𝑘ℎ 1 𝑇𝑇𝑇𝑇

b. The output of each 510-MW hydroelectric plant is calculated as follows: 𝐸𝐸𝑂𝑂 (𝑘𝑘𝑘𝑘ℎ) = 510 𝑀𝑀𝑀𝑀 × �

1,000 𝑘𝑘𝑘𝑘 � × 8,760 ℎ𝑟𝑟𝑟𝑟 × 0.40 = 1.787 × 109 𝑘𝑘𝑘𝑘ℎ 1 𝑀𝑀𝑀𝑀

e. Hence, the number of 1,300 MW hydroelectric plant is calculated as follows: # 𝑜𝑜𝑜𝑜 1,300 𝑀𝑀𝑀𝑀 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 =

1.577 × 1013 = 8,825 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 1.787 × 109

18. Calculate the number of wind turbines needed to satisfy 100% of the US end-use power of 1.8 TW in 2030 from the wind turbine alone. The average wind speed is 10 m/s. The wind turbine size is 13 MW

330

with 220 m diameter. Assume the system efficiency due to array, transmission, and distribution losses to be about 85%.

a) 590,000 b) 1,180,000 c) 271,000 d) 110,000 Solutions: a. The capacity factor for the average windspeed is as follows: 𝐶𝐶𝐶𝐶 = �0.087 × 10

𝑚𝑚 13,000 𝑘𝑘𝑘𝑘 �−� � = 0.601 (220 𝑚𝑚)2 𝑠𝑠

b. Using the wind energy equation, we have the following calculations: 𝐸𝐸𝑡𝑡 = 13,000 × 0.601 × 8,760 × 0.85 = 5.82 × 107 𝑘𝑘𝑘𝑘ℎ

c. The US end use demand in units of kWh is calculated as follows: 𝑈𝑈𝑈𝑈 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 1.8 𝑇𝑇𝑇𝑇 ×

1 × 109 𝑘𝑘𝑘𝑘 × 8,760 = 1.577 × 1013 𝑘𝑘𝑘𝑘ℎ 1 𝑇𝑇𝑇𝑇

d. Hence, the number of turbines is calculated as follows: # 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 =

(1.577 × 1013 ) = 271,000 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 5.82 × 107

19. Calculate the capacity factor for a GE Haliade-X 13 MW wind turbine with a 220-m rotor diameter operating at an annual wind speed (Vm) of 8.5 m/s.

a) 0.747 b) 0.601 c) 0.294 d) 0.360 Solutions: 331

a. The capacity factor for 10 m/s average windspeed is as follows: 𝐶𝐶𝐶𝐶 = �0.087 × 10

𝑚𝑚 13000 𝑘𝑘𝑘𝑘 �−� � = 0.601 (220 𝑚𝑚)2 𝑠𝑠

20. Calculate the number of 300-MW concentrated solar power (CSP) systems that would satisfy 100% of the US end-use power of 1.8 TW in 2030. Assume that the CSP plant has an average yearly availability of 10%.

a) 60,000 b) 40,000 c) 20,000 d) 80,000 Solutions: a. The output of the plant is calculated as follows: 𝐸𝐸𝑂𝑂 (𝑘𝑘𝑘𝑘ℎ) = 300 𝑀𝑀𝑀𝑀 × �

1,000 𝑘𝑘𝑘𝑘 � × 8,760 ℎ𝑟𝑟𝑟𝑟 × 0.10 = 2.628 × 108 𝑘𝑘𝑘𝑘ℎ 1 𝑀𝑀𝑀𝑀

b. About 100% of the US end use demand is calculated as follows: 𝑈𝑈𝑈𝑈 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 1.8 𝑇𝑇𝑇𝑇 ×

1 × 109 𝑘𝑘𝑘𝑘 × 8,760 = 1.5768 × 1013 𝑘𝑘𝑘𝑘ℎ 1 𝑇𝑇𝑇𝑇

c. Hence, the number of CSP systems of 300 MW per plant is calculated as follows: 1.5768×1013

# 𝑜𝑜𝑜𝑜 300 𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶𝐶𝐶 = 2.628×108 = 60,000 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

332