Sustainable energy edition 1st edition dunlap solutions manual

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Sustainable Energy - Chapter 8: Direct Use of Solar Energy

Chapter 8 Direct Use of Solar Energy Problem 8.1 Calculate the power radiated by a woodstove of dimensions 65 cm high by 55 cm deep by 85 cm wide with a surface temperature of 120°C. Assume that heat is radiated from all surfaces of the stove and that the stove has an emissivity of 1. Note that woodstoves are painted black because black surfaces have high absorptance, and objects with high absorptance also have high emissivity.

Solution The surface area in m2 is (from sides/front/back/top/bottom) (2×0.65 m ×0.55 m) + (2×0.65 m ×0.85 m) + (2×0.55 m ×0.85 m) = 2.755 m2 The surface temperature in K is 120°C + 273 = 393 K. The surface radiation is given by the Stefan-Boltzmann law as P = AεσT4 where ε = 1 and σ = 5.67×10-8 W m-2 K-4. Thus P = (2.755 m2)×(1.0)×(5.67×10-8 W∙m-2∙K-4)×(393 K)4 = 3.7 kW Problem 8.2 Locate information about the current cost of home heating oil or natural gas in your area (whichever is in common use) and the cost of residential electricity.

Assuming an efficiency of 85% for an oil furnace and 100% for electric heat, calculate the relative cost of electric heat compared with oil heat or natural gas if both heating systems require the same net energy to heat a house.

Solution In Halifax, Nova Scotia on 15 Sept 2011, the cost of oil is $1.08 CDN per liter and the cost of electricity is $0.139 per kWh [both including taxes]. The energy content of oil (from Appendix IV) is 38.5 MJ/L; this is for crude oil but home heating oil (basically the same as diesel fuel) is similar. Thus the cost of oil per MJ is

($1.08/L)/(38.5 MJ/L) = $0.028/MJ

At 85% efficiency, this means that the cost per MJ of heat is

($0.028/MJ)/(0.85) = $0.033/MJ

45 ©2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Sustainable Energy - Chapter 8: Direct Use of Solar Energy For electricity (at 100% efficiency) the cost per MJ of heat is ($0.139/kWh)×(0.278 kWh/MJ) = $0.039/MJ Electricity is slightly higher at this time in this location.

Problem 8.3 Consider a vertical south facing window in a house at 40ºN latitude. For an interior temperature of 20ºC make a plot of the minimum R-value as a function of outside temperature from -30ºC to 10ºC for which the passive solar heating exceeds the heat loss though the window.

Solution The heat loss through the window due to conduction per unit area per hour will be (equation (8.5)) Q T  A R where the value of R is in units of (s∙m2∙ºC)/J. The insolation per unit area per day in the winter is estimated to be (from Table 8.2) I ≈ 107 J/(m2∙d) or (107 J/(m2∙d))/(86,400 s/d) = 116 W/m2.

Equating this to the heat loss and solving for the necessary R-value gives R = ΔT/I The range of ΔT of interest (corresponding to outside temperatures of -30°C to 10°C) is ΔT = (20°C – 10°C)) = 10°C to ΔT = (20°C – (-30°C)) = 50°C Values of R calculated as above as a function of outside temperature are shown in the figure below.

0.5 0.45 0.4

R-value

0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 -30

-20

-10

0

10

outside temperature (Celsius)

46 ©2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Sustainable Energy - Chapter 8: Direct Use of Solar Energy Problem 8.4 Compare the total solar energy received at the surface of the earth in one year to the total annual global energy requirements.

Solution An average of 50% of the power given in equation (8.2) arrives at the surface of the earth. This is (0.5)×(1.73×1017 W) = 8.65×1016 W The total energy received over one year is given by (8.65×1016 W)×(3.15×107 s/y) = 2.7×1024 J From the Preface, the total annual energy used by humans at present is 5.7×1020 J. Thus the energy of sunlight incident on the earth's surface is (2.7×1024J)/(5.7×1020J) = 4700 times as much as we use.

Problem 8.5 Compare the R-values of: a. Two pieces of 3-mm thick glass in thermal contact. b. Two pieces of 3-mm thick glass with a 1-cm air space between them.

W  cm and for air it is m 2  C W  cm W  cm . Since R is defined as R = l/k where l is in cm and k is in 2 , then a 2.3 2 m  C m  C 0.3 3 mm (0.3 cm) piece of glass will have an R-value of R   0.005 and a 1-cm layer 59 1.0 of air will have an R-value of R   0.435 . Since Rtotal   R then 2.3

Solution The thermal conductivity of glass is given as 59

for (a) Rtotal=0.005+0.005=0.01 and for (b) Rtotal=0.005+0.435+0.005=0.445.

This emphasizes the importance of the air space in insulating properties.

Problem 8.6 Approximate a house as a cube with an edge length of 7 m. The house loses heat from the four walls and the roof (but not the floor). The average R-value for the walls and roof is R = 1.2 (this takes into account walls/windows/doors/etc.). Calculate the heat loss in MJ/m3 per degree day (°C) and compare this to the estimated residential heating needs as discussed in this chapter.

Solution The surface area is 5×(7 m)×(7 m) = 245 m2 47 ©2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Sustainable Energy - Chapter 8: Direct Use of Solar Energy and the volume will be (7 m)3 = 343 m3 Thus the total power loss will be P A (245 m 2 )    204 J/s per degree (°C) T R 1.2

Converting this to power per day gives (204 J/s)×(86,400 s/d) = 17.6 MJ per degree day (°C) Normalizing this to the volume of the house gives (17.6 MJ per degree day (°C))/(343 m3) = 51.4 kJ/m3 per degree day (°C) compared to the estimated value of 67 kJ/m3 per degree day (°C) given in the chapter.

Problem 8.7 A 300 liter electric hot water heater has provides 9000 W of power to heat water. If the heater is filled with water at an initial temperature is 10 °C, how long will it take for the water to reach 60 °C? Assume there are no heat losses.

Solution 300 L of water corresponds to a mass of 300 kg. The specific heat of water is 4186 J/(kg °C) and the temperature difference in °C will be (60 °C – 10 °C) = 50 °C The heat is Q = CmΔT and since the power input is constant then Q = Pt and solving for time gives t

CmT (4186J/(kg  C)  (300kg)  (50  C)   6977s P 9000J/s

or about 1h56m Problem 8.8 Compare the masses and volumes of water, concrete, sand and wood needed to store 1 GJ of heat if the operating temperatures are Tc=30 ºC and Th=80 ºC. In each case calculate the edge length of a square storage unit with a height of 2.5 m.

Solution The amount of heat (thermal energy) stored is given by equation (8.12) as Q = mCΔT 48 ©2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Sustainable Energy - Chapter 8: Direct Use of Solar Energy

Solving for mass gives m = Q/(CΔT) Using Q = 109 J, ΔT = (80°C – 30°C) = 50°C and C from Table 8.6 for the various materials of interest, the mass may be calculated. From the masses and the known densities as given in the table, the volumes are found. The length of the edge in meters is given by L = (V/2.5)1/2. The table below summarizes these results.

material water concrete sand wood

C (J/(kg°C)) 4186 653 816 2800

m (kg) 4.78×103 3.06×104 2.45×104 7.14×103

ρ (kg/m3) 1000 2300 1600 500

V (m3) 4.78 13.3 15.3 14.3

L (m) 1.38 2.31 2.47 2.39

49 ©2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Sustainable Energy - Chapter 8: Direct Use of Solar Energy

50 ©2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.