Heat and Mass Transfer Fundamentals and Applications 6th Edition Test Bank

Type:

Test Bank

Resource:

Heat and Mass Transfer Fundamentals and Applications

Edition:

6th Edition

Author(s):

Yunus A. Cengel Afshin J. Ghajar

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Solution 1-1C Thermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium state to another. Heat transfer, on the other hand, deals with the rate of heat transfer as well as the temperature distribution within the system at a specified time.

Problem 1-1C How does the science of heat transfer differ from the science of thermodynamics?

Solution 1-2C (a) The driving force for heat transfer is the temperature difference. (b) The driving force for electric current flow is the electric potential difference (voltage). (a) The driving force for fluid flow is the pressure difference.

Problem 1-2C What is the driving force for (a) heat transfer, (b) electric current flow, and (c) fluid flow?

Solution 1-3C The rating problems deal with the determination of the heat transfer rate for an existing system at a specified temperature difference. The sizing problems deal with the determination of the size of a system in order to transfer heat at a specified rate for a specified temperature difference.

Problem 1-3C How do rating problems in heat transfer differ from the sizing problems?

Solution 1-4C The experimental approach (testing and taking measurements) has the advantage of dealing with the actual physical system, and getting a physical value within the limits of experimental error. However, this approach is expensive, time consuming, and often impractical. The analytical approach (analysis or calculations) has the advantage that it is fast and inexpensive, but the results obtained are subject to the accuracy of the assumptions and idealizations made in the analysis.

Problem 1-4C What is the difference between the analytical and experimental approaches to heat transfer? Discuss the advantages and disadvantages of each approach.

Solution 1-5C Modeling makes it possible to predict the course of an event before it actually occurs, or to study various aspects of an event mathematically without actually running expensive and time-consuming experiments. When preparing a mathematical model, all the variables that affect the phenomena are identified, reasonable assumptions and approximations are made, and the interdependence of these variables are studied. The relevant physical laws and principles are invoked, and the problem is formulated mathematically. Finally, the problem is solved using an appropriate approach, and the results are interpreted.

Problem 1-5C What is the importance of modeling in engineering? How are the mathematical models for engineering processes prepared?

Solution 1-6C The right choice between a crude and complex model is usually the simplest model which yields adequate results.

Preparing very accurate but complex models is not necessarily a better choice since such models are not much use to Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

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an analyst if they are very difficult and time consuming to solve. At the minimum, the model should reflect the essential features of the physical problem it represents.

Problem 1-6C When modeling an engineering process, how is the right choice made between a simple but crude and a complex but accurate model? Is the complex model necessarily a better choice since it is more accurate?

Solution 1-7C Warmer. Because energy is added to the room air in the form of electrical work.

Problem 1-7C On a hot summer day, a student turns his fan on when he leaves his room in the morning. When he returns in the evening, will his room be warmer or cooler than the neighboring rooms? Why? Assume all the doors and windows are kept closed.

Solution 1-8C Warmer. If we take the room that contains the refrigerator as our system, we will see that electrical work is supplied to this room to run the refrigerator, which is eventually dissipated to the room as waste heat.

Problem 1-8C Consider two identical rooms, one with a refrigerator in it and the other without one. If all the doors and windows are closed, will the room that contains the refrigerator be cooler or warmer than the other room? Why? Solution 1-9C The claim is false. The heater of a house supplies the energy that the house is losing, which is proportional to the temperature difference between the indoors and the outdoors. A turned off heater consumes no energy. The heat lost from a house to the outdoors during the warming up period is less than the heat lost from a house that is already at the temperature that the thermostat is set because of the larger cumulative temperature difference in the latter case. For best practice, the heater should be turned off when no one is at home during day (at subfreezing temperatures, the heater should be kept on at a low temperature to avoid freezing of water in pipes). Also, the thermostat should be lowered during bedtime to minimize the temperature difference between the indoors and the outdoors at night and thus the amount of heat that the heater needs to supply to the house.

Problem 1-9C It is often claimed that you should leave the heating system in your house on all the time rather than turning it off when you are not home because it takes more energy to heat the house back to its normal temperature. What do you think?

Solution 1-10C No. The thermostat tells an airconditioner (or heater) at what interior temperature to stop. The air conditioner will cool the house at the same rate no matter what the thermostat setting is. So, it is best to set the thermostat at a comfortable temperature and then leave it alone. Setting the thermostat too low a home owner risks wasting energy and money (and comfort) by forgetting it at the set low temperature.

Problem 1-10C Someone claims that turning the thermostat of the central air conditioning of a warm house to the lowest level will cool the house a lot faster. Is there any truth to this claim?

Solution 1-11C No. Since there is no temperature drop of water, the heater will never kick into make up for the heat loss. Therefore, it will not waste any energy during times of no use, and there is no need to use a timer. But if the family were on a time-of-use

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tariff, it would be possible to save money (but not energy) by turning on the heater when the rate was lowest at night and off during peak periods when rate is the highest.

Problem 1-11C Consider a perfectly insulated (no heat loss from the skin) 80-L electric water heater set at 60°C. Observing that one tank of hot water is sufficient to meet the hot water needs of a family for an entire day, someone suggests using a timer to heat the water once a day to the set temperature and then turning it off to save energy and money. Do you think the family should take this advice and invest in a timer? Assume there is a single tariff for electricity for all hours of the day.

Solution 1-12C For the constant pressure case. This is because the heat transfer to an ideal gas is mcpT at constant pressure and mcvT at constant volume, and cp is always greater than cv.

Problem 1-12C An ideal gas is heated from 50°C to 80°C (a) at constant volume and (b) at constant pressure. For which case do you think the energy required will be greater? Why? Solution 1-13C The rate of heat transfer per unit surface area is called heat flux q . It is related to the rate of heat transfer by Q =

 qdA .

A

Problem 1-13C What is heat flux? How is it related to the heat transfer rate?

Solution 1-14C Energy can be transferred by heat and work to a close system. An energy transfer is heat transfer when its driving force is temperature difference.

Problem 1-14C What are the mechanisms of energy transfer to a closed system? How is heat transfer distinguished from the other forms of energy transfer?

Solution 1-15E A logic chip in a computer dissipates 3 W of power. The amount heat dissipated in 8 h and the heat flux on the surface of the chip are to be determined.

Assumptions Heat transfer from the surface is uniform.

Logic chip  Analysis (a) The amount of heat the chip dissipates during an 8-hour period is  t = (3 W)(8 h ) = 24 Wh = 0.024 kWh Q=Q

Q = 3W

(b) The heat flux on the surface of the chip is Q 3W q = = = 37.5 W/in2 A 0.08 in 2

Problem 1-15E A logic chip used in a computer dissipates 3 W of power in an environment at 120°F and has a heat transfer surface area of 0.08 in2. Assuming the heat transfer from the surface to be uniform, determine (a) the amount of heat this chip dissipates during an eight-hour workday in kWh and (b) the heat flux on the surface of the chip in W/in2.

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Solution 1-16 The filament of a 150 W incandescent lamp is 5 cm long and has a diameter of 0.5 mm. The heat flux on the surface of the filament, the heat flux on the surface of the glass bulb, and the annual electricity cost of the bulb are to be determined.

Assumptions Heat transfer from the surface of the filament and the bulb of the lamp is uniform.

Analysis (a) The heat transfer surface area and the heat flux on the surface of the filament are As = DL =  (0.05 cm)(5 cm) = 0.785 cm 2

Q 150 W q s = = = 191 W/cm 2 = 1.91 10 6 W/m 2 As 0.785 cm 2

Q Lamp 150 W

(b) The heat flux on the surface of glass bulb is As = D 2 =  (8 cm) 2 = 201.1 cm 2

qs =

Q 150 W = = 0.75 W /cm 2 = 7500 W / m 2 As 201.1 cm 2

(c) The amount and cost of electrical energy consumed during a one-year period is Electricity Consumption = Q t = (0.15 kW)(365 8 h/yr) = 438 kWh/yr Annual Cost = (438 kWh/yr)($0.08 / kWh) = $35.04/yr

Problem 1-16 Consider a 150-W incandescent lamp. The filament of the lamp is 5 cm long and has a diameter of 0.5 mm. The diameter of the glass bulb of the lamp is 8 cm. Determine the heat flux in W/m2 (a) on the surface of the filament and (b) on the surface of the glass bulb, and (c) calculate how much it will cost per year to keep that lamp on for eight hours a day every day if the unit cost of electricity is $0.08/kWh.

Solution 1-17 An aluminum ball is to be heated from 80C to 200C. The amount of heat that needs to be transferred to the aluminum ball is to be determined.

Assumptions The properties of the aluminum ball are constant.

Properties The average density and specific heat of aluminum are given to be  = 2700 kg/m3 and cp = 0.90 kJ/kgC. Analysis The amount of energy added to the ball is simply the change in its internal energy, and is determined from Metal E transfer = U = mc p (T2 − T1 ) ball Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

E

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where

m = V =

 6

D 3 =

 6

(2700 kg/m 3 )(0.15 m) 3 = 4.77 kg

Substituting, E transfer = (4.77 kg)(0.90 kJ/kg  C)(200 − 80)C = 515 kJ Therefore, 515 kJ of energy (heat or work such as electrical energy) needs to be transferred to the aluminum ball to heat it to 200C. Problem 1-17 A 15-cm-diameter aluminum ball is to be heated from 80°C to an average temperature of 200°C. Taking the average density and specific heat of aluminum in this temperature range to be ρ = 2700 kg/m3 and cp = 0.90 kJ/kg·K, respectively, determine the amount of energy that needs to be transferred to the aluminum ball.

Solution 1-18E A water heater is initially filled with water at 50F. The amount of energy that needs to be transferred to the water to raise its temperature to 120F is to be determined.

Assumptions 1 Water is an incompressible substance with constant specific. 2 No water flows in or out of the tank during heating.

Properties The density and specific heat of water at 85ºF from Table A-9E are:  = 62.17 lbm/ft3 and cp = 0.999 Btu/lbmR. 120F Analysis The mass of water in the tank is

 1 ft 3   = 498.7 lbm m = V = (62.17 lbm/ft 3 )(60 gal)   7.48 gal    Then, the amount of heat that must be transferred to the water in the tank as it is heated from 50 to 120F is determined to be

50F Water

45F Water

Q = mc p (T2 − T1 ) = (498.7 lbm)(0.999 Btu/lbm  F)(120 − 50)F = 34,874 Btu

Discussion Referring to Table A-9E the density and specific heat of water at 50ºF are:  = 62.41 lbm/ft3 and cp = 1.000 Btu/lbmR and at 120ºF are:  = 61.71 lbm/ft3 and cp = 0.999 Btu/lbmR. We evaluated the water properties at an average temperature of 85ºF. However, we could have assumed constant properties and evaluated properties at the initial temperature of 50ºF or final temperature of 120ºF without loss of accuracy.

Problem 1-18E A 60-gallon water heater is initially filled with water at 50°F. Determine how much energy (in Btu) needs to be transferred to the water to raise its temperature to 120°F. Evaluate the water properties at an average water temperature of 85°F. Solution 1-19 An electrically heated house maintained at 22°C experiences infiltration losses at a rate of 0.7 ACH. The amount of energy loss from the house due to infiltration per day and its cost are to be determined.

Assumptions 1 Air as an ideal gas with a constant specific heats at room temperature. 2 The volume occupied by the furniture and other belongings is negligible. 3 The house is maintained at a constant temperature and pressure at all times. 4 The infiltrating air exfiltrates at the indoors temperature of 22°C. Properties The specific heat of air at room temperature is cp = 1.007 kJ/kgC. Analysis The volume of the air in the house is V = (floor space)(height) = (200 m 2 )(3 m) = 600 m 3

Noting that the infiltration rate is 0.7 ACH (air changes per hour) and thus the air in the house is completely replaced by the outdoor air 0.724 = 16.8 times per day, the mass flow rate of air through the house due to infiltration is 0.7 ACH 5C

22C AIR

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m air = =

PoVair Po (ACH V house ) = RTo RTo (89.6 kPa)(16.8 600 m 3 / day) (0.287 kPa  m 3 /kg  K)(5 + 273.15 K)

= 11,314 kg/day

Noting that outdoor air enters at 5C and leaves at 22C, the energy loss of this house per day is

Q infilt = m air c p (Tindoors − Toutdoors ) = (11,314 kg/day)(1.007 kJ/kg. C)(22 − 5)C = 193,681 kJ/day = 53.8 kWh/day At a unit cost of $0.082/kWh, the cost of this electrical energy lost by infiltration is Enegy Cost = (Energy used)(Unit cost of energy) = (53.8 kWh/day)($0.082/kWh) = $4.41/day Problem 1-19 Infiltration of cold air into a warm house during winter through the cracks around doors, windows, and other openings is a major source of energy loss since the cold air that enters needs to be heated to the room temperature. The infiltration is often expressed in terms of ACH (air changes per hour). An ACH of 2 indicates that all air in the house is replaced twice every hour by the cold air outside.

Consider an electrically heated house that has a floor space of 200 m2 and an average height of 3 m at 1000 m elevation, where the standard atmospheric pressure is 89.6 kPa. The house is maintained at a temperature of 22°C, and the infiltration losses are estimated to amount to 0.7 ACH. Assuming the pressure and the temperature in the house remain constant, determine the amount of energy loss from the house due to infiltration for a day during which the average outdoor temperature is 5°C. Also, determine the cost of this energy loss for that day if the unit cost of electricity in that area is $0.082/kWh. Solution 1-20 Liquid ethanol is being transported in a pipe where heat is added to the liquid. The volume flow rate that is necessary to keep the ethanol temperature below its flashpoint is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The specific heat and density of ethanol are constant.

Properties The specific heat and density of ethanol are given as 2.44 kJ/kg∙K and 789 kg/m 3, respectively.

Analysis The rate of heat added to the ethanol being transported in the pipe is

Q = m c p (Tout − Tin ) or

Q = Vc p (Tout − Tin ) For the ethanol in the pipe to be below its flashpoint, it is necessary to keep Tout below 16.6°C. Thus, the volume flow rate should be V 

Q

c p (Tout − Tin )

=

20 kJ/s (789 kg/m 3 )(2.44 kJ/kg  K )(16.6 − 10) K

V  0.00157m 3 /s Discussion To maintain the ethanol in the pipe well below its flashpoint, it is more desirable to have a much higher flow rate than 0.00157 m3/s.

Problem 1-20 Liquid ethanol is a flammable fluid and can release vapors that form explosive mixtures at temperatures above its flash point at 16.6°C. In a chemical plant, liquid ethanol (cp = 2.44 kJ/kg·K, ρ = 789 kg/ m3) is being transported in a pipe with an inside diameter of 5 cm. The pipe is located in a hot area with the presence of an ignition source, where an estimated 20 kW Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

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of heat is added to the ethanol. Your task as an engineer is to design a pumping system to transport the ethanol safely and to prevent fire hazard. If the inlet temperature of the ethanol is 10°C, determine the volume flow rate that is necessary to keep the temperature of the ethanol in the pipe below its flash point.

Solution 1-21 A 2 mm thick by 3 cm wide AISI 1010 carbon steel strip is cooled in a chamber from 597 to 47°C to avoid instantaneous thermal burn upon contact with skin tissue. The amount of heat rate to be removed from the steel strip is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The stainless steel sheet has constant specific heat and density. 3 Changes in potential and kinetic energy are negligible.

Properties For AISI 1010 carbon steel, the specific heat of AISI 1010 steel at (597 + 47)°C / 2 = 322°C = 595 K is 682 J/kg∙K (by interpolation from Table A-3), and the density is given as 7832 kg/m3.

Analysis The mass of the steel strip being conveyed enters and exits the chamber at a rate of m = Vwt The rate of heat being removed from the steel strip in the chamber is given as Q = m c (T − T ) removed

p

in

out

= Vwtc p (Tin − Tout ) = (7832 kg/m 3 )(1 m/s)(0.030 m)(0.002 m)(682 J/kg  K)(597 − 47) K = 176 kW Discussion By slowing down the conveyance speed of the steel strip would reduce the amount of heat rate needed to be removed from the steel strip in the cooling chamber. Since slowing the conveyance speed allows more time for the steel strip to cool.

Problem 1-21

In many manufacturing plants, individuals are often working around high-temperature surfaces. Exposed hot surfaces that can cause thermal burns on human skin are considered hazards in the workplace. Metallic surfaces above 70°C can instantly damage skin that touches them. Consider an AISI 1010 carbon steel strip (ρ = 7832 kg/m3) 2 mm thick and 3 cm wide that is conveyed into a chamber to be cooled at a constant speed of 1 m/s. The steel strip enters the cooling chamber at 597°C. Determine the rate at which heat needs to be removed so that the steel strip exits the chamber at 47°C. Discuss how the conveyance speed can affect the required rate of heat removal.

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Solution 1-22 Liquid water is to be heated in an electric teapot. The heating time is to be determined.

Assumptions 1 Heat loss from the teapot is negligible. 2 Constant properties can be used for both the teapot and the water.

Properties The average specific heats are given to be 0.7 kJ/kg·K for the teapot and 4.18 kJ/kg·K for water.

Analysis We take the teapot and the water in it as the system, which is a closed system (fixed mass). The energy balance in this case can be expressed as E − E out in  Net energy transfer by heat, work, and m ass

=

E sy stem    Change in internal, kinetic, potential, etc.energies

E in = U sy stem = U water + U teapot

Then the amount of energy needed to raise the temperature of water and the teapot from 15°C to 95°C is E in = (mcT ) water + (mcT ) teapot = (1.2 kg)(4.18 kJ/kg  C)(95 − 15)C + (0.5 kg)(0.7 kJ/kg  C)(95 − 15)C = 429.3 kJ

The 1200-W electric heating unit will supply energy at a rate of 1.2 kW or 1.2 kJ per second. Therefore, the time needed for this heater to supply 429.3 kJ of heat is determined from t =

E in Totalenergy transferred 429.3 kJ = = = 358 s = 6.0 min  Rate of energy transfer E transfer 1.2 kJ/s

Discussion In reality, it will take more than 6 minutes to accomplish this heating process since some heat loss is inevitable during heating. Also, the specific heat units kJ/kg · °C and kJ/kg · K are equivalent, and can be interchanged.

Problem 1-22 1.2 kg of liquid water initially at 15°C is to be heated to 95°C in a teapot equipped with a 1200-W electric heating element inside. The teapot is 0.5 kg and has an average specific heat of 0.7 kJ/kg·K. Taking the specific heat of water to be 4.18 kJ/kg·K and disregarding any heat loss from the teapot, determine how long it will take for the water to be heated.

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A water heater uses 100 kW to heat 60 gallon (0.2271 m3) of liquid water initially at 20°C. Determine the heating duration such that the water exiting the heater would be in compliance with the ASME Boiler and Pressure Vessel Code (ASME BPVC.IV-2015) service restrictions.

Assumptions1 Heating of the heater material is negligible (i.e. 100 kW is for heating the water only). 2 Constant properties are used for the water. 3 No water flowing out of the heater during the heating.

Properties The average density and specific heat of water are given to be 970 kg/m 3 and 4.18 kJ/kg·K, respectively.

Analysis The mass of the water in the heater is m = V = (970 kg/m 3 )(0.2271m3 ) = 220.29 kg

The energy balance in this case can be expressed as Ein − Eout   

=

Net energy transfer by heat, work,and mass

Esystem    Changein internal,kinetic, potential,etc. energies

Ein = U system = U water Ein = (mcT ) water

In terms of heat transfer rate

(mcT ) water E Q = in = t t Solving for the heating duration from 20°C to 120°C, (mc) water (T2 − T1 ) t = Q

(220.29 kg)(4.18 kJ/kg  C) (100C) = 920.8 s = 15.3 min 100 kJ/s Discussion It takes about 15 minutes at 100 kW to heat 60 gallons of liquid water from 20°C to the ASME Boiler and Pressure Vessel Code service restrictions temperature of 120°C. If the heating duration is more than 15.3 minutes, then the final temperature of the water that would exit the heater would be higher than 120°C, which exceeds the code restrictions.

Problem 1-23 t =

A water heater uses 100 kW to heat 60 gallons of liquid water from an initial temperature of 20°C. The average density and specific heat of the water are 970 kg/m3 and 4.18 kJ/kg·K, respectively. According to the service restrictions of the ASME Boiler and Pressure Vessel Code (ASME BPVC.IV-2015, HG-101), water heaters should not be operating at temperatures exceeding 20°C at or near the heater outlet. Determine the heating duration such that the final water temperature that would exit the water heater would be in compliance with the ASME Boiler and Pressure Vessel Code.

Solution 1-24 A boiler (10 kg) is used to heat 20 gallon (0.07571 m 3) of liquid water with 50 kW for 30 minutes. Determine whether this operating condition would be in compliance with the ASME Boiler and Pressure Vessel Code (ASME BPVC.IV-2015) service restrictions.

Assumptions1 Heat loss from the boiler is negligible. 2 Constant properties are used for both the boiler and the water. 3 The raise in temperature for the boiler and the water is equal. 4 No water flowing out of the boiler during the heating.

Properties The average specific heats are given to be 0.48 kJ/kg·K for the boiler material and 4.18 kJ/kg·K for the water. The average density of the water is 850 kg/m3.

Analysis We take the boiler and the water in it as a closed system. The energy balance in this case can be expressed as Ein − Eout    Net energy transfer by heat, work,and mass

=

Esystem    Changein internal,kinetic, potential,etc. energies

Ein = U system = U water + U boiler Ein = (mcT ) system = (mcT ) water + (mcT ) boiler Ein = [(mc) water + (mc) boiler]T Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

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Solving for the raise in temperature T =

Ein [( V c) water + (mc) boiler]

T =

(50 kJ/s )(30  60 s) = 329C [(850 kg/m 3 )(0.07571m3 )(4.18 kJ/kg  C) + (10 kg)(0.48 kJ/kg  C) ]

The final temperature is T2 = T1 + T = 15C + 329C = 344C  120C Discussion The final temperature of the water after 30 minutes of heating at 50 kW is 224°C greater than the ASME Boiler and Pressure Vessel Code service restrictions of 120°C. Thus, the operating condition would not be in compliance with the code. A temperature control mechanism should be implemented to ensure that the water exiting the boiler stays below 120°C. Problem 1-24 A boiler with a mass of 10 kg uses 50 kW to heat 20 gallons of liquid water initially at 15°C. The average specific heats for the boiler material and the water are 0.48 kJ/kg·K and 4.18 kJ/kg·K, respectively. The average density of the water is 850 kg/m3. According to the service restrictions of the ASME Boiler and Pressure Vessel Code (ASME BPVC.IV-2015, HG101), hot water boilers should not be operating at temperatures exceeding 120°C at or near the boiler outlet. Determine whether the final water temperature that would exit the boiler after 30 min of heating at 50 kW would be in compliance with the ASME Boiler and Pressure Vessel Code.

Solution 1-25 Water is heated in an insulated tube by an electric resistance heater. The mass flow rate of water through the heater is to be determined.

Assumptions 1 Water is an incompressible substance with a constant specific heat. 2 The kinetic and potential energy changes are negligible, ke  pe  0. 3 Heat loss from the insulated tube is negligible.

Properties The specific heat of water at room temperature is cp = 4.18 kJ/kg·C. Analysis We take the tube as the system. This is a control volume since mass crosses the system boundary during the process.

We observe that this is a steady-flow process since there is no change with time at any point and thus 1 = m 2 = m  , and the tube is insulated. The energy mCV = 0 and ECV = 0 , there is only one inlet and one exit and thus m balance for this steady-flow system can be expressed in the rate form as E − E out = E sy stem0 (steady ) = 0 → E in = E out in    Rate of net energy transfer by heat, work, and m ass

Rate of changein internal, kinetic, potential, etc.energies

 h1 = m  h2 (since ke  pe  0) W e,in + m   c (T − T ) W =m e,in

p

2

WATER 15C

70C

1

Thus,

7 kW

m =

W e,in c p (T2 − T1 )

=

7 kJ/s = 0.0304 kg/s (4.18 kJ/kg  C)(70 − 15)C

Problem 1-25 Water is heated in an insulated, constant diameter tube by a 7-kW electric resistance heater. If the water enters the heater steadily at 15°C and leaves at 70°C, determine the mass flow rate of the water.

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Solution 1-26 It is observed that the air temperature in a room heated by electric baseboard heaters remains constant even though the heater operates continuously when the heat losses from the room amount to 7000 kJ/h. The power rating of the heater is to be determined.

Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ke  pe  0. 3 The temperature of the room remains constant during this process.

Analysis We take the room as the system. The energy balance in this case reduces to E −E = E sy stem inout     Net energy transfer by heat, work, and mass

Change in internal, kinetic, potential, etc.energies

AIR

We,in − Qout = U = 0 We,in = Qout since U = mcvT = 0 for isothermal processes of ideal gases. Thus,

We

 1kW   = 1.94 kW W e,in = Q out = 7000kJ/h   3600kJ/h  Problem 1-26 A room is heated by a baseboard resistance heater. When the heat losses from the room on a winter day amount to 7000 kJ/h, it is observed that the air temperature in the room remains constant even though the heater operates continuously. Determine the power rating of the heater in kW. Solution 1-27 A room is heated by an electrical resistance heater placed in a short duct in the room in 15 min while the room is losing heat to the outside, and a 300-W fan circulates the air steadily through the heater duct. The power rating of the electric heater and the temperature rise of air in the duct are to be determined.

Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ke  pe  0. 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications.

3 Heat loss from the duct is negligible. 4 The house is air-tight and thus no air is leaking in or out of the room.

Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Also, cp = 1.007 kJ/kg·K for air at room temperature (Table A-15) and cv = cp – R = 0.720 kJ/kg·K. Analysis (a) We first take the air in the room as the system. This is a constant volume closed system since no mass crosses the system boundary. The energy balance for the room can be expressed as

E −E inout 

=

Net energy transfer by heat, work, and m ass

E sy stem    Change in internal, kinetic, potential, etc.energies

We,in + Wfan,in − Qout = U  (We,in + W fan,in − Q out )t = m(u 2 − u1 )  mcv (T2 − T1 )

200 kJ/min 568 m3

The total mass of air in the room is

V = 5  6  8 m 3 = 240 m 3 m=

We

P1V (98 kPa)(240 m ) = = 284.6 kg RT1 (0.287 kPa  m 3 /kg  K)(288 K ) 3

300 W

Then the power rating of the electric heater is determined to be

W e,in = Q out − W fan,in + mcv (T2 − T1 ) / t = (200/60 kJ/s ) − (0.3 kJ/s ) + (284.6 kg)(0.720 kJ/kg  C)(25 − 15C)/(18  60 s) = 4.93 kW

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(b) The temperature rise that the air experiences each time it passes through the heater is determined by applying the energy balance to the duct,

E in = E out + m h = Q 0 + m h

W e,in + W fan,in W + W e,in

1

out

2

(since ke  pe  0)

 h = m c p T fan,in = m

Thus,

W e,in + W fan,in

T =

=

m c p

(4.93 + 0.3)kJ/s = 6.2C (50/60 kg/s )(1.007 kJ/kg  K)

Problem 1-27 A 5-m × 6-m × 8-m room is to be heated by an electrical resistance heater placed in a short duct in the room. Initially, the room is at 15°C, and the local atmospheric pressure is 98 kPa. The room is losing heat steadily to the outside at a rate of 200 kJ/min. A 300-W fan circulates the air steadily through the duct and the electric heater at an average mass flow rate of 50 kg/min. The duct can be assumed to be adiabatic, and there is no air leaking in or out of the room. If it takes 18 min for the room air to reach an average temperature of 25°C, find (a) the power rating of the electric heater and (b) the temperature rise that the air experiences each time it passes through the heater.

Solution 1-28 Air is moved through the resistance heaters in a 1200-W hair dryer by a fan. The volume flow rate of air at the inlet and the velocity of the air at the exit are to be determined.

Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ke  pe  0. 3 Constant specific heats at room temperature can be used for air. 4 The power consumed by the fan and the heat losses through the walls of the hair dryer are negligible.

Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Also, cp = 1.007 kJ/kg·K for air at room temperature (Table A-15).

Analysis (a) We take the hair dryer as the system. This is a control volume since mass crosses the system boundary during the process. We observe that this is a steady-flow process since there is no change with time at any point and thus 1 = m 2 = m  . The energy balance for this mCV = 0 and ECV = 0 , and there is only one inlet and one exit and thus m steady-flow system can be expressed in the rate form as E − E = E sy stem0 (steady ) = 0 → E in = E out inout    Rate of net energy transfer by heat, work, and mass

Rate of change in internal, kinetic, potential, etc.energies

W e,in + W fan,in 0 + m h1 = Q out 0 + m h2 (since ke  pe  0) W = m c (T − T ) e,in

p

2

1

Thus,

m = =

W e,in

c p (T2 − T1 )

P1 = 100 kPa T1 = 22C

T2 = 47C A2 = 60 cm2

1.2 kJ/s = 0.04767 kg/s (1.007 kJ/kg  C)(47 − 22)C

· We = 1200 W

Then,

v1 =

RT1 (0.287 kPa  m 3 /kg  K)(295 K) = = 0.8467 m 3 /kg P1 100 kPa

V1 = m v 1 = (0.04767 kg/s )(0.8467m 3 /kg) = 0.0404 m 3 /s (b) The exit velocity of air is determined from the conservation of mass equation,

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v2 =

RT2 (0.287 kPa  m /kg  K)(320 K) = = 0.9184 m 3 /kg P2 100 kPa

m =

1

3

v2

A2V2 ⎯ ⎯→ V2 =

m v 2 (0.04767 kg/s )(0.9184 m 3 /kg) = = 7.30 m/s A2 60 10 − 4 m 2

Problem 1-28 A hair dryer is basically a duct in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it to flow over the resistors, where it is heated. Air enters a 1200-W hair dryer at 100 kPa and 22°C and leaves at 47°C. The cross-sectional area of the hair dryer at the exit is 60 cm2. Neglecting the power consumed by the fan and the heat losses through the walls of the hair dryer, determine (a) the volume flow rate of air at the inlet and (b) the velocity of the air at the exit.

Solution 1-29E Air gains heat as it flows through the duct of an air-conditioning system. The velocity of the air at the duct inlet and the temperature of the air at the exit are to be determined.

Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 222F and 548 psia. 2 The kinetic and potential energy changes are negligible, ke  pe  0. 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications.

Properties The gas constant of air is R = 0.3704 psia·ft3/lbm·R (Table A-1E). Also, cp = 0.240 Btu/lbm·R for air at room temperature (Table A-15E).

Analysis We take the air-conditioning duct as the system. This is a control volume since mass crosses the system boundary during the process. We observe that this is a steady-flow process since there is no change with time at any point and thus 1 = m 2 = m  , and heat is lost from the system. The mCV = 0 and ECV = 0 , there is only one inlet and one exit and thus m energy balance for this steady-flow system can be expressed in the rate form as E − E out = E sy stem0 (steady ) = 0 → E in = E out in    Rate of net energy transfer by heat, work, and mass

Rate of changein internal, kinetic, potential, etc.energies

 h1 = m  h2 (since ke  pe  0) Q in + m   c (T − T ) Q =m in

p

2

1

450 ft3/min

AIR

D = 10 in

(a) The inlet velocity of air through the duct is determined from

V1 =

V1 V1 450 ft 3 /min = = = 825 ft/min A1 r 2  (5 / 12 ft)2

2 Btu/s

(b) The mass flow rate of air becomes RT1 (0.3704 psia  ft 3 /lbm  R )(510 R ) = = 12.6 ft 3 /lbm P1 15 psia V 450 ft 3 /min m = 1 = = 35.7 lbm/min = 0.595 lbm/s v 1 12.6 ft 3 /lbm

v1 =

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Then the exit temperature of air is determined to be

T2 = T1 +

Q in 2 Btu/s = 50F + = 64.0F m c p (0.595 lbm/s )(0.240 Btu/lbm  F)

Problem 1-29E Air enters the duct of an air-conditioning system at 15 psia and 50°F at a volume flow rate of 450 ft3/min. The diameter of the duct is 10 in, and heat is transferred to the air in the duct from the surroundings at a rate of 2 Btu/s. Determine (a) the velocity of the air at the duct inlet and (b) the temperature of the air at the exit.

Solution 1-30 Liquid water entering at 10°C and flowing at 10 g/s(0.01 kg/s) is heated in a circular tube by an electrical heater at 10 kW. Determine whether the water exit temperature would be below 79°C and comply with the ASME Code for Process Piping, and the minimum mass flow rate to keep the exit temperature below 79°C. Assumptions1 Water is an incompressible substance with constant properties. 2 Heat loss from the tube is negligible. 3 Steady operating conditions.

Properties The specific heat of water is given as 4.18 kJ/kg·K. AnalysisThe tube is taken as a control volume. For steady state flow, the mass flow rate at the inlet is equal to the mass flow rate at the exit: 1 = m 2 = m  m The energy balance for the control volume is 0 (steady) E in − E out = E system      Rate of net energy transfer by heat, work,and mass

Rate of changein internal,kinetic, potential,etc. energies

E in − E out = 0 E = E in

out

 h1 = m  h2 Q heater + m   c (T − T ) Q =m heater

p

2

1

Solving for the exit temperature, Q heater 10 kJ/s + T1 = + 10C = 249C  79C  cp m (0.01 kg/s)(4.18 kJ/kg  C)

T2 =

Thus, the exit temperature is not in compliance with the ASME Code for Process Piping for PVDC lining.

The minimum mass flow rate needed to keep the water exit temperature from exceeding 79°C is Q heater m  (T2 − T1 ) c p



10 kJ/s (79 − 10)C (4.18 kJ/kg  C)

 0.0347 kg/s The higher the value of mass flow rate, the lower the water exit temperature is achieved.

Discussion If the desire is to have higher exit temperature, then a different thermoplastic lining should be used.

Polytetrafluoroethylene (PTFE) lining has a recommended maximum temperature of 260°C by the ASME Code for Process Piping (ASME B31.3-2014, A323).

Problem 1-30

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Liquid water with a specific heat of 4.18 kJ/kg·K is heated in a circular tube by an electrical heater at 10 kW. The water enters the tube at 10°C with a mass flow rate of 10 g/s. The inner surface of the tube is lined with polyvinylidene chloride (PVDC) lining. According to the ASME Code for Process Piping (ASME B31.3-2014, A323), the recommended maximum temperature for PVDC lining is 79°C. Determine whether the water exiting the tube would be in compliance with the ASME Code for Process Piping. What is the minimum mass flow rate of the water needed to keep the exit temperature from exceeding the recommended maximum temperature for PVDC lining?

Solution 1-31C The thermal conductivity of a material is the rate of heat transfer through a unit thickness of the material per unit area and per unit temperature difference. The thermal conductivity of a material is a measure of how fast heat will be conducted in that material.

Problem 1-31C Define thermal conductivity, and explain its significance in heat transfer.

Solution 1-32C Diamond is a better heat conductor.

Problem 1-32C Which is a better heat conductor, diamond or silver?

Solution 1-33C The thermal conductivity of gases is proportional to the square root of absolute temperature. The thermal conductivity of most liquids, however, decreases with increasing temperature, with water being a notable exception.

Problem 1-33C How do the thermal conductivity of gases and liquids vary with temperature?

Solution 1-34C Superinsulations are obtained by using layers of highly reflective sheets separated by glass fibers in an evacuated space.

Radiation heat transfer between two surfaces is inversely proportional to the number of sheets used and thus heat loss by radiation will be very low by using this highly reflective sheets. At the same time, evacuating the space between the layers forms a vacuum under 0.000001 atm pressure which minimize conduction or convection through the air space between the layers.

Problem 1-34C Why is the thermal conductivity of superinsulation orders of magnitude lower than the thermal conductivity of ordinary insulation?

Solution 1-35C Most ordinary insulations are obtained by mixing fibers, powders, or flakes of insulating materials with air. Heat transfer through such insulations is by conduction through the solid material, and conduction or convection through the air space as well as radiation. Such systems are characterized by apparent thermal conductivity instead of the ordinary thermal conductivity in order to incorporate these convection and radiation effects.

Problem 1-35C Why do we characterize the heat conduction ability of insulators in terms of their apparent thermal conductivity instead of their ordinary thermal conductivity?

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Solution 1-36C The mechanisms of heat transfer are conduction, convection and radiation. Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles.

Convection is the mode of energy transfer between a solid surface and the adjacent liquid or gas which is in motion, and it involves combined effects of conduction and fluid motion. Radiation is energy emitted by matter in the form of electromagnetic waves (or photons) as a result of the changes in the electronic configurations of the atoms or molecules.

Problem 1-36C What are the mechanisms of heat transfer? How are they distinguished from each other?

Solution 1-37C

dT Conduction is expressed by Fourier's law of conduction as Q cond = −kA where dT/dx is the temperature gradient, k is the dx thermal conductivity, and A is the area which is normal to the direction of heat transfer.

Convection is expressed by Newton's law of cooling as Q conv = hAs (Ts − T ) where h is the convection heat transfer coefficient, As is the surface area through which convection heat transfer takes place, Ts is the surface temperature and T is the temperature of the fluid sufficiently far from the surface.

4 Radiation is expressed by Stefan-Boltzman law as Q rad =  As (Ts4 − Tsurr ) where  is the emissivity of surface, As is the surface area, Ts is the surface temperature, Tsurr is the average surrounding surface temperature and  = 5.67 10 −8 W/m 2  K 4 is the Stefan-Boltzman constant.

Problem 1-37C Write down the expressions for the physical laws that govern each mode of heat transfer, and identify the variables involved in each relation.

Solution 1-38C Convection involves fluid motion, conduction does not. In a solid we can have only conduction.

Problem 1-38C How does heat conduction differ from convection?

Solution 1-39C No. It is purely by radiation.

Problem 1-39C Does any of the energy of the sun reach the earth by conduction or convection?

Solution 1-40C In forced convection the fluid is forced to move by external means such as a fan, pump, or the wind. The fluid motion in natural convection is due to buoyancy effects only.

Problem 1-40C How does forced convection differ from natural convection?

Solution 1-41C In solids, conduction is due to the combination of the vibrations of the molecules in a lattice and the energy transport by free electrons. In gases and liquids, it is due to the collisions of the molecules during their random motion.

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Problem 1-41C What is the physical mechanism of heat conduction in a solid, a liquid, and a gas?

Solution 1-42C The parameters that effect the rate of heat conduction through a windowless wall are the geometry and surface area of wall, its thickness, the material of the wall, and the temperature difference across the wall.

Problem 1-42C Consider heat transfer through a windowless wall of a house on a winter day. Discuss the parameters that affect the rate of heat conduction through the wall.

Solution 1-43C In a typical house, heat loss through the wall with glass window will be larger since the glass is much thinner than a wall, and its thermal conductivity is higher than the average conductivity of a wall.

Problem 1-43C Consider heat loss through two walls of a house on a winter night. The walls are identical except that one of them has a tightly fit glass window. Through which wall will the house lose more heat? Explain.

Solution 1-44C The house with the lower rate of heat transfer through the walls will be more energy efficient. Heat conduction is proportional to thermal conductivity (which is 0.72 W/m.C for brick and 0.17 W/m.C for wood, Table 1-1) and inversely proportional to thickness. The wood house is more energy efficient since the wood wall is twice as thick but it has about onefourth the conductivity of brick wall.

Problem 1-44C Consider two houses that are identical except that the walls are built using bricks in one house and wood in the other. If the walls of the brick house are twice as thick, which house do you think will be more energy efficient?

Solution 1-45C The rate of heat transfer through both walls can be expressed as T − T2 T − T2 Q wood = k wood A 1 = (0.16 W/m  C) A 1 = 1.6 A(T1 − T2 ) L wood 0.1 m T − T2 T − T2 Q brick = k brick A 1 = (0.72 W/m  C) A 1 = 2.88 A(T1 − T2 ) Lbrick 0.25 m where thermal conductivities are obtained from Table A-5. Therefore, heat transfer through the brick wall will be larger despite its higher thickness.

Problem 1-45C Consider two walls of a house that are identical except that one is made of 10-cm-thick wood while the other is made of 25cm-thick brick. Through which wall will the house lose more heat in winter?

Solution 1-46C Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by a blackbody at the same temperature.

Absorptivity is the fraction of radiation incident on a surface that is absorbed by the surface. The Kirchhoff's law of radiation states that the emissivity and the absorptivity of a surface are equal at the same temperature and wavelength.

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Define emissivity and absorptivity. What is Kirchhoff’s law of radiation?

Solution 1-47C A blackbody is an idealized body which emits the maximum amount of radiation at a given temperature and which absorbs all the radiation incident on it. Real bodies emit and absorb less radiation than a blackbody at the same temperature.

Problem 1-47C What is a blackbody? How do real bodies differ from blackbodies?

Solution 1-48

The thermal conductivity of a wood slab subjected to a given heat flux of 40 W/m 2 with constant left and right surface temperatures of 40ºC and 20ºC is to be determined.

Assumptions 1 Steady operating conditions exist since the surface temperatures of the wood slab remain constant at the specified values. 2 Heat transfer through the wood slab is one dimensional since the thickness of the slab is small relative to other dimensions. 3 Thermal conductivity of the wood slab is constant.

Analysis The thermal conductivity of the wood slab is determined directly from Fourier’s relation to be L W  0.05 m  k = q = 0.10 W/mK =  40 2  T1 − T2  m  (40 − 20)C

Discussion Note that the ºC or K temperature units may be used interchangeably when evaluating a temperature difference.

T

.

q = 40 W/m2

T1 = 40oC

T2 = 20oC

T(x)

L = 0.05 m x

Problem 1-48 A wood slab with a thickness of 0.05 m is subjected to a heat flux of 40 W/m2. The left and right surface temperatures of the wood slab are kept at constant temperatures of 40° C and 20° C, respectively. What is the thermal conductivity of the wood slab?

Solution 1-49 The inner and outer surfaces of a brick wall are maintained at specified temperatures. The rate of heat transfer through the wall is to be determined.

Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Thermal Brick wall properties of the wall are constant.

Properties The thermal conductivity of the wall is given to be k = 0.69 W/mC. 0.3 m Analysis Under steady conditions, the rate of heat transfer through the wall is 5C 20C (20 − 5)C T Q cond = kA = (0.69 W/m  C)(4  7 m 2 ) = 966 W L 0.3 m Problem 1-49 The inner and outer surfaces of a 4-m × 7-m brick wall of thickness 30 cm and thermal conductivity 0.69 W/m·K are maintained at temperatures of 20° C and 5° C, respectively. Determine the rate of heat transfer through the wall in W.

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Solution 1-50E The inner and outer glasses of a double pane window with a 0.5-in air space are at specified temperatures. The rate of heat transfer through the window is to be determined Glass Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Heat transfer through the window is one-dimensional. 3 Thermal properties of the air are constant.

Properties The thermal conductivity of air at the average temperature of (60+48)/2 = 54F is k = 0.01419 Btu/hftF (Table A-15E).

Air Analysis The area of the window and the rate of heat loss through it are

Q

A = (4 ft) (4 ft) = 16 ft 2

T −T (60 − 48)F Q = kA 1 2 = (0.01419 Btu/h.ft.F)(16 ft 2 ) = 131 Btu/h L 0.25 / 12 ft

60F 48F

Problem 1-50E The inner and outer glasses of a 4-ft × 4-ft double-pane window are at 60°F and 48°F, respectively. If the 0.25-in space between the two glasses is filled with still air, determine the rate of heat transfer through the window.

Solution 1-51 The inner and outer surfaces of a window glass are maintained at specified temperatures. The amount of heat transfer through the glass in 5 h is to be determined.

Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Thermal properties of the glass are constant.

Glass Properties The thermal conductivity of the glass is given to be k = 0.78 W/mC. Analysis Under steady conditions, the rate of heat transfer through the glass by conduction is (10 − 3)C T Q cond = kA = (0.78 W/m  C)(2  2 m 2 ) = 4368 W L 0.005m

Then the amount of heat transfer over a period of 5 h becomes

10C

3C

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1-20  Q=Q cond t = (4.368 kJ/s)(5  3600 s) = 78,620 kJ

If the thickness of the glass doubled to 1 cm, then the amount of heat transfer will go down by half to 39,310 kJ. Problem 1-51 The inner and outer surfaces of a 0.5-cm thick 2-m × 2-m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass is 0.78 W/m·K, determine the amount of heat loss through the glass over a period of 5 h.

What would your answer be if the glass were 1 cm thick?

Solution 1-52

Prob. 1-51 is reconsidered. The amount of heat loss through the glass as a function of the window glass thickness is to be plotted.

Analysis The problem is solved using EES, and the solution is given below.

"GIVEN" L=0.005 [m] A=2*2 [m^2] T_1=10 [C] T_2=3 [C] k=0.78 [W/m-C] time=5*3600 [s] "ANALYSIS" Q_dot_cond=k*A*(T_1-T_2)/L Q_cond=Q_dot_cond*time*Convert(J, kJ)

L [m] 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01

Qcond [kJ] 393120 196560 131040 98280 78624 65520 56160 49140 43680 39312

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400000 350000

Qcond [kJ]

300000 250000 200000 150000 100000 50000 0 0.002

0.004

0.006

0.008

0.01

L [m] Problem 1-52

Reconsider Prob. 1–51. Using appropriate software, plot the amount of heat loss through the glass as a function of the window glass thickness in the range of 0.1 cm to 1.0 cm. Discuss the results.

Solution 1-53 Heat is transferred steadily to boiling water in the pan through its bottom. The inner surface temperature of the bottom of the pan is given. The temperature of the outer surface is to be determined.

Assumptions 1 Steady operating conditions exist since the surface temperatures of the pan remain constant at the specified values. 2 Thermal properties of the aluminum pan are constant.

Properties The thermal conductivity of the aluminum is given to be k = 237 W/mC. Analysis The heat transfer area is A =  r2 =  (0.075 m)2 = 0.0177 m2 Under steady conditions, the rate of heat transfer through the bottom of the pan by conduction is

T −T T Q = kA = kA 2 1 L L

105C

Substituting, 800 W = (237 W/m  C)(0.0177 m 2 )

T2 − 105C 0.004 m

800 W

0.4 cm

which gives T2 = 105.76C Problem 1-53 An aluminum pan whose thermal conductivity is 237 W/m·K has a flat bottom with diameter 15 cm and thickness 0.4 cm.

Heat is transferred steadily to boiling water in the pan through its bottom at a rate of 800 W. If the inner surface of the bottom of the pan is at 105°C, determine the temperature of the outer surface of the bottom of the pan.

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Solution 1-54E The inner and outer surface temperatures of the wall of an electrically heated home during a winter night are measured. The rate of heat loss through the wall that night and its cost are to be determined.

Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values during the entire night. 2 Thermal properties of the wall are constant.

Properties The thermal conductivity of the brick wall is given to be k = 0.42 Btu/hftF. Analysis (a) Noting that the heat transfer through the wall is by conduction and the surface area of the wall is A = 20 ft 10 ft = 200 ft 2 , the steady rate of heat transfer through the wall can be determined from

T − T2 (62 − 25)F Q = kA 1 = (0.42 Btu/h.ft.F)(200 ft 2 ) = 3108 Btu/h L 1 ft or 0.911 kW since 1 kW = 3412 Btu/h. (b) The amount of heat lost during an 8 hour period and its cost are

Brick Wall Q

Q = Q t = (0.911 kW)(8 h) = 7.288 kWh

Cost = (Amount of energy)(Unit cost of energy) = (7.288 kWh)($0.07/kWh)

1 ft 25F

62F

= $0.51 Therefore, the cost of the heat loss through the wall to the home owner that night is $0.51.

Problem 1-54E The north wall of an electrically heated home is 20 ft long, 10 ft high, and 1 ft thick and is made of brick whose thermal conductivity is k = 0.42 Btu/h·ft·°F. On a certain winter night, the temperatures of the inner and the outer surfaces of the wall are measured to be at about 62°F and 25°F, respectively, for a period of 8 h. Determine (a) the rate of heat loss through the wall that night and (b) the cost of that heat loss to the homeowner if the cost of electricity is $0.07/kWh. Solution 1-55 The thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.

Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Heat losses through the lateral surfaces of the apparatus are negligible since those surfaces are well-insulated, and thus the entire heat generated by the heater is conducted through the samples. 3 The apparatus possesses thermal symmetry.

Analysis The electrical power consumed by the heater and converted to heat is Q W = VI = (110 V)(0.6 A) = 66 W e

The rate of heat flow through each sample is W 66 W Q = e = = 33 W 2 2

3 cm

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3

cm

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Then the thermal conductivity of the sample becomes

D 2  (0.04 m) 2 = = 0.001257 m 2 4 4 Q L (33 W)(0.03m) T Q = kA ⎯ ⎯→ k = = = 78.8 W/m.C L AT (0.001257 m 2 )(10C) A=

Problem 1-55 In a certain experiment, cylindrical samples of diameter 4 cm and length 7 cm are used (see Fig. 1–32). The two thermocouples in each sample are placed 3 cm apart. After initial transients, the electric heater is observed to draw 0.6 A at 110 V, and both differential thermometers read a temperature difference of 10°C. Determine the thermal conductivity of the sample.

Solution 1-56 The thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.

Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Heat losses through the lateral surfaces of the apparatus are negligible since those surfaces are well-insulated, and thus the entire heat generated by the heater is conducted through the samples. 3 The apparatus possesses thermal symmetry.

Analysis For each sample we have

Q = 25 / 2 = 12.5 W

Q

Q

A = (0.1 m)(0.1 m) = 0.01 m 2 T = 82 − 74 = 8C L

Then the thermal conductivity of the material becomes Q L (12.5 W)(0.005m) T Q = kA ⎯ ⎯→ k = = = 0.781 W/m.C L AT (0.01 m 2 )(8C)

L

A

Problem 1-56 One way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical rectangular samples of the material and to heavily insulate the four outer edges, as shown in Fig. P1-56.

Thermocouples attached to the inner and outer surfaces of the samples record the temperatures.

During an experiment, two 0.5-cm-thick samples 10 cm × 10 cm in size are used. When steady operation is reached, the heater is observed to draw 25 W of electric power, and the temperature of each sample is observed to drop from 82°C at the inner surface to 74°C at the outer surface. Determine the thermal conductivity of the material at the average temperature.

Solution 1-57 To prevent a silicon wafer from warping, the temperature difference across its thickness cannot exceed 1°C. The maximum allowable heat flux on the bottom surface of the wafer is to be determined.

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Assumptions 1 Heat conduction is steady and one-dimensional. 2 There is no heat generation. 3 Thermal conductivity is constant.

Properties The thermal conductivity of silicon at 27°C (300 K) is 148 W/m∙K (Table A-3).

Analysis For steady heat transfer, the Fourier’s law of heat conduction can be expressed as Tup − Tbot dT q = −k = −k dx L Thus, the maximum allowable heat flux so that Tbot − Tup  1C is q  k

Tbot − Tup L

= (148 W/m  K)

1K 500  10-6 m

q  2.96  10 5 W/m 2

Discussion With the upper surface of the wafer maintained at 27°C, if the bottom surface of the wafer is exposed to a flux greater than 2.96×105 W/m2, the temperature gradient across the wafer thickness could be significant enough to cause warping.

Problem 1-57 Silicon wafer is susceptible to warping when the wafer is subjected to a temperature difference across its thickness.

Thus, steps needed to be taken to prevent the temperature gradient across the wafer thickness from getting large. As an engineer in a semiconductor company, your task is to determine the maximum allowable heat flux on the bottom surface of the wafer while maintaining the upper surface temperature at 27°C. To prevent the wafer from warping, the temperature difference across its thickness of 500 μm cannot exceed 1°C.

Solution 1-58 Heat loss by conduction through a concrete wall as a function of ambient air temperatures ranging from -15 to 38°C is to be determined.

Assumptions 1 One-dimensional conduction. 2 Steady-state conditions exist. 3 Constant thermal conductivity. 4 Outside wall temperature is that of the ambient air.

Properties The thermal conductivity is given to be k = 0.75, 1 or 1.25 W/mK. Analysis From Fourier’s law, it is evident that the gradient, dT dx = − q k , is a constant, and hence the temperature distribution is linear, if q and k are each constant. The heat flux must be constant under one-dimensional, steadyCopyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

1-25

state conditions; and k are each approximately constant if it depends only weakly on temperature. The heat flux and heat rate for the case when the outside wall temperature is T2 = −15C and k = 1 W/mK are:

q = −k

T −T dT 25C − (− 15C ) = k 1 2 = (1 W m  K) = 133.3 W m 2 dx L 0.30 m

(

)(

(1)

)

 = q  A = 133.3 W m 2  20 m 2 = 2667 W Q

(2)

Combining Eqs. (1) and (2), the heat rate Q can be determined for the range of ambient temperature, −15 ≤ T2 ≤ 38°C, with different wall thermal conductivities, k .

Discussion (1) Notice that from the graph, the heat loss curves are linear for all three thermal conductivities. This is true because under steady-state and constant k conditions, the temperature distribution in the wall is linear. (2) As the value of k increases, the slope of the heat loss curve becomes steeper. This shows that for insulating materials (very low k ), the heat loss curve would be relatively flat. The magnitude of the heat loss also increases with increasing thermal conductivity. (3) At T2 = 25C , all the three heat loss curves intersect at zero; because T1 = T2 (when the inside and outside temperatures are the same), thus there is no heat conduction through the wall. This shows that heat conduction can only occur when there is temperature difference.

The results for the heat loss Q with different thermal conductivities k are tabulated and plotted as follows:

Q [W]

T2 [°C]

k = 0.75 W/m∙K

k = 1 W/m∙K

k = 1.25 W/m∙K

-15

2000

2667

3333

-10

1750

2333

2917

-5

1500

2000

2500

0

1250

1667

2083

5

1000

1333

1667

10

750

1000

1250

15

500

666.7

833.3

20

250

333.3

416.7

25

0

0

0

30

-250

-333.3

-416.7

38

-650

-866.7

-1083

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Problem 1-58 A concrete wall with a surface area of 20 m2 and a thickness of 0.30 m separates conditioned room air from ambient air. The temperature of the inner surface of the wall (T1) is maintained at 25°C. (a) Determine the heat loss Q(W) through the concrete wall for three thermal conductivity values of 0.75, 1, and 1.25 W/m·K and outer wall surface temperatures of T2 = − 15, − 10, − 5, 0, 5, 10, 15, 20, 25, 30, and 38°C (a total of 11 data points for each thermal conductivity value). Tabulate the results for all three cases in one table. Also provide a computer-generated graph [Heat loss, Q(W) vs. Outside wall temperature, T2(°C)] for the display of your results. The results for all three cases should be plotted on the same graph. (b) Discuss your results for the three cases.

Solution 1-59 A hollow spherical iron container is filled with iced water at 0°C. The rate of heat gain by the iced water and the rate at which ice melts in the container are to be determined.

Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Heat transfer through the shell is one-dimensional. 3 Thermal properties of the iron shell are constant. 4 The inner surface of the shell is at the same temperature as the iced water, 0°C. 5 Treat the spherical shell as a plain wall and use the outer area.

Properties The thermal conductivity of iron is k = 80.2 W/mC (Table A-3). The heat of fusion of water is given to be 333.7 kJ/kg.

Analysis This spherical shell can be approximated as a plate of thickness 0.4 cm and area A = D2 =  (0.2 m)2 = 0.126 m2 Then the rate of heat transfer through the shell by conduction is (5 − 0)C T Q cond = kA = (80.2 W/m  C)(0.126 m 2 ) = 25,263 W = 25.3 kW L 0.002 m

Considering that it takes 333.7 kJ of energy to melt 1 kg of ice at 0°C, the rate at which ice melts in the container can be determined from  ice = m

5C

Iced water 0C

0.2 cm

Q 25.263kJ/s = = 0.0757 kg/s hif 333.7 kJ/kg

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Discussion We should point out that this result is slightly in error for approximating a curved wall as a plain wall. The error in this case is very small because of the large diameter to thickness ratio. For better accuracy, we could use the inner surface area (D = 19.6 cm) or the mean surface area (D = 19.8 cm) in the calculations.

Problem 1-59 A hollow spherical iron container with outer diameter 20 cm and thickness 0.2 cm is filled with iced water at 0°C. If the outer surface temperature is 5°C, determine the approximate rate of heat gain by the iced water in kW and the rate at which ice melts in the container. The heat of fusion of water is 333.7 kJ/kg. Treat the spherical shell as a plain wall, and use the outer area.

Solution 1-60

Prob. 1-59 is reconsidered. The rate at which ice melts as a function of the container thickness is to be plotted.

Analysis The problem is solved using EES, and the solution is given below.

"GIVEN" D=0.2 [m] L=0.2 [cm] T_1=0 [C] T_2=5 [C] "PROPERTIES" h_if=333.7 [kJ/kg] k=k_(Iron, 25) "ANALYSIS" A=pi*D^2 Q_dot_cond=k*A*(T_2-T_1)/(L*Convert(cm, m)) m_dot_ice=(Q_dot_cond*Convert(W, kW))/h_if

mice [kg/s] 0.1515 0.07574 0.0505 0.03787 0.0303 0.02525 0.02164 0.01894 0.01683

0.16 0.14 0.12

mice [kg/s]

L [cm] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0.1 0.08 0.06 0.04

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0 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1-28

1

0.01515

Problem 1-60

Reconsider Prob. 1–59. Using appropriate software, plot the rate at which ice melts as a function of the container thickness in the range of 0.1 cm to 1.0 cm. Discuss the results.

Solution 1-61 A 5 m × 5 m concrete slab with embedded heating cable melts snow at a rate of 0.1 kg/s. The power density (heat flux) for the embedded heater is to be determined whether it is in compliance with the NFPA 70 code. Also, the temperature difference between the heater surface and the slab surface is to be determined whether it exceeds 21°C, as recommended in the ASHRAE Handbook to minimize thermal stress.

Assumptions1 Steady operating conditions. 2 Slab surface and heater surface temperatures are uniform. 3 Heat transfer through the concrete layer is one-dimensional. 3 Properties of the concrete are constant. 4 The heater heats the surface uniformly.

Properties The thermal conductivity of concrete is given as 1.4 W/m·K. The latent heat of fusion for water is 333.7 kJ/kg (Table A-2).

Snow Concrete slab

Slab surface, T2 Heater surface, T1

AnalysisThe heat rate required for melting snow at 0.1 kg/s is

Q = m ice hif = (0.1 kg/s )(333700 J/kg ) = 33370 W For a surface area of 5 m × 5 m, the power density (heat flux) is q =

Q 33370 W = = 1335 W/m2  1300 W/m 2 As 25 m 2

To determine the temperature difference between the heater surface(T1) and the slab surface(T2), we use the Fourier law of conduction:

T −T q = k 1 2 L or

q L (1335 W/m2 )(0.05 m) = = 47.7C  21C k 1.4 W/m  K DiscussionThe power density for the embedded heating cable in the concrete slab slightly exceeds the limit set by the National Electrical Code® (NFPA 70) of 1300 W/m2. The temperature difference between the heater surface and the slab surface is about 27°C higher than the recommended value by the 2015 ASHRAE Handbook—HVAC Applications, Chapter 51.

T1 − T2 =

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1-29

Problem 1-61 A heating cable is embedded in a concrete slab to melt snow on a 5-m  5-m surface area. The heating cable is heated electrically with joule heating. The concrete has a thermal conductivity of 1.4 W/m·K, and the layer between the slab surface and the heater surface is 50 mm. When the surface is covered with snow, the heat generated from the heating cable can melt snow at a rate of 0.1 kg/s. According to the National Electrical Code (NFPA 70), the power density (heat flux) for embedded snow-melting equipment should not exceed 1300 W/m2. Under the given conditions, does the concrete slab with the embedded heating cable operate in compliance with the NFPA 70 code? Also, to minimize thermal stress in the concrete, the temperature difference between the heater surface (T1) and the slab surface (T2) should not exceed 21°C (2015 ASHRAE Handbook—HVAC Applications, Chap. 51). Do the current operating conditions allow T1 − T2 ≤ 21°C?

Solution 1-62E Using the conversion factors between W and Btu/h, m and ft, and C and F, the convection coefficient in SI units is to be expressed in Btu/hft2F. Analysis The conversion factors for W and m are straightforward, and are given in conversion tables to be

1 W = 3.41214Btu/h 1 m = 3.2808ft The proper conversion factor between C into F in this case is

1C = 1.8F since the C in the unit W/m2C represents per C change in temperature, and 1C change in temperature corresponds to a change of 1.8F. Substituting, we get 1 W/m 2  C =

3.41214Btu/h (3.2808 ft)2 (1.8 F)

= 0.1761Btu/h  ft 2  F

which is the desired conversion factor. Therefore, the given convection heat transfer coefficient in English units is h = 14 W/m 2  C = 14  0.1761Btu/h  ft 2  F = 2.47 Btu/h ft 2  F Problem 1-62E

An engineer who is working on the heat transfer analysis of a house in English units needs the convection heat transfer coefficient on the outer surface of the house. But the only value he can find from his handbooks is 14 W/m2·K, which is in SI units. The engineer does not have a direct conversion factor between the two unit systems for the convection heat transfer coefficient. Using the conversion factors between W and Btu/h, m and ft, and °C and °F, express the given convection heat transfer coefficient in Btu/h·ft2·°F. Solution 1-63 The heat flux between air with a constant temperature and convection heat transfer coefficient blowing over a pond at a constant temperature is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Convection heat transfer coefficient is uniform. 3 Heat . transfer by radiation is negligible. 4 Air temperature and qconv T = 20oC the surface temperature of the pond remain constant. h = 20 W/m2.K Analysis From Newton’s law of cooling, the heat flux is T s = 40oC given as Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

1-30

q conv = h (Ts - T) W (40 – 20)C = 400 W/m2 m2  K

q conv = 20

Discussion (1) Note the direction of heat flow is out of the surface since Ts > T ; (2) Recognize why units of K in h and units of ºC in (Ts - T) cancel.

Problem 1-63 Air at 20°C with a convection heat transfer coefficient of 20 W/m2·K blows over a pond. The surface temperature of the pond is at 40°C. Determine the heat flux between the surface of the pond and the air.

Solution 1-64 A series of ASME SA-193 carbon steel bolts are bolted to the upper surface of a metal plate. The upper surface is exposed to convection with the ambient air. The bottom surface is subjected to a uniform heat flux. Determine whether the use of the bolts complies with the ASME Boiler and Pressure Vessel Code, where 260°C is the maximum allowable use temperature.

Assumptions1 Heat transfer is steady. 2 One dimensional heat conduction through the metal plate. 3 Uniform heat flux on the bottom surface. 4 Uniform surface temperature at the upper plate surface. 5 The temperature of the bolts is equal to the upper surface temperature of the plate.

Air, T∞, h

Bolt

Metal plate

Uniform heat flux

Analysis The uniform heat flux subjected on the bottom plate surface is equal to the heat flux transferred by convection on the upper surface. q 0 = q cond = q conv = 5000 W/m 2

From the Newton’s law of cooling, we have q conv = h (Ts − T ) Assuming the temperature of the bolts is equal to the upper surface temperature of the plate,

Tbolt = Ts =

q conv 5000 W/m2 + T = + 30C = 530C  260C h 10 W/m2  K

Discussion The temperature of the bolts exceeds the maximum allowable use temperature by 260°C. One way to keep the temperature of the bolts below 260°C is by increasing the convection heat transfer coefficient. Higher convection heat transfer coefficient can be achieved by having forced convection. To keep the upper surface temperature of the plate at 260°C or lower, the convection heat transfer coefficient should be higher than 21.7 W/m 2·K. h

q conv 5000 W/m 2   21.7 W/m 2  K Ts − T (260 − 30) K

Problem 1-64 A series of ASME SA-193 carbon steel bolts are bolted to the upper surface of a metal plate. The bottom surface of the plate is subjected to a uniform heat flux of 5 kW/m 2. The upper surface of the plate is exposed to ambient air with a temperature of 30°C and a convection heat transfer coefficient of 10 W/m 2·K. The ASME Boiler and Pressure Vessel Code Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

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(ASME BPVC.IV-2015, HF-300) limits the maximum allowable use temperature to 260°C for the SA-193 bolts. Determine whether the use of these SA-193 bolts complies with the ASME code under these conditions. If the temperature of the bolts exceeds the maximum allowable use temperature of the ASME code, discuss a possible solution to lower the temperature of the bolts.

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