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BASICS OF HEAT TRANSFER
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Question 1 of 20
1. Question
1 pointsA 10cm diameter copper ball is to be heated from 100°C to an average temperature of 150°C in 30 minutes (Fig. ). Taking the average density and specific heat of copper in this temperature range to be P= 8950 kg/m³ and Cp 0.395 kJ/kg · °C, respectively, determine the total amount of heat transfer to the copper ball,
ASSUMPTIONS :Constant properties can be used for copper at the average temperature
Properties: The average density and specific heat of copper are given to be P= 8950 kg/m³ and Cp 0.395 kJ/kg · °C.
Correct
SOLUTION :The copper ball is to be heated from 100°C to 150°C. The total heat transfer, the average rate of heat transfer, and the average heat flux are to be determined
ANALYSIS: The amount of heat transferred to the copper ball is simply the change in its internal energy, and is determined from Energy transfer to the system= Energy increase of the system Q=ΔU mCave (T2 T1) where m =PV =Π/6 PD³=Π/6 (8950 kg/m³ )(0.1 m)³ 4.69 kg Substituting,Q (4.69 kg)(0.395 kJ/kg · °C)(150 100)°C 92.6 kJ
Incorrect
SOLUTION :The copper ball is to be heated from 100°C to 150°C. The total heat transfer, the average rate of heat transfer, and the average heat flux are to be determined
ANALYSIS: The amount of heat transferred to the copper ball is simply the change in its internal energy, and is determined from Energy transfer to the system= Energy increase of the system Q=ΔU mCave (T2 T1) where m =PV =Π/6 PD³=Π/6 (8950 kg/m³ )(0.1 m)³ 4.69 kg Substituting,Q (4.69 kg)(0.395 kJ/kg · °C)(150 100)°C 92.6 kJ

Question 2 of 20
2. Question
1 pointsA 10cm diameter copper ball is to be heated from 100°C to an average temperature of 150°C in 30 minutes (Fig.). Taking the average density and specific heat of copper in this temperature range to be P=8950 kg/m³ and Cp= 0.395 kJ/kg · °C, respectively, determine the average rate of heat transfer to the ball
Assumptions: Constant properties can be used for copper at the average temperature.
Properties: The average density and specific heat of copper are given to be P= 8950 kg/m³ and Cp =0.395 kJ/kg · °C.
Correct
The copper ball is to be heated from 100°C to 150°C. The total heat transfer, the average rate of heat transfer, and the average heat flux are to be determined.
ANALYSIS: The rate of heat transfer normally changes during a process with time. However, we can determine the average rate of heat transfer by dividing the total amount of heat transfer by the time interval. Therefore, Qave =Q/Δt=92.6 kJ/1800 s=0.0514 KJ/S=51.4
Incorrect
The copper ball is to be heated from 100°C to 150°C. The total heat transfer, the average rate of heat transfer, and the average heat flux are to be determined.
ANALYSIS: The rate of heat transfer normally changes during a process with time. However, we can determine the average rate of heat transfer by dividing the total amount of heat transfer by the time interval. Therefore, Qave =Q/Δt=92.6 kJ/1800 s=0.0514 KJ/S=51.4

Question 3 of 20
3. Question
1 pointsA 10cm diameter copper ball is to be heated from 100°C to an average temperature of 150°C in 30 minutes (Fig.). Taking the average density and specific heat of copper in this temperature range to be P=8950 kg/m³ and Cp= 0.395 kJ/kg · °C, respectively, determine the average heat flux.
PROPERTIES: Constant properties can be used for copper at the average temperature.
ASSUMPTIONS: The average density and specific heat of copper are given to be P=8950 kg/m³ and Cp 0.395 kJ/kg · °C
Correct
The copper ball is to be heated from 100°C to 150°C. The total heat transfer, the average rate of heat transfer, and the average heat flux are to be determined
ANALYSIS:Heat flux is defined as the heat transfer per unit time per unit area, or the rate of heat transfer per unit area. Therefore, the average heat flux in this case is
Qave= Qave/A=Qave/ΠD²=51.4W/Π(0.1)m²=1636 W/m²
Incorrect
The copper ball is to be heated from 100°C to 150°C. The total heat transfer, the average rate of heat transfer, and the average heat flux are to be determined
ANALYSIS:Heat flux is defined as the heat transfer per unit time per unit area, or the rate of heat transfer per unit area. Therefore, the average heat flux in this case is
Qave= Qave/A=Qave/ΠD²=51.4W/Π(0.1)m²=1636 W/m²

Question 4 of 20
4. Question
1 points1.2 kg of liquid water initially at 15°C is to be heated to 95°C in a teapot equipped with a 1200W electric heating element inside (Fig.). The teapot is 0.5 kg and has an average specific heat of 0.7 kJ/kg · °C. Taking the specific heat of water to be 4.18 kJ/kg · °C and disregarding any heat loss from the teapot, determine how long it will take for the water to be heated (ANSWER IN ONE DIGIT)
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 · °C for the teapot and 4.18 kJ/kg · °C for waterCorrect
Liquid water is to be heated in an electric teapot. The heating time is to be determined.. 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 Ein Eout =ΔEsystem Ein =ΔUsystem =ΔUwater +ΔUteapot Then the amount of energy needed to raise the temperature of water and the teapot from 15°C to 95°C is Ein (mCΔT )water (mCΔT )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 1200W 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 Δt=TOTAL ENERGY TRANSFERRED/RATE OF ENERGY TRANSFER=Ein/ E transfer=429.3KJ/1.2KJ/s=358s=6.0 MIN
Incorrect
Liquid water is to be heated in an electric teapot. The heating time is to be determined.. 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 Ein Eout =ΔEsystem Ein =ΔUsystem =ΔUwater +ΔUteapot Then the amount of energy needed to raise the temperature of water and the teapot from 15°C to 95°C is Ein (mCΔT )water (mCΔT )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 1200W 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 Δt=TOTAL ENERGY TRANSFERRED/RATE OF ENERGY TRANSFER=Ein/ E transfer=429.3KJ/1.2KJ/s=358s=6.0 MIN

Question 5 of 20
5. Question
1 pointsA 5mlong section of an air heating system of a house passes through an unheated space in the basement (Fig.). The cross section of the rectangular duct of the heating system is 20 cm x 25 cm. Hot air enters the duct at 100 kPa and 60°C at an average velocity of 5 m/s. The temperature of the air in the duct drops to 54°C as a result of heat loss to the cool space in the basement. Determine the rate of heat loss from the air in the duct to the basement under steady conditions. Also, determine the cost of this heat loss per hour if the house is heated by a natural gas furnace that has an efficiency of 80 percent, and the cost of the natural gas in that area is $0.60/therm (1 therm 100,000 Btu 105,500 kJ).(answer upto three decimal)
Assumptions: 1 Steady operating conditions exist. 2 Air can be treated as an ideal gas with constant properties at room temperature.
Properties :The constant pressure specific heat of air at the average temperature of (54 60)/2 =57°C is 1.007 kJ/kg · °CCorrect
SOLUTION:The temperature of the air in the heating duct of a house drops as a result of heat loss to the cool space in the basement. The rate of heat loss from the hot air and its cost are to be determined.
Analysis :We take the basement section of the heating system as our system, which is a steadyflow system. The rate of heat loss from the air in the duct can be determined from Q=m· CpΔT where m is the mass flow rate and ΔT is the temperature drop. The density of air at the inlet conditions is P= P/RT=100 KPA/(0.287 KPA. m³/KG.K)(60+273)K=1.046 kg/m³ The crosssectional area of the duct is Ac= (0.20 m)(0.25 m) 0.05 m² Then the mass flow rate of air through the duct and the rate of heat loss become m·=PVAc (1.046 kg/m³)(5 m/s)(0.05 m²) 0.2615 kg/s and Qloss= m· Cp(Tin Tout) =(0.2615 kg/s)(1.007 kJ/kg · °C)(60 54)°C =1.580 kJ/s or 5688 kJ/h. The cost of this heat loss to the home owner is Cost of heat loss (Rate of heat loss)(Unit cost of energy input)/Furnace efficiency =(5688 kJ/h)($0.60/therm)0.80( 1 therm/105,500 kJ)=$0.040/hIncorrect
SOLUTION:The temperature of the air in the heating duct of a house drops as a result of heat loss to the cool space in the basement. The rate of heat loss from the hot air and its cost are to be determined.
Analysis :We take the basement section of the heating system as our system, which is a steadyflow system. The rate of heat loss from the air in the duct can be determined from Q=m· CpΔT where m is the mass flow rate and ΔT is the temperature drop. The density of air at the inlet conditions is P= P/RT=100 KPA/(0.287 KPA. m³/KG.K)(60+273)K=1.046 kg/m³ The crosssectional area of the duct is Ac= (0.20 m)(0.25 m) 0.05 m² Then the mass flow rate of air through the duct and the rate of heat loss become m·=PVAc (1.046 kg/m³)(5 m/s)(0.05 m²) 0.2615 kg/s and Qloss= m· Cp(Tin Tout) =(0.2615 kg/s)(1.007 kJ/kg · °C)(60 54)°C =1.580 kJ/s or 5688 kJ/h. The cost of this heat loss to the home owner is Cost of heat loss (Rate of heat loss)(Unit cost of energy input)/Furnace efficiency =(5688 kJ/h)($0.60/therm)0.80( 1 therm/105,500 kJ)=$0.040/h 
Question 6 of 20
6. Question
1 pointsConsider a house that has a floor space of 2000 ft² and an average height of 9 ft at 5000 ft elevation where the standard atmospheric pressure is 12.2 psia (Fig.). Initially the house is at a uniform temperature of 50°F. Now the electric heater is turned on, and the heater runs until the air temperature in the house rises to an average value of 70°F. Determine the amount of energy transferred to the air assuming the house is airtight and thus no air escapes during the heating process and(answer upto three decimal)
Assumptions: 1 Air can be treated as an ideal gas with constant properties at room temperature. 2 Heat loss from the house during heating is negligible. 3 The volume occupied by the furniture and other things is negligible.
Properties: The specific heats of air at the average temperature of (50 70)/2 60°F are Cp 0.240 Btu/lbm · °F and Cv= Cp R =0.171 Btu/lbm · °FCorrect
solution :The air in the house is heated from 50°F to 70°F by an electric heater. The amount and cost of the energy transferred to the air are to be determined for constantvolume and constantpressure cases.
Analysis :The volume and the mass of the air in the house are V =(Floor area)(Height)= (2000 ft2=²)(9 ft) 18,000 ft³ m =PV/RT=(12.2 psia)(18,000 ft³)/(0.3704 psia · ft³/lbm · R)(50+ 460)R=1162 lbmThe amount of energy transferred to air at constant volume is simply the change in its internal energy, and is determined from Ein Eout =ΔEsystem Ein, constant volume =ΔUair =mCvΔT (1162 lbm)(0.171 Btu/lbm · °F)(70 50)°F 3974 Btu At a unit cost of $0.075/kWh, the total cost of this energy is Cost of energy =(Amount of energy)(Unit cost of energy) =(3974 BTU)($0.075/kWH) (1 KWH/3412 BTU)=$0.087
Incorrect
solution :The air in the house is heated from 50°F to 70°F by an electric heater. The amount and cost of the energy transferred to the air are to be determined for constantvolume and constantpressure cases.
Analysis :The volume and the mass of the air in the house are V =(Floor area)(Height)= (2000 ft2=²)(9 ft) 18,000 ft³ m =PV/RT=(12.2 psia)(18,000 ft³)/(0.3704 psia · ft³/lbm · R)(50+ 460)R=1162 lbmThe amount of energy transferred to air at constant volume is simply the change in its internal energy, and is determined from Ein Eout =ΔEsystem Ein, constant volume =ΔUair =mCvΔT (1162 lbm)(0.171 Btu/lbm · °F)(70 50)°F 3974 Btu At a unit cost of $0.075/kWh, the total cost of this energy is Cost of energy =(Amount of energy)(Unit cost of energy) =(3974 BTU)($0.075/kWH) (1 KWH/3412 BTU)=$0.087

Question 7 of 20
7. Question
1 pointsConsider a house that has a floor space of 2000 ft² and an average height of 9 ft at 5000 ft elevation where the standard atmospheric pressure is 12.2 psia (Fig.). Initially the house is at a uniform temperature of 50°F. Now the electric heater is turned on, and the heater runs until the air temperature in the house rises to an average value of 70°F. Determine the amount of energy transferred to the air assuming some air escapes through the cracks as the heated air in the house expands at constant pressure. Also determine the cost of this heat for each case if the cost of electricity in that area is $0.075/kWh(ANSWER UPTO THREE DECIMAL)
Assumptions: 1 Air can be treated as an ideal gas with constant properties at room temperature. 2 Heat loss from the house during heating is negligible. 3 The volume occupied by the furniture and other things is negligible.
Properties: The specific heats of air at the average temperature of (50 70)/2 60°F are Cp 0.240 Btu/lbm · °F and Cv= Cp R =0.171 Btu/lbm · °FCorrect
solution: The air in the house is heated from 50°F to 70°F by an electric heater. The amount and cost of the energy transferred to the air are to be determined for constantvolume and constantpressure cases.
Analysis :The volume and the mass of the air in the house are V =(Floor area)(Height)= (2000 ft2=²)(9 ft) 18,000 ft³ m =PV/RT=(12.2 psia)(18,000 ft³)/(0.3704 psia · ft³/lbm · R)(50+ 460)R=1162 lbmThe amount of energy transferred to air at constant pressure is the change in its enthalpy, and is determined from Ein, constant pressure= ΔHai=mCpΔT =(1162 IBM)(0.240 BTU/IBM.°F)(7050) °F=5578 BTU At a unit cost of $0.075/kWh, the total cost of this energy is Cost of energy= (Amount of energy)(Unit cost of energy) =(5578 BTU)($0.075/kWH)(1 kWH/3412 BTU) =$0.123
Incorrect
solution: The air in the house is heated from 50°F to 70°F by an electric heater. The amount and cost of the energy transferred to the air are to be determined for constantvolume and constantpressure cases.
Analysis :The volume and the mass of the air in the house are V =(Floor area)(Height)= (2000 ft2=²)(9 ft) 18,000 ft³ m =PV/RT=(12.2 psia)(18,000 ft³)/(0.3704 psia · ft³/lbm · R)(50+ 460)R=1162 lbmThe amount of energy transferred to air at constant pressure is the change in its enthalpy, and is determined from Ein, constant pressure= ΔHai=mCpΔT =(1162 IBM)(0.240 BTU/IBM.°F)(7050) °F=5578 BTU At a unit cost of $0.075/kWh, the total cost of this energy is Cost of energy= (Amount of energy)(Unit cost of energy) =(5578 BTU)($0.075/kWH)(1 kWH/3412 BTU) =$0.123

Question 8 of 20
8. Question
1 pointsThe roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m thick, and is made of a flat layer of concrete whose thermal conductivity is k 0.8 W/m · °C (Fig. ). The temperatures of the inner and the outer surfaces of the roof one night are measured to be 15°C and 4°C, respectively, for a period of 10 hours. Determine the rate of heat loss through the roof that night (answer upto two decimal)
Assumptions :1 Steady operating conditions exist during the entire night since the surface temperatures of the roof remain constant at the specified values. 2 Constant properties can be used for the roof.
Properties :The thermal conductivity of the roof is given to be k 0.8 W/m · °C.
Correct
solution:The inner and outer surfaces of the flat concrete roof of an electrically heated home are maintained at specified temperatures during a night. The heat loss through the roof and its cost that night are to be determined
analysis:Noting that heat transfer through the roof is by conduction and the area of the roof is A 6 m x 8 m= 48 m², the steady rate of heat transfer through the roof is determined to be Q=KA T1T2/L=(0.8 W/M °C)(48 m²) (154)°C/0.25 m=1690 W=1.69 KW
Incorrect
solution:The inner and outer surfaces of the flat concrete roof of an electrically heated home are maintained at specified temperatures during a night. The heat loss through the roof and its cost that night are to be determined
analysis:Noting that heat transfer through the roof is by conduction and the area of the roof is A 6 m x 8 m= 48 m², the steady rate of heat transfer through the roof is determined to be Q=KA T1T2/L=(0.8 W/M °C)(48 m²) (154)°C/0.25 m=1690 W=1.69 KW

Question 9 of 20
9. Question
1 pointsThe roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m thick, and is made of a flat layer of concrete whose thermal conductivity is k 0.8 W/m · °C (Fig. ). The temperatures of the inner and the outer surfaces of the roof one night are measured to be 15°C and 4°C, respectively, for a period of 10 hours. Determine the cost of that heat loss to the home owner if the cost of electricity is $0.08/kWh.(answer upto two decimal)
Assumptions :1 Steady operating conditions exist during the entire night since the surface temperatures of the roof remain constant at the specified values. 2 Constant properties can be used for the roof.
Properties :The thermal conductivity of the roof is given to be k 0.8 W/m · °C.
Correct
solution:The inner and outer surfaces of the flat concrete roof of an electrically heated home are maintained at specified temperatures during a night. The heat loss through the roof and its cost that night are to be determined.
analysis: The amount of heat lost through the roof during a 10hour period and its cost are determined from Q=Q Δt (1.69 kW)(10 h) =16.9 kWh Cost =(Amount of energy)(Unit cost of energy =(16.9 KWh)($0.08/KWh)=$1.35
Incorrect
solution:The inner and outer surfaces of the flat concrete roof of an electrically heated home are maintained at specified temperatures during a night. The heat loss through the roof and its cost that night are to be determined.
analysis: The amount of heat lost through the roof during a 10hour period and its cost are determined from Q=Q Δt (1.69 kW)(10 h) =16.9 kWh Cost =(Amount of energy)(Unit cost of energy =(16.9 KWh)($0.08/KWh)=$1.35

Question 10 of 20
10. Question
1 pointsA common way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical samples of the material, as shown in (Fig. ) The thickness of the resistance heater, including its cover, which is made of thin silicon rubber, is usually less than 0.5 mm. A circulating fluid such as tap water keeps the exposed ends of the samples at constant temperature. The lateral surfaces of the samples are well insulated to ensure that heat transfer through the samples is onedimensional. Two thermocouples are embedded into each sample some distance L apart, and a differential thermometer reads the temperature drop ΔT across this distance along each sample. When steady operating conditions are reached, the total rate of heat transfer through both samples becomes equal to the electric power drawn by the heater, which is determined by multiplying the electric current by the voltage. In a certain experiment, cylindrical samples of diameter 5 cm and length 10 cm are used. The two thermocouples in each sample are placed 3 cm apart. After initial transients, the electric heater is observed to draw 0.4 A at 110 V, and both differential thermometers read a temperature difference of 15°C. Determine the thermal conductivity of the sample.(answer upto one decimal)
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.
Correct
solution: The thermal conductivity of a material is to be determined by ensuring onedimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.
Analysis :The electrical power consumed by the resistance heater and converted to heat is We =VI (110 V)(0.4 A) 44 W The rate of heat flow through each sample is Q=1/2We=1/2 x (44 w)=22W since only half of the heat generated will flow through each sample because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possesses thermal symmetry. The heat transfer area is the area normal to the direction of heat flow, which is the crosssectional area of the cylinder in this case: A=1/4ΠD²=1/4 Π(0.05m)²=0.00196m²
Noting that the temperature drops by 15°C within 3 cm in the direction of heat flow, the thermal conductivity of the sample is determined to be Q=KA ΔT/L→ K=QL/AΔT=(22W)(0.03 m)/(0.00196m²) (15°C)=22.4 W/M°C
Incorrect
solution: The thermal conductivity of a material is to be determined by ensuring onedimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.
Analysis :The electrical power consumed by the resistance heater and converted to heat is We =VI (110 V)(0.4 A) 44 W The rate of heat flow through each sample is Q=1/2We=1/2 x (44 w)=22W since only half of the heat generated will flow through each sample because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possesses thermal symmetry. The heat transfer area is the area normal to the direction of heat flow, which is the crosssectional area of the cylinder in this case: A=1/4ΠD²=1/4 Π(0.05m)²=0.00196m²
Noting that the temperature drops by 15°C within 3 cm in the direction of heat flow, the thermal conductivity of the sample is determined to be Q=KA ΔT/L→ K=QL/AΔT=(22W)(0.03 m)/(0.00196m²) (15°C)=22.4 W/M°C

Question 11 of 20
11. Question
1 pointsAn engineer who is working on the heat transfer analysis of a brick building in English units needs the thermal conductivity of brick. But the only value he can find from his handbooks is 0.72 W/m · °C, which is in SI units. To make matters worse, the engineer does not have a direct conversion factor between the two unit systems for thermal conductivity. Can you help him out?(answer upto two decimal)
Correct
solution :The situation this engineer is facing is not unique, and most engineers often find themselves in a similar position. A person must be very careful during unit conversion not to fall into some common pitfalls and to avoid some costly mistakes. Although unit conversion is a simple process, it requires utmost care and careful reasoning. The conversion factors for W and m are straightforward and are given in conversion tables to be 1 W=3.41214 Btu/h 1 m= 3.2808 ft But the conversion of °C into °F is not so simple, and it can be a source of error if one is not careful. Perhaps the first thought that comes to mind is to replace °C by (°F 32)/1.8 since T(°C) [T(°F) 32]/1.8. But this will be wrong since the °C in the unit W/m · °C represents per °C change in temperature. Noting that 1°C change in temperature corresponds to 1.8°F, the proper conversion factor to be used is
1°C =1.8°F
Substituting, we get
1 W/m · °C=3.41214 Btu/h/(3.2808 ft)(1.8°F)=0.5778 Btu/h · ft · °F which is the desired conversion factor. Therefore, the thermal conductivity of the brick in English units is Kbrick=0.72 w/m.°C =0.72x(0.5778 BTU/h. ft.°F =0.42 BTU/h.ft.°F
Incorrect
solution :The situation this engineer is facing is not unique, and most engineers often find themselves in a similar position. A person must be very careful during unit conversion not to fall into some common pitfalls and to avoid some costly mistakes. Although unit conversion is a simple process, it requires utmost care and careful reasoning. The conversion factors for W and m are straightforward and are given in conversion tables to be 1 W=3.41214 Btu/h 1 m= 3.2808 ft But the conversion of °C into °F is not so simple, and it can be a source of error if one is not careful. Perhaps the first thought that comes to mind is to replace °C by (°F 32)/1.8 since T(°C) [T(°F) 32]/1.8. But this will be wrong since the °C in the unit W/m · °C represents per °C change in temperature. Noting that 1°C change in temperature corresponds to 1.8°F, the proper conversion factor to be used is
1°C =1.8°F
Substituting, we get
1 W/m · °C=3.41214 Btu/h/(3.2808 ft)(1.8°F)=0.5778 Btu/h · ft · °F which is the desired conversion factor. Therefore, the thermal conductivity of the brick in English units is Kbrick=0.72 w/m.°C =0.72x(0.5778 BTU/h. ft.°F =0.42 BTU/h.ft.°F

Question 12 of 20
12. Question
1 pointsA 2mlong, 0.3cmdiameter electrical wire extends across a room at 15°C, as shown in (Fig.). Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be 152°C in steady operation. Also, the voltage drop and electric current through the wire are measured to be 60 V and 1.5 A, respectively. Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room.( answer upto one decimal)
Assumptions :1 Steady operating conditions exist since the temperature readings do not change with time. 2 Radiation heat transfer is negligible
Correct
solution :The convection heat transfer coefficient for heat transfer from an electrically heated wire to air is to be determined by measuring temperatures when steady operating conditions are reached and the electric power consumed.
. Analysis :When steady operating conditions are reached, the rate of heat loss from the wire will equal the rate of heat generation in the wire as a result of resistance heating. That is,
Q=Egenerated =VI=(60 V)(1.5 A)=90 W The surface area of the wire is As=π(0.003 m)(2m)=0.01885 m² Newton’s law of cooling for convection heat transfer is expressed as Qconv=hAs(Ts T∞)
Disregarding any heat transfer by radiation and thus assuming all the heat loss from the wire to occur by convection, the convection heat transfer coefficient is determined to be h=Qconv/As(TsT∞)= 90w/(0.01885 m²) ( 152 – 15) °C=34.9 W/m².°C
Incorrect
solution :The convection heat transfer coefficient for heat transfer from an electrically heated wire to air is to be determined by measuring temperatures when steady operating conditions are reached and the electric power consumed.
. Analysis :When steady operating conditions are reached, the rate of heat loss from the wire will equal the rate of heat generation in the wire as a result of resistance heating. That is,
Q=Egenerated =VI=(60 V)(1.5 A)=90 W The surface area of the wire is As=π(0.003 m)(2m)=0.01885 m² Newton’s law of cooling for convection heat transfer is expressed as Qconv=hAs(Ts T∞)
Disregarding any heat transfer by radiation and thus assuming all the heat loss from the wire to occur by convection, the convection heat transfer coefficient is determined to be h=Qconv/As(TsT∞)= 90w/(0.01885 m²) ( 152 – 15) °C=34.9 W/m².°C

Question 13 of 20
13. Question
1 pointsIt is a common experience to feel “chilly” in winter and “warm” in summer in our homes even when the thermostat setting is kept the same. This is due to the so called “radiation effect” resulting from radiation heat exchange between our bodies and the surrounding surfaces of the walls and the ceiling. Consider a person standing in a room maintained at 22°C at all times. The inner surfaces of the walls, floors, and the ceiling of the house are observed to be at an average temperature of 10°C in winter and 25°C in summer. Determine the rate of radiation heat transfer between this person and the surrounding surfaces if the exposed surface area and the average outer surface temperature of the person are 1.4 m² and 30°C, respectively (Fig)
Assumptions :1 Steady operating conditions exist. 2 Heat transfer by convection is not considered. 3 The person is completely surrounded by the interior surfaces of the room. 4 The surrounding surfaces are at a uniform temperature.
Properties :The emissivity of a person is 0.95
Correct
solution :The rates of radiation heat transfer between a person and the surrounding surfaces at specified temperatures are to be determined in summer and winter.
Analysis :The net rates of radiation heat transfer from the body to the surrounding walls, ceiling, and floor in winter and summer are Qrad, winter =€as (T 4s T 4surr, winter) =(0.95)(5.67 x 10^8 W/m² · K4)(1.4 m²)x [(30+ 273)^4 (10+ 273)^4] K4 =152 W Qrad, winter =€as (T 4s T 4surr, winter) =(0.95)(5.67 x 10^8 W/m² · K4)(1.4 m²)x [(30+ 273)^4 (25+ 273)^4] K4=40.9w
Incorrect
solution :The rates of radiation heat transfer between a person and the surrounding surfaces at specified temperatures are to be determined in summer and winter.
Analysis :The net rates of radiation heat transfer from the body to the surrounding walls, ceiling, and floor in winter and summer are Qrad, winter =€as (T 4s T 4surr, winter) =(0.95)(5.67 x 10^8 W/m² · K4)(1.4 m²)x [(30+ 273)^4 (10+ 273)^4] K4 =152 W Qrad, winter =€as (T 4s T 4surr, winter) =(0.95)(5.67 x 10^8 W/m² · K4)(1.4 m²)x [(30+ 273)^4 (25+ 273)^4] K4=40.9w

Question 14 of 20
14. Question
1 pointsConsider a person standing in a breezy room at 20°C. Determine the total rate of heat transfer from this person if the exposed surface area and the average outer surface temperature of the person are 1.6 m² and 29°C, respectively, and the convection heat transfer coefficient is 6 W/m² · °C (Fig.)
Assumptions 1 Steady operating conditions exist. 2 The person is completely surrounded by the interior surfaces of the room. 3 The surrounding surfaces are at the same temperature as the air in the room. 4 Heat conduction to the floor through the feet is negligible. Properties The emissivity of a person is 0.95
Correct
solution :The total rate of heat transfer from a person by both convection and radiation to the surrounding air and surfaces at specified temperatures is to be determined.
Analysis: The heat transfer between the person and the air in the room will be by convection (instead of conduction) since it is conceivable that the air in the vicinity of the skin or clothing will warm up and rise as a result of heat transfer from the body, initiating natural convection currents. It appears that the experimentally determined value for the rate of convection heat transfer in this case is 6 W per unit surface area (m²) per unit temperature difference (in K or °C) between the person and the air away from the person. Thus, the rate of convection heat transfer from the person to the air in the room is Qconv=hAs (Ts T∞ =(6 W/m² · °C)(1.6 m²)(29 20)°C=86.4 WThe person will also lose heat by radiation to the surrounding wall surfaces. We take the temperature of the surfaces of the walls, ceiling, and floor to be equal to the air temperature in this case for simplicity, but we recognize that this does not need to be the case. These surfaces may be at a higher or lower temperature than the average temperature of the room air, depending on the outdoor conditions and the structure of the walls. Considering that air does not intervene with radiation and the person is completely enclosed by the surrounding surfaces, the net rate of radiation heat transfer from the person to the surrounding walls, ceiling, and floor is Qrad= ∈aAs (T 4s T 4surr)=(0.95)(5.67 x 10^8 W/m² ^ K4)(1.6 m²) x [(29 +273)^4 (20+ 273)^4] ^K4 =81.7w
Note that we must use absolute temperatures in radiation calculations. Also
note that we used the emissivity value for the skin and clothing at room temperature since the emissivity is not expected to change significantly at a slightly higher temperature. Then the rate of total heat transfer from the body is determined by adding these two quantities: Qtotal =Qconv +Qrad (86.4 +81.7) W =168.1 WIncorrect
solution :The total rate of heat transfer from a person by both convection and radiation to the surrounding air and surfaces at specified temperatures is to be determined.
Analysis: The heat transfer between the person and the air in the room will be by convection (instead of conduction) since it is conceivable that the air in the vicinity of the skin or clothing will warm up and rise as a result of heat transfer from the body, initiating natural convection currents. It appears that the experimentally determined value for the rate of convection heat transfer in this case is 6 W per unit surface area (m²) per unit temperature difference (in K or °C) between the person and the air away from the person. Thus, the rate of convection heat transfer from the person to the air in the room is Qconv=hAs (Ts T∞ =(6 W/m² · °C)(1.6 m²)(29 20)°C=86.4 WThe person will also lose heat by radiation to the surrounding wall surfaces. We take the temperature of the surfaces of the walls, ceiling, and floor to be equal to the air temperature in this case for simplicity, but we recognize that this does not need to be the case. These surfaces may be at a higher or lower temperature than the average temperature of the room air, depending on the outdoor conditions and the structure of the walls. Considering that air does not intervene with radiation and the person is completely enclosed by the surrounding surfaces, the net rate of radiation heat transfer from the person to the surrounding walls, ceiling, and floor is Qrad= ∈aAs (T 4s T 4surr)=(0.95)(5.67 x 10^8 W/m² ^ K4)(1.6 m²) x [(29 +273)^4 (20+ 273)^4] ^K4 =81.7w
Note that we must use absolute temperatures in radiation calculations. Also
note that we used the emissivity value for the skin and clothing at room temperature since the emissivity is not expected to change significantly at a slightly higher temperature. Then the rate of total heat transfer from the body is determined by adding these two quantities: Qtotal =Qconv +Qrad (86.4 +81.7) W =168.1 W 
Question 15 of 20
15. Question
1 pointsConsider steady heat transfer between two large parallel plates at constant temperatures of T1 =300 K and T2= 200 K that are L= 1 cm apart, as shown in Fig.. Assuming the surfaces to be black (emissivity∈= 1), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is filled with atmospheric air,
Assumptions :1 Steady operating conditions exist. 2 There are no natural convection currents in the air between the plates. 3 The surfaces are black and thus ∈= 1.
Properties: The thermal conductivity at the average temperature of 250 K is k =0.0219 W/m · °C for air , 0.026 W/m · °C for urethane insulation , and 0.00002 W/m · °C for the superinsulation.Correct
solution :The total rate of heat transfer between two large parallel plates at specified temperatures is to be determined for four different cases.
Analysis : The rates of conduction and radiation heat transfer between the plates through the air layer are Qcond=KA =T1T2/L=( 0.0219 W/M°C)( 1 m²) (300200)°C/0.01 m=219 W and Qrad=€aA(T 41 T 42)=(1)(5.67 x 10^8 W/m² · K4)(1 m²)[(300 K)^4 (200 K)^4] =368 W
Therefore,
Qtotal =Qcond+Qrad= 219+ 368= 587 W The heat transfer rate in reality will be higher because of the natural convection currents that are likely to occur in the air space between the plates.
Incorrect
solution :The total rate of heat transfer between two large parallel plates at specified temperatures is to be determined for four different cases.
Analysis : The rates of conduction and radiation heat transfer between the plates through the air layer are Qcond=KA =T1T2/L=( 0.0219 W/M°C)( 1 m²) (300200)°C/0.01 m=219 W and Qrad=€aA(T 41 T 42)=(1)(5.67 x 10^8 W/m² · K4)(1 m²)[(300 K)^4 (200 K)^4] =368 W
Therefore,
Qtotal =Qcond+Qrad= 219+ 368= 587 W The heat transfer rate in reality will be higher because of the natural convection currents that are likely to occur in the air space between the plates.

Question 16 of 20
16. Question
1 pointsConsider steady heat transfer between two large parallel plates at constant temperatures of T1 =300 K and T2= 200 K that are L= 1 cm apart, as shown in Fig.. Assuming the surfaces to be black (emissivity∈= 1), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates isevacuated
Assumptions :1 Steady operating conditions exist. 2 There are no natural convection currents in the air between the plates. 3 The surfaces are black and thus ∈= 1.
Properties: The thermal conductivity at the average temperature of 250 K is k =0.0219 W/m · °C for air , 0.026 W/m · °C for urethane insulation , and 0.00002 W/m · °C for the superinsulation.Correct
solution: The total rate of heat transfer between two large parallel plates at specified temperatures is to be determined for four different cases. When the air space between the plates is evacuated, there will be no conduction or convection, and the only heat transfer between the plates will be by radiation. Therefore, ·Qtotal= Qrad= 368 W
Incorrect
solution: The total rate of heat transfer between two large parallel plates at specified temperatures is to be determined for four different cases. When the air space between the plates is evacuated, there will be no conduction or convection, and the only heat transfer between the plates will be by radiation. Therefore, ·Qtotal= Qrad= 368 W

Question 17 of 20
17. Question
1 pointsConsider steady heat transfer between two large parallel plates at constant temperatures of T1 =300 K and T2= 200 K that are L= 1 cm apart, as shown in Fig.. Assuming the surfaces to be black (emissivity∈= 1), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is filled with urethane insulation, and
Assumptions :1 Steady operating conditions exist. 2 There are no natural convection currents in the air between the plates. 3 The surfaces are black and thus ∈= 1.
Properties: The thermal conductivity at the average temperature of 250 K is k =0.0219 W/m · °C for air , 0.026 W/m · °C for urethane insulation , and 0.00002 W/m · °C for the superinsulation.Correct
solution :The total rate of heat transfer between two large parallel plates at specified temperatures is to be determined for four different cases. An opaque solid material placed between two plates blocks direct radiation heat transfer between the plates. Also, the thermal conductivity of an insulating material accounts for the radiation heat transfer that may be occurring through the voids in the insulating material. The rate of heat transfer through the urethane insulation is Qtotal= Qcond=KA T1T2/L=(0.026 W/m.°C)(1 m²) (300 200)°C/0.01 m=260 W
Incorrect
solution :The total rate of heat transfer between two large parallel plates at specified temperatures is to be determined for four different cases. An opaque solid material placed between two plates blocks direct radiation heat transfer between the plates. Also, the thermal conductivity of an insulating material accounts for the radiation heat transfer that may be occurring through the voids in the insulating material. The rate of heat transfer through the urethane insulation is Qtotal= Qcond=KA T1T2/L=(0.026 W/m.°C)(1 m²) (300 200)°C/0.01 m=260 W

Question 18 of 20
18. Question
1 pointsConsider steady heat transfer between two large parallel plates at constant temperatures of T1 =300 K and T2= 200 K that are L= 1 cm apart, as shown in Fig.. Assuming the surfaces to be black (emissivity∈= 1), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is filled with superinsulation that has an apparent thermal conductivity of 0.00002 W/m · °C.
Assumptions :1 Steady operating conditions exist. 2 There are no natural convection currents in the air between the plates. 3 The surfaces are black and thus ∈= 1.
Properties: The thermal conductivity at the average temperature of 250 K is k =0.0219 W/m · °C for air , 0.026 W/m · °C for urethane insulation , and 0.00002 W/m · °C for the superinsulation.Correct
solution : The total rate of heat transfer between two large parallel plates at specified temperatures is to be determined for four different cases. The layers of the superinsulation prevent any direct radiation heat transfer between the plates. However, radiation heat transfer between the sheets of superinsulation does occur, and the apparent thermal conductivity of the superinsulation accounts for this effect. Therefore, Qtotal=KA T1T2/L=(0.00002 W/m°C) ( 1 m²) (300 200)°C/ 0.01 m=0.2 W
Incorrect
solution : The total rate of heat transfer between two large parallel plates at specified temperatures is to be determined for four different cases. The layers of the superinsulation prevent any direct radiation heat transfer between the plates. However, radiation heat transfer between the sheets of superinsulation does occur, and the apparent thermal conductivity of the superinsulation accounts for this effect. Therefore, Qtotal=KA T1T2/L=(0.00002 W/m°C) ( 1 m²) (300 200)°C/ 0.01 m=0.2 W

Question 19 of 20
19. Question
1 pointsA thin metal plate is insulated on the back and exposed to solar radiation at the front surface (Fig. ). The exposed surface of the plate has an absorptivity of 0.6 for solar radiation. If solar radiation is incident on the plate at a rate of 700 W/m² and the surrounding air temperature is 25°C, determine the surface temperature of the plate when the heat loss by convection and radiation equals the solar energy absorbed by the plate. Assume the combined convection and
radiation heat transfer coefficient to be 50 W/m2 · °C.Assumptions: 1 Steady operating conditions exist. 2 Heat transfer through the insulated side of the plate is negligible. 3 The heat transfer coefficient remains constant. Properties: The solar absorptivity of the plate is given to be ∝=0.6.
Correct
solution :The back side of the thin metal plate is insulated and the front side is exposed to solar radiation. The surface temperature of the plate is to be determined when it stabilizes.
Analysis The absorptivity of the plate is 0.6, and thus 60 percent of the solar radiation incident on the plate will be absorbed continuously. As a result, the temperature of the plate will rise, and the temperature difference between the plate and the surroundings will increase. This increasing temperature difference will cause the rate of heat loss from the plate to the surroundings to increase. At some point, the rate of heat loss from the plate will equal the rate of solar energy absorbed, and the temperature of the plate will no longer change. The temperature of the plate when steady operation is established is determined from Egained =Elost or ∝As q incident, solar =hcombined As (Ts T∞) Solving for Ts and substituting, the plate surface temperature is determined to be Ts=T∞+∝ qincident, solar/ hcombined= 25°C+0.6 x (700 W/m²)/50 W/m² · °C=33.4°CIncorrect
solution :The back side of the thin metal plate is insulated and the front side is exposed to solar radiation. The surface temperature of the plate is to be determined when it stabilizes.
Analysis The absorptivity of the plate is 0.6, and thus 60 percent of the solar radiation incident on the plate will be absorbed continuously. As a result, the temperature of the plate will rise, and the temperature difference between the plate and the surroundings will increase. This increasing temperature difference will cause the rate of heat loss from the plate to the surroundings to increase. At some point, the rate of heat loss from the plate will equal the rate of solar energy absorbed, and the temperature of the plate will no longer change. The temperature of the plate when steady operation is established is determined from Egained =Elost or ∝As q incident, solar =hcombined As (Ts T∞) Solving for Ts and substituting, the plate surface temperature is determined to be Ts=T∞+∝ qincident, solar/ hcombined= 25°C+0.6 x (700 W/m²)/50 W/m² · °C=33.4°C 
Question 20 of 20
20. Question
1 pointsThe difference of two numbers is 4, and the sum of the squares of these two numbers is equal to the sum of the numbers plus 20. Determine these two numbers.
Correct
solution :Relations are given for the difference and the sum of the squaresof two numbers. They are to be determined.
Analysis :We start the EES program by doubleclicking on its icon, open a new file, and type the following on the blank screen that appears:xy=4
x^2+y^2=x+y+20
which is an exact mathematical expression of the problem statement withx and y denoting the unknown numbers. The solution to this system of two nonlinear equations with two unknowns is obtained by a single click on the“calculator” symbol on the taskbar. It gives
x=5 and y=1Incorrect
solution :Relations are given for the difference and the sum of the squaresof two numbers. They are to be determined.
Analysis :We start the EES program by doubleclicking on its icon, open a new file, and type the following on the blank screen that appears:xy=4
x^2+y^2=x+y+20
which is an exact mathematical expression of the problem statement withx and y denoting the unknown numbers. The solution to this system of two nonlinear equations with two unknowns is obtained by a single click on the“calculator” symbol on the taskbar. It gives
x=5 and y=1