Thermodynamics An Engireeng Approach 8Th edition By SI Units – Test Bank

Thermodynamics An Engineering Approach 8Th edition By SI Units – Test Bank

Multiple-Choice Test Problems

Chapter 8: Exergy: A Measure of Work Potential

Çengel/Boles – Thermodynamics: An Engineering Approach, 8^th Edition

(Numerical values for solutions can be obtained by copying the EES solutions given and pasting them on a blank EES screen, and pressing the Solve command. Similar problems and their solutions can be obtained easily by modifying numerical values.)

Chap8-1 Exergy Destruction for Heat Conduction

Heat is transferred through a plane wall steadily at a rate of 750 W. If the inner and outer surface temperatures of the wall are 30°C and 15°C, respectively, and the environment temperature is 15°C, the rate of exergy destruction within the wall is

(a) 37 W (b) 731 W (c) 7200 W (d) 9600 W (e) 1.9 W

Answer  (a) 37 W

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

Q=750 “W”

T1=30 “C”

T2=15 “C”

To=15 “C”

“Entropy balance S_in – S_out + S_gen= DS_system for the wall for steady operation gives”

Q/(T1+273)-Q/(T2+273)+S_gen=0   “W/K”

X_dest=(To+273)*S_gen “W”

“Wrong Solutions:”

Q/T1-Q/T2+Sgen1=0; W1_Xdest=(To+273)*Sgen1   “Using C instead of K in Sgen”

Sgen2=Q/((T1+T2)/2); W2_Xdest=(To+273)*Sgen2    “Using avegage temperature in C for Sgen”     

Sgen3=Q/((T1+T2)/2+273); W3_Xdest=(To+273)*Sgen3    “Using avegage temperature in K”

W4_Xdest=To*S_gen   “Using C for To”

Chap8-2 Exergy Destrustion During Liquid Flow

Kerosene enters an adiabatic piping system at 20°C at a rate of 12 kg/s. It is observed that the temperature rises by 0.2°C in the pipe due to friction. If the environment temperature is also 20°C, the rate of exergy destruction in the pipe is

(a) 70 kW (b) 4.8 kW (c) 0.40 kW (d) 5.8 kW (e) 24 kW

Answer  (b) 4.8 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

Cp=2.0 “kJ/kg.K”

m=12 “kg/s”

T1=20 “C”

T2=20.2 “C”

To=20 “C”

S_gen=m*Cp*ln((T2+273)/(T1+273)) “kW/K”

X_dest=(To+273)*S_gen “kW”

“Wrong Solutions:”

W1_Xdest=(To+273)*m*Cp*ln(T2/T1) “Using deg. C in Sgen”

W2_Xdest=To*m*Cp*ln(T2/T1) “Using deg. C in Sgen and To”

W3_Xdest=(To+273)*Cp*ln(T2/T1) “Not using mass flow rate with deg. C”

W4_Xdest=(To+273)*Cp*ln((T2+273)/(T1+273)) “Not using mass flow rate with K”

Chap8-3 2nd-Law Efficiency of a Heat Engine

A heat engine receives heat from a source at 1200 K and rejects the waste heat to a sink at 400 K. If the thermal efficiency of the engine is 0.40, the second-law efficiency of this heat engine is

(a) 40% (b) 27% (c) 60% (d) 67% (e) 83%

Answer  (c) 60%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

TL=400 “K”

TH=1200 “K”

Eta_th=0.40

Eta_rev=1-TL/TH

Eta_II=Eta_th/Eta_rev

“Wrong Solutions:”

W1_Eta_II=Eta_rev1/Eta_th; Eta_rev1=TL/TH “Using wrong relation for reversible efficiency”

W2_Eta_II=Eta_th “Taking second-law efficiency to be thermal efficiency”

W3_Eta_II=Eta_rev “Taking second-law efficiency to be reversible efficiency”

W4_Eta_II=Eta_th*Eta_rev “Multiplying thermal and reversible efficiencies instead of dividing”

Chap8-4 Work potential of water reservoir

A pool contains 800 tons of water at an average elevation of 75 m. The maximum amount of electric power that can be generated from this water is

(a) 16,700 kWh (b) 164 kWh (c) 163,500 kWh (d) 16,300 kWh (e) 588,600 kWh

Answer  (b) 164 kWh

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

m=800000 “kg”

h=75 “m”

g=9.81 “m/s^2”

“Maximum power is simply the potential energy change,”

W_max=m*g*h/1000 “kJ”

W_max_kWh=W_max/3600 “kWh”

“Some Wrong Solutions with Common Mistakes:”

W1_Wmax =m*g*h/3600 “Not using the conversion factor 1000”

W2_Wmax =m*g*h/1000 “Obtaining the result in kJ instead of kWh”

W3_Wmax =m*g*h*3.6/1000 “Using worng conversion factor”

W4_Wmax =m*h/3600“Not using g and the factor 1000 in calculations”

Chap8-5 Second-Law Efficiency of Resistance Heater

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