Question 1

R134a is condensed at a constant temperature of 30°C as it flows through the condenser of a refrigerator by rejecting heat at a rate of 6 kW. The rate of entropy change of R134a as it flows through the condenser is
A)-0.0198 kW/K
B)0 kW/K
C)0.20 kW/K
D)-0.567 kW/K
E)0.0198 kW/K

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
T1=30 "C"
Q_out=6 "kW" 
S_change=-Q_out/(T1+273) "kW/K" 
"Some Wrong Solutions with Common Mistakes:"
W1_S_change=0 "Assuming no change"
W2_S_change=Q_out/T1 "Using temperature in C"
W3_S_change=Q_out/(T1+273) "Wrong sign"
W4_S_change=-s_fg "Taking entropy of vaporization"
s_fg=(ENTROPY(R134a,T=T1,x=1)-ENTROPY(R134a,T=T1,x=0))

Question 2

Steam is compressed from 4 MPa and 300SYMBOL 176 \f "Symbol"C to 9 MPa isentropically. The final temperature of the steam is
A)300°C
B)418°C
C)303°C
D)362°C
E)446°C

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
P1=4000 "kPa"
T1=300 "C"
P2=9000 "kPa"
s2=s1
s1=ENTROPY(Steam_IAPWS,T=T1,P=P1)
T2=TEMPERATURE(Steam_IAPWS,s=s2,P=P2)
"Some Wrong Solutions with Common Mistakes:"
W1_T2=T1 "Assuming temperature remains constant"
W2_T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2) "Saturation temperature at P2"
W3_T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2) "Saturation temperature at P1"

Question 3

The temperature of a cold orange with an average mass of 0.28 kg and average specific heat of 3.42 kJ/kg.SYMBOL 176 \f "Symbol"C rises from 7°C to 25°C as a result of heat transfer from the suroundign air. The entropy change of the orange is
A)0.213 kJ/K
B)1.22 kJ/K
C)0.0597 kJ/K
D)0 kJ/K
E)0.154 kJ/K

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
C=3.42 "kJ/kg.K"
m=0.28 "kg"
T1=7 "C"
T2=25 "C"
S_change=m*C*ln((T2+273)/(T1+273))
 
"Some Wrong Solutions with Common Mistakes:"
W1_S_change=C*ln((T2+273)/(T1+273)) "Not using mass"
W2_S_change=m*C*ln(T2/T1) "Using C"
W3_S_change=m*C*(T2-T1)/(T1+273) "Using Q/T_ave"

Question 4

Argon gas is compressed from 150 kPa nd 25°C to a pressure of 700 kPa adiabatically. The lowest temperature of argon after compression is
A)25°C
B)46°C
C)117°C
D)279°C
E)472°C

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
k=1.667
P1=150 "kPa"
T1=25 "C"
P2=700 "kPa"
"s2=s1"
"The exit temperature will be lowest for isentropic compression," 
T2=(T1+273)*(P2/P1)^((k-1)/k) "K"
T2_C= T2-273 "C"
"Some Wrong Solutions with Common Mistakes:"
W1_T2=T1 "Assuming temperature remains constant"
W2_T2=T1*(P2/P1)^((k-1)/k) "Using C instead of K"
W3_T2=(T1+273)*(P2/P1)-273 "Assuming T is proportional to P"
W4_T2=T1*(P2/P1) "Assuming T is proportional to P, using C"

Question 5

Steam expands in an adiabatic turbine from 8 MPa and 450°C to a pressure of 50 kPa at a rate of 1.8 kg/s. The maximum power output of the turbine is
A)1129 kW
B)995 kW
C)2136 kW
D)718 kW
E)1791 kW

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
P1=8000 "kPa"
T1=450 "C"
P2=50 "kPa"
m=1.8 "kg/s"
h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1)
s1=ENTROPY(Steam_IAPWS,T=T1,P=P1)
h2=ENTHALPY(Steam_IAPWS, s=s1,P=P2)
W_out=m*(h1-h2) 
"Some Wrong Solutions with Common Mistakes:"
h22=ENTHALPY(Steam_IAPWS,x=1,P=P2) "Assuming sat. vapor at turbine exit"
W1_Wout=m*(h1-h22)
W2_Wout=h1-h2 "Not using mass flow rate"

Question 6

Helium gas expands in an adiabatic turbine from 2 MPa and 750°C to 0.18 MPa at a rate of 1.5 kg/s. Assuming constant specific heats at room temperature, the maximum power output of the turbine is
A)1384 kW
B)830 kW
C)3613 kW
D)5316 kW
E)923 kW

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
Cp=5.1926
k=1.667
P1=2000 "kPa"
T1=750 "C"
m=1.5 "kg/s"
P2=180 "kPa"
"s2=s1"
T2=(T1+750)*(P2/P1)^((k-1)/k)
W_max=m*Cp*(T1-T2)
"Some Wrong Solutions with Common Mistakes:"
Cv=3.1156 "kJ/kg.K"
W1_Wmax=m*Cv*(T1-T2) "Using Cv"
T22=T1*(P2/P1)^((k-1)/k) "Using C instead of K"
W2_Wmax=m*Cp*(T1-T22)
W3_Wmax=Cp*(T1-T2) "Not using mass flow rate"
T24=T1*(P2/P1) "Assuming T is proportional to P, using C"
W4_Wmax=m*Cp*(T1-T24)

Question 7

A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from volume V₁ to volume V₂ while loosing heat to the surroundings at temperature T in the amount of q. If the gas constant of the gas is R, the entropy change of the gas $\Delta$s during this process is A)$\Delta$s =R ln(V₂/V₁)-q/T B)$\Delta$s = R ln(V₂/V₁) C)$\Delta$s =R ln(V₁/V₂) D)$\Delta$s =R ln(V₁/V₂)-q/T E)$\Delta$s= 0

Accepted Answer

For an isothermal reversible process, the entropy change is given by $\Delta$s = R ln(V₂/V₁).

Question 8

Air is compressed from 30°C and 4.2 L to a volume of 0.50 m³ in a reversible adiabatic manner. The air temperature after compression is A)30°C B)70°C C)437°C D)252°C E)635°C "

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
fluid$='R134a'
rho1=density(Fluid$, P=100, x=0)
k=1.40
v1=4.2 "L"
T1=30 "C"
v2=0.50 "L"
"s2=s1"
"The exit temperature is determined from isentropic compression relation," 
T2=(T1+273)*(v1/v2)^(k-1) "K"
T2_C= T2-273 "C"
 
"Some Wrong Solutions with Common Mistakes:"
W1_T2=T1 "Assuming temperature remains constant"
W2_T2=T1*(v1/v2)^(k-1) "Using C instead of K"
W3_T2=(T1+273)*(v1/v2)-273 "Assuming T is proportional to v"
W4_T2=T1*(v1/v2) "Assuming T is proportional to v, using C"

Question 9

Heat is lost through a plane wall steadily at a rate of 750 W. If the inner and outer surface temperatures of the wall are 50°C and 15°C, respectively, the rate of entropy generation within the wall is
A)0.0289 W/K
B)2.10 W/K
C)2.08 W/K
D)0.111 W/K
E)48.0 W/K

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
Q=600 "W"
T1=20+273 "K"
T2=5+273 "K"
"Entropy balance S_in - S_out + S_gen= DS_system for the wall for steady operation gives" 
Q/T1-Q/T2+S_gen=0 "W/K" 
"Some Wrong Solutions with Common Mistakes:"
Q/(T1+273)-Q/(T2+273)+W1_Sgen=0 "Using C instead of K"
W2_Sgen=Q/((T1+T2)/2) "Using avegage temperature in K" 
W3_Sgen=Q/((T1+T2)/2-273) "Using avegage temperature in C" 
W4_Sgen=Q/(T1-T2+273) "Using temperature difference in K"

Question 10

Air is compressed steadily and adiabatically from 15°C and 90 kPa to 310°C and 600 kPa. Assuming constant specific heats for air at room temperature, the isentropic efficiency of the compressor is
A)0.038
B)0.64
C)0.86
D)0.92
E)0.73

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
Cp=1.005 "kJ/kg.K"
k=1.4
P1=90 "kPa"
T1=15 "C"
P2=600 "kPa"
T2=300 "C"
T2s=(T1+273)*(P2/P1)^((k-1)/k)-273
Eta_comp=(Cp*(T2s-T1))/(Cp*(T2-T1))
"Some Wrong Solutions with Common Mistakes:"
T2sW1=T1*(P2/P1)^((k-1)/k) "Using C instead of K in finding T2s"
W1_Eta_comp=(Cp*(T2sW1-T1))/(Cp*(T2-T1)) 
W2_Eta_comp=T2s/T2 "Using wrong definition for isentropic efficiency, and using C"
W3_Eta_comp=(T2s+273)/(T2+273) "Using wrong definition for isentropic efficiency, with K"

Question 11

Helium gas expands in an adiabatic turbine steadily from 700°C and 900 kPa to 70 kPa at a rate of 0.15 kg/s. For an isentropic efficiency of 82%, the power produced by the turbine is
A)398 kW
B)286 kW
C)592 kW
D)2650 kW
E)239 kW

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
Cp=5.1926 "kJ/kg-K"
k=1.667
m=0.15 "kg/s"
T1=700 "C"
P1=900 "kPa"
P2=70 "kPa"
Eta_turb=0.82 
T2s=(T1+273)*(P2/P1)^((k-1)/k)-273
Eta_turb=(Cp*(T2-T1))/(Cp*(T2s-T1))
W_out=m*Cp*(T1-T2)
"Some Wrong Solutions with Common Mistakes:"
T2sW1=T1*(P2/P1)^((k-1)/k) "Using C instead of K to find T2s"
Eta_turb=(Cp*(T2W1-T1))/(Cp*(T2sW1-T1))
W1_Wout=m*Cp*(T1-T2W1)
Eta_turb=(Cp*(T2s-T1))/(Cp*(T2W2-T1)) "Using wrong definition for isentropic efficiency, and using C"
W2_Wout=m*Cp*(T1-T2W2)
W3_Wout=Cp*(T1-T2) "Not using mass flow rate"
Cv=3.1156 "kJ/kg.K"
W4_Wout=m*Cv*(T1-T2) "Using Cv instead of Cp"

Question 12

Water enters a pump steadily at 180 kPa at a rate of 0.012 m³/s and leaves at 1200 kPa. If the changes in kinetic and potential energies are negligible, the minimum power input to the pump is A)37 kW B)40 kW C)12.2 kW D)52 kW E)74 kW

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
V=0.012 "m^3/s"
g=9.81 "m/s^2"
P1=180 "kPa"
P2=1200 "kPa"
"Pump power input is minimum when compression is reversible and thus W=m*v(*P2-P1)=V*(P2-P1)"
W_min=V*(P2-P1) "kPa.m^3/s=kW"
"Some Wrong Solutions with Common Mistakes:"
W1_Win=V*(P2+P1) "Adding pressures instead of subtracting"

Question 13

Water enters a pump steadily at 180 kPa at a rate of 0.012 m³/s and leaves at 1200 kPa. If the pump efficiency is 0.75 and the changes in kinetic and potential energies are negligible, the power input to the pump is A)9.2 kW B)16.3 kW C)22.1 kW D)12.2 kW E)28.5 kW

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
V=0.012 "m^3/s"
g=9.81 "m/s^2"
P1=180 "kPa"
Eff=0.75
P2=1200 "kPa"
"Pump power input is minimum when compression is reversible and thus W=m*v(*P2-P1)=V*(P2-P1)"
W_min=V*(P2-P1) "kPa.m^3/s=kW"
W_act=W_min/Eff
"Some Wrong Solutions with Common Mistakes:"
W1_Wact=W_min*Eff "Multiplying by efficiency"
W2_Wact=W_min "Disregarding efficiency"
W3_Win=V*(P2+P1)/Eff "Adding pressures instead of subtracting"

Question 14

Air at 900 kPa and 25°C is allowed to expand steadily and isothermally to 100 kPa at a rate of 1.4 kg/s. The maximum power output of the turbine is
A)22 kW
B)188 kW
C)263 kW
D)137 kW
E)94 kW

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
Cp=1.005 "kJ/kg.K"
R=0.287 "kJ/kg.K"
Cv=0.718 "kJ/kg.K"
k=1.4
P1=900 "kPa"
T=25 "C"
m=1.4 "kg/s"
P2=100 "kPa"
Wout=m*R*(T+273)*ln(P1/P2)
"Some Wrong Solutions with Common Mistakes:"
W1_Wout=m*R*T*ln(P1/P2) "Using C instead of K"
W2_Wout=m*T*(P1-P2) "Using wrong relation"
W3_Wout=R*(T+273)*ln(P1/P2) "Not using mass flow rate"

Question 15

Air is to be compressed steadily and isentropically from 2 atm to 18 atm by a two-stage compressor. To minimize the total compression work, the intermediate pressure between the two stages must be
A)4 atm
B)12 atm
C)10 atm
D)6 atm
E)18 atm

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
P1=2 "atm"
P2=18 "atm"
P_mid=SQRT(P1*P2)
"Some Wrong Solutions with Common Mistakes:"
W1_P=(P1+P2)/2 "Using average pressure"
W2_P=P1*P2/2 "Half of product"

Question 16

Argon gas enters an adiabatic nozzle steadily at 800°C and 700 kPa with a low velocity, and exits at a pressure of 120 kPa. The highest possible velocity of helium gas at the nozzle exit is
A)649 m/s
B)503 m/s
C)888 m/s
D)1070 m/s
E)752 m/s

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
k=1.667
Cp=0.5203 "kJ/kg.K"
Cv=0.3122 "kJ/kg.K"
T1=800 "C"
P1=700 "kPa"
Vel1=0
P2=120 "kPa"
"s2=s1 for maximum exit velocity"
"The exit velocity will be highest for isentropic expansion," 
T2=(T1+273)*(P2/P1)^((k-1)/k)-273 "C"
"Energy balance for this case is h+ke=constant for the fluid stream (Q=W=pe=0)"
(0.5*Vel1^2)/1000+Cp*T1=(0.5*Vel2^2)/1000+Cp*T2
 
"Some Wrong Solutions with Common Mistakes:"
T2a=T1*(P2/P1)^((k-1)/k) "Using C for temperature" 
(0.5*Vel1^2)/1000+Cp*T1=(0.5*W1_Vel2^2)/1000+Cp*T2a
T2b=T1*(P2/P1)^((k-1)/k) "Using Cv" 
(0.5*Vel1^2)/1000+Cv*T1=(0.5*W2_Vel2^2)/1000+Cv*T2b
T2c=T1*(P2/P1)^k "Using wrong relation" 
(0.5*Vel1^2)/1000+Cp*T1=(0.5*W3_Vel2^2)/1000+Cp*T2c

Question 17

Combustion gases with a specific heat ratio of 1.34 enter an adiabatic nozzle steadily at 720°C and 400 kPa with a low velocity, and exit at a pressure of 100 kPa. The lowest possible temperature of combustion gases at the nozzle exit is
A)426°C
B)507°C
C)699°C
D)112°C
E)363°C

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
k=1.34
T1=720 "C"
P1=400 "kPa"
P2=100 "kPa"
"Nozzle exit temperature will be lowest for isentropic operation"
T2=(T1+273)*(P2/P1)^((k-1)/k)-273
"Some Wrong Solutions with Common Mistakes:"
W1_T2=T1*(P2/P1)^((k-1)/k) "Using C for temperature"
W2_T2=(T1+273)*(P2/P1)^((k-1)/k) "Not converting the answer to C"
W3_T2=T1*(P2/P1)^k "Using wrong relation"

Question 18

Refrigerant-134a enters an adiabatic turbine steadily at 40°C and 0.9 MPa, and leaves at 140 kPa. The highest possible percentage of mass of R134a that condenses at the turbine exit and leaves the turbine as a liquid is
A)2%
B)2%.
C)8%
D)12%
E)18%

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
P1=900 "kPa"
T1=40 "C"
P2=140 "kPa"
s2=s1
s1=ENTROPY(R134a,T=T1,P=P1)
x2=QUALITY(R134a,s=s2,P=P2)
misture=1-x2

Question 19

Liquid kerosene with a specific heat of 2.0 kJ/kg.°C enters an adiabatic piping system at 18° at a rate of 3 kg/s. If the water temperature rises by 0.4°C during flow due to friction, the rate of entropy generation in the pipe is
A)132 W/K
B)44 W/K
C)8.2 W/K
D)2.7 W/K
E)89 W/K

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
Cp=2000 "J/kg.K"
m=3 "kg/s"
T1=18 "C"
T2=18.4 "C"
S_gen=m*Cp*ln((T2+273)/(T1+273)) "W/K"
"Some Wrong Solutions with Common Mistakes:"
W1_Sgen=m*Cp*ln(T2/T1) "Using deg. C"
W2_Sgen=Cp*ln(T2/T1) "Not using mass flow rate with deg. C"
W3_Sgen=Cp*ln((T2+273)/(T1+273)) "Not using mass flow rate with deg. K"

Question 20

Refrigerant-134a enters an adiabatic compressor as saturated vapor at 0.18 MPa at a rate of 1.6 kg/s, and exits at 1 MPa and 60°C. The rate of entropy generation in the turbine is
A)0 kW/K
B)0.47 kW/K
C)3.34 kW/K
D)1.26 kW/K
E)14.1 kW/K "

Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
P1=180 "kPa"
x1=1
m=1.6 "kg/s"
P2=200 "kPa"
T2=300 "C"
s1=ENTROPY(R134a,x=x1,P=P1)
s2=ENTROPY(R134a,T=T2,P=P2)
S_gen=m*(s2-s1) "kW/K"
"Some Wrong Solutions with Common Mistakes:"
W1_Sgen=0 "Assuming isentropic expansion"

Quiz 7: Entropy

R134a is condensed at a constant temperature of 30°C as it flows through the condenser of a refrigerator by rejecting heat at a rate of 6 kW. The rate of entropy change of R134a as it flows through the condenser is

Steam is compressed from 4 MPa and 300SYMBOL 176 \f "Symbol"C to 9 MPa isentropically. The final temperature of the steam is

The temperature of a cold orange with an average mass of 0.28 kg and average specific heat of 3.42 kJ/kg.SYMBOL 176 \f "Symbol"C rises from 7°C to 25°C as a result of heat transfer from the suroundign air. The entropy change of the orange is

Argon gas is compressed from 150 kPa nd 25°C to a pressure of 700 kPa adiabatically. The lowest temperature of argon after compression is

Steam expands in an adiabatic turbine from 8 MPa and 450°C to a pressure of 50 kPa at a rate of 1.8 kg/s. The maximum power output of the turbine is

Helium gas expands in an adiabatic turbine from 2 MPa and 750°C to 0.18 MPa at a rate of 1.5 kg/s. Assuming constant specific heats at room temperature, the maximum power output of the turbine is

A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from volume V₁ to volume V₂ while loosing heat to the surroundings at temperature T in the amount of q. If the gas constant of the gas is R, the entropy change of the gas $\Delta$ s during this process is

Air is compressed from 30°C and 4.2 L to a volume of 0.50 m³ in a reversible adiabatic manner. The air temperature after compression is

Heat is lost through a plane wall steadily at a rate of 750 W. If the inner and outer surface temperatures of the wall are 50°C and 15°C, respectively, the rate of entropy generation within the wall is

Air is compressed steadily and adiabatically from 15°C and 90 kPa to 310°C and 600 kPa. Assuming constant specific heats for air at room temperature, the isentropic efficiency of the compressor is

Helium gas expands in an adiabatic turbine steadily from 700°C and 900 kPa to 70 kPa at a rate of 0.15 kg/s. For an isentropic efficiency of 82%, the power produced by the turbine is

Water enters a pump steadily at 180 kPa at a rate of 0.012 m³/s and leaves at 1200 kPa. If the changes in kinetic and potential energies are negligible, the minimum power input to the pump is

Water enters a pump steadily at 180 kPa at a rate of 0.012 m³/s and leaves at 1200 kPa. If the pump efficiency is 0.75 and the changes in kinetic and potential energies are negligible, the power input to the pump is

Air at 900 kPa and 25°C is allowed to expand steadily and isothermally to 100 kPa at a rate of 1.4 kg/s. The maximum power output of the turbine is

Air is to be compressed steadily and isentropically from 2 atm to 18 atm by a two-stage compressor. To minimize the total compression work, the intermediate pressure between the two stages must be

Argon gas enters an adiabatic nozzle steadily at 800°C and 700 kPa with a low velocity, and exits at a pressure of 120 kPa. The highest possible velocity of helium gas at the nozzle exit is

Combustion gases with a specific heat ratio of 1.34 enter an adiabatic nozzle steadily at 720°C and 400 kPa with a low velocity, and exit at a pressure of 100 kPa. The lowest possible temperature of combustion gases at the nozzle exit is

Refrigerant-134a enters an adiabatic turbine steadily at 40°C and 0.9 MPa, and leaves at 140 kPa. The highest possible percentage of mass of R134a that condenses at the turbine exit and leaves the turbine as a liquid is

Liquid kerosene with a specific heat of 2.0 kJ/kg.°C enters an adiabatic piping system at 18° at a rate of 3 kg/s. If the water temperature rises by 0.4°C during flow due to friction, the rate of entropy generation in the pipe is

Refrigerant-134a enters an adiabatic compressor as saturated vapor at 0.18 MPa at a rate of 1.6 kg/s, and exits at 1 MPa and 60°C. The rate of entropy generation in the turbine is