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Complete Test Bank With Answers
Sample Questions Posted Below
Multiple-Choice Test Problems
Chapter 5: Mass and Energy Analysis of Control Volumes
Çengel/Boles – Thermodynamics: An Engineering Approach, 8th 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.)
Chap5-1 Steam Nozzle m_dot
Steam is accelerated by a nozzle steadily from a low velocity to a velocity of 220 m/s at a rate of 1.2 kg/s.
If the steam at the nozzle exit is at 300C and 2 MPa, the exit area of the nozzle is
(a) 6.8 cm2 (b) 7.2 cm2 (c) 3.8 cm2 (d) 54.6 cm2 (e) 22.8 cm2
Answer (a) 6.8 cm2
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
Vel_1=0 “m/s”
Vel_2=220 “m/s”
m=1.2 “kg/s”
T2=300 “C”
P2=2000 “kPa”
“The rate form of energy balance is E_dot_in – E_dot_out = DELTAE_dot_cv”
v2=VOLUME(Steam_IAPWS,T=T2,P=P2)
m=(1/v2)*A*Vel_2 “A in m^2”
“Some Wrong Solutions with Common Mistakes:”
R=0.4615 “kJ/kg.K”
P2*v2ideal=R*(T2+273)
m=(1/v2ideal)*W1_A*Vel_2 “assuming ideal gas”
P2*v3ideal=R*T2
m=(1/v3ideal)*W2_A*Vel_2 “assuming ideal gas and using C for temperature”
m=W3_A*Vel_2 “not using specific volume”
Chap5-2 R134a Diffuser m_dot
Refrigerant 134a enters a diffuser steadily at 0.5 MPa, 50C, and 120 m/s at a rate of 1.2 kg/s. The inlet
area of the diffuser is
(a) 0.81 cm2 (b) 4.8 cm2 (c) 26 cm2 (d) 5.3 cm2 (e) 100 cm2
Answer (b) 4.8 cm2
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
Vel_1=120 “m/s”
m=1.2 “kg/s”
T1=50 “C”P1=500 “kPa”
“The rate form of energy balance is E_dot_in – E_dot_out = DELTAE_dot_cv”
v1=VOLUME(R134a,T=T1,P=P1)
m=(1/v1)*A*Vel_1 “A in m^2”
“Some Wrong Solutions with Common Mistakes:”
R=0.08149 “kJ/kg.K”
P1*v1ideal=R*(T1+273)
m=(1/v1ideal)*W1_A*Vel_1 “assuming ideal gas”
P1*v2ideal=R*T1
m=(1/v2ideal)*W2_A*Vel_1 “assuming ideal gas and using C”
m=W3_A*Vel_1 “not using specific volume”
Chap5-3 Adiabatic Air-Water HX T_cold_out
An adiabatic heat exchanger is used to heat cold water at 8C entering at a rate of 3 kg/s by hot air at
150C entering also at rate of 3 kg/s. If the exit temperature of hot air is 30C, the exit temperature of
cold water is (use constant specific heats at room temperature)
(a) 150C (b) 30C (c) 36.9C (d) 28.6C (e) 128C
Answer (c) 36.9C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
C_w=4.18 “kJ/kg-C”
Cp_air=1.005 “kJ/kg-C”
Tw1=8 “C”
m_dot_w=3 “kg/s”
Tair1=150 “C”
Tair2=30 “C”
m_dot_air=3 “kg/s”
“The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out”
m_dot_air*Cp_air*(Tair1-Tair2)=m_dot_w*C_w*(Tw2-Tw1)
“Some Wrong Solutions with Common Mistakes:”
(Tair1-Tair2)=(W1_Tw2-Tw1) “Equating temperature changes of fluids”
Cv_air=0.718 “kJ/kg.K”
m_dot_air*Cv_air*(Tair1-Tair2)=m_dot_w*C_w*(W2_Tw2-Tw1) “Using Cv for air”
W3_Tw2=Tair1 “Setting inlet temperature of hot fluid = exit temperature of cold fluid”
W4_Tw2=Tair2 “Setting exit temperature of hot fluid = exit temperature of cold fluid”
Chap5-4 Air-Water HX w/Heat Loss T_cold
A heat exchanger is used to heat cold water at 8C entering at a rate of 1.2 kg/s by hot air at 90C
entering at rate of 2.5 kg/s. The heat exchanger is not insulated, and is loosing heat at a rate of 28 kJ/s. If
the exit temperature of hot air is 20C, the exit temperature of cold water is (use constant specific heats
at room temperature)
(a) 43.1C (b) 48.6C (c) 78.0C (d) 37.5C (e) 27.5C
Answer (d) 37.5CSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
C_w=4.18 “kJ/kg-C”
Cp_air=1.005 “kJ/kg-C”
Tw1=8 “C”
m_dot_w=1.2 “kg/s”
Tair1=90 “C”
Tair2=20 “C”
m_dot_air=2.5 “kg/s”
Q_loss=28 “kJ/s”
“The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out”
m_dot_air*Cp_air*(Tair1-Tair2)=m_dot_w*C_w*(Tw2-Tw1)+Q_loss
“Some Wrong Solutions with Common Mistakes:”
m_dot_air*Cp_air*(Tair1-Tair2)=m_dot_w*C_w*(W1_Tw2-Tw1) “Not considering Q_loss”
m_dot_air*Cp_air*(Tair1-Tair2)=m_dot_w*C_w*(W2_Tw2-Tw1)-Q_loss “Taking heat loss
as heat gain”
(Tair1-Tair2)=(W3_Tw2-Tw1) “Equating temperature changes of fluids”
Cv_air=0.718 “kJ/kg.K”
m_dot_air*Cv_air*(Tair1-Tair2)=m_dot_w*C_w*(W4_Tw2-Tw1)+Q_loss “Using Cv for air”
Chap5-5 Adiabatic HX of Water T_cold_out
An adiabatic heat exchanger is used to heat cold water at 12C entering at a rate of 4 kg/s by hot water
entering at 95C at rate of 2.5 kg/s. If the exit temperature of hot water is 50C, the exit temperature of
cold water is
(a) 90C (b) 50C (c) 95C (d) 57C (e) 40C
Answer (e) 40C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
C_w=4.18 “kJ/kg-C”
Tcold_1=12 “C”
m_dot_cold=4 “kg/s”
Thot_1=95 “C”
Thot_2=50 “C”
m_dot_hot=2.5 “kg/s”
Q_loss=0 “kJ/s”
“The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out”
m_dot_hot*C_w*(Thot_1-Thot_2)=m_dot_cold*C_w*(Tcold_2-Tcold_1)+Q_loss
“Some Wrong Solutions with Common Mistakes:”
Thot_1-Thot_2=W1_Tcold_2-Tcold_1 “Equating temperature changes of fluids”
W2_Tcold_2=90 “Taking exit temp of cold fluid=inlet temp of hot fluid”
Chap5-6 Adiabatic Mixing of Water T_mix
In a water heating system, cold water at 5C flowing at a rate of 5 kg/s is mixed adiabatically with hot
water at 80C flowing at a rate of 2 kg/s. The exit temperature of the mixture is
(a) 26.4C (b) 42.5C (c) 40.0C (d) 64.3C (e) 55.2CAnswer (a) 26.4C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
C_w=4.18 “kJ/kg-C”
Tcold_1=5 “C”
m_dot_cold=5 “kg/min”
Thot_1=80 “C”
m_dot_hot=2 “kg/min”
“The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out”
m_dot_hot*C_w*Thot_1+m_dot_cold*C_w*Tcold_1=(m_dot_hot+m_dot_cold)*C_w*Tmix
“Some Wrong Solutions with Common Mistakes:”
W1_Tmix=(Tcold_1+Thot_1)/2 “Taking the average temperature of inlet fluids”
Chap5-7 Adiabatic Mixing of Air T_mix
In a heating system, cold outdoor air at 10C flowing at a rate of 6 kg/min is mixed adiabatically with hot
air at 50C flowing at a rate of 1.5 kg/min. Assuming constant specific heats at room temperature, the exit
temperature of the mixture is
(a) 30C (b) 18C (c) 45C (d) 22C (e) 38C
Answer (b) 18C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
C_air=1.005 “kJ/kg-C”
Tcold_1=10 “C”
m_dot_cold=6 “kg/min”
Thot_1=50 “C”
m_dot_hot=1.5 “kg/min”
“The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out”
m_dot_hot*C_air*Thot_1+m_dot_cold*C_air*Tcold_1=(m_dot_hot+m_dot_cold)*C_air*T
mix
“Some Wrong Solutions with Common Mistakes:”
W1_Tmix=(Tcold_1+Thot_1)/2 “Taking the average temperature of inlet fluids”
Chap5-8 Gas Turbine wHeat Loss
Hot combustion gases (assumed to have the properties of air at room temperature) enter a gas turbine at
0.8 MPa and 1500 K at a rate of 2.1 kg/s, and exit at 0.1 MPa and 800 K. If heat is lost from the turbine
to the surroundings at a rate of 150 kJ/s, the power output of the gas turbine is
(a) 1477 kW (b) 1677 kW (c) 1327 kW (d) 1124 kW (e) 872 kW
Answer (c) 1327 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.Cp_air=1.005 “kJ/kg-C”
P1=800 “kPa”
T1=1500 “K”
T2=800 “K”
m_dot=2.1 “kg/s”
Q_dot_loss=150 “kJ/s”
“The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out”
W_dot_out+Q_dot_loss=m_dot*Cp_air*(T1-T2)
“Alternative: Variable specific heats – using EES data”
W_dot_outvariable+Q_dot_loss=m_dot*(ENTHALPY(Air,T=T1)-ENTHALPY(Air,T=T2))
“Some Wrong Solutions with Common Mistakes:”
W1_Wout=m_dot*Cp_air*(T1-T2) “Disregarding heat loss”
W2_Wout-Q_dot_loss=m_dot*Cp_air*(T1-T2) “Assuming heat gain instead of loss”
Chap5-9 Adiabatic R134a Turbine
Refrigerant-134a expands in an adiabatic turbine from 1 MPa and 120C to 0.10 MPa and 50C at a rate
of 0.8 kg/s. The power output of the turbine is
(a) 72.5 kW (b) 58.0 kW (c) 43.5 kW (d) 46.4 kW (e) 54.4 kW
Answer (d) 46.4 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
P1=1000 “kPa”
T1=120 “C”
P2=100 “kPa”
T2=50 “C”
m_dot=0.8 “kg/s”
Q_dot_loss=0 “kJ/s”
h1=ENTHALPY(R134a,T=T1,P=P1)
h2=ENTHALPY(R134a,T=T2,P=P2)
“The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out”
-W_dot_out-Q_dot_loss=m_dot*(h2-h1)
“Checking using properties from tables:”
h11=332.47
h22=295.45
-Wtable-Q_dot_loss=m_dot*(h22-h11)
“Some Wrong Solutions with Common Mistakes:”
-W1_Wout-Q_dot_loss=(h2-h1)/m_dot “Dividing by mass flow rate instead of
multiplying”
-W2_Wout-Q_dot_loss=h2-h1 “Not considering mass flow rate”
u1=INTENERGY(R134a,T=T1,P=P1)
u2=INTENERGY(R134a,T=T2,P=P2)
-W3_Wout-Q_dot_loss=m_dot*(u2-u1) “Using internal energy instead of enthalpy”
-W4_Wout-Q_dot_loss=u2-u1 “Using internal energy and ignoring mass flow rate”Chap5-10 Adiabatic Steam Turbine
Steam expands in an adiabatic turbine from 4 MPa and 500C to 0.5 MPa and 250C at a rate of 1740
kg/h. The power output of the turbine is
(a) 1004 kW (b) 485 kW (c) 182 kW (d) 377 kW (e) 235 kW
Answer (e) 235 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
P1=4000 “kPa”
T1=500 “C”
P2=500 “kPa”
T2=250 “C”
m_dot=1740/3600 “kg/s”
Q_dot_loss=0 “kJ/s”
h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1)
h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2)
“The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out”
-W_dot_out-Q_dot_loss=m_dot*(h2-h1)
“Some Wrong Solutions with Common Mistakes:”
-W1_Wout-Q_dot_loss=(h2-h1)/m_dot “Dividing by mass flow rate instead of
multiplying”
-W2_Wout-Q_dot_loss=h2-h1 “Not considering mass flow rate”
u1=INTENERGY(Steam_IAPWS,T=T1,P=P1)
u2=INTENERGY(Steam_IAPWS,T=T2,P=P2)
-W3_Wout-Q_dot_loss=m_dot*(u2-u1) “Using internal energy instead of enthalpy”
-W4_Wout-Q_dot_loss=u2-u1 “Using internal energy and ignoring mass flow rate”
Chap5-11 Steam Turbine wHeat Loss
Steam expands in a turbine from 6 MPa and 500C to 0.2 MPa and 150C at a rate of 1.2 kg/s. Heat is
lost from the turbine at a rate of 34 kJ/s during the process. The power output of the turbine is
(a) 750 kW (b) 784 kW (c) 818 kW (d) 573 kW (e) 641 kW
Answer (a) 750 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
T1=500 “C”
P1=6000 “kPa”
T2=150 “C”
P2=200 “kPa”
m_dot=1.2 “kg/s”
Q_dot_loss=34 “kJ/s”
h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1)
h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2)
“The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out”
W_dot_out+Q_dot_loss=m_dot*(h1-h2)
“Some Wrong Solutions with Common Mistakes:”W1_Wout=m_dot*(h1-h2) “Disregarding heat loss”
W2_Wout-Q_dot_loss=m_dot*(h1-h2) “Assuming heat gain instead of loss”
u1=INTENERGY(Steam_IAPWS,T=T1,P=P1)
u2=INTENERGY(Steam_IAPWS,T=T2,P=P2)
W3_Wout+Q_dot_loss=m_dot*(u1-u2) “Using internal energy instead of enthalpy”
W4_Wout-Q_dot_loss=m_dot*(u1-u2) “Using internal energy and wrong direction for
heat”
Chap5-12 Adiabatic Steam Compressor
Steam is compressed by an adiabatic compressor from 0.1 MPa and 100C to 1.0 MPa and 400C at a
rate of 0.85 kg/s. The power input to the compressor is
(a) 692 kW (b) 500 kW (c) 383 kW (d) 451 kW (e) 588 kW
Answer (b) 500 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
P1=100 “kPa”
T1=100 “C”
P2=1000 “kPa”
T2=400 “C”
m_dot=0.85 “kg/s”
Q_dot_loss=0 “kJ/s”
h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1)
h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2)
“The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out”
W_dot_in-Q_dot_loss=m_dot*(h2-h1)
“Some Wrong Solutions with Common Mistakes:”
W1_Win-Q_dot_loss=(h2-h1)/m_dot “Dividing by mass flow rate instead of multiplying”
W2_Win-Q_dot_loss=h2-h1 “Not considering mass flow rate”
u1=INTENERGY(Steam_IAPWS,T=T1,P=P1)
u2=INTENERGY(Steam_IAPWS,T=T2,P=P2)
W3_Win-Q_dot_loss=m_dot*(u2-u1) “Using internal energy instead of enthalpy”
W4_Win-Q_dot_loss=u2-u1 “Using internal energy and ignoring mass flow rate”
Chap5-13 Adiabatic R134a Compressor
Refrigerant-134a is compressed by an adiabatic compressor from the saturated vapor state at 0.12 MPa
to 1.2 MPa and 70C at a rate of 0.108 kg/s. The power input to the compressor is
(a) 587 kW (b) 63.4 kW (c) 6.85 kW (d) 6.42 kW (e) 59.4 kW
Answer (c) 6.85 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
P1=100 “kPa”
x1=1
P2=1400 “kPa”
T2=60 “C”m_dot=0.15 “kg/s”
Q_dot_loss=1.80 “kJ/s”
h1=ENTHALPY(R134a,x=x1,P=P1)
h2=ENTHALPY(R134a,T=T2,P=P2)
“The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out”
W_dot_in-Q_dot_loss=m_dot*(h2-h1)
“Checking using properties from tables:”
h11=233.86
h22=298.96
Wtable-Q_dot_loss=m_dot*(h22-h11)
“Some Wrong Solutions with Common Mistakes:”
W1_Win+Q_dot_loss=m_dot*(h2-h1) “Wrong direction for heat transfer”
W2_Win =m_dot*(h2-h1) “Not considering heat loss”
u1=INTENERGY(R134a,x=x1,P=P1)
u2=INTENERGY(R134a,T=T2,P=P2)
W3_Win-Q_dot_loss=m_dot*(u2-u1) “Using internal energy instead of enthalpy”
W4_Win+Q_dot_loss=u2-u1 “Using internal energy and wrong direction for heat
transfer”
Chap5-14 R134a Compressor wCooling
Refrigerant-134a is compressed steadily from the saturated vapor state at 0.10 MPa to 1.4 MPa and 60C
at a rate of 0.15 kg/s. The refrigerant is cooled at a rate of 1.80 kJ/s during compression. The power input
to the compressor is
(a) 5.74 kW (b) 7.54 kW (c) 9.08 kW (d) 9.34 kW (e) 46.7 kW
Answer (d) 9.34 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
P1=100 “kPa”
x1=1
P2=1400 “kPa”
T2=60 “C”
m_dot=0.15 “kg/s”
Q_dot_loss=1.80 “kJ/s”
h1=ENTHALPY(R134a,x=x1,P=P1)
h2=ENTHALPY(R134a,T=T2,P=P2)
“The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out”
W_dot_in-Q_dot_loss=m_dot*(h2-h1)
“Checking using properties from tables:”
h11=233.86
h22=298.96
Wtable-Q_dot_loss=m_dot*(h22-h11)
“Some Wrong Solutions with Common Mistakes:”
W1_Win+Q_dot_loss=m_dot*(h2-h1) “Wrong direction for heat transfer”
W2_Win =m_dot*(h2-h1) “Not considering heat loss”
u1=INTENERGY(R134a,x=x1,P=P1)
u2=INTENERGY(R134a,T=T2,P=P2)W3_Win-Q_dot_loss=m_dot*(u2-u1) “Using internal energy instead of enthalpy”
W4_Win+Q_dot_loss=u2-u1 “Using internal energy and wrong direction for heat
transfer”
Chap5-15 R134a Condenser Q_out
Saturated refrigerant 134a vapor at 40C is condensed as it flows through a tube at a rate of 0.2 kg/s.
The condensate leaves the tube as saturated liquid at 40C. The rate of heat transfer from the tube is
(a) 21.0 kJ/s (b) 53.4 kJ/s (c) 162 kJ/s (d) 74.4 kJ/s (e) 32.4 kJ/s
Answer (e) 32.4 kJ/s
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
T1=40 “C”
m_dot=0.2 “kg/s”
h_f=ENTHALPY(R134a,T=T1,x=0)
h_g=ENTHALPY(R134a,T=T1,x=1)
h_fg=h_g-h_f
Q_dot=m_dot*h_fg
“Some Wrong Solutions with Common Mistakes:”
W1_Q=m_dot*h_f “Using hf”
W2_Q=m_dot*h_g “Using hg”
W3_Q=h_fg “not using mass flow rate”
W4_Q=m_dot*(h_f+h_g) “Adding hf and hg”
Chap5-16 Electric Heating of Helium
Helium gas is to be heated steadily by a 3-kW electric resistance heater as it flows through an insulated
duct. If the helium enters at 50C at a rate of 0.08 kg/s, the exit temperature of helium will be
(a) 57.2°C (b) 50.6°C (c) 62.0°C (d) 71.9°C (e) 112.0°C
Answer (a) 57.2°C
Solution 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-C”
T1=50 “C”
m_dot=0.08 “kg/s”
W_dot_e=3 “kJ/s”
W_dot_e=m_dot*Cp*(T2-T1)
“Some Wrong Solutions with Common Mistakes:”
Cv=3.1156 “kJ/kg.K”
W_dot_e=Cp*(W1_T2-T1) “Not using mass flow rate”
W_dot_e=m_dot*Cv*(W2_T2-T1) “Using Cv”
W_dot_e=m_dot*Cp*W3_T2 “Ignoring T1”Chap5-17 Throttling of Air
Air at 300 K and 200 kPa is throttled by a valve to a pressure of 100 kPa. If the valve is adiabatic and the
change in kinetic energy is negligible, the temperature of air after throttling will be
(a) 150 K (b) 200 K (c) 300 K (d) 450 K (e) 600 K
Answer (c) 300 K
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
“The temperature of an ideal gas remains constant during throttling, and thus”
T1=300 “K”
P1=200 “kPa”
P2=100 “kPa”
T2=T1 “K”
“Some Wrong Solutions with Common Mistakes:”
W1_T2=T1*P1/P2 “Assuming v=constant”
W2_T2=T1*P2/P1 “Assuming v=constant and pressures backwards”
Chap5-18 Throttling of SatLiquid Water x2
Saturated liquid water at 1.0 MPa is throttled adiabatically to a pressure of 0.4 MPa. If the change in
kinetic energy is negligible, the percentage of water that evaporates during this throttling process will be
(a) 0.0% (b) 3.8% (c) 24.8% (d) 7.4% (e) 100%
Answer (d) 7.4%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
P1=1000 “kPa”
x1=0
P2=400 “kPa”
“Quality represents the percentage of water that evaporates:”
h1=ENTHALPY(Steam_IAPWS,x=x1,P=P1)
x2=QUALITY(Steam_IAPWS,h=h1,P=P2)
“Some Wrong Solutions with Common Mistakes:”
v1=VOLUME(Steam_IAPWS,x=x1,P=P1)
W1_x2=QUALITY(Steam_IAPWS,v=v1,P=P2) “Assuming the volume to remain
constant”
u1=INTENERGY(Steam_IAPWS,x=x1,P=P1)
W2_v2=VOLUME(Steam_IAPWS,u=u1,P=P2) “Assuming u=constant”
Chap5-19 Throttling of R134a T2
Saturated Refrigerant-134a liquid at 0.8 MPa is throttled to a pressure of 0.12 MPa. The temperature of
the refrigerant after throttling is
(a) -14.6°C (b) -5.6°C (c) 0°C (d) 31.3°C (e) -22.4°C
Answer (e) -22.4°CSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
P1=800 “kPa”
x1=0
P2=120 “kPa”
“Quality represents the percentage of water that evaporates:”
h1=ENTHALPY(R134a,x=x1,P=P1)
T2=TEMPERATURE(R134a,h=h1,P=P2)
“Some Wrong Solutions with Common Mistakes:”
W1_T2=TEMPERATURE(R134a,x=0,P=P1) “Taking the temperature to be the saturation
temperature at P1″
Chap5-20 Throttling of Steam v2
Steam at 4 MPa and 400C is throttled adiabatically to a pressure of 1 MPa. If the change in kinetic
energy is negligible, the specific volume of the steam after throttling will be
(a) 0.2952 m3/kg (b) 0.2327 m3/kg (c) 0.3749 m3/kg (d) 0.5165 m3/kg (e) 0.3066 m3/kg
Answer (a) 0.2952 m3/kg
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines
on a blank EES screen.
P1=4000 “kPa”
T1=400 “C”
P2=1000 “kPa”
h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1)
v2=VOLUME(Steam_IAPWS,h=h1,P=P2)
“Some Wrong Solutions with Common Mistakes:”
W1_v2=VOLUME(Steam_IAPWS,T=T1,P=P2) “Assuming the volume to remain
constant”
u1=INTENERGY(Steam,T=T1,P=P1)
W2_v2=VOLUME(Steam_IAPWS,u=u1,P=P2) “Assuming u=constant”
W3_v2=VOLUME(Steam_IAPWS,T=T1,P=P2) “Assuming T=constant”
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