EME2007
Engineering Thermodynamics
Spring 2020 Prof. C Ryu
Solutions for Mid-term Exam
Points : 1. 2/2/2/2 2. 2/2/2 3. 3 4. 3 5. 2/2/2/4
6. 2/1/2/1
7. 2/4
8. 2/2
1. A combustion gas leaving a boiler consists of 20% CO2, 10% H2O, and 70% N2 by volume at 80°C,
1 bar. Assuming an ideal gas mixture, determine
a) average molecular weight [kg/kmol]
b) density [kg/m3]
c) relative humidity [%]
d) dew point [°C]
a) M=0.2*44.010+0.1*18.015+0.7*28.013 = 30.213 kg/kmol
b) Density=P/RT=1e5/(8314.5/30.213)/(273.15+80)=1.029 kg/m3
c) Relative humidity = 0.1*100/47.39 = 0.2110 or 21.10%
c) Dew point for Psat=PH2O=10 kPa: 45.81 °C
2. A 1-m3 rigid tank has air at 100 kPa, 300 K and connected by a valve to another
tank of 0.5 m3 with air at 250 kPa, 500 K. The valve is opened, and the two tanks
come to a uniform state without any heat transfer. Determine
a) total mass b) final temperature
c) final pressure
a) PV=mRT, R=0.287; uA=214.36 kJ/kg, uB=359.84 kJ/kg from Table A.7
mA=(PV/RT)A =100*1/(0.287*300)=1.1614 kg
mB=(PV/RT)B =250*0.5/(0.287*500) =0.8711 kg
Mass conservation: m2=mA+mB=2.0325 kg
b) Energy conservation: m2u2=mAuA+mBuB
u2=(1.1614*214.36+0.8711*359.84)/2.0325=276.71
T2=380+(276.71-271.99)/(286.49-271.99)*20 = 386.51 K
c) P =mRT/V =2.0325*0.287*386.51/1.5 = 150.30 kPa
3. Ideal gas air at 1800 K is supplied to a boiler and leaves at 600 K in a steam power plant. Inside
the boiler, water at 80 oC, 40 bar enters and is superheated to 450 oC, 40 bar. If the flow rate of
water is 10 kg/s, what is the flow rate of air?
Energy conservation: msteam(h2-h1) = mair(hin-hout)
For air, hin = 2002.99, hout = 607.32 kJ/kg
For water, h1 = 334.88, h2=3330.23 kJ/kg
mair = msteam(h2-h1)/ (hin-hout) =10*(3330.23-334.88)/(2002.99-607.32) = 21.46 kg/s
4. The steam at 450 oC, 40 bar enters a turbine to produce specific work of 900 kJ/kg and leaves at
50 kPa. Assuming no heat transfer and no changes in kinetic energy, determine the quality of
steam at the turbine exit.
Energy conservation: w = h2-h3
h2=3330.23 kJ/kg, h3=3330.23-900 =2430.23 kJ/kg
x = (2430.23-340.47)/2305.40 = 0.9065
5. A rigid tank initially contains a 1-kg mixture of 60% water and 40% steam by volume at 1 bar
(state 1). It is heated until the pressure reaches 10 bar (state 2) and then the safety value opens
to discharge the saturated steam from the top. While maintaining the same pressure, the heat is
transferred until the mass inside the tank becomes 0.5 kg (state 3). Determine
a) quality at state 1
b) volume of the tank
c) quality at state 3
d) total heat transfer between states 1 and 3 [kJ]
Sat. steam at 10 bar
Safety valve
Valve opens
1. Water/Steam
m=1 kg
1 bar
2. Water/Steam
10 bar
3. Water/Steam
m=0.5 kg
10 bar
Heat
Heat
Heat
a) at 1 bar, vf=0.001043, vg=1.694
(eq.1) 0.4v1=x1vg (eq.2) 0.6v1=(1-x1)vf
From (eq.1): v1=2.5x1vg → (eq.2): 0.6*2.5x1vg=(1-x1)vf
x1=0.001043/(0.6*2.5*1.694+0.001043) = 4.103e-4
b) From eq.1: v1=x1vg/0.4 =4.103e-4*1.694/0.4 = 0.0017376 m3/kg
or v1=0.001043+(1.694-0.001043)*4.103e-4
V=m*v1= 0.001738 m3 or 1.738 L
c) at state 3, v3=2v1=0.0017376*2 =0.0034752
x3 =(0.0034752-0.001127)/0.19332 = 0.01215
d) Mass conservation: m3-m1 = -me or me = m1-m3 =0.5 kg
Energy conservation: m3u3-m1u1 = Q-mehe
u1 =417.33+4.103e-4*2088.72 =418.19
u3 =761.67+0.01215*1821.97 =783.81
he=hg=2778.08
Q =0.5*783.81-1*418.19+0.5*2778.08 = 1362.75 kJ
6. The refrigerator shown in the figure operates using R-134a. The states at each point are given in
the table. Determine the following:
a) specific work required for the compressor (ignore the heat loss) [kJ/kg]
b) heat leaving the condenser [kJ/kg]
c) quality at the evaporator inlet
d) heat absorbed in the evaporator [kJ/kg]
a) h1=387.22, By interpolation: h2 =435.11+50/200*(431.24-435.11)=434.14 kJ/kg
wcomp =h1-h2=387.22-434.14=-46.92 or 46.92 kJ/kg
b) The state 3 was wrongly taken.
At 800 kPa, 35°C, R134a is superheated vapor,
h3=415.72+(424.86-415.72)*(35-31.3)/(40-31.3)=419.61
qcond =h3-h2 =434.14-419.61 = -14.53 or 14.53 kJ/kg
c) This will not be marked:
At 133.7 kPa, hf=173.74, hfg=212.34, hg=386.08 → h4~h3=419.6>> hg
x cannot be determined
d) This will not be marked
7. Helium gas at 135 kPa, 350 K expands from 1 m3 to 2 m3 in a polytropic process with n=1.667.
Find
a) boundary work [kJ]
b) heat transfer [kJ]
a) P1V11.667 = P2V21.667 V1 =1 m3 à P2 = P1(V1/V2)1.667 = 135*0.5^1.667 = 42.5125 kPa
Wb = (P2V2- P1V1)/(1-n) = (42.5125*2-135*1)/(1-1.667) = 74.925 kJ
b) Energy conservation: m(u2-u1) = Q-Wb
From PV = mRT, m = PV/RT = 135*1/(2.0771*350) = 0.1857 kg
T2=P2V2/mR =42.5125*2/(0.1857*2.0771) = 220.43 K
Helium is a noble gas having a constant Cv of 3.116 kJ/kg.K (Table A.5)
Q = m(u2-u1)+Wb =0.1857*3.116*(220.43-350)+74.925 =-0.004954 kJ
8. Water in a piston-cylinder system at 20 °C, 1 bar is expanded to 300 °C at a constant pressure.
Determine
a) boundary work [kJ/kg]
b) heat transfer [kJ/kg]
a) v1=0.001002, v2=2.63876 m3/kg
wb=P(v2-v1) =100*(2.63876-0.001002) = 263.78 kJ/kg
b) h1=83.94, h2=3074.28 kJ/kg
q=(h2-h1) =(3074.28-83.94)= 2990.34 kJ/kg