Engineering Thermodynamics – CHE/CEEN/MEEN 2300
Test 4
Chapter 7
100 pts.
You may use a calculator, the Properties Table booklet and a 3”x5” index card of
formulas. Please show your work. Answers without supporting calculations will be
counted wrong.
1. 20 pts A rigid tank contains an ideal gas at 300C that is being stirred by a paddle
wheel. The paddle wheel does 100 kJ of work on the ideal gas. It is observed that
the temperature of the ideal gas remains constant during this process as a result of
heat transfer between the system and the surroundings at 200C. Use the table to
organize your data
State 1
State 2
T1 = 30 C
T2 = 30 C
V1 = V2 therefore v1 = v2
v2 = v1
a. 4 pts Determine the change in internal energy of the system
Both the temperature and the specific volume are the same at State 1 and State
2, therefore the two states are identical, and all of the physical properties are
the same at both states. Therefore, there is no change in internal energy.
b. 4 pts Determine how much heat was transferred to the surroundings
Perform an energy balance on the system (which is a closed system)
Qin – Wout = ΔU = 0 therefore Qin must equal Wout and conversely, Qout must equal Win.
Thus 100 kJ was transferred to the surroundings.
c. 6 pts Determine the change in entropy of the system
As shown in part a, the two states are identical. Therefore there is no change in entropy of
the SYSTEM. This does not mean that the process does not result in a change in entropy
– since the surroundings were affected by the process. You may wonder how there can
be no change in entropy in the system, if there was heat transfer. The paddle wheel
caused an increase in entropy, as it interacted with the gas in the system. Heat transfer
caused a reduction in entropy. The two processes cancelled out!!
d. 4 pts Determine the change in entropy of the surroundings
The surroundings remained at constant temperature, therefore
ΔS = Qsurroundings/T = 100 kJ/(20+273) K = 0.34 kJ/K Several students forgot that the
temperature must be expressed in Kelvin.
e. 2 pts Determine the change in entropy of the universe (Sgen)
ΔSuniverse = Sgen = ΔSsystem + ΔSsurroundings = 0 + 0.34 kJ/K = 0.34 kJ/K
2. 20 pts A 0.5 m3 rigid tank contains R-134a initially at 200 kPa and 40% quality.
Heat is transferred now to the refrigerant from a source at 350C until the pressure
rises to 400 kPa. Use the table below to keep track of your properties
State 1
P1 = 200 kPa
x1 = 0.4
v1 = 0.0402 m3/kg
u1 = 110.586 kJ/kg
s1 = 0.459 kJ/kg K
State 2
P2 = 400 kPa
x2 = .786
v2 = 0.0402 m3/kg
u2 = 195.56 kJ/kg
s2 = 0.7701 kJ/kg K
a. 4pts Find the mass of refrigerant in the system
m = V/v
v = vf + x vfg = 0.0007532 + 0.4 * (0.0993 – 0.0007532) = 0.0402 m3/kg
m = 0.5 m3/0.0402 m3/kg = 12.44 kg
b. 4 pts Find the total change in internal energy of the system
ΔU = m Δu = m(u2 – u1)
u1 = uf + x ufg = 36.69 + 0.4 * (221.43 – 36.69) = 110.586 kJ/kg
Now we need to find u2, but we don’t know the value of x2. We can use the specific
volume to help us
v2 = vf +x2 vfg = 0.0007907 + x2 * (0.051201 – 0.0007907) = 0.0402 m3/kg
x2 = 0.78
u2 = uf + x ufg = 61.69 + 0.786 * (231.9 – 61.69) = 195.56 kJ/kg
ΔU = 12.44kg * (195.56 – 110.586) = 12.44 kg * (84.96 kJ/kg) = 1057 kJ
c. 2 pts Find the amount of heat added to the system
Perform an energy balance.
Q – W = ΔU
W=0
Q = ΔU = 1057 kJ
d. 4 pts Find the change in entropy of the system
ΔS = m Δs = m(s2 – s1)
s1 = sf + x sfg = 0.1481 + 0.4 * (0.9253 – 0.1481) = 0.459 kJ/kg K
s2 = sf + x sfg = 0.2399 + 0.786 * (0.9145 – 0.2399) = 0.7701 kJ/kgK
ΔS = 12.44 kg * (0.7701 – 0.459) kJ/kg K = 12.44 kg * 0.3111 kJ/kg K = 3.87 kJ/K
e. 4 pts Find the change in entropy of the heat source (which is large enough to stay
at constant temperature)
ΔS = Q/T = - 1057/(35 + 273) = -3.432 kJ/K
f. 2 pts Find the change in entropy of the universe due to this process (Sgen)
ΔSuniverse = Sgen = ΔSsystem + ΔSsurroundings = 3.87 – 3.432 = .44 kJ/K
3. 20 pts A 25 kg iron block initially at 3500C is quenched in an insulated tank that
contains 100 kg of water at 180C. Assuming that the water that vaporizes during the
process condenses back into the tank, determine the total entropy change during this
process. Use the table to organize your data.
water m = 100 kg
iron m = 25 kg
State 1
State 2
State 1
State 2
T1 = 18 C
Tf = 26.66 C
T1 = 350 C
Tf = 26.66 C
C = 4.2 kJ/kg K
C = 0.45 kJ/kg C
First, solve for the final temperature
mH2O CH2O ΔTH2O = - miron Ciron ΔTiron
100 kg * 4.2 kJ/kg K * (Tf – 18) = - 25 kg * 0.45 kJ/kg K * (Tf – 350)
Tf = 26.66 C
For solids and liquids (incompressible materials)
ΔS = m C ln(T2/T1) and
ΔS system = ΔSH2O + ΔSiron
ΔSH2O = 100 kg* 4.2 kJ/kg K[ (26.66 + 273)/(18 + 273)] = 12.31 kJ/K
ΔSiron = 25 kg * 0.45 kJ/kg K ln[ 26.66 + 273)/(350 + 273)] = - 8.24 kJ/K
ΔS system = 4.07 kJ/K
4. Air is compressed steadily by a 5 kW compressor from 100 kPa and 170C to 600
kPa and 1670C at a rate of 1.6 kg/min. During this process, some heat transfer takes
place between the compressor and the surrounding medium at 170C. Use the table to
keep track of your properties
Inlet conditions
Exit conditions
Pi = 100 kPa
Pe = 600 kPa
Ti = 17 C + 273 = 290 K
Te = 167 C + 273 = 440 K
0
si = 1.66802 kJ/kg K
se0 = 2.08870 kJ/kg K
hi = 290.16 kJ/kg
he = 441.61 kJ/kg
a. 8 pts Determine the rate of entropy change of air during this process
Δs = cp avg ln(TeTi) - R ln(Pe/Pi) if you assume that the heat capacity is constant.
A more accurate approach is:
Δs = se0 – si0 - R ln(Pe/Pi)
Δs = 2.08870 – 1.66802 - 0.2870 * ln(600/100) = -0.0935 kJ/kg K
ΔS = m Δs = 1.6 kg/min * (-0.0935 kJ/kg K) = -0.1497 kJ/K min = -.0025 kJ/K sec
b. 8 pts Determine the rate of entropy change of the surroundings due to heat
transfer with the compressor
First determine the rate of heat transfer into the system
Q-W = m(Δh) (I’ve left off the dot to make it easier to type, but these are rates)
Q = W + m(Δh) = - 5kJ/sec + 1.6 kg/min* min/60 sec*(441.61 – 290.16)kJ/kg = -.96 kJ/s
Therefore the rate of heat transfer into the surroundings is + 0.96 kJ/sec
ΔSsurroundings = Q/T = 0.96 kJ/s / (17 + 273) K = 0.0033 kJ/K sec
c. 4 pts Find the rate of change of entropy of the universe (Sgen) due to this
process.
ΔSuniverse = Sgen = ΔSsystem + ΔSsurroundings = -0.0025 + 0.0033 = .0008 kJ/Ksec
5. 20 pts Liquid water enters a 25 kW pump at 100 kPa pressure at a rate of 5 kg/sec.
Determine the highest pressure the liquid water can have at the exit of the pump.
(Neglect the kinetic and potential energy change of water, and take the specific volume of
water to be 0.001 m3/kg
w = Wdot/mdot = 25 kJ/sec / 5 kg/sec = 5 kJ/kg
w = v ΔP = 5 kJ/kg = .001 m3/kg (Pe – 100 kPa)
Pe = 5100 kPa
6. 20 pts Helium gas is compressed from 14 psia and 700F to 120 psia at a rate of 5 ft3/s.
Determine the power input to the compressor, assuming the compression process to be
a) isentropic
W  mw
k 1




kRT1  P2 k

w
   1

k  1  P1 


The first thing we’ll need to do is find the mass flow rate, using the ideal gas law. The
mass was worth 4 pts.
PV  mRT
m
PV

RT
14 psia *5 ft 3 / sec
 0.493lbm / s
psia * ft 3
2.6809
*  70  460  R
lbmR
Now find the work per unit mass (4 pts)
k 1
1.667 1




1.667
1.667
*0.4961
BTU
/
lbm
*
R
*
70

460
R


kRT1  P2  k
120




w
* 
 1
   1 
 14 

k  1  P1 
1.667  1




 895.2 BTU / lbm
Total work = 0.493 lbm/sec * 895.2 BTU/lbm = 44.1 Btu/sec
b) polytropic with n = 1.2 (4 pts)
Same as a, but substitute n for k, which gives
w = 679.3 BTU/lbm
Total work = 33.5 Btu/sec
c) isothermal (4 pts)
w = RT ln(P2/P1) = 0.4961 Btu/lbm R * (70 + 460)R * ln(120/14) = 564.9 Btu/ lbm
Total work = 27.85 Btu/sec
The trend of these numbers make sense, since the most efficient compressors are
isothermal, the least are isentropic, and the more realistic polytropic case should fall
between the two values.
d) ideal two stage polytropic with n=1.2 (4 pts)
Px  P2 P1  120*14  41 psia is the intermediate pressure. Each of the two
compressors carries half the load, which is:
n 1
1.2 1




1.2
1.2*0.4961
BTU
/
lbm
*
R
*
70

460
R


nRT1  Px  n
41




w
*  
 1
   1 
 14 

n  1  P1 
1.2  1




 309.31BTU / lbm
The total for the two compressors is twice this value, 618.6 Btu/ lbm, and the power
required is:
W  mw  0.0493lbm / sec*618.6 Btu / lbm  30.49 Btu / sec
this makes sense, since it is not as good as isothermal, but better than a single stage
polytropic case.