ME 354 Tutorial, Week#3 – Exergy: Control Mass Analysis
An insulated piston-cylinder device contains 2L of saturated liquid water at a
pressure of 150kPa – which is constant throughout the process. An electric
resistance heater inside the cylinder is turned on, and electrical work is done on
the water in the amount of 2200kJ. Assuming the surroundings to be at 25C and
100kPa, determine:
a) The minimum work with which this process could be accomplished and
b) The exergy destroyed during the process
Step 1: Draw a diagram to represent the system (show control mass of
interest)
The control mass boundary encloses the water in the cylinder
State1
Intermediate State
Surroundings
T0 =25°C
P0 =100 kPa
Wb
Sat. Liquid H 2O
P1 =150 kPa
V1 =2L
State 2
Surroundings
T0 =25°C
P0 =100 kPa
Q=0
H20
P2 =150 kPa
We =
2200 kJ
Step 2: Property table
State
T [°C]
1 (sat liq)
2
P [kPa]
150
150
Property
h [kJ/kg] s [kJ/kg*K]
v [m3/kg]
V [m3]
0.002
Step 3: Assumptions
Assumptions:
1) Insulated piston-cylinder (Q=0)
2) KE, PE  0
3) Compression/expansion processes are in quasi-equilibrium
Step 4: Calculations (usually start by writing First and Second Laws)
Part a)
We can determine the properties at state 1 using the information given in the
problem (saturated liquid H2O at P = 150 kPa) with Table A-5.
From Table A-5
 s1 = sf@150kPa = 1.4336 [kJ/kg*K]
 v1 = vf@150kPa = 0.001053 [m3]
 h1 = hf@150kPa = 467.11[kJ/kg]
From the specific volume at state 1, v1, and the volume of state 1, V1, we can
determine the mass of H20 in the cylinder.
1
 mH 2O
V
 1 
v1
 
0.002 m3
 1.9kg
 m3 
0.001053 
 kg 
We still must determine the properties at state 2 (we DO know the pressure is
150 kPa because this is a constant pressure process). We can write an energy
balance on the system to link state 2 to state 1, as shown in Eq1.
E1  We  Wb  E2
(Eq1)
Using the assumption KE, PE  0, we and noting that W b = P(V2-V1), Eq1 can
be re-expressed as Eq2.
U 2  U1  We  P(V2  V1 )
(Eq2)
Eq2 can be rearranged into the form presented in Eq3.
U 2  PV2   U1  PV1   We
(Eq3)
Recognizing that U+PV = H, and that H = mH20h we can rewrite Eq3 as Eq4.
mH 2O h2  h1   We
(Eq4)
Isolating h2 in Eq4, we can determine h2.
 h2 
 kJ 
 kJ 
We
2200kJ 
 h1 
 467.11   1625 
mH 2O
1.9kg
 kg 
 kg 
Using h2 = 1625 kJ/kg and P2 = 150kPa with Table A-5, we find that we are in the
vapour dome, in between the saturated liquid and vapour states on the 150
isobar line. We can determine the properties at state 2 by first finding the quality,
and then using this to interpolate in Table A-5.
x
h2  h f
hg  h f

1625  467.11
 0.52
2693.6  467.11
 s2= sf@150kPa + xsfg@150kPa = 1.4336 + 0.52(5.7897) = 4.44 [kJ/kg*K]
 v2= vf@150kPa + x(vg@150kPa -vf@150kPa) = 0.001053 + 0.52(1.1582) = 0.6033 [m3]
2
The minimum work that needs to be provided to accomplish this process
represents the reversible work input (Sgen = 0  Xdestroyed = 0). We can use the
general relation for reversible work as expressed in Eq5.
Wmin  Wrev  (U 2  U1 )  T0 (S 2  S1 )  P0 (V2  V1 )
(Eq5)
Substituting Eq2 into Eq5 for internal energy and bringing the m H2O outside the
bracket to convert the properties to specific properties, Eq5 can be written as
Eq6.
(Eq6)
Wmin  We  mH 2O P  P0 v2  v1   T0 (s 2  s1 )
Substituting the known property values into Eq6, we can calculate the minimum
work input.

 kJ 
 kJ  
 Wmin  2200[kJ ]  1.9[kg](150  100)(0.6033  0.001053)    (298)( 4.44  1.4336)   
 kg 
 kg  

= 440.6 kJ
Answer a)
Part b)
The exergy destroyed can be calculated from the expression Xdestroyed = T0Sgen.
To find the entropy generated during the process we can perform an entropy
balance on the system as shown in Eq7.
S in  S out  S gen  S system
(Eq7)
For a closed system, entropy can only be carried across the system boundary by
heat transfer. Since the system is insulated, Sin = 0 and Sout = 0. The change in
the system entropy is the difference in entropy between state 2 and state 1 i.e.
Ssystem = mH2O(s2-s1). Using the above information with Eq7, the Xdestroyed
(Irreversibility) can be determined.
 kJ 
X destroyed  T0 S gen  T0 mH 2O ( s 2  s1 )  (298[ K ])(1.9[kg])( 4.44  1.4336) 

 kg  K 
= 1702.2 kJ
Answer b)
Step 5: Concluding Statement and Remarks
The minimum work with which this process could be accomplished is 440.6kJ.
The amount of exergy destroyed during this process is 1702.2 kJ.
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