ME 354 Tutorial 2B – Exergy – Control Volume Analysis
Winter 2001
In the boiler of a typical power plant, H2O flows inside the tubes lining the walls
of the combustion chamber. Consider a case in which the H20 is brought from 0.8
Mpa & 150C to 250C at essentially constant pressure while the combustion
gases passing over the tubes cool from 1100C to 500C at essentially constant
pressure. The gases can be modeled as nitrogen (N2). There is no significant heat
transfer from the combustion chamber to the surroundings.
Find:
a) the mass flow rate of combustion gases per kg of H20 flowing inside the
tubes
b) the rate of exergy destruction per kg of H20 flow
Step 1: Draw a diagram to represent the system (show control volume of
interest)
The first control volume encloses the H2O pipes as shown below
The second control volume encloses all the N2 in the combustion chamber.
Step 2: Prepare a property table
H2O
N2
State
1
2
3
4
T [K]
523.15
343.15
1373.15
773.15
Property
P [MPa] h [kJ/kg] s [kJ/kg*K]
0.8
0.8
P3
P3
Step 3: State your assumptions
Assumptions:
1) SSSF (steady state/steady flow)
2) Combustion gases modeled as N2 – Ideal Gas
3) No heat transfer from system to surroundings
4) ke, pe  0
5) Non-constant specific heats for N2
Step 4: Calculations (usually start by writing First and Second Laws)
Part a)
Writing the First Law for the control volume around the H20 tubes gives Eq1




dEcv
 m1 (e  Pv)1  m 2 (e  Pv) 2  Q W
dt
(Eq1)
Expanding the energy terms (e = u + ke + pe), realizing that there is no shaft
work
into
the
pipes,
and
applying
the
SSSF
assumption



dE
( cv  0, m1  m 2  m H2O), Eq1 can be written as Eq2.
dt
.
.
0  m H 2O (u  ke  pe  Pv)1  (u  ke  pe  Pv) 2   Q12
(Eq2)
Using the assumption that ke and pe are approximately zero, and the fact the
property enthalpy, h, is defined as u + Pv, Eq2 can re-written as Eq3.
.
.
Q 12  m H 2O h2  h1 
(Eq3)
Applying the above analysis to the N2 control volume (noting that the heat
transfer is now out of the system) yields Eq4.


- Q 34  m N 2 h4  h3 
(Eq4)
From examination of the control volume diagrams we can see that the heat
transfer from the N2 is equal to the heat transfer into the H20. Eq5 expresses this
relation.


(Eq5)
Q 12  Q 34
Relating Eq3 & Eq4 through Eq5 gives


m H 2O h2  h1   m N 2 h3  h4 
(Eq6)

We need to find the ratio of the mass flow rate of combustion gases, m N 2 , to

mass flow rate of H20, m H 2 O . Rearranging Eq6 into the form of Eq7.

mN2

m H 20

h2  h1
h3  h4
(Eq7)
To determine the enthalpy of N2 at state 3 and 4, we can use Table A-18 (since
there is a 600 temperature difference between state 3 and 4, the constant specific heat assumption
could lead to considerable error. This is why we are not using the relation h 4 - h3 = cp(T4-T3)).
Table
A-18 gives enthalpy on a kmol basis. To convert to a kg basis, we can use Table
A-1, which gives the molar mass of N2 as 28.02 kg/kmol.
From Table A-18, interpolating between T=1360K and T=1380K for T3=
1373.15K (1100C)
_
_
1360  1373.15
42227  h 3

 h 3  42679.4 kJ/kmol
1360  1380
42227  42915
 h3 = (42679.4 kJ/kmol)/(28.02 kg/kmol) = 1523.2 kJ/kg
From Table A-18, interpolating between T=770K and T=780K for T4=
773.15K (500C)
_
_
770  773.15
22772  h 4

 h 4  22870.6 kJ/kmol
770  780
22772  23085
 h4 = (22870.6 kJ/kmol)/(28.02 kg/kmol) = 816.2 kJ/kg
To determine the enthalpy of H20 at state 1 and 2, we can use the steam tables.
At state 1, the temperature of the H20 is 150C @ 0.8MPa. Looking first in Table
A-5 at 0.8MPa we find that the corresponding saturated temperature is 170.43C.
Since T1 < 170.43C, the H20 at state 1 is subcooled (compressed liquid). We can
approximate the enthalpy at state 1 using the enthalpy of saturated liquid at
150C (see p73 of text for explanation).
From Table A-4 @ 150C
 h1 = hf =632.20 kJ/kg
At state 2, the temperature of the H20 is 250C @ 0.8MPa. Looking first in Table
A-5 at 0.8MPa we find that the corresponding saturated temperature is 170.43C.
Since T2 > 170.43C, the H20 at state 2 is superheated. We can use Table A-6 @
0.8MPa to determine the enthalpy of state 2.
From Table A-6 @ 250C, 0.8 MPa
 h2 = 2950.0 kJ/kg
We can now substitute these values into Eq7 to solve for the mass flow ratios.

mN2

m H 20

h2  h1 2950.0  632.20
= 3.28

h3  h4
1523.2  816.2
Answer a)
Part b)

To find the rate of exergy destruction, I , we can make use of the generalized
exergy equation (Week 3: Lecture 3) shown in Eq8 applied to the entire system –
see above diagram for new control volume boundary.



 
 
dCV
dV
T 
T 
 P0 CV  W  m  Q1  0   W  m  Q1  0 
I
dt
dt
 TTER  IN 
 TTER  OUT

(Eq8)
With our new control volume boundary, there is no work or heat transfer out of
the system. Also, with the steady state assumption, the time derivatives in Eq8 go
to zero. Eq8 can be reduced to Eq9.

 
 
I  m   m 

 IN 
 OUT
(Eq9)
The flow exergy terms in Eq9 can be expanded in terms of the individual flow
exergy terms of the system as shown in Eq10.




 

I   m H 2 O  1  mN 2  3    m H 2 O  2  mN 2  4 

 

(Eq10)
The question has asked for the exergy destruction per kg of H2O flow. We can

obtain an expression for this by dividing Eq10 through by m H 2 O to obtain Eq11.

I


  1   2  
m H 2O
mN 2

 3   4 
(Eq11)
m H 2O

We determined
mN 2

in part a, so the question has been reduced to finding the
m H 2O
differences in the flow exergies. Referring to Week 3: Lecture 3, the flow exergy
can be determined from Eq12.
  h  h0   T0 s  s0  

    g  z  z 
1 2
v  v0
2
2
0
(Eq12)
The difference in flow exergy can now be expressed as shown in Eq13.
 1   2  h1  h2   T0 s1  s 2  

1 
v1
2
  v   g z
2
 2
2
1
 z2 
(Eq13)
Using the assumption that ke & pe  0, Eq13 reduces to Eq14.
1   2  h1  h2   T0 s1  s2 
Similarly, applying the same steps for 3 – 4 we obtain Eq15.
(Eq14)
 3   4  h3  h4   T0 s3  s4 
(Eq15)
We determined the enthalpies in part a) we now must similarly determine the
entropies.
From Table A-18, interpolating between T=1360K and T=1380K for T3=
1373.15K (1100C)
_
_
1360  1373.15
238.376  s 3

 s 3  238.706 kJ/kmol*K
1360  1380
238.376  238.878
 s3 = (238.706 kJ/kmol*K)/(28.02 kg/kmol) = 8.519 kJ/kg*K
From Table A-18, interpolating between T=770K and T=780K for T4=
773.15K (500C)
_
_
770  773.15
219.709  s 4

 s 4  219.836 kJ/kmol*K
770  780
219.709  220.113
 s4 = (22870.6 kJ/kmol)/(28.02 kg/kmol) = 7.846 kJ/kg*K
From Table A-4 @ 150C
 s1 = sf =1.8418 kJ/kg*K
From Table A-6 @ 250C, 0.8 MPa
 s2 = 7.0384 kJ/kg*K
Substituting in these values in to Eq13 and Eq14.
1  2  632.20  2950  25  273.151.8418  7.0384 =-768.43 kJ/kg (Eq13)
 3   4  1523.2  816.2  25  273.158.519  7.846=506.36 kJ/kg
(Eq14)
Substituting Eq13 & Eq14 into Eq11,

I

 768.43  3.28506.36 =892.43 kJ/kgH2O
m H 2O
Step 5: Concluding Statement & Remarks
Answer b)
The mass flow rate of combustion gases per kg of H20 flowing inside the tubes
was found to be 3.28. The rate of exergy destruction per kg of H 20 flow was
found to be 892.43 kJ/kgH20.