SECTION XIII - IRREVERSIBILITY
Irreversibility – a very powerful tool in the design and optimization of complex thermodynamics
systems.

Consider the application of conservation of energy to a uniform-state, uniform-flow process
such as the filling or emptying of a tank:
W12

m i h i  e Ki  e Pi

Q12
(from surroundings at T0)

m e h e  e K e  e Pe
m 2 (u 2  e K 2  e P2 )

 m 2 (u 1  e K1  e P1 )

Consider now the application of conservation of energy to a reversible uniform-state, uniformflow process:
Q0
Wrev
Wc
W12 rev

m i h i  e Ki  e Pi

Reversible heat engine to facilitate
reversible heat transfer
Q12 rev
m 2 (u 2  e P2  e K 2 )

m e h e  e Pe  e K e

 m1 (u1  e P1  e K1 )
(net work done by the reversible heat engine)
Wrev = W12 rev + WC

Irreversibility: I = Wrev – W12 (i.e. actual)

Conservation of energy:
Q12 rev  m i (h i  e Ki  e Pi )  W12 rev  m e (h e  e K e  e Pe )
 m 2 (u 2  e K e  e Pe )  m1 (u 1  e Ki  e Pi )

Reversible Heat Engine: WC = Q0 – Q12 rev
(i)
(ii)
Section XIII - Irreversibility

Second Law:
Page 62
 
tQ
Q0
   12  dt
0
T0
 T  rev
T0: temperature of surrounding
(iii)
 
tQ
 WC  T0   12  dt  Q12 rev
0
 T  rev

Second Law for the process (i.e. uniform-state, uniform-flow):
 
tQ
m 2 s 2  m1s1  m e s e  m i s i    12  dt  0
0
 T  rev
reversible
 
tQ
 m 2 s 2  m1s1  m e s e  m i s i    12  dt
0
 T  rev
(iv)
WC  T0 m 2 s 2  m1s1  m e s e  m i s i   Q12 rev
(v)



  m u
 Wrev  Q12 rev  m i h i  e Ki  e Pi  m e h i  e Ki  e Pi
 
 m 2 u 2  e K 2  e Pi 2

1
1



W12 rev [see (i)]
 e K1  e P1
 T0 m 2 s 2  m1m1   m e s e   mi s i  Q12 rev



WC [See (v)]
 Wrev   m i h i  e K i  e Pi  T0 s i   m e h e  e K e  e Pe  T0 s i
 



 m 2 u 2  e K 2  e Pi 2  T0 s 2  m1 u 1  e K1  e P1  T0 s1


(vi)
For a nonflow process (i.e. a closed system):
mi = me = 0; m1 = m2 = m  (vi) becomes


 


Wrev  m u 1  u 2   e K1  e K 2  e P1  e P 2  T0 s1  s 2 
w rev 

Wrev
m
u
1

 
(vii)


 u 2   e K1  e K 2  e P1  e P 2  T0 s1  s 2 
(vii a)
For a flow process (i.e. an open system:



  m
 i h i  e K i  e Pi  T0 s i  m
 e h e  e K e  e Pe  T0 s e
W
rev

 



d
m 2 u 2  e K 2  e P2  T0 s 2  m1 u 1  e K1  e P1  T0 s1
dt

(viii)
Section XIII - Irreversibility

Page 63
d
0
dt
i m
e m
 and
If the flow process is steady, then m


 


 m
 h i  h e   e Ki  e Ke  e Pi  e Pe  T0 s i  s e 
W
rev
w rev 

W
rev


 
(ix)


 h i  h e   e K i  e K e  e Pi  e Pe  T0 s i  s e 
m
I = Wrev – W12
Wrev




  m u
 m i h i  e Ki  e Pi  T0 s i  m e h e  e K e  e Pe  T0 s e
 
 m 2 u 2  e K 2  e P2  T0 s 2
1
1
 e K1  e P1  T0 s1

W12

  




 m i h i  e K i  e Pi  m e h e  e K e  e Pe  m 2 u 2  e K 2  e P2  m1 u 1  e K1  e P1
 I   me T0 s e   mi T0 s i  m 2 T0 s 2  m1T0 s1  Q12
I  T0 m 2 s 2  m1s1   T0  m 2 s 2   m i s i   T0
Q12
T0
T0dstot

 e= 0
For a nonflow process: m1 = m2 = m, mi = m


I  T0 m(s 2  s1   Q12
i

I
 T0 s 2  s1   q 12
m
For a flow process:

Q
I  T d m s  m s    m



s

m
s

T
 i i 0 T12
0
2 2
1 1
e e
dt
0

For a stead-flow process:
i m
e m
 ,
m
d
0
dt
 dSsys dSsurr 
In general, I  T0 


dt 
 dt

Q12
I  T m

0 s e  s i   T0
T0

(x)
Section XIII - Irreversibility
Page 64
6.14 A small storage tank with a volume of 0.12 m3 initially contains 0.9 kg of argon at 40C.
Heat is transferred to the argon from a reservoir whose temperature is 1300C until the
entropy of the argon has increased by 0.46 kJ/kg K. Determine the amount of heat transferred,
the final pressure of the argon, the reversible work, and the irreversiblity of the process.
(Assume T0 = 25C, P0 = 100 k Pa)
Q12
Argon
m = 0.9 kg
 = 0.12 m3
T1 = 40C=313 K
Tres = 1300C = 1573 K
system boundary
T1 = 313 K > Tcrit argon = 151 K
P1 
mRT1 0.9  0.2081  313

 488.5 KPa  Pcrit arg on  4.86 MPa

0.12
 treat argon as ideal

Assume tank rigid  1 = 2 =  = 0.12 m3

Change in entropy
s 2  s1  c v n
T2
v
 R n 2  0.46 kJ / kg K
T1
v1
(v1=v2)
T2  T1e (s2 s1 ) c v  313 e 0.46 0.3122  1366 K
T 
 1366 
P1  mRT1 , P2   mRT 2  P2  P1  2   488.5

 313 
 T1 
P2  2.13 MPa
Q12  W12  ( U 2  U1   mc v (T2  T1 )  (0.9)(0.3122)(1366  313)  295.9 kJ
 tan k

 assumed 
 rigid



Q
298(295.9)
I  67.3 kJ
I  T0 ms 2  s1   T0 12  298(0.9)(0.46) 
Tres
1573
I  Wrev  Wactual 
Wrev  67.3 kJ
Section XIII - Irreversibility
Page 65
Steam at 1.4 MPa and 260°C enters an adiabatic nozzle with negligible velocity and exits at a
pressure of 150KPa and a quality of 96 percent. Calculate the exit velocity, the adiabatic nozzle
efficiency and the irreversibility per unit mass. (To = 25°C)
inlet
exit
 Flow is uniform at the inlet and exit
 ep  h
 Nozzle walls are rigid
Adiabatic nozzle
Steam
Pi = 1.4 MPa
Ti = 260°C
Vi  0
System
boundary
Assumptions
 Flow is steady
i m
e m

 m
Conservation of Energy
Pe = 150KPa
Xe = 0.96
Ve = ?
2
v
 W
 m
 (h e  e  hi )
Q
ie
ie
2000
(adiabatic)
V
 hi = he + e
2000

 Walls



 assumed

 rigid  d  0 


2
Steam is superheated at the inlet since Ti + 260° > Tsat = 195.07°C corresponding to
Pi = 1.4MPa
hi = 2927.2 +
3040.4  2927.2 (260  250)  2949.8 KJ
300  250
Kg
he = 467.11 + (0.96)(2226.5) = 2604.6
Exit Velocity: Ve =
KJ
Kg
2000(hi  he )  2000(2949.8  2604.6)
Ve = 830.9 m/s
Adiabatic nozzle efficiency
N 
hi  he ,a
hi  he ,s

2949.8  2604.6  0.816 (81.6%)
2949.8  2526.6
Section XIII - Irreversibility
si = 6.7467 +
se, s  seg
Page 66
(6.9534  6.7467)( 260  250)
kJ
 6.7880
 se
(300  250)
KgK
Pe0.150MPa
, se,s  se f
Pe0.150MPa
 Saturated liquid-vapour mixture at
xe,s =
se , ,s  se f
S efg

(6.7880  1.4336)
 0.925
5.7897
he,s = 467.11 +0.925 (2226.5) = 2526.6
Irreversibility:
e,s
kJ
Kg
i = To (se –si) + q12
adiabatic
se = 1.4336 + (0.96) (5.7897) = 6.9917
i = 298 (6.9917 – 6.7880) = 60.7
KJ
KgK
KJ
Kg
Steam at 4MPa and 450°C enters an adiabatic turbine at a rate of 12 Kg/s. Ten percent of the mass
flow is extracted at a point where the temperature and pressure of the steam are 300°C and 1.2 MPa
respectively, for use in an industrial process. The remainder of the steam is further expanded and
exhausts from the turbine at 75 KPa with a quality of 98 percent. Determine the adiabatic efficiency
of the turbine , the irreversibility rate and the reversible power. (Assume To = 25°C)
Pi = 4MPa
Ti = 450°C
inlet
Adiabatic Steam Turbine
Intermediate
exit
Pint = 1.2 MPa
Tint = 300°C
System
boundary
Exit
Pe = 75KPa
Xe = 0.98
m i 12 Kg / s
m int  0.1 m i 1.2 Kg / s
Assumptions
 Flow is steady
 Fluid is at steady state
 Flow is uniform at inlet
and exits
 ep & ek << h
Section XIII - Irreversibility
Page 67
Conservation of Mass
m i  m int  m e  m e  m int  12  1.2
m e  10.8Kg / s
Conservation of energy
Q ie  W ie  m e he  m int hint  m i hi
 adiabatic 


 turbine 

Ti = 450°C > Tsat = 250.4°C corresponding to Pi = 4Mpa
 Steam is superheated at the inlet
hi = 3330.3

kJ
kJ
, si = 6.9363
Kg
KgK
Tint = 300°C > Tsat = 187.99°C corresponding to Pint = 1.2 MPa  steam is superheated at the
intermediate exit
hint = 3045.8
kJ
kJ
, sint = 7.0317
Kg
KgK

he = 384.39 + 0.98 (2278.6) = 2617.4
kJ
Kg

se = 1.2130 + 0.98 (6.2434) = 7.3315
kJ
KgK

If process is turbine is reversible, it will be isentropic since it is also adiabatic
 si = sint = se = 6.9363
sint,  sintg
se,s < seg
p  2 MPa
p 75MPa
kJ
KgK
 6.5233
 6.5233 
se,s > sef P75MPa  1.2130
kJ
 superheated also at
KgK
int,s
kJ
saturated liquid-vapour mixture at
KgK
kJ
KgK
e,s
Section XIII - Irreversibility
xe,s =
Page 68
6.9363  1.2130
 0.917
6.2434
he,s = 384.39 + (0.917)(2278.6) = 2473.9
hint,s = 2935.0 +
hint,s = 2993.5
kJ
kg
(3045.8  2935.0)(6.9363  6.8294)
(7.0317  6.8294)
kJ
kg
Actual turbine power:
 i hi  m
 e he  m
 int hint
Wie,a = m
= 12(3330.5) –10.8 (2617.4)-1.2 (3045.8)
Wie,a = 8040.7 kw
Turbine power under isentropic flow conditions
Wie,s = m i hi  m e he, s  m int hint,s
= 12(3330.3) – 10.8(2473.9)-1.2 (2993.5)
Wie,s = 9653.3 kw
Adiabatic Efficiency
T 
W ie,a 8040.7

 0.833(83.3%)
W ie,s 9653.3
Irreversibility Rate
dSys 
I  To (m e s e  m int sint  m i si )  To
 Qie
dt
 steady  adiabatic 



 state  turbine 
I  29810.8(7.3315)  1.2(7.0317)  12(6.9363)
I  1306 KW
Section XIII - Irreversibility
Page 69
Reversible Power
I  W
 W
 W
  I  W

rev
ie,a
rev
ie,a
W rev  1306  8040.7  9364.7 KW
Air at 275 KPa and 360k enters an internally reversible, isothermal turbine and exhausts at 105 KPa.
Heat is transferred to the air in the turbine from a reservoir whose temperature is 550k, and To =
300k. Determine the work done per kilogram of air, assuming that changes in kinetic and potential
energies are negligible. Compare the turbine efficiency and turbine effectiveness. Which of these
performance characteristics is more meaningful?
Q ie
inlet
Isothermal
Turbine air
Wie
System
boundary
exit
Ti  Te  360 K  Tcritair  l 33K
Pi  275 KPa
  Pcritair
Pe  105KPa 
Given: Process is
isothermal and internally
reversible
ek <<0, ep <<0


  air can be treated as an ideal gas
 3.76 MPa 

P 
Entropy change: se  si  se  si  R ln  e 
 Pi 
(Te = Ti)
105
kJ
 0.2763
 se-si = -0.287 ln
275
KgK
Heat Transfer: 9ie = T (se-si) = 360 (0.2763)
 qie= 99.47
T= 550k
Assumptions
 Flow through turbine is
steady
 Flow is uniform at the
inlet and exit
 System at steady state
kJ
kg
Conservation of Energy
qie-Wie = he-hi = Cpave (Te-Ti)
Section XIII - Irreversibility
 Wie= qie = 99.47
Page 70
kJ
kg
For an isentropic process from inlet to exit
se,s–si = s e, s  si  R ln
Pe
P
 se,s  si  R ln e
Pi
Pi
se, s  6.8852  0.287 ln
he,s = 270.12 +
= 273.63
T 
Wie,a
Wie, s

105
kJ
 6.6089
275
KgK
(280.14  270.12)(6.6089  6.5961)
(6.6326  6.5961)
kJ
kg
99.47
99.47

 1.144
(360.58  273.63) 86.95
 T is not a good measure of performance for an isothermal turbine since the isentropic
process is vastly different from the actual isothermal process
W
Effectiveness :
  ie,a
(or sec ond law efficiency )
Wie,rev
Irreversibility
dS sys Q ie 

I  To m ( se  si ) 


dt
Tres 

i=

q 
I
 To ( se  si )  ie 
m
Tres 

99.47 
kJ

i  300  (0.2763) 
 28.62

550 
kg

i = wrev - wactual
wrev = i + wactual = (28.62 + 99.47) = 128.09
 
99.47
 0.777 (77.7%)
128.09
kJ
kg
Section XIII - Irreversibility
Page 71
 is a better indicator of turbine performance under isothermal conditions than T since Wrev is
the maximum work that should be produced for the given conditions.
Refrigerant –12 enters a steady-flow adiabatic compressor as a saturated vapour at -10°C at a rate of
18 kg/min. At the exit of the compressor the temperature and pressure of the refrigerant are 30°C
and 500KPa. Calculate the power input, the compressor efficiency, and the irreversibility rate if To =
25°C.
  18 kg / min
m
 0.3 kg / s
inlet
Assumptions
 Flow properties are
uniformly distributed
at the inlet and exit.
 ep & eK << h
Adiabatic
Compressor
R-12
Wie
Given: Flow is steady
exit
System
boundary
h i  h g / T 10 C 183.058
kJ
kJ
, s i  0.7014
kg
KgK
Te = 30°C > Tsat = 15.06°C corresponding to Pe = 500KPa
 R-12 is superheated at the exit
he = 203.814,
kJ
kJ
, se = 0.7230
KgK
kg
Conservation of Energy:
Q ie  W ie  m (he  hi )
 adiabatic 


 compressor 
W ie  m (hi  he )  0.3(183.058  203.814)  6.23 kW
Section XIII - Irreversibility
Page 72
Compressor Efficiency
C 


W
ie,s

W
ie,a
If process in compressor is reversible  si =se,s = 0.7014 kJ/kg K since process is also adiabatic
 R – 12 is superheated at e,s
s g P500KPa  0.6899 kJ / kg K  s e,s  s e,sg
h e,s  196.935 
C 
203.814  196.9350.7014  0.6999  197.382 kJ / kg
0.7230  0.6999
197.382  183.058  0.69 (69%)
203.814  183.058
Irreversibility Rate
I  T (m
 se  m
 s i )  298(0.3)(0.7230  0.7014)  1.93 kW
0
P = 500 kPa
T
e

e
T = 303 K

e,s
T = 288.64 K
T = 263 K
ie


si
6.7

sp
s
Two kilograms of steam expands isentropically in a closed, adiabatic system from 2 MPa and
350C to a final temperature of 90C. Sketch the T-s and P-v diagrams, and calculate the
work and the irreversibility for this process. (Assume T0 = 25C, P0 = 100 kPa)

T1 = 350C > Tsat = 212.42C corresponding to P1 = 2 MPa  steam is superheated at state (1)
 s1 = s2 = 6.9563 kJ/kg K

s 2g
T  90 C
 7.4791 kJ / kg K , s 2f
T  90 C
 1.1925 kJ / kg K
s 2f  s 2  s 2g  state (2) is in the liquid-vapour saturation region
Section XIII - Irreversibility
x2 
s 2  s f2
s fg 2

Page 73
6.9563  1.1925
 0.917
6.2866
T
T1 = 350C
P


T1 = 350C
P = 2.0 MPa
P1 = 2MPa


Tsat = 212.42C
S = Const


T2 = 90C
Psat = 0.07 MPa
s

system is expanding  Pdv work is done

system adiabatic  Q12 = 0

Process is isentropic, since adiabatic  internally reversible
 Actual work done = Reversible Work  Irreversibility = 0

First Law: Q12  W12  U 2  U1
W12  U1  U 2
u1  2859.8 kJ / kg  U1  2(2758.8)  5718.6 kJ
U 2  2(2318.8)  4637.6 kJ
W12  5719.6  4637.6  1082 kJ


T2 = 90C
v