Chapter Two
Reversibility
2-1 Reversible work
In thermodynamics, a quasi-static change is an ideal process. A quasi-static process is
approximately realized by making the system change very slowly. The quasi-static
process is referred to as an internally reversible process or as a reversible process.
In a closed system, the reversible work is given by
2
w rev   Pdv
(2-1)
1
Different process is between the same two states will results in different amount of
work.
Isothermal process Pv = const.
2
v
(2-2)
w rev   Pdv  P1 v1 ln 2
1
v1
Polytropic process Pvn = const.
w rev  c 
2
1
dv P2 v 2  P1v1

vn
1 n
(2-3)
Adiabatic process Pvk = const.
dv P2 v 2  P1 v1

1 vk
1 k
The reversible work for an open system is given by
w rev  c
2
V1  V 2
 g ( Z1  Z 2 )
1
2
In this process, the change in KE and PE are negligible, Eq.(2-5) is reduced to
2
w rev    vdP 
2
(2-4)
2
2
w rev   vdP
1
(2-5)
(2-6)
Example 2-1 Air at 100kPa and 298K enters a compressor at a velocity of 40m/s and
discharges at 1MPa and 90m/s. In this process, the airflow rate is 15k/s. Determine the
power input required for the compressor.
Sol.
1st law for an adiabatic process:
2
2
 
 
V 
V  
m i  h i  i   W rev  m e  h e  e 
2 
2 




m(Vi  Ve )
 m c p (Ti  Te ) 
2

W rev
Te  Pe

Ti  Pi



n 1
n
P
, Te   e
 Pi
2



n 1
n
2
 1 
Ti  298

 0.1 
1.4 1
1.4
 575.35K
Reversible work

W rev  (.)(   .) 
(     )
  
 990.51kJ / s
2-2 Available work
The system and environment are in equilibrium, no further change of state of the
system can occur spontaneously and hence no further work is performed. Hence the
above described process leads to the maximum reversible work or work potential
associated with the static state of a system.
When a system and its environment are in equilibrium with each other, the system is
said to be in its dead state. A system in a dead state is in mechanical and thermal
equilibrium with environment at P0 and T0. The numerical values of (P0, T0) for dead
state are those of the standard environment state, namely, 298.15K and 1.01325 bars (1
atm). Additional requirements for the dead state are that the velocity of a charged
system or fluid system be zero. Also, the gravitational potential energy is zero. The
work potential of a system relative to its dead state, which exchanges heat solely with
the environment, is called the thermo-mechanical availability or energy of that state. A
reversible heat transfer process of the system is defined as the process in which the heat
is transferred through an infinitesimal temperature difference. The process of
infinitesimal temperature difference will require an infinite amount of time or infinite
heat transfer area. Fig.2-1 shows the difference between the reversible and irreversible
heat transfer from bodyⅠto bodyⅡ. The actual heat transfer process is characterized by
a finite temperature difference, the reversible heat transfer process is characterized by
an infinite number of fictitious bodies between the two actual bodiesⅠandⅡ. In this
case, each fictitious body performs heat source and sink functions and it reduces the
temperature by an infinitesimal amount in the direction of heat flow.
Ⅰ
Heat source
at T
(a) Irreversible heat transfer
Ⅱ
Heat sink
at To
Fictitious heat sources and sinks
Heat source
at T
(b) Reversible heat transfer process
Heat sink
at To
Fig. 2-1 Reversible and irreversible heat
transfer process
Consider the two control volumes A and B in Fig. 2-2. Control volume A undergoes
an internally reversible process. Thus, control volume A’s surface temperature is
different from the temperature is the environment, the process is not externally
reversible. Therefore, the work of control volume A is not the maximum quantity of
work that could be produced (engine) or the minimum quantity of work that must be
supplied (compressor). To make the process externally reversible, one or more
reversible that engines such as Carnot engine must be attached to control volume A. In
this heat engine operates two temperature ranges between given surface temperature (T)
and the environmental temperature (T0). As shown in Fig. 2-2, control volume A and
Carnot engine are combined to form a new control volume B. In this case, the control
volume B undergoes an internally and externally reversible change.
c.v B
c.v A


me
mi
T

W av
Rev.
HE

Qo
Environment
To, Po
Fig. 2-2 An internally and externally reversible process
for the control volume B
Thus, the total work produced by control volume B must be sum of the reversible
work by control volume A and Carnot engine HE. This is the work one would have
obtained from an ideal process between two given stats.
The work from the internally and externally reversible process is defined as the
available work, Wava.
Considering the 1st law eq. to control volume B in Fig. 2-2 gives

2

2


Vi
Ve
 gZ i )  m e (h e 
 gZ e )  Q o
2
2
The 2nd law eq. of thermodynamics for the control volume B is given by
W ava  m i (h i 

(2-7)


Q
mi si  me se  o  0
To
(2-8)
Combining these two equations yields the general expression for the available work
per unit mass of working substance:

w ava 
W ava

mi
 Vi 2 Ve 2 
  g( Z i  Z e )
 (h i  h e )  To (s i  s e )  

2 
 2
Neglecting the changes in kinetic and potential energies, wava is given by
(2-9)
w ava = (h i  h e )  To (s i  s e )
(2-
10)
The available work by a closed system can be derived in similar manner. The 1st law
and 2nd law equations are combined to yield to available work as
(2-10)
w ava = (u i  u e )  To (s i  s e )
There is another approach in developing eq. (2-10) from Fig. 2-3

W rev
q rev
Rev.
HE

We
qo
Environment
To, Po
Fig. 2-3 Available work by a closed system
We start with the fact that available work is equal to the reversible work produced by
the closed system plus the work by Carnot engine.
δ Wava  δ Wrev  δ Wc
(2-11)
The work per unit mass produced by the Carnot engine is given by
 T 
δ w c  δ q rev   o 
T

(2-12)
where (-) sign for δ q rev means the direction of heat transfer from the system to the
Carnot engine.
Combining these two equations yields
 T 
δ w ava  δ w rev  δ q rev   o 
T

 w rev  q rev  To
q rev
T
q rev
 du rev  To ds
T
Integrating eq. (2-13) between state 1 and state 2 is given as
w ava = (u   u  )  To (s  s  )
=  du rev  To
(2-13)
(2-14)
This equation is identical eq. (2-10) and is use to determine the available work by a
closed system.
A more general equation for the available work is given by
Wava =  mi (h i  s i )   me (h e  To s e )  m (h  To s )  m  (h   To s  )
(2-15)
For a steady-state, steady-flow process, the difference between the state 1 and state 2 is
m u  Ts   m u   Ts  cv  
Thus, available work for a SSSF process become
Wava   mi h i  T s i    m e h e  T s e 
Wava  H i  H e   T Si  Se 
For a closed system, the available work is given by
Wava  m u   T s   m  u   T s  cv
= (U1 – U2) – T0 (S1 – S2)
Example 2-2 Air enters a turbine at 10 bar and 980 K and expands to 1 bar and 500 K.
The heat loss from the turbine is approximately 2% of the turbine output. Calculate the
actual work and available work for turbine process.
Sol.
T
1
·
Actual work wact
q + h1 = wact + h2
-0.02 wact + h1 = wact + h2
1.02 wact = cp (T1 - T2)
·2s ·2
w av 
s
Fig. 2-4
0.24(980  500)
1.02
= 112.94 kJ/kg
wact = (h1 – h2 ) – T0 (s1 – s2)
= cp (T1 –T2) – T0 ( cp ln
T1
P
– R ln 1 )
T2
P2
980
10 

 0.287 ln 
wava = 0.24 (980 – 500) – 273  0.24 ln
500
1

Problem 2-1 Air is compressed adiabatically in a reciprocating compressor. For initial
conditions of 100 kPa and 20℃ and final conditions of 300kPa and 130℃, calculate
the actual work and the available work in the compression process.
2-3 Unavailable energy
The difference between the reversible work and actual work is defined as the lost
work. Lost work is the work, which could have been obtained but was not actually
produced because of internal irreversibility.
The lost work Wlost is given by
(a)
Wlost = Wrev  Wact
For closed system, Wrev is given as
Wrev = Q rev  dU rev
Also,  W act is expressed as
Wact = Q act  dU act
(b)
(c)
Substituting Equations (b) and (c) into equation (a) and taking into consideration
dU rev = dU act , the lost work is
Wlost = ( Q rev  dU rev ) – ( Q act  dU act )
= Q rev  Q act
(d)
For a reversible process Q is expressed as
Q rev  TdS
(e)
Combining equations (d) and (e) yields
Q act Wlost
(2-16)

T
T
Equation (2-16) shows that the entropy change is caused by the heat transfer between
the system and its surroundings and by the lost work within the system. The heat
transfer can increase or decrease the entropy, depending on the heat-flow direction. The
lost work is always positive and therefore will always results in an increase of entropy.
The entropy induced by the lost work is referred to as the entropy production.
dS 
If the system undergoes the internally reversible process, there will be no lost work,
and Eq. (2-16) will reduce
Q 
dS 

T  rev
Equation of the (2-16) can be written in a rate form and expressed as
W 
 lost 
T 
dS Q act

 
dt
T
dt
The second law eq. for a closed system is

(2-17)

dS Q act 


dt
T
where  
Wlost
T
(2-18)
The lost work is caused by internal irreversibility Carnot work (Fig. 2-3,HE) is the
work lost due to external irreversibility. Since there is no Carnot engine, the work
expected from it is imaginary and does not exist at all. Therefore the Carnot work
δ Wlost,c is lost due to an external irreversibility.
We have the equation for the Carnot work :
 T 
Wlost,c  δ Wlost,c     c  δ Q rev
T

(2-19)
The value of Wlost and Carnot work are dependent on the process. For given initial
and final states the lost work and Carnot work will be determined by the nature of
process.
In this chapter, the combination of lost work and Carnot work ( Wlost  Wlost,c ) is
referred to as the unavailable energy or unavailable work. Using the definition of
available work, δ Wava is
Wava = Wrev + Wlost,c
where Wlost  Wrev  Wact
Wava  Wact  Wlost  Wlost,c
or
 ( Wlost + Wlost,c ) = Wava  Wact
(2-20)
Eq. (2-20) means that the unavailable energy in the process is the difference between
the available work and the actual work. In other words, it is the amount of work that is
not available for work conversion during the process.
For a closed system, the second law eq. is given by
 ( Wlost + Wlost,c )  dU rev  Q rev  Q act  dU act
 To dS  δ Q act
(2-21)
Eq. (2-21) can be written as
dS 
( Wlost  Wlost,c )
To

Q act
To
(2-22)
Eq. (2-22) can be expressed in a rate form as
dS 1  ( Wlost  Wlost,c )  


 Q act 
dt To 
t

(2-23)
The second law eq. for an open system can be expressed as


dScv
1
  mi s i   me s e 
dt
To

 
 1 
W

W
lost
lost
,
c

   Qact

 To
(2-24)
Example 2-5 Air enters a turbine at 1 MPa and 1093 C and the air first expands
isentropically to 0.1MPa and then change to final state by a reversible and isobaric
process. Determine for the entire turbine process. Final temp. is 540 C and heat loss
from the turbine is approximately 2.5 % of the turbine output calculate for this turbine
process. (a) the reversible work, (b) the Carnot work, (c) the lost work, and (d) the
unavailable energy
Sol.
T-s diagram for the turbine process is shown in Fig. 2-5.
The process 1-2s is isentropic process,
and the process 2s-2 is reversible and
isobaric process.
1MPa
T
1
·
0.1MPa
·
·2s 2
(a) The reversible work for the entire
process is
w rev = w rev , s + w rev , s 
s
= cp(T1 – T2 s ) + 0
Fig. 2-5
The temperature T2s is calculated by the following isentropic P-T relationship
T2s  P2s 


T1  P1 
P
T2s  T1  2
 P1
k 1
k



k 1
k
0.4
 0.1  1.4
 (1093  273)

 1 
 707.52K
w rev = 1.005  1093  273  707.52
= 661.77 kJ/kg
(b) Carnot work
w c  w c1 2 s  w c 2 s  2

    ( 
s
T
)δ q
T

To
TdS
s T
T
 c p (T  Ts )  To c p ln 
Ts
 c p (T  Ts )  
 813 
=-1.005  540  273  707.52 + (273+20)  1.005 ln 

 707.52 
= -65.087 kJ/kg
Checking:
w ava = ( h 1 - h 2 ) - To( s 1 - s 2 )

T
P 
= cp ( T 1 - T 2 ) - To  c p ln 1  R ln 1 
T2
P2 

1366
1 

 0.287 ln
= 1.005 (1093  273)  813  293  1.005 ln
813
0.1

=596.590 kJ/kg
w rev = cp ( T 1 - T 2 s )
= 1.005(1366-707.52)
=661.77 kJ/kg
w ava = w rev + wc
∴ wc = w ava  w rev
 5 9 6 . 5 9 60 6 1 . 776 5 . 0k6J / k g
The results of both methods are in good agreement.
(c) The lost work due to the internal irreversibility is the difference between the
reversible work and the actual work.
w lost  w rev  w act
The actual work is given by the first law eq.
q  h1  ke1  pe1  h 2  ke 2  pe 2  w
Neglecting the difference of ke and pe , the 1st law equation is given by
q + h1 = h 2 + w
W act = q + (h 1  h 2 )
= ( h1  h2 ) 
2.5
+( h 1  h 2 )
100
= ( h 1  h 2 ) (0.25+1)
= cp(T 1  T2 ) (1.025)
=1.005(1366-707.52) 1.025
= 678.320 kJ/kg
The lost work w lost is
w lost = w rev  w act = 661.77  678.320
=  16.55 kJ/kg
(c) The unavailable energy for the turbine process is expressed as
w lost + w lost,c =  16.55 – 65.087 =  81.64 kJ/kg
The unavailable work equals the available work minus actual work.
Check: lost work = w ava  w act
= 596.596 – 678.320
=  81.72
The results by two methods are in good agreement.
Example 2-6 Air is compressed in a cylinder of compressor. For initial conditions of 0.1
MPa and 25oC, and final conditions of 0.35MPa and 160oC. In this problem, the
compression process is assumed that the air is first compressed isentropically and then
expanded reversibly and isothermally to the final state.
Calculate (a) the reversible work, (b) the lost work (c) the Carnot work, and (d) the
unavailable energy.
Sol.
0.35MPa
T
(a) Reversible work is given by
2
·
wrev = wrev,1-2s + wrev,2s-2
0.1MPa
= cp(T2S – T1) +T2s (S2 –S2s)
T2s = 160+273 = 433K
T1 =298K
·
where, wrev,2s-2 is
1
q+h1=h2+w
s
Fig. 2-6 T - s diagram
q=w
 wrev,2s-2 = q = T2(s2-s2s) = T2 (s2-s1)
w rev , 2s  2  T2 ( 
2
1
2 vdP
dh

)
1
T
T
T2
P
443
0.35
 R ln 2 )  443(1.005 ln
 0.287 ln
)
T1
P1
298
0.1
 17.2396kJ/kg
 T2 (c P ln
w rev ,s  c v (TS  T )  .(   )  .kJ/kg
c p  c v  R ,  c v  c p  R  1.005  0.287  0.718kJ/kg
 w rev  w rev ,s  w rev , s  .  .  .kJ/kg
(b) The lost work is the reversible work minus the actual work.
w lost  w rev  w act ,
w act  c v (T2S  T1 )
 0.718(443  298)  104.11kJ/kg
w lost  86.77  (104.11)  17.34kJ/kg
(c) The Carnot work for the entire process is
w C  w C12S  w C2S2
2
 0   (1 
2S
T0
)q
T2
q  T2 ds
 (T2  T0 )(s 2  s1 )
w C  (T2  T0 )(S2  S2S )
 (T2  T0 )(c p ln
T2
P
 R ln 2 )
T1
P1
 (443  298)(1.005 ln
443
0.35
 0.287 ln
)
298
0.1
 5.643kJ/kg
(d) The unavailable energy can be determined by the sum of lost work, Wlost  Wlost,c
2-5. Total entropy change
For an open system the 2nd law eq. has been developed and expressed as eq. (2-24).
dScv
1
1 
 W

 isi   m
 e s e   (W
(2-25)
 m
Q CV
lost
lost,c ) 
dt
To
T0
i
e
When eq. (2-25) is applied to the surrounding of the open system, the second law eq. is
dSsur
1 
 es e   m
 isi  Q
 m
(2-26)
cv
dt
T0
e
i
Combining eqs (2-25) and (2-26) will produce
dScv dSsur 1
 W


  (W
lost
lost,c )
dt
dt
T0 j
or,
dS tot 1

dt
T0
 (W
lost

W
lost,c )
(2-27)
(2-28)
j
The above equations indicates that the rate of total entropy change in the system and
its surrounding is equal to the rate of change in the unavailable entropy divided by the
surrounding temp To.
Equation (2-28) can be expressed as
dS tot
(2-29)
0
dt
where the equality holds for reversible process and the inequality for irreversible
change.
The rate of entropy production is referred to as the principle of entropy production.
 W

dS tot W
lost,c
 lost
dt
T0
(2-30)
The followings are two examples for the total change of entropy. The surrounding
temperature T0 is assumed to be 25oC under unless otherwise indicated.
Example 2-7 Calculate the total entropy change of air the expanding process of turbine.
Air enters to turbine at 1MPa and 1100oC and expands to 0.1MPa and 530oC. The heat
loss from the turbine is 2.5% of the turbine output.
Sol.
The total change of entropy in the turbine process is
s tot  s cv  s sur
Because of the steady-state and steady-flow condition, the term Scv must be vanished.
s tot  s sur
 (s 2  s1 ) 
 c p ln
q0
T0
T2
P 0.025Wact
 R ln 2 
T1
P1
T0
s tot  1.005 ln
530  273
0.1 (558.88)  0.025
 0.287 ln

1100  273
1
298
 0.1685kJ/kgK
 q  h1  h 2  w act
w act  h 1  h 2  0.025w act
w act (1  0.025)  h1  h 2
w act 
c p (T1  T2 )
1.025

1.005(1373  803)
 558.88kJ/kg
1.025
Example 2-8 Superheated vapor at 15MPa and 550℃ is throttled to 12.5MPa by the
control valve. Calculate the net change of entropy and unavailable energy in the
throttling process.
Sol.
The process is adiabatic, steady-state, and steady-flow, the equation is reduced to
s tot  s e  s i
w lost  w lost,c  To (s e  s i )
From steam table, hi =
si =
se =
The second law equation involved the lost work is expressed as

dS cv
Q
 isi   m
 e s e   i  
 m
dt
i
e
j Tj
(a)
For the surrounding the second law eq. is

Q
dSsur
j


  mes e   misi  
dt
e
i
j T0
Combining equations (a) and (b), it gives the rate of total change in entropy as
dScv dSsur
 (1  1)

   Q
j
dt
dt
Tj T0
(b)
(2-31)
In above equation (2-31), the term  is always positive and would become zero
 and ( 1  1 ) are positive or
when process is internally reversible. The second term Q
j
Tj T0
negative at the same time.
Eq. (2-31) is rearranged in term of internal and external irreversibility as follows.

dS 
 dS
 T 
(2-32)
T  cv  sur   T σ       Q j
dt 
T
 dt
j 
Let the rate of internal irreversibility and external irreversibility be


I int  T σ

 T 
I ext       Q j
T
j 
Then the equation of irreversibility becomes

dS  
 dS
T  cv  sur   I int  I ext
dt 
 dt
(2-34)
(2-35)
This equation indicates that the rate of process irreversibility is related to the rate of
total entropy change in the system and environment.
Let us consider the heat reservoir, which is connected to the system. The heat reservoir
is a body of infinite capacity, when the heat reservoir under goes a thermodynamic
process, it has a constant temperature. Also, the heat reservoir produces no work and it
exchanges heat only with surroundings.
δ Wava  dU  T dS
(2-36)
Since the heat reservoir undergoes internal reversible process and produces no work,
the available work of the heat reservoir is given by
dU  δ Qr and dS 
δ Qr
Tr
(2-37)
Substituting eq. (2-37) into the eq. (2-36), we have the expression for the available
work by heat reservoir.
 T 
δ Wava     δ Q r
 Tr 
Integrating this equation is
 T 
Wava     Q r
 Tr 
(2-38)
where Tr : temperature of the reservoir
Qr : heat transfer from the reservoir
The available work of heat reservoir represents the maximum amount of work
that could be produced by the heat reservoir when it releases the heat.
For an open system, we have
 T 
Wava  ( H  H  )  T (S  S )     Q r
 Tr 
(2-39)
For a closed system, we have
 T 
Wava  ( U  U  )  T (S  S )     Q r
 Tr 
(2-40)
(2) Heat exchanger
cold fluid
in
hot fluid
out
in
out
Fig. 2-7
If the heat exchanger is well insulated and two streams (hot and cold stream) are
treated as two individual systems, the hot or cold system exchanges heat with another
system.
We first calculate the available work for the hot and cold streams as
Wava,h  m h (h i  h e ) h  m h T (s i  s e ) h
Wava,c  m c (h i  h e ) c  m c T (s i  s e ) c
The available work for the heat transfer process in the heat exchanger is
Wava  Wava ,h  Wava ,c
(2-39)