1. Thermodynamics
Chapter 3
Diploma in Engineering
Mechanical Engineering Division, Ngee Ann Polytechnic

2. Chapter 3
Steady Flow Processes with Steam
• Introduction
• Steam boiler
• Steam turbine
• Steam condenser
• Mixing chamber

3. Introduction
• This chapter deals with steady flow processes in open
systems.
• Limited to steam and water in the following devices:
 Steam boiler
 Steam turbine
 Condenser
 Mixing chamber
2 2
1 2
1 21 1 2 2( ) ( )
2 2
in outin out
c c
Q W m h gZ Q W m h gZ        
g g g g g g

4. Steam boiler
• Steam boiler is a device used in a steam power plant. Its
function is to generate steam at constant pressure.
Boiler
steam
feed water
furnace
inQ
g
lossQ
g
system
boundary
1
2
Heat energy is supplied to
convert the water into steam

5. Steam boiler
• kg/s of feed water enter
the boiler at point 1 with
specific enthalpy, h1, the
velocity, c1 and vertical
distance above the datum,
Z1
• kg/s of steam leave the
boiler at point 2 with specific
enthalpy, h2, the velocity, c2
and vertical distance above
the datum, Z2
• be the rate of heat
supplied from the furnace
into the boiler and be
the rate of heat loss from
the boiler to its surrounding.

6. Steam boiler
• Apply the continuity flow equation
• Apply the steady flow energy
• Since
• hence
2 2
1 2
1 21 1 2 2( ) ( )
2 2
in outin out
c c
Q W m h gZ Q W m h gZ        
g g g g g g
1 2
in out
m m
m m m

  
 
g g
g g g
0
0
in
out
W
W


g
g
2 2
1 2
1 21 1 2 2( ) ( )
2 2
in out
c c
Q m h gZ Q m h gZ      
g g g g

7. Steam boiler
Example
• A boiler operates at a constant pressure of 20 bar
conditions. Steam is produced at the rate of 0.5 kg/s with
a dryness fraction of 0.98. Feed water enters the boiler
at a temperature of 60°C. Assuming the heat lost to
surrounding, the change in kinetic and the change in
potential energy are negligible. Determine the rate of
heat energy supplied to the boiler.

8. Steam boiler
Solution:
Applying the continuity flow equation
At point 1, the feed water in a compressed liquid,
1 2 0.5 /
in out
m m
m m kg s

  
 
g g
g g
1 60
251.1 /
sf at t t C
h h kJ kg 
 o

9. Steam boiler
At point 2, the steam is a wet steam with dryness fraction,
x2=0.98
At p2=20 bar,
Applying
2 20
2 20
909 /
2799 /
f f at p bar
g g at p bar
h h kJ kg
h h kJ kg


 
 
2 2 2 2 2
(1 )
(1 ) 0.98 2799 (1 0.98) 909 2761.2 /
x g f
g f
h xh x h
h x h x h kJ kg
  
         

10. Steam boiler
Applying the steady flow energy equation
Since
Hence
2 2
1 2
1 21 1 2 2( ) ( )
2 2
in outin out
c c
Q W m h gZ Q W m h gZ        
g g g g g g
1 21 2
2 2
1 21 2
0
0
0
1 1
2 2
in
out
out loss
W
W
Q Q
m gZ m gZ
m c m c


 


g
g
g g
g g
g g
1 21 2
3 3 3
2 12 1 0.5 (2761.2 10 252.1 10 ) 1255.05 10 /
in
in
Q m h m h
Q m h m h J s
 
        
g g g
g g g

11. Steam turbine
• Steam turbine is a device used in a steam power plant.
Its function is to produce work output.
lossQ
g
outW
g
(power output)
1
2
high pressure
steam
low pressure
steam
1
1
1
1
m
c
h
z
g
2
2
2
2
m
c
h
z
g
During the expansion process work is produced by the
turbine and heat energy may be lost from the turbine
to its surrounding at a steady rate.

12. Steam turbine
• kg/s of steam enter the
turbine at point 1 with
specific enthalpy, h1, the
velocity, c1 and vertical
distance above the datum,
Z1
• kg/s of steam enter the
boiler at point 2 with specific
enthalpy, h2, the velocity, c2
and vertical distance above
the datum, Z2
• be the power output of
the turbine and be the
rate of heat loss from the
turbine to its surroundings.

13. Steam turbine
• Apply the continuity flow equation
• Apply the steady flow energy
• Since
• hence
2 2
1 2
1 21 1 2 2( ) ( )
2 2
in outin out
c c
Q W m h gZ Q W m h gZ        
g g g g g g
1 2
in out
m m
m m m

  
 
g g
g g g
0
0
in
in
W
Q


g
g
2 2
1 2
1 21 1 2 2( ) ( )
2 2
outout
c c
m h gZ Q W m h gZ      
g g g g

14. Steam turbine
Example
• In a steam power plant as shown, 3.5 kg/s of
superheated steam at pressure 20 bar and temperature
450 °C enters the turbine. It then expends and leaves
the turbine at pressure 0.12 bar and dryness fraction
0.92. The heat loss, the change in kinetic energy and the
change in potential energy are assumed to be negligible.
Determine the power output of the turbine.

15. Steam turbine
Solution:
Applying the continuity flow equation
Refer to the superheated steam table
1 2 3.5 /
in out
m m
m m kg s

  
 
g g
g g
1 20 , 450
3357 /at p bar t C
h h kJ kg 
 o

16. Steam turbine
At point 2, the steam is a wet steam with dryness fraction,
x2=0.92
At p2=0.12 bar,
Applying
2 0.12
2 0.12
207 /
2590 /
f f at p bar
g g at p bar
h h kJ kg
h h kJ kg


 
 
2 2 2 2 2
(1 )
(1 ) 0.92 2590 (1 0.92) 207 2399.4 /
x g f
g f
h xh x h
h x h x h kJ kg
  
         

17. Steam turbine
Applying the steady flow energy equation
Since
Hence
2 2
1 2
1 21 1 2 2( ) ( )
2 2
in outin out
c c
Q W m h gZ Q W m h gZ        
g g g g g g
1 21 2
2 2
1 21 2
0
0
0
1 1
2 2
in
in
out loss
W
Q
Q Q
m gZ m gZ
m c m c


 


g
g
g g
g g
g g
1 21 2
3 3 3
1 21 2 3.5 (3357 10 2399.4 10 ) 3351.6 10 /
out
out
m h W m h
W m h m h J s
 
        
g g g
g g g

18. Steam condenser
• Steam condenser is a device used in a steam power
plant. It normally has a low pressure and it condenses
the steam into water by taking away heat from the
steam.
Condenser
1
2
steam
condensate
rejectQ&

19. Steam condenser
• kg/s of steam enter the
boiler at point 1 with specific
enthalpy, h1, the velocity, c1
and vertical distance above
the datum, Z1
• kg/s of condensate leave
the condenser at point 2
with specific enthalpy, h2,
the velocity, c2 and vertical
distance above the datum,
Z2
• be the rate of heat
rejected from the condenser
to its surroundings.

20. Steam condenser
Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
2 2
1 2
1 21 1 2 2( ) ( )
2 2
in outin out
c c
Q W m h gZ Q W m h gZ        
g g g g g g
1 2
in out
m m
m m m

  
 
g g
g g g
0
0
0
in
out
in
out rejected
W
W
Q
Q Q




g
g
g
g g
2 2
1 2
1 21 1 2 2( ) ( )
2 2
rejected
c c
m h gZ Q m h gZ     
g g g

21. Steam condenser
Example
• In a steam plant as shown, 3.5 kg/s of steam at pressure
0.12 bar and dryness fraction 0.92 enters the condenser.
It then condenses and leaves the condenser as
saturated water. Assuming the changes in kinetic energy
and potential energy are negligible. Determine the rate of
heat rejected from the condenser to its surroundings.

22. Steam condenser
Solution:
Applying the continuity flow equation
1 2 0.35 /
in out
m m
m m kg s

  
 
g g
g g
Condenser
1
2
steam
condensate
rejectQ&
1
1
1
1
3.5 /
0.12 1
0.92
m kg s
P bar
x
h



2
2
h
m

23. Steam condenser
At point 1, the steam is a wet steam with dryness fraction,
x1=0.92
At p1=0.12 bar,
Applying
At point 2, the condensate is a saturated water at p2=0.12
bar
2 0.12
2 0.12
207 /
2590 /
f f at p bar
g g at p bar
h h kJ kg
h h kJ kg


 
 
1 1 1 1 1
(1 )
(1 ) 0.92 2590 (1 0.92) 207 2399.4 /
x g f
g f
h xh x h
h x h x h kJ kg
  
         
2 0.12 207 /f at p barh h kJ kg 

24. Steam condenser
Applying the steady flow energy equation
Since
Hence
2 2
1 2
1 21 1 2 2( ) ( )
2 2
in outin out
c c
Q W m h gZ Q W m h gZ        
g g g g g g
1 21 2
2 2
1 21 2
0
0
0
1 1
2 2
in
out
in
out rejected
W
W
Q
Q Q
m gZ m gZ
m c m c






g
g
g
g g
g g
g g
1 21 2
3 3 3
1 21 2 3.5 (3399.4 10 207 10 ) 7673.4 10 /
rejected
rejected
m h Q m h
Q m h m h J s
 
        
g g g
g g g

25. Mixing chamber
• Mixing chamber is a device used to mix high
temperature fluid and low temperature fluid, and produce
fluid at the required temperature. Hot fluid and cold fluid
enter the mixing chamber at point 1 and 2 respectively,
Fluid at the required temperature leaves the mixing
chamber at point 3. During the mixing heat energy may
be lost from the chamber to its surrounding.
Mixing chamber
1
2
3
High temp.
fluid
Low temp.
fluid
Fluid at the required
temperature
lossQ&

26. Mixing chamber
• kg/s of hot fluid enter the chamber at point 1 with
specific enthalpy, h1, the velocity, c1 and vertical distance
above the datum, Z1
• kg/s of cold fluid enter the chamber at point 2 with
specific enthalpy, h2, the velocity, c2 and vertical distance
above the datum, Z2
• kg/s of warm fluid leave the chamber at point 3 with
specific enthalpy, h3, the velocity, c3 and vertical distance
above the datum, Z3
• be the rate of heat loss from the chamber to its
surroundings.

27. Mixing chamber
Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
22 2
31 2
1 2 31 1 2 2 3 3( ) ( ) ( )
2 2 2
in outin out
cc c
Q W m h gZ m h gZ Q W m h gZ           
g g g g g g g
1 2 3
in out
m m
m m m

  
 
g g
g g g
0
0
0
in
out
in
out loss
W
W
Q
Q Q




g
g
g
g g
22 2
31 2
1 2 31 1 2 2 3 3( ) ( ) ( )
2 2 2
loss
cc c
m h gZ m h gZ Q m h gZ        
g g g g

28. Mixing chamber
Example
• In a food processing industry, steam and water are
mixed to produce continuous supply of hot water. 2.5
kg/s of dry saturated steam at pressure 1.2 bar and 60
kg/s of water at temperature 30 °C enter the mixing
chamber. If the heat loss from the chamber to its
surrounding is 500 W and the changes in kinetic energy
and potential energy are negligible. Determine the mass
flow rate and the specific enthalpy of the hot water.

29. Mixing chamber
Solution:
Applying the continuity flow equation
At point 1, the dry saturated steam at p1=1.2 bar
At point 2, the compressed water at t=30 °C
1 2 3
3 2.5 60 62.5 /
in out
m m
m m m
m kg s

  
  
 
g g
g g g
g
1 1.2 2683 /g at p barh h kJ kg 
2 30
125.7 /
sf at t C
h h kJ kg
 o

30. Mixing chamber
Applying the steady flow energy equation
Since
1 2 31 2 3
2 2 2
1 2 31 2 3
0
0
0
1 1 1
2 2 2
in
out
in
W
W
Q
m gZ m gZ m gZ
m c m c m c



 
 
g
g
g
g g g
g g g
22 2
31 2
1 2 31 1 2 2 3 3( ) ( ) ( )
2 2 2
in outin out
cc c
Q W m h gZ m h gZ Q W m h gZ           
g g g g g g g

31. Mixing chamber
Hence
1 2 31 2 3
3 1 23 1 2
1 21 2
3
3
3 3
3
2.5 2683 10 60 125.7 10 500
62.5
227.98 10 /
loss
loss
loss
m h m h Q m h
m h m h m h Q
m h m h Q
h
m
J kg
  
  
 
 
     

 
g g g g
g g g g
g g g
g

32. Thank you
Q & A