Thermodynamics chapter 3
Transcript of Thermodynamics chapter 3
Thermodynamics
Chapter 3
Diploma in EngineeringMechanical Engineering Division, Ngee Ann Polytechnic
Chapter 3Steady Flow Processes with
Steam
• Introduction• Steam boiler• Steam turbine• Steam condenser• Mixing chamber
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 21 2
1 21 1 2 2( ) ( )2 2
in outin outc cQ W m h gZ Q W m h gZ
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
furnaceinQ
lossQ
system boundary
1
2
Heat energy is supplied to convert the water into steam
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.
Steam boiler• Apply the continuity flow equation
• Apply the steady flow energy
• Since
• hence
2 21 2
1 21 1 2 2( ) ( )2 2
in outin outc cQ W m h gZ Q W m h gZ
1 2
in outm m
m m m
0
0
in
out
W
W
2 21 2
1 21 1 2 2( ) ( )2 2in outc cQ m h gZ Q m h gZ
Steam boilerExample• 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.
Steam boilerSolution:Applying the continuity flow equation
At point 1, the feed water in a compressed liquid,
1 2 0.5 /
in outm m
m m kg s
1 60251.1 /
sf at t t Ch h kJ kg
Steam boilerAt 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
Steam boilerApplying the steady flow energy equation
Since
Hence
2 21 2
1 21 1 2 2( ) ( )2 2
in outin outc cQ W m h gZ Q W m h gZ
1 21 2
2 21 21 2
0
0
0
1 12 2
in
out
out loss
W
W
Q Q
m gZ m gZ
m c m c
1 21 2
3 3 32 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
Steam turbine• Steam turbine is a device used in a steam power
plant. Its function is to produce work output.
lossQ
outW
(power output)
1
2
high pressure steam
low pressure steam
1
1
1
1
mchz
2
2
2
2
mchz
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.
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.
outW
Steam turbine• Apply the continuity flow equation
• Apply the steady flow energy
• Since
• hence
2 21 2
1 21 1 2 2( ) ( )2 2
in outin outc cQ W m h gZ Q W m h gZ
1 2
in outm m
m m m
0
0
in
in
W
Q
2 21 2
1 21 1 2 2( ) ( )2 2outoutc cm h gZ Q W m h gZ
Steam turbineExample• 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.
Steam turbineSolution:Applying the continuity flow equation
Refer to the superheated steam table
1 2 3.5 /
in outm m
m m kg s
1 20 , 4503357 /
at p bar t Ch h kJ kg
Steam turbineAt 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
Steam turbineApplying the steady flow energy equation
Since
Hence
2 21 2
1 21 1 2 2( ) ( )2 2
in outin outc cQ W m h gZ Q W m h gZ
1 21 2
2 21 21 2
0
0
0
1 12 2
in
in
out loss
W
Q
Q Q
m gZ m gZ
m c m c
1 21 2
3 3 31 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
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
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.
rejectedQ
Steam condenserApply the continuity flow equation
Apply the steady flow energy
Since
Hence
2 21 2
1 21 1 2 2( ) ( )2 2
in outin outc cQ W m h gZ Q W m h gZ
1 2
in outm m
m m m
0
0
0
in
out
in
out rejected
W
W
Q
Q Q
2 21 2
1 21 1 2 2( ) ( )2 2rejectedc cm h gZ Q m h gZ
Steam condenserExample• 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.
Steam condenser
Solution:Applying the continuity flow equation
1 2 0.35 /
in outm m
m m kg s
Condenser
1
2
steam
condensate
rejectQ
1
1
1
1
3.5 /0.12 10.92
m kg sP barxh
2
2
h
m
Steam condenserAt point 1, the steam is a wet steam with dryness
fraction, x1=0.92At 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
Steam condenserApplying the steady flow energy equation
Since
Hence
2 21 2
1 21 1 2 2( ) ( )2 2
in outin outc cQ W m h gZ Q W m h gZ
1 21 2
2 21 21 2
0
0
0
1 12 2
in
out
in
out rejected
W
W
Q
Q Q
m gZ m gZ
m c m c
1 21 2
3 3 31 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
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
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.
Mixing chamberApply the continuity flow equation
Apply the steady flow energy
Since
Hence
22 231 2
1 2 31 1 2 2 3 3( ) ( ) ( )2 2 2
in outin outcc cQ W m h gZ m h gZ Q W m h gZ
1 2 3
in outm m
m m m
0
0
0
in
out
in
out loss
W
W
Q
Q Q
22 231 2
1 2 31 1 2 2 3 3( ) ( ) ( )2 2 2loss
cc cm h gZ m h gZ Q m h gZ
Mixing chamberExample• 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.
Mixing chamberSolution: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 outm m
m m m
m kg s
1 1.2 2683 /g at p barh h kJ kg
2 30125.7 /
sf at t Ch h kJ kg
Mixing chamberApplying the steady flow energy equation
Since
1 2 31 2 3
2 2 21 2 31 2 3
0
0
0
1 1 12 2 2
in
out
in
W
W
Q
m gZ m gZ m gZ
m c m c m c
22 231 2
1 2 31 1 2 2 3 3( ) ( ) ( )2 2 2
in outin outcc cQ W m h gZ m h gZ Q W m h gZ
Mixing chamberHence
1 2 31 2 3
3 1 23 1 2
1 21 23
3
3 3
3
2.5 2683 10 60 125.7 10 50062.5
227.98 10 /
loss
loss
loss
m h m h Q m h
m h m h m h Q
m h m h Qh
m
J kg
Thank youQ & A