Superheating, Reheating, and Regeneration

advertisement
EGR 334 Thermodynamics
Chapter 8: Sections 3-4
Lecture 32:
Superheat, Reheat, and
Cogeneration
Quiz Today?
Today’s main concepts:
• Explain how a superheater changes the Rankine cycle model
• Explain how a reheat line changes the Rankine cycle model.
• Explain how regenerative heating changes the Rankine cycle
model.
• Explain how cogeneration and process steam lines change
the Rankine cycle model.
• Be able to use mass balance, energy balance, and entropy
balance to solve systems with modified Rankine cycle
models.
Reading Assignment:
Read Chapter 9, Sections 1-2
Homework Assignment:
Problems from Chap 8: 21,29, 49, 60
Sec 8.1 : Modeling Vapor Power Systems
Vapor Power System Model
3
Sec 8.2 : Rankine Cycle
4
Tools: Conservation of mass
Conservation of energy (First Law)
Entropy balance (Second Law)
Thermodynamic properties
Energy balances:
For the turbine:
W
 h1  h 2

m
For the condenser:
Q
m
Q
 h 2  h3
m c .w .
 h 6  h5
For the pump:
W
m
 h 4  h3

For the boiler:
Q
m
 h1  h 4

4
3
 WP 
vd P  
 v3  p 4  p 3 

 m  in t . rev
Sec 8.2 : Rankine Cycle
5
Process 1-2: Isentropic expansion
of the fluid through the turbine
from saturated vapor to the
condenser pressure.
Process 3-4: Isentropic
compression of fluid to compressed
liquid.
Process 2-3: Heat transfer at
constant pressure through the
condenser to saturated liquid.
Process 4-1: Heat transfer at
constant pressure through the
boiler to saturated vapor
 
W cycle
Q in
W T  W P

Q
in
Sec 8.2.2 : Modified Rankine Cycle
In addition to being able to work with the
Ideal Rankine Cycle, you should also be
prepared to evaluate performance that
includes the following modifications
-- Superheating: Heating of the fluid past a
saturated vapor and well into the
superheated region by the steam generator.
-- Reheating: Reheating fluid after it has
passed through one turbine and then
passing through a second turbine at lower
pressure.
-- Regenerative Feedwater Heating: A
process where some of the steam is bled off
and used to partially reheat the condensate
from the condenser.
6
Sec 8.2.2 : Modified Rankine Cycle
In addition to being able to work with the
Ideal Rankine Cycle, you should also be
prepared to evaluate performance that
includes the following modifications
-- Non-Ideal Rankine (with irreversibilities):
Similar to the Ideal Rankine cycle except
the processes through the turbine and
pump are not isentropic. Turbine and pump
efficiencies need to be used to determine
work in each of the components.
-- Cogeneration: A process where some of
the steam is bled off for use by other parts
of the plant, such as for heating, cleaning,
etc.
7
Sec 8.3 : Superheat and Reheat
8
How can the efficiency of the Rankine cycle be increased?
Raise the Average High Temperature.
As the temperature is raised, note that the sat. vapor location
also shifts to the left, so that when the fluid expands through the
turbine, a lower quality of the fluid is obtained. Why is this bad?
Net work is decreased as area under
the curve get narrower and there are
physical effects on the turbine that
involve more blade wear. It is more
desirable to have the exit quality of
the steam, x2, be above 90%
How can the cycle maintain high
boiler Pressure and Temperature
and still maintain a higher exit
quality?
Superheat the steam.
9
How is steam superheated?
Sec 8.3 : Superheat and Reheat
10
Superheating:
It is not necessary to have the exit from the boiler be a
saturated vapor (xboiler exit = 1).
Operate the boiler, such that the exit of the boiler is a
superheated vapor.
The only difference in calculating cycle performance from the Ideal
Rankine Cycle, is that state 3 is no longer expected to fall along the
saturated vapor curve, but will be found on the superheated steam table.
Sec 8.3 : Superheat and Reheat
When Reheating is implemented, there is a multistage stage turbine
(often identified as a High Pressure and Low Pressure turbine.)
Before vapor starts to condense (reaches saturation), the vapor is
sent back to the boiler and reheated.
11
Sec 8.3 : Superheat and Reheat
12
This affects the Ideal Rankine Cycle model because there are additional states
to determine properties for, a second pass through the turbine, and a second
heat exchange with the boiler. Therefore, identify additional properties
states (1, 2, 3, and 4) and then rework the 1st law equations as applied to the
turbine and boiler.
0   W H P _ turbine  m ( h1  h 2 )
and
0   W L P _ turbine  m ( h3  h 4 )
Boiler Reheat Equation
0  Q reheat  m ( h 2  h3 )
Sec 8.4 : Regenerative Vapor Power Cycle
13
Regenerative Feedwater Heating. (regeneration)
Steam is diverted to use in a mixing chamber to preheat the
condensate prior to entering boiler. This is useful since water is
typically cooled below saturated level out of the condensor.
Note: This rerouting will diminish the net work output. The
reduction in Qboiler should be less than the reduction of Wcycle.
Sec 8.4 : Regenerative Vapor Power Cycle
14
Open feedwater analysis:
#2
Mass Balance: m 1  m 2  m 3
#1
Often will use the term “blend fraction”
y
m 2
consequently, 1  y 
m 1
Energy Balance: 0 
#3
m 3
m 1
 m  h 
i
i
0  m 2 h 2  m 3 h3  m 1 h1  m 2 h2  m 3 h5  m 1 h6
0
m 2
m 1
h2 
m 3
m 1
h5 
m 1
m 1
h 6  yh 2  1  y h5  h 6
Thus:
y
h6  h5
h 2  h5
Sec 8.4 : Regenerative Vapor Power Cycle
15
Energy balances:
For the turbines:
W T 1
 h1  h 2
m
and
For the condenser:
Q out

m
 1  y  h3  h 4 
For the pump:
W P
m
  h 7  h 6   1  y  h5  h 4 
For the boiler:
Q in
m
 h1  h 7
W T 2
m
 1  y  h 2  h3 
16
Example (8.44): An Ideal Rankine cycle with the state properties given
below includes one open feedwater heater operating at 100 psi. Saturated
liquid exits the open feedwater heater at 100 psi. The mass flow rate of
steam into the first turbine state is 1.4 x 106 lbm/hr. Determine
(a) The net power developed, in Btu/hr.
(b) The thermal efficiency.
(c) The mass flow rate of cooling water,
if T = 20°F.
State
1
T (°F)
1100
p (psi)
1600
x
h (Btu/lbm)
s (Btu/lbm K)
2
3
4
5
6
7
100
1
1
100
100
1600
17
Example (8.44) continued:
Start by finding the state properties
State 1: Using Table A4E with T = 1100 F and p = 1600 psi,
find h1 = 1547.7
and s1 = 1.6315
State 2: Using Table A4E with s2=s1 and p = 100 psi,
find h2 = 1210.7
State 3: Using Table A3E with s3=s1, and p = 1 psi,
find x3 = 81.2% and then, find h3 = 911.2
State 4: Using Table A3E with sat. liquid and p = 1 psi,
find v4 = vf = 0.01614, h4 = hf = 69.74, and s4 = sf = 0.1327
State
1
T (°F)
1100
p (psi)
1600
x
h (Btu/lbm)
s (Btu/lbm K)
2
3
4
5
6
7
100
1
1
100
100
1600
18
Example (8.44) continued:
Start by finding the state properties
State 5: Using h5 - h4 =v4 (p5 - p4)
find h5 = 70.04
State 6: Using Table A4E with sat. liq. and p = 100 psi,
find v6 = 0.01774
h6 = 298.6, and s6 = 0.4744
State 7: Using h7 - h6 =v6 (p7 - p6)
find h7 = 303.5
State
1
2
3
4
5
6
7
T (°F)
1100
1100
p (psi)
1600
1600
100
100
1
1
100
100
100
100
1600
1600
s.h.
s.h.
0.812
0.00
liq.
0.00
liq.
h (Btu/lbm)
1547.7
1210.7
911.2
69.74
70.04
298.6
303.5
s (Btu/lbm K)
1.6315
1.6315
1.6315
0.1327
x
0.4744
19
Example (8.44) continued:
Find heat in provided in boiler:
0  Q in  m 1 ( h 7  h1 )
then
Q in  m 1 ( h1  h 7 )
 (1.4  10 lb m / hr )(1547.7  303.5) B tu / lb m
6
 1.74  10 B tu / hr
9
State
1
2
3
4
5
6
7
T (°F)
1100
p (psi)
1600
100
1
1
100
100
1600
s.h.
s.h.
0.812
0.00
liq.
0.00
liq.
h (Btu/lbm)
1547.7
1210.7
911.2
69.74
70.04
298.6
303.5
s (Btu/lbm K)
1.6315
1.6315
1.6315
0.1327
x
0.4744
20
Example (8.44) continued:
Now find how the mass flow splits from the turbine
y
h 6  h5
h 2  h5

298.6  70.04
1210.7  70.04
 0.200 c
therefore:
m 2  y m1
 (0.200)(1.4  10 lb m / hr )  0.28  10 lb m / hr
6
6
m 3  (1  y ) m 1  (1  0.200)(1.4  10 lb m / hr )  1.12  10 lb m / hr
6
State
1
T (°F)
1100
p (psi)
6
2
3
4
5
6
7
1600
100
1
1
100
100
1600
s.h.
s.h.
0.812
0.00
liq.
0.00
liq.
h (Btu/lbm)
1547.7
1210.7
911.2
69.74
70.04
298.6
303.5
s (Btu/lbm K)
1.6315
1.6315
1.6315
0.1327
x
0.4744
21
Example (8.44) continued:
Find work from turbine
0   W turbine  m 2 ( h1  h 2 )  m 3 ( h1  h3 )
W turbine  m 2 ( h1  h 2 )  m 3 ( h1  h3 )
 (0.28  10 lb m / hr )(1547.7  1210.7 ) B tu / lb m
6
 (1.12  10 lb m / hr )(1547.7  911.2) B tu / lb m
6
 94.36  10  712.88  10  807.24  10 B tu / hr
6
6
State
1
T (°F)
1100
p (psi)
6
2
3
4
5
6
7
1600
100
1
1
100
100
1600
s.h.
s.h.
0.812
0.00
liq.
0.00
liq.
h (Btu/lbm)
1547.7
1210.7
911.2
69.74
70.04
298.6
303.5
s (Btu/lbm K)
1.6315
1.6315
1.6315
0.1327
x
0.4744
22
Example (8.44) continued:
Find heat provided to pumps:
Pump 1: 0   W pum p 1  m 3 ( h 4  h5 )
W pum p 1  m 3 ( h 4  h5 )
 (1.12  10 lb m / hr )(69.74  70.04) B tu / lb m   0.336  10 B tu / hr
Pump 2:
6
6
W pum p 2  m 1 ( h 6  h 7 )
 (1.4  10 lb m / hr )(298.6  303.5) B tu / lb m   6.860  10 B tu / hr
6
6
State
1
2
3
4
5
6
7
T (°F)
1100
p (psi)
1600
100
1
1
100
100
1600
s.h.
s.h.
0.812
0.00
liq.
0.00
liq.
h (Btu/lbm)
1547.7
1210.7
911.2
69.74
70.04
298.6
303.5
s (Btu/lbm K)
1.6315
1.6315
1.6315
0.1327
x
0.4744
23
Example (8.44) continued:
therefore the Net Work
W net  W turbine  W pum p 1  W pum p 2
 807.24  10  6.860  10  0.336  10  800.0  10 B tu / hr
6
6
6
6
Plant thermal efficiency:
 
W net
800  10 B tu / hr
6

Q in
1740  10 B tu / hr
6
State
1
T (°F)
1100
p (psi)
 45.98%
2
3
4
5
6
7
1600
100
1
1
100
100
1600
s.h.
s.h.
0.812
0.00
liq.
0.00
liq.
h (Btu/lbm)
1547.7
1210.7
911.2
69.74
70.04
298.6
303.5
s (Btu/lbm K)
1.6315
1.6315
1.6315
0.1327
x
0.4744
24
Example (8.44) continued:
for the condenser:
0  m 3 ( h3  h 4 )  m cw ( h cw In  h cw O ut )
m cw   m 3
h3  h 4
h cw In  hcw O ut
  m3
  (1.4  10 lb m / hr )
6
State
1
T (°F)
1100
p (psi)
h3  h 4
c H 2 0 (T cw In  Tcw O ut )
(911.2  69.74) B tu / lb m
(1 B tu / lbm  R )(  20 R )
 58.9  10 lb m / hr
6
2
3
4
5
6
7
1600
100
1
1
100
100
1600
s.h.
s.h.
0.812
0.00
liq.
0.00
liq.
h (Btu/lbm)
1547.7
1210.7
911.2
69.74
70.04
298.6
303.5
s (Btu/lbm K)
1.6315
1.6315
1.6315
0.1327
x
0.4744
Sec 8.2.4 : Principal Irreversibilities and Losses
25
NonIdeal Rankine Cycle.
For real processes, there is some
heat transfer with the surroundings,
and thus an increase in entropy. In
the turbine and pump, this can be
quantified using the isentropic
efficiencies.
For Turbine:
W T m real
h1  h 2
T 


W T m S
h1  h 2 S



For Pump:
P 
W
P

m S
W P m real

h 4 S  h3
h 4  h3
But, for a real process, the inefficiencies associated with combustion are
more significant. These, however, are external to the power cycle, so are
not analyzed as part of the Rankine Cycle analysis.
Cogeneration Systems
Cogeneration systems provide both electrical work out and process steam
or hot water for use by an end user which is often a commerical,
industrial, or governmental user.
Electricity provided
to the community
Steam exported to
the community
Cogeneration Systems
--Exporting useful steam to the community limits the electricity that also
can be provided from a given fuel input, however.
--For instance, to produce saturated vapor at 100oC (1 atm) for export to
the community water circulating through the power plant will condense
at a higher temperature and thus at a higher pressure.
--In such an operating mode thermal efficiency is less than when
condensation occurs at a pressure below 1 atm, as in a plant fully
dedicated to power production.
T > 100oC
p > 1 atm
Sec 8.4.3 : Multiple Feedwater Heaters
As closing comment: You now have most of the tools you would need
to analyze some relatively complicated vapor power generation units
and be able to discuss their respective efficiencies or how to suggest
improvements in their performance... (not that you would want to).
28
29
End of Lecture 32 slides
So ends Chap 8!
Download
Related flashcards

Thermodynamics

57 cards

Create Flashcards