Vapor and Combined Power Cycles
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Vapor power cycles
A gas power cycle considers air as the working fluid → single phase
A vapor power cycle considers considers steam as the working fluid fluid → might be two phases
Carnot Vapor Cycle
The use of steam as working fluid • Low cost • Avai Availa labi bilility ty • High High enthalp enthalpy y of vapori vaporizat zation ion
Carnot cycle • Give Give maximum maximum thermal thermal effici efficiency ency • But not not be suitabl suitable e to use use for analyzing analyzing the efficiency of vapor power cycle
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Objectives • Analyze Analyze vapor power power cycles in in which the working working fluid fluid is alternatel alternately y vaporized and condensed. • Analyze power power generation generation coupled coupled with process process heating, heating, called called cogeneration. • Investigate Investigate methods methods to modify modify the basic Rankine Rankine vapor vapor power cycle cycle to increase the cycle thermal efficiency. • Analyze Analyze the reheat reheat and regenerative regenerative vapor vapor power power cycles. cycles. • Analyze Analyze power cycles cycles that consist consist of two separate separate cycles cycles known as combined cycles and binary cycles.
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Carnot vapor cycle The working fluid, steam (water), undergoes a thermodynamic cycle from 1-2-3-4-1.
Carnot Vapor Cycle Using Steam
700
600
500 6000 kPa
400
] C [ T 300
2
100 kPa
3
200
2
3
4
1
100
0 0 .0
1 .0
2 .0
3 .0
4 .0
5 .0
6 .0
7 .0
8 .0
9 .0
1 0 .0
s [kJ/kg-K [kJ /kg-K]]
1
4
ηth, Carnot = Carnot
Wnet Qin
= 1−
Qout Qin
= 1−
T L T H
•The larger the T H the larger the ηth, Carnot •The smaller the T L the larger the ηth, Carnot • To increase increase the thermal thermal efficiency efficiency in any power power cycle, we we try to increase increase4 the maximum temperature at which heat is added.
Reasons why the Carnot cycle is not used! • Pumping Pumping process 1-2 requires requires the pumping pumping of a mixture mixture of saturated saturated liquid liquid and saturated vapor at state 1 and the delivery of a saturated liquid at state 2. • To superheat superheat the steam steam to take advantage advantage of a higher higher temperature, temperature, elaborate controls are required to keep T H constant while the steam expands and does work. Carnot Vapor Cycle Using Steam
700
600
2
500
3
6000 kPa
400
] C [ T 300
2
100 kPa
3
200
1
4
1
100
4
0 0 .0
1 .0
2 .0
3 .0
4 .0
5 .0
6 .0
7 .0
8 .0
9 .0
1 0 .0
s [kJ/kg-K [kJ /kg-K]]
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Rankine Cycle •The simple Rankine cycle continues the condensation process 4-1 until the saturated liquid line is reached. Carnot Vapor Cycle Cycle Using Ste Steam am
700
Rankine Vapor Power Cycle 500
6000 kPa
600 400
500
3 300
6000 kPa
400
] C [ T 300
2
] C [ T
100 kPa
3
200
10 kPa
200
2
100
4
1
100
0
0 0 .0
1 .0
2 .0
3 .0
4 .0
5 .0
6 .0
s [kJ /kg-K] /kg-K]
Process 1-2 2-3 3-4 4- 1
7 .0
8 .0
9 .0
1 0 .0
4
1 0
2
4
6
8
10
1 12 2
s [kJ/kg-K] [kJ/kg-K]
Ideal Rankine Cycle Processes Description Isentropic compression in pump Constant pressure heat addition in boiler Isentropic expansion in turbine Constant pressure heat rejection in condenser
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Example Compute the thermal efficiency of an ideal Rankine cycle for which steam leaves the boiler as superheated vapor at 6 MPa, 350oC, and is condensed at 10 kPa. Use the power system and T-s diagram. P 2 = P 3 = 6 MPa = 6000 kPa T 3 = 350oC Rankine Vapor Power Cycle P 1 = P 4 = 10 kPa 50 0
6000 kPa 40 0
3 30 0
] C [ T
20 0
10 kPa
2
10 0
4
1
0 0
2
4
6
8
10
1 12 2
s [kJ/kg-K [k J/kg-K]]
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Pump - Assume Assume steadysteady-flo flow w - Ne Negl glec ecti ting ng ΔPE & ΔKE - Assume adiabatic adiabatic and reversible reversible processe processes s
2
3
= m 2 = m 1h1 + W pump = m 2 h2 m 1 m
W pump
1
4
= m (h2 − h1 )
Since the pumping process involves an incompressible liquid, state 2 is in the compressed liquid region
Rankine Vapor Power Cycle Cycle 50 0
6000 kPa 40 0
Recall the property relation:
3 30 0
dh = T ds + v dP
Since the ideal pumping process 1-2 is isentropic, ds = 0.
] C [ T
20 0
10 kPa
2
10 0
4
1
0 0
2
4
6
s [kJ/kg-K]
8
10
12 12
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The incompressible liquid assumption allows Rankine Vapor Power Cycle Cycle
v ≅ v1
= const . h2 − h1 ≅ v1 ( P2 − P1 )
50 0
6000 kPa 40 0
3 30 0
] C [ T
The pump work : W pump w pump
20 0
10 kPa
= m (h2 − h1 ) ≅ m v1 ( P2 − P1 )
2
10 0
=
pump W m
= v1 ( P2 − P1 )
4
1
0 0
2
4
6
8
10
12 12
s [kJ/kg-K]
Use the steam tables
⎧h = h = 191.81 kJ 1 f kg P1 = 10 kPa ⎫ ⎪ ⎪ ⎬⎨ 3 Sat Sat. liquid liquid ⎭ ⎪ m v = v = 0.00101 ⎪⎩ 1 f kg w pump = v1 ( P2 − P1 ) = 0.00101
m
3
kg
(6000 − 10) kPa
kJ m3 kPa
= 6.05
kJ kg
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h 2 is found from
h2
= w pump + h1 = 6.05
kJ kg
+ 191.81
kJ kg
= 197.86
kJ kg
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Boiler • To find find the heat heat supplied supplied in the boiler, boiler, assume assume • Conservation Conservation of of mass and and energy energy for steady steady flow flow • Neglect ect ΔPE & ΔKE • No work work is done on the steam steam in the boiler boiler
= m 3 = m 2 h2 + Q in = m 3h3 m (h3 − h2 ) Q in = m 2 m
• Find the properti properties es at state 3 from the superheated tables ⎧h = 3043.9 kJ 3 P3 = 6000 kPa ⎫ ⎪ kg ⎪
T3
= 350o C
⎬⎨ ⎭ ⎪ s3 = 6.3357 kJ ⎪⎩ kg ⋅ K
• The heat heat trans transfer fer per per unit unit mass mass :
qin
=
Q in
m
= h3 − h2 = (3043.9 − 197.86) kJ = 2845.1 kJ kg
kg
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Turbine To find turbine work assume - Conservatio Conservation n of mass mass and energy energy for steady steady flow. flow. - The process process is adiabati adiabatic c and reversible reversible Rankine Vapor Power Cycle - Neg egllect ΔPE & ΔKE 500
= m 4 = m 3h3 = Wturb + m 4 h4 m (h3 − h4 ) Wturb = m
6000 kPa
3 m
400
3 300
] C [ T
200
10 kPa
Find the properties at state 4 from the steam tables by noting s 4 = s 3 = 6.3357 kJ/kg-K
at P4
= 10kP kPa : s f = 0.6492 s4 x4
kJ kg ⋅ K
; sg
2
100
0
=
s fg
=
2
4
6
8
10
12 12
s [kJ/kg-K [kJ /kg-K]]
= 8.1488
= s f + x4 s fg s4 − s f
4
1
0
kJ
kg ⋅ K
6.3357 − 0.6492 7.4996
= 0.758
12
h4
The net work done by the cycle :
= h f + x4 h fg = 191.81 = 2005.0
kJ kg
wnet
+ 0.758(2392.1)
kJ
= wturb − wpump = (1038.9 − 6.05)
kg
kJ
= 1032.8
kg
kJ kg
kJ kg
Rankine Vapor Power Cycle
T
500
he turbine work per unit mass
6000 kPa 400
wturb
= h3 − h4
3 300
= (3043.9 − 2005.0) = 1038.9
kJ
kJ kg
] C [ T
200
10 kPa
2
100
0
kg
4
1
0
2
4
6
s [kJ /kg-K] /kg-K]
8
10
12 12
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The thermal efficiency
η th
=
wnet qin
1032.8
= 2845.1
= 0.363 or
kJ kg kJ kg
36.3% Rankine Vapor Power Cycle 50 0
6000 kPa 40 0
3 30 0
2
3
] C [ T
20 0
10 kPa
2
10 0
1
4
4
1
0 0
2
4
6
s [kJ/kg-K]
8
10
1 12 2
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Previous lecture Ideal Ideal Rank Rankin ine e cycl cycle e Analyzing the efficiency of components
Today lecture Deviation of actual vapor cycles Way to impro improve ve the Rank Rankine ine cycle cycle Ideal Ide al Rankin Rankine e cycle cycle with with reheat reheat Ideal Rankine Rankine cycle with regeneration regeneration
Next lecture Isentropic efficiency Cogeneration 15
Deviation of actual vapor power cycles from idealized cycles
Fluid friction → pressure drops in boiler, condenser, and piping Irreversibili Irreversibility: ty: heat loss loss from steam steam to surrounding surroundings s Isentropic efficiency
η P
=
ws wa
− h1 = h2 a − h1 h2 s
η T
=
wa ws
=
h3 − h4 a h3 − h4 s
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Example A steam power plant operates on the cycle shown in Fig. If the isentropic efficiency of turbine is 87 percent and the isentropic efficiency of the pump is 85 percent, percent, determine determine (a) the thermal effici efficiency ency of the cycle cycle and (b) the net power output of the plant for a mass flow rate of 15 kg/s
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Ways to improve the simple Rankine cycle efficiency Increase the average temperature at which heat is transferred to the working fluid in the boiler, or decrease the average temperature at which heat is rejected from the working fluid in the condenser.
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Ways to improve the simple Rankine cycle efficiency •
Lower condenser pressure - Less energy is lost lost to to surroundin surroundings. gs. - Moisture Moisture is is increased increased at turbine turbine exit. - Min. temp is limited limited by cooling cooling temperature temperature - Lower temperatur temperature e gives lower lower pressure, and and this might might cause the problem of air leakage into condenser. - Too much moisture moisture gives gives lower turbine turbine efficiency efficiency and erosion erosion of turbine blade.
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Ways to improve the simple Rankine cycle efficiency
Superheat the vapor (without increasing the boiler pressure) Average temperature is higher during heat addition. - Moisture Moisture is reduced reduced at turbine turbine exit. exit. - Max. temp temp is limited limited by metallurgical metallurgical property
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Ways to improve the simple Rankine cycle efficiency • Increase boiler pressure (for fixed maximum temperature) - Availabili Availability ty of steam is is higher at higher higher pressures. pressures. - Moisture Moisture is is increas increased ed at turbine turbine exit exit (Side (Side effect effect and improved improved by reheating).
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Example Consider a steam power plant operating on the ideal Rankine cycle. Steam enters ent ers the the turbi turbine ne at 3 MPa MPa and 3 350 50 OC and is condensed in the condenser at a pressure of 10 kPa. Determine (a) the thermal efficiency of this power plant, (b) the thermal efficiency if steam is superheated to 600OC instead of 350OC, and (c) the thermal efficiency if the boiler pressure is raised to 15 MPa while while the turbine turbine inlet temperatu temperature re is maintained maintained at at 600OC.
(a)
(b)
(c)
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ηth = qin
wnet qin
=
qin
− qou o ut qin
= 1−
qin qout
= h3 − h2
qout = h4 − h1
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(a)
(b)
(c)
Tmax
350 oC
600 oC
600oC
Pmax
3 MPa
3 MPa
15 MPa
x
0.8128
0.915
0.804
qin
2921.3
3488
3376.2
qout
1944.3
2188.5
1923.5
ηth
33.4%
37.3%
43%
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Ideal Reheat Rankine Cycle As the boiler boiler pressure is increase increased d in the simple Rankine Rankine cycle, cycle, not only does the thermal efficiency increase, but also the turbine exit moisture increases. 1. Superheat Superheat the steam to very very high temperature temperature before before it enters the turbine (limited by metallurgical property) 2. Expand the steam in the turbine in two states, and reheat it in between.
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Ideal Reheat Rankine Cycle wturb qin
= wturb,1 + wturb,2 = ( h3 − h4 ) + ( h5 − h6 )
= q primary + qreheat = ( h3 − h2 ) + ( h5 − h4 )
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Ideal Reheat Rankine Cycle - The average average temperature temperature during the the reheat process process can be increased increased by increasing the number of expansion and reheat stages. - As the number number of stages stages is increased, increased, the the expansion expansion and reheat reheat processes approach an isothermal process at the max. temp. - In theory, effici efficiency ency from the second second reheat reheat is about half half of that from a single reheat. And if the turbine inlet pressure is not high enough, double reheat would result in superheated exhaust (increase the average temperature for heat rejection) → super critical pressure P > 22.06 MPa
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Rankine Cycle with Reheat Component Boiler Turbine Condenser Pump
Process Const. P Isentropic Const. P Isentropic
First Law Result = (h 3 - h 2) + (h 5 - h 4) q in = (h 3 - h 4) + (h 5 - h 6) w out q out = (h 6 - h 1) w in = (h 2 - h 1) = v 1(P 2 - P 1)
The thermal efficiency is given by
η th
=
wnet
=
( h3 - h4 ) + (h5 - h6 ) - (h2 - h1 )
qin
= 1−
( h3 - h2 ) + (h5 - h4 ) h6 − h1
( h3 - h2 ) + (h5 - h4 ) 28
Example Consider a steam power plant operating on the ideal reheat Rankine cycle. cycle. Steam enters enters the high-press high-pressure ure turbine turbine at 15 MPa and 600OC and is condensed in the condenser at a pressure of 10 kPa. If the moisture content of the steam at the exit of the low-pressure turbine is not to exceed 10.4 percent, determine (a) the pressure at which the steam should be reheated and (b) the thermal efficiency of the cycle. Assume the steam is reheated to the inlet temperature of the high-pressure turbine.
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Assignment 5 Compare the thermal efficiency and turbine-exit quality at the condenser pressure for a simple Rankine cycle and the reheat cycle when the boiler pressure is 4 MPa, the boiler exit temperature is 400oC, and the condenser pressure is 10 kPa. The reheat takes place at 0.4 MPa and the steam leaves the reheater at 400oC. No Reheat With Reheat
ηth
35.3% 35.9%
xturb exit 0.8159 0.9664
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Ideal Regenerative Rankine Cycle • To improve improve the cycle thermal thermal efficiency efficiency,, the average temperatu temperature re at which heat is added must be increased. • Allow the the steam leaving leaving the boiler boiler to expand the steam steam in the turbine turbine to an intermediate pressure. - Some Some of stea steam m from from turbin turbine e is sent to a regenerative heater to preheat the condensate before entering the boiler.→ increase feeding water (boiler) temp. - How Howeve ever, r, this this reduces reduces the the mass of of steam expanding in the lowerpressure stages of the turbine, and, thus, the total work done by the turbine. The work that is done is done more efficiently. 36
The preheating of the condensate is done in a combination of open and closed heaters. - In the open open feed water water heater, heater, the extracted extracted steam steam and the condensate are physically mixed. - In the closed closed feed feed water heater, heater, the the extracted extracted steam steam and the condensate are not mixed.
Open feed water heater
Closed feed water heater
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Cycle with an open feedwater heater
= h5 − h4 qout = (1 − y ) ( h7 − h1 ) qin
wturb,out = ( h5 − h6 ) + (1 − y ) ( h6 − h7 ) w pump ,in
= (1 − y ) wpump ,1,in + w pump,2,in
6 / m 5 y = m w pump ,1,in
= v1 ( P2 − P1 )
w pump ,2,in
= v3 ( P4 − P3 )
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6 / m 5 be the fraction of mass extracted from the turbine for the Let y = m feedwater heater.
= m out 6 + m 2 = m 3 = m 5 m 2 = m5 − m 6 = m 5 (1 − y) m in m
Conservation of energy for the open feedwater heater:
= E out 6h6 + m 2 h2 = m 3h3 m 5h6 + (1 − y ) m 5h2 = m 5h3 ym h − h2 y = 3 h6 − h2 E in
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Cycle with a closed feedwater heater with pump to boiler pressure
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Example Consider Consider a steam power power plant operating operating on the ideal regenerativ regenerative e Rankine Rankine cycle with one open feed water heater. Steam enters the turbine at 15 MPa and 600OC and is condensed in the condenser at a pressure of 10 kPa. Some steam steam leaves leaves the turbine turbine at a pressure pressure of 1.2 MPa MPa and enters enters the open feed water heater. Determine the fraction of steam extracted from the turbine and the thermal efficiency of the cycle.
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43
44
45
%
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Example: Example: Closed Closed feed feedwate waterr heater heater A steam power power plant operates operates on an ideal regenerative Rankine Rankine cycle. Steam O enters enters the turbine turbine at 6 MPa 450 C and is condensed in the condenser at 20 kPa. Steam is extracte extracted d from the the turbine turbine at 0.4 0.4 MPa to heat heat the feedwate feedwaterr in a closed closed feedwater feedw ater heater. heater. Water Water leaves leaves the heater heater at the conden condensatio sation n temperatur temperature e of the extracted steam and that the extracted steam leaves the heater as a saturated liquid and is pumped to the line carrying the feedwater. Determine (a) the net work output per kilogram of steam flowing throught the boiler and (b) the thermal efficiency of the cycle.
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50
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Deviation from Actual Cycles • Piping losses--fr losses--friction ictional al effects reduce reduce the available available energy content content of the steam. •Turbine losses--turbine isentropic (or adiabatic) efficiency. P3
T 3
η turb
=
wactual wisentropic
=
h3 − h4 a h3 − h4 s
4a
P4
4s
s
The actual enthalpy at the turbine exit (needed for the energy analysis of the next component) is
h4 a
= h3 − η turb (h3 − h4 s ) 53
•Pump losses--pump isentropic (or adiabatic) efficiency. 2a T
η pump
=
wisentropic wactual
2s
P2
− h1 = h2 a − h1 h2 s
P1
1 s
The actual enthalpy at the pump exit (needed for the energy analysis of the next component) is
h2 a
= h1 +
1
η pump
( h2 s
− h1 )
•Condenser losses--relatively small losses that result from cooling the condensate below the saturation temperature in the condenser.
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Second law analysis of vapor power cycles - Ideal Carnot Carnot cycle cycle is a totally totally reversible reversible - Ide Ideal al Ra Ranki nkine ne cycles cycles may may invol involve ve irreve irreversi rsibil biliti ities es externa externall to the the syste system, m, e.g. heat transfer through a finite temperature difference - Second law analysi analysis s of these cycles cycles is used to to reveal reveal where where the the largest largest irreversibil irreversibilities ities occur and and what their their magnitude magnitude are.
= T0 Sgen = T0 ( Sout − S in )
X destruction
⎛ Q out Q in ⎞ = T0 ⎜⎜ ∑ m s + − ∑ m s − ⎟⎟ Tb ,out T b ,in ⎠ in ⎝ out Or on a unit mass basis for a one-inlet, one-exit, steady flow device
xdestruction
= T0 sgen
⎛ qout qin ⎞ = T0 ⎜⎜ se − si + − ⎟⎟ Tb ,out T b ,in ⎠ ⎝
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Second law analysis of vapor power cycles - Exergy destruction destruction associ associated ated with with a cycle cycle depends depends on on the magnitude magnitude of of the heat heat transfe transferr with the the highhigh- and lowlow- tem tempera peratur ture e reservoi reservoirs. rs.
⎛ qout qin ⎞ xdestruction = T 0 ⎜ ∑ ⎜ Tb,out − ∑ T b,in ⎟⎟ ⎝ ⎠ For a cycle that involves heat transfer only with a source at TH and a sink at TL, the e exer xergy gy destru destructi ction on are
⎛ qout qin ⎞ − ⎟ xdestruction = T 0 ⎜ ⎝ T L T H ⎠
Exergy Exergy of a fluid fluid strea stream m φ at any states is 2
ϕ = ( h − h0 ) − T0 ( s − s0 ) +
V
2
+ gz 56
Example Determine Determine the exergy destruction destruction associated associated with with the the Ra Rankine nkine cycle (all four processes as well as the cycle), assuming that heat is transferred to the steam in a furnace at 1600 K and heat is rejected to a cooling medium at 290 K and 100 kPa. kPa. Also, Also, determine determine the exergy exergy of the steam steam leaving leaving the turbine
100 kPa
100 kPa
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Cogeneration Cogeneration is the production of more than one useful form of energy (such as process heat and electric power) from the same energy source
Q in 120kW Turbine
Boiler
10kW
Process heater
Boiler Process heater
Q p
100kW
pump pump A simple process-heating plant
An ideal cogeneration plant
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Cogeneration Cogeneration is the production of more than one useful form of energy (such as process heat and electric power) from the same energy source
Steam turbine
Producing EE
5 ~ 7 at atm, m, 15 150 0 – 20 200 0OC
Process heat
Chemical, pulp and paper, oil production
Heat transfer to the steam in boiler
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Cogeneration
63
Utilization factor,
u
Waste heat rejection from turbine transfers to the steam in boiler and is utilized as either process heat or electric power. Utilization factor, u
εu =
=
Net work output + Process heat delivered Total heat input
Wnet
+ Q p
Q in
Q out
εu = 1 − Qin
Q out = presents the heat rejected in the condenser.
εu = 100% when no any heat losses in system
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Example Steam Steam enters enters the the turbine turbine at 7 MPa MPa and 500 500OC. Some steam is extracted from the turbine turbine at 500 kPa for process process heating. heating. The remaining remaining steam steam continues continues to expand to 5 kPa. Steam is then condensed at constant pressure and pumped to the boiler pressure of 7 MPa. At times of high demand for process heat, some steam steam leaving leaving the boiler boiler is throttled throttled to 500 500 kPa and is routed to to the process heater. The extraction fraction are adjusted so that steam leaves the process heater as a saturated liquid at 500 kPa. It is subsequently pumped to 7 MPa. The mass flow rate of steam through the boiler is 15 kg/s. Disregarding any pressure drops and heat losses in the piping and assuming the turbine and the pump to be isentropic, determine (a) the maximum rate at which process heat can be supplied, (b) the power produced and the utilization utilization factor factor when no process process heat is supplied, supplied, and (c) the rate of process process heat supply when 10 percent of the steam is extracted before it enters the turbine and 70 percent of the steam is extracted from the turbine at 500 kPa for process heating.
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Example A textile plane require 4 kg/s of saturated steam at 2 MPa, which is extracted from the turbine of a cogeneration plant. Steam enters the turbine at 8 MPa and 500OC at a rate of 11 kg/s and leaves at 20 kPa. The extracted steam leaves the process heater as a saturated liquid and mixes with the feed water at constant pressure. The mixture is pumped to the boiler pressure. Assuming and isentropi isentropic c efficiency efficiency of 88 percent percent for both the turbine and the pumps, pumps, determine determine (a) the rate rate of process process heat heat supply, supply, (b) the the net power power output, and (c) the utilization factor of the plant.
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