notes 23 - ceprofs
TRANSCRIPT
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Lec 23: Brayton cycle regeneration, Rankine cycle
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• For next time:– Read: § 8-11 to 8-13, 9-1 to 9-2.– HW12 due Wednesday, November 19, 2003
• Outline:– Rankine steam power cycle– Cycle analysis– Example problem
• Important points:– Know what assumptions you can make about
the points along the cycle path– Know how to analyze a pump– Know how all the isentropic efficiencies are
defined
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Improved Brayton Cycle: Add a Heat Exchanger (Regenerator)
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Ts Diagram for Brayton Cycle with Regeneration
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Analysis with regeneration
Heat input to the cycle:
53 hhqinput
Net work of the cycle:
)()( 1243 hhhhwww compturbinenet
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Analysis with regeneration
53
1243 )()(hh
hhhhqw
in
netregen
Efficiency of the cycle:
For the cold-air assumption:
kkpregen r
TT /)1(
3
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Regenerator in cycle
Regenerator heat transfer:25 hhqregen
Maximum heat transfer in regenerator:
242'5max, hhhhqregen
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Regenerator Effectiveness
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25
max,
,
hhhh
regen
actregen
For the cold-air assumption:
24
25
TTTT
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Efficiency with regeneration
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TEAMPLAYTEAMPLAY
Problem 8-82
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Vapor Power Cycles
• We’ll look specifically at the Rankine cycle, which is a vapor power cycle.
• It is the primary electrical producing cycle in the world.
• The cycle can use a variety of fuels.
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TEAMPLAYTEAMPLAY
How much does it cost to operate a gas fired 120 MW(output) power plant with a 35% efficiency for 24 hours/day for a full year if fuel costs are $6.00 per 106 Btu?
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Vapor Power Cycles• The Carnot cycle is still important as a
standard of comparison.• However, just as for gas power cycles, it
cannot be practically achieved in useful, economical systems.
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BOILERTURBINE
PUMP
CONDENSER
q inw out
qoutw in
1
3
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We’ll simplify the power plant
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Ideal power plant cycle is called the Rankine Cycle
• 1-2 reversible adiabatic (isentropic) compression in the pump
• 2-3 constant pressure heat addition in the boiler.
• 3-4 reversible adiabatic (isentropic) expansion through turbine
• 4-1 constant pressure heat rejection in the condenser
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Rankine cycle power plant• The steady-state first law applied to open
systems will be used to analyze the four major components of a power plant– Pump– Boiler (heat exchanger)– Turbine– Condenser (heat-exchanger)
• The second law will be needed to evaluate turbine performance
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Vapor-cycle power plants
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What are the main parameters we want to describe the cycle?
=> Net power or work output
pumpturout WWW Power
pumpturout www Work
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Main parameters….
=> Efficiency
in
net
QW
in
net
qw
Or
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General comments about analysis• Typical assumptions…
– Steady flow in all components– Steady state in all components– Usually ignore kinetic and potential
energy changes in all components– Pressure losses are considered
negligible in boiler and condenser– Power components are isentropic for
ideal cycle
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Start our analysis with the pump
pump Pump 1 20 Q W m(h h KE PE)
)pp(hhw 2121pump
The pump is adiabatic, with no kinetic or potential energy changes. The work per unit mass is:
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Pump Analysis
This expression gives us negative value for wp. It is standard practice in dealing with cycles to express works as positive values, then add or subtract depending on whether they’re in or out.
21Pump hhw
This gives us a positive value for work.
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Boiler is the next component.
]PEKEh[hmWQ0 32boilerboiler
•Boilers do no work. In boilers, heat is added to the working fluid, so the heat transfer term is already positive.•So
23boilerboiler hhqm
Q
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Proceeding to the Turbine
]PEKEh[hmWQ0 43turbineturbine
43turbine hhwm
W
Turbines are almost always adiabatic. In addition, we’ll usually ignore kinetic and potential energy changes:
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Last component is the Condenser
]PEKEh[hmWQ0 14condcond
41condcond hhqm
Q
Condensers do no work (they are heat exchangers),and if there is no KE and PE,
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More condenser...
What is the sign of qcond ?
As with work, we’re going to want the sign of all the heat transfer terms positive.
41condcond hhqm
Q
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Ideal Rankine Cycle• The pump work, because it is reversible
and adiabatic, is given by
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2
1
hhvdpwP
)( 12 PPvwP
• and
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Ideal Rankine Cycle on a p-v diagram
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Efficiency
in
out
qw
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1243
h-h)P-(P-h-h v
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Example Problem
A Rankine cycle has an exhaust pressure from the turbine of 10 kPa. Determine the quality of the steam leaving the turbine and the thermal efficiency of the cycle which has turbine inlet pressure of 15 MPa and 600C.
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Start an analysis:
Assumptions:
pump and turbine are isentropic P2 = P3 = 15 MPa T3 = 600C P4 = P1 = 10 kPA Kinetic and potential energy changes are zero
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Draw diagram of cycle
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Some comments about working cycle problems
• Get the BIG picture first - where’s the work, where’s the heat transfer, etc.
• Tables can useful - they help you put all the data you will need in one place.
• You will need to know how to look up properties in the tables!
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Put together property data
State T (C) P(kPa) v(m3/kg) h(kJ /kg) s(kJ /kgK) x
1 10 0
2 15000 n.a.
3 600 15000 ----
4 10 ----
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Property data
• h1=191.83 kJ/kg is a table look-up, as is h3=3582.3 kJ/kg.
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Let start with pump work
Pump work:
2121pump hh)p(pw
15)MPa - (0.01kgm)00101.0(w
3
pump
kgkJ1.15w pump
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More calculations...
Enthalpy at pump outlet:
pump12 whh
kgkJ)1.1583.191(h2
Plugging in some numbers:
kgkJ93.206h2
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Calculate heat input and turbine work..
kgkJ)93.2063.3582(hhq 23boiler
Boiler heat input:
kgkJ4.3375qboiler
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Property data
• Because s3= s4, we can determine that x4=0.803 and thus h4=2114.9 kJ/kg
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Turbine work
kgkJ)9.21143.3582(hhw 43turbine
kgkJ4.1467w turbine
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We’ve got the exit quality, now we need efficiency
Cycle efficiency:
in
out
qw
Substituting for net work:
in
pumpturbine
qww
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Overall thermal efficiency
in
pumpturbine
qww
kgkJ3375.4
kgkJ)1.15(1467.4
430.0
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Some general characteristics of the Rankine cycle
• Low condensing pressure (below atmospheric pressure)
• High vapor temperature entering the turbine (600 to 1000C)
• Small backwork ratio (bwr)
01.0hhhh
ww
bwr43
21pump
turbine
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TEAMPLAYTEAMPLAY
Work Problem 8-102