gas turbine 1
TRANSCRIPT
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Applied Thermodynamics
1
Gas Turbine Cycles
N S Senanayake
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Air standard cycles• Air standard cycles refers to thermodynamic cycle
with certain assumptions so as to use the principles of thermodynamics conveniently.
• Assumptions– Air is the working fluid and behaves as a perfect gas– Mass and composition of the working fluid will not
change in the cycle– Processes are reversible– Specific heat capacity of the working fluid does not
change
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Otto cycle (air standard)
1 – 2 Adiabatic compression
2 – 3 Constant volume heat addition
3– 4 Adiabatic expansion
4– 1 Constant volume heat rejection
Spark Ignition (SI) engines are based on this cycle
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Otto cycle …
rVV
VVn ratioCompressio
3
4
2
1
)( , 23 TTcqaddedHeat vin
)()( , 1441 TTcTTcqrejectedHeat vvout
)()( , 1423 TTcTTcwworkNet vvnet
inputHeatworkNetEfficiencyThermal th
,
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Otto cycle…
)()(1
)()}(){(
23
14
23
1423
TTTT
TTcTTTTc
v
vth
Consider process 1 – 2
CpV CVpV 1)( CTV 1
122
111
VTVT 11
2
1
1
2
rVV
TT
Consider process 3 – 4
144
133
VTVT 11
3
4
4
3
rVV
TT
2
3
1
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Otto cycle…From equations 2 and 3
4
1
3
2
4
3
1
2
TT
TT
TT
TT
4
1
3
2 11TT
TT
4
14
3
23
TTT
TTT
1
4
3
14
23 r
TT
TTTT
From equation 1
123
14 11)()(1
rTT
TTth
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Otto cycle efficiency vs. compression ratio
0 5 10 15 20 25 30 35 400
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
compression ratio (r)
Eff
icienc
y
( )
= 1.2
= 1.4
123
14 11)()(1
rTT
TTth
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Mean effective pressure (MEP)- Otto cycle
This is the mean pressure which is developed in the cylinder. Defined as the ratio of net work done to the displacement of volume of the piston.
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Mean effective pressure (MEP)- Otto cycle
ngeVolume chaNetMEP work
11
11224433
VpVpVpVpworkNet
( ) ( ) 1122443311 VpVpVpVpworkNet
11
11
11
2211
44
3344 Vp
VpVpVpVpVpworkNet
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Mean effective pressure (MEP)- Otto cycle
1111
11
1
211
4
344 rp
pVprp
pVpworkNet
rVV
ppVpVp
2
1
1
22211
rVV
pp
3
4
4
3 Similarly
( ) ( ) 111
1 111
114
rVprVpworkNet
Consider process 1 – 2
( ) 14
11
11 pprVworkNet
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Terminology : Reciprocating Engine
volume swept out by the piston when it moves from TDC to BDC is called the displacement volume.
distance from TDC to BDC is called stroke
• The piston is said to be at the top dead center (TDC) when it has moved to a position where the cylinder volume is minimum. This volume is called a clearance volume.
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Spark Ignition vs Compression Ignition
Spark-ignition engines: mixture of fuel and air are ignited by a spark plug.
Compression ignition engines: Air is compressed to high enough pressure and temperature that combustion occurs spontaneously when fuel is injected.
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Air-Standard Diesel Cycle The Air-Standard Diesel Cycle is the ideal cycle that approximates the compression ignition engine i.e. Compression Ignition (CI) engines are based on this cycle Process Description 1-2 Isentropic Compression 2-3 Constant Pressure Heat Addition 3-4 Isentropic Expansion 4-1 Constant Volume Heat Rejection
P
V
T
S
2 3
4
1 1
4
2
3P = P2 3
P 1
S = S1 2 S = S3 4
P 4
V =V4 1O O
T1
T4
T2
T3
y
x
V2
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Diesel cycle)( , 2323 TTcqaddedHeat p
)( , 4141 TTcqrejectedHeat v
2
1 , VVrrationCompressio
2
3 , VVratiooffCut
23
4123
qqq
addedheatheatnet
th
)(
)()(
23
1423
TTcTTcTTc
p
vpth
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Diesel cycle
)()(11
23
14
TTTT
th
( ) 112
11
2
1
1
2
rTTrVV
TT
Process 1 - 2
Process 2 - 3
11323
2
3
2
3 rTTTTTT
VV
1
2
3
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Diesel cycleProcess 3 - 4
11
3
2
2
1
1
3
2
2
4
1
3
4
4
3 ..
r
VV
VV
VV
VV
VV
TT
4( )
1
11
1
1
34 Tr
rTr
TT
Substituting from eq. 3
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Diesel cycleSubstituting for all Ts in equation 1.
)1()1(11
)()(11 11
11
1
11
rrTrTTT
th
)1()1(
11 1
rth
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Diesel cycle efficiency vs. compression ratio
0 5 10 15 20 25 300.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
4
Effic
ienc
y (
)
Compression ratio (r)
=2 = 1.4
)1()1(
11 1
rth
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Gas turbines
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Elements of simple gas turbine Power Plant
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The simple gas turbine power plant mainly consists of a gas turbine coupled to a rotary type air compressor and combustion chamber which is placed between the compressor and turbine in the fuel circuit.
Auxiliaries, such as cooling fan, water pumps, etc. and the generator itself, are also driven by the turbine.
Other auxiliaries are starting device, lubrication system, duct system, etc.
A modified plant may have in addition to the above, an inter-cooler, a regenerator and a re-heater
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Flow diagram – Gas turbine power plant
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Gas turbine cycle• Gas-turbines usually operate on an open
cycle
• A compressor takes in fresh ambient air (state 1), compresses it to a higher temperature and pressure (state 2).
• Fuel and the higher pressure air from compressor are sent to a combustion chamber, where fuel is burned at constant pressure. The resulting high temperature gases are sent to a turbine (state 3).
• The high temperature gases expand to the ambient pressure (state 4) in the turbine and produce power.
• The exhaust gases leave the turbine.
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Brayton cycleBy using the air-standard assumptions, replacing the combustion process by a constant pressure heat addition process, and replacing the exhaust discharging process by a constant pressure heat rejection process, the open cycle described above can be modeled as a closed cycle, called ideal Brayton cycle.
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Open Cycle Gas Turbine
Air
GCompressor
Turbine
CombustorFuel
Generator
50 – 70 % of turbine power
Pressure ratio: usually about 15, but up to 40 and moreTurbine inlet temperature (TIT): 900° - 1700°CTurbine exit temperature (TET): 400° - 600°CPower: 100 kW – 300 MW
Exhaust
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Closed Cycle Gas TurbineHeat Source
G
Compressor TurbineGenerator
Condensate from Process
Steam to Process
Heat Exchanger
2
1
3
4
Working fluid circulates in a closed circuit and does not cause corrosion or erosion
Any fuel, nuclear or solar energy can be used
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The ideal Brayton cycle is made up of four internally reversible processes.
1-2 Isentropic compression
2-3 Constant pressure heat addition
3- 4 Isentropic expansion
4-1 Constant pressure heat rejection
Brayton cycle
Steady Flow Energy Equation
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Efficiency of Brayton Cycle
( )232323 TTchhqq pin
( ) ( )414141 TTchhqq pout
23
41
23
4123 1
qqq
addedheatworknet
th
1
111
2
3
1
4
2
1
23
14
TTTT
TT
TTTT
th 1
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Efficiency of Brayton CycleConsider process 1 – 2, Isentropic compression
Cp
TCpTC
pTpCpV
11
1
1
2
1
2
1 1
prpp
TT
ratiopressurerpp
p 1
2
Consider process 3 – 4, Isentropic expansion
1
1
3
4
3
4 1
prpp
TT
2
3
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Efficiency of Brayton CycleFrom equations 2 and 3:
Substituting equations 2, 3 and 4 in equation 1
2
3
1
4
3
4
2
1 TT
TT
TT
TT
4
1
2
3
2
3
1
11 1
111
pp
th
rTTTT
r 1
11
p
th
r5
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Work ratio
)()()(
outputWork Net Work
34
3412
34
3412
TTTTTT
wwwrw
( ) )1(
3
11
pw rTTr
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Equation shows that the work ratio increases in direct proportion to the ratio T3 /T1 and inversely with a power of the pressure ratio.
On the other hand, thermal efficiency equation shows that thermal efficiency increases with increased pressure ratio.
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Compressor work:w12 = - (h2 – h1 ) = -Cp(T2 – T1)
Heat supplied during the cycle:q23 = (h3 – h2) = Cp(T3 – T2)
23
3412
inputHeat Net Work
qww
Summary of Equations
)1(
11
pr
Turbine work:w34 = (h3 – h4) = Cp(T3 – T4)
34
3412
outputWork Net Work
wwwrw
( )
)1(
3
11
pw rTT
r
Work ratio
Efficiency
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Improving cycle efficiency and work ratio
According to the equation (5) above, the efficiency of Brayton cycle depends only on the pressure ratio. To maximize the efficiency the pressure ratio has to be increased. This is done by compressing air isentropically from 1 to 2 to the maximum possible pressure. When pressure is increased the temperature also rises proportionately to the order of (-1)/.
The maximum temperature (T3) is fixed by the metallurgical properties, ability to withstand the high temperatures by the turbine materials. In practice the minimum temperature (T1) is limited to the atmospheric temperature.
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Isentropic compression process to reach the maximum temperature T3 is shown by 1-2. What is shown in the figure is the compression close to T3. The temperature T3 is obtained by heat addition form 2 to 3. If we reach T3 only by compression the points 2 and 3 will coincide. Under this condition the compression work and turbine work becomes equal giving zero net work. Therefore the maximum theoretical pressure ratio is obtained when work ratio is zero
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1
1
3
1
2max)(
TT
pp
rp
Therefore, for zero net work
( ) )1(
3
110
prTT
It is clear from the cycle 1-2’-3’-4’ of figures, the net work also becomes zero when pressure ratio is unity i.e. p2 = p1. Here also the figures show a pressure ratio very close to 1 for illustration. Here the maximum temperature T3 is achieved by heat supply at the same pressure.
Therefore, the pressure ratio (rp) has to be in between (rp)max and 1.
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Pressure ratio for maximum net work
The net work out put per unit mass is given by the following equation.
)()( 3412 TTcTTcw pp
Since
1/)1(
/)1(
1
212 Tr
pp
TT p
3/)1(4
1 Tr
Tp
and
111 /)1(3/)1(
1
pppp rTcrTcw
01111/)12(3/11
pp
pp
p rTc
rTc
drdw
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)1(2/
1
3
TT
rp
( ))1/(
1
3max
TT
rp
Since
max)( pp rr
The maximum net work is obtained when the pressure ratio equals the square root of maximum theoretical pressure ratio.
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Variation of efficiency and the net work out put with rp and T3
We will see how the efficiency and net work output vary with the pressure ratio and the maximum temperature.
Let us assume the following data are available
Min. temperature, T1 = 30oC = 303K
Isentropic eff. for compressor, comp = 0.85
Isentropic eff. for the turbine, Tub = 0.90
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Isentropic efficiency
Performance of turbines/compressors are measured by isentropic efficiencies.
The actual work input to the compressor is more and the actual work output from the turbine is more because of irreversibility.
Isentropic efficiencies involve a comparison between the actual performance of a device and the performance that would be achieved under idealized circumstances for the same inlet state and the same exit pressure.
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Isentropic efficiency - Turbine
The desired output from a turbine is the work output. Hence,
the definition of isentropic efficiency of a turbine is the ratio of
the actual work output of the turbine to the work output of the
turbine if the turbine undergoes an isentropic process between
the same inlet and exit pressures.
WorkIsentropicWorkTurbineActual
T
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The isentropic efficiency of turbine can be written as
12
12
hhhh
s
aT
h1 = enthalpy at the inlet
h2a = enthalpy of actual process at the exit
h2s = enthalpy of isentropic process at the exit
12
12
TTTT
s
aT
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Isentropic efficiency - compressor
The isentropic efficiency of a compressor or pump is defined as
the ratio of the work input to an isentropic process, to the work
input to the actual process between the same inlet and exit
pressures.
WorkActualWorkCompressorIsentropic
C
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The isentropic efficiency of compressor can be written as
12
12
hhhh
a
sC
h1 = enthalpy at the inlet
h2a = enthalpy of actual process at the exit
h2s = enthalpy of isentropic process at the exit
12
12
TTTT
a
sC
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The Back Work Ratio
Therefore, the turbine used in gas-turbine power plants are larger than those used in steam power plants of the same net power output, P. Usually more than half of the turbine work output is used to drive the compressor.
turbine
comp
WW
ratioworkBack
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Deviation of Actual Gas-Turbine Cycles from Ideal One
Pressure drop Isentropic efficiency
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Example 1A four stroke SI engine has the compression ratio of 6 and swept volume of 0.15m3. Pressure and temperature at the beginning of compression are 98kPa and 60oC respectively. Heat supplied in
the cycle is 150kJ. cp = 1kJ/kgK, cv = 0.71kJ/kgK
Determine
(i) the pressure , volume and temperature at all main state points
(ii) Efficiency
(iii) Mean effective pressure
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Example 2
An ideal diesel cycle using air as working fluid has a compression ratio of 16 and a cut off ratio of 2. The intake conditions are 100kPa, 20oC, and 2000cm3.Determine (a) Temperature and pressure at the end of each process(b) Net work output(c) Thermal efficiency(d) Mean effective pressure
cp = 1.0045kJ/kgK, cv 0.7175kJ/kgK
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Example 3
In an air standard Brayton cycle the minimum and maximum temperature are 300K and 1200K respectively. The pressure ratio is 10.
(i) Find out temperatures after compression and expansion
(ii) Calculate the compressor and turbine work, each in kJ/kg of air, and thermal efficiency of the cycle.
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Example 4
A gas turbine receives air at 1bar, 300K and compresses it adiabatically to 6.2bar. The isentropic efficiency of compressor is 0.88. The fuel has a heating value of 44186kJ/kg and the fuel –air ratio is 0.017kg fuel/kg of air. The turbine efficiency is 0.9. Calculate the work of turbine and compressor per kg of air compressed and the thermal efficiency.For products of combustion cp = 1.147kJ/kgK, g = 1.33.
For air cp = 1.005kJ/kgK, g = 1.4.
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The ideal air-standard Brayton cycle operates with air entering the compressor at 95 kPa, 22oC. The pressure ratio rp is 6:1 and the air leaves the heat addition process at 1100 K. Determine • the compressor work • the turbine work per unit mass flow, • the cycle efficiency,• the back work ratio, and compare the compressor exit
temperature to the turbine exit temperature.
Assume constant properties.
Example 5
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Example 6 In a gas turbine plant, working on the Brayton cycle, helium at 30° C and 22 bar is compressed to a pressure of 64 bar and then heated to a temperature of 1200 °C. After expansion in the turbine, the gas is cooled to initial pressure and temperature.Assume the following:Isentropic efficiency of the compressor – 0.85Isentropic efficiency of the turbine – 0.8Pressure loss in the combustion chamber – 1.2 barPressure loss in the cooler – 0.5 barSpecific heat (Cp) of the products of combustion is the same as that of helium and it is equal to 5.1926 kJ/kg K. Ratio of specific heats of helium – 1.667Determine the following;• Temperature at the end of compression and expansion.• Heat supplied, heat rejected and the net work per kg of helium.• Thermal efficiency of the plant• Flow rate of helium required to give an output of 12 MW.