gtg
TRANSCRIPT
Effi ciency of GTGEffi ciency of GTG
Presented by:Sanjay Kumar
IIT Kanpur
TATA CHEMICAL LTD., BabralaFertilizer Division
Gas Turbine Generator
The Gas Turbine is a power plant which produces a great amount of energy.
Basic Component of GTG
Compressor CombustorTurbine
Brayton Cycle
INLET AIRFILTER
C T
AIR COMPRESSOR TURBINE
GENERATOR
C.C.
LOAD GEAR
COMBUSTIONCHAMBER
FUEL(NG/HSD/NAP.)
KS STEAM
FEED WATER
STAGE-2
STAGE-1
HEATRECO-VERYUNIT(HRU)
D
G
(AUX.FIRING)
Brayton Cycle P-V Diagram:
P
V
Theoretical Cycle
Real Cycle
Stages in GTG
Compression 1-2 Combustion 2-3Expansion 3-4
Why there is diffrence in real and theoretical cycle???
Compression Stage
Volume decreasesPressure IncreasesTemprature IncreasesAvg. Inlet Temp=27oCInlet Pressure=1 Bar (atm)Outlet Temp= 310oCOutlet press=6.7bar17 stage Compressor
Compression Stage Cont….
Efficiency of the compressor=80% (but we assume it to be 100% while calculation.
Work done on Comp Wc = maCpa(T2-T1)
This is also equal to Wc=-nkRT1((P2/P1)(k-1)/nk-1)/.8(k-1))
n=No. of Stages , R=8.314J/Kmol K
Losses in Compressor
Designing Lossa) Rotar Lossb) Stator Loss
Losses are expressed as “Loss of heat and Enthalpy.” qtot=qth + qdf
qtot=Head Loss From disc Friction
qth=Heat Available from energy equation
Various Loss
qin=Heat loss from shock of the rotor
qsh=incidence loss
qbc=Profile Loss
qc=Clearance Loss (rotor and shroud)
qsf=Secondary Loss encountered in flow purge
qia= qin+ qsh+ qbc+ qc+ qsf
Efficiency Nimp= qia/ qtot
Compressor Stage cont...
qoa= qia- qin- qsh- qbc- qsf
qw=Head loss encountered in the stator from wake caused by impeller. qosf= Frictional Loss qex=Loss of Kinetic Head at the exit of stator.
Nstage= qia/ qtot
Compressor Stage Cont..
Due to all these factor the efficiency of compressor comes down to 80%. And thus the compression's not fully adiabatic.
P
V
Isotherm Line on Actual Curve
P
V
Shaded area= Extra Work that we have to done on Compressor have same pressure level.
Combustion Stage
The Fuel entered at this stageFuel Used:
a) Naphthab) Natural Gas
Reaction Occurred:CH4 + 2O2 ==> CO2 + 2H2O + Heat
Combustor Performance
Combustor performance is measured by:1) Efficiency decrease
encountered in the combustor2) Evenness of the outlet temperature
profile.
Ncomp=(ma+ mf)h3- mah2
(mf * CV)
Combustor Performance Cont.
Heat added to Combustor:
Q=(ma+ mf)h3- mah2
= (ma+ mf)CpgT3- maCpaT2
= (ma+ mf)Cpg(T3-T2)
Loss is Pressure is due to :
Acceleration of the fluid.Pressure drop due to aerodynamic drag:-
a) Friction in the Passage.b) Mixing Losses of the air jets.
As volume increases==>the temperature rises ==> Presure decreases.
Combustor stage cont.
P-V diagram
P
V
Pressure Loss
Expansion Stage
Gas does work in this stage using a turbine.About 25% of the gas is used.Rest is sent to make KS/LS steam in HRU.Average Inlet Temperature=1022oCInlet Pressure= 6.7 bar.Outlet Temp= 520oCOutlet Pressure=1bar (or >1bar)
Expansion Stage Cont.
Work Done by Turbine=Wc
Wc=(ma+mf)(h3-h4) = (ma+mf)Cpg(T3-T4)
Expansion is not ideally adiabatic.
Overall Efficiency of GTG
Three methods to calculate efficiency (as I used):-
a) Theoretical (Temp method)b) Actual (I/O method)c) Graphical (Using graphs)
*Graphical method is not so accurate as it is a averaged graph.
Theoretical Method
Ngtg= Wt - Wc
QNgtg=1- (T4-T1) (T3-T2)
Assumptions: 1) ma>>mf
2) Cair= Cgas
3) P2/P1= P3/P4
GTG Cont.
Known Quantity=T1, T2, T4
Unknown Quantity= T3
T3= T4( P3/ P4)((k-1)/k)
K=1.33 for gas.Approximated efficiency= 28-29 %
Input/Output Method
Efficiency=Ngtg=Power Produced Fuel Energy
GTG
Fuel=Mf*CVCV=Calorific Value
Power
Graphical Method
By graph we generate equations between various parameters.
Efficiency of GTG is Calculated.
Heat Recovery Unit
Input : 1) Feed Water 2) Steam entering inHRU from
GTG Output :1) KS Steam Efficiency of HRU :
Nhru=(mkshks - mfwhfw – mgtghgtg) (mf * CV)
GTG + HRU efficiency
Overall efficiency of GTG + HRU:
Noverall =Power Produced +Steam Generation Fuel Energy (HRU+GTG)
Steam Generation = msteam x hsteam
hsteam =0.7 according to the data in TCL
Efficiency chart
GTG + HRU= 92.7 %
GTG30.2%
62.5%
7.3%
HRU
Power
Losses
Effects of Different Parameters
Parameter which can affect:1) Inlet Temperature of Compressor
2) T3 (Temperature at Combustor outlet) 3) Relative Humidity 4) Compression Ratio
Main Points
Efficiency can be increased if:a) Compression ratio Increasesb) T1 decreases.c) T3 increases.
Presented By:Sanjay Kumar IIT Kanpur