Steam injected gas turbine, Calculation Exercise
Date: 2005-04-28
This exercise will give You 2 bonus credits for the exam in the Applied Heat and Power Technology course, spring 2005, provided that You hand it in not later than Friday, April 29, at 9.30 to the reception on Brinellvägen 68.
Clamp/join Your solution papers and mark with course number and teacher name (= Catharina Erlich)
A small gas turbine unit working with light oil as fuel has turbine, generator and compressor on a single shaft. Following data is given:
Air mass flow30 kg/s
Pressure ratio compressor10
Isentropic efficiency compressor0.82
Ambient temperature15ºC
Ambient pressure1.013 bar
Isentropic exponent, airκ = 1.4
Turbine inlet temperature800ºC
Isentropic efficiency turbine0.88
Pressure loss in combustor2%
Mechanical efficiency99%
Generator efficiency98%
Isentropic exponent, flue gasκ = 1.33
The gas turbine is to be modified with steam injection in the combustion chamber to recover hot exhaust gas from the turbine and too increase the power output. The steam is produced in a common single-pressure heat recovery steam generator (HRSG), consisting of economizer, evaporator and superheater.
Steam-injected modification, Data:
Steam mass flow5% of air mass flow
Steam superheat temperature320ºC
Steam pressure12 bar
Feed water temperature35ºC.
- Calculate the power output and efficiency of the gas turbine without steam injection
- Calculate the power output and efficiency of the gas turbine with steam injection
- Estimate the stack temperature of exhaust after HRSG.
HINTS
Follow this model for steam-injected gas turbine calculations:
a)The compressor remains unaffected by the steam injection
b)The isentropic exponent for flue gas remains the same as without steam injection, as well as the isentropic efficiency for the turbine
c)The fuel control system maintains the turbine inlet temperature at a constant value (which means that the fuel flow must increase a little to heat the injected steam up to the turbine inlet temperature)
d)Assume that the gas turbine outlet pressure remains the same as without the HRSG.
e)The heat capacity of the mixture combustion gas and steam may be represented by
CPAVE = CPG + (msteam/mair)*CPsteam
SOLUTION
- Simple gas turbine without steam injection
The electrical power output for the gas turbine is:
PGT= ηG * (PT* ηm – PC)
where
We start with the compressor:
This gives us that t2 = 15+327 °C = 342°C
The enthalpies for air in point 1 and 2 are (from table at x=0)
h1 = (15°C, x=0) = 15 kJ/kg
h2 = (342°C, x =0) = 349.6 kJ/kg (interpolated in the table)
The compressor thus consumes
Pc = 30*(349.6-15) kJ/s = 10 038 kW
For the turbine we have to take the fuel flow into consideration as well as the gas content in flue gas for the enthalpies. We need thus to calculate the specific fuel consumption β. From HEAT BALANCE OVER THE COMBUSTOR:
Point 3 is at 800°C in our case. The LHV for light oil is 42.3 MJ/kg.
Looking in the enthalpy table for air and flue gas from light oil we find that
h3,AIR = (800°C, x=0) = 856 kJ/kg
DH3 = (800°C) = 76 kJ/kg
The specific fuel consumption becomes:
The fuel flow becomes: mFUEL = 0.0126*30 = 0.38 kg/s
Now we can also calculate the gas content x:
The enthalpy in point 3 for flue gas becomes:
h3, GAS = h3, AIR + x*DH3 = 856 + 0.193*76 = 870.7 kJ/kg
The last enthalpy needed to calculate the power output is in point 4 (= turbine outlet). In order to get this enthalpy we need the temperature.
The temperature is calculated from:
We have given the temperature t3, we need first to calculate the pressure ratio.
The pressure after the compressor, p2 = 10*1.013 = 10.13 bars.
We loose 2% of this pressure in the combustion chamber. Then
p3 = 0.98*10.13 = 9.927
As the gas turbine is open, p4 = p1 and the pressure ratio p3/p4 thus becomes:
p3/p4 = 9.927/1.013 = 9.8
The temperature drop in the turbine part thus becomes
The temperature in the turbine outlet is:
t4 = 800 – 408 = 392°C
Now we can find the enthalpy in the turbine outlet:
h4, GAS = h4, AIR + x*DH4
h4,AIR = (392°C, x = 0) = 402.5 kJ/kg
DH4 = (392°C) = 30.4 kJ/kg
Thus
h4, GAS = 402.5 + 0.193*30.4 = 408.4 kJ/kg
The turbine output becomes:
PT = (30 + 0.38) * (870.7 – 408.4) kW = 14 045 kW
The electrical power output of the gas turbine is:
PGT= ηG * (PT* ηm – PC) = 0.98*(14 045*0.99 – 10 038) kW
PGT = 3789 kW ≈ 3.79 MW
Efficiency
ηGT = 23.6% (it is low because we have a small gas turbine with relatively combustion temperature )
- Steam-injected gas turbine
The power output for the steam injected gas turbine is calculated the same way as previous, we just have to consider that we have also steam into the gases, that will expand in the turbine part.
PGT = ηG * (PT* ηm – PC)
Reading the hints carefully we could assume that the compressor remains unaffected, i.e. we have the same temperature increase, same airflow and same isentropic efficiency and isentropic exponent. This gives us that
PC = 10 038 kW as in the previous task.
For the turbine output we have to make some new calculations. Steam is introduced in the combustion chamber, but will not take any active part in the combustion process, it just needs heat to increase the temperature from 320°C up to 800°C; therefore we have to add up some more fuel.
To calculate this NEW fuel flow, we have to make a heat balance over the combustion chamber.
Energy in = Energy out
mair * h2, AIR + mfuel * LHV + msteam * hSteam,320 = (mair + mfuel ) * h3, GAS + msteam * hSteam,800
The parameters written in bold, which are the steam parameters, have been added to the heat balance for the simple gas turbine.
Comparing the balance above with the simple gas turbine (see the Solution guide for gas turbines), it is identical except from the steam parameters!
As the steam is not taking active part into the combustion we can treat it separately as above. The enthalpy in point 3 considers the enthalpy for flue gas WITHOUT the steam, i.e. same enthalpy expression as for the simple gas turbine:
h3, GAS = h3, AIR + x*DH3
where x =
As the steam flow is very small compared to the airflow, we can neglect the pressure increase in the combustion chamber when steam is injected. The compressor determines the pressure in the combustion chamber. The enthalpies for steam are:
hSteam,320 = (320°C, 12 bar) = 3087.8 kJ/kg (there is a very little difference if
determining this enthalpy for
10 bars instead)
hSteam,800 = (800°C, 10 bar) = 4155.0 kJ/kg
Now considering that:
β = fuel flow (kg/s) / air flow (kg/s) = mfuel /mair
and that msteam /mAIR = 0.05
we obtain:
Which is identical to the simple gas turbine calculation except from that we have added up the steam heating term.
As we are keeping 800°C as turbine inlet temperature, h3,AIR and DH3 are the same as in the calculation for the simple gas turbine without the injection. As we also should take into consideration that the compressor remains unchanged we obtain:
Thus the NEW fuel flow becomes:
mFUEL = 30*0.0139 = 0.417 kg/s
We also get a new gas content in the gases (not considering the steam)
The enthalpy in point 3 for flue gas without steam thus becomes:
h3, GAS = h3, AIR + x*DH3 = 856 + 0.213*76 = 872.2 kJ/kg
According to the hints we could assume that the isentropic efficiency, the isentropic exponent and the outlet pressure remain the same as for the simple gas turbine without steam injection. This means that the temperature drop over the turbine will be the same as in task no 1.
This gives us that: t4 = 392°C.
The enthalpy for flue gas (without steam) in point 4 is:
h4, GAS = 402.5 + 0.213*30.4 = 409.0 kJ/kg
The enthalpy for steam in point 4 is:
hSTEAM, 392 = (1.013 bar, 392°C) = 3261 kJ/kg (interpolated in steam table)
The turbine power output is composed by expansion of the flue gas (without steam) and the expansion of the steam:
PT = (mair + mfuel)*(h3,GAS – h4,gas) + mSTEAM*(hSteam, 800 – hSteam,392)
The steam flow is: mSTEAM = 0.05* 30 kg/s = 1.5 kg/s
Which gives:
PT = (30+0.417)*(872.2 –409.0) + 1.5*(4155-3261) kW = 15 430 kW
The electrical power output becomes:
PGT = ηG * (PT* ηm – PC) = 0.98*(15 430*0.99 – 10 038) kW
PGT = 5133 kW ≈ 5.13 MW
We have thus increased the power output with 1.34 MW by injecting the steam
Efficiency
ηGT = 29.1%
We have thus increased the efficiency with 5.5% by recovering some heat from the exhaust to the steam injection
- Calculation of the stack temperature
Heat balance over the HRSG gives
hSH = hSteam,320 = 3087.8 kJ/kg
hFW = 4.18*35 kJ/kg = 146 kJ/kg
mtot = mair + mfuel + mSteam
CPAVE = CPG + (msteam/mair)*CPsteam
Assume that stack temperature is around 150°C, then the average temperature in the HRSG from inlet to outlet is:
tave = (392+150)/2 = 270°C
CPG = (CP diagram for flue gas at 270°C, x=0.213) = 1.060 kJ/kg K
Cpsteam = (CP diagram for steam at 270°C, 1bar) = 2.0 kJ/kg K
CPAVE = 1.060 + 0.05*2 = 1.16 kJ/kg
We have made an estimate. If we want a more exact value we iterate once again with a new average temperature: (392+273)/2 = 330°C
This gives CPG = 1.075 kJ/kgK and Cpsteam = 2.03 kJ/kg K and thus
CPAVE = 1.177 kJ/kg K and tG4 = 274.5 °C
As seen there is still much heat left in the HRSG that can be recovered for use as process heat.