Question 11.15: A 4500 kW gas turbine generating set operates with two compr...
A 4500 kW gas turbine generating set operates with two compressor stages ; the overall pressure ratio is 9 : 1. A high pressure turbine is used to drive the compressors, and a low-pressure turbine drives the generator. The temperature of the gases at entry to the high pressure turbine is 625°C and the gases are reheated to 625°C after expansion in the first turbine. The exhaust gases leaving the low-pressure turbine are passed through a heat exchanger to heat air leaving the high pressure stage compressor. The compressors have equal pressure ratios and intercooling is complete between the stages. The air inlet temperature to the unit is 20°C. The isentropic efficiency of each compressor stage is 0.8, and the isentropic efficiency of each turbine stage is 0.85, the heat exchanger thermal ratio is 0.8. A mechanical efficiency of 95% can be assumed for both the power shaft and compressor turbine shaft. Neglecting all pressure losses and changes in kinetic energy calculate :
(i) The thermal efficiency ; (ii) Work ratio of the plant ;
(iii) The mass flow in kg/s.
Neglect the mass of the fuel and assume the following :
For air : c_{pa} = 1.005 kJ/kg K and γ = 1.4
For gases in the combustion chamber and in turbines and heat exchanger, c_{pg} = 1.15 kJ/kg K and γ = 1.333.
Refer Fig. 33
Given : T_{1} = 20 + 273 + 293 K, T_{6} = T_{8} = 625 + 273 = 898 K
Efficiency of each compressor stage = 0.8
Efficiency of each turbine stage = 0.85, η_{mech.} = 0.95, ε = 0.8
(i) Thermal efficiency, η_{thermal} :
Since the pressure ratio and the isentropic efficiency of each compressor is the same then the work input required for each compressor is the same since both compressors have the same air inlet temperature i.e., T_{1} = T_{3} and T_{2}′ = T_{4}′ .
Also, \frac{ T_{2} }{ T_{1} } = (\frac{p_{2}}{p_{1}})^{\frac{γ – 1}{γ }} and \frac{p_{2}}{p_{1}} = \sqrt{9} = 3
∴ T_{2} = (20 + 273) × (3) ^{\frac{1.4 – 1}{1.4 }} = 401 K
Now, η_{compressor} (L.P.) = \frac{T_{2} – T_{1}}{T_{2}′ – T_{1}}
0.8 = \frac{401 – 293}{T_{2}′ – 293}
i.e., T_{2}′ = \frac{401 – 298}{0.8} + 293 = 428 K
Work input per compressor stage
= c_{pa} (T_{2}′ – T_{1}) = 1.005 (428 – 293) = 135.6 kJ/kg
The H.P. turbine is required to drive both compressors and to overcome mechanical friction.
i.e., Work output of H.P. turbine = \frac{2 × 135.6 }{0.95} = 285.5 kJ/kg
∴ c_{pg} (T_{6} – T_{7}′) = 285.5
i.e., 1.15 (898 – T_{7}′) = 285.5
∴ T_{7}′ = 898 – \frac{285.5}{1.15} = 650 K
Now, η_{turbine (H.P.)} = \frac{T_{6} – T_{7}′}{T_{6} – T_{7}} ; 0.85 = \frac{898 – 650}{898 – T_{7}}
∴ T_{7} = 898 – (\frac{898 – 650}{0.85}) = 606 K
Also, \frac{ T_{6} }{ T_{7} } = (\frac{p_{6}}{p_{7}})^{\frac{γ – 1}{γ }}
or \frac{p_{6}}{p_{7}} = (\frac{ T_{6} }{ T_{7} } )^{\frac{γ }{γ – 1}} = (\frac{ 898 }{ 606 } )^{\frac{1.333 }{0.333}} = 4.82
Then, \frac{p_{8}}{p_{9}} = \frac{9}{4.82} = 1.86
Again, \frac{ T_{8} }{ T_{9} } = (\frac{p_{8}}{p_{9}})^{\frac{γ – 1}{γ }} = (1.86)^{\frac{1.333 – 1}{1.333 }} = 1.16
∴ T_{9} = \frac{ T_{8} }{1.16} = \frac{ 898 }{1.16} = 774 K
Also, η_{turbine (L.P.)} = \frac{T_{8} – T_{9}′}{T_{8} – T_{9}} ; 0.85 = \frac{898 – T_{9}′}{898 – 774}
∴ T_{9} ′ = 898 – 0.85 (898 – 774) = 792.6 K
∴ Network output = c_{pg} (T_{8} – T_{9}′) × 0.95
= 1.15 (898 – 792.6) × 0.95 = 115.15 kJ/kg
Thermal ratio or effectiveness of heat exchanger,
ε = \frac{T_{5} – T_{4}′}{T_{9}′ – T_{4}′ } = \frac{T_{5} – 428}{792.6 – 428}
i.e., 0.8 = \frac{T_{5} – 428}{792.6 – 428}
∴ T_{5} = 0.8 (792.6 – 428) + 428 = 719.7 K
Now, Heat supplied = c_{pg} (T_{6} – T_{5}) + c_{pg} (T_{8} – T_{7}′)
= 1.15 (898 – 719.7) + 1.15 (898 – 650) = 490.2 kJ/kg
∴ η_{thermal} = \frac{Network output}{Heat supplied} = \frac{115.15}{490.2}
= 0.235 or 23.5%.
(ii) Work ratio :
Gross work of the plant = W_{turbine (H.P.)} + W_{turbine (L.P.)}
= 285.5 + \frac{115.15}{0.95} = 406.7 kJ/kg
∴ Work ratio = \frac{Network output}{Gross work output} = \frac{115.15}{406.7} = 0.283.
(iii) Mass flow in \dot{m} :
Let the mass flow be \dot{m}, then
\dot{m} × 115.15 = 4500
∴ \dot{m} = \frac{4500}{115.15} = 39.08 kg/s
i.e., Mass flow = 39.08 kg/s.
