A gas turbine employs a heat exchanger with a thermal ratio of 72%. The turbine operates between the pressures of 1.01 bar and 4.04 bar and ambient temperature is 20°C. Isentropic efficiencies of compressor and turbine are 80% and 85% respectively. The pressure drop on each side of the heat

Question

A gas turbine employs a heat exchanger with a thermal ratio of 72%. The turbine operates between the pressures of 1.01 bar and 4.04 bar and ambient temperature is 20°C. Isentropic efficiencies of compressor and turbine are 80% and 85% respectively. The pressure drop on each side of the heat exchanger is 0.05 bar and in the combustion chamber 0.14 bar. Assume combustion efficiency to be unity and calorific value of the fuel to be 41800 kJ/kg.
Calculate the increase in efficiency due to heat exchanger over that for simple cycle.
Assume [latex] c_{p} [/latex] is constant throughout and is equal to 1.024 kJ/kg K, and assume γ = 1.4.
For simple cycle the air-fuel ratio is 90 : 1, and for the heat exchange cycle the turbine entry temperature is the same as for a simple cycle.

Accepted Answer

Simple Cycle. Refer Fig. 31.

[latex]\frac{ T_{2} }{ T_{1} } = (\frac{p_{2}}{p_{1}})^{\frac{γ - 1}{γ }} = (\frac{4.40}{1.01})^{\frac{1.4 - 1}{1.4 }} [/latex] = 1.486

∴ [latex]T_{2}[/latex] = 293 × 1.486 = 435.4

Also, [latex] η_{compressor} = \frac{T_{2} - T_{1}}{T_{2}′ - T_{1}} [/latex]

0.8 = [latex] \frac{435.4 - 293}{T_{2}′ - 293} [/latex]

∴ [latex]T_{2}′ = \frac{435.4 - 293}{0.8}[/latex] + 293 = 471 K

Now, [latex]m_{f} × C = (m_{a} + m_{f}) × c_{p} × (T_{3} – T_{2}′)[/latex]
[latex] [m_{a} = mass of air, m_{f} [/latex] = mass of fuel]

∴ [latex]T_{3} = \frac{m_{f} × C }{c_{p} (m_{a} + m_{f})} + T_{2}′ = \frac{1 × 41800}{1.024 (90 + 1) } [/latex] + 471 = 919.5 K

Also, [latex]\frac{ T_{4} }{ T_{3} } = (\frac{p_{4}}{p_{3}})^{\frac{γ - 1}{γ }} [/latex]

or [latex]T_{4} = T_{3} × (\frac{p_{4}}{p_{3}})^{\frac{γ - 1}{γ }} = 919.5 × (\frac{1.01}{3.9})^{\frac{1.4 - 1}{1.4 }} [/latex] = 625 K

Again, [latex] η_{turbine } = \frac{T_{3} - T_{4}′}{T_{3} - T_{4}} [/latex]

∴ 0.85 = [latex] \frac{919.5 - T_{4}′}{919.5 - 625} [/latex]

∴ [latex]T_{4}′ [/latex] = 919.5 – 0.85(919.5 – 625) = 669 K

[latex] η_{thermal} = \frac{(T_{3} - T_{4}′) - (T_{2}′ - T_{1})}{(T_{3} - T_{2}′)} [/latex]

= [latex] \frac{(919.5 - 669) - (471 - 293)}{(919.5 - 471)} = \frac{72.5}{448.5} [/latex] = 0.1616 or 16.16%.

Heat Exchanger Cycle. Refer Fig. 32 (a, b)
[latex]T_{2}′ = 471 K (as for simple cycle) ; T_{3} [/latex]= 919.5 K (as for simple cycle)
To find [latex]T_{4}′ [/latex] :

[latex]p_{3} = 4.04 – 0.14 – 0.05 = 3.85 bar ; p_{4} [/latex] = 1.01 + 0.05 = 1.06 bar

∴ [latex]\frac{ T_{4} }{ T_{3} } = (\frac{p_{4}}{p_{3}})^{\frac{γ - 1}{γ }} = (\frac{1.06}{3.85})^{\frac{1.4 - 1}{1.4}} [/latex] = 0.69

i.e., [latex]T_{4} [/latex] = 919.5 × 0.69 = 634 K

[latex] η_{turbine } = \frac{T_{3} - T_{4}′}{T_{3} - T_{4}} ; 0.85 = \frac{919.5 - T_{4}′}{919.5 - 634} [/latex]

∴ [latex]T_{4}′ [/latex] = 919.5 – 0.85 (919.5 – 634) = 677 K

To find [latex]T_{5} [/latex] :
Thermal ratio (or effectiveness),

ε = [latex] \frac{T_{5} - T_{2}′}{T_{4}′ - T_{2}′} ∴ 0.72 = \frac{T_{5} - 471}{677 - 471} [/latex]

∴ [latex]T_{5} [/latex] = 0.72 (677 – 471) + 471 = 619 K

[latex] η_{thermal} = \frac{(T_{3} - T_{4}′) - (T_{2}′ - T_{1})}{(T_{3} - T_{5})} [/latex]

= [latex] \frac{(919.5 - 677) - (471 - 293)}{(919.5 - 619)} = \frac{64.5}{300.5} [/latex] = 0.2146 or 21.46%

∴ Increase in thermal efficiency = 21.46 – 16.16 = 5.3%.

Question 11.14: A gas turbine employs a heat exchanger with a thermal ratio ...

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