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Open cycle gas turbine with intercooling

Short answer

Intercooling splits compression into LP and HP stages with cooling between them, reducing compressor work because work is proportional to inlet temperature. Optimum intermediate pressure is sqrt(p1 p2). It increases net work and work ratio; combined with regeneration it noticeably raises thermal efficiency.

Draw a sketch of a simple open cycle constant pressure gas turbine power plant with intercooling and describe the method to improve its thermal efficiency considering intercooling and T-s diagram.

A simple open-cycle constant-pressure gas turbine draws atmospheric air, compresses it, burns fuel in it at constant pressure, expands the hot gas in a turbine and exhausts it to the atmosphere (Brayton/Joule cycle). With intercooling, compression is split into LP and HP stages with an air cooler between them.

Sketch of the plant

             water in/out
                 ||
 Air->[LPC]-2->[ IC ]-3->[HPC]-4->[CC]-5->[ T ]-6->
  1     ||                 ||      ^       ||  exh
        ++======shaft======++======|=======++==>Gen
                                  fuel

LPC = low-pressure compressor, IC = intercooler, HPC = high-pressure compressor, CC = combustion chamber, T = turbine.

T–s diagram

 T
 |                   5
 |                  /|
 |                 / |
 |         4     /   |
 |         |---'     |
 |         |   2     6
 |         |  /|    /
 |         | / |   /
 |         3   1--'
 +----------------------- s
 1-2 LPC, 2-3 intercooling at p_i,
 3-4 HPC, 4-5 combustion at p_2,
 5-6 turbine, 6-1 exhaust (p_1)

How intercooling improves the plant

  • Compressor work per kg is W=cpTin[r(γ−1)/γ−1]W = c_p T_{in}\left[r^{(\gamma-1)/\gamma} - 1\right], i.e. proportional to inlet temperature. Cooling the air from T2T_2 back to T3≈T1T_3 \approx T_1 at the intermediate pressure makes the HP stage work much smaller.
  • Total compressor work is minimum when intercooling is perfect (T3=T1T_3 = T_1) and the stage pressure ratios are equal: pi=p1p2p_i = \sqrt{p_1 p_2}.
  • Turbine work is unchanged, so net work and work ratio increase; a smaller plant gives the same output.
  • Because air now leaves the HPC colder (T4T_4 lower), more fuel is needed in 4–5, so intercooling alone gives only a small change in efficiency. When combined with a regenerator, the larger gap between exhaust temperature T6T_6 and T4T_4 allows more heat recovery, and thermal efficiency rises noticeably.
  • Other ways to raise efficiency: regeneration, reheating, higher turbine inlet temperature, optimum pressure ratio and better component efficiencies.

✓ Answer checked by Claude Opus 5.5 (high effort)

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