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Two-shaft gas turbine, pressure ratio 7:1

Short answer

HP turbine work equals compressor work, giving T4 = 654.75 K and p4 = 1.64 bar. LP turbine expansion to 100 kPa gives net work 74.38 kJ/kg, so power is 74.38 kW per kg/s. Thermal efficiency is 19.38%.

Air is drawn in a gas turbine unit at 15°C, 100 kPa and pressure ratio is 7:1. The compressor is driven by the H.P turbine and L.P turbine drives a separate power shaft. The isentropic efficiencies of compressor and the H.P and L.P turbines are 0.82, 0.85 and 0.85 respectively. If the maximum cycle temperature is 610°C, Calculate a) The pressure and temperature of the gases entering the low pressure turbine. b) The net power developed by the unit per kg/sec mass flow. c) The thermal efficiency of the unit. (For compression process cp = 1.005 kJ/kg.K and γ = 1.4 and for combustion and expansion process cp = 1.15 kJ/kg.K and γ = 1.333)

Arrangement: compressor (C) driven by the HP turbine; the LP (power) turbine drives the generator on a separate shaft.

 1        2         3        4         5
 --> [C] ---> [CC] ---> [HPT] ---> [LPT] -->
      \__________________/           |
         (HP shaft)                  +==> Gen

Data: T1=288T_1 = 288 K, p1=100p_1 = 100 kPa, rp=7r_p = 7, ηc=0.82\eta_c = 0.82, ηHP=ηLP=0.85\eta_{HP} = \eta_{LP} = 0.85, T3=610+273=883T_3 = 610 + 273 = 883 K. Air: cpa=1.005c_{pa} = 1.005, γ=1.4\gamma = 1.4 (γ−1γ=0.2857\frac{\gamma-1}{\gamma} = 0.2857). Gas: cpg=1.15c_{pg} = 1.15, γ=1.333\gamma = 1.333 (γ−1γ=0.2498\frac{\gamma-1}{\gamma} = 0.2498).

Compressor

T2s=288×70.2857=502.17 KT2=288+502.17−2880.82=549.18 KWc=1.005(549.18−288)=262.49 kJ/kg\begin{aligned} T_{2s} &= 288 \times 7^{0.2857} = 502.17\ \text{K} \\ T_2 &= 288 + \frac{502.17 - 288}{0.82} = 549.18\ \text{K} \\ W_c &= 1.005(549.18 - 288) = 262.49\ \text{kJ/kg} \end{aligned}

(a) State at entry to the LP turbine

HP turbine work = compressor work (neglecting fuel mass and mechanical losses):

cpg(T3−T4)=Wc⇒T4=883−262.491.15=654.75 KT4s=T3−T3−T4ηHP=883−228.250.85=614.47 Kp3p4=(T3T4s)1/0.2498=(883614.47)4.003=4.269p4=7004.269=163.98 kPa\begin{aligned} c_{pg}(T_3 - T_4) &= W_c \Rightarrow T_4 = 883 - \frac{262.49}{1.15} = 654.75\ \text{K} \\ T_{4s} &= T_3 - \frac{T_3 - T_4}{\eta_{HP}} = 883 - \frac{228.25}{0.85} = 614.47\ \text{K} \\ \frac{p_3}{p_4} &= \left(\frac{T_3}{T_{4s}}\right)^{1/0.2498} = \left(\frac{883}{614.47}\right)^{4.003} = 4.269 \\ p_4 &= \frac{700}{4.269} = 163.98\ \text{kPa} \end{aligned}

Gas enters the LP turbine at 1.64 bar and 654.75 K (381.8 °C).

(b) Net power per kg/s

LP turbine expands from 1.64 bar to 1 bar:

T5s=T4(p5p4)0.2498=654.75(100163.98)0.2498=578.65 KT5=T4−ηLP(T4−T5s)=654.75−0.85(76.10)=590.07 KWnet=WLP=1.15(654.75−590.07)=74.38 kJ/kg\begin{aligned} T_{5s} &= T_4\left(\frac{p_5}{p_4}\right)^{0.2498} = 654.75\left(\frac{100}{163.98}\right)^{0.2498} = 578.65\ \text{K} \\ T_5 &= T_4 - \eta_{LP}(T_4 - T_{5s}) = 654.75 - 0.85(76.10) = 590.07\ \text{K} \\ W_{net} &= W_{LP} = 1.15(654.75 - 590.07) = 74.38\ \text{kJ/kg} \end{aligned}

Net power = 74.38 kW per kg/s of air.

(c) Thermal efficiency

Qs=cpg(T3−T2)=1.15(883−549.18)=383.89 kJ/kgηth=74.38383.89=0.1938=19.38%\begin{aligned} Q_s &= c_{pg}(T_3 - T_2) = 1.15(883 - 549.18) = 383.89\ \text{kJ/kg} \\ \eta_{th} &= \frac{74.38}{383.89} = 0.1938 = 19.38\% \end{aligned}

Answer: (a) p4p_4 = 1.64 bar, T4T_4 = 654.75 K; (b) 74.38 kW per kg/s; (c) η\eta = 19.38 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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