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Gas turbine with intercooling, reheat and regeneration

Short answer

With 80% isentropic efficiencies, two-stage compression with intercooling, two-stage turbine with reheat, compressor work 293.70 kJ/kg, turbine work 630.86 kJ/kg, net 337.16 kJ/kg. Regenerator effectiveness 0.8 gives heat input 760.36 kJ/kg. Thermal efficiency 44.3%, back work ratio 0.466, net power 1958 kW.

A regenerative gas turbine with intercooling and reheat operates at steady state. Air enters the compressor at 100 kPa, 300 K with a mass flow rate of 5.807 kg/s. The pressure ratio across the two-stage compressor is 10. The pressure ratio across the two-stage turbine is also 10. The intercooler and reheater each operate at 300 kPa. At the inlets to the turbine stages, the temperature is 1400 K. The temperature at the inlet to the second compressor stage is 300 K. The isentropic efficiency of each compressor and turbine stage is 80%. The regenerator effectiveness is 80%. Determine: (a) the thermal efficiency, (b) the back work ratio, (c) the net power developed, in kW. Take cp = 1.005 kJ/kg K and γ = 1.4.

Assumptions: air-standard analysis with constant cp=1.005c_p = 1.005 kJ/kg K, γ=1.4\gamma = 1.4, γ−1γ=0.2857\frac{\gamma-1}{\gamma} = 0.2857; no pressure drops in intercooler, reheater, regenerator or combustor; steady flow; kinetic and potential energy changes neglected.

        LPC        HPC            regen          CC
 1 ->[ C1 ]-2->[IC]-3->[ C2 ]-4->[ R ]-5->[ CC ]-6
                                  ^ |
                         exhaust  | v 10 (to stack)
 6 ->[ T1 ]-7->[ RH ]-8->[ T2 ]-9-+

Pressures: 100 kPa (1), 300 kPa (2, 3, 7, 8), 1000 kPa (4, 5, 6), 100 kPa (9). Temperatures given: T1=T3=300T_1 = T_3 = 300 K, T6=T8=1400T_6 = T_8 = 1400 K.

Compressor stages (ηc=0.8\eta_c = 0.8)

Stage 1 (pressure ratio 3):

T2s=300×30.2857=410.62 KT2=T1+T2s−T1ηc=300+110.620.8=438.28 K\begin{aligned} T_{2s} &= 300 \times 3^{0.2857} = 410.62\ \text{K} \\ T_2 &= T_1 + \frac{T_{2s} - T_1}{\eta_c} = 300 + \frac{110.62}{0.8} = 438.28\ \text{K} \end{aligned}

Stage 2 (pressure ratio 1000/300 = 3.333):

T4s=300×3.3330.2857=423.17 KT4=300+123.170.8=453.96 K\begin{aligned} T_{4s} &= 300 \times 3.333^{0.2857} = 423.17\ \text{K} \\ T_4 &= 300 + \frac{123.17}{0.8} = 453.96\ \text{K} \end{aligned}

Turbine stages (ηt=0.8\eta_t = 0.8)

Stage 1 (1000 to 300 kPa):

T7s=14003.3330.2857=992.51 KT7=1400−0.8(1400−992.51)=1074.01 K\begin{aligned} T_{7s} &= \frac{1400}{3.333^{0.2857}} = 992.51\ \text{K} \\ T_7 &= 1400 - 0.8(1400 - 992.51) = 1074.01\ \text{K} \end{aligned}

Stage 2 (300 to 100 kPa):

T9s=140030.2857=1022.84 KT9=1400−0.8(1400−1022.84)=1098.27 K\begin{aligned} T_{9s} &= \frac{1400}{3^{0.2857}} = 1022.84\ \text{K} \\ T_9 &= 1400 - 0.8(1400 - 1022.84) = 1098.27\ \text{K} \end{aligned}

Regenerator (ε=0.8\varepsilon = 0.8)

T5=T4+ε(T9−T4)=453.96+0.8(1098.27−453.96)=969.41 K\begin{aligned} T_5 &= T_4 + \varepsilon (T_9 - T_4) \\ &= 453.96 + 0.8(1098.27 - 453.96) = 969.41\ \text{K} \end{aligned}

Work and heat per kg

WC=cp[(T2−T1)+(T4−T3)]=1.005(138.28+153.96)=293.70 kJ/kgWT=cp[(T6−T7)+(T8−T9)]=1.005(325.99+301.73)=630.86 kJ/kgWnet=630.86−293.70=337.16 kJ/kgQs=cp[(T6−T5)+(T8−T7)]=1.005(430.59+325.99)=760.36 kJ/kg\begin{aligned} W_C &= c_p[(T_2 - T_1) + (T_4 - T_3)] \\ &= 1.005(138.28 + 153.96) = 293.70\ \text{kJ/kg} \\ W_T &= c_p[(T_6 - T_7) + (T_8 - T_9)] \\ &= 1.005(325.99 + 301.73) = 630.86\ \text{kJ/kg} \\ W_{net} &= 630.86 - 293.70 = 337.16\ \text{kJ/kg} \\ Q_s &= c_p[(T_6 - T_5) + (T_8 - T_7)] \\ &= 1.005(430.59 + 325.99) = 760.36\ \text{kJ/kg} \end{aligned}

(a) Thermal efficiency

ηth=WnetQs=337.16760.36=0.4434\eta_{th} = \frac{W_{net}}{Q_s} = \frac{337.16}{760.36} = 0.4434

Answer (a): ηth≈44.3%\eta_{th} \approx 44.3\%

(b) Back work ratio

BWR=WCWT=293.70630.86=0.4656\text{BWR} = \frac{W_C}{W_T} = \frac{293.70}{630.86} = 0.4656

Answer (b): back work ratio ≈0.466\approx 0.466 (46.6%)

(c) Net power

P=m˙ Wnet=5.807×337.16=1957.9 kWP = \dot m\, W_{net} = 5.807 \times 337.16 = 1957.9\ \text{kW}

Answer (c): net power ≈1958\approx 1958 kW (about 1.96 MW)

Note: with 80% component efficiencies almost half of the turbine work is used by the compressor, which is why intercooling (less compressor work) and regeneration (less fuel) are both needed to get a reasonable efficiency.

✓ Answer checked by Claude Opus 5.5 (high effort)

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