--- title: "Gas turbine with intercooling, reheat and regeneration" subject: "Power Plant Equipment" chapter: 6 key: "semester_7/power_plant_equipment:gt-intercool-reheat-regen" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/gt-intercool-reheat-regen site: BCT Engineering --- # Gas turbine with intercooling, reheat and regeneration Power Plant Equipment · Chapter 6: Gas Turbine Power Plant - Asked in: 2069 Chaitra, 12 marks - Video: [Regenerative Gas Turbines with Reheat and Intercooling](https://www.youtube.com/watch?v=7R9DU-JariM) — Ahsan - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/gt-intercool-reheat-regen - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-6 ## Question 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. ## 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. ## Answer **Assumptions:** air-standard analysis with constant $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$, $\frac{\gamma-1}{\gamma} = 0.2857$; no pressure drops in intercooler, reheater, regenerator or combustor; steady flow; kinetic and potential energy changes neglected. ```text 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: $T_1 = T_3 = 300$ K, $T_6 = T_8 = 1400$ K. #### Compressor stages ($\eta_c = 0.8$) Stage 1 (pressure ratio 3): $$ \begin{aligned} T_{2s} &= 300 \times 3^{0.2857} = 410.62\ \text{K} \\ T_2 &= T_1 + \frac{T_{2s} - T_1}{\eta_c} = 300 + \frac{110.62}{0.8} = 438.28\ \text{K} \end{aligned} $$ Stage 2 (pressure ratio 1000/300 = 3.333): $$ \begin{aligned} T_{4s} &= 300 \times 3.333^{0.2857} = 423.17\ \text{K} \\ T_4 &= 300 + \frac{123.17}{0.8} = 453.96\ \text{K} \end{aligned} $$ #### Turbine stages ($\eta_t = 0.8$) Stage 1 (1000 to 300 kPa): $$ \begin{aligned} T_{7s} &= \frac{1400}{3.333^{0.2857}} = 992.51\ \text{K} \\ T_7 &= 1400 - 0.8(1400 - 992.51) = 1074.01\ \text{K} \end{aligned} $$ Stage 2 (300 to 100 kPa): $$ \begin{aligned} T_{9s} &= \frac{1400}{3^{0.2857}} = 1022.84\ \text{K} \\ T_9 &= 1400 - 0.8(1400 - 1022.84) = 1098.27\ \text{K} \end{aligned} $$ #### Regenerator ($\varepsilon = 0.8$) $$ \begin{aligned} T_5 &= T_4 + \varepsilon (T_9 - T_4) \\ &= 453.96 + 0.8(1098.27 - 453.96) = 969.41\ \text{K} \end{aligned} $$ #### Work and heat per kg $$ \begin{aligned} W_C &= c_p[(T_2 - T_1) + (T_4 - T_3)] \\ &= 1.005(138.28 + 153.96) = 293.70\ \text{kJ/kg} \\ W_T &= c_p[(T_6 - T_7) + (T_8 - T_9)] \\ &= 1.005(325.99 + 301.73) = 630.86\ \text{kJ/kg} \\ W_{net} &= 630.86 - 293.70 = 337.16\ \text{kJ/kg} \\ Q_s &= c_p[(T_6 - T_5) + (T_8 - T_7)] \\ &= 1.005(430.59 + 325.99) = 760.36\ \text{kJ/kg} \end{aligned} $$ #### (a) Thermal efficiency $$ \eta_{th} = \frac{W_{net}}{Q_s} = \frac{337.16}{760.36} = 0.4434 $$ **Answer (a): $\eta_{th} \approx 44.3\%$** #### (b) Back work ratio $$ \text{BWR} = \frac{W_C}{W_T} = \frac{293.70}{630.86} = 0.4656 $$ **Answer (b): back work ratio $\approx 0.466$ (46.6%)** #### (c) Net power $$ P = \dot m\, W_{net} = 5.807 \times 337.16 = 1957.9\ \text{kW} $$ **Answer (c): net power $\approx 1958$ 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With overall ratio 10 and intercooler/reheater at 300 kPa, the compressor stage ratios are: A. 3 and 3.333 B. 5 and 2 C. 10 and 1 D. 3 and 10 Answer: A. Stage 1 goes 100 to 300 kPa (ratio 3), stage 2 goes 300 to 1000 kPa (ratio 3.333). (Key point: Staging pressure) 2. For each compressor stage with eta = 0.8, the actual exit temperature is: A. T1 + (T2s - T1) x 0.8 B. T1 + (T2s - T1)/0.8 C. T2s x 0.8 D. T2s/0.8 Answer: B. The actual rise is the isentropic rise divided by 0.8, e.g. 300 + 110.62/0.8 = 438.28 K. (Key point: Non-isentropic stage) 3. Which expression gives heat supplied? A. cp[(T6 - T5) + (T8 - T7)] B. cp(T6 - T4) C. cp(T6 - T7) D. cp(T8 - T9) Answer: A. Heat is added in the combustor (T5 to T6) and the reheater (T7 to T8), total 760.36 kJ/kg. (Key point: Regenerator inlet and heat supplied) 4. Which set is correct? A. Efficiency 44.3 %, back work ratio 0.466, power about 1958 kW B. Efficiency 60 %, BWR 0.2, power 5807 kW C. Efficiency 44.3 %, BWR 2.1, power 337 kW D. Efficiency 46.6 %, BWR 0.443, power 760 kW Answer: A. Wnet = 337.16 kJ/kg, eta = 337.16/760.36 = 44.3 %, BWR = 293.70/630.86 = 0.466, P = 5.807 x 337.16 = 1958 kW. 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