--- title: "Gas turbine 800 kW with regenerator" subject: "Power Plant Equipment" chapter: 6 key: "semester_7/power_plant_equipment:gt-800kw-regenerator" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/gt-800kw-regenerator site: BCT Engineering --- # Gas turbine 800 kW with regenerator Power Plant Equipment · Chapter 6: Gas Turbine Power Plant - Asked in: 2081 Bhadra, 8 marks; 2071 Chaitra, 10 marks (asked 2 times) - Video: [Regeneration, Intercooling, and Reheating in 13 Minutes!](https://www.youtube.com/watch?v=EBQJv8pry2k) — Less Boring Lectures - 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-800kw-regenerator - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-6 ## Question A gas turbine plant of 800 kW capacity takes the air at 100 kPa and 288K with a mass flow rate of 6 kg/s. The pressure ratio of the cycle is 6 and the maximum temperature is limited to 900 K. A regenerator of 80% effectiveness is added in the plant to increase the overall efficiency. Assuming the isentropic efficiency of the compressor and turbine as 85%, determine the plant thermal efficiency and the net power developed. Take γ =1.4, Cp =1.005 kJ/kg.K. ## Short answer Compute compressor and turbine exit temperatures using isentropic efficiencies, then regenerator exit temperature with 80% effectiveness. Heat input is cp(T3 - Tx), net work is turbine work minus compressor work. Thermal efficiency is 24.82% and net power is 482.4 kW. ## Answer **Cycle:** 1–2 compressor, 2–x regenerator (air side), x–3 combustion chamber, 3–4 turbine, 4–y regenerator (gas side) to exhaust. **Data:** $p_1 = 100$ kPa, $T_1 = 288$ K, $\dot m = 6$ kg/s, $r_p = 6$, $T_3 = 900$ K, $\varepsilon = 0.8$, $\eta_c = \eta_t = 0.85$, $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$, so $\frac{\gamma-1}{\gamma} = 0.2857$ and $r_p^{0.2857} = 6^{0.2857} = 1.6685$. ```text +----------- Regenerator ---------+ | x (hot air) y (exhaust) | 1 |2 4 | -> [C] ------> [CC] -3-> [T] ---------+ \=========shaft=====/ ==> Gen ``` #### Compressor $$ \begin{aligned} T_{2s} &= T_1 r_p^{0.2857} = 288 \times 1.6685 = 480.53\ \text{K} \\ T_2 &= T_1 + \frac{T_{2s}-T_1}{\eta_c} = 288 + \frac{192.53}{0.85} = 514.51\ \text{K} \\ W_c &= c_p (T_2 - T_1) = 1.005 \times 226.51 = 227.64\ \text{kJ/kg} \end{aligned} $$ #### Turbine $$ \begin{aligned} T_{4s} &= \frac{T_3}{r_p^{0.2857}} = \frac{900}{1.6685} = 539.40\ \text{K} \\ T_4 &= T_3 - \eta_t (T_3 - T_{4s}) = 900 - 0.85 \times 360.60 = 593.49\ \text{K} \\ W_t &= c_p (T_3 - T_4) = 1.005 \times 306.51 = 308.04\ \text{kJ/kg} \end{aligned} $$ #### Regenerator and heat supplied Since $T_4 > T_2$, regeneration is possible: $$ \begin{aligned} T_x &= T_2 + \varepsilon (T_4 - T_2) = 514.51 + 0.8(593.49 - 514.51) = 577.70\ \text{K} \\ Q_s &= c_p (T_3 - T_x) = 1.005(900 - 577.70) = 323.92\ \text{kJ/kg} \end{aligned} $$ #### Efficiency and power $$ \begin{aligned} W_{net} &= W_t - W_c = 308.04 - 227.64 = 80.40\ \text{kJ/kg} \\ \eta_{th} &= \frac{W_{net}}{Q_s} = \frac{80.40}{323.92} = 0.2482 = 24.82\% \\ P_{net} &= \dot m W_{net} = 6 \times 80.40 = 482.4\ \text{kW} \end{aligned} $$ For comparison, without the regenerator $Q_s = 1.005(900 - 514.51) = 387.42$ kJ/kg and $\eta = 20.75\%$, so the regenerator raises efficiency by about 4.1 percentage points. **Answer:** Plant thermal efficiency = **24.82 %**, net power developed = **482.4 kW** (with the given data the plant cannot actually deliver its rated 800 kW at 6 kg/s). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the actual compressor outlet temperature found? A. T2 = T1 + (T2s - T1) x eta_c B. T2 = T2s x eta_c C. T2 = T1 x rp D. T2 = T1 + (T2s - T1)/eta_c Answer: D. Actual rise is the isentropic rise divided by efficiency. (Key point: Compressor outlet) 2. Tx is found from A. Tx = T2 + 0.8(T4 - T2) B. Tx = T4 - 0.8 T2 C. Tx = 0.8 T3 D. Tx = T2 + 0.8(T3 - T2) Answer: A. Effectiveness is actual rise over maximum possible rise to T4. (Key point: Regenerator effectiveness) 3. With regeneration, heat supplied per kg is A. cp(T3 - T2) B. cp(T3 - Tx) = 323.9 kJ/kg C. cp(T4 - T2) D. cp(T3 - T4) Answer: B. The regenerator already heats air to Tx. (Key point: Heat supplied) 4. Net power and efficiency are A. 800 kW and 20.75% B. 482.4 kW and 20.75% C. 482.4 kW and 24.82% D. 80.4 kW and 24.82% Answer: C. Wnet = 80.4 kJ/kg x 6 kg/s; without regenerator the efficiency would be 20.75%. (Key point: Result) ## Related in Power Plant Equipment - [Two-shaft gas turbine, pressure ratio 7:1](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-two-shaft-7-1) — Power Plant Equipment, chapter 6, asked 2× - [Gas turbine with reheat and 90% regenerator](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-reheat-regenerator-1400k) — Power Plant Equipment, chapter 6 - [Gas turbine with 78% effective regenerator](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-regenerator-78) — Power Plant Equipment, chapter 6 - [Gas turbine with intercooling, reheat and regeneration](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-intercool-reheat-regen) — Power Plant Equipment, chapter 6 - [Ideal gas turbine with regenerator at 1400 K](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-ideal-regenerator-1400k) — Power Plant Equipment, chapter 6 - [Gas turbine 5:1: power, efficiency, work ratio](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-5-1-work-ratio) — Power Plant Equipment, chapter 6 ## Related in other subjects - [Ideal gas turbine 100 kW: pressure ratio, work, efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aideal-gas-turbine-100kw-pressure-ratio) — Fundamental of Thermodynamics & Heat Transfer, chapter 6 - [Brayton cycle 80 MW: mass flow, efficiency, compressor fraction](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Abrayton-80mw-300k-1500k-fraction) — Fundamental of Thermodynamics & Heat Transfer, chapter 6 - [Brayton cycle pressure ratio 12: mass flow for 90 MW](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Abrayton-pressure-ratio-12-90mw) — Fundamental of Thermodynamics & Heat Transfer, chapter 6 - [Actual gas turbine cycle with component efficiencies](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aactual-brayton-cycle-numerical) — Turbo Machines, chapter 3