--- title: "Gas turbine with 78% effective regenerator" subject: "Power Plant Equipment" chapter: 6 key: "semester_7/power_plant_equipment:gt-regenerator-78" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/gt-regenerator-78 site: BCT Engineering --- # Gas turbine with 78% effective regenerator Power Plant Equipment · Chapter 6: Gas Turbine Power Plant - Asked in: 2081 Baisakh, 10 marks - Video: [Numerical of Gas Turbine - Numerical of Gas Turbine Power Plant](https://www.youtube.com/watch?v=L7zjYk0Dm6Y) — BEST MECHANICAL ENGINEERING - 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-regenerator-78 - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-6 ## Question In a gas turbine, the compressor takes in air at a temperature of 15°C and compresses it to four times the initial pressure with an isentropic efficiency of 82%. The temperature at the inlet of the turbine is 600°C and the isentropic efficiency of the turbine is 70%. A regenerator having an effectiveness of 78% is also incorporated in the cycle. Determine the thermal efficiency of the cycle. Take cp = 1.005 kJ/kg K and γ = 1.4. ## Short answer Find T2 and T4 using compressor and turbine isentropic efficiencies, then regenerator exit Tx = T2 + 0.78(T4 - T2). Net work is 29.32 kJ/kg and heat input is 248.27 kJ/kg. Thermal efficiency is 11.81%. ## Answer **Data:** $T_1 = 15 + 273 = 288$ K, $r_p = 4$, $\eta_c = 0.82$, $T_3 = 600 + 273 = 873$ K, $\eta_t = 0.70$, $\varepsilon = 0.78$, $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$; $r_p^{(\gamma-1)/\gamma} = 4^{0.2857} = 1.4860$. ```text +------ Regenerator ------+ | x y | 1 | 2 4 | --> [C] --> [CC] --3--> [T] ----+ ``` #### Compressor $$ \begin{aligned} T_{2s} &= 288 \times 1.4860 = 427.97\ \text{K} \\ T_2 &= 288 + \frac{427.97 - 288}{0.82} = 458.69\ \text{K} \\ W_c &= 1.005(458.69 - 288) = 171.54\ \text{kJ/kg} \end{aligned} $$ #### Turbine $$ \begin{aligned} T_{4s} &= \frac{873}{1.4860} = 587.49\ \text{K} \\ T_4 &= 873 - 0.70(873 - 587.49) = 673.14\ \text{K} \\ W_t &= 1.005(873 - 673.14) = 200.86\ \text{kJ/kg} \end{aligned} $$ #### Regenerator $T_4 = 673.14$ K $> T_2 = 458.69$ K, so heat can be recovered: $$ T_x = T_2 + \varepsilon(T_4 - T_2) = 458.69 + 0.78(214.45) = 625.96\ \text{K} $$ #### Thermal efficiency $$ \begin{aligned} W_{net} &= 200.86 - 171.54 = 29.32\ \text{kJ/kg} \\ Q_s &= c_p (T_3 - T_x) = 1.005(873 - 625.96) = 248.27\ \text{kJ/kg} \\ \eta_{th} &= \frac{29.32}{248.27} = 0.1181 = 11.81\% \end{aligned} $$ (Without the regenerator: $Q_s = 1.005(873 - 458.69) = 416.38$ kJ/kg, $\eta = 7.04\%$.) **Answer:** Thermal efficiency of the cycle = **11.81 %**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. T2s for rp=4 is A. T1 x 4 B. T1 / 1.486 C. T1 + 4 D. T1 x 4^0.2857 = 427.97 K Answer: D. Isentropic relation with (gamma-1)/gamma. (Key point: Isentropic temperature) 2. Actual turbine exit temperature A. T4 = T3 - eta_t(T3 - T4s) = 673 K B. T4 = T4s C. T4 = T3 - (T3 - T4s)/eta_t D. T4 = T3 x eta_t Answer: A. Turbine efficiency multiplies the ideal drop. (Key point: Turbine exit) 3. Air leaving the regenerator Tx is A. T4 B. T2 + 0.78(T4 - T2) = 626 K C. T2 + 0.78(T3 - T2) D. 0.78 T4 Answer: B. Effectiveness is applied between T2 and T4. (Key point: Regenerator) 4. Efficiency equals A. Wnet / cp(T3 - T2) B. Wt / Wc C. Wnet / cp(T3 - Tx) = 11.81% D. Wc / Wt Answer: C. Without regeneration it would be 7.04%. 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