--- title: "Actual gas turbine cycle with component efficiencies" subject: "Turbo Machines" chapter: 3 key: "semester_7/turbo_machines:actual-brayton-cycle-numerical" source: https://www.bctengineer.com/q/semester_7/turbo_machines/actual-brayton-cycle-numerical site: BCT Engineering --- # Actual gas turbine cycle with component efficiencies Turbo Machines · Chapter 3: Gas Turbine - Asked in: year not given, 8 marks - Video: [Turbines, Compressors, and Pumps - ISENTROPIC EFFICIENCY in 8 Minutes!](https://www.youtube.com/watch?v=K2A5gdROVZE) — Less Boring Lectures - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/turbo_machines/actual-brayton-cycle-numerical - Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-3 ## Question A gas turbine plant takes in air at 100 kPa and 288 K. The compressor pressure ratio is 6 and its isentropic efficiency is 80%. The pressure loss in the combustion chamber is 5% of the compressor delivery pressure and the turbine inlet temperature is 1073 K. The turbine isentropic efficiency is 85% and it exhausts to 100 kPa. For air $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$; for the gases $c_p = 1.148$ kJ/kg K, $\gamma = 1.333$. Neglect the mass of fuel. Find (a) the compressor delivery temperature, (b) the turbine exit temperature, (c) the net specific work, (d) the thermal efficiency, (e) the work ratio, and (f) the power for a flow of 15 kg/s. ## Short answer Considering compressor and turbine efficiencies and 5% pressure loss, T2=528.7 K, T4=751.4 K; net specific work=127.3 kJ/kg; thermal efficiency=20.4%; work ratio=0.345; power=1910 kW. ## Answer **Data:** $T_1 = 288$ K, $r_p = 6$, $\eta_c = 0.80$, $\eta_t = 0.85$, $T_3 = 1073$ K, $\dot m = 15$ kg/s. #### (a) Compressor $$ T_{2s} = 288 \times 6^{0.2857} = 480.5\ \text{K} $$ $$ w_c = \frac{c_p (T_{2s} - T_1)}{\eta_c} = \frac{1.005 \times 192.5}{0.80} = 241.9\ \text{kJ/kg}, \qquad T_2 = 288 + \frac{241.9}{1.005} = 528.7\ \text{K} $$ #### (b) Turbine Pressure at turbine inlet $p_3 = 0.95 \times 600 = 570$ kPa, so the expansion ratio is $570/100 = 5.70$. $$ \frac{\gamma - 1}{\gamma} = 0.2498, \quad T_{4s} = \frac{T_3}{r_t^{0.2498}} = \frac{1073}{1.5446} = 694.7\ \text{K} $$ $$ w_t = \eta_t c_p (T_3 - T_{4s}) = 0.85 \times 1.148 \times 378.3 = 369.2\ \text{kJ/kg} $$ $$ T_4 = T_3 - \frac{w_t}{c_p} = 1073 - \frac{369.2}{1.148} = 751.4\ \text{K} $$ #### (c) Net work $$ w_{net} = w_t - w_c = 369.2 - 241.9 = 127.3\ \text{kJ/kg} $$ #### (d) Thermal efficiency $$ q_{in} = c_{p,g} (T_3 - T_2) = 1.148 \times (1073 - 528.7) = 624.9\ \text{kJ/kg} $$ $$ \eta_{th} = \frac{w_{net}}{q_{in}} = \frac{127.3}{624.9} = 0.204 $$ #### (e) Work ratio and (f) power $$ \text{Work ratio} = \frac{w_{net}}{w_t} = \frac{127.3}{369.2} = 0.345 \quad (\text{back work ratio} = 0.655) $$ $$ P = \dot m w_{net} = 15 \times 127.3 = 1910\ \text{kW} $$ **Answer:** $T_2 = 528.7$ K; $T_4 = 751.4$ K; $w_{net} = 127.3$ kJ/kg; $\eta_{th} = 20.4\%$; work ratio = 0.345; $P = 1910$ kW. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the actual compressor delivery temperature found? A. T2 = T2s B. T2 = T1 + (T2s - T1)/eta_c = 528.7 K C. T2 = T1 + eta_c (T2s - T1) D. T2 = T2s x eta_c Answer: B. Actual rise is the isentropic rise divided by eta_c: 192.5/0.8 = 240.6 K, so T2 is about 528.7 K. (Key point: Compressor delivery temperature) 2. What is the turbine expansion ratio? A. 6 B. 5.7, since turbine inlet pressure is 0.95 x 600 kPa C. 6.3 D. 0.95 Answer: B. The 5% combustor loss lowers turbine inlet pressure to 570 kPa, so expansion ratio is 570/100 = 5.70. (Key point: Combustor pressure loss) 3. Which gas properties are used for the turbine? A. Air values cp = 1.005, gamma = 1.4 B. Gas values cp = 1.148, gamma = 1.333 C. Mean of the two D. No properties needed Answer: B. Compressor uses air values; the turbine and heat addition use combustion gas values. (Key point: Property use) 4. What is the work ratio and its meaning? A. 0.655; net work over compressor work B. 0.345; net work over turbine work, small because compressor takes much C. 0.204; efficiency D. 1.0; no losses Answer: B. Work ratio = 127.3/369.2 = 0.345; irreversibilities make the compressor use 65.5% of turbine work. 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