--- title: "Two-shaft gas turbine, pressure ratio 7:1" subject: "Power Plant Equipment" chapter: 6 key: "semester_7/power_plant_equipment:gt-two-shaft-7-1" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/gt-two-shaft-7-1 site: BCT Engineering --- # Two-shaft gas turbine, pressure ratio 7:1 Power Plant Equipment · Chapter 6: Gas Turbine Power Plant - Asked in: 2080 Bhadra, 10 marks; 2079 Baisakh, 12 marks (asked 2 times) - Video: [Numerical of Gas Turbine - Numerical 4](https://www.youtube.com/watch?v=Fs9Kr9ZCJr4) — 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-two-shaft-7-1 - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-6 ## Question Air is drawn in a gas turbine unit at 15°C, 100 kPa and pressure ratio is 7:1. The compressor is driven by the H.P turbine and L.P turbine drives a separate power shaft. The isentropic efficiencies of compressor and the H.P and L.P turbines are 0.82, 0.85 and 0.85 respectively. If the maximum cycle temperature is 610°C, Calculate a) The pressure and temperature of the gases entering the low pressure turbine. b) The net power developed by the unit per kg/sec mass flow. c) The thermal efficiency of the unit. (For compression process cp = 1.005 kJ/kg.K and γ = 1.4 and for combustion and expansion process cp = 1.15 kJ/kg.K and γ = 1.333) ## Short answer HP turbine work equals compressor work, giving T4 = 654.75 K and p4 = 1.64 bar. LP turbine expansion to 100 kPa gives net work 74.38 kJ/kg, so power is 74.38 kW per kg/s. Thermal efficiency is 19.38%. ## Answer **Arrangement:** compressor (C) driven by the HP turbine; the LP (power) turbine drives the generator on a separate shaft. ```text 1 2 3 4 5 --> [C] ---> [CC] ---> [HPT] ---> [LPT] --> \__________________/ | (HP shaft) +==> Gen ``` **Data:** $T_1 = 288$ K, $p_1 = 100$ kPa, $r_p = 7$, $\eta_c = 0.82$, $\eta_{HP} = \eta_{LP} = 0.85$, $T_3 = 610 + 273 = 883$ K. Air: $c_{pa} = 1.005$, $\gamma = 1.4$ ($\frac{\gamma-1}{\gamma} = 0.2857$). Gas: $c_{pg} = 1.15$, $\gamma = 1.333$ ($\frac{\gamma-1}{\gamma} = 0.2498$). #### Compressor $$ \begin{aligned} T_{2s} &= 288 \times 7^{0.2857} = 502.17\ \text{K} \\ T_2 &= 288 + \frac{502.17 - 288}{0.82} = 549.18\ \text{K} \\ W_c &= 1.005(549.18 - 288) = 262.49\ \text{kJ/kg} \end{aligned} $$ #### (a) State at entry to the LP turbine HP turbine work = compressor work (neglecting fuel mass and mechanical losses): $$ \begin{aligned} c_{pg}(T_3 - T_4) &= W_c \Rightarrow T_4 = 883 - \frac{262.49}{1.15} = 654.75\ \text{K} \\ T_{4s} &= T_3 - \frac{T_3 - T_4}{\eta_{HP}} = 883 - \frac{228.25}{0.85} = 614.47\ \text{K} \\ \frac{p_3}{p_4} &= \left(\frac{T_3}{T_{4s}}\right)^{1/0.2498} = \left(\frac{883}{614.47}\right)^{4.003} = 4.269 \\ p_4 &= \frac{700}{4.269} = 163.98\ \text{kPa} \end{aligned} $$ Gas enters the LP turbine at **1.64 bar and 654.75 K (381.8 °C)**. #### (b) Net power per kg/s LP turbine expands from 1.64 bar to 1 bar: $$ \begin{aligned} T_{5s} &= T_4\left(\frac{p_5}{p_4}\right)^{0.2498} = 654.75\left(\frac{100}{163.98}\right)^{0.2498} = 578.65\ \text{K} \\ T_5 &= T_4 - \eta_{LP}(T_4 - T_{5s}) = 654.75 - 0.85(76.10) = 590.07\ \text{K} \\ W_{net} &= W_{LP} = 1.15(654.75 - 590.07) = 74.38\ \text{kJ/kg} \end{aligned} $$ Net power = **74.38 kW per kg/s** of air. #### (c) Thermal efficiency $$ \begin{aligned} Q_s &= c_{pg}(T_3 - T_2) = 1.15(883 - 549.18) = 383.89\ \text{kJ/kg} \\ \eta_{th} &= \frac{74.38}{383.89} = 0.1938 = 19.38\% \end{aligned} $$ **Answer:** (a) $p_4$ = 1.64 bar, $T_4$ = 654.75 K; (b) 74.38 kW per kg/s; (c) $\eta$ = 19.38 %. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What determines the temperature at HP turbine exit? A. LP turbine work B. Fuel heating value C. Net output D. HP turbine work equals compressor work Answer: D. The HP turbine drives only the compressor, so cp_g(T3-T4) = Wc. (Key point: HP turbine work) 2. How is p4 found? A. From isentropic exit T4s via (T3/T4s)^(gamma/(gamma-1)) with the HP efficiency B. From T4 directly with the actual ratio C. Using gamma 1.4 D. By equating it to atmospheric pressure Answer: A. T4s = T3 - (T3-T4)/eta_HP, then the pressure ratio is 4.27, giving 1.64 bar. (Key point: Pressure at LPT inlet) 3. Net output of the unit is A. HP minus LP work B. LP turbine work only (74.38 kW per kg/s) C. Compressor work D. HP + LP work Answer: B. HP turbine work is consumed entirely by the compressor. (Key point: Net power) 4. Heat added per kg is A. cp_a(T3 - T2) B. cp_g(T3 - T4) C. cp_g(T3 - T2) = 383.9 kJ/kg D. cp_g(T4 - T5) Answer: C. Combustion uses the gas cp, giving efficiency 19.38%. 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