Compressor and Turbine: 25-Nav-B5 Marine Control Systems, May 2017, Q1

Question 1: Compressor and Turbine (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Pressure ratio r_p = 12.0; p₁ = 100 kPa; T₁ = 288 K; T₃ = 1350 K; η_c = 0.86; η_t = 0.89; η_m = 0.98; air/fuel ratio = 50; air properties k = 1.4, c_p = 1.005 kJ/kg·K.

Find. The four cycle temperatures, the net specific work per kg of gas, and the specific fuel consumption.

Fig. Q1 - T-s diagram, simple open-cycle gas turbine. 1→2 compression, 2→3 constant-pressure combustion, 3→4 turbine expansion, 4→1 exhaust to ambient.

Approach. Use the isentropic temperature ratio $r_p^{(k-1)/k}$ with the component efficiencies to get the actual compressor-exit and turbine-exit temperatures, then form the net work per unit mass of gas (turbine work degraded by mechanical efficiency, less compressor work) and divide the fuel flow by the net work for the specific fuel consumption.

Compressor exit (points 1→2). The isentropic ratio is $r_p^{(k-1)/k}=12^{0.2857}=2.034$, so $T_{2s}=T_1\,r_p^{(k-1)/k}=288(2.034)=585.8\ \text{K}$. The compressor efficiency gives the actual exit temperature,$$T_2=T_1+\frac{T_{2s}-T_1}{\eta_c}=288+\frac{585.8-288}{0.86}=634\ \text{K}.$$
Turbine exit (points 3→4). The isentropic exhaust temperature is $T_{4s}=T_3/r_p^{(k-1)/k}=1350/2.034=663.7\ \text{K}$, and the turbine efficiency gives$$T_4=T_3-\eta_t(T_3-T_{4s})=1350-0.89(1350-663.7)=739\ \text{K}.$$$$\boxed{T_1=288,\;T_2=634,\;T_3=1350,\;T_4=739\ \text{K}}$$
Specific work per kg of gas. Take a basis of 1 kg of gas leaving the turbine. With an air/fuel ratio of 50 the gas is 50/51 kg air + 1/51 kg fuel, so the compressor (which handles only air) does $w_c=\tfrac{50}{51}c_p(T_2-T_1)=0.9804(1.005)(634-288)=341\ \text{kJ}$, while the turbine expands the full gas stream, $w_t=c_p(T_3-T_4)=1.005(1350-739)=614\ \text{kJ}$. The net work at the coupling (mechanical losses on the turbine output) is$$w_{net}=\eta_m w_t-w_c=0.98(614)-341=260\ \text{kJ per kg gas}.$$$$\boxed{w_{net}=260\ \text{kW}\!\cdot\!\text{s/kg}}$$
Specific fuel consumption. Per kg of gas the fuel burned is 1/51 = 0.01961 kg. Converting the net work to a kWh basis ($1\ \text{kWh}=3600\ \text{kJ}$),$$\text{SFC}=\frac{m_{fuel}}{w_{net}}\times3600=\frac{0.01961}{260}\times3600=0.271\ \text{kg/kWh}.$$$$\boxed{\text{SFC}=0.271\ \text{kg/kWh}}$$

Check: Only air property data (c_p = 1.005 kJ/kg·K, k = 1.4) are supplied, so a cold-air-standard analysis is used throughout with the fuel mass accounted for via the air/fuel ratio. Using hotter combustion-gas properties (c_p ≈ 1.15) would raise the turbine work modestly; the method is unchanged.

Results — Question 1
Quantity	Value
Compressor exit temperature T₂	634 K
Turbine exit temperature T₄	739 K
Net specific work	260 kW·s/kg of gas
Specific fuel consumption	0.271 kg/kWh

Concept & Theory

Topic. Open-cycle (Brayton/Joule) gas turbine with component efficiencies.

Key relations. Isentropic $T_{2s}/T_1=r_p^{(k-1)/k}$; component efficiencies $\eta_c=(T_{2s}-T_1)/(T_2-T_1)$, $\eta_t=(T_3-T_4)/(T_3-T_{4s})$; net work $w_{net}=\eta_m w_t-w_c$; $\text{SFC}=\dot m_{fuel}/\dot W_{net}$.

Why this works. On a T–s diagram the compression and expansion are the two nearly vertical legs; irreversibility spreads them to the right, so the actual state points sit at higher entropy than the ideal ones. Because both turbine and compressor work scale with $c_p\Delta T$, the actual temperature rise/drop (found from the efficiencies) is all that is needed for the specific work.

Common pitfall. Applying the mechanical efficiency to the wrong stream, or omitting the fuel mass so the turbine and compressor are both taken at 1 kg — the air/fuel ratio is given precisely so that the turbine mass flow exceeds the compressor mass flow.

Source. Cohen, Rogers & Saravanamuttoo, Gas Turbine Theory, ch. 2–3; Çengel & Boles, Thermodynamics, ch. 9 (Brayton cycle).

Question 1 of 8: Compressor and Turbine

Question 1: Compressor and Turbine (10 marks)