Question 1 of 8: Turbojet gas turbine

Question 1: Turbojet gas 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. A stationary (static, sea-level) turbojet, so the flight speed is zero and there is no ram compression in the intake.

Find. The pressure and the actual temperature at the five stations of the engine — compressor inlet (1), compressor delivery (2), combustor exit (3), turbine exit (4) and nozzle exit (5) — together with the T-s diagram that locates them.

Approach. Work station by station along the gas path: isentropic relations plus the component efficiency fix each temperature, the combustor energy balance fixes the turbine inlet temperature, the condition that the turbine exactly drives the compressor fixes the turbine temperature drop (and hence, through the isentropic relation, the turbine exit pressure), and the nozzle expands what is left down to ambient.

1. Part (b) — station 1: the compressor inlet. The engine is stationary, so the air enters the intake at rest and station 1 is simply the ambient state: $$p_1 = 100\ \text{kPa}, \qquad T_1 = 27 + 273.15 = 300.15\ \text{K}$$ Because the aircraft speed is zero there is no ram rise; on a flying engine station 1 would be the stagnation state behind the intake and would be warmer and at higher pressure than ambient.
2. Station 2: compressor delivery, isentropic first. The delivery pressure follows straight from the pressure ratio, and the ideal delivery temperature from the isentropic relation on the reference sheet, $T_2/T_1 = (p_2/p_1)^{(k-1)/k}$: $$p_2 = r\,p_1 = 16 \times 100 = 1600\ \text{kPa}$$ $$T_{2s} = 300.15 \times 16^{0.4/1.4} = 300.15 \times 2.2082 = 662.78\ \text{K}$$ so the ideal temperature rise is $\Delta T_{s} = 662.78 - 300.15 = 362.63\ \text{K}$.
3. Station 2: the actual delivery state. A real compressor needs more temperature rise than the ideal one for the same pressure ratio, in the ratio of the isentropic efficiency $\eta_c = \Delta T_{s} / \Delta T_{\text{actual}}$: $$\Delta T = \frac{362.63}{0.80} = 453.29\ \text{K}, \qquad T_2 = 300.15 + 453.29 = \boxed{753.44\ \text{K} = 480.3\ {}^\circ\text{C}}$$ at $p_2 = 1600$ kPa. The power the compressor absorbs is therefore $$W_c = G\,c_p\,\Delta T = 70 \times 1.005 \times 453.29 = 31\,889\ \text{kW}$$
4. Station 3: the combustor exit. Combustion is taken at constant pressure, so $p_3 = p_2 = 1600$ kPa. The fuel releases $f \times CV = 1 \times 40\,000 = 40\,000$ kW into a gas stream that now carries both the air and the fuel, $G + f = 71$ kg/s: $$q_{23} = \frac{f\,CV}{G+f} = \frac{40\,000}{71} = 563.38\ \text{kJ/kg}$$ $$T_3 = T_2 + \frac{q_{23}}{c_p} = 753.44 + \frac{563.38}{1.005} = \boxed{1314.0\ \text{K} = 1040.9\ {}^\circ\text{C}}$$ which is a realistic turbine entry temperature for an engine of this pressure ratio.
5. Station 4: the turbine exit temperature. The question states that all of the turbine work drives the compressor, so the turbine temperature drop is set by the compressor power spread over the larger gas flow: $$\Delta T_{34} = \frac{W_c}{(G+f)\,c_p} = \frac{31\,889}{71 \times 1.005} = 446.91\ \text{K}$$ $$T_4 = 1314.0 - 446.91 = \boxed{867.1\ \text{K} = 594.0\ {}^\circ\text{C}}$$ Note the turbine drop is smaller than the compressor rise even though the two powers are equal, purely because one extra kilogram per second of gas is flowing through the turbine.
6. Station 4: the turbine exit pressure. The pressure has to be found through the isentropic end state, not the actual one, because it is the isentropic drop that the pressure ratio controls: $$\Delta T_{34,s} = \frac{\Delta T_{34}}{\eta_t} = \frac{446.91}{0.85} = 525.78\ \text{K}, \qquad T_{4s} = 1314.0 - 525.78 = 788.25\ \text{K}$$ $$p_4 = p_3\left(\frac{T_{4s}}{T_3}\right)^{k/(k-1)} = 1600 \times \left(\frac{788.25}{1314.0}\right)^{3.5} = \boxed{267.5\ \text{kPa}}$$ Using the actual temperature here instead of the isentropic one is the single most common error in this question; it would give about 214 kPa and would corrupt the whole nozzle calculation.
7. Station 5: the nozzle exit. The nozzle expands the gas from 267.5 kPa to the stated 100 kPa. Working the isentropic end state first, $$T_{5s} = T_4\left(\frac{p_5}{p_4}\right)^{(k-1)/k} = 867.1 \times \left(\frac{100}{267.5}\right)^{0.2857} = 654.61\ \text{K}$$ $$\Delta h_s = c_p\,(T_4 - T_{5s}) = 1.005 \times 212.50 = 213.57\ \text{kJ/kg}$$ and the real nozzle recovers only 95 % of that, $$\Delta h = 0.95 \times 213.57 = 202.89\ \text{kJ/kg}, \qquad T_5 = 867.1 - \frac{202.89}{1.005} = \boxed{665.2\ \text{K} = 392.1\ {}^\circ\text{C}}$$ at $p_5 = 100$ kPa.
8. Check the answer against the thrust the engine ought to make. The recovered enthalpy appears as jet kinetic energy, $\Delta h = V_5^2/2$ from the nozzle equation on page 17: $$V_5 = \sqrt{2 \times 202.89 \times 10^3} = 637.0\ \text{m/s}$$ and, with the aircraft stationary, the thrust is $$F = (G+f)\,V_5 = 71 \times 637.0 = 45\,227\ \text{N} = 45.2\ \text{kN}$$ A 70 kg/s engine of this pressure ratio does produce roughly 45 kN of static thrust, and the jet power $ (G+f)V_5^2/2 = 14\,405$ kW is 36.0 % of the fuel power — both figures are exactly what a turbojet of this class delivers, so the temperature and pressure chain above is sound.