Question

An aircraft engine operates on a simple ideal Brayton cycle with a pressure ratio of \(10 .\) Heat is added to the cycle at a rate of \(500 \mathrm{kW} ;\) air passes through the engine at a rate of \(1 \mathrm{kg} / \mathrm{s} ;\) and the air at the beginning of the compression is at \(70 \mathrm{kPa}\) and \(0^{\circ} \mathrm{C}\). Determine the power produced by this engine and its thermal efficiency. Use constant specific heats at room temperature.

Short answer

Answer: The power produced by the engine is 270 kW, and its thermal efficiency is 54%.

Step-by-step solution

Find the relevant values for the ideal Brayton cycle

To start, we need to find the necessary values for the Brayton cycle. We are given the following information: - Pressure ratio (rp) = 10 - Heat addition rate (Q_in) = 500 kW - Air mass flow rate (m_dot) = 1 kg/s - Initial air pressure (P1) = 70 kPa - Initial air temperature (T1) = 0 °C (273 K) Using constant specific heat at room temperature, we take the specific heat at constant pressure cp = 1005 J/kgK and the specific heat at constant volume cv = 717 J/kgK. We can calculate the specific heat ratio (γ) using the formula: γ = cp/cv = 1005 / 717 = 1.4 Now let's find the pressure and temperature values at the other cycle states given the Brayton cycle pressure ratio. P2 = P1 * rp = 70 kPa * 10 = 700 kPa P3 = P4 = P2 P1 = P4 Now, let's find T2 using the isentropic relation: T2 = T1 * (rp)^((γ-1)/γ) = 273 * (10)^((1.4-1)/1.4) = 542.3 K

Calculate the power output of the engine

Now, we will calculate the heat addition per unit mass (q_in) using the formula: q_in = cp * (T3 - T2) We are given Q_in = 500 kW, so we can calculate T3 from this equation. T3 = T2 + Q_in / (m_dot * cp) = 542.3 + 500000 / (1 * 1005) = 1042.3 K We can now find T4, using the isentropic relation once again for the expansion process: T4 = T3 / (rp)^((γ-1)/γ) = 1042.3 / (10)^((1.4-1)/1.4) = 772.3 K Using these temperatures, we can determine the net work done per unit mass, which is equal to the power produced by the engine, using the equation: w_net = cp * (T3 - T4) - cp * (T2 - T1) = 1005 * (1042.3 - 772.3) - 1005 * (542.3 - 273) = 270000 J/kg The power output of the engine is given by: Power_output = m_dot * w_net = 1 kg/s * 270000 J/kg = 270 kW

Calculate the thermal efficiency of the engine

Thermal efficiency (η) is calculated using the following relation: η = w_net / q_in = (cp * (T3 - T4) - cp * (T2 - T1)) / (cp * (T3 - T2)) η = (1005 * (1042.3 - 772.3) - 1005 * (542.3 - 273)) / (1005 * (1042.3 - 542.3)) = 270000 / 500000 = 0.54 So the thermal efficiency of this aircraft engine is 54%. In conclusion, the power produced by this engine is 270 kW, and its thermal efficiency is 54%.