Physics Answer Note #72

Question 1 Determine the specific work output, specific fuel consumption and cycle efficiency for a simple cycle gas turbine with a free power turbine as shown in Fig. 1 with the following specifications. Compressor pressure ratio (rp) = 10 Turbine inlet temperature, T3 = 1300 K Isentropic efficiency of the compressor, ηc = 0.85 Isentropic efficiency of each turbine, ηt = 0.90 Mechanical efficiency of each shaft, ηm = 0.99 Combustion efficiency, ηcom = 0.95 Combustion chamber pressure loss 8% compressor delivery pressure Exhaust pressure loss is 0.03 bar Ambient conditions 1 bar, 293 K Answer The simple cycle gas turbine with a free power turbine is widely used in power generation due to its high efficiency and simplicity in design. In order to determine the specific work output, specific fuel consumption, and cycle efficiency, we will need to use the Brayton cycle analysis. To begin, let us first define the Brayton cycle and its four processes: 1. Isentropic compression: Air is drawn into the compressor and compressed isentropically. 2. Constant pressure heat addition: Fuel is added in the combustion chamber, and the resulting hot gases expand in the turbine. 3. Isentropic expansion: The hot gases exiting the combustion chamber expand isentropically in the turbine. 4. Constant pressure heat rejection: The hot gases are cooled at a constant pressure in the heat exchanger, and the cycle is completed. Using the specifications provided, let us now calculate the specific work output and specific fuel consumption. Specific work output is defined as the net work output per unit mass flow rate of air. This can be calculated using the following equation: W = cp * T3 * [(rp)^((k-1)/k) -1] / (ηc * (ηt)^2 * (ηm)^2) Where: - W is the specific work output - cp is the specific heat capacity at constant pressure - T3 is the turbine inlet temperature - k is the specific heat ratio of air - ηc is the isentropic efficiency of the compressor - ηt is the isentropic efficiency of the turbine - ηm is the mechanical efficiency of the shaft - rp is the compressor pressure ratio (rp) Using the given values, we can substitute them into the equation and calculate the specific work output: W = (1005 J/kg K) * (1300 K) * [(10)^((1.4-1)/1.4) -1] / (0.85 * (0.90)^2 * (0.99)^2) = 2074169.93 J/kg Next, let us calculate the specific fuel consumption, which is defined as the amount of fuel consumed per unit of work output. This can be calculated using the following equation: SF = (ηcom * LHV) / (ηt * ηm * W) Where: - SF is the specific fuel consumption - ηcom is the combustion efficiency - LHV is the lower heating value of fuel - ηt is the isentropic efficiency of the turbine - ηm is the mechanical efficiency of the shaft - W is the specific work output Using the given values, we can calculate the specific fuel consumption: SF = (0.95 * 42736 kJ/kg) / (0.90 * 0.99 * 2074169.93 J/kg) = 0.0105 kg/kJ Lastly, let us calculate the cycle efficiency, which is defined as the ratio of the net work output to the heat added during the cycle. This can be calculated using the following equation: η = W / Q1 Where: - η is the cycle efficiency - W is the specific work output - Q1 is the heat added during the cycle = cp * T3 * (rp)^((k-1)/k) * [1 - (p2/p1)^((k-1)/k)] Using the given values, we can calculate the cycle efficiency: η = 2074169.93 J/kg / [1005 J/kg K * 1300 K * (10)^((1.4-1)/1.4) * [1 - (0.92/10)^((1.41)/1.4)]] = 0.357 or 35.7% Therefore, the specific work output is 2074169.93 J/kg, the specific fuel consumption is 0.0105 kg/kJ, and the cycle efficiency is 35.7%.