Air at temperature 300 K is compressed isentropically from a pressure of 1 bar to 10 bar in a compressor. Eighty percent of the compressed air is supplied to a combustor. In the combustor, 0.88 MJ of heat is added per kg of air. The specific heat at constant pressure is \( C_p = 1005 \, {J/kg/K} \) and the specific heat ratio is \( \gamma = 1.4 \). The temperature of the air leaving the combustor is _______ K (rounded off to one decimal place).
Air at temperature 300 K is compressed isentropically from a pressure of 1 bar to 10 bar in a compressor. Eighty percent of the compressed air is supplied to a combustor. In the combustor, 0.88 MJ of heat is added per kg of air. The specific heat at constant pressure is \( C_p = 1005 \, {J/kg/K} \) and the specific heat ratio is \( \gamma = 1.4 \). The temperature of the air leaving the combustor is _______ K (rounded off to one decimal place).
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Solution and Explanation
We will solve the problem in two main steps: first, we calculate the temperature after compression using the isentropic relation, and then we calculate the temperature after heat addition in the combustor.
Step 1: Temperature after isentropic compression.
For an isentropic process, the relationship between temperature and pressure is given by: \[ \frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}} \] where:
\( T_1 = 300 \, {K} \) (initial temperature),
\( P_1 = 1 \, {bar} \) (initial pressure),
\( P_2 = 10 \, {bar} \) (final pressure),
\( \gamma = 1.4 \) (specific heat ratio).
Substitute the given values: \[ \frac{T_2}{300} = \left( \frac{10}{1} \right)^{\frac{1.4 - 1}{1.4}} = 10^{0.2857} \approx 1.818 \] \[ T_2 = 300 \times 1.818 = 545.4 \, {K} \] So, the temperature after isentropic compression is approximately \( T_2 = 545.4 \, {K} \).
Step 2: Temperature after heat addition in the combustor.
In the combustor, 0.88 MJ of heat is added per kg of air. The temperature increase due to heat addition is given by: \[ Q = C_p \Delta T \] where:
\( Q = 0.88 \, {MJ/kg} = 880 \, {kJ/kg} \),
\( C_p = 1005 \, {J/kg/K} \),
\( \Delta T = T_3 - T_2 \) (temperature increase).
Rearranging the equation to solve for \( T_3 \):
\[ T_3 = T_2 + \frac{Q}{C_p} \] Substitute the values: \[ T_3 = 545.4 + \frac{880 \times 10^3}{1005} \approx 545.4 + 875.2 = 1420.6 \, {K} \] Thus, the temperature of the air leaving the combustor is approximately 1420.6 K, which rounds to 1420.6 K.
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