A compressor is used to compress air from an initial state of 100 kPa and 300 K to a final state of 800 kPa. Determine the work done on the air and the final temperature if the compression process is isentropic.
Guide On Rating System
Vote
To determine the work done on the air during the isentropic compression process, we can use the equation:
$$\frac{P_1V_1^{\gamma}- P_2V_2^{\gamma}}{1- \gamma} = W$$
Where:
- $P_1$ is the initial pressure = 100 kPa
- $V_1$ is the initial volume (unknown)
- $P_2$ is the final pressure = 800 kPa
- $V_2$ is the final volume (unknown)
- $\gamma$ is the heat capacity ratio for air at constant pressure and constant volume, which is approximately 1.4 for air.
In an isentropic process, we have $P_1V_1^{\gamma} = P_2V_2^{\gamma}$
Substituting this into the work equation, we get:
$$W = \frac{P_1V_1^{\gamma}- P_2V_2^{\gamma}}{1- \gamma} = \frac{P_1V_1^{\gamma}- P_1V_1^{\gamma}}{1- \gamma} = 0$$
Therefore, the work done on the air during the isentropic compression process is 0.
To determine the final temperature, we can use the equation:
$$T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}}$$
Where:
- $T_1$ is the initial temperature = 300 K
- $P_1$ is the initial pressure = 100 kPa
- $P_2$ is the final pressure = 800 kPa
- $\gamma$ is the heat capacity ratio for air at constant pressure and constant volume, which is approximately 1.4 for air.
Substituting the given values, we get:
$$T_2 = 300 \, \text{K} \left(\frac{800 \, \text{kPa}}{100 \, \text{kPa}}\right)^{\frac{1.4-1}{1.4}}$$
Calculating this equation gives us:
$$T_2 \approx 645.6 \, \text{K}$$
Therefore, the final temperature of the air after the isentropic compression process is approximately 645.6 K.
Got a question with my answer?
Cheapmath.com is completely free!
Asking question is free.
Getting answers is free.
+99