Air is compressed in an isentropic compressor from 15 psia

Step 1 of 2

Obtain properties of air at an average temperature of \(400^{\circ} \mathrm{F}\) from table A-2E, “ideal gas specific heat of gases” in the textbook.

Specific heat at constant pressure, \(c_p=0.245\mathrm{\ Btu}/\mathrm{lbm}\cdot R)

Specific heat ratio, k = 1.389

Find the temperature at the exit of the compressor by using the following isentropic relation:

\(\frac{T_{2}}{T_{1}}=\left(\frac{P_{2}}{P_{1}}\right)^{\frac{k-1}{k}}\)

Here, \(P_{1}\) is the compressor inlet pressure, \(P_{2}\) is compressor exit pressure, \(T_{1}\) is compressor inlet temperature, and \(T_{2}\) is the compressor exit temperature.

Substitute (70 + 459.67) R for \(T_{1}\), 15 psia for \(P_{1}\), 200 psia for \(P_{2}\), and 1.389 for k. 

\(\begin{array}{l}T_2=T_1\times\left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}}\\ T_2=(70+459.67)\times\left(\frac{200}{15}\right)^{\frac{1.389-1}{1.389}}\\ T_2=1094.6\ R\\ T_2\approx1095\ R \end{array}\)

Therefore, the outlet temperature of the air is 1095 R.