Question

Helium gas is compressed by an adiabatic compressor from an initial state of 14 psia and 50°F to a final temperature of 320°F in a reversible manner. Determine the exit pressure of helium.

Answer 1

Answer:

The value is  $P_2 = 40.54 \ psla$

Explanation:

From the question we are told that

 The initial pressure is  $P_1 = 14\ psla$

  The initial temperature is  $T_1 = 50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283 \ K$

   The final temperature is  $T_2 = 320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433 \ K$

Generally the equation for adiabatic process is mathematically represented as

         $PT^{\frac{\gamma}{1- \gamma} } = Constant$

=>      $P_1T_1^{\frac{\gamma}{1- \gamma} } = P_2T_2^{\frac{\gamma}{1- \gamma} }$

Generally for a monoatomic gas  $\gamma = \frac{5}{3}$

So

           $14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }$

=>       $14 * 283^{-2.5} =P_2 * 433^{-2.5}$

=>       $P_2 = 40.54 \ psla$