Question

Air at 25°C and 5 atm is throttled by a valve to 1 atm. If the valve is adiabatic and the change in kinetic energy is negligible, determine the exit temperature of air. Solve using appropriate software.

Answer 1

Answer:

$T_{2} = 25^{\circ}C$

Solution:

As per the question:

Temperature, $T_{1} = 25^{\circ}C$

Pressure, $P_{1} = 5 atm[\tex]Now,We know that the process of throttling occurs in adiabatic conditions where no heat is transferred in between the system and the surrounding, i.e., Q = 0Thus no work is done in this process, i.e., W = 0Enthalpy also remains same from one state to the other state, i.e., [tex]\Delta h = 0$

Therefore, from the eqn:

$Q - W = \Delta h + \Delta KE$

We can write:

$\Delta h = h_{1} - h_{2} = 0$

$\Delta h = C_{p}\Delta T$

$C_{p}(T_{2} - T_{1}) = 0$

Thus

$T_{1} = T_{2} = 25^{\circ}C$

where

$T_{2}$ = Exit temperature