Answer to Question #278453 in Molecular Physics | Thermodynamics for Neilmar

Question

A gas turbine receives 1 lb/sec air at 300 atm, 450°R and expands it reversibly to 30 atm in an adiabatic steady flow manner; ∆P = 0; ∆K = 0. (a) Use the deviation charts and find the exit air temperature and the work. (b) Solve using ideal gas theory and compare with foregoing answers.

Answer 1

Solution;

Given;

$\dot{m}=1lb/s=0.454kg/s$

$P_1=300atm=303.975MPa$

$T_1=450°R=222.222K$

$P_2=30atm=303.975×10^4Pa$

Ideal gas;

$\frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{k-1}{k}}$

For air,k=1.4

$T_2=222.222(\frac{303.975×10^4}{303.975×10^6})^{\frac{0.4}{1.4}}=59.615K$

Work;

$W=mc_v(T_2-T_1)$

$W=0.454×718(222.222-59.615)=530005.2J/s=53.00kW$