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

Question #278453

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.

1
Expert's answer
2021-12-16T16:18:27-0500

Solution;

Given;

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

P1=300atm=303.975MPaP_1=300atm=303.975MPa

T1=450°R=222.222KT_1=450°R=222.222K

P2=30atm=303.975×104PaP_2=30atm=303.975×10^4Pa

Ideal gas;

T2T1=(P2P1)k1k\frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{k-1}{k}}

For air,k=1.4

T2=222.222(303.975×104303.975×106)0.41.4=59.615KT_2=222.222(\frac{303.975×10^4}{303.975×10^6})^{\frac{0.4}{1.4}}=59.615K

Work;

W=mcv(T2T1)W=mc_v(T_2-T_1)

W=0.454×718(222.22259.615)=530005.2J/s=53.00kWW=0.454×718(222.222-59.615)=530005.2J/s=53.00kW






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