During a reversible adiabatic process of 1 lbm of an ideal gas in a piston-cylin

Question

During a reversible adiabatic process of 1 lbm of an ideal gas in a piston-cylinder, the pressure decreases from 196.8 psi to 15.8 psi. The gas constant is 37.7 ft.lb/lbm. degree R and the initial temperature of the gas is 863 degree F. Determine the initial and final gas density in [lbm/ft^3], the total work produced in [ft.lb], and the total change in internal energy in [Btu]. Assume that the specific heat at constant pressure is c_p = 0.22 Btu/lbm. degree R. Rho_1 = 0.568 lbm/ft^3, rho_2 = 0.080 lbm/ft^3 W= -75,210 ft. lb DeltaU= -75,210 ft. lb

Explanation / Answer

P1=196.8psi=196.8psi*144 lb/ft2 /psi=29339.2 lb/ft2

P2=15.8psi=15.8 psi*144=2275.2 lb/ft2

T1=863oF

T2=?

R=37.7 lb.ft/lbm.oR=37.7 *0.00128 btu /lbm.R=0.048 btu/lbm.R

T(oR)=T(oF)+459.67=863+459.67=1322.67 oR

For iso entropic process,

Y=cp/cv=cp/cp-R

cp=0.22  btu/lbm oR

R=0.048 btu /R.lb.mol

Y=0.22/0.22-0.048/=0.22/0.17=1.3

T2/T1=(P2/P1)^y-1/y

T2=T1*(P2/P1)^y-1/y=1322.67 oR  * (2275.2 /29339.2)^1.3-1/1.3= 1322.67 *(0.077)^0.23=733.4 oR

V1=mRT1/P1=1lbm* 37.7 ft.lb/lbmR.*1322.67 oR/29339.2 lb/ft2=1.699 ft3

V2=mRT2/P2=1 lbm*37.7 lb.ft/lbm.oR**733.4 oR /2275.2 lb/ft2=12.15ft3

Density (initial)=mass/v1=1 lbm/1.699 ft3=0.588 lbm/ft3

Density (final)=1 lbm/12.15 ft3=0.082lbm/ft3

Work produced=W=mCv* dT

dT=change of temperature

dW=dU=mCv* dT=1lbm*0.17 lb.ft /lbm R*(-589.3)R=-100.18 ft.lb

W= -100.18 ft.lb

U=-100.18*0.00128=0.128 btu

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