Air in a piston/cylinder assembly at 300 kPa and 500 K is expanded in a constant

Question

Air in a piston/cylinder assembly at 300 kPa and 500 K is expanded in a constant-pressure process to twice the original volume, state 2, as shown in the figure.  The piston is then locked with a pin, and heat is transferred to a final temperature of 500 K.  Find P, T, and h for states 2, 3, and find the wok and heat transfer in both processes.

Air in positon/cylinder assembly at 300kpa and 500 k is expanded in a constant pressure process to twice the original volume,state 2, as shown in the figure.The piston is then locked with a pin, and heat is transferred to a final temarature of 500 k.Find P,T,and h for states 2,3, and find the work and heat transfer in both processes.

Explanation / Answer

state 1:
Pressure 1= 300 Kpa
Temperature 1=500K
Volume1 =V1
Volume1 =
For ideal gas (air) : P1*V1= m*R*T1
or : P1*v1=R*T1
v1= R*T1/P1
for air R =8314/29=286.69 j/Kg.K
v1= 286.689*500/(300*1000)=0.4778 m3/kg

State 2:
Pressure 2 = Pressure 1 =300Kpa
Volume 2= 2 *Volume1 = 2*V1
Volume 2= 2 *Volume1 = 0.955 m3/kg

For ideal gas (air) : P1*V1= m*R*T1 , P2*V2= m*R*T2
as pressure is constant :
P*V1= m*R*T1 , P*V2= m*R*T2
(Temperature 2)/(Temperature 1) =(Volume 2)/(Volume 1)
(Temperature 2)/(Temperature 1) =(2*Volume 1)/(Volume 1)
(Temperature 2)/(Temperature 1) =2
(Temperature 2) =2*(Temperature 1)=1000K

work from state1 to state 2 (work at constant pressure):
Work = P(v2-v1) = (300*1000)*(0.955-0.477)=143344.8276J/kg
Work = 143.344KJ/kg

internal energy change from state1 to state 2 (work at constant pressure):
?U= Cp*(T2-T1)
Cp= K*R /(k-1)where k=1.4 for air as ideal gas
Cp= K*R/(k-1)
Cp= 1.4*286.689/(1.4-1)=1003.41 j/Kg.K
Cp= 1.0034115Kj/Kg.K
U=1.0034115*(1000-500)=501.705 Kj/Kg
Q=W+?U
Q=143.344+501.705=645.050 Kj/Kg


State 3:
Temperature 3 =500K
For ideal gas (air) : P3*V3= m*R*T3 , P2*V2= m*R*T2
as volume is constant :
V3=V2=0.95563 m3/kg
P3*V= m*R*T3 , P2*V2= m*R*T2
(Temperature 3)/(Temperature 2) =(P3)/(P2)
(500)/(1000) =(P3)/(300)
P3= (300)*(500)/(1000)=150 Kpa

work from state2 to state 3 (work at constant volume):
Work = v2*(P3-P2) = (0.955)*(150000-300000)=-143344.8276j/Kg
Work = -143.344 Kj/Kg

internal energy change from state1 to state 2 (work at constant Volume):
?U= Cv*(T3-T2)
Cv= R/(k-1) where k=1.4 for air as ideal gas
Cv= 286.689/(1.4-1)=716.7225j/Kg.K
Cv= 0.7167225Kj/Kg.K
?U=0.7167225*(500-1000)=-358.36125Kj/Kg
Q=W+?U
Q=-143.344+(-358.316)=-501.7060776Kj/Kg

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