Question

Air, considered an ideal gas, is contained in an insulated piston-cylinder assembly outfitted with a paddle wheel. It is initially at p1 = 10 psi, T1 = 600°F, V1 = 1 ft. The paddle wheel transfers 3 Btu (by work) to the air to a final state of P2 = 5 psi, V2 = 3 ft. You may neglect potential and kinetic energy changes. Mair = 28.97 lbm/bmol. Find the mass of air in the closed chamber [lbm), the temperature at state 2 [OR], and the work done by the air to the piston (Btu).

Answer 1

Our data are,

State 1:

`P_1= 10psi=68.95kPa\\V_1 = 1ft^3=0.02831m^3\\T_1 = 100\°F = 310.93K`

State 2:

`P_2 =5psi=34.474kPa\\V_2 = 3ft^3=0.0899m^3`

We know as well that
`3BTU=3.16kJ/K`

To find the mass we apply the ideal gas formula, which is given by

`P_1V_1=mRT_1`

Re-arrange for m,

`m= (P_1V_1)/(RT_1)\\m= (68.95*0.02831)/((0.287)310.9)\\m=0.021893kg=0.04806lbm\\`

Because of the pressure, temperature and volume ratio of state 1 and 2, we have to

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

Replacing,

`T_2 = (P_2V_2)/(P_1V_1)T_1\\T_2 =(34.474*0.0844)/(68.95*0.02831)*310.93\\T_2 = 464.217K=375.5\°F`

For conservative energy we have, (Cv = 0.718)

`W = m C_v = 0.718  \Delta T +dw\\dw = W - mv\Delta T\\dw = 3.16-(0.0218*0.718)(454.127-310.93)\\dw = 0.765kJ=0.72BTU`