An insulated cylinder is divided into two parts of 10 ft^3 each by an initially locked piston. Side A has air at 2 atm, 600 R and side B has air at 10 atm, 2000 R as shown in Fig. P3.151. The piston is now unlocked so it is free to move, and it conducts heat so the air comes to a uniform

Question

An insulated cylinder is divided into two parts of 10 [latex]ft ^{3}[/latex] each by an initially locked piston. Side A has air at 2 atm, 600 R and side B has air at 10 atm, 2000 R as shown in Fig. P3.151. The piston is now unlocked so it is free to move, and it conducts heat so the air comes to a uniform temperature [latex]T_{ A }=T_{ B }[/latex]. Find the mass in both A and B and also the final T and P.

Accepted Answer

C.V. A + B . Then [latex]{ }_{1} Q _{2}=\emptyset,{ }_{1} W _{2}=0[/latex].

Force balance on piston: [latex]P _{ A } A = P _{ B } A[/latex], so final state in A and B is the same.

State 1A: [latex]u _{ A 1}=102.457 Btu / lbm[/latex];

[latex]m _{ A }=\frac{ PV }{ RT }=\frac{29.4 psi imes 10 ft ^{3} imes 144 in ^{2} / ft ^{2}}{53.34 lbf - ft / lbm - R imes 600 R }=1.323 lbm[/latex]

State 1B: [latex]u _{ B 1}=367.642 Btu / lbm[/latex];

[latex]m_{B}=\frac{P V}{R T}=\frac{147 psi imes 10 ft ^{3} imes 144 in ^{2} / ft ^{2}}{53.34 lbf - ft / lbm - R imes 2000 R }=1.984 lbm[/latex]

For chosen C.V. [latex]{ }_{1} Q _{2}=0,{ }_{1} W _{2}=0[/latex] so the energy equation becomes

[latex]m _{ A }\left( u _{2}- u _{1} ight)_{ A }+ m _{ B }\left( u _{2}- u _{1} ight)_{ B }=0[/latex]

[latex]\left(m_{A}+m_{B} ight) u_{2}=m_{A} u_{A 1}+m_{B} u_{B 1}[/latex]

= 1.323 × 102.457 + 1.984 × 367.642 = 864.95 Btu

[latex]u _{2}=864.95 Btu / 3.307 lbm =261.55 Btu / lbm \Rightarrow T _{2}=1475 R[/latex]

[latex]P = m _{ tot } RT _{2} / V _{ tot }=\frac{3.307 lbm imes 53.34 lbf - ft / lbm - R imes 1475 R }{20 ft ^{3} imes 144 in ^{2} / ft ^{2}}[/latex]

[latex]=90.34 lbf / in ^{2}[/latex]

Question 3.306E: An insulated cylinder is divided into two parts of 10 ft^3 e...

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