A closed cylinder is divided into two rooms by a frictionless piston held in pla

Question

A closed cylinder is divided into two rooms by a frictionless piston held in place by a pin, as shown in Fig. P5.138. Room A has 0.3 ft3 air at 14.7 lbf/in.2, 90 F, and room B has 10 ft3 saturated water vapor at 90 F. The pin is pulled, releasing the piston and both rooms come to equilibrium at 90 F. Considering a control mass of the air and water, determine the work done by the system and the heat transfer to the cylinder.

Explanation / Answer

consider A + B, Energy equation: mA(u2 – u1)A + mB(uB2 – uB1) = 1Q2 – 1W2 Process equation: V = C ? 1W2 = 0 T = C ? (u2 – u1)A = 0 (ideal gas) The pressure on both sides of the piston must be the same at state 2. Since two-phase: P2 = Pg H2O at 90 F = PA2 = PB2 = 4.246 kPa Air, I.G.: PA1VA1 = mARAT = PA2VA2 = Pg H2O at 90 F VA2 ? VA2 = 14.7 × 0.30/.6988 = 6.31 ft3 Now the water volume is the rest of the total volume VB2 = VA1 + VB1 - VA2 = 0.30 + 10 - 6.31 = 3.99 ft3 mB = VB1/vB1 = 10/467.7 = 0.02138 lbm ? vB2 = 186.6 ft3/lbm 186.6 = 0.016099 + xB2 × (467.7 - 0.016) => xB2 = 0.39895 uB2 = 58.07 + 0.39895 × 982.2 = 449.9 Btu/lbm; uB1 = 1040.2 1Q2 = mB(uB2 – uB1) = 0.02138 (449.9 - 1040.2) = -12.6 Btu

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