Soln_Fall2008_HW3

# Soln_Fall2008_HW3 - state points The work is the area below...

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Solution 3 1 Take C.V. as all the argon in both A and B. Boundary movement work done in cylinder B against constant external pressure of 125 kPa. Argon is an ideal gas, so write out that the mass and temperature at state 1 and 2 are the same PA1VA = P2( V2) => V2 = W2 = ⌡⌠ 1 2 PextdV = Pext(VB2 - VB1) = 112.5 kPa (0.8 - 0.4) m3 = 45 kJ 2 State 1 from B.1.2 (P, x): v1 = vg = 0.8857 m3/kg (also in B.1.3) State 2 from B.1.3 (P, T): v2 = 1.0803 m3/kg Since the mass and the cross sectional area is the same we get h2 = v2/v1 × h1 = 1.0803/0.8857 × 0.1 = 0.122 m Process: P = C so the work integral is W = PdV = P(V2 - V1) = PA (h2 - h1) W = 200 kPa × 0.2 m2 × (0.122 − 0.1) m = 0.88 kJ 3 Ideal gas PV = mRT State 1: T1, P1 ideal gas so P1V1 = mRT1 V1 = mR T1 / P1 = 2 • 0.287 • 293.15/250 = 0.6731 m3 State 2: T2, P2 = P1 and ideal gas so P2V2 = mRT2 V2 = mR T2 / P2 = 2 • 0.287 • 500/250 = 1.148 m3 W 2 = ⌡⌠ PdV = P (V2 - V1) = 250 (1.148 – 0.6731) = 118.7 kJ 4 The setup has a pressure that varies linear with volume going through the initial and the final

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Unformatted text preview: state points. The work is the area below the process curve. W = ⌡⌠ PdV = AREA = 1 / 2 (P2 + P1)(V2 - V1) = 1 / 2 (P2 + 0)( V2 - 0) = 1 / 2 P2 V2 = 1 / 2 × 500 × 0.015 = 3.75 kJ 5 The process is polytropic with exponent n = -1/3. P1 = C2 V1/3 = 120 × 1^1/3 = 120 kPa P2 = C2 V1/3 = 120 × 4^1/3 = 190.4 kPa W 2 = ∫ P dV = ( P 2 V 2 - P 1 V 1 )/( 1 – n ) ( Equation 4.4) = (158.74× 4 - 100 × 1) / ( 1 - (-1/3)) = 481.44 kJ m 2 = (P2V2) /(RT2) = ( 158.74 × 3)/( 0.287 × 298) = 8.91 kg 6 State 1: P1 = 120 kPa, T1 = 400°C = 673.2 K State 2: T2 = T0 = 20°C = 293.2 K For all states air behave as an ideal gas. a) If piston at stops at 2, V2 = V1/2 and pressure less than Plift = P1 ⇒ P2 = P1 × V1 × T2/(T1*V2) = 120 × 2 × 293.2/673.2 = 104.5 kPa < P1 ⇒ Piston is resting on stops at state 2. b) Work done while piston is moving at constant Pext = P1. W2 = ∫ Pext dV = P1 (V2 - V1) ; V2 = 1/2 V1 = 2 m RT1/P1 w2 = W2/m = RT1 (1/2 - 1 ) = -1/2 × 0.287 × 673.2 = -96.6 kJ/kg...
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Soln_Fall2008_HW3 - state points The work is the area below...

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