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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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 Spring '09
 KATZ
 argon, P1 ideal gas

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