SolHW6ME2320 - pressure of 200 psia, Since the system is a...

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3 – 29 3 – 33 Given : R134a. 1 kg; V 1 = 0.14 m 3 ; T 1 = -26.4 o C T 2 = 100 o C Find : V 2 Solution: For state 1, at T 1 = -26.4 o C and v 1 = 0.14 m 3 /kg the system is saturated vapor – saturated liquid mixture with a initial pressure P 1 = 99.9 kPa. Since the system is a piston – cylinder device the pressure during the process remains constant (P 1 = P 2 ). For the final state P 2 = 99.9 kPa and T 2 = 100 o C, the system is in the superheated region with a specific volume of v 2 = 0.3017 m 3 /kg. Therefore, the final volume of the piston – cylinder system will be V 2 = 0.3017 m 3 2/8
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3 – 35 Given : Water, m = 1 lbm; V 1 = 2.3615 ft 3 ; T 1 = 400 o F T 2 = 100 o F Find: P 2 and V 2 . Solution: For the initial state: v 1 = 2.3615 ft 3 /lbm, T 1 = 400 o F This results in a system whose state is superheated vapor at a
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Unformatted text preview: pressure of 200 psia, Since the system is a piston cylinder device the final pressure will be the same as the initial pressure, thus P 2 = 200 psia and also T 2 = 100 o F. With these two properties is possible to determine the final specific volume, which is v 2 = 0.0161 ft 3 /lbm. Therefore the final volume of the system is V 2 = 0.0161 ft 3 3 35 Given : Water, m = 1 lbm; V = 2 ft 3 ; P = 100 psia Find: U and H. Solution: For v = 2 ft 3 /lbm, P = 100 psia, the state of the system is saturated vaporsaturated liquid mixture. Thus h = 697.6 Btu/lbm and u = 660.6 Btu/lbm, which results in a total enthalpy and internal energy of H = 697.6 Btu U = 660.6 Btu 3/8 3 53 4/8 3 58 5/8 3 62 6/8 7/8 8/8...
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This note was uploaded on 04/07/2008 for the course ME 232 taught by Professor Monefort during the Spring '08 term at Western Michigan.

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SolHW6ME2320 - pressure of 200 psia, Since the system is a...

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