SolHW7ME2320

# SolHW7ME2320 - ∆ m = m 1 – m 3 ∆ m = 7.0136 × 10-3...

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3 – 77 3 – 79. Given : P 1 = 310 kPa; T 1 = 298 K; T 2 = 323 K; V = 0.025 m 3 . T 3 = 50 o C; P 3 = P 1 ; V b constant Find : P 2 ( P) and ( m) Solution: P 1 /T 1 = P 2 /T 2 b P 2 = (P 1 /T 1 ) T 2 . P 2 =336 kPa . Thus P = 26 kPa Now: T 3 = 50 o C; P 3 = P 2 ; V = 0.025 m 3 . P V = mRT, then m 1 = P 1 V /RT 1 = (310)(0.025)/[(0.287)(298)] m 1 = 0.09062 kg m 2 = m 1 and P 3 = P 1 ; Then, m 3 = P 3 V /RT 3 = (310)(0.025)/[(0.287)(323)] m 3 = 0.0836 kg;

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Unformatted text preview: ∆ m = m 1 – m 3 ∆ m = 7.0136 × 10-3 kg; P 2 =336 kPa . ∆ P = 26 kPa 3 – 83. Given : He; m = 0.1 kg; V = 0.2 m 3 . P 1 = 350 kPa; P 2 = 700 KPa; Find : ∆ T Solution: T = P v /R T 2 – T 1 = v (P 2 – P 1 )/R T 2 – T 1 =(0.2)(700 – 350) ∆ T = 337.04 K = 337.04 o C...
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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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SolHW7ME2320 - ∆ m = m 1 – m 3 ∆ m = 7.0136 × 10-3...

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