ME476_HW6.1

# ME476_HW6.1 - HOMEWORK #6 5.11 and 5.13 Helpful hints: use...

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Ψ T 1 ( 29 Ψ 325K ( ) = 70 J kgair K = Ψ T1 70 J kgair K = Ψ T2 Ψ T1 nuR bar ln 1 r c - = Ψ T2 832.688 J kgair K = From Figure 4.4 T 2 820K = T1 and T2 from Figure 4-3 gives u s.1 20 kJ kgair = u s.2 440 kJ kgair = Compression Work W c1.2 u s.2 u s.1 - 420000 J kg = To determine P2, use ideal gas equation P 2 nuR bar T 2 v 2 = need v.2, must find v.1 first v 1 nuR bar T 1 P 1 = v 1 0.47 m 3 kgair = v 2 v 1 r c = P 2 nuR bar T 2 v 2 = P 2 7065kPa = HOMEWORK #6 5.11 and 5.13 Helpful hints: use Cutoff ratio = 2.18Note that State 4 is at low temperatures (not on Fig 4-5), use isentropic relationships w/ constant avg. specific heats to find T4, P4 using gamma=1.3. To find u4, must use Fig 4-10 with Temperature to find us,4. Relate by U4=us,4 – 1320 kj/kg airFor part c, use Eq 5-17 kPa 10 3 Pa = kgair kg = kJ 10 3 J = MJ 10 6 J = 5.11 - Constant Pressure Air-Fuel Cycle. Initial Conditions P 1 200kPa = T 1 325K = φ 0.4 = x b 0.025 = r c 14 = β 2.18 = γ 1.3 = a. Determine temperature and pressure at states 2,3,4 and 5

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## This note was uploaded on 01/25/2011 for the course ME 476 taught by Professor Broch during the Spring '10 term at Nevada.

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ME476_HW6.1 - HOMEWORK #6 5.11 and 5.13 Helpful hints: use...

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