index(7)

index(7) - Typical Control Mass Problem Chart 2 (Not...

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Unformatted text preview: Typical Control Mass Problem Chart 2 (Not complete material coverage) i State 1 Normally 3 variables are needed to fully determine it. · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 2 known intensive variables (May get away with T only in some approximations, eg, when using saturated values as an approximation for compressed liquids.) ? Tables, eg B.1.1-B.1.4; pv or T v diagrams; v = vf + x (vg − vf ) and similar for u, h, and s. ? Ideal gas (applicable?) pv = RT (4 forms) Tables A5-A8 for u, h, s. ? Make do with the formulae for differences in intensive variables listed below under “Process”? ? Remaining intensive variables p, T, v, (x, )u, h, s. · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · Amount of material: v = V /m u = U/m h = H/m s = S/m Process C1: Type of process (V = C, p = C, p linear in V , pV n = C, T = C, 1 Q2 =0, reversible?) C2: Mass: Energy: C3: 1 W2 1 Q2 m1 (+madded ) = m2 E2 − E1 = 1 Q2 − 1 W2 p(V2 − V1 ) p1 + p2 (V2 − V1 ) 2 T (S2 − S1 ) (E = U + KE? + P E?) p2 V 2 − p1 V 1 1−n 1 S2,gen =0 p1 V1 ln V2 V1 other? 1 Q2 Tsurr = [0 and S2 = S1 ] other? = ∆Snet = S2 − S1 − For ideal gases: 2 2 u2 − u1 = 1 Cv dT ≈ Cv ave (T2 − T1 ) p2 p1 n n−1 h2 − h1 = 1 Cp dT ≈ Cp ave (T2 − T1 ) − R ln p2 p1 = ... n=1 ? n=k s2 − s1 = s0 (T2 ) − s0 (T1 ) − R ln T T p2 = Polytropic: p1 2 1 Q2 ≈ Cp ave ln T2 T1 v1 v2 n = T2 T1 isothermal: isentropic and k constant: For solids and compressed liquids, by approximation, best at constant pressure: =m 1 C(p) dT ≈ mC(p) ave (T2 − T1 ) i State 2 i Same procedures as state 1 s2 − s1 ≈ C(p) ave ln T2 T1 ?? ? ...
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This note was uploaded on 07/09/2011 for the course EGN 5456 taught by Professor Dommelen during the Spring '09 term at FSU.

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