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Unformatted text preview: kJ/kg.K] Figure 2: T-s diagram for R245fa. 7 Formuleblad Thermodynamica 2, versie 18-4-2011 Ideale Gaswet pV nRT en dus pv RT == Constantes R = = 8.314472 J mol −1 K −1 N AV 6.022141023 moleculen/mol 1ste HW gesloten system d= δ Q − δ W U Reversibele arbeid W = V2 ∫ pdV V1 dECV c2 c2 = QCV − WCV − m2 h2 + 2 + gz2 + m1 h1 + 1 + gz1 . Hierin wordt de 2 2 dt uitgaande stroom aangeduid met “2” en de ingaande stroom met “1”. 1ste HW open systeem δQ 2de HW gesloten systeem dS = met δ S ≥ 0 +δ S T dS QCV 2de HW open system CV = trans − m2 s2 + m1s1 + σ CV met σ CV ≥ 0 waarbij het warmtetransport dt T plaatsvindt bij temperatuur T trans Thermodynamische potentialen H =U + pV F =U − TS G =U − TS + pV Veranderingen van U, H, F, G dU = TdS − pdV + µ dN dH = TdS + Vdp + µ dN dF = − pdV + µ dN dG = + Vdp + µ dN − SdT − SdT Partiële afgeleiden voor f(x,y) ∂f ∂f ∂f df = dx + dy = + Ndy dus M = Mdx ∂x y ∂x y ∂y x ∂f en N = ∂y x Maxwell relatie voor f(x,y) ∂M ∂N ∂ ∂f ∂ ∂f Mdx + Ndy dan Indien df = = oftewel = ∂x ∂y x y ∂y ∂x y x ∂y x ∂x y Integreren x2 ∂f f= f ( x1 , y1 ) + ∫ dx x2 , y1 ) ( ∂x y x1 Min 1 regel ∂x ∂z ∂y = −1 ∂y z ∂x y ∂z x 8 Definitie warmtecapaciteit ∂U ∂S CV = = T ∂T V , N ∂T V , N ∂H ∂S = = T en C p ∂T P , N ∂T P , N Departure Gibbs energy p f µ − µ IG Z ( p) − 1 ln= ln ϕ = =∫ dp p RT p 0 Clapeyron equation sg − s f hg − h f dp = = dT sat vg − v f T ( vg − v f ) Clausius-Clapeyron equation hg − h f d ln P = RT 2 dT sat Carnot efficiency T ηCarnot = 1 − L TH Thermal power cycle efficiency (I-law efficiency) W ηth = net Q in Isentropic process for an ideal gas γ −1 v T2 P2 γ = = 1 T1 P v2 1 γ −1 = met γ CP Cv Entropy change of an ideal gas with temperature and pressure and constant cp s (T2 , p2 ) − s (T1 , p1 ) =c p ln T2 p − R ln 2 T1 p1 Isentropic efficiencies for compression and expansion h −h h1 − h2 = = 2,is 1 ηis ,exp , ηis ,compr h1 − h2,is h2 − h1 9 COP (coefficient of performance) QC TC QH TH , COPheating = COPcooling = = = QH − QC TH − TC QH − QC TH − TC Exergy: specific flow exergy and balances v2 ex f =h − h0 + + gz − T0 ( s − s0 ) 2 T dEx = Ex f ,1 − Ex f ,2 + (h0 − T0 s0 )( M 1 − M 2 ) + 1 − 0 Q + W + Exd dt Tb T M ( ex f ,1 − ex f ,2 ) =1 − 0 Q + W + Exd − Tb 10 (9) (yjlCe =j 4' / ^^^) ^3 / := (/ 1000 C r .67 -Vm ^^^^^ ^ "" ''...
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