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G Compounding two scroll expanders in series. =h1–h3=h1–h3 ;1=h1–h2=h1–h2 ;2=h2–h3=h2–h3 (118) (119) Rearranging Eq 118, the overall efficiency can be expressed as a function of h1 – h3s ⋅h1 h1 – h2s 1⋅h1 h2 – h3ss Where the factors are defined by: the individual expander efficiencies and of the factors: = 1⋅ 1 1⋅ 2 − 1⋅ 2⋅ 1⋅ 2 2⋅h2 1− =1–r i p,i (120) This efficiency is the one to be maximized. To that end, the derivatives of each factor on the right side of the equation must be calculated. The derivative of 1 is straightforward and is given by: drp,1 p,1 Recognizing that r p=r p , 1⋅r p , 2 , the derivative of 2 manner: with: d drp,1 (123) (124) (125) 1 = ∂rp,1 rp,1 i=1 j=0 The derivative of 2 can then be calculated by: d r ∂ [−1]∂ d –1 [1−2⋅] 1 =[ ]⋅r (121) is calculated in that same (122) –1 [1−] [−1] 2 =[ ]⋅r ⋅r d drp,1 p p,1 As shown in section 4.2.1 , the individual efficiencies can be expressed as a function of the pressure ratios and of the supply density in the form of: n−1 n−1 1=∑∑aij⋅lnrp,1i⋅1jan0⋅lnrp,1na0n⋅n1=frp,1 ,1 i=0 j=0 n−1 n−1 2=∑∑aij⋅lnrp,2i⋅2jan0⋅lnrp,2na0n⋅n2=frp,2,2 i=0 j=0 1 being independent of r p ,1 , the derivative of 1 is straightforward: ∂ n−1 n−1 1 = 1 ∑∑aij⋅i⋅lnr p,1i−1⋅1jan0⋅n⋅lnr p ,1n−1 On the contrary, for 2 , rp,2 and 2 both depend on rp,1 . In good approximation, the density 2 can be assimilated to the density at the same pressure calculated as if the first expansion was isentropic: ≃⋅r[−1] 2 1 p,1 (126) (127) 2 =− p 2 − su,exp⋅r ⋅ 2 drp,1 r2 ∂rp,2 p,1 ∂ p,1 XXVIPDF Image | Organic Rankine Cycles for Waste Heat Recovery and Solar Uses
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