HW10 - Homework 10 Solution #1. Solution 1) Stainless steel...

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Homework 10 Solution #1. Solution 1) Stainless steel furnace At CE=394, C_P=F_P x F_M x C_B C_B=exp(0.08505+0.766 lnQ), where Q in BTU/hr 95MW -> 3.242x10^8 C_B=exp(0.08505+0.766 ln(3.242x10^8))= $ 3.599x10^6 F_P=0.986-0.0035(P/500)+0.0175(P/500)^2, where P in psig : 1000 psig F_P=0.986-0.0035(1000/500)+0.0175(1000/500)^2=1.049 F_M=1.7 for S.S. Therefore, C_P=(1.049)(1.7)(3.599x10^6)(528/394)=$ 8.6 x 10^6 = $ 8.6 million 2) Vertical pressure vessel, SA-387B low-alloy steel Di=7 feet, shell length=39 feet, 100 psig, 750 F C_P=F_M x Cv + C_PL at CE=394 Cv = exp(6.775+.18255 ln(w)+0.02297 ln(w)^2), where w in pounds w = pi x (Di+ts) x (L+0.8 Di) x ts x rho rho = 0.284 lb/in^3 ts = (P_d x Di)/(2 x S x E - 1.2 P_d) P_d = exp(0.60608+ .91615 ln(Po) + .0015655 ln(Po)^2)= exp(0.60608+ .91615 ln(100) + .0015655 ln(100)^2) = 128.8 psi
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For SA-387 B, at operating T of 750 F --> Use design T of 800 F ==> S = 14750 psi (P.529) Assume < 1.25 in thick, use E = 0.85 (check ts afterwards) ts=(P_d x Di)/(2 x S x E - 1.2 P_d) = ((128.8)x(7x12))/(2 x (14750) x 0.85 - 1.2 x
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This note was uploaded on 05/01/2011 for the course CHBE 2120 taught by Professor Gallivan during the Spring '07 term at Georgia Institute of Technology.

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HW10 - Homework 10 Solution #1. Solution 1) Stainless steel...

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