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# 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 (128.8))=0.434 in For 7 ft used, this is greater than min thickness 3/8 (P 530) Add corrosion allowance of 1/8 in = 0.434 + 0.125 = 0.549 in. Only comes in 1/8 in. increments above 1/2 in., so => 0.625 in. = ts Thus, w = pi x (Di+ts) x (L+0.8 Di) x ts x rho = pi x (7 x 12 + 0.625) x ( 39 x 12 + 0.8 x 7 x 12) x 0.625 x 0.284 = 25256 lb Once we know w, we can get Cv as following.
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HW10 - Homework 10 Solution#1 Solution 1 Stainless steel...

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