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Unformatted text preview: ASSIGNMENT 6 6.35) Assume a common 105 yr. life and use the AW method (PW could be used if 7 life cycles of Boiler A and 3 life cycles of Boiler B are explicitly considered over a 105 year study period.) AWA(10%) = −$75,000(A/P,10%,15) + $15,000(A/F,10%,15) − $10,000 = −$19,388.42 AWB(10%) = −$150,000(A/P,10%,35) + $25,000(A/F,10%,35) − $4,000 − $350(A/G,10%,35) = −$8,511.11 or PWA(10%) = −$193,875.46 over 105 years; PWB(10%) = −$85,107.26 over 105 years. Choose B. 638 AW2 cm(15%) = −$20,000(A/P, 15%, 4) + $5,000 = −$2,006 per year AW5 cm(15%) = −$40,000(A/P, 15%, 6) + $7,500 = −$3,068 per year Therefore, the 2 cm thickness should be recommended. 640 Assume repeatability. (a) Machine A: CR cost/yr = $40,000(A/P, 15%, 10) − $4,000(A/F,15%,10) = $7,774.8 Maint. Cost/yr = $1,500 CR and maintenance cost/part = $g,ggG.¡¢$£,¤¥¥ £¥,¥¥¥ = $0.9275 Labor cost/part = $£¡ /¦§ ¨ ©ª§«¬/¦§ = $6 Total cost/part (to nearest cent) = $6.9275 Machine B:...
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 Spring '12
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