Assignment%206

# Assignment%206 - ASSIGNMENT 6 6.35 Assume a common 105 yr...

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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. 6-38 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. 6-40 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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Assignment%206 - ASSIGNMENT 6 6.35 Assume a common 105 yr...

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