Contemporary Engineering Economics, 4 th edition, © 2007 Life-Cycle Cost Analysis Lecture No.22 Chapter 6 Contemporary Engineering Economics Copyright © 2006

Contemporary Engineering Economics, 4 th edition, © 2007 Why Life-Cycle Cost (LCC) Analysis? To select from among design alternatives that fulfill the same performance requirements, but differ with respect to initial costs and operating costs To predict the most cost-effective solution

Contemporary Engineering Economics, 4 th edition, © 2007 Stages of Life-Cycle Cost

Contemporary Engineering Economics, 4 th edition, © 2007 Sketch of a Pumping System in Which the Control Valve Fails

Contemporary Engineering Economics, 4 th edition, © 2007 Engineering Solution Alternatives Option A: A new control valve can be installed to accommodate the high pressure differential. Option B: The pump impeller can be trimmed so that the pump does not develop as much head, resulting in a lower pressure drop across the current valve. Option C: A variable frequency drive (VFD) can be installed, and the flow control valve removed. The VFD can vary the pump speed and thus achieve the desired process flow. Option D: The system can be left as it is, with a yearly repair of the flow control valve to be expected.

Contemporary Engineering Economics, 4 th edition, © 2007 Life-Cycle Cost Elements

Contemporary Engineering Economics, 4 th edition, © 2007 Cost Comparison for Options A Through D

Contemporary Engineering Economics, 4 th edition, © 2007 Sample LCC Calculation for Option A

Contemporary Engineering Economics, 4 th edition, © 2007 Comparison of LCC for Option A - D

Contemporary Engineering Economics, 4 th edition, © 2007 StandardPremium MotorEfficient Motor25 HP $13,000$15,60020 Years$0 89.5%93%$0.07/kWh3,120 hrs/yr. Size Cost Life Salvage Efficiency Energy Cost Operating Hours (a) At i= 13%, determine the operating cost per kWh for each motor. (b) At what operating hours are they equivalent? Life-Cycle Cost Analysis – Standard Motor versus Premium Efficiency Motor

Contemporary Engineering Economics, 4 th edition, © 2007 Solution: (a):  Operating cost per kWh per unit Determine total input power  Conventional motor: input power = kW/ = kW  PE motor: input power = kW/ 0.93 = kW

Contemporary Engineering Economics, 4 th edition, © 2007  Determine total kWh per year with 3120 hours of operation  Conventional motor: 3120 hrs/yr ( kW) = 65,018 kWh/yr  PE motor: 3120 hrs/yr ( kW) = 62,568 kWh/yr  Determine annual energy costs at $0.07/kwh:  Conventional motor: $0.07/kwh  65,018 kwh/yr = $4,551/yr  PE motor: $0.07/kwh  62,568 kwh/yr = $4,380/yr

Contemporary Engineering Economics, 4 th edition, © 2007  Capital cost:  Conventional motor : $13,000(A/P, 13%, 12) = $1,851  PE motor : $15,600(A/P, 13%, 12) = $2,221  Total annual equivalent cost:  Conventional motor: AE(13%) = $4,551 + $1,851 = $6,402 Cost per kwh = $6,402/58,188 kwh = $0.11/kwh  PE motor: AE(13%) = $4,380 + $2,221 = $6,601 Cost per kwh = $6,601/58,188 kwh = $0.1134/kwh

Contemporary Engineering Economics, 4 th edition, © 2007 (b) break-even Operating Hours = 6,742