\"All industrial facilities have a network of piping that carries water or other
ID: 3282072 • Letter: #
Question
"All industrial facilities have a network of piping that carries water or other liquids. According to a U.S. Department of Energy study, 16% of a typical facility's electricity costs are for its pumping systems. Suppose that you are considering a pumping facility having 9,000 ft to carry 580 gpm of water continuously to storage tank. The general equation for estimating annual pumping costs is Cost($)- [(Friction factor)(Pipe length in ft)(Number of hours)(S/kWh)(Flow in gpm)3]/[1705 (Combined pump and motor efficiency)(Pipe inner diameter in inches)A5] where the friction factor is 0.0089. The electricity cost is $0.05/kWh, the pumping system operates for 9,240 hours annually, and the combined pump and motor efficiency is 0.72. The price for 9,000 feet of steel pipe is $8,200 per inch of the inner diameter. Determine the optimal size of the pipe's inner diameter in inches by minimizing the annual equivalent cost. Assume that the pumping operation will last 17 years, the salvage value of the pipe is 9% of the initial investment, and a discount rate of 12%."Explanation / Answer
Let the optimal inner dia of pipe be = d inches
Annual Cost($)= [(Friction factor)(Pipe length in ft)(Number of hours)($/kWh)(Flow in gpm)^3]/[1705*(Combined pump and motor efficiency)(Pipe inner diameter in inches)^5]
Annual Cost($), C= [0.0089*9000*9240*0.05*5803]/[1705*0.72*d5]
C= 60,497,307,870*d-5
Total Estimated Life = 17 years
Initial Investment = $8,200*d
Salvage Value = 0.09*Initial Investment = 0.09*8200*d = 73,800*d
Present Value for all Future Cash Flows,
PV = -Initial Investment - PV(Annual Operating Costs) + PV(Salvage Value)
PV = -8200d - 60,497,307,870*d-5{(1-(1+0.12)-17)*0.12} + 73800d / (1+0.12)14
= -8200d - 60,497,307,870*d-5{(1-(1+0.12)-14)*0.12} + 15,092d
= 6892d - 60,497,307,870*d-5{(1-0.1456)*10}
= 6892d - 60,497,307,870*d-5*8.544
= 6892d - 516888x1011*d-5
Let equivalent annual cost = P
Thus, P{(1-(1+0.12)-17)*0.12} = 6892d - 516888x1011*d-5
P x 8.544 = 6892d - 516888x1011*d-5
P = 806.64d - 60,497,307,870*d-5
For equivalent annual costs to be minimum, first order differential with respect to d =0
=> -806.64 - 60,497,307,870*(-5)d-6 = 0
=> 60,497,307,870*(-5)d-6 = 806.64
=> d-6 = 806.64/ 60,497,307,870*(-5)
=> d6 = 60,497,307,870*(5)/806.64
Taking 6th root on both sides,
d = 26.85 inch
Hence, the optimal size required is 26.85 inch
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