(Economic Pipe Diameter): Example 1 introduced finding the economic diameter of a pipe. This example will work the problem to completion.
Consider the following pumping situation where 160 gpm of Newtonian fluid, having a viscosity of 20.6 centistokes (cs) and a specific gravity of 0.9, is pumped from Tank A to Tank B.

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Using March 2005 costs, find the most economic pipe diameter for the following situation. Pumping occurs 10 hr/day, 5 days/wk, and 50 wk/yr. March 2005 electrical power costs are $0.047/kWh. The system is manually operated. The pipe is Schedule 40 carbon steel and the pump is a cast iron ANSI centrifugal pump. Your company’s hurdle rate is 15% and its tax rate is 35%. Use a 10-year project life.

Pipe Diameter (in)	Equivalent Length of Pipe (ft)	P (ft of fluid/ 1000 ft of Pipe)¹
2 1/2	1688	281
3	1709	98.2
4	1820	27.2
6	1967	3.9

Example 1 identified pipe diameter as the independent variable. The dependent equipment variables were pump and motor size and the dependent cost variables were capital, utilities (electrical power), maintenance, insurance and taxes, operating supplies, plant overhead, and depreciation.
To solve this problem, start by sizing the pump. (We will ignore the 2 1/2-in. pipe because its ΔP exceeds 150 psig, which significantly increases the pipe and pump costs.) The following are the calculations for the 3-in. pipe.
Size the pump:

[latex]ΔP_{ft} = ( ext{Equivalent length}) * (ΔP/1000 ft) = 98.2/1000 ft * 1709 = 168 ext{ft of head}[/latex]

Pump size: 160 gpm, required head of 168 ft
Electrical power costs (assume a pump/motor efficiency of 60%):

[latex]HP= \frac{ ext{flow * head}}{ ext{Motor Efficiency}}[/latex]

In standard units, this becomes:

[latex]HP= \frac{ ext{gpm } * ΔP_{ft} * SG}{3960 * E}= \frac{160 * 168 * 0.9}{3960 * 0.6}[/latex]

[latex]\$ / ext{yr} = 10.2 ext{HP} * 10 \frac{ ext{hr}}{ ext{day}} * 5 \frac{ ext{day}}{ ext{wk}} * 50 \frac{ ext{wk}}{ ext{yr}} * 0.746 \frac{ ext{kWh}}{ ext{HP - hr}} * \frac{\$ 0.047}{ ext{ kWh}}= \$ 894 / ext{yr}[/latex]

Capital costs:
Pipe: Use the piping page in Appendix IV. The chart is based upon a CEPI of 460, which is a March 2005 index. This chart gives the installed cost of piping as:
[latex]\$K/100 ft = (0.1F +0.924) D^{0.83}[/latex]

where F = the number of fittings/100 ft of pipe = 43 fittings/1500 ft = 2.9/100 ft
D = nominal pipe diameter = 3 in

[latex]\$_{2005} = [(0.1 * 2.9 + 0.924) * 30.83] * (1500 ft/100 ft) = \$45.3K[/latex]

Pump. Use the pump page in Appendix IV.
The “gpm * ft” for the pump = 160 gpm * 168 ft = 26.9K
[latex]\$_{2005} = 5K[/latex]

Use Equation 3.3 to calculate the capital cost.

[latex] ext{Capital cost} = \sum{ [ ext{Equipment purchase cost} imes ( ext{Hand factor} imes F_{m})]} imes F_{i} imes F_{b} imes F_{p}[/latex]
The Hand Factor (Table 3.2) is 4.

TABLE 3.2
Hand Factors (CS Equipment)

Equipment Type	Factor
Fractionating columns	4
Pressure vessels/tanks	4
Heat exchangers	3.5*
Fired heaters	2
Pumps	4
Compressors	2.5
Instruments	4
Miscellaneous equipment	2.5
* The Hand factor for plate and frame heat exchangers with SS plates is also 3.5.
Source: Brown, T.R., Capital Cost Estimating, Hydrocarbon Processing, October 2000, 93–100. With permission.

[latex]F_{m} = 1[/latex] because the pump is cast iron
[latex]F_{i} = 1[/latex] because the pump is manually operated
[latex]F_{b} = 1.06[/latex] — assume the system will be part of an expansion in an existing plant
[latex]F_{p} = 1[/latex] — assume the system will be in the U.S.

[latex]\$_{2005} = 5K * 4 * 1 * 1 * 1.06 * 1 = \$21.2K[/latex]

Total capital = 45.3 + 21.2 = $66.5K
Cash flows:
BT cash flow = Utilities + (Maintenance, etc.) + Depreciation
= –$0.9K + (0.11 * –$66.5K) + (–$66.5K/10) = –$14.9K
AT cash flow = (BT cash flow)(1 – Tax rate) + Depreciation
= (–$14.9K) (1– 0.35) + $6.09K = –$3.04K

NPV:
[latex] ext{NPV} = P_{ ext{capital}} + P_{ ext{AT cash flow}} = –\$66.5K + [– \$3.04K (P/A, 15 \%, 10)] = – \$66.5K + (– \$15.3) = – \$81.8K[/latex]

Similarly, one calculates the NPVs for 4-in. and 6-in. pipe:

Pipe Diameter (in)	NPV ($K)
3	–81.8
4	–86.9
6	–106.2

The economic design point: Because there are discrete options, one need not plot NPV versus diameter. The maximum NPV (least negative) occurs for the 3-in.
pipe. This is the economic diameter.

Question 10.5: (Economic Pipe Diameter): Example 1 introduced finding the e...

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