Question

1.7.4. Cost optimization problems You are ready to solve interesting optimization problems in engineering. Here are a few useful relationships for the cost analysis of hydraulic projects. Let's define P the present value, F the future value and the end of year repeated uniform annual amount. The annual interest rate is i and is the number of years. present valve P LERR F = $1,200/(1+0.05)20 - 11, $200 F = P(1 + i)", and P = F(1 + i)- and PER [ 41.9 ") and R =P () For instance, you invest $1,200 per year at i-5% for 20 years, the future value will be 1.-10.05(1 + 0.05)20 (1 + 0.05)20 = $39,679 *** Optimization Problem Example You live in a city of 50,000 people near sea level and the water consumption including industrial needs is about 200 liters per person per day. This corresponds to a need for Q-0.125m/s, or 4.42 ft/s of treated water on a continuous basis. Your company (JJ Engineering) is asked to design the pipe diameter (single value) and pump size. In this case, you need to pump water through a L-2,500 ft long pipe into an elevated tank at H - 120 ft above the point source. Your supervisor Jan tells you to neglect minor losses but assume f-0.02 for friction. You figured out that small pipes would have large losses and would require a bigger pump, so you are eager to find the optimal pump and pipe size to minimize the cost of your system. Jan gives you a 2014 table for the cost of pumps and galvanized iron pipe. The cost of pipes is 2" (6$/ft), 3" (10/ft), 4" (255/ft). 6 (355/1), 8" (60$/ft), 10" (905/ft), 12" (125 total / s/ft), 15" (200 S/ft), 18" (320$/ft), 20" (360S/A), 24" (500$/ft), 30" (720 S/A), 36" Cost cosi I cost (900S/ft), 42" (1,250$/ft), and 48" (1,600$/ft). The cost of pumps is -$200/HP, and there is a fixed base construction cost of $150,000. Which combination of pipe diameter and pump size will minimize the cost? Diameter oplim total / Lowest cost Cost

Answer 1

We need to find the optimum cost by considering two criteria.

1. The diameter of pipe- As dia increases, friction loss decreases and the Total head required decreases. Which in turn decreases the power required to pump the water => Low HP=> Low cost.

Also If the diameter of the pipe increases, the cost of pipe increases.

So to find the balance we need to find the intersection point of graphs for cost due to Power(HP) and cost due to Pipe diameter. The intersection diameter gives the optimal diameter for design.

Chart 3 А Data- в Q(m3/s) 0.125 0.02 762 - 0.078781939 36.57 L(m) Hstat(m) 5 5 D(inch) D(m) 2 0.0508001 3 0.0762002 4 0.1016002 6 0.1524003 8 0.2032004 10 0.2540005 12 0.3048006 HE 0.143 0.132 0.124 0.115 0.108 0.104 0.100 H Total 36.713 36.702 36.694 36.685 36.678 36.674 36.670 Power(kW) 45.019 45.006 44.997 44.985 44.977 44.971 44.966 Power(HP) 60.348 60.329 60.317 60.301 60.291 60.283 60.277 Cost(HP) Cost(Dia) 12070 4572 12066 7620 12063 19050 12060 26670 12058 45720 12057 68580 12055 95250 Ovo AWN-OUW

Cost diagrams 80000 70000 60000 50000 Cost ($) 40000 30000 20000 10000 12 Pipe Diameter(Inch) • Cost(HP) Cost(Dia)

By looking into the diagram, it looks like selecting a 4inch pipe and 60 HP pump would be ideal considering cost optimization.