Water at 40 degree F is pumped from a reservoir to a farm 50

Water at 40 degree F is pumped from a reservoir to a farm 5,000 ft away. The farm is on a high plain at 4,000 ft above the level in the reservoir. The water flow through a 6 in cast iron pipe schedule 80 at an average velocity of 10 ft/s. if the energy company at the farm charges electric energy at a rate of 5 cents per kilowatt hour, how much per hour it will cost to pump the water to the farm? Assume that the overall efficiency of the pump is 80%. If the pipe line is not a single pipe line but consist of pipes joined by some elbows gauges and so on, would you expect the cost to pump the water to the farm increases or decreases from your above calculates in section (1)? Explain your answer.

Solution

For water at 40 F, we have density rho = 1000 kg/m³ and dynamic viscosity meu = 1.519*10^-3 Pa-s

Length of pipe L = 5000 ft = 1524 m

Elevation change z1 - z2 = 4000 ft = 1219.2 m

Dia of pipe = 6\" = 0.1524 m

For cast iron schedule 80 pipe, roughness e = 0.0102\" = 0.00026 m

Relative roughness e/D = 0.00026 / 0.1524 = 0.0017

Velocity V = 10 ft/s = 3.048 m/s

Reynolds number Re = rho*V*D / meu

= 1000*3.048*0.1524 / (1.519*10^-3)

= 3.05*10⁵

From Moody diagram for Re = 3.05*10⁵ and e/d = 0.0017, we get friction factor f = 0.023

1)

Head loss h = fL/D * V² / (2g)

= 0.023*1524 / 0.1524 * 3.048² / (2*9.81)

= 108.9 m

Required pump head = Elevation change + head loss

= 1219.2 + 108.9

= 1328.1 m

Flow rate Q = A*V = 3.14 / 4 * 0.1524² * 3.048 = 0.0555 m³/s

Pump power = rho*g*Q*H

= 1000*9.81*0.0555*1328.1

= 724026 W

= 724 kW

Power required = 724 / 0.8 = 905 kW

Cost = 5*905 = 4525 cents = $45.25

2)

If there are additional fittings, the head loss will increase further and this will lead to higher power requirement and hence higher cost.