Question

Water at 60°F is pumped from a reservoir to the top of a mountain through in 6-in. Schedule 120 pipe at an average velocity of 12ft/s. The pipe discharges into the atmosphere at a level 3000ft above the level in the reservoir. The pipeline itself is 4500ft long. Of overall efficiency to the pump and the motor driving it is 70% and the cost of electric energy to the motor is 4 cents per kilowatt hour, what is the hourly energy cost for pumping this water? (In fluid mechanics)

Answer 1

Mass of water flown per second$=$ Volume Flow rate$\times$ Density$=Av\times \rho$

Diameter of pipe$=6\ in=0.5\ ft=0.5\times 0.305\ m=0.1525\ m$

Area of pipe$=\frac{\pi d^2}{4}=0.018256 m^2$

Density of water$=1000Kg/m^3$

Velocity of water$=12ft/sec=0.305\times 12m/sec=3.66m/sec$

.Mass flow rate$(m_r)=0.018256\times3.66\times1000=66.81Kg/sec$

Total energy per second required to reach water upto height of $3000\ ft \ or\ 3000\times 0.305\ m(915\ m)=m_rgh=$ $66.81\times 9.81\times 915$ $=599.15\ KW$

As per question,motor has efficiency of $70$ % and motor power be $P.$

So,$\frac{70}{100}\times P=599.15\ KW$

$\implies P=855.932\ KW$

Cost is $4$ cents per $KW\ h$ .

Hourly cost will be $4\times 855.932=3423.73$ cents