A. An 80-percent efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at ar ate of 1.5 ft3 /s through a constant-diameter pipe. The free surface of the pool is 80 ft above that of the lake. Determine the mechanical power used to overcome frictional effects in piping. b. Water is pumped from a lake to a storage tank 15 m above at a rate of 70 L/s while consuming 15.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy. Determine:

1. The overall efficiency of the pump–motor unit.
b. The pressure difference between the inlet and the exit of the pump.

A) mechanical power used to overcome frictional effects in piping = 2.373 HP

B) efficiency = 66.82%

Pressure difference = 147 KPa

Explanation:

We are given;

efficiency of pump; η = 80% = 0.8

power input of pump;P_pump,in = 20 hp

rate being pumped; Q = 1.5 ft³/s = 0.04248 m³/s

free surface of pool; Δz = 80 ft = 24.384 m

mechanical pumping power used to deliver water is;

W_u = η × P_pump,in

W_u = 0.8 × 20

W_u = 16 Hp

Now, change in total mechanical energy of water will be equal to the change in potential energy.

Thus, it is expressed as;

ΔE_mech = ρ × Q × g × Δz

Where ρ is density of water = 1000 kg/m³

Thus;

ΔE_mech = 1000 × 0.04248 × 9.81 × 24.384 = 10,161.5150592 J/s

Converting to HP gives;

ΔE_mech = 13.627 HP

Now, mechanical power used to overcome frictional effects is given by;

W_frict = W_u - ΔE_mech

W_frict = 16 - 13.627

W_frict = 2.373 HP

B) We are now given;

W_elect,in = 15.4 KW

Volumetric flow rate; q = 70 l/s = 0.07 m³/s

Height; h = 15 m

Mass flow rate; m' = qρ

Where ρ is density of water = 1000 kg/m³.

m' = 0.07 × 1000

m' = 70 kg/s

Now,

ΔE_mech = workdone through height of 15m

Thus;

ΔE_mech = m'gh

ΔE_mech = 70 × 9.8 × 15

ΔE_mech = 10,290 W = 10.29 Kw

efficiency of the pump–motor unit = ΔE_mech/(W_elect,in) = 10.29/15.4 = 0.6682 or 66.82%

Pressure difference = ρgh = 1000 × 9.8 × 15 = 147,000 Pa = 147 KPa

answer: 500 wph.

explanation: