Answer:

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