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
