Question

A 60% efficient pump is run by a 50-kW electric motor for lifting 200 liters/sec of oil (SG =0.85) from ground level to an elevated storage tank through a constant diameter pipeline. How high can the tank be above the pump location to limit the head loss to 4.12 m in the system? Pressure of oil remains constant during the process.

Answer 1

13.88 m

Step-by-step explanation:

Given:

Efficiency of the pump, η = 60% = 0.6

Power of the motor, P = 50 kW = 50 × 10³ W

Discharge, Q = 200 L/s = 0.2 m³/s

Specific gravity of the oil, SG = 0.85

Limit for the head loss,
`h_L` = 4.12 m

Now, applying the Bernoulli's theorem between the entry and the exit points.

we get,

`(P_1)/(\rho g)+(V_1^2)/(2g)+Z_1+h_P=(P_2)/(\rho g)+(V_2^2)/(2g)+Z_2+h_L`

where, P is the respective pressure head

V is the velocity

`h_P` is the head loss by the pump

also,

P₁ = P₂ [since, there is atmospheric pressure at both the ends]

V₁ = V₂ [Cross-section of the pipe is constant]

thus, we get

`Z_1-Z_2 + h_P=h_L` ..........(1)

thus,

- ΔZ +
`h_P` = 4.12

now,

Power = ρghQ

ρ = SG × 1000 kg/m³ = 0.85 × 1000 = 850 kg/m³

h =
`h_P`

on substituting the values, we have

0.6 × 50× 10³ = 850 × 9.8 ×
`h_P` × 0.2

or

`h_P` = 18.007 m

now, from 1 we get

- ΔZ +
`h_P` = 4.12

or

- ΔZ +18.007 = 4.12

or

ΔZ = 13.88 m

hence, the tank should be 13.88 m above the pump