Respuesta :
Answer:
13.88 m
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, [tex]h_L[/tex] = 4.12 m
Now, applying the Bernoulli's theorem between the entry and the exit points.
we get,
[tex]\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+h_P=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+h_L[/tex]
where, P is the respective pressure head
V is the velocity
[tex]h_P[/tex] 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
[tex]Z_1-Z_2 + h_P=h_L[/tex] ..........(1)
thus,
- ΔZ + [tex]h_P[/tex] = 4.12
now,
Power = ρghQ
ρ = SG × 1000 kg/m³ = 0.85 × 1000 = 850 kg/m³
h = [tex]h_P[/tex]
on substituting the values, we have
0.6 × 50× 10³ = 850 × 9.8 × [tex]h_P[/tex] × 0.2
or
[tex]h_P[/tex] = 18.007 m
now, from 1 we get
- ΔZ + [tex]h_P[/tex] = 4.12
or
- ΔZ +18.007 = 4.12
or
ΔZ = 13.88 m
hence, the tank should be 13.88 m above the pump
