Question

A 7-cm diameter horizontal pipe connected to the side of a tank at 4 m below the water surface in the tank discharges water to the atmosphere. A pump located in the pipe can add head to the flow by the relation, hp (m) = 8 - 200 Q^2 (Q in m^3/s) If the losses in the system amounts to 5 m, determine the outflow velocity of the pipe.

Answer 1

Answer:

47.10 m/s

Explanation:

Diameter of the horizontal pipe, d = 7 cm = 0.07 m

Area of the horizontal pipe, A = $\frac{\pi}{4}d^2$

or

A =  $\frac{\pi}{4}0.07^2$ = 0.00384 m²

Head between the water surface and the pipe level, z = 4 m

Head added by the pump, hp = 8 - 200Q²

where, Q is the discharge

losses, hL = 5 m

now,

applying the Bernoulli's theorem between the water surface and the pipe outlet,

we have

$\frac{P_1}{\rho}+\frac{V_1^2}{2g}+z+h_P=\frac{P_2}{\rho}+\frac{V_2^2}{2g}+h_L$

where,

P₁ = pressure at the water surface = 0 (as atmospheric pressure only)

V₁ = Velocity at the free surface = 0

ρ is the density of the water

P₂ = pressure at the outlet = 0 (as atmospheric pressure only)

V₂ = Velocity at the outlet

on substituting the values, we get

$z+h_P=\frac{V_2^2}{2g}+h_L$

or

$4+(8-200Q^2)=\frac{V_2^2}{2\times9.81}+5$

or

$7-200Q^2=\frac{V_2^2}{2\times9.81}$

also,

Q = A × V₂

thus,

$7-200\times(0.00384\times\ V_2)^2=\frac{V_2^2}{2\times9.81}$

or

V₂ = 47.10 m/s