Answer:
47.10 m/s
Explanation:
Diameter of the horizontal pipe, d = 7 cm = 0.07 m
Area of the horizontal pipe, A = [tex]\frac{\pi}{4}d^2[/tex]
or
A = [tex]\frac{\pi}{4}0.07^2[/tex] = 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
[tex]\frac{P_1}{\rho}+\frac{V_1^2}{2g}+z+h_P=\frac{P_2}{\rho}+\frac{V_2^2}{2g}+h_L[/tex]
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
[tex]z+h_P=\frac{V_2^2}{2g}+h_L[/tex]
or
[tex]4+(8-200Q^2)=\frac{V_2^2}{2\times9.81}+5[/tex]
or
[tex]7-200Q^2=\frac{V_2^2}{2\times9.81}[/tex]
also,
Q = A × V₂
thus,
[tex]7-200\times(0.00384\times\ V_2)^2=\frac{V_2^2}{2\times9.81}[/tex]
or
V₂ = 47.10 m/s
