Respuesta :
Answer:
[tex]P_E=46.2778\ kW[/tex]
[tex]v=1.804\ m.s^{-1}[/tex]
Explanation:
Given:
- flow rate of water, [tex]\dot{V}=3400\ L.min^{-1}=3.4\ m^3.min^{-1}[/tex]
∵Density of water is 1 kg per liter
∴mass flow rate of water, [tex]\dot{m}=3400\ kg.min^{-1}[/tex]
- height of pumping, [tex]h=75\ m[/tex]
- efficiency of motor drive, [tex]\eta=0.9[/tex]
- diameter of pipe, [tex]D =0.2\ m[/tex]
Now the power required for pumping the water at given conditions:
[tex]P=\dot{m}.g.h[/tex]
[tex]P=\frac{3400}{60} \times 9.8\times 75[/tex]
[tex]P=41650\ W[/tex]
Hence the electric power required:
[tex]P_E \times \eta=P[/tex]
[tex]P_E \times 0.9=41650[/tex]
[tex]P_E=46.2778\ kW[/tex]
Flow velocity is given as:
[tex]v=\dot{V}\div a[/tex]
where: a = cross sectional area of flow through the pipe
[tex]v=\frac{3.4}{60}\div (\pi.\frac{0.2^2}{4} )[/tex]
[tex]v=1.804\ m.s^{-1}[/tex]
