3 years ago

URGENT

Physics

1 answer:

ella [17]3 years ago

4 0

Answer:

the is metallurgy .....

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You work for the city water department and need to pump 3400 liters/minute of water from a tank at ground level into a vented (i

umka21 [38]

Answer:

$P_E=46.2778\ kW$

$v=1.804\ m.s^{-1}$

Explanation:

Given:

flow rate of water, $\dot{V}=3400\ L.min^{-1}=3.4\ m^3.min^{-1}$

∵Density of water is 1 kg per liter

∴mass flow rate of water, $\dot{m}=3400\ kg.min^{-1}$

height of pumping, $h=75\ m$

efficiency of motor drive, $\eta=0.9$

diameter of pipe, $D =0.2\ m$

Now the power required for pumping the water at given conditions:

$P=\dot{m}.g.h$

$P=\frac{3400}{60} \times 9.8\times 75$

$P=41650\ W$

Hence the electric power required:

$P_E \times \eta=P$

$P_E \times 0.9=41650$

$P_E=46.2778\ kW$

Flow velocity is given as:

$v=\dot{V}\div a$

where: a = cross sectional area of flow through the pipe

$v=\frac{3.4}{60}\div (\pi.\frac{0.2^2}{4} )$

$v=1.804\ m.s^{-1}$

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