Question #eae28

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Question #eae28

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Answer 1 · 2017-08-27T06:13:05

I get this.

Explanation:

The problem consists of sum of two pressures which the pump must create

Pressure due to height of $40 m$ $40 m$ of water column
Pressure due to jet of water at a rate of $5 m s^{- 1}$ $5ms^-1$

For 1. Assuming unit area, Force $=$ $=$ Pressure

Pressure due $40 m$ $40m$ water column $P_{1} = m g = ρ \times volume \times g$ $P_1=mg=rhoxx"volume"xxg$
$P_{1} = 1000 \times (40 \times 1) \times 9.81 = 392400 p a$ $P_1=1000xx(40xx1)xx9.81=392400pa$

For 2. Assumed area of the jet be unit area.
Speed of water outlet from jet $= 5 m s^{- 1}$ $=5 ms^-1$ .
Total mass of water thrown out per sec $= ρ \times 1 \times 5 = 5000 k g$ $= rhoxx1xx5=5000kg$

Pump must exert pressure to support weight of this additional column of water of unit area $= 5000 \times g = 49050 N$ $=5000xxg=49050N$

For unit area of jet pressure $P_{2} = 49050 p a$ $P_2=49050pa$

Total pressure of the pump $P = P_{1} + P_{2} = 392400 + 49050 = 441450 p a$ $P=P_1+P_2=392400+49050=441450pa$

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Question #eae28

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