# Question #eae28

Aug 27, 2017

I get this.

#### Explanation:

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

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

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

Pressure due $40 m$ water column ${P}_{1} = m g = \rho \times \text{volume} \times g$
${P}_{1} = 1000 \times \left(40 \times 1\right) \times 9.81 = 392400 p a$

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

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

For unit area of jet pressure ${P}_{2} = 49050 p a$

Total pressure of the pump$P = {P}_{1} + {P}_{2} = 392400 + 49050 = 441450 p a$