Please I want a solution of this question with all steps on an urgent basis. Many Thanks FEDWater TowerHydraulicsPumpAFigure 1 Water Supply and Distribution System¢ ¢ ¢ ¢ ¢B.¢ ¢ ¢ ¢ ¢ Q3.A pump is required to pump water from a large reservoir to the water tower located 20 m above thereservoir (Figure 1). If 0.03 m³/s of water having a density of 1000 kg/m3 is pumped through a 30-mmpipe.а.How much power is required to be delivered to the water by the pump? Consider the flow lossesin the pipe; do the necessary assumptions.b. Determine the efficiency of the pump

a) The expression for energy between velocity 1 and 2 is given as P1γ+V122g+z1+HP=P2γ+V222g+z2…

Answered: Q3. A pump is required to pump water…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

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Please I want a solution of this question with all steps on an urgent basis. Many Thanks

F
E
D
Water Tower
Hydraulics
Pump
A
Figure 1 Water Supply and Distribution System
¢ ¢ ¢ ¢ ¢
B.
¢ ¢ ¢ ¢ ¢

Q3.
A pump is required to pump water from a large reservoir to the water tower located 20 m above the
reservoir (Figure 1). If 0.03 m³/s of water having a density of 1000 kg/m3 is pumped through a 30-mm
pipe.
а.
How much power is required to be delivered to the water by the pump? Consider the flow losses
in the pipe; do the necessary assumptions.
b. Determine the efficiency of the pump

Expert Solution

Step 1

The expression for energy between velocity 1 and 2 is given as

$\frac{P_{1}}{γ} + \frac{V_{1}^{2}}{2 g} + z_{1} + H_{P} = \frac{P_{2}}{γ} + \frac{V_{2}^{2}}{2 g} + z_{2}$

$\frac{P_{1}}{γ} + \frac{0}{2 g} + 0 + H_{P} = \frac{P_{1}}{γ} + \frac{V_{2}^{2}}{2 g} + Z_{2} H_{P} = \frac{V_{2}^{2}}{2 g} + Z_{2}$

Substitute the value in above expression

$H_{P} = \frac{\frac{0.03}{π [{(0.03)}^{2} / 4]}}{2 \times 9.81} + 20 H_{P} = 125.341 m$

The expression for horse power is given as

$P o w e r = H_{P} \times Q \times γ$

Substitute the value in above expression

$P o w e r = 125.341 \times 0.03 \times 1000 \times 9.81 P o w e r = 36888 watts P o w e r = 36.888 kW$

Therefor 36.88 kW power is required to be delivered to the water.

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