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BaLLatris [955]

3 years ago

8

g The pump inlet is located 1 m above an arbitrary datum. The pressure and velocity at the inlet are 100 kPa and 2 m/s, respecti

vely. The pump exit is located 4 m above the same datum. The pressure and velocity are 500 kPa and 3 m/s, respectively. How much power is required to drive this pump assuming and efficiency of 75%

1 answer:

oee [108]3 years ago

4 0

Answer:

The power needed to drive this pump assuming and efficiency of 75% is 1874.0 watts (W)

Explanation:

Solution:

Given that:

Velocity = 2 m/s

Pressure = 100 kPa

The pump exit is =4 m

Efficiency 75%

Thus,

We apply the method called the Bernoulli's equation between two reservoirs

p₁ / ps + v₁²/2g + z₁

=p₂/ps/v₂²/2g +z₂ + hL

The density of gasoline (pg) is = 680 kg m³

The gravity of acceleration is known to be =9.81 m/s²

So,

100/680 *9.81 + 2²/2 *9.81 + 1 = 500/ 680 * 9.81 + 3²/2 * 9.81 +4 + hL

16. 2 = 79.41 + hL

hl =79.41 - 16.2

hL =63.21 m

the unit weight of gasoline is(γ) = 680 * 9.81 = 6670.8 m/s

Now we find the efficiency

The efficiency (η) = The output power /Input power

Where hL =H

The input power = γ * Q * H/0.75

=6670.8 *12/3600 * 63.21

=6670.8 * 0.333 * 63.21

=6670.8 *0.2107

=1405.5/0.75

The input power =1874.0 Watts

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<u>Explanation</u>:

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A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco

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Answer:

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For pipe 2

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We know that Q=AV

$Q=A_1V_1=A_2V_2$

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$h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}$

Now putting the all values

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So h=0.23 m

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