Answer:
Input power given to the pump is 4.09 kw
Explanation:
the pictures below shows the full explanation
The required electric power input to the pump is 3.82 kW
In fluid mechanics, the required electric power input to the pump is the mechanical power taken by the pump shaft and can be calculated by going through the following process.
The first process is to determine the velocity of the water by using the formula:
[tex]\mathbf{V = \dfrac{Q}{A}}[/tex]
where;
[tex]\mathbf{V = \dfrac{Q}{\dfrac{\pi}{4}\times D^2 }}[/tex]
[tex]\mathbf{V = \dfrac{1.5}{\dfrac{\pi}{4}\times (\dfrac{5}{12})^2 }}[/tex]
V = 11.0 ft/s
The next process is to find the Reynolds number by using the following equation;
[tex]\mathbf{Re= \dfrac{\rho VD}{\mu}}[/tex]
From the parameters given:
∴
[tex]\mathbf{Re= \dfrac{62.30 \ lbm.ft^3 \times 11.0 \ ft/s \times (\dfrac{5}{12}) ft}{6.556 \times 10^{-4} \ lbm /ft.s}}[/tex]
Re = 435542.5
We will realize that the Reynolds number is greater than 4000, as a result of that, we need to determine the friction factor by using the Colebrook equation which can be expressed as:
[tex]\mathbf{\dfrac{1}{\sqrt{f}} = -2log \Bigg [ \dfrac{\varepsilon/D}{3.7}+ \dfrac{2.51}{Re \sqrt{f}} \Bigg]}[/tex]
From the above equation:
∴
[tex]\mathbf{\dfrac{1}{\sqrt{f}} = -2log \Bigg [ \dfrac{0/(5/12)}{3.7}+ \dfrac{2.51}{435542.5 \sqrt{f}} \Bigg]}[/tex]
[tex]\mathbf{\dfrac{1}{\sqrt{f}} = -2log \Bigg [ \dfrac{2.51}{435542.5 \sqrt{f}} \Bigg]}[/tex]
Making friction factor (f) the subject of the formula and solving for (f), we have:
Friction factor f = 0.01349
The next process involves estimating the sum of loss coefficients by using the formula:
[tex]\mathbf{\sum K_L = K_{L. entrance}+3K_{L. elbow}+ K_{L. exit}}[/tex]
For a turbulent flow, using the table of loss coefficients of various pipe parts:
where;
[tex]\mathbf{\sum K_L = 0+3(0.3)+1}[/tex]
[tex]\mathbf{\sum K_L = 1.90}[/tex]
From this, the total head loss [tex]\mathbf{h_L}[/tex] is computed as:
[tex]\mathbf{h_L = \Big[f \dfrac{L}{D} + \sum K_L \Big] \dfrac{V^2}{2g}}[/tex]
[tex]\mathbf{h_L = \Big[0.01349 ( \dfrac{125}{5/12} )+ 1.90 \Big] \dfrac{11.0^2}{2(32.2)}}[/tex]
[tex]\mathbf{h_L = 11.17 \ ft}[/tex]
However, we need to apply Bernoulli's equation to estimate the pump head input because we require the head input to overpower the potential head and velocity loss.
Now, using Bernoulli's equation, we have:
[tex]\mathbf{\dfrac{P_1}{\rho g}+ \dfrac{V_1^2}{2g}+ Z_1 + h_p = \dfrac{P_2}{\rho g} + \dfrac{V_2^2}{2g}+Z_2 + h_L}[/tex]
where;
From above equation, we have:
[tex]\mathbf{\dfrac{P_{atm}}{\rho g}= \dfrac{6}{2 \times 32.2 } + 0 + h_p = \dfrac{P_{atm}}{\rho g} + \dfrac{0^2}{2g} +12 + 11.17 }[/tex]
∴
The head pump input [tex]\mathbf{h_p}[/tex] is:
[tex]\mathbf{h_p = 12 + 11.17 - \dfrac{6^2}{2 \times 32.2}}[/tex]
[tex]\mathbf{h_p = 22.61 ft}[/tex]
Finally, using the formula for calculating the required electric power input to the pump, we have:
[tex]\mathbf{P = \dfrac{\rho \times g \tmes Q h_p}{\eta}}[/tex]
where;
∴
[tex]\mathbf{P = \dfrac{62.3 \times 32.2 \times 1.5 \times 22.61}{0.75}\times \dfrac{1}{32.2} \times \dfrac{1}{737}}[/tex]
P = 3.82 kW
Learn more about electric power input here:
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