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Sagot :
Answer:
Hp = 23.86 m
Hence, 23.86 m head pump is required for this flow.
Explanation:
Solution:
Data Given:
Elevation = Z = 14.9 m
Friction of the ductile iron pipe = f = 0.019
Length of the ductile iron pipe connecting the reservoirs = 22.4 m
Diameter of the Pipe = D = 0.5 cm
Radius of the Pipe = r = D/2 = 0.25 cm
Performance Curve of the Pump = Hp = 23.9 - 7.59[tex]Q^{2}[/tex] , where Hp is in meters.
[tex]Q^{}[/tex] is in Liters/second and equation is valid for 1.5 L/s
We are asked to find out the pump head required for this flow.
We need to apply the energy equation for this problem:
Energy Equation:
[tex]\frac{P1}{\beta 1} + \frac{v1^{2} }{2g} + Z1 + Hp = \frac{P2}{\beta 2} + \frac{v2^{2} }{2g} + Z2 + H_{L}[/tex]
where, [tex]\beta[/tex] = specific weight of the fluid.
v1 = 0
P1 = P2 = P
Z1 = 0
Z2 = 14.9
Hence,
New equation is:
[tex]Hp = \frac{v2^{2} }{2g} + Z2 + H_{L}[/tex]
And
[tex]H_{L}[/tex] = [tex]\frac{fLv^{2} }{2gD}[/tex]
So,
[tex]Hp = \frac{v2^{2} }{2g} + Z2 + \frac{fLv^{2} }{2gD}[/tex]
Now, converting formula of v into terms of Q, we get
Hp = 23.9 - 7.59[tex]Q^{2}[/tex]
[tex]Hp = \frac{Q2^{2} }{2(9.81)(\frac{\pi }{4}(0.05^{2}))^{2} } + 14.9 + \frac{(0.09)(22.4)Q^{2} }{12.1(0.05)^{2} }[/tex]
23.9 - 7.59[tex]Q^{2}[/tex] = [tex]\frac{Q2^{2} }{2(9.81)(\frac{\pi }{4}(0.05^{2}))^{2} } + 14.9 + \frac{(0.09)(22.4)Q^{2} }{12.1(0.05)^{2} }[/tex]
Solving for Q, we will get
9 - 7.59[tex]Q^{2}[/tex] = 13233.71[tex]Q^{2}[/tex] + 112555.37[tex]Q^{2}[/tex]
[tex]Q^{}[/tex] = 7.154 x [tex]10^{-5}[/tex] [tex]m^{3}/s[/tex]
Converting it into L/s
Q = 0.0715 L/s
Putting this value of Q into the Hp equation to get the required answer:
Hp = 23.9 - 7.59[tex]Q^{2}[/tex]
Hp = 23.9 - 7.59 x [tex]0.0715^{2}[/tex]
Hp = Pump head
Hp = 23.86 m
Hence, 23.86 m head pump is required for this flow.
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