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A centrifugal pump is installed in a pipeline to raise water14.9 m into an elevated holding tank. The length of the ductileiron pipeline (f=0.019) connecting the reservoirs is 22.4 m. Thepipe is 5.0 cm in diameter and the performance curve of the pump isgiven by Hp =23.9?7.59Q2 where Hp is in meters, Q is in liters persecond, and the equation is valid for flows up to 1.5 L/s. Usingthis pump, what flow do you expect in the pipeline if minor lossesare ignored? What pump head is required for this flow?

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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