Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
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.
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.