Calculating Maximum Pump Height for Given Flow Velocity | Pipe Friction Ignored

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The discussion focuses on calculating the maximum height a pump can achieve for a specified flow velocity, specifically 1.5 m/s, while ignoring pipe friction. The user has a pump with a maximum flow rate of 12 l/min and is trying to understand the relationship between static head, total head, and losses due to friction. They confirm that the equation used is a variation of Bernoulli's principle, which incorporates static head and frictional losses, defined as a constant multiplied by the square of the flow rate. The user concludes that for their maximum flow rate, the static head can be calculated as approximately 3.0196 m after accounting for frictional losses. This discussion highlights the importance of understanding fluid dynamics principles in pump system design.
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Hi

I have a pump which for 0 head will provide a max flow of 12 l/min (0.0002 m^3/s). For the piping I am using (3/8"), by my calculations this gives a theoretical MAX mass flow rate of 0.2 m/s and a MAX velocity of 2.8 m/s.

As i understand it, these values are not taking into account (1) frictional effects in the piping and (2) static head involved in raising the fluid.

Ignoring the frictional effects in the piping for now, how would I calculate the max height this pump could pump water for a given flow velocity, say 1.5 m/s.

I'm sure it is Bernoulli's but there seems to be too many variables. I don't have access to the performance curves of the pump.

Thanks for your help.
 
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Hm = Hst + losses
losses is defined as constant * Q^2
and u have Q = 12 L/S
AND d = 3/7 " so sub in this
o.8 * f * L * Q2/gD^5 you will find it a very small term because Q is small
so finally u can consider that H pump = H st + 3 or 4 meters
but H static should be taken into considerations
 
Hi

Thanks for the reply. Can you elaborate on this please. I don't understand some of what you have done.

Is Hm=Total head, Hst=Static head (elevation head) and losses=frictional effects

Losses (frictional effects) defined as a constant * Q^2

Is this correct? I'm with you up until this point.

The equation you've written now, is it some varation of Darcy's. Can you explain please.

Thanks
 
yes man its correct and its a form of bernoulis equation
 
and remeber also this equation is also used to draw the system curve of the pumping system and the intersection between it and the pump curve is the operating point
 
maxx_payne said:
and remeber also this equation is also used to draw the system curve of the pumping system and the intersection between it and the pump curve is the operating point

ok...so I also found out that the max head for 0 flow is 4.3 psi (3.02m) so I can write hw=H (static) + constant*Q^2.

Therefore for the max flow rate the pump can produce (12 l/min) and assuming frictional loss in pipe is constant at 1 for now, I can write:

H(static)=3.02-(1*0.0002^2 (m^3/s))=3.0196m

Is this right?
 
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