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Let's say I have a pump pushing fluid through a 250 ft pipe, 1.5'' diameter, at a rate of 20 gallons per minute. I would like to calculate the delta in energy required to pump this fluid with and without a restriction at the end of the 250 ft line. The restriction results in 0.01 psi pressure drop.

What equation is needed to calculate this energy?

Update:

Based on comments below, I was able to figure this out. First, I need to convert my $\Delta P$ to pressure head using the following formula:

$\Delta P = \rho_{fluid} * g * h$

where

$g =\text{ acceleration due to gravity}$

$h =\text{ pressure head}$

This formula itself is a simplified form of what can be found here

$P_2 - P_1 = \gamma (h_2 - h_1)$

where

$\gamma = \rho_{fluid} * g$

This gives us all the ammunition we need with the help of the pump power equation found here

$P_{\text{pump_power}} = q * \rho_{fluid} * g * h$

where

$q =\text{ fluid flow, volume per time}$

If you plug in our $\Delta P$ formula from above (you have to rearrange to get the h on it's own), you get the following:

$P_{\text{pump_power}} = q*\rho_{fluid}*g*\frac{\Delta P}{\rho_{fluid} * g}$

which, with handy cancellations, leads to:

$P_{\text{pump_power}} = q * \Delta P$

Nice! Dimensionnally, if your flow is in gallon per minute (which it was for me) you'll want to convert to cubic inches per second - this way, if your $\Delta P$ is in psi (which again, mine was) some of your inches will cancel out. You'll then have to convert BACK to feet from inches, but then you can easily go from $\frac{\text{ft-lb}}{s}$ to horsepower or kilowatts or whatever you want.

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    $\begingroup$ Good question, what have you tried so far? And do you have a pump curve? $\endgroup$
    – morristtu
    Commented Jan 27, 2016 at 18:00
  • $\begingroup$ I do not have the pump curve. At first I did some dimensional gymnastics to try to get to the right units, but that ended up being dumb. Eventually, I took my 20 gpm (mass flow) and multiplied by my delta P to get the extra work needed to achieve that delta P. The units worked out just right. I'm still not sure which equation I should have started with (I haven't used this stuff in years) - is it energy balance equation from thermo? $\endgroup$
    – wesanyer
    Commented Jan 27, 2016 at 18:10
  • $\begingroup$ Are you extending the pipeline, or raising/lowering the height of it at all? If so, you'll need to take these additional factors into account, and it becomes more complex than just using the pump power equation mentioned by Morrisstu below. $\endgroup$
    – Jim Hargreaves
    Commented Jan 28, 2016 at 11:46
  • $\begingroup$ nope, everything else is the same, only the restriction at the end of the line changes. $\endgroup$
    – wesanyer
    Commented Jan 28, 2016 at 11:50

1 Answer 1

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The pump power equation can be used to calculate the energy required in both scenarios. Convert the additional $\Delta P$ to head, then calculate the difference in required hydraulic pump power. Both of these equations are on Engineering Toolbox.

Converting head (ft or m) to pressure (psi or bar, kg/cm2) and vice versa

Calculate pump hydraulic and shaft power

For the question in your comment about the energy balance equation, Bernoulli's Equation is an energy balance for fluids.

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  • $\begingroup$ right on. I was completely blanking on what "head" was so I glossed over that engineering toolbox link when I came across it. it is all coming back now though, I remember that bernoullis equation was simplified from energy balance. thanks! as a side note, I wish those ET equations didn't always have a conversion factor builtin, it makes it more confusing imo $\endgroup$
    – wesanyer
    Commented Jan 28, 2016 at 11:56
  • $\begingroup$ Alright, I hate to keep beating a dead horse, put point in case: EngineeringToolbox gives all these equations for converting pressure head, for SI it is $P_{pressure}=0.433*\rho*h$. However, this is the same as $P_{pressure}=\rho*g*h$, with fancy conversions built-in. Using the second equation, though, I can plug in to $P_{power}=q*\rho*g*h$ and simplify into $P_{power}=q*P_{pressure}$. It's just easier this way! $\endgroup$
    – wesanyer
    Commented Jan 28, 2016 at 15:01

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