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 $\Delta P$ formular 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!

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.

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?

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 $\Delta P$ formular 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!