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Power of water pump

Consider a water pump pushing water through a pipe:

We wish to find the energy which the water pump has to expend to keep the water flowing through the pipe. Writing Newton's second law for water flowing out of a cylindrical cross section of it,

$$ F= v \frac{dm}{dt}$$

Now, $$ P = F \cdot v$$

$$ P= v \frac{dm}{dt} v$$

$$ \frac{dm}{dt} = \rho A v$$

$$ P= v^3 \rho A $$

Now the real answer has a factor of half, where in this derivation have I gone wrong?

Answer

The question seeks to evaluate the pump power for a very special case: steady flow in a horizontal pipe with no friction losses where the entrance velocity is small, and where all the pump work goes into creating a constant pressure difference regardless of the flow rate.  The pump power for this special case is not the pump power for a flow network in general, especially considering the change in pump power with flow rate behavior for the commonly used centrifugal pump. Therefore, I provide a general discussion of "flow work" and the application of the Bernoulli equation. Then, at the end of this response I will use the general results to address this special case and reach the same conclusion as other responders to this question; specifically, to explain the ${1 \over 2}$ in the pump power. I include the rather lengthy general discussion to emphasize that the result for the pump power for the special case is not valid in general.

At a position in the pipe where the cross sectional area is $A$ and the fluid velocity is $V$, the rate of energy flow per unit mass is $e_f \dot m_f$ where $e_f$ (J/kg) is the energy per unit mass of the flowing mass $m_f$ (kg), and $\dot m_f$ (kg/sec) is the rate at which mass flows past the cross sectional area $A$. As shown in the figure below, $\dot m_f$ equals $\rho AV$ where $\rho$ is density; $A$ and $V$ can be functions of position. $e_f = h + {V^2 \over 2} + zg$ (J/kg) where $h$ is the enthalpy per unit mass, $V$ the velocity, $z$ the elevation, and $g$ the acceleration of gravity.

There is "flow work" associated with pushing mass across a boundary. The enthalpy per unit mass is defined as $h = u + p/\rho$ where $u$ is the internal energy per unit mass and $p$ is pressure. The $p/\rho$ term is sometimes called "flow work" since it is the work per unit mass done at the boundary by the fluid following $\Delta m_f$ to push $\Delta m_f$ through the distance $V\Delta t$.

For details see a good book on basic thermodynamics, such as Thermodynamics by Obert, where the first law of thermodynamics is developed for an open system, defined as one where mass can enter and leave the system.

[![area flow][1]][1]

It is important to understand the previous discussion is for mass flowing across a fixed boundary. In most applications, we are concerned with the change in energy of a system (an open thermodynamic system), defined as a region between boundaries through which mass can flow. Consider a system with a boundary 1 where mass enters and a boundary 2 where mass exits.  The "flow work" per unit mass done by the surroundings on the system due to fluid in the surroundings pushing mass into the system at boundary 1 is $p_1/\rho_1$. The "flow work" per unit mass done by the system on the surroundings due to fluid in the system pushing mass into the surroundings as boundary 2 is $p_2/\rho_2$.  The energy per unit mass entering the system at boundary 1 is $e_{f_\enspace 1}$ and the energy per unit mass exiting the system at boundary 2 is $e_{f_\enspace 2}$.

The power from the pump is the rate of work done by the pump on a system.  The pump power depends on both the pump head characteristics, and the system head for the flow elements (piping and fittings) that the pump supplies.  The pump head can depend on the flow rate and the flow rate depends on the system head.  Assuming steady state flow for an incompressible liquid, the first law of thermodynamics for a steady flow system can be expressed using the Bernoulli equation. Head has units of distance (meters) and is simply each term in the Bernoulli equation expressed in meters.

Consider the pumping system shown in the figure below using a centrifugal pump.  The pump operates at an operating point determined by where the pump head curve intersects the system head curve.  For a centrifugal pump, pump head $h_{pump}$ decreases with volumetric flow rate $Q$ (cubic meters per sec) and the system head increases with $Q$ due to friction losses in the flow network.  The work per unit mass $W_{mass}$ (J/kg) done by the pump equals $gh_{pump}$ and $h_{pump} = h_{system}$ where $g$ is the acceleration of gravity, $h_{pump}$ is the pump head at the operating point and $h_{system}$ is the system head at the operating point. The power from the pump is $P_{pump} = \rho Qgh_{pump} = \rho Qgh_{system}$ (J/sec) where $\rho$ is the liquid density.  So to evaluate the pump power the operating point is required and that depends on the pump characteristics and the flow system that the pump supplies.  The "brake" power is the power supplied to the pump by the driver (for example, a motor or a turbine) and is equal to the pump power divided by the efficiency of the pump.  For more details, see the Pump Handbook, edited by Karassik et al.

[![flow operating point][2]][2]

Now consider the special case of a system defined as steady flow in a horizontal pipe with no friction losses where the entrance velocity is small.  To avoid issues with the performance of a real pump, the "pump" here is a head of liquid in a large tank with the pipe at the bottom of the tank.  This example is based on the response by @knzhou to this question. 1 is bottom of the tank and 2 is the end of the pipe.

[![enter image description here][3]][3]

The Bernoulli equation for the system is ${p_1 \over \rho} + {V_1^2 \over 2} = {p_2 \over \rho} + {V_2^2 \over 2}$. For negligible velocity $V_1$ we have ${(p_1 - p_2) \over \rho} = {V_2^2 \over {2}}$ (J/kg).  The mass flow rate is $\rho AV_2$ (kg/sec) so the power is ${V_2^2 \over {2}} \rho AV_2 = {1 \over 2} \rho A V_2^3$  (J/sec or Watts) due to the pressure difference $p_1 - p_2$. This pressure difference is created by the head of liquid. This is the same approach used by other responders to this question.  As previously discussed, this is the power for this special case, and is not in general the pump power for a flow network.

The power can also be evaluated considering the "flow work". The net flow work per unit mass done on the fluid in the pipe (flow in minus flow out) is ${(p_1 - p_2) \over \rho}$ (J/kg), and as previously shown, for the special case under consideration, this results in the pump power being ${1 \over 2} \rho A V_2^3$.

[1]: https://i.sstatic.net/0iOpg.jpg
  [2]: https://i.sstatic.net/7P3rJ.jpg
  [3]: https://i.sstatic.net/FnMz6.jpg

Edit Summary

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$\begingroup$ On topic point: This Q discusses the same point as this one $\endgroup$
– Cathartic Encephalopathy
Commented Mar 27, 2022 at 5:18
$\begingroup$ Why did my original method fail? $\endgroup$
– Cathartic Encephalopathy
Commented Mar 27, 2022 at 12:17
$\begingroup$ You need to consider the change in kinetic energy for the special case of: a horizontal pipe with no friction loss and negligible inlet velocity, with a pump that only increases pressure and the increase in pressure is constant regardless of the flow rate. $\endgroup$
– John Darby
Commented Mar 27, 2022 at 19:59
$\begingroup$ For the special case summarized in my above comment, pump power is ${1 \over 2} \rho A V^3$. I address this special case after a general discussion. It took me a while to see the special conditions where this is the pump power. So I agree with other responders, but I want to provide the more general approach, then apply it to this special case, to emphasize this special result is not in general the pump work. Sorry it took me a while to understand the special case. Answer updated. $\endgroup$
– John Darby
Commented Mar 27, 2022 at 20:01
$\begingroup$ Hi, we've noticed that you have made a large number of minor edits to this post. Please be mindful that every edit bumps the post in the "active" tab of the site and try to make your edits substantial. If you foresee improving this post repeatedly, maybe collect several edits and make them in one go instead of submitting them individually. $\endgroup$
– Chris ♦
Commented Mar 28, 2022 at 3:12

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