Power in Fluid Flow

Power is the rate at which work is done or energy in any form is used or supplied. To see the relationship of power to fluid flow, consider Bernoulli’s equation:

\(P+\cfrac{1}{2}{\mathrm{\rho v}}^{2}+\rho \text{gh}=\text{constant}\text{.}\)

All three terms have units of energy per unit volume, as discussed in the previous section. Now, considering units, if we multiply energy per unit volume by flow rate (volume per unit time), we get units of power. That is, \((E/V)(V/t)=E/t\). This means that if we multiply Bernoulli’s equation by flow rate \(Q\), we get power. In equation form, this is

\((P+\cfrac{1}{2}{\mathrm{\rho v}}^{2}+\rho \text{gh})Q=\text{power}\text{.}\)

Each term has a clear physical meaning. For example, \(\text{PQ}\) is the power supplied to a fluid, perhaps by a pump, to give it its pressure \(P\). Similarly, \(\cfrac{1}{2}{\mathrm{\rho v}}^{2}Q\) is the power supplied to a fluid to give it its kinetic energy. And \(\rho \text{ghQ}\) is the power going to gravitational potential energy.

Making Connections: Power

Power is defined as the rate of energy transferred, or \(E/t\). Fluid flow involves several types of power. Each type of power is identified with a specific type of energy being expended or changed in form.

Example: Calculating Power in a Moving Fluid

Suppose the fire hose in the previous example is fed by a pump that receives water through a hose with a 6.40-cm diameter coming from a hydrant with a pressure of \(0\text{.}\text{700}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\). What power does the pump supply to the water?

Strategy

Here we must consider energy forms as well as how they relate to fluid flow. Since the input and output hoses have the same diameters and are at the same height, the pump does not change the speed of the water nor its height, and so the water’s kinetic energy and gravitational potential energy are unchanged. That means the pump only supplies power to increase water pressure by \(0\text{.}\text{92}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\) (from \(0.700×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\) to \(1.62×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\)).

Solution

As discussed above, the power associated with pressure is

\(\begin{array}{lll}\text{power}& =& \text{PQ}\\ & =& (\text{0.920}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2})(\text{40}\text{.}0×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\text{/s})\text{.}\\ \text{}& =& 3\text{.}\text{68}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{W}=\text{36}\text{.}8\phantom{\rule{0.25em}{0ex}}\text{kW}\end{array}\)

Discussion

Such a substantial amount of power requires a large pump, such as is found on some fire trucks. (This kilowatt value converts to about 50 hp.) The pump in this example increases only the water’s pressure. If a pump—such as the heart—directly increases velocity and height as well as pressure, we would have to calculate all three terms to find the power it supplies.

Summary

Power in fluid flow is given by the equation \(({P}_{1}+\cfrac{1}{2}{\mathrm{\rho v}}^{2}+\rho \text{gh})Q=\text{power}\text{,}\) where the first term is power associated with pressure, the second is power associated with velocity, and the third is power associated with height.

Power in Fluid Flow