What is volume flow rate?

You might hear the term volume flow rate and think it sounds boring, but volume flow rate keeps you alive. I'll tell you how in a second, but first we should define volume flow rate. The volume flow rate $Q$Q of a fluid is defined to be the volume of fluid that is passing through a given cross sectional area per unit time. The term cross sectional area is just a fancy term often used to describe the area through which something is flowing, e.g., the circular area inside the dashed line in the diagram below.

Since volume flow rate measures the amount of volume that passes through an area per time, the equation for the volume flow rate looks like this:

In S.I. units (International System of Units), volume flow rate has units of meters cubed per second, $\dfrac{\text m^3}{\text s}$start fraction, start text, m, end text, cubed, divided by, start text, s, end text, end fraction, since it tells you the number of cubic meters of fluid that flow per second.

So how does volume flow rate keep you alive? Your heart pumps a volume of blood roughly equal to the volume of a can of soda every four seconds.

It turns out there's a useful alternative to writing the volume flow rate as $Q=\dfrac{V}{t}$Q, equals, start fraction, V, divided by, t, end fraction.

The volume of a portion of the fluid in a pipe can be written as $V=Ad$V, equals, A, d, where $A$A is the cross sectional area of the fluid and $d$d is the width of that portion of fluid, see the diagram below. We can substitute this formula for volume $V$V into the volume flow rate to get the following:

But the term $\dfrac{d}{t}$start fraction, d, divided by, t, end fraction is just the length of the volume of fluid divided by the time it took the fluid to flow through its length, which is just the speed of the fluid. So we can replace $\dfrac{d}{t}$start fraction, d, divided by, t, end fraction with $v$v in the previous equation and get

$A$A is the cross sectional area of a section of the pipe, and $v$v is the speed of the fluid in that section. So, we get a new formula for the volume flow rate $Q=Av$Q, equals, A, v that is often more useful than the original definition of volume flow rate because the area $A$A is easy to determine. Most pipes are cylindrical—which means the area can be found with $A=\pi r^2$A, equals, pi, r, squared—and the speed $v$v of the fluid is a quantity that is of particular interest in many situations.

Be careful though, we're now dealing with two terms that look very similar. The volume is represented with a capital letter $V$V, and the speed is represented with a lowercase letter $v$v. People often mix up the notation for volume, $V$V, and speed, $v$v, since they look so similar.

It turns out that most liquids are nearly incompressible. This means that a gallon of milk can be put into a differently shaped gallon-sized container, but you wouldn't be able to squeeze that entire gallon of milk into a half-gallon-sized container no matter how hard you squeeze.

Because liquids are incompressible, any portion of liquid flowing through a pipe could change shape, but it must maintain the same volume. This is true even if the pipe changes diameter. In the diagram below, the volume, $V$V, of liquid on the left changes shape as it enters a narrow section of pipe, but it maintains the same volume since liquids are incompressible.

Liquids must maintain their volume as they flow in a pipe since they are nearly incompressible. This means that the volume of liquid that flows into a pipe in a given amount of time must equal the volume of liquid that flows out of a pipe in the same amount of time. For instance, if in one hour you pump 2 m$^3$cubed of water into a pipe that is already full of water, 2 m$^3$cubed has to flow out of that pipe during that same hour. The only alternatives would be for the liquid to compress inside the pipe—which shouldn't happen—or the pipe balloons in size—which we assume doesn't happen if the pipe is rigid. Remember, you're not confined to considering points only at the beginning or end of the pipe, this argument works just as well for water entering and exiting any two sections of the pipe.

So, the volume flow rate $Q$Q for an incompressible fluid at any point along a pipe is the same as the volume flow rate at any other point along a pipe.

This can be represented mathematically with the formula $Q=constant$Q, equals, c, o, n, s, t, a, n, t, or—choosing any two points in the pipe—we can state mathematically that the volume flow rate is the same at any two points by writing

Now if we substitute the formula $Q=\dfrac{V}{t}$Q, equals, start fraction, V, divided by, t, end fraction, we get

Alternatively, we could plug in the alternative form of the volume flow rate, $Q=Av$Q, equals, A, v, into the formula, $Q_1=Q_2$Q, start subscript, 1, end subscript, equals, Q, start subscript, 2, end subscript, which would give us

This equation is known as the equation of continuity for incompressible fluids—the previous two equations are also sometimes referred to as the equation of continuity. The equation isn't really as mysterious as the name suggests since we found it simply by requiring that volumes be incompressible as they flow through a pipe.

The equation is quite useful though, particularly in this form, since it says that the value of $Av$A, v has a constant value throughout the pipe. In other words, no matter where in the pipe you choose to find $Av$A, v, the value will always come out to be the same number for a given pipe, if the fluid is incompressible.

So, if the area, $A$A, of a section of pipe decreases, the speed, $v$v, of the liquid there must increase so that the product, $Av$A, v, remains the same. This means that fluids speed up when they reach a narrow section of a pipe and slow down when they reach a wider section of a pipe. This matches everyday experience—think about what happens if you block a portion of the water hose with your thumb, effectively reducing its area, $A$A. The water must come out with higher speed, $v$v, to ensure the volume flow rate, $Av$A, v, remains the same. This is why narrow nozzles, which reduce the area ($A$A), attached to water hoses cause a significant increase in the speed, $v$v, of the fluid at that point.

A very wealthy woman who loves soda builds her house with a cylindrical pipe that transports Mountain Dew from downstairs to her upstairs bedroom. The Mountain Dew enters the house downstairs via a pipe with a cross sectional area of 0.0036 m$^2$squared where it is traveling with a speed of 0.48 meters per second. At the wealthy lady's bedroom, the faucet pipe through which the Mountain Dew exits has an area of 0.0012 m$^2$squared.

Note: We could have also solved this problem just by noticing that the area, $A_2$A, start subscript, 2, end subscript, of the pipe in the bedroom was $\dfrac{1}{3}$start fraction, 1, divided by, 3, end fraction the area of the pipe downstairs, $A_1$A, start subscript, 1, end subscript. This means that the speed of the Mountain Dew has to be going three times as fast in the bedroom pipe, compared to the downstairs pipe, in order for the factor $Av$A, v to remain the same.

A chef wants to make sure he always has coconut milk ready for all his cupcake recipes, so he creates a cylindrical pipe that goes from the storeroom to the kitchen. The pipe at the storeroom has a radius of 4 cm where the coconut milk has a speed of 0.25 meters per second. The coconut milk exits the tube in the kitchen with a speed of 1 meter per second.

Note: We plugged in our radius, $r_1=4\text{ cm}$r, start subscript, 1, end subscript, equals, 4, start text, space, c, m, end text, in units of centimeters, which just means that our answer came out in units of centimeters.