What is volume flow rate?

You know all about the motion of individual objects. Now, let's talk about how to analyze the motion of a fluid.

What does volume flow rate mean?

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 QQ 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.
This is a great question. I don't know and it doesn't seem to be discussed anywhere I could find. So here's a challenge for you. If you figure out why physicists use the letter Q for volume flow rate as opposed to some other letter please leave an explanation with a reference in the comments below. I, for one, would love to know. Thanks!
Possible explanation from users in comment section. The latin word for quantity is Quantitatum.

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:
Q=Vt=Volumetime\Large Q=\dfrac{V}{t}=\dfrac{Volume}{time}
In S.I. units (International System of Units), volume flow rate has units of meters cubed per second, m3s\dfrac{\text m^3}{\text s}, 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.
The average resting human heart pumps roughly 330 ml of blood every 4 seconds. So the volume flow rate can be found with
Q=Vt=330 mL4 s=83 mLs=0.083LsQ=\dfrac{V}{t}=\dfrac{ 330 \text{ mL}}{4 \text{ s}}=83 \dfrac{\text{ mL}}{\text{s}} = 0.083 \dfrac{\text{L}}{\text{s}}
We can convert this to conventional S.I. units of meters cubed per second, m3s\dfrac{\text m^3}{s}, by converting liters to meters cubed as follows:
Q=0.083Ls×(1 m31,000L)=0.000083m3sQ=0.083 \dfrac{\text L}{s} \times (\dfrac{1 \text { m}^3}{1,000 \text L}) = 0.000083 \dfrac{\text{m}^3}{\text{s}}
When expressed in meters cubed per second this volume flow rate seems really small, but that's just because a meter cubed is such an enormous volume. A meter cubed is the volume enclosed by a cube with side lengths of 1 m x 1 m x 1 m. That's equivalent to the volume of over 2,850 cans of soda.
Note that volume flow rates are going to be very small numbers when expressed in m3s\dfrac{\text{m}^3}{\text{s}} because the volumes we deal with in many situations are only a small fraction of a meter cubed. For this reason, many people choose to use alternative units for volume flow rate like mLs\dfrac{\text{mL}}{\text{s}} or cm3s\dfrac{\text{cm}^3}{\text{s}} so that the numbers don't come out so annoyingly small.

Is there another formula for volume flow rate?

It turns out there's a useful alternative to writing the volume flow rate as Q=VtQ=\dfrac{V}{t}.
The volume of a portion of the fluid in a pipe can be written as V=AdV=Ad, where AA is the cross sectional area of the fluid and dd is the width of that portion of fluid, see the diagram below. We can substitute this formula for volume VV into the volume flow rate to get the following:
No, the pipe does not have to be cylindrical for this derivation to work. Regardless of the shape of the pipe, you could just consider a portion of volume that has a width small enough that the shape and size of the cross sectional area doesn't change by much. That way the volume could be still be represented approximately by the area, AA, times that very small width, dd. But since the distance dd ends up being divided by the time, tt, in this derivation, we can make the width infinitesimally small which would give us the instantaneous speed, dt\dfrac{\text d}{\text t}, of the fluid at that point.
Q=Vt=Adt=AdtQ=\dfrac{V}{t}=\dfrac{Ad}{t}=A\dfrac{d}{t}
But the term dt\dfrac{d}{t} 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 dt\dfrac{d}{t} with vv in the previous equation and get
Q=Av\Large Q=Av
AA is the cross sectional area of a section of the pipe, and vv is the speed of the fluid in that section. So, we get a new formula for the volume flow rate Q=AvQ=Av that is often more useful than the original definition of volume flow rate because the area AA is easy to determine. Most pipes are cylindrical—which means the area can be found with A=πr2A=\pi r^2—and the speed vv 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 VV, and the speed is represented with a lowercase letter vv. People often mix up the notation for volume, VV, and speed, vv, since they look so similar.

Incompressibility of liquids

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.
No, not really. To be completely honest, liquids are not absolutely incompressible. They are just extremely hard to compress by any noticeable amount. For instance, it takes a pressure of around 400 times atmospheric pressure just to compress 1 meter cubed of water down to a volume of 0.98 meters cubed. So essentially, unless you are considering a situation where the pressures are staggeringly large, it's usually pretty safe to just assume that liquids are completely incompressible.
Gases on the other hand are much more easily compressible than liquids, so neglecting the compression of a gas is more likely to cause significant errors in the description of its behavior.
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, VV, 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.

What is the equation of continuity?

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 m3^3 of water into a pipe that is already full of water, 2 m3^3 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 QQ 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=constantQ=constant, 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
Q1=Q2\Large Q_1=Q_2
Now if we substitute the formula Q=VtQ=\dfrac{V}{t}, we get
V1t1=V2t2\Large \dfrac{V_1}{t_1}=\dfrac{V_2}{t_2}
Alternatively, we could plug in the alternative form of the volume flow rate, Q=AvQ=Av, into the formula, Q1=Q2Q_1=Q_2, which would give us
A1v1=A2v2\Large A_1 v_1 =A_2 v_2
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 AvAv has a constant value throughout the pipe. In other words, no matter where in the pipe you choose to find AvAv, the value will always come out to be the same number for a given pipe, if the fluid is incompressible.
So, if the area, AA, of a section of pipe decreases, the speed, vv, of the liquid there must increase so that the product, AvAv, 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, AA. The water must come out with higher speed, vv, to ensure the volume flow rate, AvAv, remains the same. This is why narrow nozzles, which reduce the area (AA), attached to water hoses cause a significant increase in the speed, vv, of the fluid at that point.

What do solved examples involving volume flow rate look like?

Example 1: Mountain Dew dream house

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 m2^2 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 m2^2.
What is the speed of the Mountain Dew as it exits the faucet pipe in the lady's bedroom?
A1v1=A2v2(Start with the equation of continuity since liquids are incompressible.)A_1 v_1 =A_2 v_2 \quad(\text{Start with the equation of continuity since liquids are incompressible.})
v2=(A1A2)v1(Solve symbolically for the speed of the liquid at the bedroom.)v_2 = (\dfrac{A_1}{A_2})v_1 \quad(\text{Solve symbolically for the speed of the liquid at the bedroom.})
v2=0.0036m20.0012m2(0.48 m/s)(Plug in the values for the areas and speed.)v_2 = \dfrac{0.0036 \text m^2}{0.0012 \text m^2}(0.48 \text{ m/s}) \quad(\text{Plug in the values for the areas and speed.})
v2=1.44 m/s(Calculate and celebrate!)v_2=1.44 \text{ m/s} \quad(\text{Calculate and celebrate!})
Note: We could have also solved this problem just by noticing that the area, A2A_2, of the pipe in the bedroom was 13\dfrac{1}{3} the area of the pipe downstairs, A1A_1. 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 AvAv to remain the same.

Example 2: Coconut-milk cupcakes

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.
What is the radius of the tube at the kitchen through which the coconut milk exits?
A1v1=A2v2(Start with the equation of continuity since liquids are incompressible.)A_1 v_1 =A_2 v_2 \quad(\text{Start with the equation of continuity since liquids are incompressible.})
π(r1)2v1=π(r2)2v2(Plug in the formula  for the cross sectional area of the cylindrical pipe.πr2)\pi (r_1)^2 v_1 =\pi (r_2)^2 v_2 \quad(\text{Plug in the formula $\pi r^2$ for the cross sectional area of the cylindrical pipe.})
(r1)2v1=(r2)2v2(Cancel the common factor of .π)(r_1)^2 v_1 = (r_2)^2 v_2 \quad(\text{Cancel the common factor of $\pi$.})
(r2)2=(r1)2v1v2(Symbolically solve for the square of the radius of the pipe at the kitchen.)(r_2)^2 = (r_1)^2 \dfrac{v_1}{v_2} \quad(\text{Symbolically solve for the square of the radius of the pipe at the kitchen.})
r2=r1v1v2(Take a square root of both sides.)r_2 = r_1 \sqrt{\dfrac{v_1}{v_2}} \quad(\text{Take a square root of both sides.})
r2=(4 cm)0.25 m/s1.0 m/s(Plug in values for the radius and speeds.)r_2 = (4 \text{ cm}) \sqrt{\dfrac{0.25 \text{ m/s}}{1.0 \text{ m/s}}} \quad(\text{Plug in values for the radius and speeds.})
r2=2 cm or 0.02 mr_2 = 2\text{ cm} \text{ or } 0.02\text { m}(Calculate and celebrate!)\quad(\text{Calculate and celebrate!})
Note: We plugged in our radius, r1=4 cmr_1=4\text{ cm}, in units of centimeters, which just means that our answer came out in units of centimeters.
Nave, R. "Bulk Elastic Properties." HyperPhysics. http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html.
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