\(\def \u#1{\,\mathrm{#1}}\) \(\def \abs#1{\left|#1\right|}\) \(\def \ast{*}\) \(\def \deg{^{\circ}}\) \(\def \ten#1{\times 10^{#1}}\) \(\def \redcancel#1{{\color{red}\cancel{#1}}}\) \(\def \BLUE#1{{\color{blue} #1}}\) \(\def \RED#1{{\color{red} #1}}\) \(\def \PURPLE#1{{\color{purple} #1}}\) \(\def \th#1,#2{#1,\!#2}\) \(\def \lshift#1#2{\underset{\Leftarrow\atop{#2}}#1}}\) \(\def \rshift#1#2{\underset{\Rightarrow\atop{#2}}#1}}\) \(\def \dotspot{{\color{lightgray}{\circ}}}\)
Chapter 8: Fluids
10.

Flow Rate

Flow Rate
\(\Phi\)
m3/s
The flow rate of a fluid is the total volume $V$ of fluid that travels past a particular point in a given time $\Delta t$:
$$\Phi={V\over \Delta t}$$

Its SI units are cubic meters per second, although this is a very large flow; a more practical unit is the liter per second: $$1{\rm m^3\over s}=1000{\rm L\over s}$$

(no alternate text)
The cross-sectional area of a circular pipe is the area of a circle.
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Rivers have cross-sectional areas too.

The flow rate through a pipe (or channel, or river, or whatever) depends on two things: how big the pipe is (bigger pipes mean greater flow) and how fast the water runs through the pipe. In fact, the flow rate can be found using the formula

$$\Phi=Av$$

where $v$ is the speed of the water, and $A$ is the cross-sectional area of the pipe.

When the fluid is a liquid, so that its volume remains constant, then the total flow rate into an object (a pipe, a box, whatever) must be equal to the total flow rate out of the object: if not, then either there will be a buildup of fluid inside the object (if $\Phi_{\rm in}$ is bigger) or there would have to be a secret cache of fluid inside the object which is being depleted (if $\Phi_{\rm out}$ is bigger). In either case, the situation would not be able to go on for very long. In the case of steady flow, then, or flow which lasts for a long time, we must have $$\Phi_{\rm in}=\Phi_{\rm out}$$

For example, if water flows through a pipe that changes diameter, then we can write $$A_1v_1=A_2v_2$$ If the pipe gets narrower, so that $A_1>A_2$, then the final speed $v_2$ must be greater than the initial speed $v_1$ in order to compensate for the smaller area. Water speeds up when the pipe narrows (picture a hose when you place a finger over the end) and slows down when the pipe (or a river) widens (such as a river which goes from narrow rapids to become wider, deeper, and calmer later on).