Put your thumb over the end of a garden hose and the water shoots out faster. Nothing new pushed it; the water simply had to get through a smaller opening. This lesson turns that into a rule, using the pressure picture of a liquid and a proportion.
Why it works
A liquid barely compresses. In a pipe that is full, water cannot pile up anywhere, so every second the same volume must pass every cross-section. In a time $\Delta t$ the water moves a distance $v\,\Delta t$, so the volume passing is $A v\,\Delta t$, and the volume per second is $$Q = Av.$$ If $Q$ is the same at two places, then $A_1 v_1 = A_2 v_2$. Halve the area and the speed must double.
Key idea
Flow rate $Q = Av$ (in $\text{m}^3\text{/s}$; $1\,\text{L} = 10^{-3}\,\text{m}^3$). For a liquid filling a pipe, $A_1 v_1 = A_2 v_2$: the liquid moves fastest where the pipe is narrowest.
Worked examples
Worked example: A nozzle
Water moves at $1.5\,\text{m/s}$ in a hose of area $2.0\,\text{cm}^2$. The nozzle opening is $0.50\,\text{cm}^2$. How fast does the water leave?
- $A_1 v_1 = A_2 v_2$, so $v_2 = v_1 \dfrac{A_1}{A_2}$.
- $v_2 = (1.5\,\text{m/s}) \dfrac{2.0\,\text{cm}^2}{0.50\,\text{cm}^2} = (1.5)(4) = 6.0\,\text{m/s}$.
- The areas form a ratio, so they can stay in $\text{cm}^2$.
Answer: $6.0\,\text{m/s}$
Worked example: Filling a bucket
Water flows at $2.5\,\text{m/s}$ through a pipe of area $4.0\,\text{cm}^2$. What is the flow rate, and how long does a $10\,\text{L}$ bucket take to fill?
- Convert the area: $4.0\,\text{cm}^2 = 4.0 \times 10^{-4}\,\text{m}^2$.
- $Q = Av = (4.0 \times 10^{-4}\,\text{m}^2)(2.5\,\text{m/s}) = 1.0 \times 10^{-3}\,\text{m}^3\text{/s} = 1.0\,\text{L/s}$.
- $t = V/Q = (10\,\text{L})/(1.0\,\text{L/s}) = 10\,\text{s}$.
Answer: $1.0\,\text{L/s}$; $10\,\text{s}$
Worked example: Half the diameter
A round pipe narrows from a diameter of $4.0\,\text{cm}$ to $2.0\,\text{cm}$. Water moves at $0.50\,\text{m/s}$ in the wide part. How fast in the narrow part?
- Area goes as diameter squared ($A = \pi d^2/4$), so $\dfrac{A_1}{A_2} = \left(\dfrac{4.0}{2.0}\right)^2 = 4$.
- $v_2 = (0.50\,\text{m/s})(4) = 2.0\,\text{m/s}$. Half the diameter means four times the speed.
Answer: $2.0\,\text{m/s}$
Common mistake
Thinking water slows down where a pipe narrows, as if it were squeezed or held back. In a full pipe it must speed up, or the water behind it would have nowhere to go. Rivers show it too: the same water that drifts through a wide, deep pool races through narrow rapids. Check: does $A \times v$ come out the same at both places?
Air compresses, so $A_1 v_1 = A_2 v_2$ holds for a gas only when its density stays about the same, as in slow airflow.