Visual walkthrough — Applications — Pitot tube, Venturi meter, orifice flow

This page derives ONE result in pictures: the Venturi flow rate, the formula that lets a pinched pipe measure how much water is racing through it. We start from a picture of water in a pipe and end at a boxed equation, one small step at a time. Every symbol is drawn before it is used.

Parent: Applications — Pitot / Venturi / Orifice.

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Step 1 — Draw the pipe and name what we see

WHAT. Picture a horizontal pipe that is wide on the left and pinched in the middle. Water flows left to right. We mark two spots along one imaginary line the water follows — call that line a streamline (the path one water particle traces).

• Spot 1 sits in the wide part. It has cross-section area $A_1$ (how big the circle of pipe is there, in square metres) and the water there moves at speed $v_1$ (metres per second).
• Spot 2 sits in the narrow throat. Its area is $A_2$ and speed is $v_2$.

WHY. Before any equation, we must know what every letter points at in the real picture. $A$ is an area, $v$ is a speed, the little number is which spot. Nothing else enters the whole derivation.

PICTURE. Look at the two dashed circles below: the big one (blue) is $A_1$, the small one (yellow) is the throat $A_2$. The arrow shows flow direction.

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Step 2 — The narrow part MUST run faster (Continuity)

WHAT. The same amount of water that enters the wide part per second has to leave the throat per second — water is not created or destroyed and (for our purposes) not squashed. The amount crossing a spot each second is $\text{area}\times \text{speed}$. Call it $Q$, the volume flow rate (cubic metres per second). So:

$\underset{\text{wide area}}{\underbrace{A_1}}\underset{\text{wide speed}}{\underbrace{v_1}}=\underset{\text{throat area}}{\underbrace{A_2}}\underset{\text{throat speed}}{\underbrace{v_2}}=\underset{\text{same for all spots}}{\underbrace{Q}}$

WHY. This is the Continuity Equation. We use it because it is the only rule that connects the two unknown speeds through the pipe geometry. Without it, $v_1$ and $v_2$ would be two independent mysteries.

Since $A_2<A_1$ (throat is smaller) and their product must stay equal, the throat speed $v_2$ is forced to be larger. Picture a wide slow river squeezing into a narrow fast rapid.

PICTURE. Same water per second through a smaller window ⇒ it has to go faster. The short thick arrow (slow, wide) becomes a long thin arrow (fast, narrow).

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Step 3 — Faster water pushes less: Bernoulli's balance

WHAT. Along that same streamline, Bernoulli's rule says a certain sum stays constant. Because the pipe is horizontal, the height is the same at both spots, so the height term is identical on both sides and simply cancels. What remains:

$\underset{\begin{array}{c}\text{pressure}\\ \text{at spot 1}\end{array}}{\underbrace{P_1}}+\underset{\begin{array}{c}\text{"motion pressure"}\\ \text{at spot 1}\end{array}}{\underbrace{\frac{1}{2}\rho v_1^2}}=\underset{\begin{array}{c}\text{pressure}\\ \text{at spot 2}\end{array}}{\underbrace{P_2}}+\underset{\begin{array}{c}\text{"motion pressure"}\\ \text{at spot 2}\end{array}}{\underbrace{\frac{1}{2}\rho v_2^2}}$

Here $P$ is pressure (push per unit area, in pascals) and $\rho$ (Greek "rho") is the fluid's density (kilograms per cubic metre — how heavy the fluid is per box of it). The term $\frac{1}{2}\rho v^2$ is the dynamic pressure: the part of the push that comes purely from motion.

WHY. We use Bernoulli's Equation because it is the one law that trades speed against pressure. Step 2 told us the throat is faster; Bernoulli now converts "faster" into "lower pressure". We saw motion goes up at the throat, so pressure must come down to keep the sum fixed — see Dynamic vs Static vs Stagnation Pressure.

PICTURE. The bar chart below: at each spot, a green pressure bar plus a red motion bar reach the same total height. Where the red (motion) bar is tall (throat), the green (pressure) bar is short.

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Step 4 — Rearrange to isolate the pressure difference

WHAT. Move both pressures to one side and both motion terms to the other:

$\underset{\text{the drop we can measure}}{\underbrace{P_1-P_2}}=\underset{\text{extra motion gained at throat}}{\underbrace{\frac{1}{2}\rho \left(v_2^2-v_1^2\right)}}$

WHY. The left side, $P_1-P_2$, is exactly what an instrument can read — the pressure is higher in the wide part and lower in the throat, and a manometer reports that difference directly. We are organizing the equation so the measurable thing sits alone.

Notice $v_2>v_1$ from Step 2, so $v_2^2-v_1^2>0$, so $P_1-P_2>0$: the wide part really is at higher pressure. The sign comes out correct automatically.

PICTURE. A see-saw: on the left pan sits the pressure gap; on the right pan sits the motion gain. They balance.

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Step 5 — Kill one unknown using Continuity again

WHAT. We still have two speeds. From Step 2, $A_1{v}_1=A_2{v}_2$, so

$v_1=\frac{A_2}{A_1}{v}_2.$

Substitute this into $v_2^2-v_1^2$:

$v_2^2-v_1^2=v_2^2-{\left(\frac{A_2}{A_1}\right)}^2{v}_2^2=v_2^2\left(1-\frac{A_2^2}{A_1^2}\right).$

So Step 4 becomes:

$P_1-P_2=\frac{1}{2}\rho v_2^2\left(\underset{\text{shrink factor}}{\underbrace{1-\frac{A_2^2}{A_1^2}}}\right)$

WHY. An equation with two unknowns ($v_1$ and $v_2$) can't be solved. Continuity lets us write $v_1$ in terms of $v_2$, leaving a single unknown $v_2$. That is why we invoke continuity a second time — not to find speeds yet, but to eliminate one of them.

PICTURE. Two speed-boxes ($v_1$, $v_2$) merge into one, with the small area-ratio $A_2/A_1$ acting as a shrinking dial.

Recall Why is

$1-A_2^2/A_1^2$ between 0 and 1? Because $A_2<A_1$, the fraction $A_2^2/A_1^2$ is a positive number less than 1, so subtracting it from 1 leaves a positive number less than 1. It never turns negative. Meaning of the shrink factor ::: how much of the throat's motion actually shows up as a readable pressure drop.

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Step 6 — Solve for the throat speed

WHAT. Divide both sides by $\frac{1}{2}\rho \left(1-A_2^2/A_1^2\right)$ and take the positive square root (speed is positive):

$v_2=\sqrt{\frac{2(P_1-P_2)}{\rho \left(1-\frac{A_2^2}{A_1^2}\right)}}$

WHY. The square root appears because speed sat squared in the dynamic-pressure term $\frac{1}{2}\rho v^2$. Undoing "squared" means taking a root — that is the only tool that answers "what number, squared, gives this?" We keep the positive root because water in the throat has a real forward speed, not a negative one.

PICTURE. Turning $v_2^2$ back into $v_2$: the parabola-to-line "undo squaring" arrow.

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Step 7 — Turn speed into flow rate $Q$

WHAT. The quantity we actually want is $Q$, the litres-per-second of water. From Step 2, $Q=A_2{v}_2$. Multiply the Step-6 speed by $A_2$ and tidy the algebra (multiply top and bottom inside the root by $A_1^2$ so the messy fraction disappears):

$Q=A_2{v}_2=\boxed{A_1{A}_2\sqrt{\frac{2(P_1-P_2)}{\rho (A_1^2-A_2^2)}}}$

Term by term:

• $A_1{A}_2$ — the two areas set the overall scale (bigger pipe, bigger flow).
• $2(P_1-P_2)$ — the measured pressure drop; doubling it multiplies $Q$ by only $\sqrt{2}$.
• $\rho$ — a heavier fluid flows slower for the same push.
• $A_1^2-A_2^2$ — the geometry of the squeeze; if the pipe barely narrows ($A_2\to A_1$) this vanishes.

WHY. Pressure differences are what we measure; flow rate is what we want. This last multiply by $A_2$ bridges the two.

PICTURE. The finished formula tree — measured $\Delta P$ enters, $Q$ comes out.

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Step 8 — The degenerate cases (never leave the reader stranded)

WHAT & WHY, case by case:

• No squeeze ($A_2=A_1$). Then $A_1^2-A_2^2=0$, the denominator is zero, and the formula blows up — unless $P_1-P_2$ is also zero. A pipe of constant width has no pressure drop and gives no reading: nothing narrows, nothing speeds up. The meter needs a real pinch.
• Zero flow ($Q=0$). Then $P_1-P_2=0$: still water shows equal pressure at both spots. Correct — no motion, no dynamic dip.
• Tiny throat ($A_2\ll A_1$). Then $A_1^2-A_2^2\approx A_1^2$, and $Q\approx A_2\sqrt{2(P_1-P_2)/\rho }$ — the wide-side speed becomes negligible, so it looks like a plain Pitot reading of the throat jet.
• Sign check. $P_1-P_2$ is always positive (wide side higher), so the number under the root is positive and $Q$ is real. If a manometer ever said $P_2>P_1$, you have swapped the ports.

PICTURE. Three mini-pipes: no-pinch (flat pressure), normal pinch (dip), extreme pinch (deep dip). The flat one has no dip to read.

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The one-picture summary

The whole chain in one frame: Continuity forces the throat to speed up → Bernoulli turns that speed-up into a pressure dip → algebra swaps the readable pressure drop for the flow rate $Q$.

Recall Feynman retelling — say it to a friend

Water runs through a fat pipe with a pinched middle. Because the same water-per-second has to get through the skinny part, it has to rush there — like cars merging into one lane. But rushing water leans less on the pipe walls, so the pinched spot has lower pressure than the fat spot. We stick two pressure gauges in, read how much the pressure dropped, and run it backwards: a bigger drop means the water was rushing harder, which means more water per second. We wrote that "backwards" recipe as one boxed formula, $Q=A_1{A}_2\sqrt{2\Delta P/[\rho (A_1^2-A_2^2)]}$. And we checked the weird cases: a pipe that never narrows shows no drop and measures nothing, while dead-still water shows equal pressure everywhere — exactly as common sense demands.

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Connections

• Bernoulli's Equation — Steps 3–4, the speed-for-pressure trade.
• Continuity Equation — Steps 2 and 5, links the two speeds.
• Dynamic vs Static vs Stagnation Pressure — what $\frac{1}{2}\rho v^2$ means.
• Manometers and Pressure Measurement — how $P_1-P_2$ is actually read.
• Torricelli's Law — the same energy idea for a draining tank.
• Projectile Motion — companion for the orifice jet.