Fluids in Motion: Continuity and Bernoulli - Physics for Pilot Assessments

Fluids in motion behave in a way that is genuinely counter-intuitive, and the test banks know it: where a fluid speeds up, its pressure falls, and almost everybody's first instinct is the opposite. By the end of this lesson you should be able to work out how a pipe's width changes the speed of the flow through it, split a moving fluid's pressure into the part that pushes sideways and the part that comes from its motion, and explain why the narrow throat of a venturi is the point of lowest pressure.

Flow rate: what is being conserved

Start with the thing that cannot change. Consider water flowing steadily through a pipe, steadily meaning that the picture is not changing with time — the same amount passes any given point every second. The natural measure of that is the volume flow rate: the volume of fluid crossing a cross-section each second, in cubic metres per second. If the fluid in a pipe of cross-sectional area A is moving at speed v, then in one second every particle advances a distance v, so the fluid that crosses the cross-section during that second is a slug of length v and area A — a volume of A multiplied by v.

Volume flow rate

Q = Av

$Q$: the volume crossing a cross-section each second, in cubic metres per second (m³/s)
$A$: the cross-sectional area of the pipe, in square metres (m²)
$v$: the speed of the flow, in metres per second (m/s)

Cross-sectional area means the area of the slice you would see if you cut the pipe straight across — for a circular pipe, the area of that circle. It is worth reading the formula in units to see that it makes sense: square metres multiplied by metres per second gives cubic metres per second, which is a volume per unit of time, exactly as claimed.

Continuity: narrower means faster

Now take a pipe that is wide at one end, narrows in the middle and widens again, completely full of water, with no leaks and nothing being added along the way. Whatever volume enters each second must leave each second. It cannot pile up inside, because the water is incompressible — it cannot be squeezed into a smaller space to make room for more, as the hydraulics lesson established — and it has nowhere else to go. So the volume flow rate is the same at every cross-section along the pipe, however the pipe's width changes. Since that rate is the area multiplied by the speed, and it cannot change, a smaller area forces a larger speed. This is the equation of continuity, and it is why a river runs fastest where its channel narrows and why putting your thumb over the end of a hose makes the water shoot further.

Continuity

A_1 v_1 = A_2 v_2

$A_1$: the cross-sectional area at the first point, in square metres (m²)
$v_1$: the flow speed at the first point, in metres per second (m/s)
$A_2$: the cross-sectional area at the second point, in square metres (m²)
$v_2$: the flow speed at the second point, in metres per second (m/s)

Read it as a see-saw: the product of the two cannot change, so if the area halves the speed must double to compensate, and a third of the area gives three times the speed. Nothing about the fluid changed and nothing pushed harder — the same amount of water simply has less room to pass through, and passing the same amount through less room can only be done faster.

The same volume passes every second, so the flow is fastest where the pipe is narrowest. The tapping above the throat reads the lowest static pressure of the three, and the pressure recovers as the pipe widens again.

Example 1 — the pipe narrows

Water flows at 3 metres per second through a pipe of cross-sectional area 0.02 square metres. The pipe then narrows to 0.01 square metres. How fast does the water move in the narrow part?

Find the flow rate in the wide part: $Q = A \times v = 0.02 \times 3 = 0.06\ \text{m}^3/\text{s}$
That same 0.06 cubic metres per second must pass through the narrow part, because there is nowhere else for it to go.
Rearrange Q = A × v for the speed by dividing both sides by the area, which leaves $v = \frac{Q}{A}$ : $v = \frac{0.06}{0.01} = 6\ \text{m/s}$
The area halved and the speed doubled, which is the entire content of continuity — you could have written the answer down without calculating the flow rate at all.

Three pressures, not one

A moving fluid has more than one pressure worth naming, and most confusion about Bernoulli is really confusion about which one is meant. Static pressure is the pressure the fluid exerts on a surface parallel to the flow — the sideways push on the pipe wall, the pressure that a small hole drilled into the wall measures, and the pressure that would still be there if the flow stopped and the fluid simply sat in the pipe. Dynamic pressure is the extra pressure that comes from the fluid's motion: the amount by which the push rises on a surface that faces directly upstream and so brings the flow to a halt. Total pressure, also called stagnation pressure, is the sum of the two, and it is what a forward-facing open tube held in the flow measures, because the fluid entering that tube is brought to rest.

How big is the dynamic pressure? It depends on how much fluid is being stopped and on how fast it was going, so density and speed are its two ingredients — and the speed enters squared, so doubling the speed quadruples the dynamic pressure. There is also a factor of one half, which comes out of the mechanics of bringing a moving mass to rest.

Dynamic pressure

q = \tfrac{1}{2}\rho v^2

$q$: dynamic pressure — the pressure due to the motion, in pascals (Pa)
$\rho$: the density of the fluid, in kilograms per cubic metre (kg/m³)
$v$: the speed of the flow, in metres per second (m/s)

The squared term deserves respect, because it means dynamic pressure grows far faster than speed does: three times the speed is nine times the dynamic pressure. It also means that in slow flows the dynamic pressure is a small addition to the static pressure, while in fast ones it dominates. Adding the two gives the total.

Total pressure

p_\text{total} = p_\text{static} + q

$p_\text{total}$: total (stagnation) pressure, in pascals (Pa)
$p_\text{static}$: static pressure — the sideways push on the wall, in pascals (Pa)
$q$: dynamic pressure, in pascals (Pa)

That sum is the piece of bookkeeping everything else rests on, because for a steady flow of an incompressible fluid along a level pipe, with friction ignored, the total stays the same from one point to another. Which is what the next section is about.

Why speeding up costs static pressure

Here is the mechanism, and it is worth going through properly because the result is the one people get wrong. Follow a small parcel of water along the pipe as it approaches the narrowing. Continuity says it must be moving faster once it is in the throat than it was in the wide section, so it has to accelerate on the way in. From the mechanics module, accelerating anything requires a net force in the direction of the acceleration, and a parcel of water is no exception. What could supply that forward force? Not gravity, in a level pipe. The only candidate left is the fluid around it — that is, a difference in pressure between the parcel's back and its front. For the parcel to be pushed forwards, the pressure behind it must exceed the pressure ahead of it. So the pressure ahead, in the throat, must be lower than the pressure behind, in the wide section. The fall in static pressure is not some curiosity that happens to accompany the speed-up; it is what causes it.

Run the same argument on the far side of the throat and it works in reverse. The flow has to slow down again as the pipe widens, which needs a net backward force, which needs a higher pressure ahead than behind — so the static pressure recovers as the pipe opens out. That is why the third tapping in the figure stands as high as the first. In a real pipe it does not recover quite fully, because friction takes a small permanent share.

Bernoulli's principle, along a level pipe

p_\text{static} + \tfrac{1}{2}\rho v^2 = \text{constant}

$p_\text{static}$: static pressure at the point, in pascals (Pa)
$\rho$: the density of the fluid, in kilograms per cubic metre (kg/m³)
$v$: the flow speed at the point, in metres per second (m/s)

In words: static pressure and dynamic pressure trade against one another and their sum is fixed, so speeding up raises the dynamic pressure and the static pressure must fall by exactly the same number of pascals, while slowing down does the reverse. The conditions attached to that are real and worth stating rather than hiding: the flow must be steady, the fluid effectively incompressible, the friction negligible, and the pipe level — a change of height brings the fluid's own ρgh into the accounting as well. Every venturi item you will meet is arranged so that those conditions hold.

Example 3 — dynamic pressure, both directions

Air of density 1.25 kilograms per cubic metre flows at 20 metres per second. What is its dynamic pressure? And at what speed would the dynamic pressure be 1000 pascals?

Dynamic pressure is a half times the density times the speed squared. Square the speed first: $v^2 = 20 \times 20 = 400$
Half of 1.25 is 0.625, and 0.625 × 400 = 250 pascals.
For the reverse, start from q = ½ρv² and get v² by itself. Multiply both sides by 2, giving 2q = ρv², then divide both sides by ρ, giving $v^2 = \frac{2q}{\rho}$ .
So $v^2 = \frac{2 \times 1000}{1.25} = \frac{2000}{1.25} = 1600$ .
Now take the square root of 1600, which means finding the number that multiplied by itself gives 1600. Since 40 × 40 = 1600, the speed is 40 metres per second.
Sanity-check the relationship: the speed went from 20 to 40, a doubling, and the dynamic pressure went from 250 to 1000, a quadrupling — which is what the square demands.

Example 4 — the pressure drop at a venturi throat

Water of density 1000 kilograms per cubic metre flows along a level pipe at 2 metres per second, where the static pressure is 30 000 pascals. The pipe then narrows to half its cross-sectional area. What is the static pressure in the throat?

Continuity first: half the area means twice the speed, so the water moves at 4 metres per second in the throat.
Dynamic pressure in the wide section: $q = \tfrac{1}{2} \times 1000 \times 2^2 = 500 \times 4 = 2000\ \text{Pa}$
Total pressure there = static + dynamic = 30 000 + 2000 = 32 000 pascals.
Dynamic pressure in the throat: $q = \tfrac{1}{2} \times 1000 \times 4^2 = 500 \times 16 = 8000\ \text{Pa}$
The total is the same all along the pipe, so the static pressure in the throat = 32 000 − 8000 = 24 000 pascals.
Doubling the speed therefore cost 6000 pascals of static pressure. Note that the static pressure fell while the total did not — two statements that are only compatible if you keep the two pressures apart, which is the whole reason for the vocabulary.

Where you meet it

A venturi — a deliberate narrowing in a pipe or duct — is the standard way of putting the effect to work. Tap the wall at the throat and you have a source of reduced pressure that can draw something in: a scent atomiser and a garden-hose fertiliser mixer both work by letting the low pressure at a throat pull liquid up a side tube into the passing stream. Tap it in two places instead, one in the wide section and one at the throat, and the difference between the two static pressures tells you the speed of the flow and therefore the flow rate — which is exactly what a venturi flow meter is. The same principle explains why a shower curtain swings inwards towards a fast jet of water, and why two boats running close and parallel are drawn together: fast flow between them, reduced static pressure between them, and the ordinary pressure outside pushes them in.

One caution, because this is the most commonly overstated result in the course. The static pressure falls where the fluid speeds up along its own path; it does not follow that any fast-moving fluid anywhere is at a low pressure compared with any other fluid somewhere else. A fire hose at high pressure delivers fast water — high pressure and high speed together — because the comparison being made there is with the still air outside, not with a point further back along the same flow. Bernoulli compares two points on one flow, and the moment you start comparing across unrelated flows the statement stops meaning anything.

Why it matters in aviation

This lesson appears twice in flight: once in the air passing over the wing, and once inside the instruments that measure the flight. An aerofoil works by changing the speed of the air over each of its surfaces, and the accompanying difference in static pressure between the upper and lower surfaces is what the structure carries as lift — which is also why the same wing produces very different lift at different airspeeds, since the dynamic pressure available grows with the square of the speed.

The instruments use the same bookkeeping deliberately. A pitot tube faces into the airflow, so the air entering it is brought to rest and the tube senses total pressure, while static ports set flush with the fuselage sense the static pressure alone. An airspeed indicator is a device that takes the difference between those two and displays the remaining dynamic pressure as a speed, which is why a blocked or iced pitot or static opening produces a characteristically wrong reading rather than no reading. The temperature side of the same effect matters in carburetted piston engines: air accelerating and expanding through a venturi cools sharply, so moisture can freeze in the induction system while the outside air is still well above freezing.

What the test asks

The single most examined misconception in this module is the intuition that faster flow must mean higher pressure. It feels right, because fast water hurts more — but what you feel when a jet hits you is dynamic pressure, and the quantity Bernoulli says falls is the static pressure, the sideways push on the wall. Expect an item showing a pipe with a narrow section and asking where the pressure is lowest: the answer is the narrow section, where the flow is fastest, and the confident-sounding distractor says the opposite. Expect a continuity item too, where the trap is the proportionality the wrong way round, with halving the area offered as halving the speed. And watch for items that say only "the pressure" without saying which one: if it were the total pressure that changed along the pipe, nothing in this lesson would work.