Topic 07 · 13 min · 5 figures

The Bernoulli equation

One equation gets asked to do more work, in more places, than almost anything else in this course — and gets misapplied more often, because its assumptions are usually printed as a footnote instead of priced as a cost. Written honestly, it is four conditions and one line of algebra, and the line of algebra is the easy part.

Nothing about the Bernoulli equation is difficult to write down. The difficulty, and the reason it deserves a topic of its own rather than a paragraph in fluid kinematics, is that it looks like a universal law of fluid flow and is nothing of the sort. It is a special case, purchased with four specific assumptions, and every one of those assumptions is a decision to ignore something that is usually present in a real flow. Used where it fits, it replaces a differential equation with arithmetic. Used where it does not, it gives a confident, wrong number, which is the more dangerous failure of the two.

Deriving it from Newton's second law

Take a small cylindrical sliver of fluid, aligned along a streamline, with cross-sectional area \(dA\) and length \(ds\), and apply \(\sum F_s = \delta m\, a_s\) to it in the direction of the streamline. Two things act on it along \(s\): the pressure difference between its two faces, and the component of its own weight resolved along the streamline's slope. Pressure acts inward on both faces, but not equally — if the streamline is speeding the fluid up, the pressure pushing from behind must exceed the pressure resisting from ahead, giving a net force of \(-\,dA\,dP\). The weight contributes \(-\rho g \, dA \, ds \, \sin\theta\), where \(\sin\theta\,ds = dz\) is simply how much the streamline has climbed. Divide through by the mass \(\rho\, dA\, ds\) and what remains is Euler's equation for steady, inviscid flow along a streamline:

\[\frac{dP}{\rho} + V\,dV + g\,dz = 0\]

which is already, in differential form, the whole of the physics. Every term has the same units — energy per unit mass — and the equation says that along this one curve, whatever a fluid particle gains in one form it loses in the sum of the others. Integrate it between any two points on the same streamline, holding \(\rho\) constant so it can be pulled outside the integral, and the differential statement becomes an algebraic one:

\[\frac{P}{\rho} + \frac{V^2}{2} + gz = \text{const along a streamline}\]

or, comparing two stations 1 and 2 directly on the same streamline,

\[\frac{P_1}{\rho} + \frac{V_1^2}{2} + gz_1 = \frac{P_2}{\rho} + \frac{V_2^2}{2} + gz_2\]

This is nothing more exotic than \(F = ma\) integrated once along a path, with the acceleration replaced by the material derivative met in kinematics. What makes it powerful is not the derivation, which is short, but how much can be read off from the result without solving a differential equation ever again — provided the next section's conditions actually hold.

F = ma, along s — step 1 of 4

Take a sliver of fluid riding its own streamline and apply Newton's second law along it.

\(\sum F_s = \delta m\, a_s\)
\(-\frac{dP}{\rho}\)
\(-g\,dz\)
\(\frac{P}{\rho} + \frac{V^2}{2} + gz = \text{const}\)
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The assumptions

Four words did a great deal of quiet work in the derivation above, and each one is a place where the equation can fail if it is applied without checking. Treating them as footnotes to memorise rather than costs to weigh is the single most common reason Bernoulli's equation gets used where it should not.

Steady means the local acceleration term, \(\partial \mathbf{V}/\partial t\), was dropped from the material derivative before the derivation even started. For a genuinely unsteady flow — a valve closing, a pump starting up — that term does not vanish, and an extra integral, \(\int \partial V/\partial t\, ds\), belongs in the balance. It is rarely small next to the other terms, which is why water hammer in a suddenly closed pipe produces pressure spikes nothing in the steady equation predicts.

Incompressible is what let \(\rho\) come outside the integral. For a liquid this costs almost nothing; for a gas it is a genuine restriction, reasonable below roughly Mach 0.3 and increasingly wrong above it, where density changes become part of the problem rather than a detail to neglect.

Inviscid is the most expensive assumption and the one nature grants least willingly. No real fluid has zero viscosity, and next to any solid wall a boundary layer forms in which friction is not a small correction but the dominant physics — velocity falls from the free-stream value to exactly zero at the wall, and Bernoulli's equation, which has no mechanism for spending energy irreversibly, has nothing to say inside that layer. It remains fair game in the core of a flow away from walls and wakes, which is most of what a short, well-rounded intake or nozzle contains, and is the reason those devices are the equation's best-behaved customers.

Along a streamline is the condition most often forgotten entirely, because the algebra never seems to need it. Two points on different streamlines are not guaranteed the same constant \(H\) unless the flow is also irrotational — zero vorticity everywhere — in which case a stronger version of the theorem applies and the same \(H\) holds across the whole flow field at once. Away from walls, in a flow that started from a uniform, irrotational upstream condition, this usually holds and is exactly why engineers reach for Bernoulli across an entire flow field so readily; it is not a licence granted for free, only one that is very often already paid for by the upstream conditions.

Four clauses, four costs — step 1 of 4

Steady: nothing at any fixed point is allowed to change with time.

\(\frac{\partial \mathbf{V}}{\partial t} = 0\)
\(\rho = \text{const}\)
\(\mu \to 0\)
\(H(\psi_1) \ne H(\psi_2)\)
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Three forms of head

Divide the energy form of the equation through by \(g\) and every term acquires units of length:

\[\frac{P}{\rho g} + \frac{V^2}{2g} + z = H\]

This is the same equation, rewritten in what fluid engineers call head — energy per unit weight rather than energy per unit mass — and the three terms earn names of their own because each reads directly off a physical instrument. \(z\) is elevation head, simply the height above whatever datum was chosen; it is the same term gravitational potential energy has always contributed, and choosing the datum is bookkeeping, not physics. \(P/\rho g\) is pressure head, and it is literally readable: it is the height a column of the fluid itself would need to stand to produce that pressure at its base, which is exactly what a piezometer tube — an open tube stood vertically in the flow — shows you. \(V^2/2g\) is velocity head, the kinetic energy written in the same length units, and it is what a Pitot tube adds on top of the piezometer reading.

Writing the equation this way earns more than tidiness. Because all three terms share one unit, they can be drawn on the same vertical scale and compared directly, which is exactly what the energy grade line does in the next topic, and it makes the trade-off between the terms visible rather than algebraic: narrow a duct and velocity head must rise, so — with elevation fixed and the total constant — pressure head must fall to pay for it. That single trade is the entire mechanism behind both instruments in the last section of this article.

It is worth being clear that \(H\) being constant is not a new piece of physics beyond the energy form; it is the same three quantities under a different name, chosen because a length is easier to picture, measure and draw than an energy per unit mass ever is.

One currency, three forms — step 1 of 4

Elevation head: height above a chosen datum, in metres of the fluid itself.

\(z\)
\(\frac{P}{\rho g}\)
\(\frac{V^2}{2g}\)
\(z + \frac{P}{\rho g} + \frac{V^2}{2g} = H\)
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Where it breaks down

Every one of the assumptions in the second section can fail, but the one that matters for almost every piece of real pipework is viscosity, and it is worth following through what actually happens when it is switched back on. A real pipe, run at steady, incompressible conditions, still does not hold \(H\) constant along its length. Plot the total head against distance and the ideal prediction is a flat line; the measured one slopes steadily downward, and the gap between the two grows the further downstream you look.

That gap is not a modelling error to be corrected by a better Bernoulli equation — it is real energy, permanently converted to heat by viscous shearing against the pipe wall, and it is called head loss:

\[\frac{P_1}{\rho g} + \frac{V_1^2}{2g} + z_1 = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + z_2 + h_L\]

The sign convention is not optional. \(h_L\) is added to the downstream side and is never negative, because friction only ever removes useful mechanical energy from a flow — it cannot, by the second law, hand any back. This single added term is the difference between the Bernoulli equation as derived above and the mechanical energy equation used for real pipe systems, and it is worth remembering that the correction is entirely bookkeeping: nothing about the derivation changes, one honest term is restored to an equation that dropped it on purpose.

What the loss actually costs in engineering terms — how much of it comes from the pipe wall itself against how much comes from fittings, bends and valves — is deliberately not this article's subject. That detail belongs to flow in pipes and head losses, where it gets the full treatment it needs; here the point is narrower and more important to establish first — that the term exists, that it only ever subtracts, and that the equation derived in the first section was never meant to survive contact with a wall on its own.

The bill for friction — step 1 of 4

Bernoulli's promise is a flat total head, all the way along the streamline.

\(H = \text{const}\)
\(H(x) < H(0)\)
\(h_L \ge 0\)
\(\frac{P_1}{\rho g} + \frac{V_1^2}{2g} + z_1 = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + z_2 + h_L\)
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Applications: the Pitot tube and the venturi

Two instruments exist almost entirely because this one equation holds, and both work by deliberately forcing a change in one term of it so that another term can be read off.

A Pitot tube is a probe with a small hole facing directly into the oncoming flow. The streamline that arrives exactly at that hole has nowhere to go, so its velocity there is forced to zero — the stagnation point — and every bit of velocity head it carried upstream reappears as pressure head instead. Apply Bernoulli between a point in the free stream, at pressure \(P\) and speed \(V\), and the stagnation point, at pressure \(P_0\) and zero speed, both at the same elevation:

\[\frac{P}{\rho} + \frac{V^2}{2} = \frac{P_0}{\rho} \;\;\Longrightarrow\;\; V = \sqrt{\frac{2(P_0 - P)}{\rho}}\]

Measure \(P_0 - P\) with a manometer — the stagnation pressure against the ordinary static pressure taken from a hole in the side of the same probe, undisturbed by the flow — and the local velocity comes out directly, with nothing else needed. It is the working principle behind airspeed indicators on most aircraft in service today, unchanged since Henri Pitot built the first one to measure the flow of the Seine.

A venturi meter runs the same trade the other way. A pipe is narrowed smoothly to a throat and widened back out, and continuity forces the velocity up at the throat: \(A_1V_1 = A_2V_2\) with \(A_2 < A_1\) means \(V_2 > V_1\). Apply Bernoulli between the wide section and the throat, at the same elevation, and the rise in velocity head must be paid for by a fall in pressure head:

\[\frac{P_1}{\rho} + \frac{V_1^2}{2} = \frac{P_2}{\rho} + \frac{V_2^2}{2}\]

Combine that with continuity to eliminate \(V_1\) and solve for the throat velocity in terms of a pressure difference that a manometer can measure directly:

\[V_2 = \sqrt{\dfrac{2(P_1 - P_2)}{\rho\left(1 - (A_2/A_1)^2\right)}}\]

Multiply by the throat area and the volume flow rate follows, \(\dot V = A_2 V_2\), from nothing more than a pressure gauge and two known areas. Both instruments, in the end, are the same trick: force the flow into a state where one term of Bernoulli's equation is either zero or forced to a known ratio, and let the equation hand back whatever is left.

Two instruments, one equation — step 1 of 4

A probe facing upstream brings the streamline that hits it to a dead stop.

\(\frac{P}{\rho} + \frac{V^2}{2} = \frac{P_0}{\rho}\)
\(V = \sqrt{\frac{2(P_0 - P)}{\rho}}\)
\(A_1 V_1 = A_2 V_2\)
\(V_2 = \sqrt{\dfrac{2(P_1 - P_2)}{\rho\left(1 - (A_2/A_1)^2\right)}}\)
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Every number above came from one arrangement of a venturi. The model below is the same equation, live: change the inlet speed, the throat's area, or its elevation, and watch the flow, the pressure and the total head respond exactly as the algebra says they must.

12PRESSURE HEAD
Interactive model

Bernoulli's principle: a venturi

StationVP (gauge)H
1 · inlet1.50 m/s0 Pa0.11 m
2 · throat3.00 m/s-3,375 Pa0.11 m

Total head H agrees at both stations no matter where the sliders sit — that agreement, not either number on its own, is Bernoulli's principle. Narrow the throat and the flow speeds up, which is paid for by a fall in pressure head, exactly the trade worked through above.

Saved on this device only.