Identity 3.5.1.
Note the following identity:
\begin{equation}
(\bu \cdot \nabla) \bu = \nabla \left(\frac{1}{2}|\bu|^2\right) + (\nabla \times \bu)
\times \bu.\tag{3.5.1}
\end{equation}
Section 3.5 Bernoulli’s equation
The Euler equations are a system of nonlinear partial differential equations, in which the velocity \(\bu\) and the pressure \(p\) are coupled together. Solving them in general is difficult. In many problems of interest, however, the velocity can be found by other means. For example:
-
in flow through a pipe or out of a tank, mass conservation often determines the speed;
-
in Chapter 4, the velocity follows from solving Laplace’s equation;
-
in Chapter 5, the motion of the free surface is linked to the velocity by a kinematic condition.
In each case, the remaining task is to find the pressure, for instance to calculate the force on a body, or to impose that the pressure is atmospheric at a free surface.
Bernoulli’s equation does exactly this. It is a first integral of the momentum equation: rather than a differential equation, it is an algebraic relation between the pressure, the speed and the height of the fluid,
\begin{equation*}
B = \frac{p}{\rho} + \frac{1}{2}|\bu|^2 + \chi.
\end{equation*}
In this section, we first rewrite the momentum equation in a form from which \(B\) emerges naturally. We then show that, under different assumptions, \(B\) is constant along streamlines, constant everywhere, or a function of time only. Finally, we apply these results to some practical problems.
Subsection 3.5.1 A first integral of the momentum equation
We will first need a vector identity.
Proof.
The proof follows from direct expansion of both sides.
The second term on the right-hand side involves the curl of the velocity, which has its own name.
Definition 3.5.2 Vorticity.
The vorticity, \(\bomega\text{,}\) of the velocity field \(\bu\) is defined by
\begin{equation}
\bomega \equiv \nabla \times \bu.\tag{3.5.2}
\end{equation}
The vorticity is a measure of the local rotation of the flow.
Definition 3.5.3 Irrotational flows.
A flow is said to be irrotational if the vorticity is identically zero:
\begin{equation}
\textrm{flow is irrotational} \Longleftrightarrow \bomega = \nabla \times \bu \equiv
\mathbf{0}.\tag{3.5.3}
\end{equation}
In Chapter 4, we will show that an irrotational flow in a simply connected region can always be written as \(\bu = \nabla\phi\) for some scalar function \(\phi\text{,}\) called the velocity potential (Theorem 4.1.1). We use this fact below.
Let us recall the definition of a conservative force.
Definition 3.5.4 Conservative forces.
A force \(\bF\) is a conservative force if and only if there exists a potential \(\chi\text{,}\) such that
\begin{equation*}
\bF = -\nabla \chi,
\end{equation*}
in a simply connected region where the quantities are defined.
Note the restriction to a simply connected region. The above is not quite the usual definition of a conservative force (typically defined to be a force for which the work done on an object between two points is independent of path). For now, this is not an important distinction since we will focus on fluid regions that are free of holes. Until told otherwise, we will always assume that the fluid is simply connected and therefore the above serves as a definition of conservative forces.
Remark 3.5.5.
For the typical gravity force directed in the \(-z\) direction, we can write \(\bg =
-g[0, 0, 1] = -g \be_z\text{.}\) So in this case, the (gravitational) potential is \(\chi = gz\text{.}\)
Theorem 3.5.6 The momentum equation in Bernoulli form.
Consider an inviscid, incompressible fluid of constant density \(\rho\text{,}\) subject to a conservative body force \(\bg = -\nabla\chi\text{.}\) Define the Bernoulli function
\begin{equation}
B(\bx, t) \equiv \frac{p}{\rho} + \frac{1}{2}|\bu|^2 + \chi.\tag{3.5.4}
\end{equation}
Then the momentum equation (3.4.5) can be written as
\begin{equation}
\pd{\bu}{t} + \bomega \times \bu = -\nabla B.\tag{3.5.5}
\end{equation}
Proof.
Start from the momentum equation (3.4.5), with \(\bg =
-\nabla \chi\text{,}\)
\begin{equation*}
\pd{\bu}{t} + (\bu \cdot \nabla)\bu = - \frac{1}{\rho}\nabla p - \nabla \chi.
\end{equation*}
Using the vector identity (3.5.1), and noting that \(\frac{1}{\rho}\nabla
p = \nabla(p/\rho)\) since \(\rho\) is constant, gives
\begin{equation}
\pd{\bu}{t} + (\nabla \times \bu) \times \bu = - \nabla \left(\frac{p}{\rho} +
\frac{1}{2} |\bu|^2 + \chi\right).\tag{3.5.6}
\end{equation}
Every version of Bernoulli’s equation follows from (3.5.5) by removing terms from the left-hand side.
Aside: Removed in 2026.
Subsection 3.5.2 Three forms of Bernoulli’s equation
Table 3.4 summarises the three forms of Bernoulli’s equation derived in this subsection, together with where each is used in the course. In each case, the assumptions of Theorem 3.5.6 also apply.
Theorem 3.5.7 Bernoulli’s equation for steady flow.
For a steady flow,
\begin{equation}
B(\bx) = \frac{p}{\rho} + \frac{1}{2} |\bu|^2 + \chi = \textrm{constant along
streamlines}.\tag{3.5.7}
\end{equation}
Proof.
Since the flow is steady, the first term of (3.5.5) vanishes. This leaves
\begin{equation*}
\bomega \times \bu = -\nabla B.
\end{equation*}
We now take the dot product of both sides of the equation with \(\bu\text{.}\) We use the fact that,
\begin{equation*}
\bu \cdot (\bomega \times \bu ) = 0,
\end{equation*}
since it is a triple scalar product with two repeated entries. It therefore results in the fact that
\begin{equation*}
\bu \cdot \nabla B = 0.
\end{equation*}
Since \(B\) does not depend on time, \(\DD{B}{t} = \bu \cdot \nabla B\text{,}\) so this states that \(\DD{B}{t} = 0\text{.}\) Hence \(B\) is constant along streamlines of the flow (which, for steady flow, coincide with particle paths).
Theorem 3.5.8 Bernoulli’s equation for steady irrotational flow.
For a steady and irrotational flow,
\begin{equation}
B(\bx) = \frac{p}{\rho} + \frac{1}{2} |\bu|^2 + \chi = \textrm{constant
everywhere}.\tag{3.5.8}
\end{equation}
Proof.
For steady flow, the first term of (3.5.5) vanishes. If, in addition, the flow is irrotational, then \(\bomega = \mathbf{0}\text{,}\) and therefore
\begin{equation*}
\nabla B = \mathbf{0}.
\end{equation*}
Since all spatial derivatives of \(B\) are zero, \(B\) is constant throughout the (connected) fluid domain.
Theorem 3.5.9 Bernoulli’s equation for unsteady potential flow.
For an irrotational, possibly unsteady, flow with velocity potential \(\phi\text{,}\) so that \(\bu = \nabla\phi\text{,}\)
\begin{equation}
\pd{\phi}{t} + \frac{1}{2}|\nabla\phi|^2 + \frac{p}{\rho} + \chi = F(t),\tag{3.5.9}
\end{equation}
for some function \(F(t)\text{.}\)
Proof.
Set \(\bomega = \mathbf{0}\) and \(\bu = \nabla\phi\) in (3.5.5). Since space and time derivatives commute, \(\pd{}{t}(\nabla
\phi) = \nabla\left(\pd{\phi}{t}\right)\text{,}\) and so
\begin{equation*}
\nabla\left( \pd{\phi}{t} + \frac{1}{2}|\nabla\phi|^2 + \frac{p}{\rho} + \chi
\right) = \mathbf{0}.
\end{equation*}
The quantity in brackets therefore has no spatial dependence, i.e. it is a function of \(t\) alone.
Remark 3.5.10 The function \(F(t)\).
The function \(F(t)\) can be absorbed into the potential. Replacing \(\phi\) by \(\phi
- \int F \, \de{t}\) does not change the velocity \(\bu = \nabla\phi\text{,}\) but sets \(F
\equiv 0\text{.}\) Equivalently, \(F(t)\) may be chosen for convenience, as is done in Chapter 5.
Remark 3.5.11 Constant \(B\) on every streamline.
In the steady case, \(B\) can be constant everywhere even when the flow is not irrotational. This happens in many practical problems, where every streamline starts in a large reservoir in which the fluid is at rest, so that \(B\) takes the same value on every streamline.
Subsection 3.5.3 Applications
In each example below, we state which form of Bernoulli’s equation is used and why its assumptions hold.
Example 3.5.12 Hydrostatics and Pascal’s bursting barrel.
For a fluid at rest, \(\bu = \mathbf{0}\text{.}\) The flow is then trivially steady and irrotational, so Theorem 3.5.8 gives
\begin{equation*}
\frac{p}{\rho} + gz = \textrm{constant everywhere}.
\end{equation*}
If the free surface is at \(z = H\text{,}\) where \(p = p_\text{atm}\text{,}\) then
\begin{equation*}
p = p_\text{atm} + \rho g (H - z).
\end{equation*}
The pressure at depth \(d = H - z\) is \(p_\text{atm} + \rho g d\text{.}\) It depends only on depth, not on the shape of the container or the amount of fluid above.
This explains Pascal’s bursting barrel. A long thin vertical tube is attached to a sealed barrel full of water, and the tube is then filled. Although the column in the tube holds very little water, it raises the pressure throughout the barrel by \(\rho g
h\text{.}\) For a \(10\) m column, that is about \(10^5\) Pa, roughly one atmosphere. Acting on a lid of diameter \(0.5\) m, this is a force of about \(2 \times 10^4\) N, the weight of roughly two tonnes.
Watch this short video demonstration from Princeton, showing the power of hydrostatic pressure.
Example 3.5.13 Torricelli’s law.
A large open tank of cross-sectional area \(A\) holds water of depth \(h(t)\) above a small hole of area \(a \ll A\) in its base.
-
Find the speed \(u\) of the jet leaving the hole.
-
If the initial depth is \(h_0\text{,}\) find the time for the tank to drain.
Solution.
(a) The free surface falls at speed \(U\text{,}\) where mass conservation gives \(AU = au\text{,}\) so \(U/u = a/A \ll 1\text{.}\) The flow therefore changes slowly, and we treat it as steady. Taking a streamline from the free surface (\(z = h\)) to the jet (\(z = 0\)), where the pressure is atmospheric at both ends, Theorem 3.5.7 gives
\begin{equation*}
\frac{p_\text{atm}}{\rho} + \frac{1}{2}U^2 + gh = \frac{p_\text{atm}}{\rho} +
\frac{1}{2}u^2.
\end{equation*}
Using \(U = (a/A)u\text{,}\)
\begin{equation*}
u = \sqrt{\frac{2gh}{1 - (a/A)^2}} \approx \sqrt{2gh}.
\end{equation*}
This is Torricelli’s law (1643): the jet leaves at the speed of a particle falling freely from height \(h\text{.}\)
(b) Mass conservation gives
\begin{equation*}
A \dd{h}{t} = -a\sqrt{2gh}.
\end{equation*}
Separating variables and using \(h(0) = h_0\) gives
\begin{equation*}
\sqrt{h} = \sqrt{h_0} - \frac{a}{A}\sqrt{\frac{g}{2}} \, t,
\end{equation*}
so the tank empties at time
\begin{equation*}
T = \frac{A}{a}\sqrt{\frac{2h_0}{g}}.
\end{equation*}
In practice, the jet from a sharp-edged hole contracts just after leaving it (the vena contracta), and the measured flow rate is about \(0.6\) of this prediction.
You will practise an application of Bernoulli’s equation in Exercise 3.6.6, which combines the calculation of Example 3.5.13 with a choice of jar shape that makes the water level fall at a constant rate.
Example 3.5.14 Urination across species.
Watch this short video describing the work by Yang et al. on "Duration of urination does not change with body size".
Yang et al. model the bladder as a pressurised reservoir that drains through a long, narrow tube, the urethra. Applying Bernoulli’s equation between the bladder and the exit of the urethra, gravity acting over the length of the urethra adds to the bladder pressure, much as the depth \(h\) drives the jet in Torricelli’s law (Example 3.5.13). Larger animals have longer urethras, so urine leaves faster, which compensates for their larger bladders. As a result, the duration of urination is roughly the same, about \(21\) s, for animals heavier than about \(3\) kg. For smaller animals, viscosity and surface tension dominate and urine is released as drops, so this argument no longer applies.
Example 3.5.15 The limits of Bernoulli’s equation: airflow into or out of a room.
Watch this short video by Matthias Wandel about airflow into and out of a room.
Bernoulli’s equation treats flow into and out of an opening in the same way, but in practice the two look very different. Air pushed out through an opening forms a narrow jet that travels a long way, whereas air drawn in through an opening arrives from all directions. The difference comes from the flow separating at the edges of the opening, a viscous effect that inviscid theory cannot capture. We meet viscosity in Chapter 6.
