Irodov 1.325: Euler's Equation Becomes Bernoulli's Law

Problem Irodov 1.325

The equation of motion of an ideal (inviscid, incompressible) fluid element moving under gravity is

\rho\,\frac{d\vect v}{dt} = \rho\vect g - \nabla p,

where $\vect v$ is the fluid’s velocity, $p$ its pressure, $\rho$ its (constant) density, and $\vect g$ the gravitational acceleration. Demonstrate that when the flow is steady, this equation reduces, along any single streamline, to Bernoulli’s equation,

p + \frac{1}{2}\rho v^2 + \rho gz = \text{const (along that streamline)},

with $z$ the height measured upward against $\vect g$ .

Two points, 1 and 2, on one streamline of a steady flow through a rising, narrowing conduit. A manometer column of height p_i/ρ g reads each section's pressure; the arrows mark the flow, longer in the faster narrow bore. — Two points, $1$ and $2$ , on one streamline of a steady flow through a rising, narrowing conduit. A manometer column of height $p_i/\rho g$ reads each section’s pressure; the arrows mark the flow, longer in the faster narrow bore.

Given

$\rho\,d\vect v/dt = \rho\vect g-\nabla p$ (steady flow: $\partial\vect v/\partial t = 0$ )

Find

Show

p+\rho v^2/2+\rho gz = \text{const}

along a streamline

Solution

Step 1Expand the convective derivative and restrict to steady flow

The material derivative $d\vect v/dt$ splits into a local (time) part and a convective part carried by the flow itself,

\frac{d\vect v}{dt} = \frac{\partial\vect v}{\partial t}+(\vect v\cdot\nabla)\vect v.

For a steady flow the velocity at each fixed point of space never changes, $\partial\vect v/\partial t=0$ , so only the convective part survives and the equation of motion becomes

\rho(\vect v\cdot\nabla)\vect v = \rho\vect g-\nabla p. \tag{1}

Step 2Rewrite the convective term as a gradient plus a term orthogonal to

\vect v

The vector identity $(\vect v\cdot\nabla)\vect v = \nabla(v^2/2)-\vect v\times(\nabla\times\vect v)$ replaces the left side, giving

\rho\nabla\!\left(\dfrac{v^2}{2}\right)-\rho\,\vect v\times(\nabla\times\vect v) = \rho\vect g-\nabla p

(2)

Step 3Collect every term into one gradient

With $z$ measured upward against $\vect g$ , the uniform gravitational acceleration is itself a gradient, $\vect g=-\nabla(gz)$ . Substituting into (2) and moving every gradient to one side, using here for the first and only time that $\rho$ is constant, which is what allows it to be carried inside both gradients,

\rho\nabla\left(\frac{v^2}{2}\right)-\rho\,\vect v\times(\nabla\times\vect v) = -\rho\nabla(gz)-\nabla p,

and gathering the three gradients on the left,

\nabla\!\left(p+\dfrac{1}{2}\rho v^2+\rho gz\right) = \rho\,\vect v\times(\nabla\times\vect v)

(3)

Step 4Project along a streamline

Let $d\vect r$ be an elementary displacement along a streamline, so by definition $d\vect r$ is parallel to $\vect v$ at every point. Dotting (3) with $d\vect r$ , the left side becomes the total differential of the bracketed quantity along the streamline, while the right side vanishes identically, since $\vect v\times(\nabla\times\vect v)$ is perpendicular to $\vect v$ by the definition of a cross product, and so perpendicular to $d\vect r$ too:

d\vect r\cdot\nabla\left(p+\frac{1}{2}\rho v^2+\rho gz\right) = \rho\,d\vect r\cdot\bigl[\vect v\times(\nabla\times\vect v)\bigr] = 0.

The left side is exactly $d(p+\rho v^2/2+\rho gz)$ evaluated along the streamline, so

d\left(p+\frac{1}{2}\rho v^2+\rho gz\right) = 0 \quad\text{along the streamline.}

Step 5Integrate along the streamline

A quantity whose differential vanishes along a curve is constant on that curve. Integrating from any one point of the streamline to any other,

\left(p+\frac{1}{2}\rho v^2+\rho gz\right)_{\text{point 2}} - \left(p+\frac{1}{2}\rho v^2+\rho gz\right)_{\text{point 1}} = 0.

p+\dfrac{1}{2}\rho v^2+\rho gz = \text{const along a streamline}

(4)

Check

Every term reduces to the same units of pressure, kg/(m s $^2$ ), matching $p$ . And in the limiting case $\vect v=\vect 0$ everywhere, the equation of motion (1) reduces to $\vect 0 = \vect g-\nabla p/\rho$ , ordinary hydrostatic equilibrium; (3) then has a vanishing right side and gives $\nabla(p+\rho gz)=\vect 0$ , that is $p+\rho gz=\text{const}$ throughout the fluid, the hydrostatic pressure law. That last step has to be taken from (3) and not from the box: where $\vect v$ vanishes there are no streamlines to be constant along, and the projection’s requirement that $d\vect r$ be parallel to $\vect v$ says nothing at a stagnation point either.

Remark

Nowhere did this derivation assume the flow is irrotational ( $\nabla\times\vect v=\vect 0$ ); only that it is steady. That is why the boxed constant is generally different on different streamlines of the same flow, and Bernoulli’s equation as proved here may only be applied point to point along one streamline, never between two arbitrary points of the fluid. The stronger, single whole-flow constant that many textbook applications quietly use is a separate fact, and (3) says exactly what it costs: the bracket is uniform throughout the fluid precisely when $\vect v\times(\nabla\times\vect v)=\vect 0$ everywhere, rather than merely along each streamline’s own direction. Irrotational flow, $\nabla\times\vect v=\vect 0$ , is the familiar way to secure that, but it is not the only one: a Beltrami flow, whose vorticity is everywhere parallel to its own velocity (helical flows are the standard example), satisfies it just as well.