advanced34 min studyLesson 173 in Physics

Pressure forces and Bernoulli’s qualified constant

Derive an inviscid momentum balance and use Bernoulli only along the paths and under the assumptions that support it.

Before you begin

On this page

The question for today

What you’ll understand

Interpret Euler’s equation as acceleration caused by pressure gradients and body forces.
Apply steady Bernoulli along an appropriate streamline.
Explain why different streamlines need not share a Bernoulli constant.

Pressure is a local compressive stress, not a stored reserve that must point along the flow. Its spatial gradient can accelerate fluid, curve its motion, or balance gravity in still water. Euler’s equation keeps these roles within one momentum balance after viscous stresses have been neglected. Bernoulli’s familiar relation follows from that balance under additional assumptions, so using it starts with an assumption check rather than a numerical substitution.

A small parcel feels the pressure difference across it

Pressure pushes inward on every face of a fluid element. A uniform pressure produces no net force on a closed element because opposing face forces cancel. A gradient produces a resultant toward lower pressure. Dividing by mass gives pressure-gradient force per unit mass. Gravity contributes as a body force throughout the volume. This reasoning assumes continuum scales and an inertial frame; rotating-frame forces would need additional terms.

Euler’s equation drops the divergence of viscous stress, not the acceleration or pressure field. It is a useful outer-flow approximation when viscous forces are small on the scales considered, and a model for ideal fluids. It cannot in general impose both no penetration and viscous no slip at a wall. Static equilibrium is a useful sign check: an upward pressure-gradient force must oppose the downward gravitational force.

\rho\left(\partial_t\mathbf u+(\mathbf u\cdot\nabla)\mathbf u\right)=-\nabla p-\rho\nabla\Phi

Here Φ is a conservative body-force potential per unit mass; uniform downward gravity has Φ = gz. Pressure p is force per area and −∇p/ρ has acceleration units.↔ Scroll wide equations horizontally.

Integrate along the direction that the fluid takes

For steady motion, take the component of Euler’s equation tangent to a streamline. The tangential acceleration is the directional derivative of one half the speed squared. With constant density, the pressure term integrates to pressure divided by density. The sum of kinetic energy per unit mass, pressure divided by density, and conservative potential is then constant along that streamline. It is a statement about specified points in one modeled flow.

Different streamlines can have different constants in a rotational flow. If the steady flow is irrotational and the relevant connected region satisfies the same constant-density conservative-force assumptions, the constant can be common throughout that region. A pump, viscous loss, or shock changes the mechanical-energy account and invalidates an unmodified two-point formula across that device or region. Compressible barotropic flow instead uses an integral of dp divided by density.

B=\frac{|\mathbf u|^2}{2}+\frac{p}{\rho}+\Phi,\qquad \mathbf u\cdot\nabla B=0

This constant-density form assumes steady inviscid flow and a conservative body force. B may vary from one streamline to another; pressure is not generally uniform across a curved stream.↔ Scroll wide equations horizontally.

Use both continuity and a boundary condition

A constricting horizontal duct illustrates the division of labor. Mass conservation gives the relation between section mean speeds if the profile approximations are appropriate. Bernoulli then relates pressure change to speed change along a suitable streamline when losses are negligible. The duct shape does not alone give an absolute pressure. One measured pressure, a reservoir condition, or another boundary datum is still needed to fix the level.

Be careful with everyday slogans about fast flow and low pressure. Changing elevation, comparing different streamlines, or changing the upstream forcing can reverse an apparently simple comparison. Curved flow also needs a pressure gradient normal to the streamlines to provide centripetal acceleration. A successful calculation checks dimensions, signs, a static limit, and whether the measured pressure difference is large enough that neglected friction or compressibility may matter.

Work it through

A horizontal ideal constriction

A steady constant-density fluid with $\rho=1000\,\mathrm{kg\,m^{-3}}$ enters a horizontal duct section at $2\,\mathrm{m\,s^{-1}}$ . Its downstream area is half the upstream area. Neglect viscous losses and use uniform section speeds. Find the pressure change between points on one streamline.

Continuity gives $A_1U_1=A_2U_2$ . Because $A_2=A_1/2$ , the downstream speed is $U_2=4\,\mathrm{m\,s^{-1}}$ . This step uses mass balance and constant density, not Bernoulli.
Cancel equal gravitational potentials in Bernoulli and rearrange: $p_2-p_1=\rho(U_1^2-U_2^2)/2$ . The pressure difference is negative because kinetic energy per mass has increased along the chosen streamline.
Substitute to obtain $1000(4-16)/2=-6000\,\mathrm{Pa}$ . Compare the six-kilopascal drop with measured data; a larger drop can indicate losses omitted by the ideal model.

The result

The modeled downstream pressure is 6 kPa lower. Absolute pressures remain undetermined until one boundary pressure is supplied; the ideal estimate also requires the fluid to remain in the assumed single phase.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Resolve a pressure-gradient force

Treat the displayed vector as a normalized pressure-gradient force per unit mass. Predict that at an angle of 90 degrees its horizontal component vanishes, then vary magnitude and inspect the projections. The sliders change one vector’s angle and magnitude.

Vectorx / y projectionsTrace marker

x component: 1.393
y component: 0.975
Magnitude: 1.7

Angle35°Magnitude1.7

t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forVector resolution supports the component form of momentum balance. This lab does not calculate pressure, a streamline, or Bernoulli flow through a constriction.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The stated conditions permit integration along that streamline. Equal speed alone does not supply the assumptions; pumps add work, and different streamlines in rotational flow may have different Bernoulli constants.

Your turn

Try it yourself

An ideal incompressible fluid rises $2\,\mathrm m$ along a steady streamline while its speed remains constant. With density $800\,\mathrm{kg\,m^{-3}}$ and $g=10\,\mathrm{m\,s^{-2}}$ , find the pressure change. Would a different streamline at the final elevation necessarily have that pressure?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The speed terms cancel, leaving pressure per density plus gravitational potential constant along the chosen streamline.

Compare with the solution

The pressure change is $\Delta p=-\rho g\Delta z=-800(10)(2)=-16000\,\mathrm{Pa}$ . The modeled pressure falls by 16 kPa. In a rotational flow, another streamline may carry a different Bernoulli constant, so its final pressure cannot be inferred from this calculation alone. Constant density and conservative gravity do not remove that distinction.

Keep the idea

Take it with you

Euler’s equation retains pressure-gradient and inertial forces after viscous stresses are omitted.
Steady Bernoulli is a streamline result with explicit constitutive and force assumptions.
Continuity fixes speed relations; boundary data and momentum balance fix pressures.

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