Question bank — Assumptions in Bernoulli — steady, inviscid, incompressible, along streamline

False. The constant is guaranteed equal only along one streamline; different streamlines can carry different constants unless the flow is also irrotational, which promotes it to a global constant.

False. Steady kills only the $\partial v/\partial t$ term; you still need inviscid, incompressible and same-streamline. A steady viscous flow in a rough pipe still loses pressure.

False. The $P$–$v$ trade-off is a consequence of Bernoulli, so it only holds when all four assumptions do. Inside a Boundary Layer or a pulsing pipe, pressure can behave completely differently.

True. It is Newton's 2nd law integrated along the streamline, which is the work–energy theorem in disguise — every assumption is the price of one dropped energy term.

True — for ideal flow. With $v$ and $h$ fixed by continuity and geometry, Bernoulli forces $P_1=P_2$. Any measured drop signals viscosity, i.e. the inviscid assumption failing.

False. A pump injects energy, so the "constant" jumps upward across it. You must use the extended energy equation with a pump-head term $h_{\text{pump}}$.

False. Below Mach $\approx 0.3$ (i.e. $M=v/c<0.3$) density changes are negligible, so air behaves incompressibly and Bernoulli is legitimate — this is exactly the low-speed-wing case. See Mach Number and Compressibility.

False in practice. It means viscous forces are negligible for this flow region — true far from walls, false inside the thin boundary layer where shear dominates.

False. The $\rho gh$ term exists only because gravity comes from a height potential; a non-conservative body force cannot be written as a clean potential, so no single "constant" could absorb it.

In a free vortex (idealised drain) the fluid circles the centre with speed $v\propto 1/r$ — faster nearer the middle — so the two points lie on different circular streamlines; the same-streamline license doesn't apply, and unless the flow is irrotational the constants differ. (See the free-vortex sketch in the figure below.)

During start-up the flow is unsteady, $\partial v/\partial t\ne 0$ everywhere, so the dropped inertial term is nonzero — the steady assumption is violated and Bernoulli is missing the water-hammer term.

Because at high Mach ($M=v/c$ not small) $\rho$ varies with $P$, so it cannot be pulled out of the integral; the incompressible assumption fails and an internal-energy correction is needed.

The gauge is fine; the assumption is broken. A wall tapping reads static pressure, and real pipes have wall friction, so the inviscid premise fails and mechanical energy leaks into heat, dropping the static pressure downstream.

A Pitot tube facing the flow reads total (stagnation) pressure $P+\frac{1}{2}\rho v^2$, while a sidewall tapping reads static pressure $P$; the difference is the dynamic pressure, so they should not agree wherever $v\ne 0$.

Right at the surface you are inside the boundary layer, where viscosity dominates — inviscid fails there. Lift-explaining Bernoulli applies to the streamlines outside that thin layer.

Steady only kills $\partial v/\partial t$. The convective term survives — water in a narrowing pipe still accelerates in perfectly steady flow because it moves along $s$ into a faster region.

Viscosity does irreversible work converting mechanical energy into heat, so $P+\frac{1}{2}\rho v^2+\rho gh$ steadily decays downstream instead of staying fixed. See Viscosity and Poiseuille Flow.

A parcel accelerates either because the field changes in time ($\partial v/\partial t$) or because it moves along $s$ into a faster region ($v\partial v/\partial s$). Steady deletes the first, the local term; see Material Derivative.

Newton's law was written for the force component along $s$, so only motion along the streamline appears; the perpendicular balance is a separate equation, which is why the constant is only guaranteed along one streamline.

For Irrotational Flow the cross-streamline pressure balance forces every streamline to share the same constant, so the equation relates any two points in the field.

For an isentropic gas the fractional density change grows like $\frac{\Delta \rho }{\rho }\approx \frac{M^2}{2}$ (leading term), so at $M=0.3$ it is only about $0.3^2/2\approx 4.5\%$ — small enough to ignore, but it climbs steeply as $M$ rises. The related pressure-change estimate carries a factor of $\gamma$ (the adiabatic index) since pressure and density are linked by $P\propto {\rho }^{\gamma }$ in a gas. See Mach Number and Compressibility.

The constant is a closed energy balance for a passive parcel; a pump is an external energy source (its pump-head $h_{\text{pump}}$) that breaks that closure, so its head must be added explicitly.

Only a conservative force can be written as a potential (here $gh$), which is what lets its work integrate into the single clean term $\rho gh$; otherwise the work would depend on the path and no fixed "constant" could exist.

Yes — it reduces to $P+\rho gh=\text{const}$, the hydrostatic pressure law, which is the zero-velocity limit of the same equation.

To $P+\frac{1}{2}\rho v^2=\text{const}$, so pressure and speed trade off directly — the classic Venturi/wing result, valid only if all four assumptions hold.

Yes. Incompressibility only requires $\rho$ constant, not small; the derivation integrates $dP/\rho$ cleanly regardless of the value of $\rho$.

The incompressible step: $\int dP/\rho$ no longer equals $P/\rho$, so you need the compressible form of Bernoulli, not the standard one.

Only a good approximation — the density has already shifted a few percent ($\approx M^2/2\approx 4.5\%$), so the incompressible result carries a small but real error that grows fast above $0.3$.

Not quite "exact" — even with a negligible boundary layer, curved core streamlines carry a cross-stream pressure gradient (so different streamlines hold different constants), and entrance/exit and area-change losses still occur; Bernoulli is only exact along a single straight core streamline far from those effects. See Boundary Layer.