Exercises — Assumptions in Bernoulli — steady, inviscid, incompressible, along streamline

Constant area ⇒ by continuity $A_1{v}_1=A_2{v}_2$ with $A_1=A_2$ gives $v_1=v_2$. Horizontal ⇒ $h_1=h_2$. So Bernoulli's $\frac{1}{2}\rho v^2$ and $\rho gh$ terms are equal at both ends and it forces $P_1=P_2$. Reality: $P$ dropped by $40\text{kPa}$. The broken license is inviscid. Wall friction (viscosity) converts mechanical energy irreversibly into heat, so the "constant" decays downstream. The fix is the extended equation with a head-loss term $h_L$.

Where the missing term comes from. The parent derived Bernoulli from Newton's 2nd law on a parcel — the "Euler equation" along the streamline coordinate $s$: $\rho \left(\underset{\text{unsteady}}{\underbrace{\frac{\partial v}{\partial t}}}+\underset{\text{convective}}{\underbrace{v\frac{\partial v}{\partial s}}}\right)=-\frac{\partial P}{\partial s}-\rho g\frac{\partial h}{\partial s}.$ Integrating this along $s$ from point 1 to point 2 gives ordinary Bernoulli plus one extra piece from the unsteady term: $\underset{\text{unsteady / inertial term}}{\underbrace{\rho {\int }_1^2\frac{\partial v}{\partial t}ds}}+\left(P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2\right)-\left(P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1\right)=0.$ Here the integral is taken along the same streamline (that is what "$ds$" means — the streamline-odometer step defined above), not over time. (In smooth irrotational flow this same term is often written $\rho \partial \varphi /\partial t$ using a velocity potential $\varphi$; the two are equivalent forms.) Answer. The local (unsteady) term $\frac{\partial v}{\partial t}\ne 0$: the velocity field is changing in time at every fixed point. That kills the steady assumption, and it is exactly this $\rho {\int }_1^2\partial v/\partial tds$ (the "water-hammer / inertial" term) that plain Bernoulli drops. Once flow settles, $\partial v/\partial t\to 0$ and Bernoulli is valid again.

Mach number $=\frac{v}{a}=\frac{30}{340}\approx 0.088$. Since $0.088<0.3$, density change is negligible: the incompressible assumption holds, and Bernoulli may be used.

Step 1 — continuity (WHY: in a steady, incompressible flow the same mass of water must cross every window each second, so a smaller window forces a faster speed — mass conservation). $A_1{v}_1=A_2{v}_2\Rightarrow v_2=\frac{A_1}{A_2}{v}_1=\frac{8}{2}(1.5)=6\text{m/s}.$ Step 2 — Bernoulli, horizontal so $h$ drops out (WHY: same streamline, all licenses assumed valid). $P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$ $P_1-P_2=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)=\frac{1}{2}(1000)\left(6^2-1.5^2\right)=\frac{1}{2}(1000)(36-2.25).$ $P_1-P_2=16,875\text{Pa}\approx 16.9\text{kPa}.$ Faster narrow section ⇒ lower pressure, exactly as the throat gauge would read.

Pick a streamline from the free surface (point 1) down to the hole (point 2). Both are exposed to atmosphere, so $P_1=P_2=P_{\text{atm}}$ — the pressure terms cancel. $\underset{\approx 0}{\underbrace{\frac{1}{2}\rho v_1^2}}+\rho gH=\frac{1}{2}\rho v_2^2+0.$ $v_2=\sqrt{2gH}=\sqrt{2(9.8)(1.8)}=\sqrt{35.28}\approx 5.94\text{m/s}.$ This is Torricelli's law — free-fall speed from height $H$, because pressure energy plays no net role here.

Extended (viscous) Bernoulli along the streamline: $P_1+\frac{1}{2}\rho v^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v^2+\rho g{h}_2+\underset{\text{friction}}{\underbrace{\rho g{h}_L}}.$ Constant area + horizontal ⇒ $v$ and $h$ terms cancel, leaving $P_1-P_2=\rho g{h}_L$: $h_L=\frac{\Delta P}{\rho g}=\frac{40,000}{1000\times 9.8}=4.08\text{m}.$ So the viscous friction "eats" $4.08\text{m}$ of head — mechanical energy gone to heat. This is precisely the term plain Bernoulli omits by assuming inviscid flow.

The failing region is the boundary layer — the paper-thin film right against the skin where velocity drops from the free-stream value to zero at the wall. There inviscid fails: shear (viscosity) dominates. The bulk flow outside this layer is effectively frictionless, steady, and low-Mach (incompressible), so along each outer streamline Bernoulli is legitimate and the faster-top ⇒ lower-$P$ ⇒ lift argument stands. The two regions are simply different streamlines with different physics.

(a) Ideal drop. Continuity first (WHY: mass conservation in steady incompressible flow — the same volume per second squeezes through the smaller window, so it must move faster): $v_2=\frac{A_1}{A_2}{v}_1=\frac{10}{4}(2)=5\text{m/s}.$ $\Delta P_{\text{ideal}}=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)=\frac{1}{2}(1000)(25-4)=10,500\text{Pa}=10.5\text{kPa}.$ (b) Loss. The manometer reads a smaller drop ($8.0\text{kPa}$) than the ideal ($10.5\text{kPa}$) — friction spoils the pressure recovery, so the measured mechanical-energy drop between the tappings falls short of ideal by $\Delta P_{\text{loss}}=\Delta P_{\text{ideal}}-\Delta P_{\text{measured}}=10.5-8.0=2.5\text{kPa}.$ This $2.5\text{kPa}$ is the viscous cost. (c) Held: steady (fixed flow), incompressible (water), along a streamline (centre-line). Failed: inviscid — the $2.5\text{kPa}$ is friction into heat.

Energy balance with a pump term (plain Bernoulli is illegal — the pump makes the "constant" jump): $\underset{\text{lower surface}}{\underbrace{P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1}}+\rho g{h}_P=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2.$ WHY $P_1=P_2$ cancels: both the reservoir surface and the outlet are open to the same atmosphere, so $P_1=P_2=P_{\text{atm}}$. Equal numbers on opposite sides of the equation subtract to zero — the atmosphere pushes on both ends alike and does no net work, so it simply drops out. With $v_1\approx 0$ (wide reservoir), $h_1=0,h_2=12$: $\rho g{h}_P=\frac{1}{2}\rho v_2^2+\rho g{h}_2\Rightarrow h_P=\frac{v_2^2}{2g}+h_2=\frac{3^2}{2(9.8)}+12.$ $h_P=0.459+12=12.46\text{m}.$ The velocity head ($0.459\text{m}$) plus the lift ($12\text{m}$) is exactly what the pump supplies.

(a) $v_1=\frac{0.6}{0.1}=6\text{m/s}$, $v_2=\frac{0.6}{0.4}=1.5\text{m/s}$. $\Delta P=\frac{1}{2}\rho \left(v_1^2-v_2^2\right)=\frac{1}{2}(1000)(36-2.25)=16,875\text{Pa}\approx 16.9\text{kPa}.$ (b) Points $r_1$ and $r_2$ sit on different streamlines (concentric circles). Bernoulli's constant is guaranteed equal only along one streamline, so equating across them is normally illegal. It is rescued only because a free vortex ($v\propto 1/r$) is irrotational — for irrotational flow the Bernoulli constant is global (the same on every streamline). Hence the along a streamline license is upgraded to "anywhere," and the number stands. A forced vortex ($v\propto r$, rotational) would make this equation nonsense.

Threshold speed: $\text{Mach}=v/a=0.3\Rightarrow v=0.3(340)=102\text{m/s}.$ Ideal pressure drop accelerating from rest to $v$ (incompressible Bernoulli, still just valid at the edge): $\Delta P=\frac{1}{2}\rho v^2=\frac{1}{2}(1.2)(102{)}^2=\frac{1}{2}(1.2)(10404)=6242.4\text{Pa}\approx 6.24\text{kPa}.$ Above $102\text{m/s}$ the incompressible license fails: $\int dP/\rho$ no longer equals $\Delta P/\rho$ and a compressible correction is required.

(a) $\frac{\partial v}{\partial t}=(2+0.5s)(0.1)$. At $s=4$: $=(2+2)(0.1)=0.4\ne 0$. Not steady — Bernoulli is not licensed. (b) Convective part: $\frac{\partial v}{\partial s}=0.5(1+0.1t)$; at $t=0$, $=0.5$. Speed at $(s=4,t=0)$: $v=(2+2)(1)=4\text{m/s}$. $v\frac{\partial v}{\partial s}=4(0.5)=2{\text{m/s}}^2,\frac{\partial v}{\partial t}=0.4{\text{m/s}}^2.$ $\frac{Dv}{Dt}=\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial s}=0.4+2.0=2.4{\text{m/s}}^2.$ (c) Plain Bernoulli silently sets $\partial v/\partial t=0$, so it would ignore the $0.4{\text{m/s}}^2$ unsteady piece and mis-predict the pressure field by exactly that contribution.

One-line diagnosis for each broken license: Constant-area pipe, pressure still drops ::: inviscid fails (friction → heat) Valve just opened, everything accelerating ::: steady fails ($\partial v/\partial t\ne 0$) Mach $>0.3$ ::: incompressible fails ($\int dP/\rho \ne \Delta P/\rho$) Comparing points on different streamlines ::: along-a-streamline fails (unless irrotational) A pump or fan sits between the two points ::: energy injected, "constant" jumps