Exercises — Bernoulli's equation — derivation from F = ma along streamline

• $P$ = static pressure — energy stored per unit volume in the pressure of the fluid.
• $\frac{1}{2}\rho v^2$ = dynamic pressure — kinetic energy per unit volume.
• $\rho gy$ = gravitational (hydrostatic) term — potential energy per unit volume.

All three share the unit pascal, and $1\text{Pa}=1{\text{J/m}}^3$ — energy per volume. That equality is why the three can be added: they are the same kind of quantity.

(a) Honey is thick — inviscid fails (large viscosity, energy lost to internal friction). See Viscosity and Poiseuille Flow. (b) The velocity field changes in time — steady fails ($\partial /\partial t\ne 0$). (c) Near/above the speed of sound air's density changes a lot — incompressible fails ($\rho$ not constant).

Lower. Horizontal ⇒ the $\rho gy$ term is the same at both points, so $P+\frac{1}{2}\rho v^2$ is constant. Bigger $v$ forces smaller $P$. (This is the Venturi Meter principle.)

Streamline from the free surface (1) to the hole (2). Both open to atmosphere so $P_1=P_2$; the wide surface barely moves so $v_1\approx 0$; drop in height is $h$. $\rho gh=\frac{1}{2}\rho v_2^2\Rightarrow v_2=\sqrt{2gh}=\sqrt{2(9.8)(5.0)}\approx 9.9\text{m/s}.$ Same as an object dropped from $5\text{m}$ — see Torricelli's Law.

At the stagnation mouth the air is brought to rest ($v=0$); horizontal, so height cancels. $P+\frac{1}{2}\rho v^2=P_0\Rightarrow v=\sqrt{\frac{2(P_0-P)}{\rho }}=\sqrt{\frac{2(3000)}{1.2}}\approx 70.7\text{m/s}.$

Horizontal, so height cancels: $P_2=P_1+\frac{1}{2}\rho (v_1^2-v_2^2)=150000+\frac{1}{2}(1000)(2^2-6^2).$ $P_2=150000+500(4-36)=150000-16000=134000\text{Pa}.$ Pressure drops by $16\text{kPa}$ where the flow speeds up.

Step 1 — link the speeds by continuity (see Equation of Continuity). Same volume per second passes both areas — in the figure, the same red streamline threads the wide part and the throat: $A_1{v}_1=A_2{v}_2\Rightarrow v_2=\frac{A_1}{A_2}{v}_1=\frac{8}{2}{v}_1=4v_1.$ Step 2 — Bernoulli, horizontal (the streamline stays at one height, so $\rho gy$ cancels between points 1 and 2): $P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}\rho ((4v_1{)}^2-v_1^2)=\frac{1}{2}\rho (15v_1^2).$ Step 3 — solve: $6000=\frac{1}{2}(1000)(15)v_1^2=7500v_1^2\Rightarrow v_1=\sqrt{0.8}\approx 0.894\text{m/s}.$ (Then $v_2=4v_1\approx 3.58\text{m/s}$ if asked — the "fast" label on the throat in the figure.)

Same width ⇒ same speed ⇒ the dynamic terms cancel; only the height term shifts: $P_2=P_1-\rho g(y_2-y_1)=200000-(1000)(9.8)(4.0)=200000-39200=160800\text{Pa}.$ Rising fluid "spends" pressure to buy potential energy — this is just Hydrostatic Pressure reappearing inside a moving fluid.

First, gauge vs absolute. Gauge pressure is measured relative to the atmosphere — it is what a tyre gauge reads, and it can be negative (below atmospheric). Absolute pressure is measured from a perfect vacuum ($P_{\text{abs}}=P_{\text{gauge}}+P_{\text{atm}}$) and physically cannot go below zero, because zero absolute pressure already means "no molecules pushing at all." A predicted absolute pressure of $-5000\text{Pa}$ is therefore impossible: it would require the fluid to pull, and liquids cannot sustain much tension. In reality, as $P$ falls toward water's vapour pressure (a small positive number), the water boils/cavitates: vapour bubbles form and break the smooth streamline. Bernoulli's assumptions (incompressible, single continuous streamline) then fail. So a negative absolute pressure is a warning flag that the flow cavitates before ever reaching it — not a real pressure.

Treat top and bottom as (approximately) equal-height points on nearby streamlines, apply Bernoulli to each. The pressure difference: $P_{\text{bot}}-P_{\text{top}}=\frac{1}{2}\rho (v_{\text{top}}^2-v_{\text{bot}}^2)=\frac{1}{2}(1.2)(120^2-100^2).$ $=0.6(14400-10000)=0.6(4400)=2640\text{Pa}.$ Lift $=\Delta P\times \text{area}=2640\times 30=79200\text{N}\approx 7.9\times 10^4\text{N}$. (See Aerodynamic Lift. Real lift needs circulation theory, but the Bernoulli estimate captures the sign and rough size.)

Continuity: $A_1{v}_1=A_2{v}_2\Rightarrow v_1=\frac{A_2}{A_1}{v}_2=\frac{0.05}{0.5}{v}_2=0.1v_2$. Bernoulli (both surfaces at atmosphere, drop $h$): $\rho gh=\frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}\rho (v_2^2-(0.1v_2{)}^2)=\frac{1}{2}\rho v_2^2(1-0.01).$ $gh=0.495v_2^2\Rightarrow v_2=\sqrt{\frac{9.8\times 1.8}{0.495}}=\sqrt{35.63}\approx 5.97\text{m/s}.$ Compare naive Torricelli $\sqrt{2gh}=\sqrt{35.28}\approx 5.94\text{m/s}$ — the correction is tiny (the $0.01$), justifying why $v_1\approx 0$ usually works.

Uniform width ⇒ by continuity the speed is the same everywhere in the tube; call it $v$. (a) Outlet speed. Streamline from tank surface (1, $v\approx 0$, $P_{\text{atm}}$, $y=0$) to outlet (3, $P_{\text{atm}}$, $y=-D$) — the full red path in the figure: $P_{\text{atm}}+0+0=P_{\text{atm}}+\frac{1}{2}\rho v^2+\rho g(-D).$ $\frac{1}{2}\rho v^2=\rho gD\Rightarrow v=\sqrt{2gD}=\sqrt{2(9.8)(2.5)}\approx 7.0\text{m/s}.$ Outlet speed depends only on the drop $D$ below the source (the lower double-arrow) — the bend height cancels for speed. (b) Pressure at the top of the bend (point 2 in the figure, $y=+H$, speed $v$). From surface (1) to top (2): $P_{\text{atm}}=P_2+\frac{1}{2}\rho v^2+\rho gH.$ $P_2=P_{\text{atm}}-\frac{1}{2}\rho v^2-\rho gH=101325-\frac{1}{2}(1000)(49)-(1000)(9.8)(1.0).$ $P_2=101325-24500-9800=67025\text{Pa}.$ Positive (above vapour pressure), so the siphon works. If $P_2$ had gone toward zero, the water column would break — the max lift height of a siphon.

Step 1 — pressure difference from the manometer. In the U-tube, the two arms connect to the two pipe sections. At the level of the lower mercury surface, the pressure must match from both sides (same fluid, same height ⇒ same pressure — Hydrostatic Pressure). Going down arm 1: $P_1$ plus a water column; going down arm 2: $P_2$ plus a shorter water column plus a mercury column of height $\Delta h$. Balancing these and cancelling the common water heights leaves the mercury column standing in for the pressure difference, but we must subtract the water that mercury displaces on the other side: $P_1-P_2=({\rho }_{\text{Hg}}-{\rho }_{\text{water}})g\Delta h.$ The $-{\rho }_{\text{water}}$ appears because the $\Delta h$ of mercury on one side is balanced against $\Delta h$ of water on the other; only the density difference produces net pressure. Numerically: $P_1-P_2=(13600-1000)(9.8)(0.15)=12600\times 9.8\times 0.15=18522\text{Pa}.$ Step 2 — continuity: $v_2=\frac{A_1}{A_2}{v}_1=4v_1$. Step 3 — Bernoulli, horizontal: $P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1000)(16v_1^2-v_1^2)=7500v_1^2.$ Step 4 — solve: $v_1^2=\frac{18522}{7500}=2.4696\Rightarrow v_1\approx 1.571\text{m/s}.$ Step 5 — flow rate ($A_1=40{\text{cm}}^2=40\times 10^{-4}{\text{m}}^2$): $Q=A_1{v}_1=40\times 10^{-4}\times 1.571\approx 6.29\times 10^{-3}{\text{m}}^3/\text{s}.$ That's about $6.3$ litres per second.

At the bend, pressure from surface (1) to bend top (2) with $v\approx 0$: $P_2=P_{\text{atm}}-\rho gH.$ Set $P_2=P_{\text{vap}}$ for the failure point: $H_{\max }=\frac{P_{\text{atm}}-P_{\text{vap}}}{\rho g}=\frac{101325-2300}{(1000)(9.8)}=\frac{99025}{9800}\approx 10.1\text{m}.$ This is the same $\sim 10\text{m}$ ceiling that limits how high suction alone can raise water — a beautiful cross-check with Hydrostatic Pressure.

Horizontal ⇒ drop the $\rho gy$ term. Evaluate the Bernoulli sum (the "constant") on each line: $C_A=P_A+\frac{1}{2}\rho v_A^2=120000+\frac{1}{2}(1000)(5^2)=120000+12500=132500\text{Pa}.$ $C_B=P_B+\frac{1}{2}\rho v_B^2=120000+\frac{1}{2}(1000)(1^2)=120000+500=120500\text{Pa}.$ Since $C_A=132500\ne 120500=C_B$, the Bernoulli constant is generally different on different streamlines — exactly the parent-note warning. You may not carry a value from line A over to line B. Extra condition for equality: if the flow is additionally irrotational (no swirl anywhere), the same constant holds everywhere in the fluid, not just along each streamline. These two lines have different constants, so this flow is rotational.

Continuity ties the speeds; Bernoulli trades pressure ↔ speed ↔ height along one streamline; watch signs (negative pressure = cavitation) and never hop streamlines unless the flow is irrotational.