Applications — Pitot tube, Venturi meter, orifice flow

Three classic devices that turn Bernoulli's equation into a measuring tool. Master one idea — trade pressure for speed — and all three fall out.

0. The tools we derive everything from

We will only combine these two. Nothing else is needed.

1. Pitot tube — measuring the speed of a flow

Derivation (from scratch). Take a streamline at the same height $h$ , from a point far upstream (1) with speed $v$ and static pressure $P_0$ , to the stagnation mouth (2) where $v_2=0$ and pressure $P_s$ : $P_0 + \tfrac12\rho v^2 = P_s + 0$ Why this step? Heights are equal, so $\rho g h$ cancels; point 2 is at rest by definition of stagnation.

So the dynamic pressure is $P_s - P_0 = \tfrac12\rho v^2$ . Solve:

2. Venturi meter — measuring flow rate in a pipe

Derivation. Horizontal pipe ( $h_1=h_2$ ), wide section (1) and throat (2): $P_1+\tfrac12\rho v_1^2 = P_2+\tfrac12\rho v_2^2 \;\Rightarrow\; P_1-P_2=\tfrac12\rho(v_2^2-v_1^2)$ Continuity gives $v_1 = \dfrac{A_2}{A_1}v_2$ . Substitute: $P_1-P_2=\tfrac12\rho\,v_2^2\left(1-\frac{A_2^2}{A_1^2}\right)$ Why this step? We replaced $v_1$ to leave a single unknown $v_2$ . Solve for $v_2$ and multiply by $A_2$ to get $Q$ :

Worked example Water through a venturi

$A_1=10\ \text{cm}^2=10^{-3}\,\text{m}^2$ , $A_2=5\ \text{cm}^2=5\times10^{-4}\,\text{m}^2$ , $P_1-P_2=2000\ \text{Pa}$ , water $\rho=1000$ . $Q=(10^{-3})(5\times10^{-4})\sqrt{\dfrac{2(2000)}{1000\,(10^{-6}-2.5\times10^{-7})}}$ $=5\times10^{-7}\sqrt{\dfrac{4000}{7.5\times10^{-4}}}=5\times10^{-7}\sqrt{5.33\times10^{6}}\approx 1.15\times10^{-3}\ \text{m}^3/\text{s}$ . Why this step? Plug into boxed formula directly; $A_1^2-A_2^2=7.5\times10^{-4}\,\text{m}^4$ .

Figure — Applications — Pitot tube, Venturi meter, orifice flow

3. Orifice flow (Torricelli) — fluid draining from a tank

Derivation. Surface (1) at the top, hole (2) at depth $H$ . Both open to atmosphere so $P_1=P_2=P_{atm}$ . If hole area $\ll$ tank area, $v_1\approx 0$ : $\cancel{P_{atm}}+0+\rho g H = \cancel{P_{atm}}+\tfrac12\rho v^2+0$ Why this step? Pressures cancel; surface speed $\approx0$ by continuity ( $A_{hole}\ll A_{tank}$ ); set hole as height reference.

Forecast-then-Verify

Recall Predict before you compute

If you double the venturi pressure drop, does $Q$ double? No — $Q\propto\sqrt{\Delta P}$ , so it rises by $\sqrt2\approx1.41\times$ .
Two holes, one at depth $H$ and one at $4H$ . Speed ratio? $1:2$ (since $v\propto\sqrt H$ ).
Pitot in water vs air for same $\Delta P$ — which gives larger $v$ ? Air, because $v\propto1/\sqrt\rho$ and air is lighter.

Common mistakes (steel-manned)

80/20 — the highest-yield takeaways

Every device = Bernoulli + Continuity, then solve for a speed.
Pitot: $v=\sqrt{2\Delta P/\rho}$ . Venturi: $Q=A_1A_2\sqrt{\tfrac{2\Delta P}{\rho(A_1^2-A_2^2)}}$ . Orifice: $v=\sqrt{2gH}$ .
Faster fluid ⇒ lower pressure. Memorize that arrow and you can re-derive all three.

Recall Feynman: explain to a 12-year-old

Imagine running in a hallway. In a narrow part you have to speed up to get through (that's continuity). When you rush, you push less on the side walls (low pressure) — when you stand still you lean on the wall more (high pressure). A pitot tube is like sticking your hand straight into the wind: you feel a hard push that tells you how fast the wind is. A venturi is the narrow hallway: the squeeze makes water race and press lightly, and we read that. An orifice is a hole in a water bucket: the deeper the hole, the more "fall" the water gets, so it squirts out faster — exactly like dropping a stone from that depth.

Connections

Bernoulli's Equation — the parent principle.
Continuity Equation — supplies the $A_1v_1=A_2v_2$ link.
Torricelli's Law — special case of orifice flow.
Projectile Motion — used for the orifice jet range.
Manometers and Pressure Measurement — how $\Delta P$ is read as $\Delta h$ .
Dynamic vs Static vs Stagnation Pressure.

Flashcards

What is a stagnation point?

A point where the fluid is brought to rest (

v=0

), converting kinetic energy into extra (stagnation) pressure.

Pitot tube speed formula

v=\sqrt{2(P_s-P_0)/\rho}

, where

P_s-P_0

is the dynamic pressure.

Why is throat pressure lower in a venturi?

Continuity forces higher speed at the smaller throat; Bernoulli then requires lower pressure for constant total energy.

Venturi volume flow rate

Q=A_1A_2\sqrt{\dfrac{2(P_1-P_2)}{\rho(A_1^2-A_2^2)}}

State Torricelli's law

Efflux speed from a hole at depth

H

v=\sqrt{2gH}

, same as free fall through

H

Does orifice speed depend on hole area?

No — speed is

\sqrt{2gH}

; area only changes the flow rate

Q=A\sqrt{2gH}

Horizontal range of an orifice jet (hole depth H, height y above floor)

x=2\sqrt{Hy}

If venturi

\Delta P

doubles, how does

Q

change?

Increases by

\sqrt2\approx1.41\times

, since

Q\propto\sqrt{\Delta P}

Two equations that derive all three devices

Bernoulli (

P+\tfrac12\rho v^2+\rho gh=

const) and Continuity (

A_1v_1=A_2v_2

Converting

\Delta P

to manometer height

\Delta P=(\rho_m-\rho)g\,\Delta h

for a fluid column of density

\rho_m

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, teeno devices ka dil ek hi hai: Bernoulli + Continuity. Bernoulli kehta hai ki ek streamline pe $P+\tfrac12\rho v^2+\rho gh$ constant rehta hai — yani jahan fluid tez bhaagta hai, wahan pressure kam ho jaata hai. Bas yahi ek line yaad rakho, baaki sab isi se nikal aata hai.

Pitot tube: ek tube ko seedha flow ke saamne rakho. Wahan fluid ruk jaata hai (stagnation point, $v=0$ ), aur uski speed wali energy extra pressure ban jaati hai. Static pressure se difference lo, aur $v=\sqrt{2\Delta P/\rho}$ se speed nikal lo — yahi se plane apni airspeed maapta hai.

Venturi meter: pipe ko beech mein patla kar do. Continuity ( $A_1v_1=A_2v_2$ ) ki wajah se throat mein fluid tez ho jaata hai, aur Bernoulli ke hisaab se wahan pressure gir jaata hai. Us pressure drop ko manometer se padho aur flow rate $Q=A_1A_2\sqrt{2\Delta P/[\rho(A_1^2-A_2^2)]}$ nikal lo. Yaad rakhna: patla matlab tez aur kam pressure, na ki zyada pressure — yahan log galti karte hain.

Orifice (Torricelli): tank mein depth $H$ pe chhota chhed karo. Upar aur chhed dono jagah atmospheric pressure, to sirf gravity ki height bachti hai jo speed mein badal jaati hai: $v=\sqrt{2gH}$ — bilkul jaise koi cheez $H$ height se gir rahi ho. Jet ki horizontal range $x=2\sqrt{Hy}$ projectile motion se aati hai. Mantra: Push, Pinch, Pour — Pitot, Venturi, Orifice. Exam mein yeh teen formula aur unki derivation guaranteed marks hai.

Applications — Pitot tube, Venturi meter, orifice flow

0. The tools we derive everything from

1. Pitot tube — measuring the speed of a flow

2. Venturi meter — measuring flow rate in a pipe

3. Orifice flow (Torricelli) — fluid draining from a tank

Forecast-then-Verify

Common mistakes (steel-manned)

80/20 — the highest-yield takeaways

Connections

Flashcards

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Fluid Mechanics

Connections