Applications — Pitot tube, Venturi meter, orifice flow: Question bank

Definition Symbol glossary (skim before you start)

$P_{s}$ = stagnation pressure, read where the fluid is brought to rest at a pitot mouth.
$P_{0}$ = static pressure of the undisturbed stream (side-on port).
$P_{atm}$ = atmospheric pressure — the pressure of the open air. For a jet or a tank surface exposed to the air this is the local static pressure, so $P_{atm}$ plays the role of $P_{0}$ at any open-to-air point.
$Δ P$ = the pressure difference a device measures ( $P_{s} - P_{0}$ for pitot, $P_{1} - P_{2}$ for venturi).
$ρ$ = density of the flowing fluid (air, water, …).
$v$ = local flow speed of the fluid; $v_{1}, v_{2}$ = speeds at the two labelled points of a device (inlet 1 vs throat 2 in a venturi; surface 1 vs hole 2 in a draining tank).
$g$ = gravitational acceleration ( $\approx 9.8 m/s^{2}$ ), the strength of gravity that appears in every $ρ g h$ and $2 g H$ term.
$A_{1}, A_{2}$ = cross-section areas of the venturi's wide inlet and narrow throat.
$A_{hole}, A_{tank}$ = area of the drain hole and of the tank's surface.
$H$ = depth of the drain hole below the free surface (drives Torricelli speed).
$h$ = a point's height in Bernoulli's $ρ g h$ term (elevation on a streamline).
$Δ h$ = the manometer liquid column height used to read off $Δ P$ .
$ρ_{m}$ = density of the manometer liquid; $ρ$ (again) = density of the flowing fluid in $P_{1} - P_{2} = (ρ_{m} - ρ) g Δ h$ .
$x$ = horizontal range of a jet along the floor; $y$ = vertical fall height of the jet from the hole down to the floor (used in projectile motion).

True or false — justify

The narrow throat of a venturi meter has higher pressure because the fluid is squeezed.

False. A fluid is not compressed like a solid — by continuity it speeds up in the throat, and by Bernoulli faster flow means lower pressure. The throat is a pressure dip.

Torricelli's speed

v = 2 g H

depends on the size of the drain hole.

False. The formula contains no area at all;

v

is set purely by the depth

H

. Hole size changes the flow rate

Q = A_{hole} v

, not the speed of each escaping particle.

Stagnation pressure at a pitot mouth equals atmospheric pressure.

False. The mouth faces the stream and brings fluid to rest, so it reads static plus the dynamic term:

P_{s} = P_{0} + \frac{1}{2} ρ v^{2} > P_{0}

. That extra bit is exactly what encodes the speed.

Doubling the venturi pressure drop

Δ P

doubles the measured flow rate

Q

False. Bernoulli gives

Δ P = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})

, so

Δ P

scales with speed squared; continuity (

Q = A_{2} v_{2}

) then makes

Q \propto v_{2} \propto Δ P

. Doubling

Δ P

multiplies

Q

by only

2 \approx 1.41

For the same measured

Δ P

, a pitot tube in air reports a larger speed than one in water.

True.

v = 2Δ P / ρ

, so

v \propto 1/ ρ

; air is far lighter than water, so the same

Δ P

corresponds to a much faster flow.

In the orifice derivation we can ignore the surface speed

v_{1}

regardless of hole size.

False. We drop

v_{1}

only because

A_{hole} ≪ A_{tank}

forces

v_{1} = (A_{hole} / A_{tank}) v_{2} \approx 0

. If the hole is comparable to the tank cross-section,

v_{1}

must be kept.

Bernoulli's equation requires you to use absolute pressures, never gauge pressures.

False. All three devices use only the difference

P_{1} - P_{2}

, so any consistent reference cancels. The rule is: pick one convention and never mix Pa with atm mid-problem.

The height term

ρ g h

in Bernoulli matters for a pitot tube in air.

True in principle, negligible in practice. Over the tiny vertical span of the tube in a light gas,

ρ g h

is minuscule compared to the dynamic term, so we cancel it — but it is a size argument, not an exact zero.

Bernoulli's incompressible form is safe for a pitot tube at any airspeed.

False. Bernoulli here assumes constant density

ρ

. For gases above roughly Mach 0.3 (about

100 m/s

in air) compressibility matters and

v = 2Δ P / ρ

under-reads; a compressible correction is required. See the edge-case section.

Spot the error

"In a venturi the fluid slows down in the throat because there is less room."

The error inverts continuity. Less area means higher speed, not lower:

A_{1} v_{1} = A_{2} v_{2}

with

A_{2} < A_{1}

forces

v_{2} > v_{1}

. Same volume per second must squeeze through a smaller door.

"A pitot tube measures pressure, so its output is a pressure, not a speed."

It measures a pressure difference, but the physics converts it:

v = 2 (P_{s} - P_{0}) / ρ

. The device is calibrated to output speed because the dynamic pressure is a known function of speed.

"Torricelli says the jet leaves faster from a deeper hole because the water above weighs more."

The reasoning lands on the right answer for the wrong reason. It is not the weight pushing but the conversion of potential energy of the falling surface height

H

into kinetic energy —

v = 2 g H

is exactly a free-fall speed through

H

"For a horizontal jet, the deepest hole always throws water the farthest along the floor."

Wrong — range is

x = 2 H y

where

y

is the fall height below the hole. A deeper hole (larger

H

) has a smaller remaining fall

y

, so range peaks at the mid-height, not the bottom.

"Since both the venturi ports are on a horizontal pipe, the manometer fluid doesn't matter."

The manometer fluid density

ρ_{m}

sets the conversion

P_{1} - P_{2} = (ρ_{m} - ρ) g Δ h

, where

ρ

is the flowing fluid. A different manometer liquid gives a different

Δ h

for the same

Δ P

, so it very much matters.

"The static port of a pitot-static tube should also face into the wind."

No — the static port must be side-on (parallel to the flow) so it reads undisturbed static pressure

P_{0}

. Facing it into the wind would make it read stagnation pressure too, and the difference would vanish.

Edge cases

What does the pitot formula give if the stream is at rest (

v = 0

Then

P_{s} = P_{0}

, so

Δ P = 0

and

v = 0 = 0

— self-consistent. No flow, no dynamic pressure, no reading.

What happens to venturi

Q

if the throat area

A_{2}

approaches the pipe area

A_{1}

Q = A_{1} A_{2} 2Δ P / [ρ (A_{1}^{2} - A_{2}^{2})]

the denominator

A_{1}^{2} - A_{2}^{2} \to 0

, so the formula blows up — but physically

Δ P \to 0

too, since there is no constriction. The device stops working: no squeeze, no signal.

What is the drain speed at a hole placed exactly at the water surface (

H = 0

v = 2 g \cdot 0 = 0

. Water at the surface has no depth of potential energy to convert, so it dribbles out with essentially zero speed.

For the horizontal jet range

x = 2 H y

, what happens as the hole nears the floor (

y \to 0

Range

\to 0

: the jet has almost no vertical fall

y

, so it lands right at the base of the tank no matter how fast it leaves.

At what airspeed does the incompressible pitot formula start to fail for a gas?

Above roughly Mach 0.3 (about

100 m/s

in sea-level air), density changes are no longer negligible, so treating

ρ

as constant in

v = 2Δ P / ρ

introduces error; real airspeed indicators apply a compressible-flow correction. The same warning applies to a high-speed gas venturi.

Can the venturi throat pressure drop so low that the formula stops being trustworthy?

Yes — if the throat speed is high enough,

P_{2}

can fall to the liquid's vapor pressure and the water flashes to vapor bubbles (cavitation). Bernoulli assumes a single incompressible phase, so beyond that limit the measurement (and the pipe surface) degrades.

Does the same low-pressure danger arise in orifice flow?

It can in a submerged or venturi-nozzle orifice where the jet's high speed creates a local pressure below vapor pressure; a simple open-air tank hole stays near atmospheric and is safe. Cavitation is the real-world floor on how low pressure may drop before the ideal model fails.

What does Bernoulli predict if a venturi is tilted so the throat is higher than the inlet?

The

ρ g h

terms no longer cancel; the pressure drop then reflects both the speed increase and the height gain, so the horizontal-pipe formula must be corrected with the

ρ g (h_{2} - h_{1})

term.

Recall One-line summary of every trap

Speed comes from energy, rate from geometry. Pitot $v = 2Δ P / ρ$ powered the "air vs water" and "at rest → zero" items; Torricelli $v = 2 g H$ powered the "hole size irrelevant", " $H = 0$ ", and "range peaks mid-height" items; and $Q = A_{1} A_{2} 2Δ P / [ρ (A_{1}^{2} - A_{2}^{2})]$ (rate = area × speed) powered the " $Δ P$ " and " $A_{2} \to A_{1}$ " traps. Always trace continuity first, Bernoulli second, use pressure differences (references cancel), remember the incompressible model breaks above Mach 0.3 in gases, and that cavitation is the physical floor on low pressure.

Connections

Bernoulli's Equation — the energy balance behind every trap.
Continuity Equation — resolves the venturi "squeeze" confusion.
Torricelli's Law — the free-fall analogy.
Projectile Motion — the jet-range edge cases.
Manometers and Pressure Measurement — why manometer fluid matters.
Dynamic vs Static vs Stagnation Pressure — the static-vs-stagnation port distinction.

Question bank — Applications — Pitot tube, Venturi meter, orifice flow

True or false — justify

Spot the error

Why questions

Edge cases

Connections