Applications — Pitot tube, Venturi meter, orifice flow: Worked examples

This page is a drill. We take the three master formulas from the parent note and hit every kind of case they can produce — normal numbers, zero inputs, degenerate geometry, limiting behaviour, a real-world word problem, and an exam twist. First we map the territory, then we walk each cell.

The only tools, restated so nothing is used before it is defined:

The scenario matrix

Every problem this topic can throw is one of these cells. The examples below are labelled with the cell they cover.

Cell	Device	What makes it "a case"
A	Pitot	Standard: given $Δ P$ , find $v$
B	Pitot	Read via a manometer height $Δ h$ (indirect $Δ P$ )
C	Venturi	Standard: given $Δ P$ , find $Q$
D	Venturi	Degenerate geometry $A_{2} \to A_{1}$ (throat = pipe) — limiting behaviour
E	Orifice	Standard depth $H$ , find jet speed
F	Orifice	Zero input $H = 0$ (hole at the surface)
G	Orifice	Real-world word problem: jet range on the floor (Projectile Motion)
H	Orifice	Exam twist: which of two holes lands farther? (max range)
I	Mixed	Sanity/limiting: double $Δ P$ → $Q$ up by $2$ , not $2$

A — Pitot, the plain case

Worked example Cell A: standard airspeed

A pitot tube on a glider reads a dynamic pressure $P_{s} - P_{0} = 1800 Pa$ in air of density $ρ = 1.2 kg/m^{3}$ . Find the airspeed $v$ .

Forecast: larger $Δ P$ or lighter fluid → faster. Air is very light, so guess a "highway" speed, tens of m/s.

Write the Pitot formula. $v = \frac{2 ( P _{s} - P _{0} )}{ρ}$ . Why this step? The tube's mouth stops the flow ( $v = 0$ there), converting all kinetic energy into the extra pressure $P_{s} - P_{0}$ ; Bernoulli at equal height gives exactly this.
Plug numbers. $v = \frac{2 ( 1800 )}{1.2} = 3000$ . Why this step? Direct substitution — the height term $ρ g h$ cancels because both ports sit at the same level.
Evaluate. $3000 \approx 54.8 m/s$ .

Verify: Units: $Pa / (kg/m^{3}) = (N/m^{2}) (m^{3} / kg) = N\cdotm/kg = m^{2} / s^{2} = m/s$ . ✓ And $54.8$ m/s ≈ 197 km/h — a believable glide speed.

B — Pitot read through a manometer

Worked example Cell B: dynamic pressure hidden in a liquid height

A pitot–static tube in water ( $ρ = 1000 kg/m^{3}$ ) is connected to a mercury manometer ( $ρ_{m} = 13600 kg/m^{3}$ ) that shows a height difference $Δ h = 5 cm = 0.05 m$ . Take $g = 10$ . Find the water speed $v$ .

Forecast: mercury is 13.6× denser than water, so even a small $5$ cm column stands for a big pressure. But water is heavy too, so the speed won't be enormous — a few m/s.

Turn the height into a pressure. $P_{s} - P_{0} = ρ_{m} g Δ h$ . Why this step? A manometer balances the unknown pressure difference against a known column of liquid; the column's weight per area is $ρ_{m} g Δ h$ . (See Manometers and Pressure Measurement.)
Compute it. $P_{s} - P_{0} = 13600 \times 10 \times 0.05 = 6800 Pa$ . Why this step? This is the dynamic pressure the moving water carries.
Feed into Pitot. $v = \frac{2 ( 6800 )}{1000} = 13.6 \approx 3.69 m/s$ . Why this step? Same formula as Cell A, but $Δ P$ now comes from the manometer, not a direct gauge.

Verify: $v^{2} = 13.6$ ; re-multiply: $\frac{1}{2} ρ v^{2} = \frac{1}{2} (1000) (13.6) = 6800$ Pa = the $Δ P$ we started from. ✓ Loop closes.

C — Venturi, the plain case

Worked example Cell C: flow rate through a constriction

Water ( $ρ = 1000$ ) flows through a venturi with wide area $A_{1} = 20 cm^{2} = 2 \times 1 0^{- 3} m^{2}$ and throat $A_{2} = 8 cm^{2} = 8 \times 1 0^{- 4} m^{2}$ . The pressure drops $P_{1} - P_{2} = 1500 Pa$ into the throat. Find $Q$ .

Forecast: a real pipe carries something like litres per second, so $Q$ around $1 0^{- 3} m^{3} / s$ .

Write $Q$ . $Q = A_{1} A_{2} \frac{2 ( P _{1} - P _{2} )}{ρ ( A _{1}^{2} - A _{2}^{2} )}$ . Why this step? Continuity ( $A_{1} v_{1} = A_{2} v_{2}$ ) removes one unknown; Bernoulli supplies the pressure link. Combining gives $Q$ in terms of measurable areas and $Δ P$ only.
Areas. $A_{1} A_{2} = (2 \times 1 0^{- 3}) (8 \times 1 0^{- 4}) = 1.6 \times 1 0^{- 6}$ . And $A_{1}^{2} - A_{2}^{2} = 4 \times 1 0^{- 6} - 6.4 \times 1 0^{- 7} = 3.36 \times 1 0^{- 6} m^{4}$ . Why this step? These are the two geometry pieces the formula needs.
Inside the root. $\frac{2 ( 1500 )}{1000 \times 3.36 \times 1 0 ^{- 6}} = \frac{3000}{3.36 \times 1 0 ^{- 3}} = 8.93 \times 1 0^{5}$ .
Finish. $Q = 1.6 \times 1 0^{- 6} 8.93 \times 1 0^{5} = 1.6 \times 1 0^{- 6} \times 944.9 \approx 1.51 \times 1 0^{- 3} m^{3} / s$ . Why this step? $8.93 \times 1 0^{5} \approx 945$ ; multiply by the area product.

Verify: $1.51 \times 1 0^{- 3} m^{3} / s = 1.51$ L/s — realistic for a garden-hose-sized pipe. Units inside root: $Pa / (kg/m^{3} \cdot m^{4}) = s^{- 2} m^{- 2}$ , root gives $m^{- 1} s^{- 1}$ , times $m^{4}$ gives $m^{3} / s$ . ✓

D — Venturi with no constriction (degenerate limit)

Worked example Cell D: the throat equals the pipe,

$A_{2} \to A_{1}$ What does the venturi formula predict for $Q$ if the "throat" is not narrowed at all, i.e. $A_{2} = A_{1}$ , but you still read some pressure drop $P_{1} - P_{2} > 0$ ? Interpret it physically.

Forecast: if the pipe never narrows, continuity says $v_{1} = v_{2}$ , so Bernoulli forces $P_{1} = P_{2}$ . A nonzero $Δ P$ across a uniform pipe should be impossible — the formula ought to blow up or forbid it.

Set $A_{2} = A_{1} = A$ in the denominator. $A_{1}^{2} - A_{2}^{2} = A^{2} - A^{2} = 0$ . Why this step? The denominator $ρ (A_{1}^{2} - A_{2}^{2})$ is exactly the "how much does the fluid speed up" factor. No narrowing → it vanishes.
Read the formula. $Q = A^{2} \frac{2 ( P _{1} - P _{2} )}{ρ \cdot 0} \to \infty$ for any $Δ P > 0$ . Why this step? Dividing by zero → unbounded. Mathematically the model refuses this input.
Physical reading. A uniform pipe cannot sustain $P_{1} \neq = P_{2}$ from Bernoulli alone (heights equal, speeds equal). So $Δ P = 0$ is the only consistent value; the $\frac{0}{0}$ that results is the model saying "no information about $Q$ from an unconstricted pipe." Why this step? Degenerate geometry means the measurement principle disappears — you need the squeeze.

Verify: consistency check — if instead $Δ P = 0$ and $A_{2} = A_{1}$ , the numerator is $0$ too, giving $Q = A^{2} 0/0$ , the honest "indeterminate": the meter provides no reading, exactly right. ✓

E — Orifice, plain depth

F — Orifice with zero depth (zero input)

G — Orifice + projectile: range on the floor (word problem)

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

Jet speed (Torricelli). $v = 2 g H = 2 (10) (0.20) = 4 = 2 m/s$ . Why this step? This is the horizontal launch speed — see the red jet in the figure.

Fall time (free fall from $y$ ). The jet has no vertical speed at launch, so $y = \frac{1}{2} g t^{2} \Rightarrow t = 2 y / g = 2 (0.45) /10 = 0.09 = 0.30 s$ . Why this step? Horizontal and vertical motions are independent (Projectile Motion); vertical is pure free fall.

Range. $x = v t = 2 \times 0.30 = 0.60 m$ . Why this step? No horizontal force → constant horizontal speed, so distance = speed × time.

Verify: the shortcut $x = 2 H y = 2 0.20 \times 0.45 = 2 0.09 = 2 (0.30) = 0.60 m$ agrees. ✓

H — Exam twist: which hole lands farthest?

Write the range function. $x (y) = 2 (D - y) y = 2 D y - y^{2}$ . Why this step? Combining Torricelli speed and free-fall time gives range purely as a function of the hole height $y$ .

Maximize the inside. The parabola $f (y) = D y - y^{2}$ peaks where $f^{'} (y) = D - 2 y = 0 \Rightarrow y = D /2$ . Why this step? $x$ grows with $f$ , so maximizing $f$ maximizes $x$ ; a downward parabola peaks at its vertex.

Best height and range. $y = D /2 = 0.5 m$ (mid-height). Then $H = D - y = 0.5 m$ and $x_{m a x} = 2 0.5 \times 0.5 = 2 (0.5) = 1.0 m$ . Why this step? At $y = D /2$ the depth equals the height — the balance point the forecast predicted.

Verify: at the vertex $x_{m a x} = D = 1.0 m$ (a known result: max range equals the fill height). Check a neighbour: $y = 0.6 \Rightarrow x = 2 0.4 \times 0.6 = 2 0.24 = 0.98 m < 1.0$ . ✓ Also symmetric: $y = 0.4$ gives the same $0.98$ m as $y = 0.6$ — holes symmetric about mid-height land equally far.

I — Sanity limit: doubling the pressure drop

Recall One-line recall per cell

Pitot direct ::: $v = 2Δ P / ρ$ Pitot via manometer ::: $Δ P = ρ_{m} g Δ h$ , then Pitot Venturi ::: $Q = A_{1} A_{2} 2Δ P / [ρ (A_{1}^{2} - A_{2}^{2})]$ Venturi with $A_{2} = A_{1}$ ::: forbids $Δ P \neq = 0$ ( $Q \to \infty$ / indeterminate) Orifice speed ::: $v = 2 g H$ , no area, no density Orifice at $H = 0$ ::: $v = 0$ Jet range ::: $x = 2 H y$ Max range hole height ::: $y = D /2$ , giving $x_{m a x} = D$ Double $Δ P$ ::: $Q$ up by $2$

Connections

Bernoulli's Equation — every cell starts here.
Continuity Equation — supplies the area link in the venturi cells.
Torricelli's Law — cells E, F, G, H.
Projectile Motion — the range cells G and H.
Manometers and Pressure Measurement — cell B's $Δ h \to Δ P$ .
Dynamic vs Static vs Stagnation Pressure — the pressure vocabulary behind Pitot and Venturi.