Applications — Pitot tube, Venturi meter, orifice flow: Exercises

Symbols recap so nothing is used unexplained. Height convention: all heights are measured upward from a chosen reference level called the datum; we use different letters only to signal what the height is of, but they all mean "metres above the datum" (or a depth below a named surface where stated):

$P$ = pressure (force per area, units pascal $Pa$ ).
$v$ = fluid speed. $ρ$ = density.
$z$ = elevation of a point above the datum (used when a pipe/throat sits higher or lower).
$H$ = a depth below a free surface (used for draining tanks).
$y$ = height of a hole above the floor (used only for projectile range).
$h$ = a generic height inside the raw Bernoulli statement.
$A$ = cross-section area the fluid flows through. $Q = A v$ = volume per second flowing.
$Δ P = P_{1} - P_{2}$ , a pressure difference; $Δ h$ = a manometer height reading.

Level 1 — Recognition

L1.1

A tube is bent so its open mouth points straight into an oncoming river current. The fluid piling up at the mouth is momentarily brought to rest. Name that point and state the fluid speed there.

Recall Solution

The mouth faces the flow, so incoming fluid decelerates and stops: this is the stagnation point, and the speed there is $v = 0$ . Why: all the incoming kinetic energy $\frac{1}{2} ρ v^{2}$ has turned into extra pressure. See Dynamic vs Static vs Stagnation Pressure. Answer: stagnation point, $v = 0$ .

L1.2

In a horizontal pipe that narrows to a throat, is the pressure at the narrow throat higher or lower than in the wide section? One sentence of reasoning.

Recall Solution

Lower. By continuity the throat forces the fluid to speed up ( $A_{2} < A_{1} \Rightarrow v_{2} > v_{1}$ ); by Bernoulli, faster fluid at the same height carries lower pressure. Answer: lower pressure at the throat.

L1.3

Water drains from a tank through a small hole a depth $H$ below the surface. Write the jet speed formula and state what it does not depend on.

Recall Solution

Torricelli: $v = 2 g H$ (see Torricelli's Law). It does not depend on the hole's area or shape — only the depth $H$ . Area affects the rate $Q = A 2 g H$ , not the speed.

Level 2 — Application

L2.1

A pitot tube on a glider reads a pressure difference $P_{s} - P_{0} = 1080 Pa$ in air ( $ρ_{a} = 1.2 kg/m^{3}$ ). Find the airspeed.

Recall Solution

WHAT: apply the pitot speed formula $v = 2 (P_{s} - P_{0}) / ρ$ . WHY: the mouth faces the stream, so $P_{s} - P_{0} = \frac{1}{2} ρ v^{2}$ is exactly the dynamic pressure. Solve for $v$ . $v = \frac{2 ( 1080 )}{1.2} = 1800 \approx 42.4 m/s .$ Answer: $v \approx 42.4 m/s$ .

L2.2

A tank has a hole at depth $H = 1.25 m$ . Find the jet speed. If the hole area is $A = 4 cm^{2}$ , find the volume flow rate $Q$ .

Recall Solution

Speed: $v = 2 g H = 2 (10) (1.25) = 25 = 5 m/s$ . Rate: convert area $A = 4 cm^{2} = 4 \times 1 0^{- 4} m^{2}$ , then $Q = A v = (4 \times 1 0^{- 4}) (5) = 2 \times 1 0^{- 3} m^{3} / s .$ Answer: $v = 5 m/s$ , $Q = 2 \times 1 0^{- 3} m^{3} / s$ .

L2.3

A pitot's pressure difference is read on a mercury manometer whose two legs are tapped into the same flowing water on both sides, so the water fills the tubes down to the mercury surfaces on each leg. Mercury height difference $Δ h = 5 cm$ , mercury $ρ_{m} = 13600 kg/m^{3}$ , flowing water $ρ = 1000 kg/m^{3}$ . Find the water speed. (See Manometers and Pressure Measurement.)

Recall Solution

WHAT: turn the manometer height into a pressure. Because water sits above the mercury in both legs, the water columns partly balance each other and the net reading is driven by the excess weight of mercury over the displaced water. The correct relation is therefore $Δ P = (ρ_{m} - ρ) g Δ h .$ WHY the $(ρ_{m} - ρ)$ ? Over the height $Δ h$ where one leg has mercury and the other has water, only the density difference is unbalanced; if you forgot the water and used $ρ_{m} g Δ h$ you would over-count by $ρ g Δ h$ . (When the fluid above the mercury is a gas like air, $ρ$ is negligible and the simpler $ρ_{m} g Δ h$ is fine — that's the geometry that justifies the shortcut.) $Δ P = (13600 - 1000) (10) (0.05) = 12600 \times 0.5 = 6300 Pa .$ Then pitot speed with the water density (the fluid that's flowing): $v = \frac{2 ( 6300 )}{1000} = 12.6 \approx 3.55 m/s .$ Answer: $v \approx 3.55 m/s$ .

Level 3 — Analysis

L3.1

A horizontal venturi carries water. Wide area $A_{1} = 8 cm^{2}$ , throat $A_{2} = 2 cm^{2}$ , measured pressure drop $P_{1} - P_{2} = 3000 Pa$ . Find the flow rate $Q$ .

Figure below (s01): a lavender-filled pipe drawn wide on the left, pinched to a narrow throat in the middle, then widening again. A slow coral inlet arrow $v_{1}$ becomes a longer, faster arrow $v_{2}$ at the throat; the wide section is tagged " $P_{1}$ high" and the throat " $P_{2}$ low," with pressure taps rising off each. It shows visually why the pressure dips exactly where the fluid speeds up.

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

Recall Solution

Derive the formula first (WHY, not just plug). Horizontal pipe so $z_{1} = z_{2}$ and the $ρ g z$ terms cancel. Bernoulli between wide (1) and throat (2): $P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2} \Rightarrow P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) .$ WHY continuity now: we have one equation but two unknown speeds. Continuity $A_{1} v_{1} = A_{2} v_{2}$ lets us write $v_{1} = \frac{A _{2}}{A _{1}} v_{2}$ , killing one unknown: $P_{1} - P_{2} = \frac{1}{2} ρ v_{2}^{2} (1 - \frac{A _{2}^{2}}{A _{1}^{2}}) .$ Solve for $v_{2}$ , then multiply by $A_{2}$ to get the volume rate $Q = A_{2} v_{2}$ . Doing the algebra (multiply top and bottom inside the root by $A_{1}^{2}$ ) gives the compact form: $Q = A_{1} A_{2} \frac{2 ( P _{1} - P _{2} )}{ρ ( A _{1}^{2} - A _{2}^{2} )} .$ Now plug in. Convert areas: $A_{1} = 8 \times 1 0^{- 4} m^{2}$ , $A_{2} = 2 \times 1 0^{- 4} m^{2}$ . $A_{1}^{2} - A_{2}^{2} = 64 \times 1 0^{- 8} - 4 \times 1 0^{- 8} = 60 \times 1 0^{- 8} = 6 \times 1 0^{- 7} m^{4} .$ $Q = (8 \times 1 0^{- 4}) (2 \times 1 0^{- 4}) \frac{2 ( 3000 )}{1000 ( 6 \times 1 0 ^{- 7} )} .$ Inside the root: $\frac{6000}{6 \times 1 0 ^{- 4}} = 1 0^{7}$ , so $1 0^{7} \approx 3162$ . $Q = (1.6 \times 1 0^{- 7}) (3162) \approx 5.06 \times 1 0^{- 4} m^{3} / s .$ Answer: $Q \approx 5.06 \times 1 0^{- 4} m^{3} / s$ (about $0.5 L/s$ ).

L3.2

For the venturi in L3.1, find the throat speed $v_{2}$ and the wide-section speed $v_{1}$ , and confirm they obey continuity.

Recall Solution

$v_{2} = Q / A_{2} = \frac{5.06 \times 1 0 ^{- 4}}{2 \times 1 0 ^{- 4}} \approx 2.53 m/s$ . $v_{1} = Q / A_{1} = \frac{5.06 \times 1 0 ^{- 4}}{8 \times 1 0 ^{- 4}} \approx 0.632 m/s$ . Continuity check: $A_{1} v_{1} = (8 \times 1 0^{- 4}) (0.632) = 5.06 \times 1 0^{- 4} = Q$ , and $v_{2} / v_{1} = A_{1} / A_{2} = 4$ . ✓ The throat runs $4 \times$ faster, exactly as areas demand.

L3.3

A tank is filled so the free surface is a depth $H = 0.80 m$ above a hole in its side wall. The hole is $y = 1.25 m$ above the floor. How far from the base does the horizontal jet land?

Figure below (s02): a lavender water tank with the surface marked; a coral dot marks the hole. A mint double-arrow labels the depth $H$ from surface to hole, a butter double-arrow labels the height $y$ from hole to floor, and a coral parabola traces the jet arching down to a landing mark $x$ on the floor. It separates the two ideas visually: $H$ sets the speed, $y$ sets the fall time.

Recall Solution

WHAT: two separate physics steps — get the exit speed, then treat the jet as a projectile. WHY split it? Once fluid leaves the hole, gravity alone acts; horizontally there is no force, so it's ordinary Projectile Motion. Exit speed (Torricelli): $v = 2 g H = 2 (10) (0.80) = 16 = 4 m/s$ . Fall time from height $y$ : $y = \frac{1}{2} g t^{2} \Rightarrow t = 2 y / g = 2 (1.25) /10 = 0.25 = 0.5 s$ . Horizontal range: $x = v t = 4 \times 0.5 = 2 m$ . (Compact form: $x = 2 H y = 2 0.80 \times 1.25 = 21 = 2 m$ .) ✓ Answer: $x = 2 m$ .

Level 4 — Synthesis

L4.1

Two holes are drilled in a tall tank of total water depth $D = 1.8 m$ : hole A at depth $H_{A} = 0.2 m$ below the surface, hole B at depth $H_{B} = 1.6 m$ . Both are on the same vertical wall. Show that they land at the same spot on the floor, and find that distance. (Floor is at the tank base.)

Figure below (s03): the same tank drawn once, with two holes — a coral one high up (depth $H_{A}$ ) and a mint one low down (depth $H_{B}$ ). Two jets, coral and mint, arch out and are shown meeting the floor at the same landing mark, making the "mirror-depth" symmetry visible at a glance.

Recall Solution

Setup: a hole at depth $H$ below the surface sits a height $y = D - H$ above the floor. Its exit speed is $v = 2 g H$ and fall time $t = 2 y / g$ , so range $x = v t = 2 g H \cdot \frac{2 ( D - H )}{g} = 2 H (D - H) .$ Symmetry insight: $H (D - H)$ is unchanged when $H \to D - H$ . Hole A has $H_{A} = 0.2$ ; the mirror depth is $D - H_{A} = 1.6 = H_{B}$ . So A and B are mirror partners ⇒ equal range. Numerically: $x_{A} = 2 0.2 (1.8 - 0.2) = 2 0.2 \times 1.6 = 2 0.32 \approx 1.13 m$ . $x_{B} = 2 1.6 (1.8 - 1.6) = 2 1.6 \times 0.2 = 2 0.32 \approx 1.13 m$ . ✓ Same. Answer: both land $\approx 1.13 m$ from the wall.

L4.2

For that same tank, at what single depth $H_{⋆}$ would a hole give the maximum possible range, and what is that range?

Recall Solution

WHAT: maximise $x = 2 H (D - H)$ over $H$ . WHY calculus/symmetry? The product $H (D - H)$ is a downward parabola in $H$ ; its peak is exactly at the midpoint $H_{⋆} = D /2$ . $H_{⋆} = 1.8/2 = 0.9 m$ . Max range: $x_{m a x} = 2 0.9 (1.8 - 0.9) = 2 0.9 \times 0.9 = 2 (0.9) = 1.8 m$ . Neatly, $x_{m a x} = 2 (D /2) (D /2) = D$ — the maximum horizontal range equals the water depth. Answer: hole at mid-depth $H_{⋆} = 0.9 m$ , range $= 1.8 m$ .

Level 5 — Mastery

L5.1

A venturi is mounted vertically in a pipe carrying water upward. The wide inlet (1) is at the bottom, area $A_{1} = 12 cm^{2}$ ; the throat (2) is a height $Δ z = 0.30 m$ higher, area $A_{2} = 4 cm^{2}$ . A gauge reads the true pressure difference $P_{1} - P_{2} = 4200 Pa$ . Find the flow rate $Q$ , correctly accounting for the height difference. (Here $z_{1}$ and $z_{2}$ are the elevations of points 1 and 2 above a common datum; take the datum at the inlet so $z_{1} = 0$ and $z_{2} = Δ z = 0.30 m$ .)

Recall Solution

WHAT/WHY: now the two points are at different elevations, so the $ρ g z$ terms in Bernoulli do not cancel. Full Bernoulli between (1) and (2): $P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g z_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g z_{2} .$ Rearrange, with $z_{2} - z_{1} = Δ z = 0.30 m$ : $effective drop Δ P_{eff} (P_{1} - P_{2}) - ρ g Δ z = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) .$ WHY this grouping: part of the measured pressure difference is spent just lifting the fluid the height $Δ z$ ; only the leftover drives the speed change. Compute the lift term: $ρ g Δ z = (1000) (10) (0.30) = 3000 Pa$ , so $Δ P_{eff} = 4200 - 3000 = 1200 Pa .$ From here it is the same derivation as L3.1 (continuity substitutes out $v_{1}$ ), just with $Δ P_{eff}$ in place of $P_{1} - P_{2}$ : $Q = A_{1} A_{2} \frac{2 Δ P _{eff}}{ρ ( A _{1}^{2} - A _{2}^{2} )} .$ Areas: $A_{1} = 12 \times 1 0^{- 4}$ , $A_{2} = 4 \times 1 0^{- 4} m^{2}$ ; $A_{1}^{2} - A_{2}^{2} = (144 - 16) \times 1 0^{- 8} = 128 \times 1 0^{- 8} = 1.28 \times 1 0^{- 6} m^{4}$ . Inside the root: $\frac{2 ( 1200 )}{1000 ( 1.28 \times 1 0 ^{- 6} )} = \frac{2400}{1.28 \times 1 0 ^{- 3}} = 1.875 \times 1 0^{6}$ ; $\approx 1369$ . Prefactor: $A_{1} A_{2} = (12 \times 1 0^{- 4}) (4 \times 1 0^{- 4}) = 4.8 \times 1 0^{- 7}$ . $Q = (4.8 \times 1 0^{- 7}) (1369) \approx 6.57 \times 1 0^{- 4} m^{3} / s .$ Answer: $Q \approx 6.57 \times 1 0^{- 4} m^{3} / s$ . (Ignoring $Δ z$ would have overstated the drop by $3000 Pa$ and inflated $Q$ .)

L5.2

A pitot–static tube in a wind tunnel gives $P_{s} - P_{0} = 800 Pa$ in air ( $ρ_{a} = 1.2$ ). The same tube in a water tunnel gives the same reading $800 Pa$ . Find both speeds and the ratio, and explain physically which fluid moves faster for equal $Δ P$ .

Recall Solution

Air: $v_{a} = 2 (800) /1.2 = 1333.3 \approx 36.5 m/s$ . Water: $v_{w} = 2 (800) /1000 = 1.6 \approx 1.265 m/s$ . Ratio: $\frac{v _{a}}{v _{w}} = \frac{ρ _{w}}{ρ _{a}} = \frac{1000}{1.2} = 833.3 \approx 28.9$ . Physical why: dynamic pressure is $\frac{1}{2} ρ v^{2}$ . To reach the same pressure, the lighter fluid (air) must move much faster to make up for its small $ρ$ — speed scales as $1/ ρ$ . Answer: $v_{a} \approx 36.5 m/s$ , $v_{w} \approx 1.27 m/s$ , ratio $\approx 28.9$ .

Recall One-line self-test

Which single equation, combined with Bernoulli, converts a measured area ratio into a speed ratio? ::: The continuity equation $A_{1} v_{1} = A_{2} v_{2}$ .

Connections

2.2.16 Applications — Pitot tube, Venturi meter, orifice flow (Hinglish) · Bernoulli's Equation · Continuity Equation
Torricelli's Law · Projectile Motion · Manometers and Pressure Measurement · Dynamic vs Static vs Stagnation Pressure