MHT-CET Physics · Mechanical Properties of Fluids

Streamline Flow: Continuity, Bernoulli and Torricelli

In steady streamline flow the volume passing any section per second is the same, so the fluid speeds up where the pipe narrows (Av constant); Bernoulli's theorem then says the pressure falls where the speed rises, and Torricelli's theorem gives the speed of a jet from a hole at depth h as √(2gh).

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Why this matters

17 PYQs, 3 HARD. Six are continuity — speed in a narrower pipe, a nozzle, a sprinkler, how fast a tank empties; seven are Bernoulli — the pressure at the narrow part, the lift on a roof in a wind, the flow rate from a pressure difference, the speed of water from an opened valve; four are Torricelli — jets from holes at different depths, the time to drain, and the recoil of a tank with holes on opposite sides. Three cards.

Concept 1 of 3: Continuity: Narrow Pipe, Faster Flow

Water cannot pile up in a pipe, so the volume flow rate Q = Av is the same everywhere along it. A pipe of a third the radius has a ninth the area, so the water moves nine times faster. The same rule links a pipe to a nozzle (d√(V/V₁)), a hose to a sprinkler with n holes (v' = R²v/(nr²)), and a tap to the falling water level in a tank (dh/dt = a v/A). Streamline flow means the velocity at any fixed point does not change with time and the layers stay parallel.

Definition

$A_1v_1 = A_2v_2$ ; for a circular pipe $r_1^2v_1 = r_2^2v_2$ . Radius R → R/3 ⇒ speed × 9.
Flow rate $Q = \pi r^2 v$ : $\pi \times 10^{-1}$ m³/s through radius 0.1 m ⇒ 10 m/s.
Nozzle from diameter d at V to speed $V_1$ : $d_1 = d\sqrt{\dfrac{V}{V_1}}$ . Sprinkler with n holes of radius r: $v' = \dfrac{R^2v}{nr^2}$ .
Tank draining through a tap: $A\,\dfrac{dh}{dt} = a\,v$ .
Streamline flow: velocity at a point is constant in time, below the critical velocity, layers parallel, no random motion.

Continuity

A_1 v_1 = A_2 v_2 = Q

Worked example

A hose of radius 1 cm carries water at 2 m/s into a sprinkler with 20 holes of radius 1 mm. Speed from each hole?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q7Easy

Example 1 · Mechanical Properties of Fluids · Bernoulli, Continuity, Streamline, and Torricelli

A gardening pipe having an internal radius 'R' is connected to a water sprinkler having 'n' holes each of radius 'r'. The water in the pipe has a speed 'v'. The speed of water leaving the sprinkler is

Scaling speed with radius, not area

Q = Av and A ∝ r². A third of the radius means a ninth of the area and nine times the speed, not three.

Drill 5 more on continuity: narrow pipe, faster flow

Concept 2 of 3: Bernoulli's Theorem

Along a streamline P + ½ρv² + ρgh is constant. In a horizontal pipe the height term drops out, so wherever the fluid speeds up its pressure falls by ½ρ(v₂² − v₁²) — the narrowest part has the highest speed and the lowest pressure. Pair it with continuity: find the speeds from the areas, then the pressure change from Bernoulli. A wind over a roof is the same idea: fast air above, still air below, and a lift of ½ρv² on every square metre.

Definition

$P + \tfrac{1}{2}\rho v^2 + \rho g h =$ constant along a streamline.
Horizontal pipe: $P_1 - P_2 = \tfrac{1}{2}\rho\left(v_2^2 - v_1^2\right)$ . Speed v → 3v ⇒ pressure falls by $4\rho v^2$ .
Narrowest section: maximum speed, minimum pressure.
Roof in a wind of speed v: force $= \tfrac{1}{2}\rho v^2 A$ (50 m/s, 300 m², air 1.2 kg/m³ ⇒ $4.5 \times 10^5$ N).
Valve opened: gauge reading falls from $P_1$ to $P_2$ , so $v = \sqrt{\dfrac{2(P_1 - P_2)}{\rho}}$ .
Flow rate from a pressure difference in a tapering pipe: use $v_1 = \dfrac{A_2}{A_1}v_2$ in Bernoulli, solve for $v_2$ , then $Q = A_2v_2$ .

Bernoulli, horizontal pipe

P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2

Worked example

Water flows at 1 m/s where the area is 12 cm² and the pressure 5000 Pa. Pressure where the area is 4 cm²?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift II · Q12Moderate

Example 2 · Mechanical Properties of Fluids · Bernoulli, Continuity, Streamline, and Torricelli

A horizontal pipeline carries water in a streamline flow. At a point along the pipe, where the cross-sectional area is

10\,cm^{2}

, the velocity of water is

1\,m/s

and pressure is 2000 Pa. The pressure of water at another point where the cross-sectional area is

5\,cm^{2}

is [Given: density of water

=1000\,kg/m^{3}

]

Putting the high pressure at the narrow part

Squeezing the pipe speeds the fluid up, and faster flow means LOWER pressure. The narrowest section has maximum speed and minimum pressure.

Drill 6 more on bernoulli's theorem

Concept 3 of 3: Torricelli: Jets and Draining Tanks

Water leaving a hole at depth h below the open surface comes out as fast as if it had fallen h: v = √(2gh). The deeper the hole, the faster the jet. A full tank's gauge pressure at the bottom does the same job: v = √(2ΔP/ρ). Because v grows as √h, a tank four times as full takes twice as long to drain. Two holes on opposite sides throw jets with momentum ρAv² each; the difference is ρA × 2g × (height difference), so the tank feels a net sideways force proportional to h.

Definition

Efflux speed at depth h: $v = \sqrt{2gh}$ ; lower orifices are faster, $V_1 < V_2 < V_3$ going down.
From pressure: bottom gauge pressure $4H - H = 3H$ ⇒ $v = \sqrt{\dfrac{6H}{\rho}}$ .
Drain time $\propto \sqrt{h}$ : height × 4 ⇒ time × 2.
Two holes on opposite sides, height difference h: net thrust $= \rho A(v_1^2 - v_2^2) = 2\rho A g h \propto h$ .

Torricelli

v = \sqrt{2gh}, \qquad t_{\text{drain}} \propto \sqrt{h}

Worked example

A tank drains in 3 minutes when filled to height h. How long from 9h?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 13th May Shift 2 · Q33Moderate

Example 3 · Mechanical Properties of Fluids · Bernoulli, Continuity, Streamline, and Torricelli

There is hole of area 'A' at the bottom of a cylindrical vessel. Water is filled to a height 'h' and water flows out in 't' second. If water is filled to a height '4h', it will flow out in time (in second)

Measuring depth from the bottom

h in √(2gh) is the depth of the hole BELOW THE FREE SURFACE, not its height above the base. The hole nearest the bottom gives the fastest jet.

Drill 3 more on torricelli: jets and draining tanks

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Continuity: Narrow Pipe, Faster Flow
Continuity
$A_1 v_1 = A_2 v_2 = Q$
Bernoulli's Theorem
Bernoulli, horizontal pipe
$P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2$
Torricelli: Jets and Draining Tanks
Torricelli
$v = \sqrt{2gh}, \qquad t_{\text{drain}} \propto \sqrt{h}$

Watch out for (3)

Scaling speed with radius, not area
→ Continuity: Narrow Pipe, Faster Flow
Putting the high pressure at the narrow part
→ Bernoulli's Theorem
Measuring depth from the bottom
→ Torricelli: Jets and Draining Tanks

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Related notes

Pressure — the ρgh term →