QCC Notes
CLASS 11 · PHYSICSJEE MAIN × NEETहिंदी
§ 9.3NCERT Class 11 · Physics · Chapter 9

Fluid Dynamics: Equation of Continuity, Bernoulli's Theorem & Torricelli's Law

Fluid Dynamics (Hydrodynamics) treats fluids in motion. To analyze complex flow patterns rigorously for engineering and entrance examinations, we model fluid behavior using an ideal fluid: an idealized substance that is (1) incompressible (ρ = const), (2) non-viscous (η = 0), (3) in steady laminar flow, and (4) irrotational.

1. Streamline Flow & The Equation of Continuity

In steady (streamline) flow, every particle passing through a particular point follows the identical path and possesses the same velocity as the particle preceding it. A tangent drawn at any point along a streamline yields the direction of fluid velocity at that point. Streamlines never intersect; if they did, a particle at the junction would have two ambiguous velocities simultaneously.

               End 1 (Inlet)                                   End 2 (Constriction)
               Area A1, Velocity v1                          Area A2, Velocity v2
               +-----------------                            /----------------+
               |                                            /                 |
               |  -----> v1       _/       -----> v2 |
               |                    [ Narrow Neck: A2 < A1 ]                 |
               +-----------------/                          -----------------+
                                        v2 > v1
                        Mass Conservation: A1 * v1 = A2 * v2
      
Fig 9.3A: Flow of an incompressible fluid through a tube of varying cross-section, illustrating mass conservation.

By conservation of mass, the mass of fluid entering a tube of flow in time Δ t must equal the mass exiting in the same time interval:

Δ m1 = Δ m2 ⇒ ρ1 A1 v1 Δ t = ρ2 A2 v2 Δ t

For an incompressible fluid (ρ1 = ρ2 = ρ):

Equation of Continuity:
A1 v1 = A2 v2 = Q = dV/dt = Constant Volume Flow Rate
Physical Consequence: Fluid velocity is inversely proportional to cross-sectional area (v ∝ 1/A). Where a river or pipe narrows, the fluid speeds up; where it widens, the fluid slows down.

2. Derivation of Bernoulli's Theorem

Bernoulli's Principle (1738) represents the fundamental statement of the work-energy theorem applied to non-viscous, steady fluid flow along a streamline:

Consider an incompressible fluid flowing through a non-uniform pipe from height h1 to height h2. In a time interval Δ t:

  • Work done by pressure force at inlet: W1 = F1 Δ x1 = P1 A1 (v1 Δ t) = P1 Δ V.
  • Work done by pressure force at outlet: W2 = -F2 Δ x2 = -P2 A2 (v2 Δ t) = -P2 Δ V.
  • Net work done by pressure forces: Wnet = (P1 - P2)Δ V.
  • Change in kinetic energy: Δ K = 1/2Δ m (v22 - v12) = 1/2ρ Δ V (v22 - v12).
  • Change in gravitational potential energy: Δ U = Δ m g (h2 - h1) = ρ Δ V g (h2 - h1).

Applying the Work-Energy Theorem (Wnet = Δ K + Δ U):

(P1 - P2)Δ V = 1/2ρ Δ V (v22 - v12) + ρ g Δ V (h2 - h1)

Rearranging terms:

Bernoulli's Equation:
P1 + 1/2ρ v12 + ρ g h1 = P2 + 1/2ρ v22 + ρ g h2
P + 1/2ρ v2 + ρ g h = Constant along a streamline

Alternative "Head" Formulation

Dividing the entire equation by ρ g gives all terms in units of length (meters of fluid column):

P/(ρ g) + v2/2g + h = Total Head (Constant)
  • P/(ρ g) = Pressure Head
  • v2/2g = Velocity (Dynamic) Head
  • h = Elevation (Datum) Head

3. Torricelli's Law of Efflux & Tank Emptying Time

Consider a large open storage tank filled with liquid of density ρ to a total depthH, possessing a small orifice of areaalocated at a depthhbelow the open free surface (a ≪ A, whereA is the cross-sectional area of the tank):

       Free Surface: Level 1 (Area A, P1 = P0, v1 approx 0)
       ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
       |                                                   |
       |                   |                               |
       |                   | h                             |
       |                   v                               |
       |                Orifice: Level 2 (Area a, P2 = P0) =====> Speed of Efflux v2
       |                                                   |
       |                                      (H - h)      |
       +---------------------------------------------------+
       |<----------------- Horizontal Range R ------------>|
      
Fig 9.3B: Torricelli's tank: liquid discharges through an orifice at depth h, executing projectile motion over range R.

Applying Bernoulli's equation between the open free surface (1) and the discharge orifice (2):

P0 + 1/2ρ v12 + ρ g H = P0 + 1/2ρ v22 + ρ g (H - h)

By continuity, v1 = (a/A)v2 ≈ 0 since a ≪ A:

ρ g h = 1/2ρ v22 ⇒ v = √(2gh)

Kinematics of the Efflux Stream

  • Time of Flight to Ground: Fluid leaves horizontally from elevation (H - h):
    t = √((2(H - h))/g)
  • Horizontal Range (R):
    R = v · t = √(2gh) × √((2(H - h))/g) = 2√(h(H − h))
  • Maximum Range Condition: By differentiating R with respect to h, the range is maximized when the hole is drilled precisely at mid-depth (h = H/2):
    Rmax = 2√((H/2)(H − H/2)) = H
  • Complementary Depths: Two holes drilled at depths h and (H - h) produce identical horizontal ranges (R1 = R2).
  • Time to Empty Tank Completely:
    Tempty = (A/a)√(2H/g)

4. Applications: Venturimeter, Aerofoils & Magnus Effect

The Venturimeter

A Venturimeter measures the volumetric discharge rate Q of liquid flowing through a pipeline. It consists of a wide tube of area A1 connected to a narrow throat of area A2 (A2 < A1):

From continuity: v2 = (A1 / A2)v1. Because A2 < A1, v2 > v1. By Bernoulli's equation, this acceleration causes the pressure at the throat to drop (P2 < P1):

P1 - P2 = 1/2ρ (v22 - v12) = 1/2ρ v12 [ (A1/A2)2 - 1 ] = ρm g h

The rate of flow Q is given by:

Q = A1 v1 = A1 A2 √((2(P1 - P2))/(ρ (A12 - A22))) = A1 A2 √((2ρm g h)/(ρ (A12 - A22)))

Dynamic Aerodynamic Lift (Aerofoil)

An airplane wing (aerofoil) is sculpted with a convex upper curvature and a flat lower surface. As the wing moves forward through the air, streamlines crowd together above the curved top, forcing air to travel faster (vtop > vbottom). By Bernoulli's principle, the static pressure on top drops below that beneath the wing (Ptop < Pbottom). The resulting upward pressure difference produces aerodynamic lift:

Flift = (Pbottom - Ptop)A = 1/2ρ (vtop2 - vbottom2)A

JEE & NEET Solved Practice Problems

Problem 1: Water flows through a horizontal pipe of varying cross section. At a point where the pipe diameter is 10 cm, the velocity is 2 m/s and pressure is 2 × 105 Pa. Find the pressure at another point where the pipe diameter constricts to 5 cm. (ρ = 1000 kg/m3).
Solution:
1. From equation of continuity:
A1 v1 = A2 v2 ⇒ d12 v1 = d22 v2

v2 = v1 (d1/d2)2 = 2 × (10/5)2 = 2 × 4 = 8 m/s

2. Applying Bernoulli's equation for a horizontal pipe (h1 = h2):
P1 + 1/2ρ v12 = P2 + 1/2ρ v22

P2 = P1 - 1/2ρ (v22 - v12) = 2 × 105 - 1/2(1000)(82 - 22)

P2 = 200000 - 500(64 - 4) = 200000 - 30000 = 1.70 × 105 Pa = 170 kPa
Problem 2: A cylindrical water tank of height H = 5 m is completely full. A small orifice is opened at a depth h = 1.25 m below the water surface. Calculate: (a) the speed of efflux, (b) the horizontal range of the issuing jet on the floor, and (c) at what other depth an orifice could be drilled to yield the exact same horizontal range. Take g = 10 m/s2.
Solution:
(a) Speed of efflux from Torricelli's law:
v = √(2gh) = √(2 × 10 × 1.25) = √(25) = 5 m/s

(b) Horizontal range on the ground:
R = 2√(h(H − h)) = 2√(1.25 × (5 − 1.25)) = 2√(1.25 × 3.75) = 2√(4.6875) ≈ 4.33 m

(c) The complementary depth that produces identical horizontal range is:
h' = H - h = 5 - 1.25 = 3.75 m below the free surface

Frequently Asked Questions

Q1. What are the four defining assumptions of an ideal fluid in fluid dynamics?
An ideal fluid is assumed to be: (1) Incompressible (density rho remains constant regardless of pressure); (2) Non-viscous (zero internal fluid friction between flowing layers); (3) Steady flow (velocity, density, and pressure at any spatial point remain constant over time); and (4) Irrotational (fluid elements have zero angular momentum about their own mass centers).
Q2. What is the physical meaning of each term in Bernoulli's Equation?
In Bernoulli's equation, P + (1/2)*rho*v2 + rho*g*h = constant: P is the static pressure energy per unit volume; (1/2)*rho*v2 is the kinetic energy per unit volume; and rho*g*h is the gravitational potential energy per unit volume. Dividing by rho*g yields the head form: P/(rho*g) [Pressure Head] + v2/(2g) [Velocity Head] + h [Datum/Elevation Head] = Total Head (constant).
Q3. What is Torricelli's Law of Efflux?
Torricelli's Law states that the velocity of efflux (v) of an ideal liquid discharging through a small orifice situated at depth h below the open free surface is identical to the velocity acquired by a freely falling body dropped from height h: v = sqrt(2 * g * h).
Q4. What is the Magnus Effect, and how does it explain the curve of a spinning ball in air?
When a ball spins while translating through air, it drags surrounding air with it. On the side where surface spin reinforces oncoming airflow, airstream velocity increases, lowering local pressure (by Bernoulli's principle). On the opposing side where spin opposes airflow, velocity decreases, raising local pressure. The resulting crosswise pressure imbalance exerts a net aerodynamic lateral force (Magnus force), causing the ball's flight path to curve.
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