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NEET/Unit 7: Properties of Solids and Liquids/Bernoulli's Equation and Applications

Bernoulli's Equation and Applications

For steady, incompressible, non-viscous flow along a streamline, the sum of pressure energy, kinetic energy per unit volume and gravitational potential energy per unit volume is constant. This is Bernoulli's equation, and it underlies Venturi meters, Pitot tubes, atomisers and Torricelli's law.

Concept

The equation of continuity for an incompressible fluid is

A1v1=A2v2(volume flow rate).A_{1} v_{1} = A_{2} v_{2} \quad \text{(volume flow rate)}.

Bernoulli's theorem then states that along a streamline,

P+12ρv2+ρgh=constant.P + \tfrac{1}{2}\rho v^{2} + \rho g h = \text{constant}.

A narrower section of a horizontal pipe carries faster fluid (continuity) at lower pressure (Bernoulli). This is the principle behind aerofoils, atomisers and the Venturi meter.

Formula Derivation

Apply the work-energy theorem to a fluid element of mass Δm\Delta m moving from point 1 (cross-section A1A_{1}, speed v1v_{1}, height h1h_{1}, pressure P1P_{1}) to point 2:

Wnet=ΔKE+ΔPE.W_{\text{net}} = \Delta KE + \Delta PE.

The work done by pressure forces is (P1−P2) ΔV(P_{1} - P_{2})\,\Delta V, while the changes in KE and PE per unit volume are 12ρ(v22−v12)\tfrac{1}{2}\rho(v_{2}^{2} - v_{1}^{2}) and ρg(h2−h1)\rho g(h_{2} - h_{1}). Rearranging,

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2.P_{1} + \tfrac{1}{2}\rho v_{1}^{2} + \rho g h_{1} = P_{2} + \tfrac{1}{2}\rho v_{2}^{2} + \rho g h_{2}.

For a tank with a small hole at depth hh below the free surface, setting P1=P2=P0P_{1} = P_{2} = P_{0} and v1≈0v_{1} \approx 0 gives Torricelli's law:

v=2gh.v = \sqrt{2 g h}.

NEET-style Worked Example

Water flows through a horizontal pipe whose cross-section narrows from 20 cm220\ \text{cm}^{2} to 5 cm25\ \text{cm}^{2}. The speed at the wider section is 1 m/s1\ \text{m/s} and the pressure is 2×105 Pa2 \times 10^{5}\ \text{Pa}. Find the pressure at the constriction.

Continuity: v2=v1A1/A2=1×(20/5)=4 m/sv_{2} = v_{1} A_{1}/A_{2} = 1 \times (20/5) = 4\ \text{m/s}.

Bernoulli (horizontal pipe):

P2=P1+12ρ(v12−v22)=2×105+12(103)(1−16)=2×105−7500≈1.925×105 Pa.P_{2} = P_{1} + \tfrac{1}{2}\rho(v_{1}^{2} - v_{2}^{2}) = 2 \times 10^{5} + \tfrac{1}{2}(10^{3})(1 - 16) = 2 \times 10^{5} - 7500 \approx 1.925 \times 10^{5}\ \text{Pa}.

Common Confusions

  • Bernoulli applies to steady, incompressible, non-viscous, irrotational flow along a streamline.
  • The same equation does not hold across streamlines unless flow is irrotational.
  • Speed at a constriction increases (continuity), so pressure must decrease (Bernoulli) — common in NEET MCQs.
  • Torricelli's law uses the height of free surface above the orifice, regardless of the tank shape.

Key Takeaways

  • Continuity: Av=A v = constant for incompressible flow.
  • Bernoulli: P+12ρv2+ρgh=P + \tfrac{1}{2}\rho v^{2} + \rho g h = constant.
  • Torricelli: v=2ghv = \sqrt{2gh}.
  • Faster flow corresponds to lower pressure in a horizontal pipe.
  • Applications: Venturi meter, Pitot tube, atomiser, aerofoil lift, Magnus effect.

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