Fluid Mechanics: Common Mistakes and Fixes (3)

hard • 3 min read
Tags Fluid Mechanics

Question

A horizontal pipe has a cross-sectional area A1=4×10−2A_1 = 4 \times 10^{-2} m2^2 at the wider end and A2=1×10−2A_2 = 1 \times 10^{-2} m2^2 at the narrower end. Water flows through it. If the speed at the wider end is v1=2v_1 = 2 m/s and the pressure there is P1=2×105P_1 = 2 \times 10^5 Pa, find the speed and pressure at the narrower end. Density of water ρ=1000\rho = 1000 kg/m3^3.

Solution — Step by Step

Incompressible flow: A1v1=A2v2A_1 v_1 = A_2 v_2.

v2=A1v1A2=(4×10−2)(2)1×10−2=8 m/sv_2 = \frac{A_1 v_1}{A_2} = \frac{(4 \times 10^{-2})(2)}{1 \times 10^{-2}} = 8 \text{ m/s}

For a horizontal pipe (same height), Bernoulli simplifies:

P1+12ρv12=P2+12ρv22P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2

P2=P1+12ρ(v12−v22)P_2 = P_1 + \tfrac{1}{2}\rho(v_1^2 - v_2^2)

P2=2×105+12(1000)(4−64)P_2 = 2 \times 10^5 + \tfrac{1}{2}(1000)(4 - 64)

P2=2×105+500(−60)=2×105−30000P_2 = 2 \times 10^5 + 500(-60) = 2 \times 10^5 - 30000

P2=1.7×105 PaP_2 = 1.7 \times 10^5 \text{ Pa}

The pipe narrows, so speed increases (continuity) and pressure drops (Bernoulli). P2<P1P_2 < P_1. ✓

Final answer: v2=8v_2 = 8 m/s, P2=1.7×105P_2 = 1.7 \times 10^5 Pa.

Why This Works

Continuity comes from mass conservation in incompressible flow — the same volume per second must pass every cross-section. So if area shrinks, speed grows by the same factor.

Bernoulli’s equation is energy conservation per unit volume of fluid: P+12ρv2+ρghP + \tfrac{1}{2}\rho v^2 + \rho gh is constant along a streamline (for ideal flow). When speed goes up, kinetic energy goes up, so pressure must drop to compensate.

Continuity: A1v1=A2v2A_1 v_1 = A_2 v_2 (incompressible, no leaks)

Bernoulli: P+12ρv2+ρgh=constP + \tfrac{1}{2}\rho v^2 + \rho gh = \text{const} along a streamline

Conditions: ideal fluid (no viscosity), steady flow, incompressible, along a streamline

Alternative Method

Direct substitution into the combined form:

P2=P1+12ρv12[1−(A1A2)2]P_2 = P_1 + \tfrac{1}{2}\rho v_1^2 \left[1 - \left(\frac{A_1}{A_2}\right)^2\right]

P2=2×105+500⋅4⋅(1−16)=2×105+2000(−15)=1.7×105 Pa✓P_2 = 2 \times 10^5 + 500 \cdot 4 \cdot (1 - 16) = 2 \times 10^5 + 2000(-15) = 1.7 \times 10^5 \text{ Pa} \checkmark

Three classic fluid mechanics traps:

  1. Forgetting the ρgh\rho gh term when the pipe is not horizontal. If the wider end is at a different height, you must include ρgh\rho g h for both points.
  2. Mixing A1v1=A2v2A_1 v_1 = A_2 v_2 with A1P1=A2P2A_1 P_1 = A_2 P_2. Continuity is for AvAv, NOT for APAP. There’s no “pressure flux” conservation.
  3. Using Bernoulli across a pump or a turbulent region. Bernoulli assumes ideal, steady, non-turbulent flow. If energy is added (pump) or dissipated (turbulence), you must add/subtract those terms.

For Venturi-meter problems, this exact setup gives ΔP\Delta P, which is then read from a manometer. Practice this combination — it’s a JEE Main staple every alternate year.

Common Mistake

The most damaging mistake: treating Bernoulli as a force-balance equation. It is not. It is an energy-per-unit-volume conservation. Pressure here is the static pressure of the fluid, not a “force on the wall.”

Another sneaky one: students compute ΔP\Delta P correctly but forget which end is which. The pressure is higher at the wider, slower end. If your answer says pressure is higher at the narrow end, you’ve swapped a sign.

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