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Question

Water is flowing through a pipe under constant pressure. At some place the pipe becomes narrow. The pressure of water at this place:


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Solution

Step 1: Concept used

Principle of continuity:
According to the principle of continuity, if the fluid is in streamlined flow and is in-compressible then we can say that mass of fluid passing through different cross-sections are equal.

a1v1=a2v2

Where, a1 and a2 are the areas of cross-section, and v1 and v2 are the flow velocities at two different points as shown in the figure below.

According to Bernoulli's theorem:

P+12ρv2+ρgh=costant

Here, P is the pressure exerted by the fluid, ρ is the density of the fluid, v is the velocity of the fluid, h is the height of the container.

Step 2: Explanation

  1. In streamlined flow, the product of cross-section area and velocity remains constant (equation of continuity). So in the narrowest part of the pipe velocity is maximum.
  2. From Bernoulli's theorem, we know that the sum of potential energy, kinetic energy, and pressure energy remains constant. Since the pipe is horizontal, potential energy is equal at all the points. So, at the narrowest part of the pipe, pressure (pressure energy) will be minimum because velocity (kinetic energy) is maximum in the narrowest part.
  3. The value of pressure from the Bernoulli equation is given by,

P=costant-12ρv2+ρgh

From the above equation, it is clear that if velocity is increasing then the pressure will decrease.

Hence, the pressure of water in narrow places will decrease.


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