Physics, asked by gagan1035, 1 year ago

State and prove bernoulli theorem

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Answered by chahu2123
11

Let the velocity, pressure and area of the fluid column be v1, P1 and A1 at Q and v2, P2 and A2 at R. Let the volume bounded by Q and R move to S and T where QS = L1, and RT = L2. If the fluid is incompressible:

A1L1 = A2L2

The work done by the pressure difference per unit volume = gain in k.e. per unit volume + gain in p.e. per unit volume. Now:

Work done = force x distance = p x volume

Net work done per unit volume = P1 - P2

k.e. per unit volume = ½ mv2 = ½ Vρ v2 = ½ρv2 (V = 1 for unit volume)

Therefore:

k.e. gained per unit volume = ½ ρ(v22 - v12)

p.e. gained per unit volume = ρg(h2 – h1)

where h1 and h2 are the heights of Q and R above some reference level. Therefore:

P1 - P2 = ½ ρ(v12 – v22) + ρg(h2 - h1)

P1 + ½ ρv12 + ρgh1 = P2 + ½ ρv22 + rgh2

Therefore:

P + ½ ρv2 + ρgh is a constant

For a horizontal tube h1 = h2 and so we have:

P + ½ ρv2 = a constant

This is Bernoulli's theorem You can see that if there is a increase in velocity there must be a decrease of pressure and vice versa.

No fluid is totally incompressible but in practice the general qualitative assumptions still hold for real fluids.

Answered by Bhasksr
23

hope it helps you

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