Question

please solve numerical in question no. 25

Answer 1

Hello,
This is a easy question
Its answer:-
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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.

Important so pls read it:-No fluid is totally incompressible but in practice the general qualitative assumptions still hold for real fluids.   If you like it then don't forget to click on the thanks button!
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