Derive Bernoulli's principle.
Answers
Let the velocity, pressure and area of a fluid column at a point X be v1, p1 and A1 and at another point Y be v2, p2 and A2 . Let the volume that is bounded by X and Y be moved to M and N . let XM = L1 and YN = L2 . Now if we can compress the fluid then we have,
A1 × L1 = A2 × L2
We know that that the work done by the pressure difference per volume of the unit is equal to the sum of the gain in kinetic energy and gain in potential energy per volume of the unit.
This implies
Work done = force × distance
⇒ Work done = p × volume
Therefore, net work done per volume = p1 – p2
Also, kinetic energy per unit volume = 12 m v2 = 12 ρ v2
Therefore, we have,
Kinetic energy gained per volume of unit = 12 ρ ((v2)2 – (v1)2)
And potential energy gained per volume of unit = p g (h2 – h1)
Here, h1 and h2 are heights of X and Y above the reference level taken in common.
Finally we have
p1 – p2 = 12 ρ ((v2)2 – (v1)2) + ρ g (h2 – h1)
⇒ p1 + 12 ρ (v1)2 + ρ g h1 = p2 + 12 ρ (v2)2 + ρ g h2
⇒ p + 12 ρ v2 + ρ g h is a constant
When we have h1 = h2
Then we have, p + 12 ρ v2 is a constant.
This proves the Bernoulli’s Theorem
Statement:—
- If an ideal fluid is flowing through a tube in streamline motion then each point of its path total energy per unit volume is constant.
- P+1/2pv²+pgh= constant or
- P1+1/2pv²1+pgh1= P2+1/2pv²2+pgh2
Derivation:—
Let p = (density of fluid)
m= (mass of fluid entering at x/ leaving at y in 1 sec)
from principle of continuity
A1V1=A2V2= m/p .....(1)
Work done on fluid entering at x in 1 sec
= P1A1V1
Work done by fluid leaving at Y in 1 sec=
P2A2V2
Net work done in 1 sec = P1A1V1-P2A2V2
= P1m/p - P2m/p......(2)
K.E of fluid entering at x in 1 sec = 1/2mV²1
K.E of fluid leaving at y in 1 sec= 1/2mv²2
Change in KE = 1/2mv²2-1/2mv²1...(3)
Potential energy of fluid entering at x in 1 sec = mgh1
PE of fluid leaving at y in one sec= mgh2
Change in PE = mgh2-mgh1
now,
applying law of conservation of energy
Net work done = change in KE + change in PE
P1+1/2pv²1= P2+1/2pv²2+pgh2