Bernoulli’s Theorem States That:

As we move along a streamline, the sum of the pressure (P), the kinetic energy,

the per unit volume (ρ and the potential energy per unit volume (ρgh) remains a

constant.”

• Proof:



*Consider a fluid moving in a pipe if varying cross-sectional area.

*Let the pipe have varying heights & the fluid flowing is incompressible.

*Fluid is initially at E and flows to B. Consider the parts of the tube BC & DE

which are equal in volume, thus having equal volume of fluid.

Distance fluid travels at left end è v2 Δt

Distance fluid travels at right end è v1 Δt

Work done on fluid at left end è W=-F.d è W=-P.A.d è W2= -P2.A2.v2 Δt

Work done on fluid at left end è W=F.d è W=P.A.d è W1=P1.A1.v1 Δt

Ai.Vi Δt = ΔV

W1-W2 = (P1-P2). ΔV (Volumes are same)

Also, W= Kinetic Energy + Potential Energy.

ρ=m/v è m=ρ.v è m=ρ.ΔV

ΔKE of fluid from top to bottom = (1/2).ρ. ΔV(V22-V12)

ΔPE of fluid from top to bottom = ρ. ΔV.g.(h2-h1)

Applying Work-Energy theorem…

ΔW= (1/2).ρ. ΔV(V22-V12) + ρ. ΔV.g.(h2-h1)

Equating Eq.1 & Eq.2

(P1-P2). ΔV = (1/2).ρ. ΔV(V22-V12) + ρ. ΔV.g.(h2-h1)

(P1-P2)= (1/2).ρ.(V22-V12) + ρ.g.(h2-h1)

Re-arranging,

P1 + (1/2).ρ.V12 + ρ.g.(h1) = P2 + (1/2).ρ.V22 + ρ.g.(h2)

--> P + (1/2)ρV2 + ρgh = Constant [Cancelling ΔV