To prove Bernoulli’s theorem, we make the following assumptions:
1. The liquid is incompressible.
2. The liquid is non–viscous.
3. The flow is steady and the velocity of the liquid is less than the critical velocity for the liquid.
It states that the total energy (pressure energy, potential energy and kinetic energy) of an incompressible and non–viscous fluid in a steady flow through a pipe remains constant throughout the flow, provided there is no source or sink of the fluid along the length of the pipe. This statement is based on the assumption that there is no loss of energy due to friction.
Mathematically, for a unit mass of fluid flowing through a pipe. Pρ+g×h+12×V2=constant
Consider a fluid of negligible viscosity moving with the laminar flow, as shown in Figure 1.
Let the velocity, pressure and area of the fluid column be v1,P1
at Q and v2,P2
at R. Let the volume bounded by Q and R move to S and T where QS=L1,andRT=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.
Work done = force x distance =p×volume
Net work done per unit volume =P1−P2
K.E. per unit volume =12mv2=12Vρv2=12ρv2
(V = 1 for unit volume)
K.E. gained per unit volume =12ρ(v22−v21)
P.E. gained per unit volume =ρg(h2–h1)
are the heights of Q and R above some reference level.
is a constant.
For a horizontal tube h1=h2
and so we have: P+12ρv2
= a (constant)
This is Bernoulli's theorem. You can see that if there is an increase in velocity there must be a decrease in pressure and vice versa.