  Question

# State and prove Bernoulli's theorem.

Solution

## 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 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,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. Now: 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) Therefore: K.E.    gained per unit volume =12ρ(v22−v21) 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=12ρ(v21–v22)+ρg(h2−h1)P1+12ρv21+ρgh1=P2+12ρv22+rgh2 Therefore:  P+12ρv2+ρgh 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.  Suggest corrections   