Question

State and prove Bernoulli’s theorem for steady flow of an ideal fluid

Answer 1

Bernoulli's theorem
According to Deniel Bernoulli "the sum of pressure energy , kinetic energy and potential energy(for unit volume and mass) of an incompressible and non-viscous fluid(ideal fluid) is remain constant."
1) for unit volume
P+egh+1/2ev^2=constant
2) for unit mass
P/eg+h+v^2/2g=constant
Proof
As the fluid is incompressible so what ever mass of fluid enters the pipe at section A in unit time(t),an equal mass of fluid flow out in section B in unit time (t) then mass of ideal fluid.
mass=volume×density
At point A
m=A1 V1×e -(1).
At point B
m=A2 V2×e -(2).
(a) change in KE =KE at B - KE at A
=1/2mv2^2-1/2mv1^2
=1/2m(v2^2-v1^2) -(3)
(b) change in PE = PE at B - PE at A
=mgh2-mgh1
=mg(h2-h1) -(4)
c Net work done on the fluid= Work done on the fluid at A- Work on the fluid at B
=P1A1V1-P2A2V2
=P1A1V1-P2A1V1
=A1V1(P1-P2)
Net work done= change in KE+ change in PE
A1V1(P1-P2)=1/2m(v2^2-v1^2)+mg(h2-h1)
A1V1(P1-P2)=A1V1{e(1/2v2^2-v1^2)+eg(h2-h1)}
P1-P2=1/2ev2^2-1/2ev1^2+egh2-egh1
P1+1/2ev1^2+egh1=P2+1/2ev2^2+egh2
P+egh+1/2ev^2=constant