Question

Derive bernoulli equation​

Answer 1

Answer:

Assumptions:

The density of the incompressible fluid remains constant at both the points.

Energy of the fluid is conserved as there are no viscous forces in the fluid.

Therefore, the work done on the fluid is given as:

dW = F1dx1 – F2dx2

dW = p1A1dx1 – p2A2dx2

dW = p1dV – p2dV = (p1 – p2)dV

We know that the work done on the fluid was due to conservation of gravitational force and change in kinetic energy. The change in kinetic energy of the fluid is given as:

dK=12m2v22−12m1v21=12ρdV(v22−v21)

The change in potential energy is given as:

dU = mgy2 – mgy1 = ρdVg(y2 – y1)

Therefore, the energy equation is given as:

dW = dK + dU

(p1 – p2)dV = 12ρdV(v22−v21) + ρdVg(y2 – y1)

(p1 – p2) = 12ρ(v22−v21) + ρg(y2 – y1)

Rearranging the above equation, we get

p1+12ρv21+ρgy1=p2+12ρv22+ρgy2

Explanation:

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Answer 2

Answer:

Explanation:

Assumptions:

The density of the incompressible fluid remains constant at both the points.

Energy of the fluid is conserved as there are no viscous forces in the fluid.

Therefore, the work done on the fluid is given as:

dW = F1dx1 – F2dx2

dW = p1A1dx1 – p2A2dx2

dW = p1dV – p2dV = (p1 – p2)dV

We know that the work done on the fluid was due to conservation of gravitational force and change in kinetic energy. The change in kinetic energy of the fluid is given as:

dK=12m2v22−12m1v21=12ρdV(v22−v21)

The change in potential energy is given as:

dU = mgy2 – mgy1 = ρdVg(y2 – y1)

Therefore, the energy equation is given as:

dW = dK + dU

(p1 – p2)dV = 12ρdV(v22−v21) + ρdVg(y2 – y1)

(p1 – p2) = 12ρ(v22−v21) + ρg(y2 – y1)

Rearranging the above equation, we get

p1+12ρv21+ρgy1=p2+12ρv22+ρgy2