Question

Derive the Bernoulli's Equation step by step..No Spam Please... NB : No Need of Figure...

Answer 1

Bernoulli's Principle :

The total mechanical energy of the moving fluid comprising the gravitational potential energy of elevation, the energy associated with the fluid pressure and the kinetic energy of the fluid motion, remains constant.

Before the derivation, we have to assume that :

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

• The 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 :

$\displaystyle\sf{dW=F_1dx_1-F_2dx_2}$

$\implies\displaystyle\sf{dW=p_1A_1dx_1-p_2A_2dx_2}$

$\implies\displaystyle\sf{dW=p_1dV-p_2dV}$

$\implies\displaystyle\sf{dW=(p_1-p_2)dV}$

• 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 :

$\displaystyle\sf{dK = \frac{1}{2}\:m_{2}\:v_{2}^{2}-\frac{1}{2}\:m_{1}\:v_{1}^{2}}$

$\implies\displaystyle\sf{dK=\frac{1}{2}\:\rho \:dV\:(v_{2}^{2}-v_{1}^{2})}$

The change in potential energy is given as:

$\displaystyle\sf{dU = m\:g\:h_2 - m\:g\:h_1}$

$\implies\displaystyle\sf{dU=\rho\:dV\:g(h_2-h_1)}$

Therefore, the energy equation is given as:

$\displaystyle\sf{dW = dK + dU}$

$\implies\displaystyle\sf{(p_1-p_2)dV=\frac{1}{2}\:\rho \:dV(v_2^2-v_1^2)+\rho\:dV\:g\:(h_2-h_1)}$

$\implies\displaystyle\sf{(p_1-p_2)=\frac{1}{2}\:\rho\:(v_2^2-v_1^2)+\rho\:g\:(h_2-h_1)}$

Rearranging the above equation, we get

$\implies\displaystyle\sf{p_{1}+\frac{1}{2}\:\rho \:v_{1}^{2}+\rho \:g\:h_{1}=p_{2}+\frac{1}{2}\:\rho\: v_{2}^{2}+\rho \:g\:h_{2}}$

The Bernoulli’s equation is :

$\boxed{\displaystyle\sf{p_{1}+\frac{1}{2}\:\rho \:v_{1}^{2}+\rho \:g\:h_{1}=p_{2}+\frac{1}{2}\:\rho\: v_{2}^{2}+\rho \:g\:h_{2}}}$

$\boxed{\displaystyle\sf{p+\frac{1}{2}\:\rho\:v^2+\rho\:g\:h=constant}}$

Where,

• p is the pressure exerted by the fluid .

• v is the velocity of the fluid .

• ρ is the density of the fluid .

• h is the height of the container.

Hope You Understood ! !

Figure is Attached !!

Answer 2

Answer:

Definition

"The total mechanical energy of the moving fluid comprising the gravitational potential energy of elevation, the energy associated with the fluid pressure and the kinetic energy of the fluid motion, remains constant."

Let us consider two different regions in the above diagram. Let us name the first region as BC and the second region as DE. Now consider the fluid was previously present in between B and D. However, this fluid will move in a minute (infinitesimal) interval of time (∆t).

If the speed of fluid at point B is v1 and at point D is v2. Therefore, if the fluid initially at B moves to C then the distance is v1∆t. However, v1∆t is very small and we can consider it constant across the cross-section in the region BC.

Similarly, during the same interval of time ∆t the fluid which was previously present in the point D is now at E. Thus, the distance covered is v2∆t. Pressures, P1 and P2, will act in the two regions, A1 and A2, thereby binding the two parts.

V

Finding the Work Done

First, we will calculate the work done (W1) on the fluid in the region BC. Work done is

W1 = P1A1 (v1∆t) = P1∆V

Moreover, if we consider the equation of continuity, the same volume of fluid will pass through BC and DE. Therefore, work done by the fluid on the right-hand side of the pipe or DE region is

W2 = P2A2 (v2∆t) = P2∆V

Thus, we can consider the work done on the fluid as – P2∆V. Therefore, the total work done on the fluid is

W1 – W2 = (P1 − P2) ∆V

The total work done helps to convert the gravitational potential energy and kinetic energy of the fluid. Now, consider the fluid density as ρ and the mass passing through the pipe as ∆m in the ∆t interval of time.

Hence, ∆m = ρA1 v1∆t = ρ∆V

Hope it helps u

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