Question

derivation of bernoulis theorem

Answer 1

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

Change in Gravitational Potential and Kinetic Energy

Now, we have to calculate the change in gravitational potential energy ∆U.

Bernoulli's equation

Similarly, the change in ∆K or kinetic energy can be written as

Bernoulli's equation

Calculation of Bernoulli’s Equation

Applying work-energy theorem in the volume of the fluid, the equation will be

Bernoulli's equation

Dividing each term by ∆V, we will obtain the equation

Bernoulli's equation

Rearranging the equation will yield

Bernoulli's equation

The above equation is the Bernoulli’s equation. However, the 1 and 2 of both the sides of the equation denotes two different points along the pipe. Thus, the general equation can be written as

Bernoulli's equation

Thus, we can state that Bernoulli’s equation state that the Pressure (P), potential energy (ρgh) per unit volume and the kinetic energy (ρv2/2)   per unit volume will remain constant.

Important Points to Remember

It is important to note that while deriving this equation we assume there is no loss of energy because of friction if we apply the principle of energy conservation. However, there is actually a loss of energy because of internal friction caused during fluid flow. This, in fact, will result in the loss of some energy.

Answer 2

$\huge\mathfrak\red{Answer :}$

According to it for an stream flow of an liquid, the total energy [sum of kinetic energy, potential energy and pressure energy] per unit mass remains constant.. At every cross section throughout the liquid flow.

Consider a streamline flow of ideal liquid across a pipe AB as shown in figure....

For an ideal liquid $\rho$ = constant

$\eta$ = 0

a1, a2 be the area of cross section at A and B

v1, v2 be the volume of liquid at A and B.

P1, p2 the pressure of liquid at A and B.

$\rho$ = density of liquid.

If m is mass entering per sec.

Then;

a1v1$\rho$ = a2v2$\rho$ = m

a1v1 = a2v2 = $\dfrac{m}{\rho}$ = v

F = P1a1

W1 = P1 a1 v1 = P1 V [Work done per pressure at A]

W2 = P2 a2 v2 = P2 V [Work done per pressure at B]

Net work done = P1 V - P2 V

Kinetic Energy at A = $\dfrac{1}{2}$ m$v_{1}^{2}$

Kinetic Energy at B = $\dfrac{1}{2}$ m$v_{2}^{2}$

Potential Energy at A = mg$h_{1}$

Potential Energy at B = mg$h_{2}$

According to work energy principle...

$P_{1}V\:-\:P_{2}V$ = $\dfrac{1}{2}mv_{2}^{2}\:-\:\dfrac{1}{2}mv_{1}^{2}$ + $mgh_{2}\:-\:mgh_{1}$

$P_{1}V$ + $\dfrac{1}{2}mv_{1}^{2}$ + $mgh_{1}$ = $P_{2}V$ + $\dfrac{1}{2}mv_{2}^{2}$ + $mgh_{2}$ ....(1)

Divide eq. (1) by m

$\dfrac{P_{1}V}{{m}}$ + $\dfrac{1}{2m}$$mv_{1}^{2}$ + $\dfrac{mgh_{1}}{{m}}$ = $\dfrac{P_{2}V}{{m}}$ + $\dfrac{1}{2m}$$mv_{2}^{2}$ + $\dfrac{mgh_{1}}{{m}}$

$\dfrac{P_{1}}{\rho}$ + $\dfrac{1}{2}\:\times\:V_{1}^{2}$ + $gh_{1}$ = $\dfrac{P_{2}}{\rho}$ + $\dfrac{1}{2}\:\times\:V_{2}^{2}$ + $gh_{2}$

$\dfrac{P}{\rho}$ + $\dfrac{1}{2}{V}^{2}$ + $gh$ = Constant...

Hence, proved...