Question

expression for Bernoulli's theorm

Answer 1

In fluid dynamics, Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy. The principle is named after Daniel Bernoulli who published it in his book Hydrodynamica in 1738.

Answer 2

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


$\huge\textbf{Bernoulli's Theoram :}$


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...