Question

State and prove Bernoulli's theorem.

Answer 1

Bernoulli's Theorem:

According to Bernoulli's theorem, the sum of the energies possessed by a flowing ideal liquid at a point is constant provided that the liquid is incompressible and non-viseous and flow in streamline.

Potential energy + Kinetic energy + Pressure energy = Constant

$P+\frac{1}{2}p{v}^2+pgh=Constant$

$gh+\frac{1}{2}{v}^2+\frac{P}{p}=C$ ............(11.11)

Where C is a constant.

This relation is called Bernoulli's theorem.

Dividing eqn. (11.11) by g, we get

$h+\frac{P}{pg}+\frac{1}{2}\frac{v^2}{g}=C^{\prime }$ ............(11.12)

Where C is another constant.

For horizontal flow, h remains same throughout.

So,

$\frac{P}{pg}+\frac{v^2}{2g}=Constant$

or; $P+\frac{1}{2}p{v}^2=Constant$

P is static pressure of the liquid and $\frac{1}{2}p{v}^2$ is its dynamic and velocity pressure.

Thus, for horizontal motion, the sum of static and dynamic pressure is constant. If $p_1{v}_1$ and $p_2{v}_2$ represent pressure and velocities at two points. Then

$P_1+\frac{1}{2}p{v}_1^2=P_2+\frac{1}{2}p{v}_2^2$

Concepts : 1. In Bernoulli's theorem $P+pgh+\frac{1}{2}e{v}^2$

= constant. The term $(p+pgh)$ is called static pressure and the term $\frac{1}{2}p{v}^2$ is the dynamic pressure of the fluid.

2. Bernoulli theorem is fundamental principle of the energy.

3. The equation $\frac{P}{pg}+\frac{1}{2}\frac{v^2}{g}+h=constant$

the term $\frac{P}{pg}$ = pressure head

the term $\frac{v^2}{2g}$ = velocity head

h = gravitational head.

Derivation of Bernoulli's Theorem :

The energies possessed by a flowing liquid are mutually convertible. When one type of energy increases, the other type of energy decreases and vice-versa. Now, we will derive the Bernoulli's theorem using the work-energy theorem.

Consider the flow of liquid. Let at any time, the liquid lies between two areas of flowing liquid $A_1$ and $A_2$. In time interval $\Delta t$, the liquid displaces from $A_1$ by $\Delta x_1=v_1\Delta t$ and displaces from $A_2$ by $\Delta x_2=v_2\Delta t$. Here $v_1$ and $v_2$ are the velocities of the liquid at $A_1$ and $A_2$.

The work done on the liquid is $P_1{A}_1\Delta x_1$ by the force and $P_2{A}_2\Delta x_2$ against the force respectively.

Net work done,

$W=P_1{A}_1\Delta x_1-P_2{A}_2\Delta x_2$

$\Rightarrow W=P_1{A}_1{v}_1\Delta t-P_2{A}_2{v}_2\Delta t$

$=(P_1-P_2)\Delta V$ .........(11.13)

Here, $\Delta V\to$ the volume of liquid that flows through a cross-section is same (from equation of continuity).

But, the work done is equal to net change in energy (K.E. + P.E.) of the liquid, and

$\Delta K=\frac{1}{2}p\Delta V(v_1^2-v_2^2)$ ........ (11.14)

and $\Delta U=p\Delta V(h_2-h_1)$ ........ (11.15)

$\therefore (P_1-P_2)\Delta V=\frac{1}{2}p\Delta V(v_1^2-v_2^2)+pg\Delta V(h_2-h_1)$

$P_1+\frac{1}{2}{v}_1^2+pg{h}_1=P_2+\frac{1}{2}{v}_2^2+pg{h}_2$

or $P+\frac{1}{2}p{v}^2+pgh=constant$ .......... (11.16)

This is the required relation for Bernoulli's theorem.

$\therefore A_1{v}_1=A_2{v}_2$

So, more the cross-sectional area, lesser is the velocity and vice-versa.

So, Bernoulli's theorem,

$P_1+\frac{1}{2}{p}_1{v}_1^2=P_2+\frac{1}{2}{p}_2{v}_2^2$