# Bernoulli’s Theorem

Bernoulli’s theorem states that:
For the steady flow of a incompressible non-viscous fluid, the sum of pressure energy per unit mass, potential energy per unit mass and kinetic energy per unit mass remains constant at every point throughout the flow.

$\text{i.e. } \frac{P}{ρ} + gh +\frac{v^2}{2} = constant$ where, $\frac{P}{ρ}=\text{Pressure energy per unit mass}$ $gh=\text{Potential energy per unit mass}$ $\frac{v^2}{2}=\text{Kinetic energy per unit mass}$

This theorem is an outcome of the principle of conservation of energy.​ Consider an incompressible non-viscous liquid is flowing inside a tube $XY$ of varying cross sectional area.

​If the flow is steady, then the mass $m$ of the liquid flowing per second through any section of the tube is same. Let the density of the liquid be $ρ$.

Let, $P_1=\text{Pressue at X}$ $P_2=\text{Pressue at Y}$

$A_1=\text{Area of cross section of the tube at X}$ $A_2=\text{Area of cross section of the tube at Y}$

$v_1=\text{Velocity of liquid at X}$ $v_2=\text{Velocity of liquid at Y}$

$h_1=\text{Height of section X from ground}$ $h_2=\text{Height of section Y from ground}$

The flow work associated with the liquid at $X$, $W_1=P_1A_1v_1$ where, $v_1$ is the distance travelled per second and $F_1=P_1A_1$ $∴W_1=P_1V$ where, $V$ is volume of liquid flowing per second.

Similarly, The flow work associated with the liquid at $Y$, $W_2=P_2A_2v_2$ $∴W_2=P_2V$

Now, $\text{Total work done (W)}=\text{Sum of two flow works}$ $W=W_1+(-W_2)$ $W=W_1-W_2$

There is negative sign because the flow work at $Y$ is negative. $∴W=P_1V-P_2V$ $W=(P_1-P_2)V$ $W=(P_1-P_2)\frac{m}{ρ}\text{ ____(1) }\left[ρ=\frac{m}{V}\right]$

Potential energy increases from $X$ to $Y$ because the height is increasing. $\text{Change in P.E. } ΔP.E.=mg(h_2-h_1)$

Kinetic energy also increases from $X$ to $Y$ because the velocity increasing. $\text{Change in K.E. } ΔK.E.=\frac{1}{2}m(v_2^2-v_1^2)$

Now- $\text{Total work done (W)}=ΔP.E+ΔK.E.$ $W=mg(h_2-h_1)+\frac{1}{2}m(v_2^2-v_1^2)\text{ ____(2)}$

From (1) and (2), $(P_1-P_2)\frac{m}{ρ}=mg(h_2-h_1)+\frac{1}{2}m(v_2^2-v_1^2)$ $\frac{P_1}{ρ}-\frac{P_2}{ρ}=gh_2-gh_1+\frac{v_2^2}{2}-\frac{v_1^2}{2}$ $\frac{P_1}{ρ}+gh_1+\frac{v_1^2}{2}=\frac{P_2}{ρ}+gh_2+\frac{v_2^2}{2}$

Hence, $\frac{P}{ρ}+gh+\frac{v^2}{2}=constant\text{ ___(a)}$

This verifies Bernoulli’s theorem. Also, multiplying $(a)$ by $ρ$, we get, $P+ρ gh+\frac{ρv_1^2}{2}=constant$

Thus, the sum of pressure energy per unit volume, potential energy per unit volume and kinetic energy per unit volume remains constant.

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