Hydrodynamic

(i) Reynold's Number

According to Reynold, the critical velocity (v_c) of a liquid flowing through a long narrow tube is

(a) directly proportional to the coefficient of viscosity ($\eta$) of the liquid.

(b) inversely proportional to the density $\rho$ of the liquid and

(c) inversely proportional to the diameter (D) of the tube.

That is $v_c\propto \frac{\eta }{\rho D}or{v}_c=\frac{R\eta }{\rho D}or=\frac{v_c\rho D}{\eta }$ ... (1)

where R is the Reynold number.

If R < 2000, the flow of liquid is streamline of laminar. If R > 3000, the flow is turbulent. If R lies between 2000 and 3000, the flow is unstable and may change from streamline flow to turbulent flow.

(ii) Steady Flow (Stream Line Flow)

Then the velocity of fluid particles reaching a particular point is the same at all time. Thus, each following particle takes the same path as taken by a previous particle through that point.

(iii) Line of Flow

It is the path taken by a particle in a flowing liquid, in case of a steady flow, it is called a streamline.

Consider an area S in a fluid in steady flow. Draw streamlines from all the points of the periphery of S. These streamlines enclose a tube, of which S in a cross-section.

No fluid enters or leaves across the surface of this tube.

Consider an area S in a fluid in steady flow. Draw streamlines from all the points of the periphery of S. These streamlines enclose a tube, of which S in a cross-section.

No fluid enters or leaves across the surface of this tube.

(iv) Equation of Continuity

In a time $\Delta t$, the volume of liquid entering, the tube of flow in a steady flow is $A_1{V}_1$$\Delta t$, the same volume must flow out, as liquid is incompressible. The volume flowing out in $\Delta t$ is $A_2{V}_2\Delta t$

$\Rightarrow A_1{V}_1=A_2{V}_2$

(v) Bernoulli's Theorem

Consider a tube of flow, ABCD. In a time $\Delta$t, liquid moves and the liquid element becomes A'B'C'D'. In other words, we can also interpret that ABB'A' has gone to DCC'D'.

$\Delta$m = $\rho$A₁V₁ $\Delta$t = $\rho$ A₂V₂ $\Delta$t

Work done by fluid pressure at 1

= (P₁A₁) v₁ $\Delta$t = P₁ $\Delta$m/$\rho$

Work done by fluid pressure at 2

= - (P₂A₂) v₂ $\Delta$t = -P₂ $\Delta$m/$\rho$

Work done by gravity = -($\Delta$m).g. (h₂ - h₁)

Change in kinetic energy = 1/2 $\Delta$m [V₂² - V₁²]

Using, work energy theorem, (W = $\Delta$K)

$\begin{align}& \Rightarrow {{P}_{1}}\dfrac{\Delta m}{\rho }-{{P}_{2}}\dfrac{\Delta m}{\rho }-\Delta mg\,\,({{h}_{2}}-{{h}_{1}})=\dfrac{1}{2}\Delta m\,\,[{{V}_{2}}^{2}-{{V}_{1}}^{2}]\\&\dfrac{{{P}_{1}}}{\rho}+g{{h}_{1}}+\dfrac{{{V}_{1}}^{2}}{2}=\dfrac{{{P}_{2}}}{\rho}+g{{h}_{2}}+\dfrac{{{V}_{2}}^{2}}{2} \\& \Rightarrow {{P}_{1}}+\rho g{{h}_{1}}+\dfrac{\rho {{V}_{1}}^{2}}{2}={{P}_{2}}+\rho g{{h}_{2}}+\dfrac{\rho {{V}_{2}}^{2}}{2} \\ & P+\rho gh+\dfrac{rho {{V}^{2}}}{2}=Cons\tan t \\\end{align}$

Example : The reading of pressure-meter fitted in a closed pipe is 4.5$\times$10⁵ N/m². On opening the value of the pipe, the reading the meter reduces to 4.0$\times$10⁵N/m². Calculate the speed of water flowing in the pipe.

Solution: From Bernoulli's theorem

P₁ + $\frac{1}{2}\rho v_1^2$ = P₂ + $\frac{1}{2}\rho v_2^2$

or P₁ – P₂ = _{$\frac{1}{2}\rho \left(v_2^2-v_1^2\right)$}

Here, v₁ = 0

(Velocity of the water is zero because initially pipe is closed)

$\therefore$ $v_2^2=\frac{2}{\rho }\left(P_1-P_2\right)$

$v_2^2=\frac{2}{\left(1\times {10}^3\right)}\times \left[4.5\times {10}^3-4.0\times {10}^5\right]$

= 2 $\times$ 0.5$\times$10³ = 100m/s

v₂ = 10m/s