Subject: Physics

Poiseulle's Studied the motion of liquid Through a narrow horizontal tube. Example: (capillary tube) assuming the streamline motion of the liquid. The layer of liquid id contact with the surface of a horizontal tube is at rest velocity of layers increases as we move towards the axis of tube as shown. He found that volume of liquid flowing out per second (V) through narrow horizontal tube is

- directly propostional between two ends of a tube.

$$\text {i.e} \: V \propto p$$ - directly proportional to the fourth power of the radius of a tube.

$$\text {i.e} \: V \propto r^4$$ - inversely propostional to the length of tube.

$$\text {i.e} \: V \propto \frac 1l$$ - inversely propostional to the length of tube.

$$\text {i.e} \: V \propto \frac {1}{\eta}$$

Combining all

\begin{align*} V &\propto \frac {Pr^4}{\eta l} \\ V &= \frac {\pi}{8} \frac {Pr^4}{\eta l}\dots (V)\\ \end{align*}

The value of proportionality is found to be \(\frac {\pi }{8}\) from experiment.

Equation (v) is known as poiseuille’s formula

**Derivation of Poiseuille’s formula by dimensional analysis**

Poiseuille studied the volume of a liquid flowing out per second through a narrow horizontal tube (example: Capillary tube) assuming the motion of a liquid as stream line and is found to depend upon

- The pressure gradient of a tube.

$$\text {i.e.} \: V \propto \left ( \frac Pl \right )^q \dots (i) $$ - The radius of tube.

$$\text {i.e.} \: V \propto r^b \dots (ii) $$

- The coefficient of viscosity of a liquid

$$\text {i.e.} \: V \propto \eta ^c \dots (iii) $$

Combining all

\begin{align*} V &\propto \left (\frac {P}{ l}\right )^a r^b \eta ^c \\ V &= K\left (\frac {P}{ l}\right )^a r^b \eta ^c \dots (iv) \\ \end{align*}

Writing dimensional formula on both sides we have

$$[M^0 L^3 T^{-3}] = \left [ \frac {ML^{-1}T^{-2}}{L} \right ]^a . [L]^b . [ML^{-1}T^{-1}] $$

Here, the constant of proportionality is a dimensionless quantity.

$$[M^0 L^3 T^{-3}] =[M^{a+c} L^(-2a+b-c) T^{-2a-c}]$$

Equating power of M, L, and T on both sides.

\begin{align*} \text {Power of M} &= a + c = 0 \\ \text { Power of L} &= -2a + b – c = 3 \\ \text {Power of T} & = -2a – c = -3 \\ \end{align*}

Solving all we get a = 1, b = 4 and c = -1. Now by substituting vales of a, b and c, in equation (iv) we get

\begin{align*} V &= K\left (\frac {P}{ l}\right ) r^4 \eta ^{-1} \\ &= K \frac {Pr^4}{nl} \\ V &= \frac {\pi}{8} \frac {Pr^4}{nl} \dots (v) \\ \end{align*}

The value of constant proportionality is found to be \(\frac {\pi }{8}\) from experiment. Equation (v) is known as poiseuille’s formula.

When a small spherical body is dropped in a viscous medium, the layer in contact with it starts moving with the same velocity as that of the body whereas the layer at a considerable far distance will be at rest. In this way, a relative motion between different layers of a medium is created. When a spherical body is dropped into it, the viscous force comes into play which opposes the motion of a body.

As time passes the velocity of falling body increases and viscous force acting on it also increases in that proportion. A state will come when the body starts moving with uniform velocity called terminal velocity in the medium. Terminal velocity means the net force acting on the body becomes zero.

Stroke studied the motion of a small spherical body through a viscous medium and viscous force (F) acting on the body is found to depend upon following factors:

- The coefficient of viscosity of the medium.

$$\text {i.e.} \: V \eta ^a \dots (i) $$ - The radius of a spherical body.

$$\text {i.e.} \: V \propto r^b \dots (ii) $$ - The terminal velocity required by the body in the medium.

$$\text {i.e.} \: V \propto v^c \dots (iii) $$

Combining all

\begin{align*} F &\propto n^ar^bv^c \\ F &= k n^ar^bv^c \end{align*}

Writing dimensional formula on both sides we have,

$$[MLT^{-2}] =[ML^{-1}T^{-1}]^a[L]^b[LT^{-1}]^c $$

Here the proportionality constant is dimensionless

$$[M LT^{-2}] =[M^a L^(-a+b+c) T^{-a-c}]$$

Equating powers of M,L and T on both sides, we have

\begin{align*} a &= 1 \\ b – a + c &= 1 \\ - a - c &= -2 \\ \end{align*}

Solving all we get a = 1, b = 1 and c= 1. Substituting value of a, b, and c in equation (iv)

\begin{align*} F &= n^1r^1v^1 \\ F &= k n\: r\:v \\ \end{align*}

**Determination of Viscosity of a Liquid using Stroke's Formula**

When a spherical ball falls freely through a viscous medium such as a liquid, its velocity at first goes on increasing. Finally, a stage is reached at which the weight of the ball is just equal to the sum of upthrust due to buoyancy and the upward viscous force. So, the net force acting on the body is zero and the ball starts to fall with a constant velocity, i.e. terminal velocity

Let r be the radius of the spherical ball falling through the viscous fluid of density (σ) and coefficient of viscosity (η). Let (Ρ) be the density of the material of the ball. The various forces acting on the body are:

- Its weight, W in the downward direction,
- Upward thrust, U is equal to the weight of the displaced fluid and,
- The viscous force, F in a direction opposite to the direction of motion of the body.

So, the downward force acting on the ball = U + F

When the ball is falling with the terminal velocity v, then

\begin{align*} \text {downward force } &= \text {upward force } \\ F &= W – U \dots (i) \\ \text {According to Stroke’s law, viscous force is F} &= 6\pi \eta rv \end{align*}

Weight of body, \( W = mg = (\text {volume} \times \text {density} ) \text {of the body } \times g = \frac 43 \pi r^3 \rho g\)

Upthrust, U = weight of liquid displaced = \(\text {volume} \times \text {density} ) \text {of fluid } \times g = \frac 43 \pi r^3 \sigma g\)

Equation (i) becomes as,

\begin{align*} 6 \pi \eta rv & = \frac 43 \pi r^3 \rho g - \frac 43 \pi r^3 \sigma g \\ &= \frac 43 \pi r^3 (\rho - \sigma) g \\ \text {or,} \: \eta &= \frac {2r^2 (\rho - \sigma ) g}{9v} \dots (ii) \end{align*}

This is the expression for the coefficient of viscosity of liquid or fluid.

\begin{align*} 6 \pi \eta rv & = \frac 43 \pi r^3 \rho g - \frac 43 \pi r^3 \sigma g \\ &= \frac 43 \pi r^3 (\rho - \sigma) g \\ \text {or,} \: \eta &= \frac {2r^2 (\rho - \sigma ) g}{9v} \dots (ii) \end{align*}_{\begin{align*} 6 \pi \eta rv & = \frac 43 \pi r^3 \rho g - \frac 43 \pi r^3 \sigma g \\ &= \frac 43 \pi r^3 (\rho - \sigma) g \\ \text {or,} \: \eta &= \frac {2r^2 (\rho - \sigma ) g}{9v} \dots (ii) \end{align*}}=

43πr3ρg−43πr3σg=43πr3(ρ−σ)g=2r2(ρ−σ)g

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