Notes on Coefficient of Linear Expansion by Pullinger's Apparatus, Bimetallic Thermostat and Differential Expansion | Grade 11 > Physics > Thermal Expansion | KULLABS.COM

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#### Determination Of Coefficient Of Linear Expansion Of A Solid By Pullinger’s Apparatus

Construction

The linear expansivity of a material can be determined by the Pullinger’s apparatus. It consists of a hollow cylinder where the experimental rod is placed inside the cylinder. There are three opening in the cylinder. The upper and lower opening are used for steam inlet and steam outlet. The middle opening is used for placing thermometer which measures the temperature of rod. The spherometer is placed in a free end of the instrument which measure the increase in length of the rod. An electric circuit is connected with the instrument to find whether spherometer touches the rod or not.

Working

The experimental rod is taken and its length is measured (say) L. The rod is placed inside the cylinder and the initial temperature of the rod is taken (say)θ1. Now the spherometer is rotated downward when it just touches the rod. Initial reading on spherometer is taken and spherometer is rotated upward to give a small space for expansion of the rod. The steam is passed into the cylinder as steam is passed the reading in thermometer rise. When the thermometer shows constant reading 8-10 the final temperature of the rod is taken (say)θ2.The spherometer is rotated downward when it touches the rod and the final reading of spherometer is taken. let initial spherometer reading is R1 and final reading is R2. Thenthe increase in radius is R2 - R1.

We have

$$Linear \;expansivity = \frac{increase\;in\; length}{original\;length\times rise \;in\; temperature}$$

$$\therefore\alpha = \frac{R_2-R_1}{R_1(\theta_2-\theta_1)}$$

By finding values of R1, R2,$$\theta_2, \theta_1$$and linear expansivity is determined.

#### Force Set up Due to Expansion or Contraction

Let us consider a metal rod of length l1 is fixed at the two rigid fixed ends S1 and S2 as shown in the figure. Suppose the initial temperature be θ1.Now the rod is heated up to θ2,metal rod will try expand to the length of l2, but it cannot do so.

From linear expansion of solid, we have

$$l_2 = l_1(1+ \alpha\Delta\theta)$$

Where $$\alpha$$ is the coefficient of linear expansion of the rod, and $$\Delta\theta = \theta_1 -\theta_2$$ .So, the increase in length is

$$l_2 - l_1= l_1( \alpha\Delta\theta)$$

Due to increase in temperature, the rod tries to expand but will not be able to, to expand due to rigid ends. As a result, a force or tension is produced which would compress the rod. From the definition of Young’s modulus of elasticity, we have,

Young’s modulus, $$\Upsilon = \frac{\text {tensile stress}}{\text {tensile strain}} = \frac{tension/area}{\text{ change in length/original length}} = \frac {T/A}{l_2 - l_1/ l_1}$$

Where A is a cross-sectional area of the rod. Then,

$$\Upsilon = \frac TA\times \frac{l_1}{l_2 - l_1} = \frac{T}{A} \times\frac{l_1}{l_1\alpha\Delta\theta} = \frac {T}{A\alpha\Delta\theta}$$

Or, $$T = \Upsilon A \alpha\Delta\theta$$

Tension or force $$=\Upsilon A \alpha(\theta_1 - \theta_2)$$

#### Bimetallic Thermostat

Bimetallic thermostats are used to control higher temperature. It works as an electric contact breaker in an electrical heating circuit. It consists of two strips of metal that have different coefficients of linear expansion such as brass and steel. The two pieces are welded and riveted together. When the bimetallic strip is heated, the brass, having a greater value of, expands more than the steel. Since the two strips are bounded together, the bimetallic strip bends into the arc as shown in the figure. The metallic strip is in contact with screw S at point P and on heating, the strip curves downwards and the contact at P is broken. Thermostats are used for controlling the temperature of laundry irons, hot water storage tanks, aquaria for tropical fish and for many other purposes.

#### Differential Expansion (fig different from book)

When metal rod of different metal are heated to the same range of temperature their expansion are different i.e. differential expansion.

Let us consider two rod B and C having length and with linear expansivity and respectively. Both the rods are heated so that the final length of the rod are and respectively at the temperature.

We have,

$$l_2 = l_1(1+ \alpha\Delta\theta)$$

$$\Delta\theta = \theta_2-\theta_1$$

Increase in length of rod $$\beta$$

For equal difference in length

$$l_2 - l_1 =l_1\alpha\Delta\theta$$

or, $$l_1\alpha\Delta\theta = l_1 '\alpha '\Delta\theta$$

or, $$l_1\alpha=l_1'\alpha '$$

or, $$\frac{l_1}{l_2 '} = \frac{\alpha '}{\alpha}$$

This is the constant difference in length in all temperature.

Relation between real and cubical expansion of liquid

Liquid does not have fixed shape so, we do not take linear and superficial expansion in the case of liquid instead we take cubical expansion.

Let us consider liquid contain in a vessel having an original volume up to level A suppose the system is heated at first. After heating the expansion of vessel takes place and the level of a liquid decreases to point B. After sometimes liquid gets heated the expansion of a liquid is greater than that of solid. So, finally liquid reaches C.

Here,

AB = expansion of vessel

AC = apparent expansion of liquid

BC = real expansion of liquid

From the figure

BC = AC + AB ----------(i)

$$BC = AC + AB \dots (i)$$

real expansion of liquid = apparent expansion of liquid + expansion of a vessel

Real expansivity of liquid $$(\gamma_r)$$ is defined as real increase in volume of the liquid per unit original volume of the liquid per unit rise in temperature. .i.e

Real expansivity of liquid $$(\gamma_r) = \frac {\text{Increase in volume of the liquid}}{\text{original volume of the liquid× rise in temperatures}}$$

$$\gamma_r = \frac {BC}{V \times \Delta\theta}$$

$$\because\text{original volume of liquid} = \text{volume of vessel(V)} \times {\text{rise in temperature}\Delta\theta }$$

$$BC = \gamma_r V \times \Delta\theta\dots (ii)$$

Apparent expansivity is defined as an apparent increase in a volume of liquid per unit rise in temperature.

apparent expansivity of liquid $$(\gamma_a) = \frac{\text{apparent increase in volume of the liquid}}{\text{original volume of the liquid} \times \text{ rise in temperature}}$$

$$\gamma_a = \frac {AC}{V \times \Delta\theta}\$$

$$AC = \gamma_a V \times \Delta\theta\dots(iii)$$

Cubical expansivity is defined as increase in a volume of the vessel per unit original volume of the vessel per unit rise in temperature.

Expansion of vessel $$(\gamma) = \frac{\text{increase in volume of the vessel}} {\text{original volume of the vessel} \times \text{rise in temperature}}$$

$$\gamma = \frac {AB}{V \times \Delta\theta}$$

$$AB = \gamma V \times \Delta\theta\dots(iv)$$

Using equation (i), (ii), (iii) and (iv)

$$\gamma_r V \times \Delta\theta = \gamma_a V \times \Delta\theta + \gamma V \times \Delta\theta$$

$$\gamma_r V \times \Delta\theta = (\gamma_a + \gamma) V \times \Delta\theta$$

$$\gamma_r = \gamma_a +\gamma$$

$$\gamma_r = \gamma_a + 3\alpha [\because \gamma = 3\alpha]$$

The linear expansivity of a material can be determined by the Pullinger’s apparatus.

When the bimetallic strip is heated, the brass, having a greater value of, expands more than the steel.

When metal rod of different metal are heated to the same range of temperature their expansion are different i.e. differential expansion.

Real expansivity of liquid is defined as real increase in volume of the liquid per unit original volume of the liquid per unit rise in temperature.

Apparent expansivity is defined as an apparent increase in a volume of liquid per unit rise in temperature.

Cubical expansivity is defined as increase in a volume of the vessel per unit original volume of the vessel per unit rise in temperature.

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