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# Heat transfer, Conduction, Conduction Law: Differential Form, thermal resistance, solved examples, FAQs

You might have seen someone or used the spatula while cooking. These spatulas are usually made out of metal or wood. If you’d have observed carefully, the wooden spatulas can be used to stir the hot food for a lot longer as compared to the metal one. But why doesn’t using wooden spatulas for long burns one’s hand while metal one doesn’t take much time to produce a burning sensation? To know the reason you must be aware about an important mode of heat transfer known as conduction.

## Heat Transfer

Heat is a form of energy that is transmitted from a region of high temperature to that of a low temperature.

Heat transfer may take place in any of the following modes:

• Conduction
• Convection

## Conduction

Heat conduction is the flow of internal energy from a region of higher temperature to that of a lower temperature by the interaction of adjacent particles (atoms, molecules, ions, electrons, etc.) in the intervening space.

Consider a metal rod on which several nails are glued by wax between points A and B at certain distances. The left end of the rod, point A, is heated by an external source, and as a result, its temperature increases, and the thermal energy starts transmitting inside the rod by means of conduction. As time progresses, the temperatures of the points between A and B gradually increase, causing the wax to melt, which results in the nails to lose the required adhesive force and drop one after another, starting from point A to point B.

Points to Remember

Consider a conducting slab of length L and cross-sectional area A, as shown in the figure. Faces P and Q are set at constant temperatures T1 and T2 (T1 > T2), respectively. As a result, the heat starts flowing from face P to face Q in the slab. Consider a rectangular cross section in the slab of thickness dx at a distance of x from face P. At some time t, this cross section receives Q1 amount of heat and passes down Q2 amount of heat (Q1 > Q2) to its neighbouring cross section at a distance of + dx. The difference of heat (Q1− Q2) contributes to the increase in the temperature of the cross section. As time progresses, every cross section in the slab achieves a constant temperature ranging from T1 at face P to T2 at face Q. Therefore, the temperature gradient is generated inside the slab, following which every cross section in the slab between faces P and Q receives and releases the same amount of heat that enters face P.

The definition of the steady-state heat transfer is as follows:

If the temperature of a cross section at any position in the slab remains constant with time, it is said to have achieved a steady state. However, it is not the same as thermal equilibrium.

A metallic conductor is shown in the figure. When the steady-state heat transfer is achieved, the temperatures of cross sections a, b, and c, become constant with time. The same amount of heat, 𝛿Q, flows through all such cross sections.

## Fourier’s Law of Heat Conduction

The rate of heat transfer through a medium is directly proportional to the cross-sectional area normal to the direction of heat and the temperature gradient across the medium.

Let the rate of heat transfer be q.

According to the definition,

The minus sign in the expression of the temperature gradient proportionality comes from the fact that the temperature is decreasing in the direction of heat transfer, where ‘k’ is known as the thermal conductivity of the material.

Fourier’s law of heat conduction is applicable under the following conditions:

• A body through which the heat is being transferred has to be in a steady state.
• The thermal conductivity () of the body has to be constant.
• It is only valid for unidimensional (1D) heat transfer.

The order of thermal conductivity of different substances, in general, is as follows:

kgases < kliquids < ksolids

Among solids, metals have higher thermal conductivity than non-metals:

knon-metals < kmetals

SI unit of thermal conductivity is

Dimensions of

Combination of slabs

• Series Combination

In a series combination of slabs, there are n number of slabs of lengths L1, L2, …, Ln and the thermal conductivities k1, k2, …, kn, respectively. They are connected in series combinations as shown in the figure. The heat is supplied to the first slab, and the temperature at the cross sections are T1, T2, ….., Tn - 1, Tn (T1 > Tn). The heat current is constant through all the slabs.

Since the heat current, q, remains constant, we can write it as follows:

Where

This is the same formula as the series combination of resistors. So, the equivalent thermal resistance is given by,

If the area of cross section of all the slabs are equal, i.e., A1 = A2 = … = An,then the equivalent thermal conductivity is given by,

• Parallel combination

In a parallel combination of slabs, the change in temperature, i.e., ΔT is constant for all the slabs. However, the heat current through different slabs is different.

The following figure represents the equivalence to the parallel combination of resistors.

The net heat current for the parallel combination of the slabs is given by,

The net thermal conductivity (keq) is given by,

For the slabs having an equal area, we get the following:

I.e.

## Circuit problems

It deals with the similarity between the heat conduction formula and Ohm’s law for electric current.

where

(say)

This is analogous to the expression for electric current

Comparing both the expressions, thermal resistance can be obtained as

## Practice problems

Q1. A hollow tube has length l, inner radius r1, and outer radius r2. The material has a thermal conductivity of k. Find the heat current through the walls of the tube if the flat ends are maintained at temperatures T1 and T2. (T2 > T1)

The flat ends of the tube are maintained at temperatures T1 and T2, and after some time, a steady state has been reached.

The rate of heat transfer can be found as follows:

Q2. Consider the situation shown in the figure. The frame is made of the same material and has a uniform cross-sectional area. Calculate the amount of heat flowing per second through the cross section of the bent part if the total heat taken per second from the end at 100 °C is 130 J.

The frame is made of the same material (thermal conductivity k) and has a uniform cross-sectional area (say A). The thermal resistance, of the material varies in proportion with the length of the conductor.

Assuming that the thermal resistance of the material of length 5 cm is R, the thermal network can be pictured as shown in the figure.

Given,
The rate of heat entering the network from the end kept at 100°C is

At junction C, it gets divided into two parts.

(i) Heat H1 passing through the bent part

(ii) Heat H2 passing through the straight portion

Hence,

Since the bent part consists of three thermal resistors (R, 12R, and R) placed in series, the equivalent thermal resistance of the path is as follows:

Req = R + 12R + R = 14R

The given thermal network can be simplified into two thermal resistances, 14R and 12R, placed in a parallel combination as shown in the figure.

The temperature differences across both AB and CD are equal.

Heat passing through conductor AB per second is as follows:

Heat passing through the conductor CD per second is as follows:

From equation ,

Q3. Four identical rods, AB, CD, CF, and DE, are joined as shown in the figure. The length, cross-sectional area, and thermal conductivity of each rod are and respectively. The ends, A, E, and F, are maintained at temperatures T1, T2, and T3, respectively. Assuming no loss of heat to the atmosphere, find the temperature at B.

Answer: The rods are made of the same material (thermal conductivity ), having the same cross-sectional area, , and length .

Let the temperature of junction B be . The length of divisions BC and BD is , and they are connected in series with rods CF and DE, respectively. A simplified thermal network is shown in the figure.

where

Since the algebraic sum of heat leaving the junction is zero,

Q4. A composite slab is prepared by pasting two plates of lengths L1 and L2 and thermal conductivities and . The slabs have equal cross-sectional areas. Find the equivalent conductivity of the composite slab.

Answer: If the plates of different thermal conductivities and lengths were to be replaced by another plate of length , which would retain the same thermal resistance as earlier, the thermal conductivity of the new plate will be known as the equivalent thermal conductivity.

Therefore,

## FAQs

Q1. Why are metals good conductors of heat?
Answer: Metallic solids are good conductors of heat because their free electrons carry the heat energy, whereas in non-metallic solids and fluids, the conduction is only due to the vibrations of molecules.

Q2. What is the SI unit of thermal resistance?