Chapter 11.8 - Convection And Radiation

In the previous section we saw heat transfer by conduction. In this section, we will see convection and radiation

• First we will see convection
• Convection can occur only in fluids
• In convection, there is actual movement of particles
    ♦ The particles which are in direct contact with the heat source, become hot
    ♦ They move away towards colder parts of the fluid mass
• This process can be explained in detail, by considering the formation of sea breeze and land breeze. It can be explained in 17 steps:
1. The heat from the sun travels through the atmosphere and reaches the surface of the earth
2. The heat traveling from sun to earth is in the form of short waves
• The molecules in the atmosphere cannot absorb these short waves. They can absorb only long waves
• So, while traveling from sun to earth, the sun’s rays cannot heat up the atmosphere
• Then how does the atmosphere get it’s heat?
3. The answer can be written in 4 steps:
(i) The earth gets heated up first
(ii) The earth then emits heat back into the atmosphere
    ♦ This heat is in the form of long waves
(iii) The molecules in the atmosphere can absorb these long waves
(iv) Thus the atmosphere gets heated
4. However, our present discussion is about sea breeze and land breeze
• For that discussion, we do not need to consider the heating of atmosphere by the long waves emitted by the earth
• What we need to remember at this stage is that:
    ♦ the earth’s surface gets heated first by the suns rays
5. Now, the earth’s surface has both land and water
• The soil particles in the land has much smaller specific heat capacity when compared to water
• So land gets heated more quickly than the seas
(Recall that, water is used as a coolant in automobiles because, it can absorb large quantities of heat with out becoming hot)
6. So a difference in heating occurs:
    ♦ The atmospheric air in contact with the land
    ♦ gets heated more
    ♦ Than the atmospheric air in contact with the sea
7. Note that, this heating is achieved by conduction
    ♦ The vibrating soil particles
    ♦ set up vibrations in the air particles
    ♦ which are in contact with the soil
8. When the air is heated in this way, it expands
• Expansion means: ‘increase in volume’ and consequent ‘decrease in density'
• The 'hot air' thus has a lesser density than cold air
• The lesser dense air rises up due to buoyancy
• This upward motion is marked as ‘A’ in the fig.11.18 below:

Fig.11.18

9. So a vacant space is created above the land
• The air above the sea will move towards that vacant space
    ♦ This movement of air from sea to land is known as sea breeze
    ♦ This is marked as ‘B’ in the fig.11.18 above
10. This ‘B’ creates a vacant space above sea
• The air high up above the sea settles downwards to this vacant space
    ♦ This downward movement is marked as ‘C’
11. This ‘C’ creates a vacant space high up above the sea
• The air already present high up above the land moves horizontally towards this vacant space
    ♦ This horizontal movement is marked as ‘D’
12. Thus the process continues in a cyclic manner
• This cyclic process continues as long as the sun heats up the land
13. During night, there is no heat from the sun
• The land cools down more quickly than the sea
(Recall that, water is used in hot water bags because, it cools down slowly)
14. So a difference in heating occurs:
    ♦ The air above sea
    ♦ will be hotter
    ♦ Than the air above the land
15. The hot air above the sea rises up
This is marked as 'A' in fig.11.19 below:

Fig.11.19

• The air above land moves towards the sea
    ♦ This movement of air from land to sea is called land breeze
    ♦ This is marked as 'B' in the fig.11.19 above
16. We see that, the movements in fig.11.19 are just the opposites of the movements in the previous fig.11.18
• So we can write:
In the night, the cycle is reversed
17. Thus we see that, in ‘heat transfer by convection’, there indeed is actual movement of particles from one part of the fluid body to other parts

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Next we will see heat transfer by radiation. It can be written in steps:
1. Consider a piece of iron
• Let it be heated gradually. It will pass through several stages:
• Stage 1:
    ♦ The iron becomes a little hot
    ♦ This hotness can be detected if we move our hands closer to the iron
• Stage 2:
    ♦ As the temperature increases, the iron becomes red hot
• Stage 3:
    ♦ As the temperature increases further, the iron becomes white hot
2. For our present discussion, we are interested in stage 1
• In the stage 1, the heat reached us even though we did not touch the iron
• How is that possible?
• The following steps from (3) to (7) will give the answer:
3. When a body is heated, it's molecules, atoms, electrons, protons etc., begin to vibrate
• Due to the vibration of the charged particles, electromagnetic waves are produced
(Some basic details can be seen here)
• These waves begin to radiate out from the body
4. Such electromagnetic waves do not need a medium to travel
• They can travel through vacuum
• They travel at the speed of light
• In our present case, the electromagnetic waves radiated out from the iron and reached our hands
• Thus we felt the heat
5. The ‘frequency of the waves’ depend on the energy content of the iron piece 
    ♦ If the energy content is high, the emitted waves will have higher frequency
    ♦ If the energy content is low, the emitted waves will have lower frequency
6. In stage 1, the energy content is low. So the emitted radiation have a low frequency
• Radiations of lower frequencies are detected in the form of heat
7. In stages 2 and 3, the energy contents are high. So the emitted radiations have high frequencies
• Radiations of higher frequencies are detected in the form of light
    ♦ So they appear red and white in color
8. So now we know how the heat from the iron piece reached us
■ All bodies (solid, liquid and gas) which have a temperature above absolute zero, will emit heat radiations
■ The electromagnetic radiation emitted by a body by virtue of it’s temperature is called thermal radiation
■ The heat that we receive from the sun is thermal radiation
9. Consider the 'earth together with the atmosphere' as one unit
• Then, the space between this unit and the sun is a vacuum
• All radiations (including thermal radiation) emitted from the sun, can travel through this vacuum and reach the earth
10. Let us see the relation between black color and thermal radiation:
    ♦ A body with black color can absorb most of the thermal radiation falling on it
    ♦ A body with white color can reflect most of the thermal radiation falling on it
• So we prefer to wear:
    ♦ white or light colored cloths in summer
    ♦ black or dark colored cloths in winter
• The bottom side of cooking vessels is given a black color
    ♦ This will enable the vessels to absorb maximum heat from the flame
11. Let us see the working of a Dewar flask. It can be written in 7 steps:
(i) A Dewar flask consists of two flasks, kept one inside the other
• Some images can be seen here
(ii) The inner flask is double walled
    ♦ The space between the two walls is a vacuum
(iii) The inner surface of the inner wall is coated with silver
    ♦ So heat from the contents will be reflected back
(iv) The outer surface of the outer wall is also coated with silver
    ♦ So heat from the surroundings will be reflected back
(v) The vacuum space between the walls prevents any heat movement by conduction or convection
(vi) The inner flask is fixed to the outer flask using insulating materials like cork
(vii) All the above steps will help to keep hot contents hot and cold contents cold for a longer duration

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In the next section, we will see Newton's law of cooling

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Chapter 11.7 - Heat Transfer by Conduction

In the previous section we completed a discussion on phase diagrams. In this section we will see heat transfer by conduction

• Heat flows from a body at higher temperature to a body at lower temperature
• This flow can be through conduction, convection or radiation
• First we will see conduction. Some basics about conduction can be written in 3 steps:
1. In conduction, the molecules or atoms in a body begin to vibrate when heated
2. Consider a metallic rod. Let one end of the rod be heated
• Those molecules at the heated end will begin to vibrate
    ♦ They will collide with the adjacent cold molecules
• Those cold molecules will begin to vibrate and their temperature also increases
    ♦ Those newly heated molecules will collide with the adjacent colder molecules
• This process continues and heat reaches the other end of the rod
■ This process is called conduction
3. During this process, there is no transfer of particles
• This is because, vibration of molecules is a ‘to and fro motion’ about a 'mean point'
    ♦ Each molecule has it’s own mean point.  This mean point do not change
    ♦ So there is no actual movement of molecules along the length of the rod
• In convection, there is actual movement of molecules from one part of the body to other parts. We will see convection in the next section 

Rate of heat transfer

• When heat is transferred from one part of a body to another by conduction, we will want to know the rate at which the heat is transferred
• That means, we will want to know the ‘quantity of heat’ which flows in one second
• This quantity is known as heat current

To get a good understanding about heat current, we will see a comparison with the rate of flow of a liquid. It can be written in 8 steps:
1. In fig.11.16(a) below, two vessels contain water
• The vessels are connected by a pipe of varying cross section

Fig.11.16

2. The level difference between the two water surfaces is maintained at a constant value of ‘h’
• This is accomplished by two simple steps:
(i) The excess water reaching B over-flows through an outlet pipe
(ii) The water used up from A, is replenished by an inlet pipe
■ Since ‘h’ is constant, we will have a steady flow through the connecting pipe
3. Consider any cross section 1-1 through the pipe
• In one second, a volume of (A₁V₁) will pass the section 1-1
    ♦ Where
    ♦ A₁ is the cross sectional area at 1-1
    ♦ V₁ is the velocity of flow at 1-1
• We say that:
Rate of flow at 1-1 is A₁V₁
4. Consider any other cross section 2-2
• In one second, a volume of (A₂V₂) will pass the section 1-1
    ♦ Where
    ♦ A₂ is the cross sectional area at 2-2
    ♦ V₂ is the velocity of flow at 2-2
• We say that:
Rate of flow at 2-2 is A₂V₂
5. Just like 1-1 and 2-2, there can be a large number of cross sections along the length of the pipe
• The rate at all those sections will be the same
• That means:
A₁V₁ = A₂V₂ = A₃V₃  . . . = a constant
• That means, the same volume will be flowing per second at all seconds
• This is so because, the ‘volume entering a section’ must be equal to ‘volume leaving that section’
• Other wise, there will be accumulation of water at certain sections
    ♦ And the farther sections will not receive expected quantities of water
6. Similar is the case with flow of heat through a conductor
• Take any number of sections along the conductor
• If there is no heat loss to the surroundings, the 'quantity of heat passing through all those sections in one second' will be the same
7. We measure 'quantity of heat' in joules
• So the ‘number of joules’ passing the various sections along a conductor in any second will be the same
• If it is not the same, there will be accumulation of heat at certain sections
    ♦ And the farther sections will not receive the expected quantities of heat
8. The unit of 'rate of flow of water' is m^₃ s^_-1
• In the same way:
The unit of 'rate of flow of heat' is ^_J s^_-1
    ♦ 'Rate of flow of heat' is called heat current
    ♦ The symbol for heat current is H

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Let us see a practical way to measure H. It can be written in 9 steps:
1. In fig.11.16(b) above, a metallic bar AB  is shown in yellow color
    ♦ It’s length is L
    ♦ It has a varying cross sectional area
2. The left end A of the rod is embedded in a large heat reservoir C
(A 'heat reservoir' has a large heat capacity so that, it’s temperature remains the same even when heat is removed or added from it)
• By suitable mechanism, the temperature of C is maintained at a constant value T_C
3. The other end B is embedded in another heat reservoir D
• By suitable mechanism, the temperature of D is maintained at a constant value T_D
■ T_C is greater than T_D
4. Due to the difference in temperatures, heat begins to flow through AB
• The sides of AB are covered with insulating material
    ♦ So there is no heat loss to the surroundings
5. In such a situation, there will be a continuous flow of heat through AB
• This flow will continue as long as there is a temperature difference between the reservoirs C and D
6. If in the case of water in fig.11.16(a), the height difference h is zero, the flow of water will stop
    ♦ Such a situation will arise if the water from inlet stops
    ♦ The water level in the first vessel will gradually fall
    ♦ Soon the two water levels will become equal
    ♦ Then h will be zero
■ If in fig.b, the temperature difference (T_C - T_D) becomes zero, the flow of heat will stop
7. As long as T_C and T_D remain at constant values, the reservoir C will supply heat at a constant rate of H joules per second through AB
• The reservoir D will receive heat at the same constant rate of H joules per second
8. We must be aware about a gradual decrease in temperature along the length of the rod. It can be written in 4 steps:
(i) The end A is in contact with reservoir C
    ♦ So the temperature at end A will be T_C
(ii) The end B is in contact with reservoir D
    ♦ So the temperature at end B will be T_D
(iii) That means, the temperature at end B will be lower than the temperature at end A
(iv) That means, as we move from end A to end B, temperature of the rod gradually decreases
9. Our aim is to find an expression for H. It can be written in 6 steps:
(i) Scientists discovered that:
    ♦ H is directly proportional to the temperature difference (T_C - T_D)
    ♦ H is directly proportional to the area A
    ♦ H is inversely proportional to the length L
[The role of A and L can be compared to their ‘similar roles in electrical resistance’. We have seen it in our earlier physics classes. (See fig.8.20 here)]
(ii) So we can write: $\mathbf\small{\rm{H \propto \frac{A (T_C - T_D)}{L}}}$
Thus we get Eq.11.10: $\mathbf\small{\rm{H = KA\frac{ (T_C - T_D)}{L}}}$
    ♦ Where K is the constant of proportionality
    ♦ It is called thermal conductivity of the material
(iii) Every material has it’s own unique value of K
• K of a material represents the readiness with which that material will allow heat to pass through it
(iv) We see that K is in the numerator. So we can write:
    ♦ If K is high, more joules can pass in each second
    ♦ If K is low, only less joules can pass in each second
(v) We can obtain K of different materials from the data book
• We see that:
    ♦ Metals like silver and copper have high K values
    ♦ Non-metals like glass and wood have low K values
(vi) Let us find the unit of K:
• Substituting the units of various quantities in Eq.11.10, we get:
$\mathbf\small{\rm{J \,S^{-1} = (unit \, of \, K)\, m^2\frac{ (K)}{m}}}$
⇒ Unit of K = $\mathbf\small{\rm{J\,S^{-1} \, m^{-1} \, K^{-1}}}$
• But $\mathbf\small{\rm{J\,S^{-1}}}$ is watt. So we get:
Unit of K = $\mathbf\small{\rm{W \, m^{-1} \, K^{-1}}}$

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Now we will see four solved examples. They are given in pdf format at the link below:

Solved example 11.24 to 11.27

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Next we will see a solved example which demonstrates the heat transfer when two conductors are connected in series

Solved example 11.28
What is the temperature of the steel-copper junction in the steady state of the system shown in Fig. 11.17. Length of the steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace = 300 °C, temperature of the other end = 0 °C. The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel = 50.2 J s_^-1 m_^-1 K_^-1 ; and of copper = 385 J s_^-1 m_^-1 K_^-1)

Fig.11.17

Solution:
Data given is:
• Length of steel rod, L₁ = 0.15 m
• Length of copper rod, L₂ = 0.10 m
• Area of cross section of the copper rod = A
• Area of cross section of the steel rod = 2A
• Higher temperature T₁ = 300 °C
• Lower temperature T₂ = 0 °C
• K of steel, K_S = 50.2 J s^_-1 m^_-1 K^_-1
• K of copper, K_C = 385 J s^_-1 m^_-1 K^_-1
• Temperature at the steel-copper junction, T = ?
1. Imagine that, the system is separated into two parts at the steel-copper junction
2. Consider the left part
• A rate of flow H₁ will be set up due to the difference between T₁ and T
• We have Eq.11.10: $\mathbf\small{\rm{H = KA\frac{ (T_C - T_D)}{L}}}$
• So we get: $\mathbf\small{\rm{H_1 = K_S 2A\frac{ (300 - T)}{0.15}}}$
3. Consider the right part
• A rate of flow H₂ will be set up due to the difference between T and T₂
• Using Eq.11.10, we get: $\mathbf\small{\rm{H_2 = K_C A\frac{ (T - 0)}{0.10}}}$
4. The rate of heat flow must be the same at all points between T₁ and T₂
• Otherwise, there will be accumulation of heat at some points
    ♦ Then heat at farther points will be lower than the expected quantities
• So equating the two rates, we get: $\mathbf\small{\rm{K_S 2A\frac{ (300 - T)}{0.15}=K_C A\frac{ (T - 0)}{0.10}}}$
⇒ $\mathbf\small{\rm{\frac{ (300 - T)}{(T-0)}=\frac{ 0.15 \times K_C}{2 \times 0.10 \times K_S}}}$
• Solving this equation, we get: T = 44.43 °C

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Now we will see two solved examples which demonstrates the general case when two conductors are connected in series. The calculations are written in pdf format at the link given below:

Solved example 11.29 and 11.30

■ From the above two solved examples, we get the following details:
• Two conductors having the same area, A and length, L are connected in series
• The temperature at the junction between the two conductors will be given by:
Eq.11.11: $\mathbf\small{\rm{T=\frac{K_1 T_1 + K_2 T_2}{K_1 + K_2}}}$
• Heat current through the composite bar is given by:
Eq.11.12: $\mathbf\small{\rm{H= \frac{A(T_1 - T_2)}{L \left(\frac{1}{K_1} + \frac{1}{K_2} \right)}}}$
• Thermal conductivity of the composite bar is given by:
Eq.11.13: $\mathbf\small{\rm{K'=\frac{2K_1 K_2}{K_1 + K_2}}}$

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In the next section, we will see heat transfer by convection and radiation



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