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
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| 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 m3 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
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 TC
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 TD
■ TC is greater than TD
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 (TC - TD) becomes zero, the flow of heat will stop
7. As long as TC and TD 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 TC
(ii) The end B is in contact with reservoir D
♦ So the temperature at end B will be TD
(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 (TC - TD)
♦ 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 = $\mathbf\small{\rm{W \, m^{-1} \, K^{-1}}}$
Now we will see four solved examples. They are given in pdf format at the link below:
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, KS = 50.2 J s-1 m-1 K-1
• K of copper, KC = 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
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: $\mathbf\small{\rm{K'=\frac{2K_1 K_2}{K_1 + K_2}}}$
In the next section, we will see heat transfer by convection and radiation
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