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Thermal conductivity can be defined as the ability of a material to transfer heat energy from a region at a higher temperature to a region at a lower temperature.
- The two words “Thermal” means heat, and “Conductivity” means to conduct or convey.
- The SI unit of the compound is watts per meter kelvin (W/(m·K)).
- Suppose you are burning a potato in a campfire and accidentally touch the potato to the flame, is a real-life example of thermal conductivity.
Key Terms: Thermal Conductivity, Heat, Temperature, Thermal Conductivity Formula, Conductors, Thermal Resistivity, Conductivity
Key Highlights
- Thermal Conductivity is a property of materials that has ability to conduct heat.
- It is affected by temperature, length of the material and type of cross-section.
- The reciprocal of this form of conductivity is known as thermal resistance.
- Materials having higher thermal conductivity are considered good conductors of thermal energy.
- It is a scalar quantity that is also called Fourier's Law for heat conduction.
What is Thermal Conductivity?
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Thermal conductivity states that the rate of heat transfer through a material is proportional to the negative gradient in temperature and to the area. It is denoted by k and also λ’ and κ’ in some cases.
- If the materials have higher thermal conductivity, they are considered good conductors of heat.
- Meanwhile, materials that have lower conductivity are considered bad conductors of heat.
- The base unit for thermal conductivity is kg⋅m⋅s−3⋅K-1.
- It is a microstructure-sensitive property, which depends on the impurities and imperfections of the structure.
- The value of thermal conductivity is in the range of 20 – 400 for metals, 2 – 50 for ceramics and 0.3 for polymers.
- Reciprocal of thermal conductivity is called thermal resistivity.
Thermal Conductivity Formula
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Thermal conductivity formula for conduction and transfer of heat is explained by Joseph Fourier. It is also known as Law of Heat Conduction or Fourier's Law of Thermal Conduction.
Consider that heat flows through a uniform rod. The rate of heat flow ΔQ/ ΔT is proportional to the ' A' cross sectional area of rod and ‘ΔT/ Δx' temperature gradient.
- k is the constant of proportionality and thermal conductivity.
- Here the heat flows down a temperature gradient, so a negative sign is introduced.
ΔQ/ ΔT = -kA ΔT/ Δx
q / A = -k. dT/ dx
k = QL / A ΔT
- Where, q / A is heat flux
- K is thermal conductivity of material
- dT / dx is Temperature difference
- Q is amount of heat transferred through the material
- L is distance between two isothermal places of a material
- A is the cross sectional area of surface
- ΔT is the difference in temperature
Measurement of Thermal Conductivity
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Several methods have been adopted to measure the thermal conductivity of materials. They are classified into two types: Steady State Method and Transient State Method.
Steady State Method
The steady state method measures thermal conductivity when the materials are in equilibrium, i.e., when the temperature is constant. Searle's bar method and Lee's dice method are two examples of steady state methods used for calculating thermal conductivity.
Advantages
The advantages of the steady state method are as follows:
- The readings obtained are accurate.
- Since the material is in a steady state, it infers constant signals.
Disadvantages
The disadvantages of the steady state method are as follows:
- Materials take a longer period of time to reach the equilibrium state.
- A well-equipped set-up is required to perform the experiment.
Transient State Method
The transient state method is used to measure thermal conductivity when the materials are heated, i.e., there is a change in temperature. Since the temperature is not constant and changes, it is called the Transient State Method.
- The plane source method, transient line source method, and laser flash method are some examples of transient state methods used to calculate thermal conductivity.
Advantages
The advantages of the transient state method are as follows:
- Readings can be obtained while heating the material.
- Since the material is in a non-steady state, it infers non-constant values.
- The readings can be taken comparatively faster.
Disadvantages
The disadvantages of the transient state method are as follows:
- The readings are not accurate.
- Analysis of data is difficult.
Factors affecting Thermal Conductivity :
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Various factors affecting thermal conductivity are as follows:
Temperature
Conducting heat implies the movement of electrons within the material. If the temperature is increased, thermal conductivity in the metals decreases. So, a little variation in the temperatures exhibits a difference in thermal conductivity.
- The value of thermal conductivity ranges from 2000 to 10,000 in the case of pure metals.
- It is completely in contrast when observed in non-metals.
- If the temperature is decreased, thermal conductivity also decreases in non-metals.
Chemical Phase of a Material
A change of phase in a material implies that there is a change in thermal conductivity. So, along with the phase change, the value of thermal conductivity also rises.
Length of Material
The rate of heat flow depends on the length of the material. The longer the length, the slower the flow rate. However, the thermal conductivity increases with the increase in length at a slower pace.
Types of Cross Section:
Round, C and hollow shapes of the materials also affect the value of thermal conductivity. Thermal conductivity is higher in the C and hollow-shaped materials when compared to the round-shaped materials.
Sample Questions
Ques: The thermal conductivity of a material in the CGS system is 0.4. In steady state the rate of flow of heat 10 cals−1, then the thermal gradient will be? (4 marks)
Ans: The differential form of Fourier’s law of thermal conduction can be expressed through the following equation:
Q=−K⋅∇T
The thermal conductivity of a material is expressed using the following formula:
K=QL /AΔT
Re-writing the differential form for the law of heat conduction, we get.
ΔQ / Δt=KAΔT / Δx
Thermal gradient is given as,
ΔT/ x =ΔQ / Δt / KA
ΔT / x=1 / KA(ΔQ / Δt)
Where,
ΔQ / Δt is the rate of heat flow
KA is the thermal conductivity of the material
Putting values,
ΔQ / Δt=10 cals−1
KA=0.4
We get,
ΔT / x=100.4=25°Ccm−1
The thermal gradient of the material in steady state is 25°Ccm−1
Ques: Consider a window of width 1.3m and height 1.7m, having a thickness of 6.3mm and a thermal conductivity value of 0.28 W/m/degree C. The temperature inside the house is 22 degrees Celsius and that outside the house -5 degrees Celsius. Calculate the rate of heat transfer? (3 marks)
Ans: Surface Area of the window = length x breadth = (1.3 x 1.7) m2 = 2.21 meter square
The thickness of the window = 6.3 mm = 0.0063 m
Rate of heat transfer = (Thermal conductivity) x (Surface area) x (Difference of the temperatures) / (Thickness of the window)
= (0.28 W/m/degree C) x (2.21 meter square) x (22-(-5) degrees C) / (0.0063 m) ~ 2652 W
The rate of heat transfer for the given example is 2652 W.
Ques: The thermal conductivity of copper is four times that of brass. Two rods of copper and brass having the same length and cross-section are joined end to end. The free end of the copper is at 0°C and the free end of brass is at 100°C. The temperature of the junction is? (4 marks)
Ans: Thermal conductivity is given by,
dQ / dt=KAΔT / L
dQ / dt=KAΔT / L
Therefore, Thermal conductivity of brass can be given by,
dQ / dt=KA(100−T) / L
dQ / dt=KA(100−T) / L …(1}
Similarly, the thermal conductivity of copper can be given as:
dQ / dt=4KA(T−0)/ L
dQ / dt=4KA(T−0) / L …(2}
Equating equation. (1) and equation. (2) we get,
KA(100−T) /L=4KA(T−0) / L
KA(100−T) / L=4KA(T−0) / L
⇒100−T=4T⇒100−T=4T
⇒5T=100⇒5T=100
⇒T=20°⇒T=20°
Hence, the temperature of the junction is 20°C.
Ques: Calculate the amount of heat transfer for a material having thermal conductivity is 0.181. The cross-sectional area of the material is 1200 m2 and thickness of the material is 2m. The hot temperature is 250 ? where the cold temperature is 25?. (4 marks)
Ans: Given, thermal conductivity of the material, K =0.181
Cross Sectional Area, A = 1200
Thickness, d = 2 m
Hot side temperature, THOT = 250 degrees C
Cold side temperature, TCOLD = 25 degrees C
Now, applying the formula,
Conduction Heat Transfer (W): 24435
Q= K×A×(THOT–TCOLD) / d
Substituting the known values,
Q= 0.181×1200×(250–25) / 2
Q = K×A×(THOT–TCOLD) / d
= 0.181×1200×(250–25) / 2
= 488702
= 24435 WWat
Therefore the heat transfer will be 24435 watt
Ques: What are the applications of thermal conductivity? (3 marks)
Ans: The applications of thermal conductivity are as follows:
- Observed in aluminum sheets used in houses to block the incoming and outgoing of air to keep warmth inside.
- Materials like graphene and diamonds are used recently to transfer heat.
- Elements like copper, steel can be used in electric circuits for conductivity.
- Carbon nanotubes are found to have good thermal conductivity.
- Hence they can be used as electrodes in batteries and capacitors.
Ques: What is Fourier's law of thermal conduction? (2 marks)
Ans: Fourier's law of thermal conduction is an important principle in heat transfer. It describes the relationship between the heat transfer rate through a material and the temperature difference across that material. The law states that the heat transfer rate through a material is proportional to the negative gradient of temperature.
Ques: Consider a window of width 2.0m and height 2.5m, having a thickness of 6.3mm and a thermal conductivity value of 0.28 W/m/degree C. The temperature inside the house is 22 degrees Celsius and that outside the house -2 degrees Celsius. Calculate the rate of heat transfer? (3 marks)
Ans: Surface Area of the window = length x breadth = (2.0 x 2.5) m2 = 5 meter square
The thickness of the window = 6.3 mm = 0.0063 m
Rate of heat transfer = (Thermal conductivity) x (Surface area) x (Difference of the temperatures) / (Thickness of the window)
= (0.28 W/m/degree C) x (5 meter square) x (22-(-2) degrees C) / (0.0063 m) ~ 5333 W
The rate of heat transfer for the given example is 5333 W.
Ques: How thermal conductivity is affected by anisotropic behavior? (2 marks)
Ans: In anisotropic materials, thermal conductivity will have different values depending on the direction of heat flow. It also depends on the orientation of the reinforcing fibers or layers within the material. For example, heat might flow much more readily in one direction compared to another. The internal structure, composition, or grain boundaries within the material can cause this change in conductivity.
Ques: Hot air at 100°C flows over a flat plate maintained at 50°C. If the forced convection heat transfer coefficient is 70W/m2K, the heat gain rate by the plate through an area of 5m2will be? (2 marks)
Ans: Given that,
h = 70 W/m2K
Tp = 50°C,
T∞ = 150°C,
A = 5m2
Q = h x A x ΔT = hA (T∞–Tp)
Q = 70 × 5 × (100-50) = 17500 W
= 17.5 kW
Ques: Compute the heat transfer amount for a material in which thermal conductivity is 0.50. The cross-sectional area is 1000 square m and a thickness of 5m. The hot temperature is 200-degree c where the cold temperature is 20 degrees C? (3 marks)
Ans: Given that Thermal conductivity of material, K =0.50
Cross Sectional Area, A = 1000
Thickness, d = 5 m
Hot side temperature = 200 degrees C
Cold side temperature = 20 degrees C
Substituting the known values,
Q= 0.50×1000×(200–20)/5
= 18000 Watt
Ques: Hot air at 10°C flows over a flat plate maintained at 20°C. If the forced convection heat transfer coefficient is 60W/m2K, the heat gain rate by the plate through an area of 10m2will be? (2 marks)
Ans: Given that,
h = 60 W/m2K
Tp = 10°C,
T∞ = 20°C,
A = 10m2
Q = h x A x ΔT = hA (T∞–Tp)
Q = 60 × 10 × (20-10) = 6000 W
= 6 kW
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