Conduction Heat Transfer : Everything You need to Know

This image shows a Man wearing thermal insulating gloves to protect himself from heat due to conduction heat transfer From hot Plate

While holding hot objects, it feels hot. This phenomenon occurs because of the transfer of heat from hot objects to our body due to “Conductive heat Transfer”. This rate of conductive heat transfer depends on a material thermal properties, temperature difference and area of contact. In this article we will discuss What is conduction Heat Transfer? and how to calculate Conduction heat Transfer? We suggest you also read this article on various modes of heat transfer.

Transfer of heat within thermally conductive body or between thermally contacted bodies due to temperature difference is known as conductive heat transfer.
Example of Conductive Heat Transfer in a Conductive Material

Conduction Heat Transfer is the transfer of heat within thermally conductive or thermally contacted bodies due to temperature difference. It occurs in solid bodies due to molecular excitement or lattice vibrations of free electrons. Whereas in liquids and gases conductive heat transfer occurs due to molecular movements and collision respectively.

Examples of Conduction Heat Transfer

Following are the examples of conduction heat transfer in products that we use in our day to day life.

  1. Transfer of heat from burner to metal pan.
  2. Hands gets burned if we touch hot surfaces due to conductive heat transfer from hot surfaces to our skin.
  3. Ironing of clothes.
  4. Transfer of heat from heat dissipating electronics components to the heatsink. Click this Link!! to know more about transfer of heat in a heat sink.
  5. Car engine gets hot when turned on.
  6. When hot coffee is poured into a cup, the cup becomes hot.

Fourier’s Law (Conduction Heat Transfer Calculation)

Rate of conduction heat transfer is directly proportional to the contact area, material thermal conductivity, temperature difference and inversely proportional to the thickness.

According to Fourier’s Law: Rate of conductive heat transfer is directly proportional to the contact area, material thermal conductivity, temperature difference and inversely proportional to the thickness.

Conduction heat Transfer Equation

Conductive Heat transfer (Qc) = – K A (dT / L)

Where

Qc = Conductive heat transfer per unit time in Watt

A = cross section area in square meter

k = Material thermal conductivity (W/mK)

dT = Temperature differential

L = Thickness or length of the part in meter

Where (dT / L) is known as temperature gradient. It’s value is always negative because the temperature gradient vector is in the opposite direction of heat flow. Conduction heat transfer is more significant in solid bodies and it follows the first law of thermodynamics.

Thermal Conductivity of a Material

Rate of conduction heat transfer is directly proportional to material thermal conductivity. For example, considering all other parameters equal the rate of conductive heat transfer will be higher in copper compared to plastic material. Because copper has a higher thermal conductivity than aluminum.

Thermal Conductivity (These values are for Indicative Purpose Only)
Material Thermal Conductivity (W/mK)
Copper 384
Aluminum 205
Iron 79
Water 0.6
Air 0.026
Silicon 0.01
Diamond 895
Bakelite 0.14

Thermal Insulation

Thermal insulation is a property of a material that is inversely proportional to the thermal conductivity. In other words, higher the thermal insulation, lower will be the thermal conductivity. Therefore insulation materials are used to reduce conductive heat transfer.

Insulator material are used to hold hot objects

To reduce the transfer of heat from one material to another, Low thermal conductivity or high insulation materials are used. For example, heating pan handle is manufactured in Bakelite material. Because bakelite has high thermal insulation properties.

Relationship in thermal and electrical Conductivity of a Material

There is no direct relationship between electrical and thermal conductivity of a material. It is not mandatory a good conductor of electricity will be a good conductor of heat as well. For example, diamond is a good conductor of heat but a bad conductor of electricity.

Conductive Heat Transfer Calculator

Conductive Heat Transfer Problem Examples

Question: Calculate the difference between the rate of heat transfer from 1000 mm² area and 2mm thickness Bakelite and Aluminum wall considering 80°C temperature difference.

Answer: 

Temperature difference between two walls (dT) = 80°C

Wall thickness (L) = 2 mm = 0.002 mm

Cross Section Area (A) = 1000 mm² = 0.001 Meter²

Bakelite thermal conductivity (k1) = 0.14 W/mK

Aluminum thermal conductivity (K2) = 205 W/mK

Rate of Heat Transfer (Bakelite) = -KA (dT/L) = 0.14 × 0.001 ×  (80/0.002) = 5.6 Watt

Rate of Heat Transfer (Aluminum) = 205 × 0.001 × (80/0.002) = 8200 Watt

Therefore the rate of heat transfer in aluminum is higher than Bakelite. You can also use the above calculator for this calculation.

We will keep adding more information on conductive heat transfer and how to calculate it. Please add your suggestions, comments or questions on conduction heat transfer in the comment box. We suggest you also read these articles on convection and radiation heat transfer.

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