Thermal conductivity is the ability of a given material to conduct or transfer heat. It is generally denoted by the symbol ‘k’ or sometimes\( \lambda\). The reciprocal of this physical quantity is referred to as thermal resistivity. Materials with high thermal conductivity are used in heat sinks, on the other hand, materials with low values of \(\lambda\) used as thermal insulators. Learn the thermal conductivity formula here.
Source: Wikipedia
Thermal Conductivity Formula
What is Thermal Conductivity?
Fourier’s law of thermal conduction also known as the law of heat conduction is very relevant for heat transfer computation. This principle is applicable for heat transfer between two isothermal planes.
It states that the rate at which heat is transferred through a given material is proportional to the negative value of the temperature gradient. And it is also proportional to the area through which the heat flows, but inversely proportional to the distance between the two isothermal planes.
The Formula for Thermal Conductivity
Every substance has its own capacity for conducting and transferring the heat. The thermal conductivity of a material is explained by the following formula:
K =\( \frac{ Q d}{A \Delta T} \)
K  is the thermal conductivity in Watt \( m^{1}K^{1} \)

Q  is the amount of heat transferred through the material in Joules/second or Watts

d  is the distance between the two isothermal planes

A  is the area of the surface in square meters

\( \Delta T i.e. T_{HOT} – T_{COLD} \)  is the difference in temperature

Also, the above formula can be rearranged to give the value of transfer of heat, as follows:
Q= \( \frac {K \times A \times(T_{HOT} – T_{COLD})}{d} \)
The SI unit of this quantity is watts per meterKelvin or Wm^{1}K^{1}. These units will describe the rate of conduction of heat through the material having the unit thickness and for each Kelvin of temperature difference.
Solved Examples
Q.1: Compute the heat transfer amount for a material in which thermal conductivity is 0.181. The crosssectional area is 1200 square m and a thickness of 2m. The hot temperature is 250degree c where the cold temperature is 25 degrees C.
Solution:
As given in the problem,
Thermal conductivity of material, K =0.181
Cross Sectional Area, A = 1200
Thickness, d = 2 m
Hot side temperature, \( T_{HOT} \) = 250 degrees C
Cold side temperature, \( T_{COLD}\) = 25 degrees C
Now, applying the formula,
Conduction Heat Transfer (W): 24435
Q= \( \frac {K \times A \times(T_{HOT} – T_{COLD})}{d}\)
Substituting the known values,
Q= \( \frac {0.181 \times 1200 \times(250 – 25) }{2} \)
Q = \( \frac {K \times A \times(T_{HOT} – T_{COLD})}{d} \)
= \( \frac {0.181 \times 1200 \times(250 – 25)}{2}\)
= \( \frac {48870}{2} \)
= 24435 Watt
Therefore the heat transfer will be 24435 watt.
Q.2: Compute the heat transfer amount for a material in which thermal conductivity is 0.5. The crosssectional area is 1200 square m and a thickness of 0.2m. The hot temperature is 58degree c where the cold temperature is 12 degrees C.
Solution: Given values are,
Thermal conductivity of the material (Watt per mK)
K =0.5
CrossSectional Area (m2) A = 1200
Thickness (m) d = 0.2 m
Hot side temperature = 58 degrees C
Cold side temperature = 12 degrees C
By applying the above formula we will get Heat Transfer = 27025 Watt.
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