Thermal time constant in heat transfer: Definition, Unit, Formula [with Pdf]

What is Thermal time constant?

For the body undergoing cooling or heating, the thermal time constant is the time to reach the temperature gradient equals 63.21 % of the initial temperature gradient.

The thermal time constant is denoted by the symbol `\tau_{th}`.

E.g.:- If the object with initial temperature Ti is cooled by the surrounding medium at temperature `T_{\infty}` then at the time equals to the thermal time constant the relation between temperature is given by,

`\Delta T=0.6321\times\Delta T_{INITIAL}`


The SI unit of the thermal time constant is second.

Thermal time constant equation:

The thermal time constant is given by the following equation,

`\tau_{th}=\frac{\rho VC}{hA_{S}}`

ρ = Density of the body
V = Volume of the body
C = Specific heat of an object
h = Convective heat transfer coefficient
As = Surface area of the body

Therefore thermal time constant depends on the above factors.

By using a thermal time constant, the lumped system equation is also rewrite as


Significance of thermal time constant:

  1. It indicates the response of the system or object to the change in the surrounding temperature.
  2. The system with a lower time constant takes less time to achieve temperature change.
  3. The system with a higher value of time constant takes much time to achieve temperature change.

Relation between temperature gradient at `t=\tau_{th}`

To check the relation between temperature gradient at `t=\tau_{th}`, put the time interval equal to `\tau_{th}` in the equation of lumped system analysis.



`\therefore (T-T_{\infty})=0.367(T_{i}-T_{\infty})`

`\therefore \Delta T_{AFTER}=0.367\Delta T_{INITIAL}`

`\therefore \Delta T_{AFTER}=36.7\%\Delta T_{INITIAL}`

From the above equation, it is clear that after the interval of `t=\tau_{th}` the temperature difference between the system and surroundings is equal to 36.7% of the initial temperature difference.

Therefore at `t=\tau_{th}`, the system achieves a temperature gradient equal to 63.2% of the initial temperature gradient.

Pratik is a Graduated Mechanical engineer. He enjoys sharing the engineering knowledge learned by him with people.

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