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}`

`(T-T_{\infty})=0.6321(T_{i}-T_{\infty})`

Thermal time constant equation:

The thermal time constant is given by the following equation,

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

Where,

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.

The SI unit of the thermal time constant is second.
By using a thermal time constant the lumped system equation is also rewrite as

`\frac{T-T_{\infty}}{T_{i}-T_{\infty}}=e^{-\frac{t}{\tau}}`

Significance of thermal time constant:

  1. It indicates response of the system or object to the change in surrounding temperature.
  2. The system with lower time constant takes less time to achieve temperature change.
  3. The system with 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 equals to `\tau_{th}` in the equation of lumped system analysis.

`\frac{T-T_{\infty}}{T_{i}-T_{\infty}}=e^{-\frac{t}{\tau}}`

`\frac{T-T_{\infty}}{T_{i}-T_{\infty}}=e^{-1}`

`\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 equals 63.2% of the initial temperature gradient.

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