## What is Thermal time constant?

**The thermal time constant is the time taken by the system to undergo temperature change equal to 63.2% of the initial temperature difference between the system and its surrounding. It can be considered for the system undergoing cooling or heating.**

In other words, it is time taken by the system to attain following condition,

Temperature change in system = 63.2% (Initial difference between T_{system} & T_{surrounding})

Let’s discuss this with an example.

In the below figure, the solid body is subjected to cooling.

Consider, initially the body is at temperature T_{i} and the surrounding is at T_{∞} temperature. After the time interval τ_{th} seconds, consider the temperature of the body as T.

Thus `\tau_{th}` is the time interval, when,

`\mathbf{T_(\text{i}} – T = 63.2% \text{of} (T_{\text{i}} – T_{\infty})}`

OR

`\mathbf{T_(i} – T = 0.632\times(T_{\text{i}} – T_{\infty})}`

It means that, at τ_{th}, the system covers around 63.2% of the initial temperature difference.

The thermal time constant is important to understand the response of a system to the surrounding temperature changes.

The thermal time constant is denoted by the symbol ‘τ_{th}‘ and has an SI unit as seconds (s).

**In this article, we’re going to discuss:**

- Temperature relation at τth for cooling and heating:
- Relation between `\DeltaT_{t=\tau_{th}}` and `\DeltaT_{\text{initial}}`:

- Thermal time constant equation:
- Significance of thermal time constant:
- Relation between temperature difference at `t=\tau_{th}`:
- Solved Numerical:
- FAQ’s:

## Temperature relation at τ_{th} for cooling and heating:

In the case of cooling, the below temperature Vs time graph shows the temperature relation at a time interval of τ_{th}.

For cooling, the temperature relation after a time interval of τ_{th} is given by,

`T_{i} – T = 63.2%(T_{i} – T_{\infty})`

In the case of heating, the below temperature Vs time graph shows the temperature relation at a time interval of τ_{th}.

For heating, the temperature relation after a time interval of τ_{th} is given by,

`T – T_{i} = 63.2%(T_{\infty} – T_{i})`

### Relation between `\DeltaT_{t=\tau_{th}}` and `\DeltaT_{\text{initial}}`:

At τ_{th}, the relation between the final temperature difference `\DeltaT_{t=\tau_{th}}` (between the system and its surrounding) and the initial temperature difference `\DeltaT_{\text{initial}}` is given by,

`\DeltaT_{t=\tau_{th}} = (1 – 0.632)\DeltaT_{\text{initial}}`

`\DeltaT_{t=\tau_{th}} = 36.7%\DeltaT_{\text{initial}}`

## 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

A_{s} = Surface area of the body

The above factors decide the thermal time constant of the object or system.

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

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

## Significance of thermal time constant:

It has following significances:

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

## Relation between temperature difference at `t=\tau_{th}`:

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

`\frac{T-T_{\infty}}{T_{i}-T_{\infty}}=e^{-\frac{hA_{s}}{\rho VC}t}`

`\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_{t = \tau_{th}}=0.367\Delta T_{\text{intial}}`

`\therefore \Delta T_{t = \tau_{th}}=36.7\%\Delta T_{\text{intial}}`

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 covers a temperature difference equal to 63.2% of the initial temperature difference.

## Solved Numerical:

1] Find the thermal time constant for a spherical object with a radius of 2.5 mm. (Assume h = 350 W/m^{2}°C, K = 25 W/m°C. density = 7600 Kg/m^{3}, specific heat = 300 J/Kg°C)

**Given:**

R = 2.5 mm = `2.5 \times 10^{-3}` m

h = 350 W/m^{2}°C

K = 25 W/m°C

`\rho` = 7600 Kg/m^{3}

C = 300 J/Kg°C

**Solution:**

The surface area of the spherical object,

`A_{S} = 4\piR^{2} = 4\pi\times(2.5 \times 10^{-3})^{2}`

`A_{S} = 7.853\times10^{-5}`m^{2}

Volume of the spherical object,

`V = \frac{4}{3}.\pi.R^{3}`

V = `\frac{4}{3}.\pi.(2.5 \times 10^{-3})^{3}`

V = `6.544\times10^{-8}` m^{3}

The thermal time constant is given by,

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

`\tau_{th}=\frac{7600\times(6.544\times10^{-8})\times300}{350\times(7.853\times10^{-5})}`

`\tau_{th}` = 5.428 seconds

2] An object subjected to cooling is initially at 300°C and its surroundings are at 50°C. Find the temperature of the object at the time equal to 3τ_{th}.

**Given:**

Ti = 300°C

T_{∞} = 50°C

t = 3τ_{th}

**Solution:**

The temperature of the object after a time interval of 3τ_{th} is given by,

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

`\frac{T-50}{300-50}=e^{-\frac{3\tau}{\tau}}`

`\frac{T-50}{300-50}=e^{-3}`

T = 62.44 °C

Thus, after the interval of 3τ_{th}, the temperature of the object reaches 62.44 °C.

## FAQ’s:

Which factors affect the thermal time constant?

The thermal time constant depends on the following factors:

Density of system, Volume, Surface area, Convective heat transfer coefficient, Specific heat of the material.