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alberand
alberand
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The energy loss causecaused by Thermal Resistance

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alberand
alberand
  • 103
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The absolute thermal resistance can be expressed in the following formula:

$$R = \frac{\Delta T}{P} \left[\frac{K}{W}\right]$$

It shows how big is the temperature difference between two points in the object per watt of applied energy. So, for known R = 3 K/W and P = 2 W we will get 6 K drop.

What happens with this power loss (temperature)? In case of electric analogy the lost currentlost current voltage drop (energy loss) goes into heat, but it is not clear for me where this energy goes to. Or is it only mathematical model which is used only in calculations?

The absolute thermal resistance can be expressed in the following formula:

$$R = \frac{\Delta T}{P} \left[\frac{K}{W}\right]$$

It shows how big is the temperature difference between two points in the object per watt of applied energy. So, for known R = 3 K/W and P = 2 W we will get 6 K drop.

What happens with this power loss (temperature)? In case of electric analogy the lost current (energy) goes into heat, but it is not clear for me where this energy goes to. Or is it only mathematical model which is used only in calculations?

The absolute thermal resistance can be expressed in the following formula:

$$R = \frac{\Delta T}{P} \left[\frac{K}{W}\right]$$

It shows how big is the temperature difference between two points in the object per watt of applied energy. So, for known R = 3 K/W and P = 2 W we will get 6 K drop.

What happens with this power loss (temperature)? In case of electric analogy the lost current voltage drop (energy loss) goes into heat, but it is not clear for me where this energy goes to. Or is it only mathematical model which is used only in calculations?

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alberand
alberand
  • 103
  • 4

The energy loss cause by Thermal Resistance

The absolute thermal resistance can be expressed in the following formula:

$$R = \frac{\Delta T}{P} \left[\frac{K}{W}\right]$$

It shows how big is the temperature difference between two points in the object per watt of applied energy. So, for known R = 3 K/W and P = 2 W we will get 6 K drop.

What happens with this power loss (temperature)? In case of electric analogy the lost current (energy) goes into heat, but it is not clear for me where this energy goes to. Or is it only mathematical model which is used only in calculations?