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# How To Calculate Heat Dissipation In Watts?

Heat dissipation is one of the deciding factors in designing heat transfer components. For example, we can use the heat dissipation capabilities to determine the effectiveness of a heat exchanger.

Using CFD, how can we calculate the amount of heat dissipated from a fluid?

## Solution

To determine how much heat a fluid loses (or gains) through a system, we can use the following equation:

$$Q = m C_p \Delta T \tag{1}$$

Where $$Q$$ ($$W$$) is the heat that the fluid loses/gains, $$m$$ ($$\frac{kg}{s}$$) is the mass flow rate of the fluid, $$C_p$$ ($$\frac{J}{kg.K}$$) is the specific heat of the fluid, and $$\Delta T$$ $$(K)$$ is the temperature difference between the outlet and the inlet.

In the next section, we will show how to use this formula, using a heat exchanger as an example.

## Expected Outcome

Let’s consider a Conjugate heat transfer (CHT) simulation using the shell and tube heat exchanger from Figure 1: Figure 1: Water, in the tube side, is dissipating heat from the hot air, in the shell side of the heat exchanger.

From equation 1, we know what information is necessary to calculate the heat dissipation on the hot fluid. The total air mass flow rate $$m$$ and the inlet air temperature $$T_{Inlet}$$ are provided as a boundary condition for the simulation. Figure 2: In case you are using a fixed value or a volumetric flow rate for the inlet, you can also calculate the mass flow rate.

Under the Materials tab, we can obtain the specific heat $$C_p$$ of the fluid: Figure 3: The specific heat represents how much heat is necessary to raise the temperature of 1 kg of a given substance by 1 Kelvin.

Before calculating the amount of heat that dissipates from the hot air, we need to determine the air outlet temperature. To obtain this information, we can set an Area average result control for the air outlet, and run the CHT simulation. Figure 4: With an area average result control, we can quickly obtain all parameters on specific faces.

Using equation 1 as a reference, the total heat dissipated ($$Q$$) from the hot air in this example is:

$$Q = 0.21 \times 1004 \times (335.58\ – 573.15) =\ – 50089\ W \tag{2}$$

Note that the value of $$Q$$ in equation 2 is negative since the hot air loses heat through the system.