Question

Four power transistors, each dissipating \(12 \mathrm{~W}\), are mounted on a thin vertical aluminum plate \(22 \mathrm{~cm} \times 22 \mathrm{~cm}\) in size. The heat generated by the transistors is to be dissipated by both surfaces of the plate to the surrounding air at \(25^{\circ} \mathrm{C}\), which is blown over the plate by a fan. The entire plate can be assumed to be nearly isothermal, and the exposed surface area of the transistor can be taken to be equal to its base area. If the average convection heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the aluminum plate. Disregard any radiation effects.

Short answer

Answer: The temperature of the aluminum plate is approximately 44.84°C.

Step-by-step solution

Calculate the total heat dissipated by the transistors

There are 4 transistors, each dissipating 12 W of power. Therefore, the total heat generation (Q) can be found by multiplying the power dissipation of each transistor by the number of transistors, which is given by: $$Q = 4 \times 12 \mathrm{~W} = 48 \mathrm{~W}$$

Convert the plate dimensions to meters

The dimensions of the aluminum plate are given in centimeters. We need to convert them to meters to use in the convection heat transfer equation. The conversion factor is 1 meter = 100 centimeters. Therefore, the dimensions of the aluminum plate in meters are: $$0.22 \mathrm{~m} \times 0.22 \mathrm{~m}$$

Calculate the surface area of the plate

Since heat is dissipated by both surfaces of the plate, we need to calculate the surface area of both sides to use in the convection heat transfer equation. The surface area (A) of one side of the plate is given by: $$A = 0.22 \mathrm{~m} \cdot 0.22 \mathrm{~m} = 0.0484 \mathrm{~m}^2$$ The surface area of both sides of the plate is then: $$2 \times A = 2 \times 0.0484 \mathrm{~m}^2 = 0.0968 \mathrm{~m}^2$$

Apply the convection heat transfer equation and solve for the temperature difference

The convection heat transfer equation is given by: $$Q = h \cdot A \cdot \Delta T$$ Where Q is the heat dissipation, h is the convection heat transfer coefficient, A is the surface area, and \(\Delta T\) is the temperature difference between the plate and the surrounding air. We want to find the \(\Delta T\). Rearranging the formula gives us: $$\Delta T = \frac{Q}{h \cdot A}$$ Now, we can plug in the given values: Q = 48 W, h = 25 W/m²·K, and A = 0.0968 m². $$\Delta T = \frac{48 \mathrm{~W}}{(25 (\mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K})) \times (0.0968 \mathrm{~m}^2)} \approx 19.84 ^{\circ}\mathrm{C}$$

Calculate the aluminum plate temperature

Now that we have the temperature difference between the aluminum plate and the surrounding air, we can determine the aluminum plate temperature by adding this temperature difference to the surrounding air temperature. The surrounding air temperature is given as \(25^{\circ} \mathrm{C}\). Therefore, the aluminum plate temperature can be calculated as follows: $$T_\text{plate} = T_\text{air} + \Delta T = 25^{\circ} \mathrm{C} + 19.84^{\circ} \mathrm{C} \approx 44.84^{\circ} \mathrm{C}$$ The temperature of the aluminum plate is approximately \(44.84^{\circ} \mathrm{C}\).

Key concepts

Aluminum Plate Heat Dissipation

Heat dissipation is a critical concept in managing the temperature of electronic devices. In our scenario, a thin aluminum plate absorbs the heat generated by four power transistors. Aluminum is a popular material for heat dissipation due to its high thermal conductivity, which allows it to effectively spread heat across its surface.
This plate has dimensions of 22 cm x 22 cm, converted into meters as 0.22 m x 0.22 m for calculations. The heat dissipated by such structures is crucial for the longevity and stability of electronic components. When the plate absorbs the heat from the transistors, it transfers this energy to the surrounding air, mitigating potential overheating.
Here, understanding the surface area of both sides of the plate is vital. Calculating this gives us a total surface area of 0.0968 m². This area is critical for analyzing how effectively the plate can dissipate heat into the environment.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient, often denoted as "h," measures how efficiently heat is transferred from a surface to a fluid. In this problem, the fluid is air, and the heat is transferred from the aluminum plate to this air.
In essence, the higher the convection heat transfer coefficient, the more effective the heat dissipation from the plate. This coefficient depends on various factors, including the type of fluid (air in this case), the flow condition (forced or natural convection), and the surface characteristics of the plate.
In our scenario, the given coefficient is 25 W/m²·K. This means for every square meter of the plate, there's a potential for 25 watts of heat to be transferred for every degree of temperature difference between the plate and the air.

Power Transistor Heat Management

Power transistors are essential components in many electronic devices, often generating significant amounts of heat during operation. Effective heat management for power transistors is critical to prevent performance degradation or damage.
In the exercise, each transistor dissipates 12 W, totaling 48 W for all four transistors. Managing this heat involves transferring it away efficiently, which is achieved using the aluminum plate as a heat sink. By spreading and dissipating the heat through convection, the plate helps maintain the transistors' operating temperatures within safe limits.
Effective heat management requires understanding the balance between heat generated by the transistors and the capacity of the surrounding systems, like our aluminum plate, to dissipate that heat to the environment. This balance ensures that the device continues to operate efficiently and safely.

Isothermal Assumption in Heat Transfer

An isothermal assumption helps simplify heat transfer calculations by assuming that an object's temperature remains consistent across its surface. This is especially useful in calculations involving heat dissipation, like in the case of our aluminum plate.
In this problem, the plate's isothermal nature is assumed, meaning its temperature is uniform across its entire surface. This assumption holds mostly when the plate is thin and has a high thermal conductivity, like aluminum.
This simplification allows us to use straightforward methods to calculate the heat transfer, as applied in our convection heat transfer equation. By making these assumptions, it becomes feasible to determine the overall temperature of the plate relative to the surrounding environment efficiently.