Question

Consider a power transistor that dissipates \(0.2 \mathrm{~W}\) of power in an environment at \(30^{\circ} \mathrm{C}\). The transistor is \(0.4 \mathrm{~cm}\) long and has a diameter of \(0.5 \mathrm{~cm}\). Assuming heat to be transferred uniformly from all surfaces, determine \((a)\) the amount of heat this resistor dissipates during a \(24-h\) period, in kWh; \((b)\) the heat flux on the surface of the transistor, in \(\mathrm{W} / \mathrm{m}^{2}\); and \((c)\) the surface temperature of the resistor for a combined convection and radiation heat transfer coefficient of \(18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Short answer

Question: Calculate the surface temperature of a power transistor dissipating 0.2 W of power in an environment with a combined convection and radiation heat transfer coefficient of 18 W/m²·K. The transistor has a cylindrical shape with a length of 0.4 cm and a diameter of 0.5 cm. Answer: The surface temperature of the power transistor is approximately 30.27°C.

Step-by-step solution

Calculate the amount of heat dissipated during a 24-hour period

We are given that the power dissipation of the transistor is 0.2 W. To find the amount of heat dissipated during a 24-hour period, we can use the formula: Heat Dissipated = Power × Time We need to convert the time from hours to seconds: Time = 24 hours × 3600 seconds/hour = 86400 seconds Now, calculate the heat dissipated: Heat Dissipated = 0.2 W × 86400 seconds = 17280 Joules Now, we need to convert Joules to kWh. We know that 1 Joule = 2.778 × 10^{-7} kWh. Therefore: Heat Dissipated = 17280 Joules × 2.778 × 10^{-7} kWh/Joule ≈ 0.0048 kWh

Calculate the surface area of the transistor

The transistor has a cylindrical shape, with a length of 0.4 cm and a diameter of 0.5 cm. To find the surface area, we can use the formula for the surface area of a cylinder: Surface Area = 2πr(h + r) First, convert the length (h) and diameter to meters: Length (h) = 0.4 cm × 0.01 m/cm = 0.004 m Diameter = 0.5 cm × 0.01 m/cm = 0.005 m Now calculate the radius (r): Radius (r) = Diameter / 2 = 0.005 m / 2 = 0.0025 m Finally, find the surface area: Surface Area = 2π(0.0025 m)(0.004 m + 0.0025 m) ≈ 4.07 × 10^{-5} m²

Calculate the heat flux on the surface of the transistor

Given the power dissipation (0.2 W) and the surface area, we can now calculate the heat flux on the surface. The formula for heat flux is: Heat Flux = Power / Surface Area Substitute the values we found: Heat Flux = 0.2 W / (4.07 × 10^{-5} m²) ≈ 4915 W/m²

Calculate the surface temperature of the resistor

Now, we will find the surface temperature of the resistor using the given combined convection and radiation heat transfer coefficient (18 W/m²·K). We can use the formula: Heat Transfer = h × A × (T_surface - T_environment) Rearrange the equation to find T_surface: T_surface = Heat Transfer / (h × A) + T_environment The heat transfer is equal to the power dissipation: Heat Transfer = 0.2 W Substitute the known values: T_surface = 0.2 W / (18 W/m²·K × 4.07 × 10^{-5} m²) + 30°C ≈ 30.27°C The surface temperature of the resistor is approximately 30.27°C.

Key concepts

Power Dissipation

Power dissipation refers to the process of electrical energy being converted into heat energy in a component, such as a transistor. This is a common occurrence in electronic devices and can affect their efficiency and temperature.
For the exercise, the power dissipation of the transistor is given as 0.2 W. This means that continuously, during operation, the transistor dissipates or emits power in the form of heat at this rate. Over a 24-hour period, this amount becomes significant.
To calculate the total heat dissipated, you would use the formula: Heat Dissipated = Power × Time. With time converted into seconds (since we're often dealing with energy in Joules which is seconds-based) the calculation becomes straightforward. This result can further be converted from Joules to kilowatt-hours (kWh) for a more conventional energy unit, giving a clearer sense of energy consumption over day-long periods.

Heat Flux

When discussing heat transfer in materials, the term "heat flux" often comes up. Heat flux is the rate of heat energy transfer through a given surface per unit area. Think of it as how intensely the heat "flows" through a particular point on the surface of an object.
In the case of the power transistor, once the surface area of the cylindrical shape is determined, you can calculate the heat flux using the formula: Heat Flux = Power / Surface Area. This calculation provides a measure of how much power is being emitted per square meter, highlighting the intensity of the heat flow across the surface of the transistor.
The heat flux is a crucial piece of information when designing devices since high heat flux areas may require additional cooling solutions to prevent overheating.

Convection and Radiation

Convection and radiation are two primary modes of heat transfer that can occur between the transistor and its surrounding environment. Convection involves the transfer of heat through the movement of fluids or gases, while radiation involves the transfer of heat through electromagnetic waves.
The combined convection and radiation heat transfer coefficient in the problem is given as 18 W/m²·K. This coefficient provides a way to quantify the ability of the surroundings to absorb or dissipate heat per degree of temperature difference per square meter. It accounts for both how air moves around the transistor and how the transistor emits heat energy in the form of infrared radiation.
Understanding this coefficient helps in determining how efficiently the heat dissipated by the transistor can be removed by the environment. It's crucial for ensuring the component runs safely without excessive temperature buildup.

Surface Temperature

The concept of surface temperature is important because it determines how hot the object (like a transistor) is to the touch and can indicate potential overheating. In the given scenario, the surface temperature is calculated to approximate 30.27°C.
This calculated temperature involves understanding all parts of the heat balance equation: the power dissipation, the area through which the heat is transferred, the environmental temperature, and the heat transfer coefficient.
Using the relationship: \[ T_\text{surface} = \frac{\text{Heat Transfer}}{h \times A} + T_\text{environment} \]where \(h\) is the convection and radiation heat transfer coefficient, \(A\) is the area, and \(T_\text{environment}\) is the ambient temperature, this formula rearranges to find the surface temperature of the transistor. This value gives an insight into potential thermal stresses and the need for further cooling mechanisms.