Question

Consider a power transistor that dissipates \(0.15 \mathrm{~W}\) 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 \(24-\mathrm{h}\) period, in \(\mathrm{kWh} ;(b)\) 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

a) Heat Dissipated in 24 hours: Approximately 3.6 x 10^-3 kWh b) Heat Flux on the surface of the transistor: Approximately 2.31 x 10^4 W/m² c) Surface Temperature of the transistor: Approximately 1286 K

Step-by-step solution

Calculate the Surface Area of the Transistor

Given the dimensions of the transistor, we can model it as a cylinder. The surface area of a cylinder can be calculated as follows: Surface Area = Area of top surface + Area of bottom surface + Area of side \(r = \frac{d}{2} = \frac{0.5 \mathrm{~cm}}{2} = 0.25 \mathrm{~cm}\) (radius) \(h = 0.4 \mathrm{~cm}\) (height) Surface Area \(= 2\pi r^{2} + 2\pi rh\) Convert cm to meters: \(r = 0.0025 \mathrm{~m}\) \(h = 0.004 \mathrm{~m}\) Surface Area \(= 2\pi (0.0025)^{2} + 2\pi (0.0025)(0.004)\) Surface Area \(\approx 6.50 \times 10^{-6} \mathrm{~m}^2\)

Calculate the Amount of Heat Dissipated in 24 Hours

We are given the power dissipated as 0.15 W. To find the total heat dissipated in 24 hours, we can use: Heat Dissipated (kWh) \( = \frac{Power (W) \times Time (h)}{1000}\) Heat Dissipated (kWh) \( = \frac{0.15 \times 24}{1000}\approx 3.6 \times 10^{-3} \mathrm{~kWh}\)

Calculate the Heat Flux on the Surface of the Transistor

The heat flux can be calculated using the formula: Heat Flux (W/m²) \(=\frac{Power (W)}{Surface Area (m^2)}\) Heat Flux (W/m²) \(= \frac{0.15}{6.50 \times 10^{-6}} \approx 2.31 \times 10^{4} \mathrm{ \frac{W}{m^2}}\)

Calculate the Surface Temperature of the Resistor

Finally, we can determine the surface temperature of the resistor using the heat transfer coefficient: Heat Flux (W/m²) \(= h(T_{surface} - T_{environment})\) \(2.31 \times 10^{4} = 18(T_{surface} - 30)\) \(T_{surface} = \frac{2.31 \times 10^{4}}{18} + 30 \approx 1286 \mathrm{~K}\) The surface temperature of the resistor is approximately 1286 K.