Question

A cylindrical resistor element on a circuit board dissipates \(0.8 \mathrm{~W}\) of power. The resistor is \(2 \mathrm{~cm}\) long and has a diameter of $0.4 \mathrm{~cm}$. Assuming heat to be transferred uniformly from all surfaces, determine \((a)\) the amount of heat this resistor dissipates during a \(24-\mathrm{h}\) period, \((b)\) the heat flux, and \((c)\) the fraction of heat dissipated from the top and bottom surfaces.

Short answer

Answer: The total heat dissipated during a 24-hour period is 69,120 J, the heat flux is approximately 3634.63 W/m², and about 9.09% of the heat is dissipated from the top and bottom surfaces combined.

Step-by-step solution

Calculate the total heat dissipated during a 24-hour period

To find the total heat dissipated during a 24-hour period, we need to use the power equation, \(P = E/t\), where \(P\) is the power, \(E\) is the total energy, and \(t\) is the time. Rearranging the equation for energy, we get \(E=P \cdot t\). We are given the power as \(0.8\mathrm{~W}\), and the time is \(24\mathrm{~h}\) (converting from hours to seconds: \(24\mathrm{~h} \times 3600\mathrm{s/h} = 86,400\mathrm{~s}\)), so we have: \(E = 0.8\mathrm{~W} \cdot 86400\mathrm{~s} = 69,120\mathrm{~J}\) So, the total heat dissipated during a 24-hour period is \(69,120\mathrm{~J}\).

Calculate the surface area of the resistor

To find the surface area of the cylindrical resistor, we need to consider both the lateral (side) surface area and the top and bottom surface areas. The formula for the lateral surface area of a cylinder is \(A_\mathrm{L} = 2 \pi rh\), where \(r\) is the radius, and \(h\) is the height (length) of the cylinder. The formula for the top or bottom surface area of a cylinder is \(A_\mathrm{T} = \pi r^2\). So, the total surface area of the resistor is \(A_\mathrm{TOT} = A_\mathrm{L} + 2A_\mathrm{T}\). First, find the radius of the cylinder: \(r = \frac{0.4\mathrm{~cm}}{2} = 0.2\mathrm{~cm} = 0.002\mathrm{~m}\) The height (length) of the cylinder is: \(h = 2\mathrm{~cm} = 0.02\mathrm{~m}\) Now, calculate the surface areas: \(A_\mathrm{L} = 2 \pi (0.002\mathrm{~m})(0.02\mathrm{~m}) = 8 \times 10^{-5} \pi\mathrm{~m}^2\) \(A_\mathrm{T} = \pi (0.002\mathrm{~m})^2 = 4 \times 10^{-6} \pi\mathrm{~m}^2\) \(A_\mathrm{TOT} = 8 \times 10^{-5} \pi\mathrm{~m}^2 + 2(4 \times 10^{-6} \pi\mathrm{~m}^2) = 8.8 \times 10^{-5} \pi\mathrm{~m}^2\)

Calculate the heat flux

The heat flux is the amount of heat transferred per unit surface area. We can find the heat flux by dividing the power dissipation by the total surface area of the resistor. Since, we know that power is energy divided by time (from the first step) and total heat dissipated by the resistor in one second equals to its power (in that second), we get: \(q = \frac{P}{A_\mathrm{TOT}} = \frac{0.8\mathrm{W}}{8.8 \times 10^{-5} \pi\mathrm{~m}^2} \approx 3634.63 \mathrm{~W/m^2}\) So, the heat flux is approximately \(3634.63 \mathrm{~W/m^2}\).

Calculate the fraction of heat dissipated from the top and bottom surfaces

To find the fraction of heat dissipated from both the top and bottom surfaces, we need to compare the top/bottom surface area combined \(\left(2A_\mathrm{T}\right)\) to the total surface area (\(A_\mathrm{TOT}\)): \(f = \frac{2A_\mathrm{T}}{A_\mathrm{TOT}} = \frac{2 (4 \times 10^{-6} \pi\mathrm{~m}^2)}{8.8 \times 10^{-5} \pi\mathrm{~m}^2} \approx 0.0909\) So, approximately \(9.09\%\) of the heat is dissipated from the top and bottom surfaces combined.