Question

A 0.1-W small cylindrical resistor mounted on a lower part of a vertical circuit board is \(0.3\) in long and has a diameter of \(0.2 \mathrm{in}\). The view of the resistor is largely blocked by another circuit board facing it, and the heat transfer through the connecting wires is negligible. The air is free to flow through the large parallel flow passages between the boards as a result of natural convection currents. If the air temperature near the resistor is \(120^{\circ} \mathrm{F}\), determine the approximate surface temperature of the resistor. Evaluate air properties at a film temperature of \(170^{\circ} \mathrm{F}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption? Answer: \(211^{\circ} \mathrm{F}\)

Short answer

Answer: The approximate surface temperature of the cylindrical resistor is \(211^{\circ} \mathrm{F}\).

Step-by-step solution

Find the Nusselt number

Calculate the Nusselt number (Nu) for the given scenario using the expression for natural convection for a vertical cylinder: \[Nu = C \times (Gr \times Pr)^n\] where C and n are constants for natural convection, Gr is the Grashof number, and Pr is the Prandtl number. Using the air properties at the film temperature of \(170^{\circ} \mathrm{F}\) and \(1 \mathrm{~atm}\) pressure, we obtain: \[\textrm{Density} (\rho) = 0.075 \,\mathrm{lbm/ft^3}\] \[\textrm{Thermal conductivity} (k) = 0.0147 \, \mathrm{Btu/hr \cdot ft \cdot ^{\circ} F}\] \[\textrm{Dynamic viscosity} (\mu) = 3.28 \times 10^{-7} \, \mathrm{lbm/ft \cdot s}\] \[\textrm{Specific heat} (c_p) = 0.240 \, \mathrm{Btu/lbm \cdot ^{\circ} F}\] Now, we can calculate Pr and Gr: \[Pr = \frac{c_p \times \mu}{k} \approx 0.710\] \[Gr = \frac{g \times \beta \times L^3 \times (\Delta{T})}{\nu^2}\] where g is the gravitational acceleration, \(\beta\) is the thermal expansion coefficient, L is the length of the resistor, \(\Delta{T}\) is the temperature difference between the surface and ambient air, and \(\nu\) is the kinematic viscosity: \[g = 32.2 \, \mathrm{ft/s^2}\] \[\beta = \frac{1}{T_{film}} = \frac{1}{(170 + 459.67) \, \mathrm{R}} \approx 0.00310~\mathrm{R^{-1}}\] \[L = 0.3 \, \mathrm{in} = 0.025 \, \mathrm{ft}\] \[\nu = \frac{\mu}{\rho} \approx 4.37 \times 10^{-5} \, \mathrm{ft^2/s}\] Ignoring \(\Delta{T}\) for an initial estimation: \[Gr \approx \frac{32.2 \times 0.00310 \times (0.025)^3}{(4.37 \times 10^{-5})^2} \approx 1.09 \times 10^6\] Assuming laminar flow, we have C = 0.59 and n = 1/4. Therefore, the Nusselt number is: \[Nu = 0.59 \times (1.09 \times 10^6 \times 0.710)^{1/4} \approx 4.04\]

Calculate the convection heat transfer coefficient

With the Nusselt number, we can now calculate the convection heat transfer coefficient (h) using the equation: \[h = \frac{Nu \times k}{L} \approx \frac{4.04 \times 0.0147}{0.025} \approx 2.34 \, \mathrm{Btu/hr \cdot ft^2 \cdot ^{\circ} F}\]

Calculate the surface temperature of the resistor

The energy balance equation between the heat generated by the resistor and the heat loss due to convection at the surface is: \[\textrm{Heat generated} = \textrm{Heat loss due to convection}\] \[0.1 \, \mathrm{W} \times 3.41 \, \mathrm{Btu/hr \cdot W} = h \times A \times (\Delta{T})\] where A is the surface area of the resistor and \(\Delta{T} = T_s - T_a\), with \(T_s\) being the surface temperature and \(T_a\) being the ambient air temperature. We can calculate the surface area: \[A = \pi D L = \pi(0.2)(0.3) \, \mathrm{in^2} \approx 0.00265 \, \mathrm{ft^2}\] Now we can find the surface temperature: \[T_s = T_a + \frac{0.1 \times 3.41}{h\times A} \approx 120 + \frac{0.1 \times 3.41}{2.34 \times 0.00265} \approx 211^{\circ} \mathrm{F}\] The surface temperature of the resistor is approximately \(211^{\circ} \mathrm{F}\).

Check the assumption

We initially ignored the temperature difference between the surface and ambient air to estimate the Grashof number. The calculated surface temperature is within reasonable limits, confirming that our assumption is valid. In conclusion, the surface temperature of the resistor is approximately \(211^{\circ} \mathrm{F}\), and our assumption was reasonable.