Question

An incandescent lightbulb is an inexpensive but highly inefficient device that converts electrical energy into light. It converts about 10 percent of the electrical energy it consumes into light while converting the remaining 90 percent into heat. (A fluorescent lightbulb will give the same amount of light while consuming only one-fourth of the electrical energy, and it will last 10 times longer than an incandescent lightbulb.) The glass bulb of the lamp heats up very quickly as a result of absorbing all that heat and dissipating it to the surroundings by convection and radiation. Consider a 10 -cm-diameter, 100 -W lightbulb cooled by a fan that blows air at \(30^{\circ} \mathrm{C}\) to the bulb at a velocity of $2 \mathrm{~m} / \mathrm{s}\(. The surrounding surfaces are also at \)30^{\circ} \mathrm{C}$, and the emissivity of the glass is \(0.9\). Assuming 10 percent of the energy passes through the glass bulb as light with negligible absorption and the rest of the energy is absorbed and dissipated by the bulb itself, determine the equilibrium temperature of the glass bulb. Assume a surface temperature of \(100^{\circ} \mathrm{C}\) for evaluation of \(\mu_{x}\). Is this a good assumption?

Short answer

Answer: The estimated equilibrium temperature of the glass bulb is approximately \(193^{\circ} C\).

Step-by-step solution

Calculate the energy converted into heat

The lightbulb converts 90% of the electrical energy it consumes into heat. Thus, the amount of heat generated is: \(Q = 0.9 \times 100 \thinspace W = 90 \thinspace W\)

Calculate the convective heat transfer coefficient

The fan blows air at a velocity of \(2 \thinspace m/s\) to the bulb. We will use the correlation for the convective heat transfer coefficient for a sphere with laminar flow: \(h = Nu \frac{k}{D}\) where \(Nu\) is the Nusselt number, \(k\) is the thermal conductivity of air, and \(D\) is the diameter of the glass bulb. For our given conditions, the Nusselt number based on \(D\) can be calculated as: \(Nu = 2.0 + 0.6 Re_D^{1/2} Pr^{1/3}\) Here, \(Re_D\) is the Reynolds number based on the diameter and \(Pr\) is the Prandtl number: \(Re_D = \frac{\rho u D}{\mu} = \frac{997 \times 2 \times 0.1}{10^{-3}} = 199400\) (using the properties of air at \(30^{\circ} C\) and assuming \(\mu_{x} = 10^{-3} \thinspace Ns/m^2\)) \(Pr = \frac{\mu C_p}{k} = \frac{10^{-3} \times 10^3}{0.027} = 36.7\) (using the properties of air at \(30^{\circ} C\)) Now, we can calculate the Nusselt number: \(Nu = 2.0 + 0.6 \times 199400^{1/2} \times 36.7^{1/3} = 48.25\) Finally, we can calculate the convective heat transfer coefficient: \(h = 48.25 \frac{0.027}{0.1} = 13.05 \thinspace W/m^2 K\)

Calculate the heat loss due to convection

The heat loss due to convection can be calculated using the equation: \(Q_{conv} = h A (T_{s}-T_{\infty})\) Here, \(A\) is the surface area of the glass bulb and \(T_{s}\) and \(T_{\infty}\) are the surface and ambient temperatures. Since we want to find \(T_s\), we will write down the equation as: \(Q_{conv} = 13.05 \times \pi \times 0.1^2 \times (T_s-30)\)

Calculate the heat loss due to radiation

The heat loss due to radiation can be calculated using the equation: \(Q_{rad} = \epsilon \sigma A (T_s^4 - T_{\infty}^4)\) Here, \(\epsilon\) is the emissivity of the glass bulb and \(\sigma\) is the Stefan-Boltzmann constant. For our case, this equation can be written as: \(Q_{rad} = 0.9 \times 5.67 \times 10^{-8} \times \pi \times 0.1^2 \times (T_s^4 - 30^4)\)

Solve for the equilibrium temperature

In order to find the equilibrium temperature, we need to balance the heat generated (\(Q\)) with the sum of the heat loss due to convection and radiation: \(Q = Q_{conv}+Q_{rad}\) Combining Step 3 and Step 4 equations, we get: \(90 = 13.05 \times \pi \times 0.1^2 \times (T_s-30) + 0.9 \times 5.67 \times 10^{-8} \times \pi \times 0.1^2 \times (T_s^4 - 30^4)\) We must solve this nonlinear equation for \(T_s\) using an appropriate method, such as the bisection method or Newton-Raphson method. By solving the equation, we get the equilibrium temperature: \(T_s \approx 193^{\circ} C\)

Check the assumption

Finally, we check if the assumption of a surface temperature of \(100^{\circ} C\) for evaluation of \(\mu_{x}\) was reasonable. The equilibrium temperature we found is \(193^{\circ} C\), which is higher than the assumed \(100^{\circ} C\). It would be more accurate to re-calculate the properties of air (including \(\mu_{x}\)) at an average temperature between \(193^{\circ} C\) and \(30^{\circ} C\), and re-do the calculations outlined in Steps 2-5. However, since this problem requires an estimation, and properties of air do not change dramatically over such a temperature range, the assumption is acceptable.