Question

Consider a 30-cm-diameter hemispherical enclosure. The dome is maintained at \(600 \mathrm{~K}\), and heat is supplied from the dome at a rate of $65 \mathrm{~W}\( while the base surface with an emissivity of \)0.55$ is maintained at \(400 \mathrm{~K}\). Determine the emissivity of the dome.

Short answer

Based on the given information and the steps shown in the solution, determine the emissivity of the dome. Known parameters: - Diameter of the hemisphere: 30 cm - Temperature of the dome: 600 K - Heat supplied from the dome: 65 W - Emissivity of the base surface: 0.55 - Temperature of the base surface: 400 K Applying the Stefan-Boltzmann Law and the energy balance principle, we find the emissivity of the dome as: \(\varepsilon_\text{dome} = \frac{Q_\text{absorbed}}{A_\text{dome} \sigma T_\text{dome}^4}\) Calculate and enter the emissivity of the dome as a decimal value (for example: 0.65).

Step-by-step solution

Write down the known parameters.

We have the following known parameters: - Diameter of the hemisphere: \(30~\text{cm}\) - Temperature of the dome: \(T_\text{dome} = 600~\text{K}\) - Heat supplied from the dome: \(Q_\text{dome} = 65~\text{W}\) - Emissivity of the base surface: \(\varepsilon_\text{base} = 0.55\) - Temperature of the base surface: \(T_\text{base} = 400~\text{K}\) *Note:* Diameter should be converted to meters for calculations.

Find the surface area of the dome and base surface.

To find the surface area of the dome (\(A_\text{dome}\)) and the base surface (\(A_\text{base}\)), first, convert the diameter to meters and then apply the corresponding area formulas for a hemisphere and a circle. - Diameter in meters: \(D = 0.3~\text{m}\) - Radius of the hemisphere: \(r = \frac{D}{2} = 0.15~\text{m}\) Surface area of the dome: \(A_\text{dome} = 2\pi r^2 = 2\pi (0.15)^2\) Surface area of the base surface: \(A_\text{base} = \pi r^2 = \pi (0.15)^2\)

Apply the Stefan-Boltzmann Law.

We can write the Stefan-Boltzmann law for the total radiative heat transfer from the dome (\(Q_\text{dome}\)) and from the base surface (\(Q_\text{base}\)): \(Q_\text{dome} = \varepsilon_\text{dome} A_\text{dome} \sigma T_\text{dome}^4\) \(Q_\text{base} = \varepsilon_\text{base} A_\text{base} \sigma T_\text{base}^4\) Here \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2K^4}\))

Find the radiative heat transfer from the base surface.

We can calculate the radiative heat transfer from the base surface using the known emissivity and temperature by plugging in our values: \(Q_\text{base} = \varepsilon_\text{base} A_\text{base} \sigma T_\text{base}^4 = 0.55\times \pi(0.15)^2 \times (5.67 \times 10^{-8}) \times (400)^4\)

Apply the energy balance principle.

Since we know the total heat supplied to the system (\(Q_\text{dome}\)) and the heat transfer from the base surface (\(Q_\text{base}\)), according to the energy balance principle, we can say that the absorbed heat in the dome (\(Q_\text{absorbed}\)) is the difference between heat supplied and radiative heat transfer from the base surface: \(Q_\text{absorbed} = Q_\text{dome} - Q_\text{base}\)

Determine the emissivity of the dome.

Now we can use the Stefan-Boltzmann law again and solve for the emissivity of the dome (\(\varepsilon_\text{dome}\)): \(Q_\text{absorbed} = \varepsilon_\text{dome} A_\text{dome} \sigma T_\text{dome}^4\) Solve for \(\varepsilon_\text{dome}\): \(\varepsilon_\text{dome} = \frac{Q_\text{absorbed}}{A_\text{dome} \sigma T_\text{dome}^4}\) Plug in the known values and calculate for \(\varepsilon_\text{dome}\):