Question

Two concentric spheres of diameters \(D_{1}=0.3 \mathrm{~m}\) and \(D_{2}=0.4 \mathrm{~m}\) are maintained at uniform temperatures \(T_{1}=700 \mathrm{~K}\) and \(T_{2}=500 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.5\) and \(\varepsilon_{2}=0.7\), respectively. Determine the net rate of radiation heat transfer between the two spheres. Also, determine the convection heat transfer coefficient at the outer surface if both the surrounding medium and the surrounding surfaces are at \(30^{\circ} \mathrm{C}\). Assume the emissivity of the outer surface is \(0.35\).

Short answer

Answer: The net rate of radiation heat transfer between the two spheres is approximately 397.43 W, and the convection heat transfer coefficient at the outer surface is approximately 12.32 W/m²K.

Step-by-step solution

Calculate areas of the sphere surfaces

To calculate the heat transfer, we need to know the surface areas of both spheres. The surface area of a sphere can be calculated using the formula \(A = 4 \pi r^2\), where r is the radius. We are given the diameters of the spheres, so we can easily find the radii: \(r_1 = \frac{D_1}{2} = 0.15\) m \(r_2 = \frac{D_2}{2} = 0.20\) m Now, we can calculate the surface areas: \(A_1 = 4 \pi (0.15)^2 \approx 0.2827\) m\(^2\) \(A_2 = 4 \pi (0.20)^2 \approx 0.5027\) m\(^2\)

Calculate radiation heat transfer between the spheres

We can now calculate the rate of radiation heat transfer between the two spheres using the Stefan-Boltzmann law: \(q_{radiation} = \frac{\varepsilon_1 \varepsilon_2 A_1 A_2 \sigma (T_1^4 - T_2^4)}{A_1 + A_2}\) where \(q_{radiation}\): net rate of radiation heat transfer (W) \(\varepsilon_1\) and \(\varepsilon_2\): emissivities of the spheres (dimensionless) \(A_1\) and \(A_2\): surface areas of the spheres (m^2) \(\sigma\): Stefan-Boltzmann constant (\(5.67 \times 10^{-8}\) W/m\(^2\)K\(^4\)) \(T_1\) and \(T_2\): temperatures of the spheres (K) Substituting the values into the equation, we have: \(q_{radiation} = \frac{0.5 \times 0.7 \times 0.2827 \times 0.5027 \times 5.67 \times 10^{-8} (700^4 - 500^4)}{0.2827 + 0.5027}\) \(q_{radiation} \approx 397.43\) W

Calculate the convective heat transfer coefficient

Now we can determine the convection heat transfer coefficient at the outer surface of the second sphere using the energy balance: \(q_{total} = q_{radiation} + q_{convection} = q_{radiation} + h A_2 (T_2 - T_{\infty})\) where \(q_{total}\): total heat transfer from the second sphere (W) \(q_{convection}\): rate of convection heat transfer (W) \(h\): convection heat transfer coefficient (W/m\(^2\)K) \(T_{\infty}\): temperature of the surrounding medium and surfaces (K) Rearranging for \(h\), we get: \(h = \frac{q_{total} - q_{radiation}}{A_2 (T_2 - T_{\infty})}\) In this problem, it's assumed that the outer surface emissivity is \(0.35\), so the total heat transfer from the second sphere is: \(q_{total} = 0.35 \times A_2 \times \sigma \times T_2^4\) Now substitute the values into the equation for \(h\): \(h = \frac{(0.35 \times 0.5027 \times 5.67 \times 10^{-8} \times 500^4) - 397.43}{0.5027 \times (500 - (30 + 273))}\) \(h \approx 12.32\) W/m\(^2\)K The net rate of radiation heat transfer between the two spheres is approximately \(397.43\) W, and the convection heat transfer coefficient at the outer surface is approximately \(12.32\) W/m\(^2\)K.

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a cornerstone in understanding thermal radiation and its role in heat transfer. It states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's temperature.

The mathematical expression for this law is given by the formula:
\[\begin{equation}E = \text{\(\sigma\)} T^4\end{equation}\]where

• \text{\(E\)} is the emitted energy per unit area,
• \text{\(\sigma\)} is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8}\) W/m\text{\(^2\)}K\text{\(^4\)},
• \text{\(T\)} is the absolute temperature in kelvins (K).

For real-life materials, which are not perfect black bodies, an emissivity factor (\text{\(\epsilon\)}) is introduced, adjusting the law to account for the object's ability to emit radiation. This factor has a value between 0 and 1, where 1 represents a perfect black body. The law for real surfaces thus becomes \[\begin{equation}q = \epsilon \sigma A T^4\end{equation}\]where

• \text{\(q\)} is the rate of heat transfer due to radiation,
• \text{\(A\)} is the area of the surface.

In the given exercise, by applying the Stefan-Boltzmann Law using the given temperatures and the emissivity values for two concentric spheres, we can calculate the net radiation heat transfer rate between them. The solution uses this law to find the heat emitted and absorbed by the spheres, leading us to the correct answer.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient, symbolized as \text{\(h\)}, defines the rate at which heat is transferred between a surface and a fluid moving past it. It is a measure of the effectiveness of the convective heat transfer process and is expressed in units of W/m\text{\(^2\)}K.

The coefficient provides us with a convenient way to calculate the heat transfer due to convection by using the simple relationship:
\[\begin{equation}q_{convection} = h A \Delta T\end{equation}\]where

• \text{\(A\)} is the surface area through which convection is occurring,
• \text{\(\Delta T\)} represents the temperature difference between the surface and the fluid.

Obtaining the convection heat transfer coefficient often involves empirical correlations derived from fluid dynamics and heat transfer research. For the given problem, once the net radiation heat transfer rate was determined, the remaining heat transfer rate \text{\(assuming total heat transfer is the sum of radiation and convection\)} can be attributed to convection. The respective equation was then rearranged to find the value for \text{\(h\)}, which helps us understand how efficiently the outer sphere's surface is exchanging heat with its surroundings.

Thermal Radiation

Thermal radiation is the emission of electromagnetic waves from all matter that possesses a temperature greater than absolute zero. It is a form of heat transfer that does not require any medium for propagation, making it distinct from conduction and convection, which do require a medium.

All bodies emit thermal radiation, but the characteristics of the radiation depend on the surface temperature and the nature of the surface itself. As mentioned, a perfect black body, which fully absorbs all incident radiation, is an ideal emitter and follows the Stefan-Boltzmann Law directly.
\[\begin{equation}q = \sigma A T^4\end{equation}\]
In real situations, we adjust this for an object's emissivity \text{\(\epsilon\)}, which may depend on material properties and surface conditions. The challenge in problems involving thermal radiation is often in determining the correct emissivity values and comprehending that the heat transfer rate significantly increases with higher temperatures due to the T to the fourth power relationship. In the textbook exercise, understanding thermal radiation allows us to predict how heat is exchanged between the spheres solely due to their temperature difference.