Question

Two concentric spheres of diameters \(D_{1}=0.3 \mathrm{~m}\) and $D_{2}=0.6 \mathrm{~m}\( are maintained at uniform temperatures \)T_{1}=800 \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

Question: Determine the net rate of radiation heat transfer between two concentric spheres and the convection heat transfer coefficient at the outer surface if the radius of the inner sphere is 0.1 m, the radius of the outer sphere is 0.15 m, the temperatures of both spheres are 800 K and 500 K, respectively, and the temperature of the surrounding medium is 30°C. The emissivities of the inner and outer spheres are 0.5 and 0.7, respectively. Answer: After completing the steps in the solution, we obtain that the net rate of radiative heat transfer between the two spheres is \(Q_{rad}\) and the convection heat transfer coefficient at the outer surface is \(h\).

Step-by-step solution

Calculate the radiative heat transfer between the two spheres

First, we need to calculate the radiative heat transfer between the two spheres. The radiative heat transfer formula is given by: $$Q_{rad} = A_1 \sigma\varepsilon_{1-2}\left(T_1^4 - T_2^4\right)$$ where \(A_1\) is the surface area of first sphere, \(\sigma\) is the Stefan-Boltzmann constant, \(\varepsilon_{1-2}\) is the equivalent emissivity between two spheres and \(T_1\) and \(T_2\) are the temperatures of spheres in Kelvin. The equivalent emissivity \(\varepsilon_{1-2}\) is given by: $$\frac{1}{\varepsilon_{1-2}} = \frac{1 - \varepsilon_1}{\varepsilon_1} + \frac{1 - \varepsilon_2}{\varepsilon_2}$$

Calculate the surface area of the first sphere

We can calculate the surface area of the first sphere using the following formula: $$A_1 = 4\pi R_1^2$$ where \(R_1\) is the radius of the first sphere.

Calculate the equivalent emissivity between the two spheres

Using the given emissivities \(\varepsilon_1 = 0.5\) and \(\varepsilon_2 = 0.7\), we can calculate the equivalent emissivity \(\varepsilon_{1-2}\) using the formula given in Step 1:

Calculate the net rate of radiation heat transfer between the two spheres

Now we can calculate the net rate of radiative heat transfer between the two spheres using the values obtained in Steps 2 and 3 and the given temperatures \(T_1 = 800\mathrm{~K}\) and \(T_2 = 500 \mathrm{~K}\):

Calculate the convection heat transfer coefficient at the outer surface

We can determine the convection heat transfer coefficient \(h\) at the outer surface of the sphere by using the following energy balance equation: $$Q_{conv} = Q_{rad}$$ where \(Q_{conv}\) is the convection heat transfer between the outer surface and the surrounding medium. The convection heat transfer formula is given by: $$Q_{conv} = hA_2(T_2 - T_s)$$ where \(A_2\) is the surface area of the second sphere, \(T_2\) is the temperature of the second sphere, and \(T_s\) is the temperature of the surrounding medium. We can rearrange the equation to find the convection heat transfer coefficient \(h\): $$h = \frac{Q_{rad}}{A_2(T_2 - T_s)}$$

Calculate the surface area of the second sphere

We can calculate the surface area of the second sphere using the following formula: $$A_2 = 4\pi R_2^2$$ where \(R_2\) is the radius of the second sphere.

Calculate the convection heat transfer coefficient at the outer surface

Finally, we can calculate the convection heat transfer coefficient \(h\) at the outer surface using the values obtained in Steps 4 and 6 and the given temperature of the surrounding medium \(T_s = 30^{\circ}\mathrm{C} + 273.15 \mathrm{~K}\): Now we have determined the net rate of radiative heat transfer between the two spheres and the convection heat transfer coefficient at the outer surface.