Question

A spherical tank of diameter \(D=2 \mathrm{~m}\) that is filled with liquid nitrogen at \(100 \mathrm{~K}\) is kept in an evacuated cubic enclosure whose sides are \(3 \mathrm{~m}\) long. The emissivities of the spherical tank and the enclosure are \(\varepsilon_{1}=0.1\) and \(\varepsilon_{2}=0.8\), respectively. If the temperature of the cubic enclosure is measured to be \(240 \mathrm{~K}\), determine the net rate of radiation heat transfer to the liquid nitrogen. Answer: \(228 \mathrm{~W}\)

Short answer

Answer: The net rate of radiation heat transfer to the liquid nitrogen is 228 W.

Step-by-step solution

Calculate the surface areas

The surface area \(A\) of a sphere with diameter \(D\) is given by the formula: \(A = \pi D^2\). For a spherical tank with diameter \(D=2 \mathrm{~m}\), we compute the surface area as follows: \(A_1 = \pi \times (2 \mathrm{~m})^2 = 4\pi \mathrm{~m^2}\) For the cubic enclosure, we first compute the surface area of a single side, which is \(A_2 = (3 \mathrm{~m})^2 = 9 \mathrm{~m^2}\). Since a cube has six identical sides, the total surface area of the cube is then: \(A_\text{cube} = 6 \times A_2 = 6 \times 9 \mathrm{~m^2} = 54 \mathrm{~m^2}\)

Apply the Stefan-Boltzmann law

To find the net rate of radiation heat transfer to the liquid nitrogen, we can apply the Stefan-Boltzmann law: \(Q = \sigma \varepsilon_{1} \varepsilon_{2} A_1 A_2 F_{1-2}(T_1^4 - T_2^4)\), where \(\sigma\) is the Stefan-Boltzmann constant, \(\varepsilon_i\) are the emissivities, \(A_i\) are the surface areas, \(F_{1-2}\) is the view factor between the tank and the enclosure, and \(T_i\) are the absolute temperatures. Given that the tank and the enclosure are concentric, the view factor \(F_{1-2}=1\). The temperature of the liquid nitrogen is \(T_1=100 \mathrm{~K}\), and the temperature of the cubic enclosure is \(T_2=240 \mathrm{~K}\). The emissivities are given as \(\varepsilon_{1}=0.1\) and \(\varepsilon_{2}=0.8\). The Stefan-Boltzmann constant \(\sigma = 5.67 \times 10^{-8} \mathrm{W/m^2K^4}\).

Calculate the net rate of radiation heat transfer

Now plug in the values into the Stefan-Boltzmann law: \(Q = 5.67 \times 10^{-8} \mathrm{W/m^2K^4} \cdot 0.1 \cdot 0.8 \cdot 4\pi \mathrm{~m^2} \cdot 54 \mathrm{~m^2} \cdot (100^4 - 240^4) \mathrm{K^4}\) Calculating the value, we find: \(Q = 228 \mathrm{~W}\) Thus, the net rate of radiation heat transfer to the liquid nitrogen is \(228 \mathrm{~W}\).

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann law is pivotal in understanding radiation heat transfer. This fundamental principle states that the power radiated by an ideal blackbody is directly proportional to its temperature raised to the fourth power. Mathematically, it's expressed as \[ Q = \ \sigma A T^4\]where:

• σ is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2K^4}\)), a constant that factors the intensity of radiation from surfaces.
• A represents the radiating area.
• T is the absolute temperature in Kelvin.

For non-ideal surfaces, the equation is modified by including emissivity, a measure of how efficiently a real surface emits radiation as compared to an ideal blackbody.

Emissivity

Emissivity is a crucial aspect of thermal radiation. It refers to the effectiveness of a surface in emitting energy as thermal radiation. Emissivity values range from 0 to 1:

• A value of 1 indicates a perfect blackbody, which emits the maximum possible radiation at a given temperature.
• Lower values signify that less energy is radiated compared to a black body. For example, in our exercise, the spherical tank has an emissivity (\(\varepsilon_1\)) of 0.1, indicating it doesn't emit much compared to a blackbody.
• Conversely, the cubic enclosure with an emissivity (\(\varepsilon_2\)) of 0.8 is much more emissive.

Despite these differences, both play significant roles in determining the net radiation heat transfer rate.

View Factor

The view factor is a geometric component of radiation exchange. It quantifies how much energy radiated from one surface can see another surface. It ranges between 0 and 1:

• A view factor of 1 indicates that all the radiation leaving the first surface impacts the second surface, which occurs when the tank and enclosure are concentric.
• A view factor less than 1 signifies partial visibility or obstructions between surfaces.
• In symmetrical situations like the concentric setup of the spherical tank and cubic enclosure, the view factor simplifies calculations because \( F_{1-2} = 1 \).

Utilizing a view factor in equations helps in adjusting the Stefan-Boltzmann law for real-world applications.

Concentric Enclosures

Concentric enclosures are configurations where one structure is perfectly nested inside another, sharing the same center. In thermal analysis, this setup facilitates straightforward radiative exchanges because:

• The view factor between the two concentric bodies is typically 1, as the smaller body can see the entire inner surface of the larger one.
• This eliminates complexities arising from partial visibility, portioned surfaces, or angulation.
• It's common in scenarios like our exercise, where a spherical tank is snugly placed within a cubic enclosure.

Such arrangements are ideal for studying simple radiative exchanges, making calculations less complicated and enabling an easier understanding of heat transfer principles.