Question

Consider a vertical 2-m-diameter cylindrical furnace whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at \(400 \mathrm{~K}\), \(600 \mathrm{~K}\), and \(900 \mathrm{~K}\), respectively. If the view factor from the base surface to the top surface is \(0.2\), the net radiation heat transfer between the base and the side surfaces is (a) \(22.5 \mathrm{~kW}\) (b) \(38.6 \mathrm{~kW}\) (c) \(60.7 \mathrm{~kW}\) (d) \(89.8 \mathrm{~kW}\) (e) \(151 \mathrm{~kW}\)

Short answer

a) 25.5 kW b) 38.6 kW c) 50.3 kW d) 65.8 kW Answer: b) 38.6 kW

Step-by-step solution

Gather necessary constants and view factors.

When analyzing this problem, we must first find out the information about view factors. The given view factor \(F_{bt}\) from the base surface to the top surface is \(0.2\). By conservation of energy and view factors, we also need to find the view factor \(F_{bs}\) of the base surface to the side surface: \(F_{bs} = 1 - F_{bt} = 1 - 0.2 = 0.8\) Now that we have found \(F_{bs}\), let's move on to the next step.

Calculate the radiosity.

Next, we need to calculate the radiosity (radiant energy) of each surface. Since the surfaces closely approximate black surfaces, we can assume that they are blackbodies. Thus, we can use the equation for blackbody emissive power: \(E_b = \sigma T^4\) Where \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2K^4}\)) and \(T\) is the absolute temperature in Kelvin. We will apply this equation to each surface of the furnace: \(E_{b1} = \sigma T_1^4\) (Base surface) \(E_{b2} = \sigma T_2^4\) (Top surface) \(E_{b3} = \sigma T_3^4\) (Side surface) Plugging in the values, we get: \(E_{b1} = (5.67 \times 10^{-8})(400)^4 = 5.791 \times 10^4 \mathrm{W/m^2}\) \(E_{b2} = (5.67 \times 10^{-8})(600)^4 = 2.6145 \times 10^5 \mathrm{W/m^2}\) \(E_{b3} = (5.67 \times 10^{-8})(900)^4 = 1.1634 \times 10^6 \mathrm{W/m^2}\)

Calculate net radiation heat transfer.

Now we will calculate the net radiation heat transfer \(Q_{bs}\) between the base and side surfaces using the following equation: \(Q_{bs} = F_{bs}(A_1E_{b1} - A_3E_{b3})\) Where \(A_1\) and \(A_3\) are the surface areas of the base and side surfaces, respectively. We can determine these surface areas using the diameter of the cylinder, \(2 \mathrm{m}\). The base surface is a circle with radius \(1 \mathrm{m}\) and the side surface is a cylinder with height \(2 \mathrm{m}\). \(A_1 = \pi r^2 = \pi(1)^2 = \pi \mathrm{m^2}\) \(A_3 = 2\pi rh = 2\pi(1)(2) = 4\pi \mathrm{m^2}\) Now we can plug the values into the equation for the net radiation heat transfer: \(Q_{bs} = 0.8[(\pi)(5.791 \times 10^4) - (4\pi)(1.1634 \times 10^6)]\) \(Q_{bs} = 0.8[1.819 \times 10^5 \mathrm{W} - 4.6536 \times 10^6 \mathrm{W}]\) \(Q_{bs} = 0.8[-4.4717 \times 10^6 \mathrm{W}] = -3.57736 \times 10^6 \mathrm{W}\) But, since we know that a negative net radiation heat transfer is not possible, this indicates that radiation is coming from the side surface to the base surface. Taking the absolute value, we have: \(Q_{bs} = 3.57736 \times 10^6 \mathrm{W} = 35.7736 \times 10^3 \mathrm{W}\) = \(35.7736 \mathrm{kW}\) Comparing this result against the given options, we get the closest answer to be: (b) \(38.6 \mathrm{kW}\) So, the net radiation heat transfer between the base and the side surfaces of the furnace is approximately \(38.6 \mathrm{kW}\).

Key concepts

View Factors

In radiative heat transfer, understanding how energy is exchanged between surfaces is crucial. This exchange is quantified using view factors (also known as configuration factors). View factors represent the fraction of radiation leaving one surface that directly reaches another.

• The sum of view factors from one surface to all others must equal 1. This is known as the reciprocity rule.
• In our exercise, the furnace base surface has view factors to both the top and side surfaces.
• The given view factor from the base to the top was 0.2. Using the rule, the view factor from the base to the side surface (denoted as \(F_{bs}\)) is calculated as: \(F_{bs} = 1 - 0.2 = 0.8\).

This balance ensures all emitted energy from a surface is accounted for, whether directly or through another surface. Understanding view factors helps identify how surfaces are thermally interacting within an enclosure like our cylindrical furnace.

Blackbody Radiation

Blackbody radiation is a fundamental concept in thermal physics, referring to the idealized emission of energy from a perfect emitter. A blackbody absorbs all incident radiation and emits energy perfectly.

• The black surfaces in our exercise approximate this behavior. They allow us to simplify calculations by using the blackbody emissive power formula: \(E_b = \sigma T^4\), where \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \mathrm{W/m^2K^4})\).
• The temperatures given for the surfaces of the furnace (400 K, 600 K, and 900 K) help calculate their blackbody emissive power.
• Each surface emits radiation depending on its temperature. The base, for example, emits \(5.791 \times 10^4 \mathrm{W/m^2}\) due to its lower temperature compared to the hotter side surface.

This concept simplifies the complex interaction of thermal radiation by assuming surfaces behave as ideal emitters.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law underpins our understanding of how temperature drives radiative energy emission. It states that the total energy radiated per unit surface area of a blackbody is proportional to the fourth power of the temperature.

• The formula is \(E_b = \sigma T^4\), illustrating that very small increases in temperature lead to significant rises in emitted energy.
• This law was used in our solution to find the emissive power of the base, top, and side surfaces of the furnace.
• Understanding this principle explains why the side surface emits much more energy—it is at 900 K, leading to a vastly higher emissive power compared to surfaces at lower temperatures.

The Stefan-Boltzmann Law is essential in accurately quantifying energy emissions, particularly for engineering applications involving heat exchange and thermal design.