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Chapter 13: Problem 87

Two convex objects are inside a large vacuum enclosure whose walls are maintained at \(T_{3}=300 \mathrm{~K}\). The objects have the same area, \(0.2 \mathrm{~m}^{2}\), and the same emissivity, 0.2. The view factor from object 1 to object 2 is \(F_{12}=0_{3} 3\). Embedded in object 2 is a heater that generates 400 W. The temperature of object 1 is maintained at \(T_{1}=200 \mathrm{~K}\) by circulating a fluid through channels machined into it. At what rate must heat be supplied (or removed) by the fluid to maintain the desired temperature of object 1? What is the temperatare of object 2 ?

Short Answer

Expert verified
The rate at which heat must be supplied or removed by the fluid to maintain the desired temperature of object 1 is approximately 400 W. The temperature of object 2 is approximately 233.66 K.

Step by step solution

01

Define the given data

The following data is given: \(T_3 = 300 K\) (temperature of the enclosure) \(T_1 = 200 K\) (temperature of object 1) Area of both objects: \(A_1 = A_2 = 0.2 m^2\) Emissivity of both objects: \(\varepsilon_1 = \varepsilon_2 = 0.2\) View factor from object 1 to object 2: \(F_{12} = F_{21} = 0.3\) Heater power in object 2: \(q_{heater} = 400 W\)
02

Calculate heat transfer from object 1 to object 2

Using the radiative heat transfer equation, we can calculate the heat transfer from object 1 to object 2: \(q_{12} = A_1 F_{12} \sigma \varepsilon_1 \varepsilon_2 (T_1^4 - T_2^4)\) Replace the known quantities and keep the unknown quantities (\(T_2\) and \(q_{12}\)): \(q_{12} = 0.2 \times 0.3 \times 5.67\times 10^{-8} \times 0.2 \times 0.2 (200^4 - T_2^4)\)
03

Apply the conservation of energy

Heat transfers between object 1, object 2, and the enclosure: \(q_{heater} = q_{12} + q_{13}\) \(q_{12} + q_{13} = 400 W\) Where \(q_{13}\) is the heat transfer from object 1 to the enclosure. Similarly to step 2, the radiative heat transfer equation can be applied: \(q_{13} = A_1 (1 - F_{12}) \sigma \varepsilon_1 (T_1^4 - T_3^4)\) Replace the known quantities and simplify: \(q_{13} = 0.2 \times (1 - 0.3) \times 5.67\times 10^{-8} \times 0.2 (200^4 - 300^4)\) Calculate the value of \(q_{13}\): \(q_{13} \approx 19.65 W\) Now substitute the calculated value of \(q_{13}\) into the conservation of energy equation: \(q_{12} = 400 - 19.65\) \(q_{12} \approx 380.35 W\)
04

Solve for T2

Now we have the value of \(q_{12}\), we can substitute it back into the radiative heat transfer equation: \(380.35 = 0.2 \times 0.3 \times 5.67\times 10^{-8} \times 0.2 \times 0.2 (200^4 - T_2^4)\) Solve the equation for \(T_2\): \(T_2^4 = 32400000\) \(T_2 \approx 233.66 K\)
05

Determine the rate of heat supply or removal

Now that we have the values of both \(q_{12}\) and \(q_{13}\), we can determine the rate of heat supplied or removed by the fluid to maintain the desired temperature of object 1: Heat supplied or removed by the fluid: \(q_{fluid} = q_{12} + q_{13}\) Substitute the values of \(q_{12}\) and \(q_{13}\): \(q_{fluid} = 380.35 + 19.65\) \(q_{fluid} \approx 400 W\) #Summary# The rate at which heat must be supplied or removed by the fluid to maintain the desired temperature of object 1 is approximately 400 W. The temperature of object 2 is approximately 233.66 K.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Exchange in Vacuum
When we talk about heat exchange in vacuum, we're looking at how heat can travel without a medium, like air or water, to transfer the energy. Since convection and conduction require matter to transport heat, the only way heat can be exchanged in a vacuum is through radiation. Radiative heat transfer is governed by electromagnetic waves and doesn't need physical contact or a medium to move from one place to another.

For instance, space itself is a vacuum, and the heat from the sun reaches the Earth via radiation. In our exercise example, two objects are in a vacuum enclosure, implying that the only form of heat exchange between these objects and their environment can occur through radiation. Understanding this concept is crucial for solving problems related to space technology, thermos flasks, and solar power generation, where the vacuum plays a significant role in how heat is managed.
Stefan-Boltzmann Law
The Stefan-Boltzmann law is a principle that's key to understanding radiative 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 formula for this law is given by
\[ j^* = \'sigma T^4 \]
where \( j^* \) is the total emissive power, \( \'sigma \) is the Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \, W/m^2K^4 \) ), and \( T \) is the absolute temperature in kelvins.

The exercise uses this law to find out how much heat is exchanged between the objects and their surroundings. It's worth noting that real objects are not perfect black bodies, which is why we consider emissivity, a factor that modifies the Stefan-Boltzmann law to describe the radiative output of non-black bodies, like the objects in our example.
Conservation of Energy
The principle of conservation of energy is a universal law that states that energy cannot be created or destroyed, only transformed from one form to another. In the context of our exercise, we applied this concept to ensure that the energy supplied to the system by the heater equals the sum of energy radiated to both the second object and to the enclosure.

Breaking it down, if we have a heater inputting 400 W of power into object 2, this energy must go somewhere. It gets partly radiated to object 1 and partly to the enclosure. By calculating these heat exchanges accurately, they should add up to the heater's power, which confirms that our use of the conservation of energy law is correct. If they don't, it indicates either a miscalculation or a misunderstanding of the system's energy dynamics.
View Factor in Radiation
The concept of a view factor, also known as a configuration factor or shape factor, is crucial in radiative heat transfer calculations because it quantifies the proportion of the radiation leaving one surface that directly strikes another. In mathematical terms, it's a dimensionless number between 0 and 1. For our exercise, the view factor helps determine how much of the heat radiated from object 1 reaches object 2.

Imagine holding up a camera and taking a picture of one object from the perspective of another; the view factor is somewhat like the fraction of the camera's field of view that is filled by the object. It's an essential parameter when designing systems where radiative heat exchange is significant, such as in satellite components or in high-temperature industrial processes.

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Most popular questions from this chapter

A diffuse, gray radiation shield of \(60-\mathrm{mm}\) diameter and emissivities of \(\varepsilon_{2,1}=0.01\) and \(\varepsilon_{2,0}=0.1\) on the inner and outer surfaces, respectively, is concentric with a long tube transporting a hot process flu盲d. The tube surface is black with a diameter of \(20 \mathrm{~mm}\). The region interior to the shield is evacuated. The exterior surface of the shield is exposed to a large room whose walls are at \(17^{\circ} \mathrm{C}\) and experiences convection with air at \(27^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). Determine the operating temperature for the inner tube if the shield temperature is maintained at \(42^{\circ} \mathrm{C}\).

Options for thermally shielding the top ceiling of a large furnace include the use of an insulating material of thickness \(L\) and thermal conductivity \(k\), case (a), or an air space of equ盲valent thickness formed by installing a steel sheet above the ceiling, case \((b)\). (a) Develop mathematical models that could be used to assess which of the two approaches is better. In both cases the interior surface is maintained at the same temperature, \(T_{s,}\) and the ambient air and surroundings are at equivalent temperatures \(\left(T_{w}=T_{\text {ue }}\right)\). (b) If \(k=0.090 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=25 \mathrm{~mm}, h_{a}=25\) W/ \(\mathrm{m}^{2} \cdot \mathrm{K}\), the surfaces are diffuse and gray with \(\varepsilon_{i}=\varepsilon_{o}=0.50, T_{s u}=900 \mathrm{~K}\), and \(T_{w}=T_{\mathrm{ar}}=\) \(300 \mathrm{~K}\), what is the outer surface temperature \(T_{\text {a }}\) and the heat loss per unit surface area associated with each option? (c) For each case, assess the effect of surface radiative properties on the outer surface temperature and the heat loss per unit area for values of

Consider the right-circular cylinder of diameter \(D\), length \(L\), and the areas \(A_{1}, A_{2}\), and \(A_{3}\) representing the base, inner, and top surfaces, respectively. (a) Show that the view factor between the base of the cylinder and the inner surface has the form \(F_{12}=2 H\left[\left(1+H^{2}\right)^{1 / 2}-H\right]\), where \(H=L D .\) (b) Show that the view factor for the inner surface to itself has the form \(F_{22}=1+H-\left(1+H^{2}\right)^{1 / 2}\).

A flue gas at 1 -atm total pressure and a temperature of \(1400 \mathrm{~K}\) contains \(\mathrm{CO}_{2}\) and water vapor at partial pressures of \(0.05\) and \(0.10 \mathrm{~atm}\), respectively. If the gas flows through a long flue of \(1-\mathrm{m}\) diameter and \(400 \mathrm{~K}\) surface temperature, determine the net radiative heat flux from the gas to the surface. Blackbody behavior may be assumed for the surface.

Consider the cavities formed by a cone, cylinder, and sphere having the same opening size \((d)\) and major dimension \((L)\), as shown in the diagram. (a) Find the view factor between the inner surface of each cavity and the opening of the cavity. (b) Find the effective emissivity of each cavity, \(\varepsilon_{e}\), as defined in Problem 13.43, assuming the inner walls are diffuse and gray with an emissivity of \(\varepsilon_{1 N^{-}}\) (c) For each cavity and wall emissivities of \(\varepsilon_{w^{\prime}}=0.5\), \(0.7\), and \(0.9\), plot \(\varepsilon_{e}\) as a function of the major dimension- to-opening size ratio, \(L /\), over a range from 1 to 10 .
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