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A radiant heater, which is used for surface treatment processes, consists of a long cylindrical heating element of diameter \(D_{1}=0.005 \mathrm{~m}\) and emissivity \(\varepsilon_{1}=0.80\). The heater is partially enveloped by a long. thin parabolic reflector whose inner and outer surface emissivities are \(\varepsilon_{24}=0.10\) and \(\varepsilon_{20}=0.80\), respectively. Inner and outer surface areas per unit length of the reflector are each \(A_{2}^{\prime}=A_{20}^{\prime}=0.20 \mathrm{~m}\), and the average convection coefficient for the combined inner and outer surfaces is \(\bar{h}_{2 \dot{m}}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The system may be assumed to be in an infinite, quiescent medium of atmospheric air at \(T_{-}=300 \mathrm{~K}\) and to be exposed to large surroundings at \(T_{\text {aur }}=300 \mathrm{~K}\). (a) Sketch the appropriate radiation circuit, and write expressions for each of the network resistances. (b) If, under steady-state conditions, electrical power is dissipated in the heater at \(P_{1}^{\prime}=\) \(1500 \mathrm{~W} / \mathrm{m}\) and the heater surface temperature is \(T_{1}=1200 \mathrm{~K}\), what is the net rate at which radiant energy is transferred from the heater? (c) What is the net rate at which radiant energy is transferred from the heater to the surroundings? (d) What is the temperature, \(T_{2}\), of the reflector?

Step by step solution

01

(a) Sketch the Radiation Circuit and Write Resistance Expressions

To analyze this problem, we'll create a radiation circuit showing all the thermal resistances involved: 1. Radiative resistance from the heating element to the reflector, given by: \[R^{rad}_{1,2} = \frac{1}{\varepsilon_1 A_1'} - \frac{1}{A_1'} = \frac{1-\varepsilon_1}{\varepsilon_1 A_1'}\] 2. The combined radiative and convective resistance from the heating reflector to the surroundings, calculated as: \[R_{2,aur} = R_{2}^{rad} + R_2^{conv} = \frac{1}{\varepsilon_{20} A_{20}'} + \frac{1}{\bar{h}_{2} A_2^{'}}\] We have the circuit, and the equations for thermal resistances.

02

(b) Calculate Radiant Energy Transfer from Heater

To find the net rate of radiant energy transfer from the heater, we first need to calculate \(q^{'}_{1,2}\) (radiative heat transfer per unit length from surface 1 to surface 2): \[q^{'}_{1,2} = \frac{T^4_1 - T^4_2}{R^{rad}_{1,2}}\] Given that under steady-state conditions, the electrical power dissipated in the heater is equal to the rate at which heat is transferred from surface 1 of the heater: \[P^{'}_1 = q^{'}_{1,2}\] Plugging in the values, we have: \[1500 \mathrm{~W/m} = \frac{(1200 \mathrm{~K})^4 - T^4_2}{\frac{1 - 0.8}{0.8 \cdot A_1^{'}}}\] We need to compute the surface area of the heating element per unit length, \(A_1'\): \[A_1^{'} = \pi D_1\] Using the given diameter of the heating element, \(D_1 = 0.005 \, \text{m}\): \[A_1^{'} = \pi \cdot 0.005 \, \text{m} = 0.0157 \, \text{m}^2\] Now we can solve for \(T_2^4\): \[T^4_2 = (1200 \mathrm{~K})^4 - \frac{1500 \mathrm{~W/m} \cdot \frac{1 - 0.8}{0.8 \cdot 0.0157 \mathrm{~m^2}}} = 2.0053 \times 10^9 \mathrm{K^4}\] So, net rate at which radiant energy is transferred from the heater to surface 2: \[q^{'}_{1,2} = 1500 \mathrm{~W/m}\]

03

(c) Calculate Radiant Energy Transfer from Heater to Surroundings

Now, we'll calculate the net rate at which radiant energy is transferred from the heater to the surroundings: \[q^{'}_{2,aur} = \frac{T^4_2 - T^4_{aur}}{R_{2,aur}}\] Using values for \(T_2^4\), \(T_1\), \(\varepsilon_{20}\), \(A_{20}'\), and \(\bar{h}_{2}\): \[q^{'}_{2,aur} = \frac{2.0053 \times 10^9 \mathrm{K^4} - (300 \mathrm{~K})^4}{\frac{1}{0.8 \cdot 0.20 \mathrm{~m^2}} + \frac{1}{2 \cdot 0.20 \mathrm{~m^2}}} = 1460.37 \mathrm{~W/m}\] The net rate at which radiant energy is transferred from the heater to the surroundings is approximately: \[q^{'}_{2,aur} = 1460.37 \mathrm{~W/m}\]

04

(d) Determine the Reflector Temperature

Lastly, we'll determine the reflector temperature, \(T_2\). We know the value \(T_2^4 = 2.0053 \times 10^9 \mathrm{K^4}\). We can now find \(T_2\) by taking the fourth root of this value: \[T_2 = \sqrt[4]{2.0053 \times 10^9 \mathrm{K^4}} = 361.16 \mathrm{~K}\] The reflector's temperature is: \[T_2 = 361.16 \mathrm{~K}\]

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Key Concepts

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

Radiation Circuit Analysis

Understanding radiation circuit analysis is crucial for solving problems related to radiant energy transfer. In thermal systems, this concept is similar to electrical circuits, where temperature differences drive heat flow, and thermal resistances impede it. Just as electrical resistance opposes current, thermal resistance hinders heat transfer.

In our textbook problem, a radiation circuit was drawn to represent the radiative and convective heat transfer from a cylindrical heater to its surroundings via a parabolic reflector. Here, each resistance is derived based on known emissivities and the Stefan-Boltzmann law for radiant exchange. The unique feature of a radiation circuit is that it helps to visualize and calculate the flow of radiant energy between different surfaces, taking into account both emission and reflection characteristics of the materials involved.

Such analysis is typically used to determine the heater's power dissipation and surface temperatures within the system. By using formulas that quantify thermal resistance, students can calculate the rate at which heat is exchanged and understand how modifying surface properties or configurations will impact the overall heat transfer.

Thermal Resistance in Heat Transfer

The concept of thermal resistance plays a central role in heat transfer, particularly when dealing with steady-state conditions. Thermal resistance is a measure of a material's ability to resist the flow of heat. In the context of our exercise, both radiative and convective thermal resistances are taken into account for comprehensive heat transfer analysis.

The radiative resistance depends on surface emissivity and area, while the convective resistance is a function of the convection heat transfer coefficient and the surface area. It's akin to resistance in an electrical circuit – higher thermal resistance means less heat transfer for a given temperature difference. Let's analogize with a blanket: a thicker blanket (higher resistance) means less heat loss from your body (the heater) to the air (the surroundings).

Students should grasp that by summing these resistances, much like in series electrical circuits, they will be able to find the overall resistance which governs the heat transfer rate under steady state in our textbook problem.

Steady-State Heat Transfer

When we discuss steady-state heat transfer, we refer to a condition where the temperature distribution in a system does not change over time. This equilibrium is an assumption that simplifies the analysis of thermal systems, as it implies that all heat entering a system must also leave it at the same rate – a principle applied in our exercise as well.

In the context of our problem, steady-state conditions allow us to set the power dissipated by the heater equal to the radiant energy transfer rate. This simplification is fundamental because it negates the need to consider time-dependent changes in temperature, enabling students to focus on applying resistance concepts to the system’s current state. Think of it as water flow through a pipe: at steady state, the amount of water entering one end equals the amount of water leaving at the other, assuming there’s nowhere else for the water to go.