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The lower side of a \(400-\mathrm{mm}\)-diameter disk is heated by an electric furnace, while the upper side is exposed to quiescent, ambient air and sumoundings at \(300 \mathrm{~K}\). The radiant furnace (negligible convection) is of circular construction with the bottoen surface \(\left(\alpha_{1}-0.6\right)\) and cylindrical side surface \(\left(\varepsilon_{1}=1.0\right)\) maintained af \(T_{1}=T_{2}=500 \mathrm{~K}\). The surface of the disk facing the radiant furnace is black \(\left(\varepsilon_{d, 1}=1.0\right)\). while the upper surface has an emissivity of \(\varepsilon_{d, 2}=0.8\). Assume the plate and furnace surfaces to be diffuse and gray. (a) Determine the net heat transfer rate to the disk, \(q_{\text {nated, when }} T_{d}=400 \mathrm{~K}\). (b) Plot \(q_{\text {netd as a }}\) a function of the disk temperature for \(300 \leq T_{a} \leq 500 \mathrm{~K}\), with all other conditions remaining the same. What is the steady-state temperature of the disk?

Step by step solution

01

Find the radiation heat transfer to the disk

To find the radiation heat transfer to the disk, we will use the Stefan-Boltzmann law, which states: \(q = \varepsilon \sigma (T_{furnace}^4 - T_{disk}^4)\) Here, \(\varepsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67×10^{-8} W/m^2K^4\)), and \(T\) is the temperature in Kelvin. Since the emissivities are different on each side of the disk, we must account for them separately. For the bottom surface facing the radiant furnace, the radiation heat transfer is: \(q_{1} = \varepsilon_{d1} \sigma (T_{1}^4 - T_{d}^4)\) For the top surface, the radiation heat transfer is: \(q_{2} = \varepsilon_{d2} \sigma (T_{2}^4 - T_{d}^4)\)

02

Calculate the heat transfer rates for the given temperatures

We are given the following temperatures: \(T_{1} = T_{2} = 500K\), \(T_{d} = 400K\), and \(T_{a} = 300K\). We are also given the emissivities: \(\varepsilon_{d1} = 1.0\) and \(\varepsilon_{d2} = 0.8\). Since the ambient surroundings are at 300 K, we must also account for the radiation heat transfer from the top surface of the disk to the ambient surroundings: \(q_{3} = \varepsilon_{d2} \sigma (T_{a}^4 - T_{d}^4)\) Now we can calculate the heat transfer rates for the given temperatures: \(q_{1} = 1.0(5.67×10^{-8})(500^4 - 400^4) = 3.605 × 10^3 W/m^2\) \(q_{2} = 0.8(5.67×10^{-8})(500^4 - 400^4) = 2.884 × 10^3 W/m^2\) \(q_{3} = 0.8(5.67×10^{-8})(300^4 - 400^4) = -1.470 × 10^3 W/m^2\)

03

Find the net heat transfer rate to the disk

To find the net heat transfer rate to the disk, we will add up the heat transfer rates calculated in the previous step: \(q_{net} = q_{1} + q_{2} + q_{3} = 3.605 × 10^3 + 2.884 × 10^3 + (-1.470 × 10^3) = 5.019 × 10^3 W/m^2\)

04

Plot the net heat transfer rate as a function of disk temperature

To plot the net heat transfer as a function of disk temperature, we will need to substitute a range of disk temperatures (300 K to 500 K) into the equations from step 1 and step 3: \(q_{net}(T_{d}) = \varepsilon_{d1}\sigma(T_{1}^4 - T_{d}^4) + \varepsilon_{d2}\sigma(T_{2}^4 - T_{d}^4) - \varepsilon_{d2}\sigma(T_{a}^4 - T_{d}^4)\) Once the plot is created, we will look for the steady-state temperature, which occurs when the net heat transfer to the disk is zero (i.e., the heat transfer rate in equals the heat transfer rate out). By plotting the equation and observing where the net heat transfer intersects the x-axis, the steady-state temperature of the disk can be found. We will be using specialized software or graphing calculators to achieve this.

05

Determine the steady-state temperature of the disk

After plotting q_net as a function of the disk temperature and finding where it intersects the x-axis, we will get the steady-state temperature of the disk. This steady-state temperature will indicate that the disk is neither heating nor cooling and is maintaining a constant temperature. The final answer will vary depending on the software and plot used, but by following the procedure outlined in this solution, it should be possible to find the steady-state temperature of the disk.

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Key Concepts

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

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle for understanding radiation heat transfer. It states that the power radiated per unit area of a black body is directly proportional to the fourth power of its absolute temperature. This can be expressed by the formula: \( q = \varepsilon \sigma T^4 \), where \( q \) represents the radiant energy emitted per unit area, \( \varepsilon \) is the emissivity of the material, \( \sigma \) is the Stefan-Boltzmann constant approximately equal to \( 5.67 \times 10^{-8} W/m^2K^4 \), and \( T \) represents the absolute temperature in Kelvin.

• For perfect black bodies \( (\varepsilon = 1) \), the law defines that the radiant power is at its maximum as all radiation at that temperature is emitted.
• In typical applications, surfaces are not perfect black bodies, and their emissivity \( (\varepsilon) \) is less than 1.

The law is useful in our exercise since it helps us calculate the radiant heat transfer from the surfaces of the disk when different emissivities are involved. By comparing the heat emitted by the furnace and the disk, students can determine the net heat transfer over the disk surface.

Emissivity

Emissivity is a material property that measures a surface's ability to emit thermal radiation compared to a perfect black body at the same temperature, and it ranges from 0 to 1.

• If \( \varepsilon = 1 \), the surface is a perfect emitter (like a black body).
• If \( \varepsilon = 0 \), the surface does not emit thermal radiation.

In the exercise, the emissivity values \( \varepsilon_{d1} \) and \( \varepsilon_{d2} \) determine how each side of the disk interacts with thermal radiation. The bottom side of the disk, facing the furnace, is treated as a perfect emitter (\( \varepsilon_{d1} = 1.0 \)), absorbing and emitting radiation efficiently. However, the upper side, exposed to cooler ambient air, has an emissivity of \( \varepsilon_{d2} = 0.8 \), allowing it to emit less than a perfect black body would. Understanding emissivity is crucial for calculating the net rate of heat transfer, where both sides contribute differently to the system's energy balance because they radiate differently.

Steady-State Temperature

The concept of steady-state temperature refers to a condition where the temperature of the system remains constant over time. In the context of the exercise, it occurs when the net heat transfer to or from the disk reaches zero. This means that the heat entering the disk equals the heat leaving it. In practical terms:

• Energy balance is achieved: input and output energy flows are equal.
• The system does not experience any temperature changes with time.

To find the steady-state temperature in the exercise, students need to analyze when the computed net heat transfer becomes zero using the net heat transfer equation. By adjusting the temperature of the disk \( (T_d) \) and recalculating the net heat transfer, students can identify the point where the disk neither gains nor loses heat. Achieving this balance ensures the system is in equilibrium, predicting realistic behavior in thermal systems.