91Ó°ÊÓ

Square plates freshly sprayed with an epoxy paint must be cured at \(140^{\circ} \mathrm{C}\) for an extended period of time. The plates are located in a large enclosure and heated by a bank of infrared lamps. The top surface of each plate has an emissivity of \(\varepsilon=0.8\) and experiences convection with a ventilation airstream that is at \(T_{\infty}=27^{\circ} \mathrm{C}\) and provides a convection coefficient of \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The irradiation from the enclosure walls is estimated to be \(G_{\text {wall }}=450 \mathrm{~W} / \mathrm{m}^{2}\), for which the plate absorptivity is \(\alpha_{\text {wall }}=0.7\). (a) Determine the irradiation that must be provided by the lamps, \(G_{\text {lamp. }}\). The absorptivity of the plate surface for this irradiation is \(\alpha_{\text {Lamp }}=0.6\). (b) For convection coefficients of \(h=15,20\), and \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), plot the lamp irradiation, \(G_{\text {lamp, as a }}\) function of the plate temperature, \(T_{s}\), for \(100 \leq\) \(T_{x} \leq 300^{\circ} \mathrm{C}\). (c) For convection coefficients in the range from 10 to \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a lamp irradiation of \(G_{\text {lmp }}=\) \(3000 \mathrm{~W} / \mathrm{m}^{2}\), plot the airstream temperature \(T_{x}\) required to maintain the plate at \(T_{x}=140^{\circ} \mathrm{C}\).

Step by step solution

01

Energy balance for the plate

The energy balance equation for the plate can be written as: \[q_{conv} + q_{\text{lamp}} + q_{\text{wall}} = 0\] where \(q_{conv}\) is the convective heat transfer rate, and \(q_{\text{lamp}}\) and \(q_{\text{wall}}\) are the absorbed irradiation rates from the lamps and enclosure walls, respectively.

02

Calculate convective heat transfer rate

The convective heat transfer rate \(q_{conv}\) can be calculated as: \[q_{conv} = h A (T_s - T_\infty)\] We need to solve for \(T_s\), the plate temperature. We will get this value from the given information: the plate temperature must be \(140^{\circ}\mathrm{C}\). The convective heat transfer coefficient, h, is given as 20 \(\mathrm{W/m^2K}\), and the air stream temperature is given as \(27^{\circ}\mathrm{C}\).

03

Calculate absorbed irradiation rates

The absorbed irradiation rates from the lamps and enclosure walls can be calculated using the emissivity and absorptivity values given: \[q_{\text{lamp}} = \alpha_{\text{lamp}}G_{\text{lamp}}A\] \[q_{\text{wall}} = \alpha_{\text{wall}}G_{\text{wall}}A\] We are given the values for \(\alpha_{\text{lamp}} = 0.6\), \(\alpha_{\text{wall}} = 0.7\), and \(G_{\text{wall}} = 450\,\mathrm{W/m^2}\).

04

Solve for \(G_{\text{lamp}}\)

Now, we substitute values from steps 2 and 3 into the energy balance equation and solve for \(G_{\text{lamp}}\): \[q_{conv} + q_{\text{lamp}} + q_{\text{wall}} = 0\] \[20\,(T_s - 27) + 0.6G_{\text{lamp}} - 0.7(450) = 0\] \[20\,(140 - 27) + 0.6 G_{\text{lamp}} - 0.7(450) = 0\] Now solve for \(G_{\text{lamp}}\): \[0.6 G_{\text{lamp}} = -20(113) + 0.7(450)\] \[G_{\text{lamp}} = \frac{-20(113) + 0.7(450)}{0.6} = 1366.67\,\mathrm{W/m^2}\] So, the required lamp irradiation is \(G_{\text{lamp}} = 1366.67\,\mathrm{W/m^2}\). **Part (b)**

05

Create functions for \(G_{\text{lamp}}\) with different convection coefficients

Use the energy balance equation as a function of the plate temperature, \(T_s\), for various convection coefficients: \[q_{conv}(h) + q_{\text{lamp}}(h) + q_{\text{wall}} = 0\] Find \(G_{\text{lamp}}(h)\) as a function of \(T_s\) for each given convection coefficient: \[G_{\text{lamp}}(h) = \frac{20\,(T_s - T_\infty) + 0.7(450)}{0.6}\]

06

Plot functions for different convection coefficients

Plot \(G_{\text{lamp}}(h)\) as a function of \(T_s\) for each given convection coefficient \(h = 15, 20, 30\,\mathrm{W/m^2K}\). The plot should be within the range of \(100 \leq T_s \leq 300^{\circ}\mathrm{C}\). **Part (c)**

07

Create functions for airstream temperature as a function of convection coefficient

Use the energy balance equation as a function of the convection coefficient: \[q_{conv}(T_x) + q_{\text{lamp}} + q_{\text{wall}} = 0\] \[G_{\text{lamp}} = \frac{h\,(T_s - T_x) + q_{\text{wall}}}{0.6}\]

08

Solve for \(T_x\) and plot

For a given lamp irradiation of \(3000\,\mathrm{W/m^2}\) and convection coefficients in the range of 10 to 30 \(\mathrm{W/m^2K}\), solve for the airstream temperature \(T_x\) that maintains the plate at \(140^{\circ} \mathrm{C}\): \[T_x = T_s - \frac{0.6 G_{\text{lamp}} + q_{\text{wall}}}{h}\] \[T_x = 140 - \frac{0.6(3000) - 0.7(450)}{h}\] Plot the airstream temperature \(T_x\) as a function of the convection coefficient from 10 to 30 \(\mathrm{W/m^2K}\).

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Key Concepts

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

Infrared Heating

Infrared heating is an innovative method of heat transfer that utilizes electromagnetic radiation, specifically in the infrared spectrum. This form of heating is especially useful in applications such as curing freshly painted surfaces, as in our exercise, where heat needs to be absorbed directly by an object rather than warming the surrounding air first.
Infrared lamps, like the ones used in this situation, emit invisible wavelengths that interact directly with the surface of materials. This direct contact ensures efficient heating, as the emitted energy is absorbed locally on the plate surface. The energy from the infrared lamps is crucial in reaching the curing temperature of the epoxy paint, ensuring the process is both effective and consistent.

• Infrared heating offers rapid and localized energy transfer.
• Irrelevant of air movement, it directly heats the object's surface.
• Infrared is useful for energy-efficient heating in specific applications like curing and drying.

Convection Heat Transfer

When we talk about convection heat transfer, we're discussing the mechanism by which thermal energy is transmitted through a fluid, whether it's a gas or a liquid. In the context of the exercise, heat is transferred between the airstream and the plate through convection.
Convection involves both the bulk movement of molecules within fluids and surface interactions driven by differences in temperature. The heat transfer coefficient (\( h \) in this case, 20 \( \mathrm{W/m^2K} \)) quantifies how efficiently this exchange occurs. It varies with factors such as flow velocity and fluid properties, which is why we see different outcomes for different coefficients in the problem.

• Convection heat transfer combines fluid movement and thermal conduction.
• The efficiency is dictated by the convection coefficient.
• This process is highly dependent on fluid velocity and temperature gradients.

Emissivity

Emissivity is a measure of a surface's ability to emit thermal radiation compared to an ideal black body. An emissivity value close to 1 implies that the surface is an excellent emitter of radiation. In the problem, the plate's emissivity is 0.8, indicating it is quite efficient in radiating energy.
A surface with high emissivity will lose heat through radiation more readily than one with low emissivity. This property becomes particularly important in balancing the heat needed from the infrared lamps and the surrounding environment when attempting to reach or maintain specific temperatures.

• Emissivity indicates a surface's efficiency in thermal radiation emission.
• A value of 0.8 shows significant radiation capability compared to a black body.
• It influences how thermal and infrared energy is managed.

Energy Balance

The concept of energy balance is fundamental in thermal analyses, ensuring that the energy entering and leaving a system is accounted for. In this exercise, the total energy from convection, lamp irradiation, and wall irradiation must sum to zero to maintain a constant plate temperature.
The energy flowing into the system from the lamps and the walls must equal the energy leaving through convection and emission. This balance allows us to solve for unknowns, like the required lamp irradiation to maintain a particular temperature.

• An accurate energy balance ensures a system is not gaining or losing energy.
• It is key to calculating and predicting system behavior under thermal processes.
• It can be expressed as \( q_{conv} + q_{\text{lamp}} + q_{\text{wall}} = 0 \).

Irradiation

Irradiation refers to the process by which radiant energy is received on a surface. In this specific case, irradiation from the lamps and the walls plays a significant role in determining the surface temperature of the plate.
The amount of irradiation absorbed depends on both the intensity of the source and the absorptive properties of the receiving surface. With given absorptivity values (\( \alpha_{\text{lamp}} = 0.6 \) and \( \alpha_{\text{wall}} = 0.7 \)), one can calculate how much of the incident energy is effectively absorbed by the plate.

• Irradiation is critical in supplying heat energy to surfaces.
• The effectiveness of irradiation depends on source intensity and surface absorptivity.
• Calculating irradiation helps determine how much energy a surface can absorb effectively.