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Chapter 12: Problem 3

The top surface of an \(L=5\)-mm-thick anodized aluminum plate is irradiated with \(G=1000 \mathrm{~W} / \mathrm{m}^{2}\) while being simultaneously exposed to convection conditions characterized by \(h=40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=30^{\circ} \mathrm{C}\). The bottom surface of the plate is insulated. For a plate temperature of \(400 \mathrm{~K}\) as well as \(\alpha=0.14\) and \(\varepsilon=0.76\), determine the radiosity at the top plate surface, the net radiation heat flux at the top surface, and the rate at which the temperature of the plate is changing with time.

Short Answer

Expert verified
The radiosity at the top surface is \(1514.20 \mathrm{W/m^2}\), the net radiation heat flux at the top surface is \(-62.00 \mathrm{W/m^2}\) (indicating a net heat loss), and the rate at which the temperature of the plate is changing with time is \(0.0593 \mathrm{K/s}\).

Step by step solution

01

Determine the surface temperature in Kelvin

The given temperature is \(400 \mathrm{~K}\), which is already in Kelvin, so no conversion is needed. We will use this temperature for further calculations.
02

Calculate the Emissive Power

We will use the Stefan-Boltzmann Law to compute the emissive power of the top surface. The formula is: \[E_b = \sigma T^4\] Where \(E_b\) is emissive power, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{Wm^{-2}K^{-4}}\)), and \(T\) is the surface temperature in Kelvin. Using the given temperature: \[E_b = (5.67 \times 10^{-8} \mathrm{Wm^{-2}K^{-4}})(400 \mathrm{K})^4 = 1452.20 \mathrm{W/m^2}\]
03

Calculate the Radiosity

The radiosity (\(J\)) is the sum of the emissive power and the reflected irradiation. The reflectance (\(1 - \alpha\)) can be used to calculate the reflected irradiation fraction: \[J = E_b + (1 - \alpha)G\] Using the given values, we get: \[J = 1452.20 \mathrm{W/m^2} + (1 - 0.14)(1000 \mathrm{W/m^2}) = 1514.20 \mathrm{W/m^2}\]
04

Calculate the Net Radiation Heat Flux

The net radiation heat flux (\(q_r \)) is the difference between the emissive power and the radiosity. It can be calculated using: \[q_r = E_b - J\] Using our computed values, we get: \[q_r = 1452.20 \mathrm{W/m^2} - 1514.20 \mathrm{W/m^2} = -62.00 \mathrm{W/m^2}\] The negative value indicates that there is net heat loss from the top surface due to radiation.
05

Determine the Convective Heat Transfer

We will use the convective heat transfer equation to calculate the heat transfer due to convection: \[q_c = hA(T - T_{\infty})\] Where \(q_c\) is convective heat transfer, \(h\) is the convection heat transfer coefficient, \(A\) is the area of the plate, and \(T_{\infty}\) is the ambient temperature. Since we need to find the change in temperature with time, we don't need the heat transfer in absolute terms but only in relative terms, so we can write: \[q_c = h(T - T_{\infty})\] Using the given values, we get: \[q_c = (40 \mathrm{W/m^2K})(400 - (30 + 273)) = 6400 \mathrm{W/m^2}\]
06

Calculate the rate of change in temperature

The rate of change in temperature of the plate (\(\frac{dT}{dt}\)) can be found using: \[\frac{dT}{dt} = \frac{q_c - q_r}{mc_p}\] Where \(m\) is the mass of the plate, \(c_p\) is the specific heat of the aluminum, and \(L\) is the thickness of the plate. Considering the plate has equal unit area, we have: \[V = L \implies m = \rho L\] Where \(\rho\) is the density of aluminum. Using the known values (\(\rho = 2700 \: \mathrm{kg/m^3}\), \(c_p = 900 \: \mathrm{J/kgK}\)), we can find the rate of change in temperature: \[\frac{dT}{dt} = \frac{6400 - (-62)}{2700 \times 5 \times 10^{-3} \times 900} = 0.0593 \: \mathrm{K/s}\] So the rate at which the temperature of the plate is changing with time is approximately \(0.0593 \: \mathrm{K/s}\).

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Key Concepts

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

Radiosity Calculation
Radiosity represents the thermal radiation emitted and reflected from a surface. To visualize this, imagine a heated plate in a room; the radiosity is the total radiant energy leaving the surface, which includes the plate鈥檚 own emission plus any reflected energy from surrounding surfaces. The calculation of radiosity is a crucial step in understanding heat transfer because it accounts for both radiation emission and reflection.

In the exercise, we evaluate the radiosity of an anodized aluminum plate, using the formula \( J = E_b + (1 - \alpha)G \). Explaining what each term means, \( E_b \) is the emissive power of the plate, interpreted as the energy the plate emits due to its temperature; \( \alpha \) indicates the plate鈥檚 absorptivity; and \( G \) is the irradiation received by the surface from external sources, like the sun. The product \( (1 - \alpha)G \) symbolizes the reflected component. It's the portion of incident light that isn't absorbed but rather reflected back into the environment. In short, radiosity tells us how 'bright' thermally a surface is, considering both its temperature and how it interacts with external radiation.
Net Radiation Heat Flux
The net radiation heat flux is the difference between the thermal radiation emitted by a surface and the radiant energy incident upon it, essentially the net amount of thermal radiation that the surface gains or loses. In practical terms, if you have a hot surface facing a cooler environment, it's going to radiate some of its heat to the surroundings. But if that surface is also receiving some radiation (like sunlight or heat radiated from other objects), you have to account for that incoming energy as well.

To calculate this value, we use the formula: \( q_r = E_b - J \). The negative result in the exercise indicates that, after considering the reflected irradiation, the surface emits more energy than it receives, which translates to a net loss of energy. This information is pivotal when determining the cooling rate of an object or simulating the thermal conditions of a system such as a satellite in space or a metal plate in industrial processing.
Convective Heat Transfer
Convective heat transfer is the movement of heat through a fluid (which could be a liquid or gas) that is caused by the fluid's movement. This might seem abstract, but it happens all the time in daily life; for instance, when you put your hand above a hot stove, you feel heat not only from radiation but also from the air moving around, heated by the stove. This is convection at play.

The exercise's convective heat transfer calculation uses the coefficient of convection \( h \) and the temperature difference between the plate and its surroundings \( (T - T_{\infty}) \) via the equation \( q_c = h(T - T_{\infty}) \). The result shows how much heat energy per unit area is transferred by convection. It鈥檚 important for determining how quickly an object will equalize with the ambient temperature or how effective a heating or cooling system might be. Clearer understanding and precise calculation of convective heat transfer are essential in many fields, from designing efficient heating and cooling systems to forecasting weather patterns.

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Most popular questions from this chapter

Consider an opaque, gray surface whose directional absorptivity is \(0.8\) for \(0 \leq \theta \leq 60^{\circ}\) and \(0.1\) for \(\theta>60^{\circ}\). The surface is horizontal and exposed to solar irradiation comprised of direct and diffuse components. (a) What is the surface absorptivity to direct solar radiation that is incident at an angle of \(45^{\circ}\) from the normal? What is the absorptivity to diffuse irradiation? (b) Neglecting convection heat transfer between the surface and the surrounding air, what would be the equilibrium temperature of the surface if the direct and diffuse components of the irradiation were 600 and \(100 \mathrm{~W} / \mathrm{m}^{2}\), respectively? The back side of the surface is insulated.

An instrumentation transmitter pod is a box containing electronic circuitry and a power supply for sending sensor signals to a base receiver for recording. Such a pod is placed on a conveyor system, which passes through a large vacuum brazing furnace as shown in the sketch. The exposed surfaces of the pod have a special diffuse, opaque coating with spectral emissivity as shown. To stabilize the temperature of the pod and prevent overheating of the electronics, the inner surface of the pod is surrounded by a layer of a phase- change material (PCM) having a fusion temperature of \(87^{\circ} \mathrm{C}\) and a heat of fusion of \(25 \mathrm{~kJ} / \mathrm{kg}\). The pod has an exposed surface area of \(0.040 \mathrm{~m}^{2}\) and the mass of the PCM is \(1.6 \mathrm{~kg}\). Furthermore, it is known that the power dissipated by the electronics is \(50 \mathrm{~W}\). Consider the situation when the pod enters the furnace at a uniform temperature of \(87^{\circ} \mathrm{C}\) and all the \(\mathrm{PCM}\) is in the solid state. How long will it take before all the PCM changes to the liquid state?

A roof-cooling system, which operates by maintaining a thin film of water on the roof surface, may be used to reduce air-conditioning costs or to maintain a cooler environment in nonconditioned buildings. To determine the effectiveness of such a system, consider a sheet metal roof for which the solar absorptivity \(\alpha_{5}\) is \(0.50\) and the hemispherical emissivity \(\varepsilon\) is \(0.3\). Representative conditions correspond to a surface convection coefficient \(h\) of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), a solar irradiation \(G_{S}\) of \(700 \mathrm{~W} / \mathrm{m}^{2}\), a sky temperature of \(-10^{\circ} \mathrm{C}\), an atmospheric temperature of \(30^{\circ} \mathrm{C}\), and a relative humidity of \(65 \%\). The roof may be assumed to be well insulated from below. Determine the roof surface temperature without the water film. Assuming the film and roof surface temperatures to be equal, determine the surface temperature with the film. The solar absorptivity and the hemispherical emissivity of the film-surface combination are \(\alpha_{S}=\) \(0.8\) and \(s=0.9\), respectively.

A manufacturing process involves heating long copper rods, which are coated with a thin film, in a large furnace whose walls are maintained at an elevated temperature \(T_{w}\). The furnace contains quiescent nitrogen gas at 1-atm pressure and a temperature of \(T_{x}=T_{w}\). The film is a diffuse surface with a spectral emissivity of \(\varepsilon_{\lambda}=0.9\) for \(\lambda \leq 2 \mu \mathrm{m}\) and \(\varepsilon_{\lambda}=0.4\) for \(\lambda>2 \mu \mathrm{m}\). (a) Consider conditions for which a rod of diameter \(D\) and initial temperature \(T_{i}\) is inserted in the furnace, such that its axis is horizontal. Assuming validity of the lumped capacitance approximation, derive an equation that could be used to determine the rate of change of the rod temperature at the time of insertion. Express your result in terms of appropriate variables. (b) If \(T_{w}=T_{\mathrm{o}}=1500 \mathrm{~K}, T_{i}=300 \mathrm{~K}\), and \(D=10 \mathrm{~mm}\), what is the initial rate of change of the rod temperature? Confirm the validity of the lumped capacitance approximation. (c) Compute and plot the variation of the rod temperature with time during the heating process.

Growers use giant fans to prevent grapes from freezing when the effective sky temperature is low. The grape, which may be viewed as a thin skin of negligible thermal resistance enclosing a volume of sugar water, is exposed to ambient air and is irradiated from the sky above and ground below. Assume the grape to be an isothermal sphere of \(15-\mathrm{mm}\) diameter, and assume uniform blackbody irradiation over its top and bottom hemispheres due to emission from the sky and the earth, respectively. (a) Derive an expression for the rate of change of the grape temperature. Express your result in terms of a convection coefficient and appropriate temperatures and radiative quantities. (b) Under conditions for which \(T_{\text {sky }}=235 \mathrm{~K}, T_{\mathrm{s}}=\) \(273 \mathrm{~K}\), and the fan is off \((V=0)\), determine whether the grapes will freeze. To a good approximation, the skin emissivity is 1 and the grape thermophysical properties are those of sugarless water. However, because of the sugar content, the grape freezes at \(-5^{\circ} \mathrm{C}\). (c) With all conditions remaining the same, except that the fans are now operating with \(V=1 \mathrm{~m} / \mathrm{s}\), will the grapes freeze?
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