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The diameter and surface emissivity of an electrically heated plate are \(D=300 \mathrm{~mm}\) and \(\varepsilon=0.80\), respectively. (a) Estimate the power needed to maintain a surface temperature of \(200^{\circ} \mathrm{C}\) in a room for which the air and the walls are at \(25^{\circ} \mathrm{C}\). The coefficient characterizing heat transfer by natural convection depends on the surface temperature and, in units of \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), may be approximated by an expression of the form \(h=0.80\left(T_{s}-T_{\infty}\right)^{1 / 3}\). (b) Assess the effect of surface temperature on the power requirement, as well as on the relative contributions of convection and radiation to heat transfer from the surface.

Short Answer

Expert verified
The power requirement to maintain the surface temperature of 200°C is approximately 394.77 W. As the surface temperature increases, the power requirement also increases, and the contribution of heat transfer by radiation becomes more significant compared to convection for higher surface temperatures.

Step by step solution

01

Identify the given values and formulas

We are given: Diameter of the plate, D = 300mm = 0.3m Surface emissivity, ε = 0.80 Surface temperature, Ts = 200°C = 473K Air and wall temperature, T∞ = 25°C = 298K Heat transfer coefficient, h = 0.80*(Ts-T∞)^(1/3)
02

Calculate the heat transfer coefficient

Let's calculate the heat transfer coefficient, h, using the given formula: h = 0.80*(Ts-T∞)^(1/3) h = 0.80*(473K - 298K)^(1/3) h ≈ 3.91 W/m²K
03

Calculate heat transfer by natural convection

Heat transfer by natural convection, Qconv, can be calculated using the formula: Qconv = h*A*(Ts - T∞) First, let's find the plate area A: A = π*(D/2)² A = π*(0.3/2)² A ≈ 0.0707 m² Now, calculate the heat transfer by natural convection: Qconv = 3.91 W/m²K * 0.0707 m² * (473K - 298K) Qconv ≈ 88.68 W
04

Calculate heat transfer by radiation

Heat transfer by radiation, Qrad, can be calculated using the Stefan-Boltzmann Law: Qrad = ε*A*σ*(Ts^4 - T∞^4) The Stefan-Boltzmann constant, σ, is approximately 5.67*10^(-8) W/m²K^4 Now, calculate the heat transfer by radiation: Qrad = 0.80 * 0.0707 m² * 5.67*10^(-8) W/m²K^4 * (473K^4 - 298K^4) Qrad ≈ 306.09 W
05

Find the total power requirement

The total power requirement, P, is the sum of the heat transfers by natural convection and radiation: P = Qconv + Qrad P = 88.68 W + 306.09 W P ≈ 394.77 W Therefore, the power requirement to maintain the surface temperature of 200°C is approximately 394.77 W.
06

Assess the effect of surface temperature on the power requirement, convection, and radiation contributions

As the surface temperature increases, the heat transfer by both natural convection and radiation will also increase. As seen from the calculations, the heat transfer by radiation is generally higher than that of natural convection. Thus, an increase in surface temperature will have a larger effect on the radiation contribution to heat transfer. When the surface temperature is low, natural convection can be the dominant mode of heat transfer. But as the surface temperature increases, the difference between Ts and T∞ becomes larger, causing the radiation term become more significant. The power requirement also increases with the increase in surface temperature. In conclusion, as the surface temperature increases, the power requirement also increases. The contribution of heat transfer by radiation becomes more significant compared to convection for higher surface temperatures.

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Key Concepts

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

Surface Emissivity
Understanding surface emissivity is crucial when studying heat transfer mechanisms. Surface emissivity, denoted by the symbol \(\varepsilon\), measures a material's ability to emit energy as thermal radiation. It is a dimensionless quantity that ranges from 0 to 1, where 1 corresponds to a perfect blackbody that radiates energy most efficiently, and values less than 1 represent real materials that radiate less energy than a blackbody at the same temperature.

High-emissivity materials are effective in both absorbing and emitting radiant energy, while low-emissivity materials reflect more of the incident radiant energy. This characteristic profoundly influences the heating or cooling of objects. In the context of the provided exercise, a plate with a surface emissivity of 0.80 indicates that the plate is quite effective in radiating energy as heat. When assessing power requirements for heating, accounting for emissivity is essential, as it impacts the energy radiated away and thus the energy required to maintain a specific surface temperature.
Natural Convection
Natural convection is a heat transfer process that occurs without external forces and is driven by buoyancy forces. When a fluid such as air or water comes into contact with a surface at a different temperature, the fluid near the hot surface warms up, becomes less dense, and rises, while cooler, denser fluid moves in to replace it, creating a convective flow. The heat transfer coefficient \(h\) is a measure used to quantify the rate of heat transfer from the surface to the fluid in natural convection scenarios.

The equation \(h=0.80(T_s-T_{\infty})^{1/3}\) in the exercise demonstrates how the coefficient depends on the temperature difference between the surface and the ambient environment: as the temperature difference increases, the convective heat transfer rate also increases, but less than proportionally because of the \(1/3\) power relationship. This knowledge helps students understand the power needed to maintain a surface temperature in a room, highlighting the relationship between the plate's surface temperature, ambient conditions, and the natural convective heat transfer process.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a fundamental principle in understanding radiative heat transfer. It describes the power radiated from a blackbody in terms of its temperature, stating that the total energy radiated per unit surface area of a blackbody across all wavelengths is proportional to the fourth power of the blackbody's absolute temperature. The law is mathematically expressed as \(Q_{rad} = \varepsilon A \sigma (T_s^4 - T_{\infty}^4)\), where \(Q_{rad}\) is the radiative heat transfer, \(\sigma\) is the Stefan-Boltzmann constant (approximately \(5.67\times10^{-8} W/m^2K^4\)), and \(A\) is the area of the emitting surface.

This law becomes exponentially significant as the temperature of the radiating body increases. From the exercise, we can see that the radiative heat loss from the plate is much greater than the heat loss due to convection, indicating that radiation is the dominant heat transfer mode at higher temperatures. Understanding this law helps predict how changing temperatures will impact the power required to maintain a certain condition, and it elucidates the intense impact of radiation at higher temperatures in contrast to convection.

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Most popular questions from this chapter

The heat flux that is applied to the left face of a plane wall is \(q^{\prime \prime}=20 \mathrm{~W} / \mathrm{m}^{2}\). The wall is of thickness \(L=10\) \(\mathrm{mm}\) and of thermal conductivity \(k=12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the surface temperatures of the wall are measured to be \(50^{\circ} \mathrm{C}\) on the left side and \(30^{\circ} \mathrm{C}\) on the right side, do steady-state conditions exist?

The concrete slab of a basement is \(11 \mathrm{~m}\) long, \(8 \mathrm{~m}\) wide, and \(0.20 \mathrm{~m}\) thick. During the winter, temperatures are nominally \(17^{\circ} \mathrm{C}\) and \(10^{\circ} \mathrm{C}\) at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what is the rate of heat loss through the slab? If the basement is heated by a gas furnace operating at an efficiency of \(\eta_{f}=0.90\) and natural gas is priced at \(C_{g}=\$ 0.02 / \mathrm{MJ}\), what is the daily cost of the heat loss?

A computer consists of an array of five printed circuit boards (PCBs), each dissipating \(P_{b}=20 \mathrm{~W}\) of power. Cooling of the electronic components on a board is provided by the forced flow of air, equally distributed in passages formed by adjoining boards, and the convection coefficient associated with heat transfer from the components to the air is approximately \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Air enters the computer console at a temperature of \(T_{i}=20^{\circ} \mathrm{C}\), and flow is driven by a fan whose power consumption is \(P_{f}=25 \mathrm{~W}\). (a) If the temperature rise of the airflow, \(\left(T_{o}-T_{i}\right)\), is not to exceed \(15^{\circ} \mathrm{C}\), what is the minimum allowable volumetric flow rate \(\dot{\forall}\) of the air? The density and specific heat of the air may be approximated as \(\rho=1.161\) \(\mathrm{kg} / \mathrm{m}^{3}\) and \(c_{p}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (b) The component that is most susceptible to thermal failure dissipates \(1 \mathrm{~W} / \mathrm{cm}^{2}\) of surface area. To minimize the potential for thermal failure, where should the component be installed on a PCB? What is its surface temperature at this location?

If \(T_{s} \approx T_{\text {sur }}\) in Equation \(1.9\), the radiation heat transfer coefficient may be approximated as $$ h_{r, a}=4 \varepsilon \sigma \bar{T}^{3} $$ where \(\bar{T} \equiv\left(T_{s}+T_{\text {sur }}\right) / 2\). We wish to assess the validity of this approximation by comparing values of \(h_{r}\) and \(h_{r, a}\) for the following conditions. In each case, represent your results graphically and comment on the validity of the approximation. (a) Consider a surface of either polished aluminum ( \(\varepsilon=\) \(0.05)\) or black paint \((\varepsilon=0.9)\), whose temperature may exceed that of the surroundings \(\left(T_{\text {sur }}=25^{\circ} \mathrm{C}\right)\) by 10 to \(100^{\circ} \mathrm{C}\). Also compare your results with values of the coefficient associated with free convection in air \(\left(T_{\infty}=T_{\text {sur }}\right)\), where \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)=0.98 \Delta T^{1 / 3}\). (b) Consider initial conditions associated with placing a workpiece at \(T_{s}=25^{\circ} \mathrm{C}\) in a large furnace whose wall temperature may be varied over the range \(100 \leq\) \(T_{\text {sur }} \leq 1000^{\circ} \mathrm{C}\). According to the surface finish or coating, its emissivity may assume values of \(0.05\), \(0.2\), and \(0.9\). For each emissivity, plot the relative error, \(\left(h_{r}-h_{r, a}\right) / h_{r}\), as a function of the furnace temperature.

During its manufacture, plate glass at \(600^{\circ} \mathrm{C}\) is cooled by passing air over its surface such that the convection heat transfer coefficient is \(h=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent cracking, it is known that the temperature gradient must not exceed \(15^{\circ} \mathrm{C} / \mathrm{mm}\) at any point in the glass during the cooling process. If the thermal conductivity of the glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and its surface emissivity is \(0.8\), what is the lowest temperature of the air that can initially be used for the cooling? Assume that the temperature of the air equals that of the surroundings.

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