An electric heater with the total surface area of \(0.25 \mathrm{~m}^{2}\) and emissivity \(0.75\) is in a room where the air has a temperature of \(20^{\circ} \mathrm{C}\) and the walls are at \(10^{\circ} \mathrm{C}\). When the heater consumes \(500 \mathrm{~W}\) of electric power, its surface has a steady temperature of \(120^{\circ} \mathrm{C}\). Determine the temperature of the heater surface when it consumes \(700 \mathrm{~W}\). Solve the problem (a) assuming negligible radiation and (b) taking radiation into consideration. Based on your results, comment on the assumption made in part ( \(a\) ).

Short Answer

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1. In part (a), we neglect radiation. Calculate the heater surface temperature (T2) in this case by using the relationship: \(T_2 = T_\text{air} + \frac{P_2}{P_1} (T_1 - T_\text{air})\) 2. In part (b), we consider radiation. Solve the equation numerically to find the value of T2: \(P_2 = K (T_2 - T_\text{air}) + \epsilon \sigma A (T_2^4 - T_\text{wall}^4)\) 3. Compare the results obtained in parts (a) and (b) and comment on the differences in surface temperatures.

Step by step solution

01

Part (a): Neglecting Radiation

Let's start by analyzing the situation when we assume negligible radiation. In this case, 100% of the delivered power (P) is dedicated to heating the room's air. From the steady-state situation, we can establish a relationship between temperature and power consumption: \(P_1 = K (T_1 - T_\text{air})\) Here, K is a constant representing the heat transfer from the heater's surface to the air. Since we don't take into account the radiation, the value of K should remain constant when the power consumption increases. Now, we can use the same relationship for the problem scenario: \(P_2 = K (T_2 - T_\text{air})\) From these two relationships, we can find T2: \(\frac{P_1}{P_2} = \frac{T_1 - T_\text{air}}{T_2 - T_\text{air}}\) Solve for T2: \(T_2 = T_\text{air} + \frac{P_2}{P_1} (T_1 - T_\text{air})\) Finally, plug in the given values to determine T2.
02

Part (b): Considering Radiation

Now we will take radiation into account. The Stefan-Boltzmann law helps us determine the power radiated from the heater's surface. The Stefan-Boltzmann law states that the radiated power is given by: \(P_\text{rad} = \epsilon \sigma A (T^4 - T_\text{wall}^4)\) where \(\sigma\) is the Stefan-Boltzmann constant: \(5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}}\). As the heat given to the room's air is now decreased, we have: \(P_\text{air} = P - P_\text{rad}\) The relationships between power, temperature, and radiation can be established for the steady-state case and the problem scenario: For the initial steady-state case: \(P_1 = K (T_1 - T_\text{air}) + \epsilon \sigma A (T_1^4 - T_\text{wall}^4)\) For the problem scenario: \(P_2 = K (T_2 - T_\text{air}) + \epsilon \sigma A (T_2^4 - T_\text{wall}^4)\) There is no closed-form analytical solution for T2 in this case, so we need to solve the above equation numerically using a numerical method such as the Newton-Raphson method or a root-finding algorithm. Then, compare the results obtained in parts (a) and (b) and comment based on the differences in surface temperatures.

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Key Concepts

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

Stefan-Boltzmann Law
Understanding the Stefan-Boltzmann Law is foundational when studying heat transfer, particularly radiation. This law reveals how the amount of radiant energy emitted from a black body in space is proportional to the fourth power of the black body's temperature. Mathematically, it's represented as:
\[P_{\text{rad}} = \epsilon \sigma A (T^4 - T_{\text{environment}}^4)\]
where \(P_{\text{rad}}\) is the power radiated, \(\epsilon\) is the emissivity of the material, \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the area of the body, and \(T\) is the absolute temperature of the body's surface. Importantly, this relationship indicates an intense increase in radiated power with even a moderate rise in temperature, highlighting the significance of radiation in heat transfer scenarios at higher temperatures. For objects that are not perfect black bodies, which is most real-world cases, the emissivity factor adjusts the ratio to account for less than perfect radiation efficiency.
Radiation Heat Transfer
Radiation heat transfer is one of the three modes of heat transfer, the others being conduction and convection. What sets radiation apart is its ability to transfer energy through a vacuum—it does not require a medium. This type of heat transfer is based on electromagnetic waves and involves the emission of infrared radiation from heated objects.
In the context of our heater, radiation plays a pivotal role as the heater's temperature escalates. As the heater emits energy, some of that energy is absorbed, reflected, and transmitted by surrounding surfaces, ultimately altering the room's temperature distribution. Since heat radiation can occur without direct contact, it's crucial to consider the impact of surrounding surfaces' temperatures, such as the walls, especially when they're at different temperatures from the air.
Temperature-Heat Power Relationship
The temperature-heat power relationship is an essential concept in heat transfer, relating to how a change in power consumption affects an object's temperature. When all other factors remain constant, as power consumption increases, so does the temperature. This relationship, linear in cases without substantial radiation, becomes complex with increasing temperatures where radiation cannot be neglected.
In the electric heater scenario, the proportionality constant \(K\) represents the heater's efficiency at transferring heat to the surrounding air (assuming no radiation). When we neglect radiation, the relationship simplifies to (assuming \(K\) is constant):
\[P \propto (T - T_{\text{air}})\]
However, in real-life applications—and in our exercise—radiation's influence grows as temperatures increase, requiring the use of more nuanced calculations to accurately determine the heater's surface temperature at different power levels. This insight helps us grasp how the heater's efficiency can change with varying conditions.
Numerical Methods in Heat Transfer
Numerical methods in heat transfer are crucial for solving complicated problems that do not have straightforward analytical solutions. Such methods break down complex equations into more manageable calculations, often through iteration, to find approximate solutions.
To solve for the temperature of our electric heater when considering radiation, we turn to numerical methods because the relationship between power consumption, temperature, and radiated heat involves high-order polynomials due to the Stefan-Boltzmann Law. By implementing algorithms like the Newton-Raphson method or using computational software, we can iteratively hone in on the temperature value that balances the heat equation, accounting for both the direct heating (convection) and radiation effects.

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Most popular questions from this chapter

Consider a flat-plate solar collector placed on the roof of a house. The temperatures at the inner and outer surfaces of the glass cover are measured to be \(33^{\circ} \mathrm{C}\) and \(31^{\circ} \mathrm{C}\), respectively. The glass cover has a surface area of \(2.5 \mathrm{~m}^{2}\), a thickness of \(0.6 \mathrm{~cm}\), and a thermal conductivity of \(0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is lost from the outer surface of the cover by convection and radiation with a convection heat transfer coefficient of \(10 \mathrm{~W} /\) \(\mathrm{m}^{2} \cdot \mathrm{K}\) and an ambient temperature of \(15^{\circ} \mathrm{C}\). Determine the fraction of heat lost from the glass cover by radiation.

A 2.1-m-long, 0.2-cm-diameter electrical wire extends across a room that is maintained at \(20^{\circ} \mathrm{C}\). Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be \(180^{\circ} \mathrm{C}\) in steady operation. Also, the voltage drop and electric current through the wire are measured to be \(110 \mathrm{~V}\) and \(3 \mathrm{~A}\), respectively. Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room. Answer: \(156 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

An electronic package in the shape of a sphere with an outer diameter of \(100 \mathrm{~mm}\) is placed in a large laboratory room. The surface emissivity of the package can assume three different values \((0.2,0.25\), and \(0.3)\). The walls of the room are maintained at a constant temperature of \(77 \mathrm{~K}\). The electronics in this package can only operate in the surface temperature range of \(40^{\circ} \mathrm{C} \leq T_{s} \leq 85^{\circ} \mathrm{C}\). Determine the range of power dissipation \((\dot{W})\) for the electronic package over this temperature range for the three surface emissivity values \((\varepsilon)\). Plot the results in terms of \(\dot{W}(\mathrm{~W})\) vs. \(T_{s}\left({ }^{\circ} \mathrm{C}\right)\) for the three different values of emissivity over a surface temperature range of 40 to \(85^{\circ} \mathrm{C}\) with temperature increments of \(5^{\circ} \mathrm{C}\) (total of 10 data points for each \(\varepsilon\) value). Provide a computer generated graph for the display of your results and tabulate the data used for the graph. Comment on the results obtained.

Consider heat transfer through a windowless wall of a house on a winter day. Discuss the parameters that affect the rate of heat conduction through the wall.

Steady heat conduction occurs through a \(0.3\)-m-thick \(9 \mathrm{~m} \times 3 \mathrm{~m}\) composite wall at a rate of \(1.2 \mathrm{~kW}\). If the inner and outer surface temperatures of the wall are \(15^{\circ} \mathrm{C}\) and \(7^{\circ} \mathrm{C}\), the effective thermal conductivity of the wall is (a) \(0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (b) \(0.83 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (c) \(1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(2.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(5.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

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