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Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front side is exposed to ambient air that is at \(20^{\circ} \mathrm{C}\) and provides a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), solar irradiation of \(600 \mathrm{~W} / \mathrm{m}^{2}\), and an effective sky temperature of \(-40^{\circ} \mathrm{C}\). What is the electrical power \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\) required to maintain the plate surface temperature at \(T_{s}=60^{\circ} \mathrm{C}\) if the plate is diffuse and has the designated spectral, hemispherical reflectivity?

Short Answer

Expert verified
To maintain the plate surface temperature at \(T_{s}=60^{\circ} \mathrm{C}\), no electrical power is needed, as the solar irradiation is enough to keep the temperature steady.

Step by step solution

01

Define the known variables and parameters

First, let's break down the given information: - Plate surface temperature, \(T_s = 60^{\circ} \mathrm{C} = 333.15 \mathrm{K}\) - Ambient air temperature, \(T_a = 20^{\circ} \mathrm{C} = 293.15 \mathrm{K}\) - Convection heat transfer coefficient, \(h = 10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) - Solar irradiation, \(G_{solar} = 600 \mathrm{~W} / \mathrm{m}^{2}\) - Effective sky temperature, \(T_{sky} = -40^{\circ} \mathrm{C} = 233.15 \mathrm{K}\)
02

Calculate heat loss due to convection

To maintain the plate at a constant temperature, the total heat loss from the plate must be equal to the heat input from both the electrical heater and solar irradiation. Heat loss due to convection can be calculated using Newton's Law of Cooling: \(q_{conv} = h \cdot A \cdot (T_s - T_a)\) Since we are asked to find the electrical power per unit area (W/m²), we just need to find the term inside the brackets: \(Q_{conv} = h \cdot (T_s - T_a) = 10 \cdot (333.15 - 293.15) = 400 \mathrm{~W} / \mathrm{m}^{2}\)
03

Calculate heat loss due to radiation

Heat loss due to radiation can be calculated using Stefan-Boltzmann Law: \(q_{rad} = \sigma \cdot A \cdot (T_s^4 - T_{sky}^4)\) \(\sigma = 5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^4\) is the Stefan-Boltzmann constant. As in the previous step, we want to find the term inside the brackets: \(Q_{rad} = \sigma \cdot (T_s^4 - T_{sky}^4) = 5.67 \times 10^{-8} \cdot (333.15^4 - 233.15^4) \approx 176.82 \mathrm{~W} / \mathrm{m}^{2}\)
04

Calculate the net heat gain from solar irradiation

The net heat gain from solar irradiation can be calculated as follows: \(Q_{solar} = G_{solar} - Q_{rad} = 600 - 176.82 \approx 423.18 \mathrm{~W} / \mathrm{m}^{2}\)
05

Calculate the required electrical power per unit area

Finally, let's calculate the required electrical power per unit area to maintain the surface temperature at \(60^{\circ} \mathrm{C}\): \(Q_{elec} = Q_{conv} - Q_{solar} = 400 - 423.18 \approx -23.18 \mathrm{~W} / \mathrm{m}^{2}\) However, a negative value for the required electrical power per unit area indicates that the solar irradiation alone is enough to maintain the surface temperature at \(60^{\circ} \mathrm{C}\). Therefore, no additional electrical power is needed. In conclusion, to maintain the plate surface temperature at \(T_{s}=60^{\circ} \mathrm{C}\), no electrical power is needed, as the solar irradiation is enough to keep the temperature steady.

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Key Concepts

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

Convection Heat Transfer
Convection heat transfer is a core principle governing how heat energy transfers between a surface and a fluid in motion over the surface. This process significantly influences heat loss or gain in systems exposed to fluid environments, such as air or water. When a solid surface is at a different temperature than the surrounding fluid, heat will naturally transfer between them. This is convection.
The efficiency of this heat transfer depends on the convection heat transfer coefficient, denoted as \( h \). It is measured in \( \text{W/m}^2\cdot \text{K} \) and considers both the properties of the fluid and the characteristics of the surface.
  • The convection heat transfer equation is given by: \[ q_{conv} = h \cdot A \cdot (T_s - T_a) \] where \( q_{conv} \) is the rate of heat transfer by convection, \( A \) is the area, \( T_s \) is the surface temperature, and \( T_a \) is the ambient air temperature.
  • In this context, we're interested in heat power per unit area, simplifying the equation to: \[ Q_{conv} = h \cdot (T_s - T_a) \]
This equation helps calculate how much heat is being lost or gained due to convection, which is crucial for maintaining temperatures of surfaces like the plate from the exercise.
Radiation Heat Transfer
Radiation heat transfer pertains to the energy transfer through electromagnetic waves. This kind of heat transfer does not require a medium, meaning it can even occur in a vacuum.
Infrared radiation is the form of radiation most commonly associated with heat transfer, influencing the temperature regulation of bodies exposed to the atmosphere.In our exercise, radiation heat loss is significant when a plate is exposed to the sky because the difference between the plate's temperature and the effective sky temperature leads to heat being radiated away. To model this, we utilize the Stefan-Boltzmann Law. By employing: \[ q_{rad} = \sigma \cdot A \cdot (T_s^4 - T_{sky}^4) \] we can determine the rate of radiative heat loss, where \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the area, and \( T_s \) and \( T_{sky} \) are the respective temperatures of the surface and the sky.
  • Radiation allows heat transfer without direct contact, using waves to move energy from hot surfaces to cooler surroundings.
  • By calculating the heat lost via radiation, we can adjust other input forms like solar or electrical to keep the surface at a designated temperature.
This ability to compute radiation heat transfer helps us balance energy inputs and outputs, crucial for efficient temperature management.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a fundamental principle in thermal radiation. It states that the power radiated by a black body is directly proportional to the fourth power of the absolute temperature of the body.
The law is expressed through the equation in: \[ q = \sigma A T^4 \] where \( q \) is the power per unit area, \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\cdot \text{K}^4 \) is the Stefan-Boltzmann constant, \( A \) is the emitting surface's area, and \( T \) is the temperature in Kelvin.
  • This law is crucial in determining how much heat is being emitted from a surface at a given temperature. In our context, it helps compute the rate of radiative heat loss from the plate to the environment.
  • The calculation of \( q_{rad} \) with temperatures of the surface and surrounding medium (like the sky) informs us how energy balance can be achieved without excess electrical power.
Understanding the Stefan-Boltzmann Law enables proper accounting for radiative losses, which is essential when designing systems to maintain desired thermal conditions.
Newton's Law of Cooling
Newton's Law of Cooling provides a simplistic approach to understanding how an object's temperature changes in relation to its surroundings. This concept is fundamentally linked to convection.
According to this principle, the rate of change in temperature is proportional to the difference between the object's temperature and the ambient temperature.In mathematical form, Newton's Law of Cooling is represented as:\[ \frac{dT}{dt} = -k(T-T_a) \] where \( dT/dt \) is the rate of temperature change, \( k \) is the cooling constant, \( T \) is the object's temperature, and \( T_a \) is the ambient temperature.For steady conditions such as the exercise where a consistent temperature must be maintained, this law simplifies to help determine the convection heat loss, \( Q_{conv}\):\[ Q_{conv} = h \cdot (T_s - T_a) \] This principle is especially helpful in calibrating systems that involve maintaining constant temperatures, like the heated plate in the problem.
  • Newton's Law helps predict how rapidly an object will lose or gain heat in still air, which directly correlates to the measurement of the convection process.
  • It simplifies the complexity of surface heat interactions by focusing on temperature differences, laying a foundation for understanding convection-based systems.
Recognizing the role of Newton's Law in such systems informs the necessity of additional heat sources or adjustments to insulation to maintain thermal stability.

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Most popular questions from this chapter

A proposed method for generating electricity from solar irradiation is to concentrate the irradiation into a cavity that is placed within a large container of a salt with a high melting temperature. If all heat losses are neglected, part of the solar irradiation entering the cavity is used to melt the salt while the remainder is used to power a Rankine cycle. (The salt is melted during the day and is resolidified at night in order to generate electricity around the clock.) Consider conditions for which the solar power entering the cavity is \(q_{\mathrm{sal}}=7.50 \mathrm{MW}\) and the time rate of change of energy stored in the salt is \(\dot{E}_{\mathrm{st}}=3.45 \mathrm{MW}\). For a cavity opening of diameter \(D_{s}=1 \mathrm{~m}\), determine the heat transfer to the Rankine cycle, \(q_{R}\). The temperature of the salt is maintained at its melting point, \(T_{\text {salt }}=T_{\text {m }}=1000^{\circ} \mathrm{C}\). Neglect heat loss by convection and irradiation from the surroundings.

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?

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?

The spectral transmissivity of a \(50-\mu\) m-thick polymer film is measured over the wavelength range \(2.5 \mu \mathrm{m} \leq\) \(\lambda \leq 15 \mu \mathrm{m}\). The spectral distribution may be approximated as \(\tau_{\lambda}=0.80\) for \(2.5 \mu \mathrm{m} \leq \lambda \leq 7 \mu \mathrm{m}, \tau_{\lambda}=0.05\) for \(7 \mu \mathrm{m}<\lambda \leq 13 \mu \mathrm{m}\), and \(\tau_{\lambda}=0.55\) for \(13 \mu \mathrm{m}<\) \(\lambda \leq 15 \mu \mathrm{m}\). Transmissivity data outside the range cannot be acquired due to limitations associated with the instrumentation. An engineer wishes to determine the total transmissivity of the film. (a) Estimate the maximum possible total transmissivity of the film associated with irradiation from a blackbody at \(T=30^{\circ} \mathrm{C}\). (b) Estimate the minimum possible total transmissivity of the film associated with irradiation from a blackbody at \(T=30^{\circ} \mathrm{C}\). (c) Repeat parts (a) and (b) for a blackbody at \(T=\) \(600^{\circ} \mathrm{C}\)

The exposed surface of a power amplifier for an earth satellite receiver of area \(130 \mathrm{~mm} \times 130 \mathrm{~mm}\) has a diffuse, gray, opaque coating with an emissivity of \(0.5\). For typical amplifier operating conditions, the surface temperature is \(58^{\circ} \mathrm{C}\) under the following environmental conditions: air temperature, \(T_{\infty}=27^{\circ} \mathrm{C}\); sky temperature, \(T_{\text {sy }}=-20^{\circ} \mathrm{C} ;\) convection coefficient, \(h=\) \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\); and solar irradiation, \(G_{S}=800 \mathrm{~W} / \mathrm{m}^{2}\). (a) For the above conditions, determine the electrical power being generated within the amplifier. (b) It is desired to reduce the surface temperature by applying one of the diffuse coatings (A, B, C) shown as follows. Which coating will result in the coolest surface temperature for the same amplifier operating and environmental conditions?

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