/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 For a boiling process such as sh... [FREE SOLUTION] | 91Ó°ÊÓ

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For a boiling process such as shown in Figure \(1.5 c\), the ambient temperature \(T_{\infty}\) in Newton's law of cooling is replaced by the saturation temperature of the fluid \(T_{\text {sat }}\). Consider a situation where the heat flux from the hot plate is \(q^{\prime \prime}=20 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2}\). If the fluid is water at atmospheric pressure and the convection heat transfer coefficient is \(h_{w}=20 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the upper surface temperature of the plate, \(T_{s, w^{\circ}}\). In an effort to minimize the surface temperature, a technician proposes replacing the water with a dielectric fluid whose saturation temperature is \(T_{\text {sat,d }}=52^{\circ} \mathrm{C}\). If the heat transfer coefficient associated with the dielectric fluid is \(h_{d}=3 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), will the technician's plan work?

Short Answer

Expert verified
The surface temperature for the water case is calculated using the equation \(q^{\prime \prime} = h_w (T_{s,w} - T_{\text{sat}})\), resulting in \(T_{s,w} = 104.01^\circ\text{C}\). For the dielectric fluid, we use the equation \(q^{\prime \prime} = h_d (T_{s,d} - T_{\text{sat,d}})\), giving \(T_{s,d} = 174.33^\circ\text{C}\). Since \(T_{s,d} > T_{s,w}\), the technician's plan to replace water with the dielectric fluid will not work in minimizing the surface temperature.

Step by step solution

01

Calculate the surface temperature for water

We will use Newton's law of cooling to calculate the surface temperature of the hot plate when using water. The equation is: \[q^{\prime \prime} = h_w (T_{s,w} - T_{\text{sat}})\] Where: \(q^{\prime \prime}\) = heat flux (\(\frac{\mathrm{W}}{\mathrm{m}^2}\)) \(h_w\) = convection heat transfer coefficient of water (\(\frac{\mathrm{W}}{\mathrm{m}^2.\mathrm{K}}\)) \(T_{s,w}\) = upper surface temperature of the plate when using water (K) \(T_{\text{sat}}\) = saturation temperature of the fluid (K) Since we know \(q^{\prime \prime}\), \(h_w\), and \(T_{\text{sat}}\) for water, we can solve for \(T_{s,w}\).
02

Calculate the surface temperature for the dielectric fluid

Next, we will use Newton's law of cooling to calculate the surface temperature of the hot plate when using the dielectric fluid. The equation is: \[q^{\prime \prime} = h_d (T_{s,d} - T_{\text{sat,d}})\] Where: \(h_d\) = convection heat transfer coefficient of the dielectric fluid (\(\frac{\mathrm{W}}{\mathrm{m}^2\cdot \mathrm{K}}\)) \(T_{s,d}\) = upper surface temperature of the plate when using the dielectric fluid (K) \(T_{\text{sat,d}}\) = saturation temperature of the dielectric fluid (K) Since we know \(q^{\prime \prime}\), \(h_d\), and \(T_{\text{sat,d}}\) for the dielectric fluid, we can solve for \(T_{s,d}\).
03

Compare the surface temperatures

Now that we have calculated the upper surface temperatures for both water and the dielectric fluid, we can compare them to determine if the technician's plan will work. The plan will work if the surface temperature when using the dielectric fluid is lower than the one when using water. That is, if \(T_{s,d} < T_{s,w}\), then the plan works.

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Key Concepts

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

Boiling Process
The boiling process is a critical phase in heat transfer where a liquid changes to a vapor state. This transformation occurs when a liquid reaches its saturation temperature under a given pressure. In the context of our exercise, boiling is relevant because it changes the way heat is transferred from a hot surface to a liquid. When a liquid boils, it forms vapor bubbles, which rapidly rise to the surface and release the absorbed heat. This process helps to equalize the temperature difference between the liquid and the heating surface.
In practical terms, this is visualized in systems where a fluid is heated, such as water in a boiler. Once the saturation temperature—also called the boiling point—is reached, phase changes start occurring, and heat transfer accelerates due to the latent heat of vaporization. This makes boiling a highly efficient method of heat transfer compared to processes like conduction. Moreover, the phase change can carry heat away more efficiently, often increasing the rate of heat transfer between the surface and the fluid. So, understanding how boiling works is crucial for optimizing industrial processes, from power generation to cooling systems.
Newton's Law of Cooling
Newton's Law of Cooling provides a mathematical framework to model the cooling (or heating) of an object from a high temperature to a lower surrounding temperature. The law states that the rate of heat loss is proportional to the difference in temperature between the object and its environment. In the exercise, we apply this principle to determine the surface temperature of a plate exposed to both water and dielectric fluid at their respective saturation temperatures.
Mathematically, it is expressed as:
  • \[ q^{ ext{''}} = h(T_s - T_{ ext{sat}}) \]
Here, \( q^{ ext{''}} \) represents the heat flux, \( h \) is the heat transfer coefficient, and \( T_s \) and \( T_{ ext{sat}} \) are the surface and saturation temperatures, respectively. This equation shows that for a higher heat transfer coefficient or a larger temperature gradient, the rate of heat transfer increases, directly affecting thermal management strategies. The law underscores a fundamental aspect of heat transfer where efficient systems often hinge upon minimizing the difference between the surface and surrounding temperatures to control energy loss.
Convection Heat Transfer
Convection heat transfer is a mode of thermal energy transfer between a solid surface and a fluid in motion. It stands out from conduction and radiation as it involves the bulk movement of fluid that helps transport heat. When the fluid near the surface is heated, its density decreases, causing it to rise while cooler, denser fluid replaces it. This creates a circulation cycle known as convection.
In the scenario described in the exercise, convection plays a vital role as it dictates how heat is transferred from the hot plate to both water and dielectric fluid. The convection heat transfer coefficient \( h \) quantifies this heat transfer efficiency. If the coefficient is higher, it implies more efficient heat transfer for a given fluid. Convection can be natural, driven by buoyancy forces due to temperature differences, or forced, using external forces such as fans or pumps. Understanding this principle is vital for designing systems that require efficient cooling or heating, like HVAC systems or electronic cooling devices.
Saturation Temperature
Saturation temperature is the boiling point of a fluid at a specific pressure, which is crucial for processes involving phase changes. It is the temperature at which a liquid transforms into vapor at a constant pressure without any further temperature rise as heat is added. For example, at standard atmospheric pressure, water has a saturation temperature of 100°C. However, this temperature changes if the pressure conditions differ.
In our exercise, knowing the saturation temperature helps us understand how the fluid behaves under heat. By replacing water with a dielectric fluid with a different saturation temperature, we can influence the rate of phase change and thereby the efficiency of the heat transfer process. Lower saturation temperatures in dielectric fluids might lead to earlier boiling, which can be less efficient if the heat transfer coefficient decreases as seen with the lower value given in the scenario. Therefore, understanding this temperature ensures that we can optimize thermal systems by selecting the right fluid and conditions for maximum efficiency.

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Most popular questions from this chapter

Liquid oxygen, which has a boiling point of \(90 \mathrm{~K}\) and a latent heat of vaporization of \(214 \mathrm{~kJ} / \mathrm{kg}\), is stored in a spherical container whose outer surface is of \(500-\mathrm{mm}\) diameter and at a temperature of \(-10^{\circ} \mathrm{C}\). The container is housed in a laboratory whose air and walls are at \(25^{\circ} \mathrm{C}\). (a) If the surface emissivity is \(0.20\) and the heat transfer coefficient associated with free convection at the outer surface of the container is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the rate, in \(\mathrm{kg} / \mathrm{s}\), at which oxygen vapor must be vented from the system? (b) Moisture in the ambient air will result in frost formation on the container, causing the surface emissivity to increase. Assuming the surface temperature and convection coefficient to remain at \(-10^{\circ} \mathrm{C}\) and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, compute the oxygen evaporation rate \((\mathrm{kg} / \mathrm{s})\) as a function of surface emissivity over the range \(0.2 \leq \varepsilon \leq 0.94\).

The heat flux through a wood slab \(50 \mathrm{~mm}\) thick, whose inner and outer surface temperatures are 40 and \(20^{\circ} \mathrm{C}\), respectively, has been determined to be \(40 \mathrm{~W} / \mathrm{m}^{2}\). What is the thermal conductivity of the wood?

An aluminum plate \(4 \mathrm{~mm}\) thick is mounted in a horizontal position, and its bottom surface is well insulated. A special, thin coating is applied to the top surface such that it absorbs \(80 \%\) of any incident solar radiation, while having an emissivity of \(0.25\). The density \(\rho\) and specific heat \(c\) of aluminum are known to be \(2700 \mathrm{~kg} / \mathrm{m}^{3}\) and \(900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (a) Consider conditions for which the plate is at a temperature of \(25^{\circ} \mathrm{C}\) and its top surface is suddenly exposed to ambient air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and to solar radiation that provides an incident flux of \(900 \mathrm{~W} / \mathrm{m}^{2}\). The convection heat transfer coefficient between the surface and the air is \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the initial rate of change of the plate temperature? (b) What will be the equilibrium temperature of the plate when steady-state conditions are reached? (c) The surface radiative properties depend on the specific nature of the applied coating. Compute and plot the steady-state temperature as a function of the emissivity for \(0.05 \leq \varepsilon \leq 1\), with all other conditions remaining as prescribed. Repeat your calculations for values of \(\alpha_{S}=0.5\) and \(1.0\), and plot the results with those obtained for \(\alpha_{S}=0.8\). If the intent is to maximize the plate temperature, what is the most desirable combination of the plate emissivity and its absorptivity to solar radiation?

The inner and outer surface temperatures of a glass window \(5 \mathrm{~mm}\) thick are 15 and \(5^{\circ} \mathrm{C}\). What is the heat loss through a \(1 \mathrm{~m} \times 3 \mathrm{~m}\) window? The thermal conductivity of glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A photovoltaic panel of dimension \(2 \mathrm{~m} \times 4 \mathrm{~m}\) is installed on the roof of a home. The panel is irradiated with a solar flux of \(G_{S}=700 \mathrm{~W} / \mathrm{m}^{2}\), oriented normal to the top panel surface. The absorptivity of the panel to the solar irradiation is \(\alpha_{S}=0.83\), and the efficiency of conversion of the absorbed flux to electrical power is \(\eta=P / \alpha_{S} G_{S} A=0.553-0.001 \mathrm{~K}^{-1} T_{p}\), where \(T_{p}\) is the panel temperature expressed in kelvins and \(A\) is the solar panel area. Determine the electrical power generated for (a) a still summer day, in which \(T_{\text {sur }}=T_{\infty}=35^{\circ} \mathrm{C}\), \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and (b) a breezy winter day, for which \(T_{\text {sur }}=T_{\infty}=-15^{\circ} \mathrm{C}, h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The panel emissivity is \(\varepsilon=0.90\).

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