/*! 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

The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is \(225^{\circ} \mathrm{C}\) and the change in plate temperature with time \((d T / d t)\) is \(-0.022 \mathrm{~K} / \mathrm{s}\). The ambient air temperature is \(25^{\circ} \mathrm{C}\) and the plate measures \(0.3 \times 0.3 \mathrm{~m}\) with a mass of \(3.75 \mathrm{~kg}\) and a specific heat of \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

In considering the following problems involving heat transfer in the natural environment (outdoors), recognize that solar radiation is comprised of long and short wavelength components. If this radiation is incident on a semitransparent medium, such as water or glass, two things will happen to the nonreflected portion of the radiation. The long wavelength component will be absorbed at the surface of the medium, whereas the short wavelength component will be transmitted by the surface. (a) The number of panes in a window can strongly influence the heat loss from a heated room to the outside ambient air. Compare the single- and double-paned units shown by identifying relevant heat transfer processes for each case. (b) In a typical flat-plate solar collector, energy is collected by a working fluid that is circulated through tubes that are in good contact with the back face of an absorber plate. The back face is insulated from the surroundings, and the absorber plate receives solar radiation on its front face, which is typically covered by one or more transparent plates. Identify the relevant heat transfer processes, first for the absorber plate with no cover plate and then for the absorber plate with a single cover plate. (c) The solar energy collector design shown in the schematic has been used for agricultural applications. Air is blown through a long duct whose cross section is in the form of an equilateral triangle. One side of the triangle is comprised of a double-paned, semitransparent cover; the other two sides are constructed from aluminum sheets painted flat black on the inside and covered on the outside with a layer of styrofoam insulation. During sunny periods, air entering the system is heated for delivery to either a greenhouse, grain drying unit, or storage system. Identify all heat transfer processes associated with the cover plates, the absorber plate(s), and the air. (d) Evacuated-tube solar collectors are capable of improved performance relative to flat-plate collectors. The design consists of an inner tube enclosed in an outer tube that is transparent to solar radiation. The annular space between the tubes is evacuated. The outer, opaque surface of the inner tube absorbs solar radiation, and a working fluid is passed through the tube to collect the solar energy. The collector design generally consists of a row of such tubes arranged in front of a reflecting panel. Identify all heat transfer processes relevant to the performance of this device.

Pressurized water \(\left(p_{\text {in }}=10\right.\) bar, \(\left.T_{\text {in }}=110^{\circ} \mathrm{C}\right)\) enters the bottom of an \(L=10\)-m-long vertical tube of diameter \(D=100 \mathrm{~mm}\) at a mass flow rate of \(\dot{m}=1.5 \mathrm{~kg} / \mathrm{s}\). The tube is located inside a combustion chamber, resulting in heat transfer to the tube. Superheated steam exits the top of the tube at \(p_{\text {out }}=7\) bar, \(T_{\text {out }}=600^{\circ} \mathrm{C}\). Determine the change in the rate at which the following quantities enter and exit the tube: (a) the combined thermal and flow work, (b) the mechanical energy, and (c) the total energy of the water. Also, (d) determine the heat transfer rate, \(q\). Hint: Relevant properties may be obtained from a thermodynamics text.

A wall is made from an inhomogeneous (nonuniform) material for which the thermal conductivity varies through the thickness according to \(k=a x+b\), where \(a\) and \(b\) are constants. The heat flux is known to be constant. Determine expressions for the temperature gradient and the temperature distribution when the surface at \(x=0\) is at temperature \(T_{1}\).

Three electric resistance heaters of length \(L=250 \mathrm{~mm}\) and diameter \(D=25 \mathrm{~mm}\) are submerged in a 10 -gal tank of water, which is initially at \(295 \mathrm{~K}\). The water may be assumed to have a density and specific heat of \(\rho=990 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (a) If the heaters are activated, each dissipating \(q_{1}=500 \mathrm{~W}\), estimate the time required to bring the water to a temperature of \(335 \mathrm{~K}\). (b) If the natural convection coefficient is given by an expression of the form \(h=370\left(T_{s}-T\right)^{1 / 3}\), where \(T_{s}\) and \(T\) are temperatures of the heater surface and water, respectively, what is the temperature of each heater shortly after activation and just before deactivation? Units of \(h\) and \(\left(T_{s}-T\right)\) are \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\mathrm{K}\), respectively. (c) If the heaters are inadvertently activated when the tank is empty, the natural convection coefficient associated with heat transfer to the ambient air at \(T_{\infty}=300 \mathrm{~K}\) may be approximated as \(h=0.70\) \(\left(T_{s}-T_{\infty}\right)^{1 / 3}\). If the temperature of the tank walls is also \(300 \mathrm{~K}\) and the emissivity of the heater surface is \(\varepsilon=0.85\), what is the surface temperature of each heater under steady-state conditions?

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