/*! 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 83 Highly reflective aluminum coati... [FREE SOLUTION] | 91Ó°ÊÓ

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Highly reflective aluminum coatings may be formed on the surface of a substrate by impacting the surface with molten drops of aluminum. The droplets are discharged from an injector, proceed through an inert gas (helium), and must still be in a molten state at the time of impact. \(V=3 \mathrm{~m} / \mathrm{s}\), and \(T_{i}=1100 \mathrm{~K}\), respectively, traverse a stagnant layer of atmospheric helium that is at a temperature of \(T_{\infty}=300 \mathrm{~K}\). What is the maximum allowable thickness of the helium layer needed to ensure that the temperature of droplets impacting the substrate is greater than or equal to the melting point of aluminum \(\left(T_{f} \geq T_{\text {mp }}=933 \mathrm{~K}\right)\) ? Properties of the molten aluminum may be approximated as \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3}, c=\) \(1200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
The maximum allowable thickness of the helium layer, x, can be found using the equation: \[x = \frac{k(mc)}{A} \ln \frac{T_{mp} - T_{\infty}}{T_i - T_{\infty}}\] where \(k\) is the thermal conductivity of aluminum, \(m\) is the droplet's mass, \(c\) is the specific heat of aluminum, \(A\) is the surface area of the droplet, \(T_{mp}\) is the melting point of aluminum, \(T_i\) is the initial temperature of the droplet, and \(T_{\infty}\) is the surrounding temperature. This equation relates the maximum thickness of the helium layer to the other parameters in the problem. In practice, one can calculate the specific values of m and A for a particular droplet and use this relationship to find the maximum thickness x.

Step by step solution

01

Identify the heat transfer mode

The molten aluminum droplets are cooling down as they travel through the helium atmosphere. So, the heat transfers from aluminum droplets to the surrounding helium. Since the droplets are in motion and the helium is still, the heat transfer will occur due to convection. We will use Newton's Law of Cooling to describe this process.
02

Calculate the convective heat transfer coefficient, h

First, we will find the convective heat transfer coefficient, h, using the equation: \[h = \frac{k}{x}\] where k is the thermal conductivity of aluminum and x is the thickness of the helium layer. We will use this equation later to find the required maximum thickness of the helium layer.
03

Apply Newton's Law of Cooling

Newton's Law of Cooling states that the rate of heat loss is proportional to the difference between the temperature of the body and the surroundings: \[\frac{dQ}{dt} = hA(T_i - T_{\infty})\] Where: \(dQ/dt\) is the rate of heat loss from the droplet \(h\) is the convective heat transfer coefficient \(A\) is the surface area of the droplet \(T_i\) is the initial temperature of the droplet \(T_{\infty}\) is the surrounding temperature We know that \(T_f \geq T_{mp}\). So, we will use this fact and the melting point to find the maximum thickness of the helium layer. To relate heat loss with the temperature change of the droplet, we next use:
04

Relate heat transfer with temperature change

We know that the heat lost must be equal to the heat required to decrease the droplets' temperature: \[\Delta Q = mc\Delta T\] Therefore, \[\frac{dQ}{dt} = mc\frac{dT}{dt}\] Now, we substitute the Newton's Law of Cooling equation here: \[hA(T_i - T_{\infty}) = mc\frac{dT}{dt}\]
05

Integrate the equation and solve for x

Now, we will integrate the equation to find the temperature change as a function of time. Then, we will substitute the melting point of aluminum to find the maximum thickness of the helium layer: \[\int_{T_i}^{T_f} \frac{dT}{T_i - T_{\infty}}=\int_0^t \frac{hA}{mc} dt\] To find the integration on the left-hand side of the equation, we will use the following formula: \[ \int \frac{dy}{y} = \ln|y|\] Integrating and solving for x, we obtain: \[x = \frac{k(mc)}{A} \ln \frac{T_{mp} - T_{\infty}}{T_i - T_{\infty}}\] Now plug in the known values and calculate the maximum thickness of the helium layer x: \(k = 200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) \(c = 1200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) \(T_{mp} = 933 \mathrm{~K}\) \(T_i = 1100 \mathrm{~K}\) \(T_{\infty} = 300 \mathrm{~K}\) We do not know the exact value of (mc)/A for this problem. However, we have found the relationship between the thickness of the helium layer and the other parameters. In practice, one can calculate the specific values of m and A for a particular droplet and use this relationship to find the maximum thickness x.

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Key Concepts

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

Understanding Newton's Law of Cooling
Newton's Law of Cooling is central to understanding how temperature changes over time due to heat loss. This law provides a simple yet powerful way to relate the rate at which an object loses heat to its immediate surroundings. Specifically, it tells us that the rate of heat loss from an object is proportional to the difference in temperature between the object and its environment.

The equation we often use to express Newton's Law of Cooling is: \[\frac{dQ}{dt} = hA(T - T_{\text{environment}})\]
where \(dQ/dt\) is the rate of heat loss, \(h\) is the convective heat transfer coefficient, \(A\) is the surface area of the object, \(T\) is the temperature of the object, and \(T_{\text{environment}}\) is the ambient temperature. In practical terms, this law assists engineers and scientists in predicting how quickly a hot object will cool down in a given environment, which is a critical consideration in many industrial processes.

For instance, in the exercise, molten aluminum droplets must remain above a critical temperature to ensure proper adherence to a substrate. By applying Newton's Law of Cooling, we can predict whether the aluminum droplets will maintain the required temperature during their travel through a layer of helium, thus ensuring the effectiveness of the coating process.
The Role of Convective Heat Transfer Coefficient
The convective heat transfer coefficient, denoted as \(h\), is a measure of the convective heat transfer per unit area and temperature difference. It's expressed in watts per square meter per Kelvin (\(W/m^2K\)) and plays a pivotal role in calculations involving heat transfer between a surface and a fluid in motion.

To clarify, the value of \(h\) characterizes how well a particular fluid—such as air, water, or helium in our example—can absorb heat from a surface. It is influenced by various factors, including the properties of the fluid, the velocity of fluid motion, and the surface geometry. In the exercise provided, finding \(h\) using the equation \(h = k/x\) allows us to compute the maximum allowable thickness of the helium layer to ensure that the aluminum droplets don't cool below their melting point upon impact with the substrate.

It’s important to note that determining \(h\) in real-world situations might involve more complex calculations or empirical correlations rather than a straightforward formula, as each scenario's specifics can greatly influence convective heat transfer.
Thermal Conductivity and its Impact on Heat Transfer
Thermal conductivity, symbolized by \(k\), is an intrinsic property of a material that quantifies its ability to conduct heat. Represented in units of watts per meter-Kelvin (\(W/mK\)), a higher thermal conductivity means that the material can transfer heat more effectively. This property is a crucial factor when analyzing heat transfer in solids and fluids alike.

In the context of our exercise, the thermal conductivity of molten aluminum determines how rapidly heat can travel through the droplets before they make contact with the substrate. We use the known thermal conductivity value along with the thickness of the helium layer to ascertain the heat transfer coefficient (\(h\)), which then feeds into the application of Newton’s Law of Cooling.

Understanding thermal conductivity is exceptionally important in material science for selecting materials based on the desired thermal properties for applications including insulation, heating, electronics, and, as demonstrated, industrial coating processes. Materials with high thermal conductivity are selected for components that require efficient heat dissipation, while those with low thermal conductivity are suited for insulation purposes.

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Most popular questions from this chapter

A spherical thermocouple junction \(1.0 \mathrm{~mm}\) in diameter is inserted in a combustion chamber to measure the temperature \(T_{\infty}\) of the products of combustion. The hot gases have a velocity of \(V=5 \mathrm{~m} / \mathrm{s}\). (a) If the thermocouple is at room temperature, \(T_{i}\), when it is inserted in the chamber, estimate the time required for the temperature difference, \(T_{\infty}-T\), to reach \(2 \%\) of the initial temperature difference, \(T_{\infty}-T_{i}\). Neglect radiation and conduction through the leads. Properties of the thermocouple junction are approximated as \(k=100 \mathrm{~W} / \mathrm{m}+\mathrm{K}, c=385 \mathrm{~J} / \mathrm{kg}+\mathrm{K}\), and \(\rho=8920 \mathrm{~kg} / \mathrm{m}^{3}\), while those of the combustion gases may be approximated as \(k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\nu=50 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\), and \(\operatorname{Pr}=0.69\). (b) If the thermocouple junction has an emissivity of \(0.5\) and the cooled walls of the combustor are at \(T_{c}=400 \mathrm{~K}\), what is the steady- state temperature of the thermocouple junction if the combustion gases are at \(1000 \mathrm{~K}\) ? Conduction through the lead wires may be neglected. (c) To determine the influence of the gas velocity on the thermocouple measurement error, compute the steady-state temperature of the thermocouple junction for velocities in the range \(1 \leq V \leq 25 \mathrm{~m} / \mathrm{s}\). The emissivity of the junction can be controlled through application of a thin coating. To reduce the measurement error, should the emissivity be increased or decreased? For \(V=5 \mathrm{~m} / \mathrm{s}\), compute the steadystate junction temperature for emissivities in the range \(0.1 \leq \varepsilon \leq 1.0\).

The roof of a refrigerated truck compartment is of composite construction, consisting of a layer of foamed urethane insulation \(\left(t_{2}=50 \mathrm{~mm}, k_{i}=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) sandwiched between aluminum alloy panels \(\left(t_{1}=5 \mathrm{~mm}\right.\), \(\left.k_{p}=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). The length and width of the roof are \(L=10 \mathrm{~m}\) and W \(=3.5 \mathrm{~m}\), respectively, and the temperature of the inner surface is \(T_{s, i}=-10^{\circ} \mathrm{C}\). Consider conditions for which the truck is moving at a speed of \(V=105 \mathrm{~km} / \mathrm{h}\), the air temperature is \(T_{\infty}=32^{\circ} \mathrm{C}\), and the solar irradiation is \(G_{S}=750 \mathrm{~W} / \mathrm{m}^{2}\). Turbulent flow may be assumed over the entire length of the roof. (a) For equivalent values of the solar absorptivity and the emissivity of the outer surface \(\left(\alpha_{S}=\varepsilon=0.5\right)\), estimate the average temperature \(T_{s, o}\) of the outer surface. What is the corresponding heat load imposed on the refrigeration system? (b) A special finish \(\left(\alpha_{S}=0.15, \varepsilon=0.8\right)\) may be applied to the outer surface. What effect would such an application have on the surface temperature and the heat load? (c) If, with \(\alpha_{S}=\varepsilon=0.5\), the roof is not insulated \(\left(t_{2}=0\right)\), what are the corresponding values of the surface temperature and the heat load?

Motile bacteria are equipped with flagella that are rotated by tiny, biological electrochemical engines which, in turn, propel the bacteria through a host liquid. Consider a nominally spherical Escherichia coli bacterium that is of diameter \(D=2 \mu \mathrm{m}\). The bacterium is in a water-based solution at \(37^{\circ} \mathrm{C}\) containing a nutrient which is characterized by a binary diffusion coefficient of \(D_{\mathrm{AB}}=0.7 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\) and a food energy value of \(\mathcal{N}=16,000 \mathrm{~kJ} / \mathrm{kg}\). There is a nutrient density difference between the fluid and the shell of the bacterium of \(\Delta \rho_{\mathrm{A}}=860 \times 10^{-12} \mathrm{~kg} / \mathrm{m}^{3}\). Assuming a propulsion efficiency of \(\eta=0.5\), determine the maximum speed of the E. coli. Report your answer in body diameters per second.

A long, cylindrical, electrical heating element of diameter \(D=10 \mathrm{~mm}\), thermal conductivity \(k=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), density \(\rho=2700 \mathrm{~kg} / \mathrm{m}^{3}\), and specific heat \(c_{p}=900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) is installed in a duct for which air moves in cross flow over the heater at a temperature and velocity of \(27^{\circ} \mathrm{C}\) and \(10 \mathrm{~m} / \mathrm{s}\), respectively. (a) Neglecting radiation, estimate the steady-state surface temperature when, per unit length of the heater, electrical energy is being dissipated at a rate of \(1000 \mathrm{~W} / \mathrm{m}\). (b) If the heater is activated from an initial temperature of \(27^{\circ} \mathrm{C}\), estimate the time required for the surface temperature to come within \(10^{\circ} \mathrm{C}\) of its steady-state value.

Cylindrical dry-bulb and wet-bulb thermometers are installed in a large- diameter duct to obtain the temperature \(T_{\infty}\) and the relative humidity \(\phi_{\infty}\) of moist air flowing through the duct at a velocity \(V\). The dry-bulb thermometer has a bare glass surface of diameter \(D_{\mathrm{db}}\) and emissivity \(\varepsilon_{g}\). The wet-bulb thermometer is covered with a thin wick that is saturated with water flowing continuously by capillary action from a bottom reservoir. Its diameter and emissivity are designated as \(D_{\text {wb }}\) and \(\varepsilon_{w}\). The duct inside surface is at a known temperature \(T_{s}\), which is less than \(T_{\infty}\). Develop expressions that may be used to obtain \(T_{\infty}\) and \(\phi_{\infty}\) from knowledge of the dry-bulb and wet-bulb temperatures \(T_{\mathrm{db}}\) and \(T_{\mathrm{ub}}\) and the foregoing parameters. Determine \(T_{\infty}\) and \(\phi_{\infty}\) when \(T_{\mathrm{db}}=45^{\circ} \mathrm{C}, T_{\mathrm{wb}}=25^{\circ} \mathrm{C}, T_{s}=35^{\circ} \mathrm{C}, p=1 \mathrm{~atm}\), \(V=5 \mathrm{~m} / \mathrm{s}, D_{\mathrm{db}}=3 \mathrm{~mm}, D_{\mathrm{wb}}=4 \mathrm{~mm}\), and \(\varepsilon_{\mathrm{g}}=\varepsilon_{w}=\) \(0.95\). As a first approximation, evaluate the dry- and wet-bulb air properties at 45 and \(25^{\circ} \mathrm{C}\), respectively.

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