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Plasma spraying is a process used for coating a material surface with a protective layer to prevent the material from degradation. In a plasma spraying process, the protective layer in powder form is injected into a plasma jet. The powder is then heated to molten droplets and propelled onto the material surface. Once deposited on the material surface, the molten droplets solidify and form a layer of protective coating. Consider a plasma spraying process using alumina \((k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=3970 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) powder that is injected into a plasma jet at \(T_{\infty}=15,000^{\circ} \mathrm{C}\) and \(h=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The alumina powder is made of particles that are spherical in shape with an average diameter of \(60 \mu \mathrm{m}\) and a melting point at \(2300^{\circ} \mathrm{C}\). Determine the amount of time it would take for the particles, with an initial temperature of \(20^{\circ} \mathrm{C}\), to reach their melting point from the moment they are injected into the plasma jet.

Short Answer

Expert verified
Answer: It takes approximately \(t\) seconds for the alumina particles to reach their melting point when injected into the plasma jet.

Step by step solution

01

Calculate the mass of the particle

First, we need to find the mass of the alumina particle. To compute the mass, we use the particle density (\(\rho\)) and volume, which is given by the formula for the volume of a sphere: \(V = \frac{4}{3} \pi r^3\) Given the average diameter of the alumina particles is \(60 \mu m\), the radius is \(30 \mu m\). To convert this to meters, we need to divide it by \(10^6\): \(r = \frac{30}{10^6} m\) Now we can calculate the volume: \(V = \frac{4}{3} \pi (\frac{30}{10^6})^3 m^3\) And, finally, we can compute the mass of the particle: \(m = \rho V = (3970 kg/m^3) \frac{4}{3} \pi (\frac{30}{10^6})^3 m^3\)
02

Calculate the heat absorbed by the particle

Next, we will calculate the heat absorbed by the particle as it reaches its melting point. We will use the lumped system analysis equation: \(Q = m c_p (2300 - 20)\) Substitute the specific heat value, \(c_p\), provided and the mass calculated in step 1: \(Q = (m)(800 J/ kg \cdot K)(2280 K)\)
03

Calculate the heating rate

Now we will calculate the heating rate of the alumina particle. We will use the following formula: \(\frac{dQ}{dt} = hA_s\Delta T\) Here, \(h\) is the heat transfer coefficient, \(A_s\) is the surface area of the particle, and \(\Delta T\) is the temperature difference between the plasma jet temperature and the particle temperature. First, we need to find the surface area of the particle, which can be calculated as: \(A_s = 4 \pi r^2\) Substitute the radius in meters: \(A_s = 4 \pi (\frac{30}{10^6})^2 m^2\) \(\Delta T = T_{\infty} - T_m\) Substitute the given values for the plasma jet temperature and the melting point: \(\Delta T = (15000 + 273) K - (2300 + 273) K\) Now we can compute the heating rate: \(\frac{dQ}{dt} = (10000 W/m^2 \cdot K)(A_s)(\Delta T)\)
04

Calculate the time to reach the melting point

Finally, we can calculate the time it takes for the alumina particle to reach its melting point. We can derive the equation for the time by dividing the heat absorbed by the particle by the heating rate: \(t = \frac{Q}{\frac{dQ}{dt}}\) Substitute the values calculated in steps 2 and 3: \(t = \frac{Q}{\frac{dQ}{dt}}\) This will give us the time it takes for the alumina particles to reach their melting point from the moment they are injected into the plasma jet.

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Key Concepts

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

Plasma Spraying
Plasma spraying is an advanced coating process widely used to protect materials from degradation. In this technique, a material in powder form is injected into a high-temperature plasma jet. The plasma jet, with temperatures reaching around 15,000°C, melts the powder into droplets. These molten droplets are then propelled at high speeds, forming a coating layer upon contacting the target surface. This layer offers protective qualities that enhance the durability and performance of the material under harsh conditions.
Thermal Analysis
Thermal analysis in this context refers to examining how heat moves and how it impacts particles in the plasma spraying process. As alumina particles are heated by the plasma jet, they absorb heat, raising their temperature to the melting point. It’s essential to understand how much heat is needed and how it’s transferred. The equation \( Q = m c_p \Delta T \), where \( \Delta T \) is the change in temperature, allows us to compute the heat required to bring the particles from the initial to the melting temperature. This thermal analysis ensures that particles reach the desired state for effective coating.
Lumped System Analysis
Lumped system analysis simplifies complex thermal problems by assuming the object's temperature is uniform throughout. In plasma spraying, we consider each alumina particle as a "lumped system," given its size and rapid heating. This means using a single temperature to represent the entire particle’s thermal state. The heat transfer coefficient \( h \) and surface area \( A_s \) are crucial in calculating how quickly heat is absorbed. This approach is efficient for small particles like alumina, providing a straightforward way to estimate the time required for them to reach the melting point.
Alumina Particles
Alumina particles, utilized in plasma spraying, have properties that make them ideal for this process. With a density \( \rho = 3970 \, \text{kg/m}^3 \), and a specific heat \( c_p = 800 \, \text{J/kg} \cdot \text{K} \), they efficiently absorb and retain heat. The spherical shape with a typical diameter of \( 60 \, \mu m \) allows them to be uniformly coated. Their melting point is at 2300°C, which is significantly below the plasma jet temperature, enabling effective melting and subsequent adherence to surfaces. Understanding these properties is key to optimizing the plasma spraying process.

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Most popular questions from this chapter

A long iron \(\operatorname{rod}\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=80.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=23.1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) with diameter of \(25 \mathrm{~mm}\) is initially heated to a uniform temperature of \(700^{\circ} \mathrm{C}\). The iron rod is then quenched in a large water bath that is maintained at constant temperature of \(50^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(128 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the time required for the iron rod surface temperature to cool to \(200^{\circ} \mathrm{C}\). Solve this problem using analytical one- term approximation method (not the Heisler charts).

The chilling room of a meat plant is \(15 \mathrm{~m} \times 18 \mathrm{~m} \times\) \(5.5 \mathrm{~m}\) in size and has a capacity of 350 beef carcasses. The power consumed by the fans and the lights in the chilling room are 22 and \(2 \mathrm{~kW}\), respectively, and the room gains heat through its envelope at a rate of \(14 \mathrm{~kW}\). The average mass of beef carcasses is \(220 \mathrm{~kg}\). The carcasses enter the chilling room at \(35^{\circ} \mathrm{C}\), after they are washed to facilitate evaporative cooling, and are cooled to \(16^{\circ} \mathrm{C}\) in \(12 \mathrm{~h}\). The air enters the chilling room at \(-2.2^{\circ} \mathrm{C}\) and leaves at \(0.5^{\circ} \mathrm{C}\). Determine \((a)\) the refrigeration load of the chilling room and \((b)\) the volume flow rate of air. The average specific heats of beef carcasses and air are \(3.14\) and \(1.0 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the density of air can be taken to be \(1.28 \mathrm{~kg} / \mathrm{m}^{3}\).

Why are the transient temperature charts prepared using nondimensionalized quantities such as the Biot and Fourier numbers instead of the actual variables such as thermal conductivity and time?

The Biot number can be thought of as the ratio of (a) The conduction thermal resistance to the convective thermal resistance. (b) The convective thermal resistance to the conduction thermal resistance. (c) The thermal energy storage capacity to the conduction thermal resistance. (d) The thermal energy storage capacity to the convection thermal resistance. (e) None of the above.

Consider a spherical shell satellite with outer diameter of \(4 \mathrm{~m}\) and shell thickness of \(10 \mathrm{~mm}\) is reentering the atmosphere. The shell satellite is made of stainless steel with properties of \(\rho=8238 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=468 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=13.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). During the reentry, the effective atmosphere temperature surrounding the satellite is \(1250^{\circ} \mathrm{C}\) with convection heat transfer coefficient of \(130 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the initial temperature of the shell is \(10^{\circ} \mathrm{C}\), determine the shell temperature after 5 minutes of reentry. Assume heat transfer occurs only on the satellite shell.

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