91Ó°ÊÓ

A vertical slab of Woods metal is joined to a substrate on one surface and is melted as it is uniformly irradialed by a laser souree on the opposite surface. The metal is initially at its fusion temperature of \(T_{f}=72^{\circ} \mathrm{C}\). and the melt rans off by gravity as soon as it is formed. The abserptivity of the metal to the laser radiation is \(\alpha_{1}=0.4\), and its latent heat of fusion is \(h_{y}=33 \mathrm{~kJ} / \mathrm{kg}\). (a) Neglecting heat transfer from the irradiated surface by convection or radiation exchange with the surroundings, determine the instantanecus rate of melting in \(\mathrm{kg} / \mathrm{s}-\mathrm{m}^{2}\) if the laser irradiation is \(5 \mathrm{~kW} / \mathrm{m}^{2}\). How much material is removed if irradiation is maintained for a period of \(2 \mathrm{~s}\) ? (b) Allowing for convection to ambient air, with \(T_{\mathrm{w}}=20 r \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and radiation exchange with large surroundings \(\left(\varepsilon=0.4, T_{\text {max }}=\right.\) \(20^{\circ} \mathrm{C}\), determine the instantaneous rate of melting during imadiation.

Step by step solution

01

Calculate Absorbed Irradiation

Firstly, determine the power absorbed by the metal from laser irradiation. Since the laser radiation is given as \( G = 5 \, \text{kW/m}^2 \) and the absorptivity \( \alpha_1 = 0.4 \), the absorbed power can be calculated as: \[ q_\text{absorbed} = \alpha_1 \times G = 0.4 \times 5000 \, \text{W/m}^2 = 2000 \, \text{W/m}^2 \] .

02

Calculate Rate of Melting with No Losses

Use the absorbed power to calculate the instantaneous rate of melting by dividing it by the latent heat of fusion \( h_f = 33 \, \text{kJ/kg} \). Convert \( h_f \) to \( \text{J/kg}\) for consistency in units: \[ h_f = 33,000 \, \text{J/kg} \] Hence, \[ \dot{m} = \frac{q_{absorbed}}{h_f} = \frac{2000}{33000} \, \text{kg/s/m}^2 \approx 0.0606 \, \text{kg/s/m}^2 \] .

03

Calculate Total Material Removed (No Convection/Radiation)

To determine the total amount of material removed over a period of 2 seconds, multiply the instantaneous rate by time: \[ m_\text{total} = \dot{m} \times \text{time} = 0.0606 \, \text{kg/s/m}^2 \times 2 \, \text{s} = 0.1212 \, \text{kg/m}^2 \] .

04

Calculate Convection and Radiation Losses

When accounting for convection and radiation loss, calculate each: Convection loss is given by \( q_\text{conv} = h(T_f - T_\infty) \). Using \( h = 15 \, \text{W/m}^2\cdot\text{K} \), \( T_f = 72^{\circ} \text{C} \), and \( T_\infty = 20^{\circ} \text{C} \): \[ q_\text{conv} = 15 \times (72 - 20) = 780 \text{W/m}^2 \]. Radiation loss can be computed with \[ q_\text{rad} = \varepsilon \sigma (T_f^4 - T_\infty^4) \]. \( \varepsilon = 0.4 \), \( \sigma = 5.67 \times 10^{-8} \text{W/m}^2 \cdot \text{K}^4 \), \( T_f = 345 \text{K} \), \( T_\infty = 293 \text{K} \). Then, compute \( q_\text{rad} \): \[ q_\text{rad} = 0.4 \times 5.67 \times 10^{-8} \times (345^4 - 293^4) \approx 143.3 \text{W/m}^2 \].

05

Calculate Net Power Absorbed with Losses

Sum convection and radiation losses and subtract from the absorbed power to find net absorbed power: \[ q_\text{losses} = q_\text{conv} + q_\text{rad} = 780 + 143.3 = 923.3 \text{W/m}^2 \] \[ q_\text{net} = q_\text{absorbed} - q_\text{losses} = 2000 - 923.3 = 1076.7 \text{W/m}^2 \] .

06

Calculate Rate of Melting with Losses

Finally, calculate the new rate of melting with the adjusted net power: \[ \dot{m}' = \frac{q_\text{net}}{h_f} = \frac{1076.7}{33000} \, \text{kg/s/m}^2 \approx 0.0326 \, \text{kg/s/m}^2 \] .

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Key Concepts

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

Laser Irradiation

Laser irradiation involves the application of laser light onto a material, in this case, a slab of Woods metal. The laser provides energy that can affect the material's state or properties. Here it's used to melt the metal. The rate at which energy is supplied by the laser is measured in power per unit area, often kilowatts per square meter (kW/m²).
Laser light contains immense energy, and when it strikes the surface of a material, it can lead to heating and, subsequently, melting. The efficiency of this process depends on the material's absorptivity, which tells us how much of the laser energy the material can absorb. For Woods metal, the absorptivity is given as 0.4 or 40%. This means 40% of the laser's energy is absorbed, directly influencing the melting process.
Understanding laser irradiation helps in calculating how much energy is actually being used for melting, compared to what is potentially wasted or reflected. This is a critical step in determining the melting rate for practical applications such as manufacturing or scientific experiments.

Latent Heat of Fusion

The latent heat of fusion is a pivotal concept when studying phase changes in materials. It refers to the amount of energy required to change a substance from solid to liquid at its melting point without changing its temperature. The energy breaks the bonds within the solid structure, leading to a phase change.
For Woods metal, the latent heat of fusion is given as 33 kJ/kg. This value indicates that 33,000 Joules are needed to melt 1 kilogram of this metal, starting at its fusion temperature. This property helps us determine how much metal will melt under a specific amount of absorbed power.

• High latent heat means more energy is required to melt a given mass.
• This property is intrinsic and varies between different materials.

In practical scenarios, understanding latent heat is crucial when predicting and calculating the energy efficiency and feasibility of processes like welding or metal casting. Without adequate energy input, the desired melting won’t occur.

Convection and Radiation Losses

Heat loss through convection and radiation significantly impacts the efficiency of heat transfer operations. Convection involves the transfer of heat through the movement of fluids like air or liquid around the heated material. This process can remove energy from the surface, reducing the rate at which a material melts.
In our problem, convection is quantified by a coefficient, here denoted by 15 \text{W/m}^2\cdot\text{K}. It considers the temperature difference between the metal's surface and the surrounding air to calculate the energy lost through convective currents.

• Convection is influenced by fluid speed and temperature difference.
• It often accounts for a substantial portion of heat loss in many industrial processes.

Radiation loss, on the other hand, occurs through electromagnetic waves. This is independent of any medium between the surface and surroundings, as opposed to convection. The loss depends on the temperature difference to the fourth power, and the material's emissivity, denoted by \( \varepsilon \), which is 0.4 for Woods Metal
The combined convection and radiation losses can greatly reduce the net energy available for the melting process. Accurate calculations in such scenarios can help in optimizing industrial processes for better energy use and cost-efficiency.