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A vertical slab of Wood's metal is joined to a substrate on one surface and is melted as it is uniformly irradiated by a laser source on the opposite surface. The metal is initially at its fusion temperature of \(T_{f}=72^{\circ} \mathrm{C}\), and the melt runs off by gravity as soon as it is formed. The absorptivity of the metal to the laser radiation is \(\alpha_{1}=0.4\), and its latent heat of fusion is \(h_{s f}=33 \mathrm{~kJ} / \mathrm{kg}\). (a) Neglecting heat transfer from the irradiated surface by convection or radiation exchange with the surroundings, determine the instantaneous rate of melting in \(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m}^{2}\) 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 \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and radiation exchange with large surroundings \((\varepsilon=0.4\), \(T_{\text {sur }}=20^{\circ} \mathrm{C}\) ), determine the instantaneous rate of melting during irradiation.

Step by step solution

01

Calculate the absorbed heat energy

We are given that the laser irradiation is \(5 kW/m^2\) and the absorptivity is \(0.4\). Calculate the absorbed heat energy: \[E_{abs} = Irradiation \times Absorptivity\] \[E_{abs}=5\, kW/m^{2}\times0.4=2 kW/m^{2}\]

02

Determine the rate of melting

To find the rate of melting (in kg/s.m虏), divide the absorbed energy by the latent heat of fusion: \[Rate_{melting}= \frac{E_{abs}}{h_{sf}}\] \[Rate_{melting}= \frac{2000\, \cancel{J/s} \cdot m^{2}}{33000\, \cancel{J}/kg}= 0.0606\,kg/s\cdot m^{2}\]

03

Calculate the material removed in 2 seconds

Multiply the rate of melting by the irradiation time (2 s) to find the total material removed: \[Material_{removed}= Rate_{melting} \times Time\] \[Material_{removed}= 0.0606\, kg/s\cdot m^{2}\times 2s= 0.1212\, kg/m^{2}\] (a) The instantaneous rate of melting is \(0.0606\, kg/s\cdot m^{2}\) and \(0.1212\, kg/m^{2}\) of material is removed after 2 seconds. (b) Considering heat transfer:

04

Calculate the convective heat transfer

Determine the convective heat transfer using the given temperature and heat transfer coefficient: \[Q_{conv} = h(T_{\infty} - T_{f})\] \[Q_{conv} = 15\,W/m^{2}K(20^{\circ} C - 72^{\circ} C) = -780\,W/m^2\]

05

Calculate the radiative heat transfer

Calculate the radiative heat transfer using the given emissivity and surrounding temperature: \[Q_{rad}=\varepsilon\sigma(T_{f}^{4} - T_{sur}^{4})\] \[Q_{rad} = 0.4(5.67\times10^{-8} W/m^2 K^4)((345 K)^4 -(293)^4) \approx -350.724\,W/m^2\]

06

Calculate the remaining energy for melting

To find the remaining energy for melting, sum the energy absorbed and heat transfer energies: \[E_{remaining}= E_{abs} - Q_{conv} - Q_{rad}\] \[E_{remaining}= 2000\,W/m^{2}+780\,W/m^2+350.724\,W/m^2 \approx 3130.724\,W/m^2\]

07

Calculate the instantaneous rate of melting with heat transfer

Divide the remaining energy by the latent heat of fusion to find the rate of melting with heat transfer: \[Rate_{melting\_heat\_transfer}=\frac{E_{remaining}}{h_{sf}}\] \[Rate_{melting\__ht\_transfer}=\frac{3130.712\,\cancel{J/s}\cdot m^{2}}{33000\,\cancel{J}/kg}\approx0.0948\,kg/s\cdot m^{2}\] (b) The instantaneous rate of melting considering heat transfer is \(0.0948\, kg/s\cdot m^{2}\).

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Key Concepts

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

Latent Heat of Fusion

Latent heat of fusion refers to the amount of heat required to change a substance from a solid to a liquid at constant temperature and pressure. In the context of our original exercise, the Wood's metal slab is subjected to this process. The metal starts melting at its fusion temperature of 72掳C.

When the laser irradiates the metal at a power of 5 kW/m虏, the absorbed heat energy is calculated by multiplying this power by the metal's absorptivity, which is 0.4:

• Absorbed Energy: 2 kW/m虏

To determine how fast the metal melts, you can divide the absorbed heat energy by the latent heat of fusion, given as 33 kJ/kg. This calculation gives you the melting rate, which shows how much material turns from solid to liquid per square meter every second:

• Instantaneous Melting Rate: 0.0606 kg/s路m虏

Understanding latent heat is essential for calculating heat required in phase transitions without temperature changes.

Convective Heat Transfer

Convective heat transfer is the transfer of heat between a solid surface and a fluid due to fluid motion. In the exercise, convection occurs between the surface of the Wood's metal and ambient air.

The convective heat loss is determined by the surrounding air temperature (20掳C), the metal's fusion temperature (72掳C), and a heat transfer coefficient (15 W/m虏路K). You utilize the formula:

• \[Q_{conv} = h(T_{f} - T_{ ext{ambient}})\]

The result of this calculation is a heat loss of -780 W/m虏, meaning heat is leaving the metal. This needs to be subtracted from the absorbed energy to find out how much energy actually aids the melting process. This accounts for the real-world scenario where not all energy results in melting because of surrounding heat exchanges.

Taking convective heat loss into account helps in identifying less apparent influences on thermal systems, allowing for more precise engineering calculations.

Radiative Heat Transfer

Radiative heat transfer is the process where heat is emitted as electromagnetic waves. This method does not require a medium and can occur across a vacuum, unlike conduction and convection.

In the problem, the metal also exchanges heat through radiation with its large surroundings maintained at 20掳C. The emissivity of the surface is given as 0.4 and the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \) W/m虏K鈦�) is used in the formula:

• \[Q_{rad}= \varepsilon \sigma(T_{f}^{4} - T_{sur}^{4})\]

This gives a radiative heat loss of approximately -350.724 W/m虏.

Again, this amount must also be deducted from the absorbed energy. By considering radiation, you assess how much heat is lost to large, cool surroundings through emission. Understanding radiative effects is critical for designing systems involved with high temperatures or in space, where radiation is a dominant form of heat transfer.