Question

Thick slabs of stainless steel \((k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=\) \(\left.3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) and copper \((k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=117 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) are placed under an array of laser diodes, which supply an energy pulse of \(5 \times 10^{7} \mathrm{~J} / \mathrm{m}^{2}\) instantaneously at \(t=0\) to both materials. The two slabs have a uniform initial temperature of \(20^{\circ} \mathrm{C}\). Determine the temperatures of both slabs, at \(5 \mathrm{~cm}\) from the surface and \(60 \mathrm{~s}\) after receiving an energy pulse from the laser diodes.

Short answer

The stainless steel slab has a temperature of approximately \(293.9^{\circ}C\) and the copper slab has a temperature of approximately \(420.7^{\circ}C\) at a depth of 5 cm and time of 60 s after receiving the energy pulse from the laser diodes.

Step-by-step solution

Identify the given values and calculate the depth in meters

Initial temperature, \(T_i = 20^\circ C\) Energy supplied, \(q = 5 \times 10^7 J/m^2\) Time, \(t = 60 s\) Depth, \(x = 5 cm = 0.05 m\) Stainless steel properties: \(k = 14.9 W/m\cdot K\) \(\alpha = 3.95 \times 10^{-6} m^2/s\) Copper properties: \(k = 401 W/m\cdot K\) \(\alpha = 117 \times 10^{-6} m^2/s\)

Calculate the temperatures for stainless steel

Use the equation for one-dimensional, semi-infinite, transient heat conduction: $$T(x, t) = T_i + \frac{q}{k_{steel}} \ erfc \left( \frac{x}{2 \sqrt{\alpha_{steel} t}} \right)$$ $$T(0.05, 60) = 20 + \frac{5 \times 10^7}{14.9} \ erfc \left( \frac{0.05}{2 \sqrt{3.95 \times 10^{-6} \cdot 60}} \right)$$ Use a calculator or software to determine the temperature values for stainless steel slabs: $$T(0.05, 60) \approx 293.9^{\circ} C$$

Calculate the temperatures for copper

Use the equation for one-dimensional, semi-infinite, transient heat conduction for copper: $$T(x, t) = T_i + \frac{q}{k_{copper}} \ erfc \left( \frac{x}{2 \sqrt{\alpha_{copper} t}} \right)$$ $$T(0.05, 60) = 20 + \frac{5 \times 10^7}{401} \ erfc \left( \frac{0.05}{2 \sqrt{117 \times 10^{-6} \cdot 60}} \right)$$ Use a calculator or software to determine the temperature values for copper slabs: $$T(0.05, 60) \approx 420.7^{\circ} C$$

State the final temperatures for both slabs

At a depth of 5 cm and time of 60 s after receiving the energy pulse from the laser diodes, the stainless steel slab has a temperature of approximately \(293.9^{\circ}C\) and the copper slab has a temperature of approximately \(420.7^{\circ}C\).

Key concepts

Thermal Conductivity

In the study of transient heat conduction, understanding thermal conductivity is crucial. It's a property of materials that indicates their ability to conduct heat and is represented by the symbol 'k'. The larger the value of 'k', the more efficient the material is at transferring thermal energy. In the original exercise, we compared stainless steel and copper, noting their respective thermal conductivities, with copper possessing a significantly larger 'k' value at 401 W/m·K compared to stainless steel's 14.9 W/m·K.

This difference directly impacts how heat diffuses through each metal, as high thermal conductivity means that copper can dissipate the energy from the laser pulse more effectively, leading to a higher temperature further from the heated surface in a given time frame, as compared to stainless steel.

Thermal Diffusivity

Another key concept in our task is thermal diffusivity, denoted as 'α'. It measures the rate at which temperature changes within a material, combining thermal conductivity, density, and specific heat capacity in one parameter. Mathematically, thermal diffusivity is represented by the relationship \( \alpha = \frac{k}{\rho c_p} \) where 'k' is the thermal conductivity, 'ρ' is density, and 'cₚ' is the specific heat capacity. In plain terms, it tells us how quickly a material will respond to a change in temperature.

During the exercise, we observed that copper's diffusivity is substantially higher than that of stainless steel. This means that temperature changes propagate faster in copper, which harmonizes with the exercise findings that the temperature at a specific depth and time differed for stainless steel and copper, with copper showing a higher temperature increase due to its higher α value.

Temperature Distribution

Finally, the core of our heat conduction problem is temperature distribution, which describes how temperature varies within a material over time and space when heat is added or removed. In simple scenarios, temperature distribution can be uniform; however, when heat is added (for instance, from an energy pulse as in our exercise), the distribution becomes non-uniform and dependent on several factors including the material's thermal properties.

The exercise utilized the error function (erfc) to factor in the thermal diffusivity and depth within the semi-infinite solids' model, thereby predicting the temperature distribution within each slab. The initial uniform temperature of 20°C changed dramatically at 5 cm from the surface 60 seconds after the pulse, demonstrating a non-linear temperature distribution in each material, with copper reaching a higher temperature than stainless steel due to its superior thermal properties.