A stainless steel slab \((k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=\) \(\left.3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) and a copper slab \((k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) 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 \(20^{\circ} \mathrm{C}\). Using \(\mathrm{EES}\) (or other) software, investigate the effect of time on the temperatures of both materials at the depth of \(5 \mathrm{~cm}\) from the surface. By varying the time from 1 to \(80 \mathrm{~s}\) after the slabs have received the energy pulse, plot the temperatures at \(5 \mathrm{~cm}\) from the surface as a function of time.

Short Answer

Expert verified
Answer: To determine the temperature changes of stainless steel and copper at a depth of 5 cm, we can utilize the heat equation and consider the given initial and boundary conditions. By applying Laplace Transforms and using computational software like EES, we can calculate the temperature at the depth of 5 cm over a time range of 1 to 80 seconds. Finally, plotting the results will provide insights into the temperature changes for both materials at the given depth due to their differing thermal properties.

Step by step solution

01

Prepare the Heat Equation

The heat equation can be written as: \[u_t = \alpha u_{xx}\] where \(u_t\) is the time derivative of the function \(u(x,t)\), \(u_{xx}\) is the second spatial derivative of the function \(u(x,t)\), u(x,t) represents the temperature at position x and time t, and \(\alpha\) is the thermal diffusivity. In this problem, we have to solve this equation for both materials with the given thermal properties.
02

Apply Boundary Conditions and Initial Conditions

At t=0, both slabs receive an energy pulse, so we set the initial condition as the uniform temperature of 20°C: \[u(x, 0) = 20\] Boundary conditions: At the surface(x=0), the energy pulse is supplied which must be taken into account. At the depth of interest (x=5cm) we don't have any boundary condition, so we have to solve the equation for both materials at the depth of 5 cm. The best approach to solve this type of problem with an initially known temperature is applying Laplace Transforms on both sides of the heat equation regarding the time variable 't'. It will simplify the integration and makes solving the problem easier.
03

Use EES software or other available tools to solve the transformed equation with given conditions

We will use EES (Engineering Equation Solver) or any other computational software to solve the transformed equation with given initial conditions and material properties. Carry out the computations for the temperature at depth 5 cm and vary the time from 1 to 80 s.
04

Plot the results

After obtaining the results, plot the temperature at 5 cm depth as a function of time for both materials, stainless steel, and copper. This plot will show the effect of time on the temperatures of both materials at the depth of 5 cm from the surface. By following these steps, you can investigate the effect of time on the temperatures of stainless steel and copper at the depth of 5 cm. The plot generated will give you insight into the temperature changes over time for both materials, considering their different thermal properties.

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Most popular questions from this chapter

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