91Ó°ÊÓ

A very thick slab with thermal diffusivity \(5.6 \mathrm{x}\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) and thermal conductivity \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is ini tially at a uniform temperature of \(325^{\circ} \mathrm{C}\). Sudklenly, tte surface is exposed to a ccolant at \(15^{\circ} \mathrm{C}\) for which the convection heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine temperatures at the surface and at a depth of \(45 \mathrm{~mm}\) after 3 min have elapsed. (b) Compute and plot temperature histories \((0 \leq t \leq\) \(3(x)\) s) at \(x=0)\) and \(x=45 \mathrm{~mm}\) for the following parametric variations: (i) \(\alpha=5.6 \times 10^{-7} .5 .6 \times\) \(10^{-6}\), and \(5.6 \times 10^{-3} \mathrm{~m}^{2} / \mathrm{s}\); and (ii) \(k=2,20\), and \(200 \mathrm{~W} / \mathrm{m}-\mathrm{K}\).

Step by step solution

01

Determine the Block Parameters

We have thermal diffusivity \(\alpha = 5.6 \times 10^{-6} \text{ m}^2/\text{s}\), thermal conductivity \(k = 20 \text{ W/m} \cdot \text{K}\), initial slab temperature \(T_i = 325^\circ \text{C}\), surface coolant temperature \(T_\infty = 15^\circ \text{C}\), and convection heat transfer coefficient \(h = 100 \text{ W/m}^2 \cdot \text{K}\). The slab thickness \(L\) is very large, and we need to find temperatures after \(t = 3 \times 60 = 180\) seconds.

02

Calculate Biot Number

The Biot number \(Bi\) determines the appropriateness of different heat transfer solutions. Use \(Bi = \frac{h L}{k}\). Since \(L\) is large, the assumption of a semi-infinite wall is applicable, especially since \(Bi \gg 0.1\), enabling use of the error function solution for semi-infinite solids.

03

Use the Error Function Solution

For a semi-infinite solid, the temperature \(T(x, t)\) at depth \(x\) and time \(t\) is given by:\[ T(x,t) = T_\infty + (T_i - T_\infty) \text{erf}\left( \frac{x}{2\sqrt{\alpha t}} \right) \]where the error function is based on depth \(x\) and time \(t\): \(\text{erf}(z) = \text{error function}(z)\).

04

Calculate Surface Temperature

Set \(x=0\) for the surface temperature calculation:\[ T(0, 180) = 15 + (325 - 15) \times \text{erf}(0) \]Since \(\text{erf}(0) = 0\), it simplifies to:\[ T(0, 180) = 15^\circ \text{C} \].

05

Calculate Temperature at 45mm Depth

For \(x = 0.045 \text{ m}\), use:\[ T(0.045, 180) = 15 + (325 - 15) \times \text{erf}\left( \frac{0.045}{2\sqrt{5.6 \times 10^{-6} \times 180}} \right) \]Calculate the argument for the error function and determine \(\text{erf}\left(\frac{0.045}{2 \times 0.01}\right)\) from standard tables.

06

Vary Parameters and Plot

For different values of \(\alpha\) and \(k\), repeat steps 3 to 5. Consider new scenarios with parameters: (i) \(\alpha = 5.6 \times 10^{-7}, 5.6 \times 10^{-6}, 5.6 \times 10^{-3} \text{ m}^2/\text{s}\)(ii) \(k = 2, 20, 200 \text{ W/m} \cdot \text{K}\)Use these values to recompute and plot \(T(0, t)\) and \(T(0.045, t)\) from \(t = 0\) to \(t = 180\) seconds using the same formula structure for error function temperatures.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Thermal Diffusivity

Thermal diffusivity is a property of materials that measures how quickly heat spreads through an object. In simpler terms, it determines the speed at which a material heats up or cools down. It is particularly crucial when analyzing the heat conduction of a material. The units of thermal diffusivity are square meters per second (m²/s). Thermal diffusivity is calculated using the formula \( \alpha = \frac{k}{\rho c_p} \), where:

• \( k \) is the thermal conductivity
• \( \rho \) is the density of the material
• \( c_p \) is the specific heat capacity

In practical applications, a high thermal diffusivity means a material will rapidly adjust its temperature. Meanwhile, a low thermal diffusivity indicates a slow temperature change. Understanding thermal diffusivity helps us determine how a material reacts to temperature changes, which is essential in designing heating and cooling systems.

Biot Number

The Biot number (Bi) is a dimensionless number that compares internal conduction resistance within a body to the convection resistance at the surface of the body. It helps us decide the type of heat conduction problem we're dealing with. The Biot number is given by the formula: \( Bi = \frac{hL}{k} \), where:

• \( h \) is the convection heat transfer coefficient
• \( L \) is the characteristic length of the object
• \( k \) is the thermal conductivity

A small Biot number (\( Bi < 0.1 \)) suggests that temperature within the object is uniform and can be treated as a lumped system, which means simpler mathematical models can be used. For large objects or when \( Bi > 0.1 \), the temperature gradient within the body is significant enough to consider more detailed models like semi-infinite solid approximations.

Error Function Solution

The error function solution is a crucial part of solving heat conduction problems, especially when dealing with semi-infinite solids, like very thick slabs. This function helps us determine the temperature at a specific depth and time after a sudden change at the surface.The error function, denoted as \( \text{erf}(z) \), is a special mathematical function often used in diffusion problems. For heat conduction, the temperature \( T(x,t) \) at depth \( x \) and time \( t \) can be calculated by:\[ T(x, t) = T_{\infty} + (T_i - T_{\infty}) \times \text{erf}\left( \frac{x}{2\sqrt{\alpha t}} \right) \]Where:

• \( T_{\infty} \) is the coolant temperature
• \( T_i \) is the initial temperature of the body
• \( \alpha \) is the thermal diffusivity

The error function value varies between -1 and 1, reflecting how far the temperature has reached through the material at any given time.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient (h) is a measure of how effective a fluid is at transferring heat through convection. It describes the thermal boundary layer between a solid surface and a moving fluid.The value of \( h \) depends on several factors:

• Type of fluid (air, water, etc.)
• Flow characteristics (laminar or turbulent)
• Surface roughness
• Fluid velocity

In many practical applications, determining the correct \( h \) value is crucial for designing energy-efficient systems. It is also used to calculate the Biot number, which, as mentioned earlier, helps in analyzing heat transfer problems. The unit is typically \( \text{W/m}^2 \cdot \text{K} \), describing energy transfer per unit area per degree of temperature difference.