91Ó°ÊÓ

The top surface of a plate, including its grooves, is maintained at a uniform temperature of \(T_{1}=200^{\circ} \mathrm{C}\). The lower surface is at \(T_{2}=20^{\circ} \mathrm{C}\), the thermal conductivity is \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and the groove spacing is \(0.16 \mathrm{~m}\). (a) Using a finite-difference method with a mesh size of \(\Delta x=\Delta y=40 \mathrm{~mm}\), calculate the unknown nodal temperatures and the heat transfer rate per width of groove spacing \((w)\) and per unit length normal to the page. (b) With a mesh size of \(\Delta x=\Delta y=10 \mathrm{~mm}\), repeat the foregoing calculations, determining the temperature field and the heat rate. Also, consider conditions for which the bottom surface is not at a uniform temperature \(T_{2}\) but is exposed to a fluid at \(T_{\infty}=20^{\circ} \mathrm{C}\). With \(\Delta x=\Delta y=10 \mathrm{~mm}\), determine the temperature field and heat rate for values of \(h=5,200\), and 1000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), as well as for \(h \rightarrow \infty\).

Step by step solution

01

Part (a): Calculation of Nodal Temperatures with Mesh Size 40 mm

First, create a grid using the given mesh size(\( \Delta x = \Delta y = 40 \mathrm{~mm} \)). Next, apply the following finite-difference equation for conduction in two dimensions, at each node: \[ \frac{T_{i+1,j} - 2T_{i,j} + T_{i-1,j}}{\Delta x^2} + \frac{T_{i,j+1} - 2T_{i,j} + T_{i,j-1}}{\Delta y^2} = 0 \] Since \( \Delta x = \Delta y \), the equation can be simplified as follows: \[ T_{i+1,j}+T_{i-1,j}+T_{i,j+1}+T_{i,j-1}-4T_{i,j} = 0 \] Now substitute the known boundary and initial conditions, solve for the unknown nodal temperatures using a suitable numerical technique (e.g. Gauss-Seidel iteration) and calculate the heat transfer rate per width, w, and per unit length.

02

Part (b): Calculation of Nodal Temperatures with Mesh Size 10 mm

Create a new grid, using the given mesh size (\( \Delta x = \Delta y = 10 \mathrm{~mm} \)). Apply the finite-difference equation for conduction in two dimensions, as discussed in Part (a). Solve for the unknown nodal temperatures and calculate the heat transfer rate per width, w, and per unit length. Next, consider the additional condition for this part: the bottom surface is exposed to a fluid at \(T_{\infty}=20^{\circ} \mathrm{C}\). So the boundary condition at the bottom surface will be updated: \[ q_{i,1} = -k\left(\frac{T_{i,1}-T_{i,2}} {\Delta y}\right) = h(T_{i,1}-T_\infty) \] Now, solve the finite-difference equation with the updated boundary condition at the bottom surface for h = 5, 200, and 1000, and for \(h \rightarrow \infty\). Calculate the nodal temperatures and the heat transfer rate per width, w, and per unit length for each of these scenarios. Remember to use proper numerical techniques, like Gauss-Seidel iteration, to solve for the unknown nodal temperatures in both parts (a) and (b).

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.

Heat Conduction

Heat conduction is a mode of heat transfer within a body or between two bodies in direct contact. In this exercise, we're dealing with a plate that has different temperatures on its top and bottom surfaces. The heat will naturally flow from the hot surface at 200°C on the top to the cooler bottom surface at 20°C.

The thermal conductivity of the material, given as 15 W/m·K, is crucial as it dictates how easily heat flows through the material. Higher thermal conductivity means the material can conduct heat more effectively. In essence, a good conductor will allow heat to transfer quickly, while a poor conductor will resist the flow of heat.

• The temperature gradient drives the heat flow, with heat moving from areas of high temperature to areas of low temperature.
• The thermal conductivity is constant in this exercise, simplifying calculations as it does not vary with temperature.

Nodal Temperatures

Nodal temperatures are the unknown temperatures at specific points (nodes) within a domain. This exercise uses the finite-difference method to estimate these temperatures. We divide the plate into smaller sections using a grid, specifying node points at each intersect.

Using the finite-difference equations, we compute temperatures at these nodes. This approach converts the continuous problem of heat conduction into a discrete set of algebraic equations. By solving these equations, we find the temperatures at each node.

• The method uses boundary conditions, such as the temperatures at the surfaces, to solve for internal node temperatures.
• Nodes at the borders of the grid have known temperatures based on boundary conditions, simplifying calculations at these points.
• Interior nodes are computed iteratively using techniques like Gauss-Seidel iteration, adjusting each node temperature until the system stabilizes to a solution.

Heat Transfer Rate

The heat transfer rate quantifies the amount of heat energy transferred per unit time. For this exercise, we look at the rate across a unit width of the groove spacing. This is calculated after establishing the nodal temperatures.

The rate of heat transfer can be computed using Fourier's Law of Heat Conduction, which states that the rate is proportional to the thermal conductivity, the area through which heat flows, and the temperature gradient.

• For a flat plate, the typical process involves integrating over the domain, accounting for the conduction at each node.
• Calculating the heat rate requires knowledge of each nodal temperature, ensuring the integration reflects gradients between adjacent nodes accurately.
• The results tell us how much heat is being conducted through the material in a given direction.

Numerical Techniques

Numerical techniques play a crucial role in solving complex engineering problems like heat conduction described here. Unlike analytical solutions, numerical methods handle irregular geometries, variable properties, and complex boundary conditions effectively.

The finite-difference method, used in this problem, involves discretizing the domain into a grid of nodes, then applying difference equations to approximate derivatives. This replaces continuous partial differential equations with algebraic equations.

• Techniques such as the Gauss-Seidel iteration simplify solving these equations by iteratively adjusting estimates of nodal temperatures until they converge to a stable solution.
• This method is preferable in engineering applications since it allows flexibility in accommodating various boundary conditions and material properties.
• Adjusting mesh sizes alters precision and computational resources needed; finer meshes like 10 mm offer more accuracy but require more computing power.