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Consider a stainless steel spoon $(k=15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, \)\varepsilon=0.6$ ) that is partially immersed in boiling water at \(100^{\circ} \mathrm{C}\) in a kitchen at \(32^{\circ} \mathrm{C}\). The handle of the spoon has a cross section of about $0.2 \mathrm{~cm} \times 1 \mathrm{~cm}\( and extends \)18 \mathrm{~cm}$ in the air from the free surface of the water. The spoon loses heat by convection to the ambient air with an average heat transfer coefficient of $h=13 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ as well as by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}=295 \mathrm{~K}\). Assuming steady one- dimensional heat transfer along the spoon and taking the nodal spacing to be \(3 \mathrm{~cm},(a)\) obtain the finite difference formulation for all nodes, \((b)\) determine the temperature of the tip of the spoon by solving those equations, and (c) determine the rate of heat transfer from the exposed surfaces of the spoon.

Short Answer

Expert verified
Based on the given information about the stainless steel spoon submerged in boiling water, we first used the finite difference method to approximate the heat transfer equation in the spoon handle. Next, we applied the boundary conditions for the nodes in contact with boiling water and the tip of the spoon, considering convection and radiation heat transfer. By solving the system of equations, we obtained the tip temperature and calculated the rate of heat transfer from the exposed surfaces of the spoon.

Step by step solution

01

Obtain the finite difference formulation for all nodes

Let's divide the spoon handle into segments of 3 cm length, which will give us 6 nodes from the free surface of the water to the tip of the spoon. We will use the finite difference method to approximate the heat transfer equation at each node. The general equation for steady-state, one-dimensional heat conduction is given by: $$\frac{d}{dx}\left(k\frac{dT}{dx}\right) - q_{generation} = 0$$ Since there is no heat generation within the spoon, the equation simplifies to: $$\frac{d}{dx}\left(k\frac{dT}{dx}\right) = 0$$ Using the central difference approximation for the spatial derivatives, we get the finite difference equation for each node: $$\frac{k}{\Delta x^2} (T_{i-1} - 2T_i + T_{i+1}) = 0$$ where \(T_i\) is the temperature at node \(i\), and \(\Delta x = 0.03\) m is the distance between nodes.
02

Apply boundary conditions and solve for the tip temperature

We need to apply boundary conditions at the first node (in contact with boiling water) and at the last node (the tip of the spoon). Boundary condition at node 1 (in contact with boiling water): $$T_1 = 100^{\circ} \mathrm{C}$$ Boundary condition at node 6 (the tip of the spoon): Here we have heat transfer by convection and radiation: $$q_{conv} = h \cdot A_{cs} \cdot (T_6 - T_\infty)$$ $$q_{rad} = \varepsilon \cdot \sigma \cdot A_{cs} \cdot (T_6^4 - T_{surr}^4)$$ The sum of these heat fluxes must be equal to the heat conduction flux through the spoon at node 5: $$k \cdot \frac{A_{cs}}{\Delta x} \cdot (T_5 - T_6) = q_{conv} + q_{rad}$$ Substituting the expressions for \(q_{conv}\) and \(q_{rad}\) and rearranging for \(T_6\), we get: $$T_6 = \frac{k \cdot (T_5 - T_6) \cdot \Delta x + h \cdot (T_6 - T_\infty) + \varepsilon \cdot \sigma \cdot (T_6^4 - T_{surr}^4)}{h + \varepsilon \cdot \sigma}$$ Now, we have finite difference equations for nodes 2 to 5, and boundary conditions for nodes 1 and 6. We can solve this system of equations using any numerical method (e.g., Gauss-Seidel or Newton-Raphson method) to obtain the temperatures for all nodes, including the tip temperature \(T_6\).
03

Determine the rate of heat transfer from the exposed surfaces of the spoon

Now that we have the temperatures at all nodes, we can calculate the rate of heat transfer from the exposed surfaces of the spoon by summing the convection and radiation heat transfer rates at each node: $$Q = \sum_{i=1}^{6} [q_{conv}(T_i) + q_{rad}(T_i)] \cdot A_{cs} \cdot \Delta x$$ Calculate the total heat transfer by plugging in the known values and the calculated temperatures at each node. This will give us the rate of heat transfer from the exposed surfaces of the spoon.

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Most popular questions from this chapter

Consider a medium in which the finite difference formulation of a general interior node is given in its simplest form as $$ \frac{T_{m-1}-2 T_{m}+T_{m+1}}{\Delta x^{2}}+\frac{\dot{e}_{m}}{k}=0 $$ (a) Is heat transfer in this medium steady or transient? (b) Is heat transfer one-, two-, or three-dimensional? (c) Is there heat generation in the medium? (d) Is the nodal spacing constant or variable? (e) Is the thermal conductivity of the medium constant or variable?

Consider a large plane wall of thickness \(L=0.3 \mathrm{~m}\), thermal conductivity \(k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=24 \mathrm{~m}^{2}\). The left side of the wall is subjected to a heat flux of \(\dot{q}_{0}=350 \mathrm{~W} / \mathrm{m}^{2}\), while the temperature at that surface is measured to be \(T_{0}=60^{\circ} \mathrm{C}\). Assuming steady one-dimensional heat transfer and taking the nodal spacing to be $6 \mathrm{~cm},(a)$ obtain the finite difference formulation for the six nodes and \((b)\) determine the temperature of the other surface of the wall by solving those equations.

Using appropriate software, solve these systems of algebraic equations. (a) \(\quad 3 x_{1}-x_{2}+3 x_{3}=0\) $$ \begin{array}{r} -x_{1}+2 x_{2}+x_{3}=3 \\ 2 x_{1}-x_{2}-x_{3}=2 \end{array} $$ (b) $$ \begin{aligned} 4 x_{1}-2 x_{2}^{2}+0.5 x_{3} &=-2 \\ x_{1}^{3}-x_{2}+x_{3} &=11.964 \\ x_{1}+x_{2}+x_{3} &=3 \end{aligned} $$ Answers: \((a) x_{1}=2, x_{2}=3, x_{3}=-1,(b) x_{1}=2.33, x_{2}=2.29\), \(x_{3}=-1.62\)

Starting with an energy balance on the volume element, obtain the steady three-dimensional finite difference equation for a general interior node in rectangular coordinates for \(T(x, y, z)\) for the case of constant thermal conductivity and uniform heat generation.

Consider steady two-dimensional heat conduction in a square cross section $(3 \mathrm{~cm} \times 3 \mathrm{~cm}, k=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=6.694 \times\( \)10^{-6} \mathrm{~m}^{2} / \mathrm{s}$ ) with constant prescribed temperature of \(100^{\circ} \mathrm{C}\) and $300^{\circ} \mathrm{C}$ at the top and bottom surfaces, respectively. The left surface is exposed to a constant heat flux of \(1000 \mathrm{~W} / \mathrm{m}^{2}\), while the right surface is in contact with a convective environment $\left(h=45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\( at \)20^{\circ} \mathrm{C}\(. Using a uniform mesh size of \)\Delta x=\Delta y\(, determine \)(a)$ finite difference equations and \((b)\) the nodal temperatures using the Gauss- Seidel iteration method.

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