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Chapter 5: Problem 34

A 1-m-long and 0.1-m-thick steel plate of thermal conductivity \(35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is well insulated on its both sides, while the top surface is exposed to a uniform heat flux of \(5500 \mathrm{~W} / \mathrm{m}^{2}\). The bottom surface is convectively cooled by a fluid at \(10^{\circ} \mathrm{C}\) having a convective heat transfer coefficient of \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming one dimensional heat conduction in the lateral direction, find the temperature at the midpoint of the plate. Discretize the plate thickness into four equal parts.

Short Answer

Expert verified
The cooling fluid near the bottom surface has a temperature of 10°C. Answer: To calculate the midpoint temperature of the steel plate, follow the steps in the solution provided. First, calculate the conduction and convection thermal resistances, then find the equivalent thermal resistance. Next, find the heat transfer rate and the temperature difference between the bottom surface and the midpoint. Finally, calculate the midpoint temperature using the cooling fluid temperature and the temperature difference.

Step by step solution

01

Calculate the conduction thermal resistance for each discretized section of the plate

For a 0.1 m thick steel plate discretized into 4 equal parts, each section's thickness will be 0.025 m. We have the thermal conductivity (\(k\)) as 35 W/m·K. The conduction resistance for each section (\(R_c\)) can be calculated as: \[R_c = \frac{L}{k \cdot A} = \frac{0.025}{35 \cdot 1}\]
02

Calculate the convection thermal resistance

Now, we have to calculate the convection resistance (\(R_{conv}\)) which is given by the following formula: \[R_{conv} = \frac{1}{h \cdot A}\] Here, using the given convective heat transfer coefficient value, \(h\) = 150 W/m²·K, and area A = 1 m².
03

Calculate the equivalent thermal resistance

We know that the middle point is at the interface between the two internal discretized sections. First, we have to find the equivalent thermal resistance considering both conduction and convection, which can be calculated as: \[ R_{eq} = R_{conv} + 2 \cdot R_c \]
04

Calculate the heat transfer rate

As the problem states, the top surface is exposed to a uniform heat flux of 5500 W/m². We can calculate the heat transfer rate (\(q\)) using the formula: \[q = q'' \cdot A\] where \(q''\) is the given heat flux as 5500 W/m², and A is the area as 1 m².
05

Calculate the temperature difference between the bottom surface and the midpoint

Using the equivalent thermal resistance and heat transfer rate, we can find the temperature difference (\(\Delta T\)) between the cooling fluid temperature at the bottom surface and the midpoint temperature using the formula: \[\Delta T = q \cdot R_{eq}\]
06

Calculate the midpoint temperature

Finally, we can calculate the midpoint temperature (\(T_{midpoint}\)) using the cooling fluid temperature (\(T_{fluid}\)) and the temperature difference: \[T_{midpoint} = T_{fluid} + \Delta T\] where \(T_{fluid}\) is 10°C.

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Most popular questions from this chapter

A hot brass plate is having its upper surface cooled by impinging jet of air at temperature of \(15^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(220 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The 10 -cm-thick brass plate \(\left(\rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=380 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\), and \(\alpha=33.9 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) had a uniform initial temperature of \(650^{\circ} \mathrm{C}\), and the lower surface of the plate is insulated. Using a uniform nodal spacing of \(\Delta x=2.5 \mathrm{~cm}\) and time step of \(\Delta t=10 \mathrm{~s}\) determine \((a)\) the implicit finite difference equations and \((b)\) the nodal temperatures of the brass plate after 10 seconds of cooling.

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?

Explain how the finite difference form of a heat conduction problem is obtained by the energy balance method.

Why do the results obtained using a numerical method differ from the exact results obtained analytically? What are the causes of this difference?

Design a fire-resistant safety box whose outer dimensions are \(0.5 \mathrm{~m} \times 0.5 \mathrm{~m} \times 0.5 \mathrm{~m}\) that will protect its combustible contents from fire which may last up to \(2 \mathrm{~h}\). Assume the box will be exposed to an environment at an average temperature of \(700^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the temperature inside the box must be below \(150^{\circ} \mathrm{C}\) at the end of \(2 \mathrm{~h}\). The cavity of the box must be as large as possible while meeting the design constraints, and the insulation material selected must withstand the high temperatures to which it will be exposed. Cost, durability, and strength are also important considerations in the selection of insulation materials.
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