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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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Key Concepts

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

Thermal Resistance
Thermal resistance is a crucial concept when dealing with heat conduction. It's a measure of how well an object resists the flow of heat, much like electrical resistance works in the case of electric current. The higher the thermal resistance, the more difficult it is for heat to pass through the material.
In the context of the given problem, the steel plate's thermal resistance comes from its structure and material properties. When we discretize the plate into smaller sections for analysis, we can calculate the thermal resistance for each section using the formula:\[ R_c = \frac{L}{k \cdot A} \]
where:
  • \(L\) is the thickness of the section
  • \(k\) is the thermal conductivity
  • \(A\) is the area through which heat is conducted
Understanding thermal resistance helps us predict how heat flows through different materials, which is vital for engineering applications like the cooling of electronic devices and building insulation.
Heat Flux
Heat flux refers to the rate of heat energy transfer through a given surface. In the exercise, the top surface of the steel plate experiences a uniform heat flux of 5500 W/m². This represents how much heat is being applied to each square meter of the plate.
The heat transfer rate \(q\), which signifies the total amount of heat energy transferred per unit time, is calculated using the formula:\[ q = q'' \cdot A \]
where:
  • \(q''\) is the heat flux
  • \(A\) is the area of the surface
Heat flux is an essential variable in many thermal analyses as it directly affects the temperature distribution within materials and is often controlled to achieve desired thermal conditions.
Convective Heat Transfer
Convective heat transfer occurs when heat is transferred between a surface and a fluid flowing across it. It is a combination of conduction and the heat carried by the fluid itself. The rate at which heat is transferred by convection is often determined by the convective heat transfer coefficient \(h\).
In the given scenario, the steel plate's bottom surface is convectively cooled by a fluid at 10°C, with a heat transfer coefficient of 150 W/m²·K. To consider this in the calculations, we use the convection resistance \(R_{conv}\):\[ R_{conv} = \frac{1}{h \cdot A} \]
This formula helps us understand how efficiently heat is transferred from the plate to the cooling fluid. Convective cooling is widely used in many systems, such as in radiators, heat sinks, and various HVAC systems.
Temperature Calculation
Calculating temperature in thermodynamics, especially in conduction problems, involves understanding how heat moves through materials and how it affects their temperatures.
The exercise asks for the temperature at the midpoint of the steel plate. Using the equivalent thermal resistance \(R_{eq}\) and heat transfer rate \(q\), the temperature difference \(\Delta T\) between the cooling fluid's temperature and the midpoint is calculated with:\[ \Delta T = q \cdot R_{eq} \]
Finally, the midpoint temperature \(T_{midpoint}\) is found by adding this temperature difference to the fluid temperature \(T_{fluid}\):\[ T_{midpoint} = T_{fluid} + \Delta T \]
where the initial cooling fluid temperature is 10°C.
This approach to temperature calculation is frequently employed in engineering tasks to ensure that devices and structures are operating within safe temperature ranges, thereby preventing overheating and potential failure.

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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?

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 an aluminum alloy fin \((k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of triangular cross section whose length is \(L=5 \mathrm{~cm}\), base thickness is \(b=1 \mathrm{~cm}\), and width \(w\) in the direction normal to the plane of paper is very large. The base of the fin is maintained at a temperature of \(T_{0}=180^{\circ} \mathrm{C}\). The fin is losing heat by convection to the ambient air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}=290 \mathrm{~K}\). Using the finite difference method with six equally spaced nodes along the fin in the \(x\)-direction, determine \((a)\) the temperatures at the nodes and \((b)\) the rate of heat transfer from the fin for \(w=1 \mathrm{~m}\). Take the emissivity of the fin surface to be \(0.9\) and assume steady one-dimensional heat transfer in the fin.

A DC motor delivers mechanical power to a rotating stainless steel shaft ( \(k=15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with a length of \(25 \mathrm{~cm}\) and a diameter of \(25 \mathrm{~mm}\). The DC motor is in a surrounding with ambient air temperature of \(20^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the base temperature of the motor shaft is \(90^{\circ} \mathrm{C}\). Using a uniform nodal spacing of \(5 \mathrm{~cm}\) along the motor shaft, determine the finite difference equations and the nodal temperatures by solving those equations.

Consider steady two-dimensional heat transfer in a long solid bar \((k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of square cross section \((3 \mathrm{~cm} \times 3 \mathrm{~cm})\) with the prescribed temperatures at the top, right, bottom, and left surfaces to be \(100^{\circ} \mathrm{C}, 200^{\circ} \mathrm{C}, 300^{\circ} \mathrm{C}\), and \(500^{\circ} \mathrm{C}\), respectively. Heat is generated in the bar uniformly at a rate of \(\dot{e}=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Using a uniform mesh size \(\Delta x=\Delta y=1 \mathrm{~cm}\) determine \((a)\) the finite difference equations and \((b)\) the nodal temperatures with the Gauss-Seidel iterative method.

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