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Starting with an energy balance on a volume element, derive the two- dimensional transient heat conduction equation in rectangular coordinates for \(T(x, y, t)\) for the case of constant thermal conductivity and no heat generation.

Short Answer

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Question: Derive the two-dimensional transient heat conduction equation in rectangular coordinates with constant thermal conductivity and no heat generation. Answer: The two-dimensional transient heat conduction equation in rectangular coordinates with constant thermal conductivity and no heat generation is given by: $$\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = \frac{\rho c_p}{k} \frac{\partial T}{\partial t}$$

Step by step solution

01

Define the Problem Variables

First, let us define the variables that we will use to describe the heat conduction in two-dimensional rectangular coordinates. - \(T(x, y, t)\): Temperature (K) at position \((x, y)\) and time \(t\) - \(k\): Thermal conductivity (W/m K) – constant - \(q\): Heat transfer rate per unit volume (W/m³) – no heat generation, so \(q=0\)
02

Write the energy balance equation for a volume element

We will now write the energy balance equation for a volume element \(\delta x \delta y \delta z\) located at position \((x, y)\). The energy balance equation states that the rate of heat entering the volume element must be equal to the rate of change of energy stored in the element. $$\frac{\partial}{\partial x} \left( k \frac{\partial T}{\partial x} \right) \delta x \delta y \delta z + \frac{\partial}{\partial y} \left( k \frac{\partial T}{\partial y} \right) \delta x \delta y \delta z = \rho c_p \frac{\partial T}{\partial t} \delta x \delta y \delta z$$ Here, \(\rho\) is the density (kg/m³) and \(c_p\) is the specific heat capacity at constant pressure (J/kg K).
03

Simplify the energy balance equation

As we know that \(k\) (thermal conductivity) is constant, we can take it out of the partial derivatives, and then we divide the entire equation by \(\delta x\delta y\delta z\) to simplify. $$k \frac{\partial^2 T}{\partial x^2} + k \frac{\partial^2 T}{\partial y^2} = \rho c_p \frac{\partial T}{\partial t}$$
04

The final two-dimensional transient heat conduction equation

Finally, we divide both sides of the equation by \(k\) and obtain the desired two-dimensional transient heat conduction equation in rectangular coordinates for \(T(x, y, t)\), with constant thermal conductivity and no heat generation: $$\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = \frac{\rho c_p}{k} \frac{\partial T}{\partial t}$$

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Key Concepts

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

Energy Balance Equation
When analyzing heat conduction, an essential part of the process is setting up the energy balance equation. This equation helps us understand how energy moves in and out of a particular volume element. To picture this, imagine that we are looking at a small section within a solid, where energy can both enter and exit.
Typically, the rate of heat entering the element must match the rate of energy change inside the element. In mathematical terms, this is expressed by balancing the heat flow in each direction with the energy stored or lost in the volume. For steady-state conditions, this means that if heat is flowing uniformly through a solid, the energy within the unit remains unchanged.
In the case of transient (or time-dependent) heat conduction, our interest lies in how quickly the energy inside the volume element is changing over time. This is where time derivatives come into play. The classic energy balance equation takes into account variables such as:
  • Temperature changes over time: expressed as \( \frac{\partial T}{\partial t} \)
  • Spatial temperature gradients in each direction: represented by \( \frac{\partial T}{\partial x} \) and \( \frac{\partial T}{\partial y} \)
During the derivation, density (\( \rho \)) and specific heat capacity (\( c_p \)) determine how much energy is stored per degree of temperature change. The energy balance equation serves as the foundation for more detailed heat conduction equations.
Rectangular Coordinates
In the study of heat conduction, choosing the right coordinate system is crucial for simplifying the problem and making the math more manageable. Rectangular coordinates are typically used when dealing with problems involving objects shaped like rectangular prisms or blocks.
Rectangular coordinates, denoted by \( (x, y, z) \), help break down the problem into three perpendicular spatial directions. This makes it simpler to describe how temperature changes along these axes over time. In two-dimensional problems, we often use only \( x \) and \( y \), assuming there is no variation in the third dimension. This is particularly true for flat surfaces or thin objects.
Using rectangular coordinates, we express spatial temperature gradients as partial derivatives. These derivatives, such as \( \frac{\partial T}{\partial x} \) and \( \frac{\partial T}{\partial y} \), are crucial for analyzing how heat flows in the system. They allow us to look at how the temperature changes at specific points within a structure, thus providing insight into the overall heat conduction process.
Overall, rectangular coordinates are practical and intuitive, especially when applying mathematical models to real-world engineering problems where objects have straight edges and flat surfaces.
Thermal Conductivity
Thermal conductivity, denoted as \( k \), is a material-specific property that defines how well a material can conduct heat. It represents the ability of a substance to transfer thermal energy through itself. High thermal conductivity materials, like metals, easily pass heat from one side to the other, whereas low thermal conductivity materials, like wood or foam, resist the flow of heat.
When considering thermal processes, knowing the thermal conductivity helps predict how fast and efficiently heat can travel through an object. In our two-dimensional heat conduction problem, we assume thermal conductivity is constant, which simplifies the math. This assumption means that \( k \) does not change with temperature, time, or place in the material.
Thermal conductivity factors directly into the equations for heat conduction. It appears in the energy balance equations because it defines the rate of heat flow due to temperature gradients, ensuring that heat can spread from hotter regions to cooler ones. This property also affects how quickly the system stabilizes to a uniform temperature when heat is applied or removed.
Understanding and using thermal conductivity correctly allows engineers to design systems where heat transfer is optimized. It is critical in contexts like building insulation, electronic cooling, and even clothing manufacturing to control thermal comfort. The simplified equations used in our derivation rely on this constant property to represent real life scenarios where detailed temperature profiles are calculated accurately.

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Most popular questions from this chapter

Consider a 20-cm-thick large concrete plane wall \((k=0.77 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) subjected to convection on both sides with \(T_{\infty 1}=27^{\circ} \mathrm{C}\) and \(h_{1}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the inside, and \(T_{\infty 2}=8^{\circ} \mathrm{C}\) and \(h_{2}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the outside. Assuming constant thermal conductivity with no heat generation and negligible radiation, (a) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, (b) obtain a relation for the variation of temperature in the wall by solving the differential equation, and \((c)\) evaluate the temperatures at the inner and outer surfaces of the wall.

Hot water flows through a PVC \((k=0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) pipe whose inner diameter is \(2 \mathrm{~cm}\) and outer diameter is \(2.5 \mathrm{~cm}\). The temperature of the interior surface of this pipe is \(50^{\circ} \mathrm{C}\) and the temperature of the exterior surface is \(20^{\circ} \mathrm{C}\). The rate of heat transfer per unit of pipe length is (a) \(77.7 \mathrm{~W} / \mathrm{m}\) (b) \(89.5 \mathrm{~W} / \mathrm{m}\) (c) \(98.0 \mathrm{~W} / \mathrm{m}\) (d) \(112 \mathrm{~W} / \mathrm{m}\) (e) \(168 \mathrm{~W} / \mathrm{m}\)

The variation of temperature in a plane wall is determined to be \(T(x)=52 x+25\) where \(x\) is in \(\mathrm{m}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). If the temperature at one surface is \(38^{\circ} \mathrm{C}\), the thickness of the wall is (a) \(0.10 \mathrm{~m}\) (b) \(0.20 \mathrm{~m}\) (c) \(0.25 \mathrm{~m}\) (d) \(0.40 \mathrm{~m}\) (e) \(0.50 \mathrm{~m}\)

A large steel plate having a thickness of \(L=4\) in, thermal conductivity of \(k=7.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\), and an emissivity of \(\varepsilon=0.7\) is lying on the ground. The exposed surface of the plate at \(x=L\) is known to exchange heat by convection with the ambient air at \(T_{\infty}=90^{\circ} \mathrm{F}\) with an average heat transfer coefficient of \(h=12 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) as well as by radiation with the open sky with an equivalent sky temperature of \(T_{\text {sky }}=480 \mathrm{R}\). Also, the temperature of the upper surface of the plate is measured to be \(80^{\circ} \mathrm{F}\). Assuming steady onedimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the plate, \((b)\) obtain a relation for the variation of temperature in the plate by solving the differential equation, and \((c)\) determine the value of the lower surface temperature of the plate at \(x=0\).

The outer surface of an engine is situated in a place where oil leakage can occur. Some oils have autoignition temperatures of approximately above \(250^{\circ} \mathrm{C}\). When oil comes in contact with a hot engine surface that has a higher temperature than its autoignition temperature, the oil can ignite spontaneously. Treating the engine housing as a plane wall, the inner surface \((x=0)\) is subjected to \(6 \mathrm{~kW} / \mathrm{m}^{2}\) of heat. The engine housing \((k=13.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) has a thickness of \(1 \mathrm{~cm}\), and the outer surface \((x=L)\) is exposed to an environment where the ambient air is \(35^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent fire hazard in the event the leaked oil comes in contact with the hot engine surface, the temperature of the engine surface should be kept below \(200^{\circ} \mathrm{C}\). Determine the variation of temperature in the engine housing and the temperatures of the inner and outer surfaces. Is the outer surface temperature of the engine below the safe temperature?

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