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Does a heat flux vector at a point \(P\) on an isothermal surface of a medium have to be perpendicular to the surface at that point? Explain.

Short Answer

Expert verified
Explain your answer. Answer: Yes, the heat flux vector at a point P on an isothermal surface must be perpendicular to the surface. This is because heat transfer within the isothermal surface does not occur due to the constant temperature, and any existing heat transfer must happen in the direction perpendicular to the surface where a temperature gradient may exist.

Step by step solution

01

Understanding Heat Flux

Heat flux is the rate at which heat energy is transferred through a unit area of a material's surface perpendicular to the flow of heat. It is a vector quantity and is commonly denoted by \(q\). Heat transfer occurs from regions of high temperature to regions of low temperature. The heat flux vector determines the direction and magnitude of the heat transfer.
02

Isothermal Surfaces

An isothermal surface is a two-dimensional surface in a medium where the temperature is constant at every point on that surface. Due to this constant temperature property, there is no heat transfer happening within the isothermal surface.
03

Relating Heat Flux to Isothermal Surface

Given that there is no temperature variation along the isothermal surface, any heat transfer that may occur will happen in a direction perpendicular to that surface. Because heat flux is a vector quantity, the direction of the heat flux vector (the direction of heat transfer) will be directly related to the surface properties.
04

Heat Flux on Point P

Now let's consider the point P on the isothermal surface. If there were a heat transfer happening at that point, it would not occur along the surface since the temperature is constant there. Instead, the direction of the heat flux vector must be perpendicular to the surface because heat transfer only occurs in directions with temperature gradients.
05

Conclusion

The heat flux vector at a point P on an isothermal surface does indeed have to be perpendicular to the surface. This is because heat transfer within the isothermal surface does not exist due to the constant temperature, and any existing heat transfer must occur in the direction perpendicular to the surface where a temperature gradient may exist.

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Key Concepts

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

Heat Transfer
The concept of heat transfer lies at the core of thermodynamics and is central to understanding how energy moves from one place to another. It is a process where thermal energy is transported due to a temperature difference. There are three main modes of heat transfer: conduction, convection, and radiation.

Conduction occurs through direct contact, where heat is passed from molecule to molecule. Convection is the movement of heat by the physical movement of fluid, encompassing both liquids and gases. Lastly, radiation is the transfer of heat through electromagnetic waves, where no medium is necessary for the process to occur.

To further clarify, thermal energy moves spontaneously from a warmer to a cooler region. This transfer continues until thermal equilibrium is reached - a state where temperature is uniform throughout the system. In exercises like the one provided, understanding the mechanisms and directions of heat transfer is crucial in predicting the behavior of the heat flux vector.
Isothermal Surfaces
An isothermal surface is characterized by a uniform temperature at every point on the surface. This means that, if you were to measure the temperature at different locations on an isothermal surface, the readings would be identical. A common example of such a surface could be the walls of a refrigerator, which strive to maintain a constant internal temperature.

In thermodynamics, isothermal surfaces are important because they help us understand where heat transfer is, and is not, occurring. As mentioned in the solution, our exercise highlights that within the bounds of an isothermal surface, there's no heat transfer — the temperature is the same throughout. This concept is essential when analyzing thermal systems and predicting how and where heat will flow.
Temperature Gradient
The temperature gradient is a way to mathematically describe the rate at which temperature changes in space. This gradient has both magnitude and direction, and it can be thought of as an arrow pointing from warm to cool regions, with its length related to how quickly temperature changes occur.

This concept is pivotal when considering heat transfer via conduction. According to Fourier's law, the rate of heat flow through a material is proportional to the temperature gradient and the cross-sectional area through which heat is flowing. A steep temperature gradient implies a high rate of heat transfer, while a shallow one indicates a slower rate. Understanding temperature gradients is key to optimizing systems for efficient thermal management.
Vector Quantity in Heat Transfer
Heat transfer is inherently a vectorial phenomenon, meaning it has both magnitude and direction. The heat flux vector is a fundamental concept that represents the direction and rate of heat transfer per unit area. Like any vector, it is depicted graphically by an arrow: the length indicates the magnitude of heat transferring through a given area, and the direction shows where the heat is moving towards.

As discussed in the exercise, the heat flux vector will always be perpendicular to isothermal surfaces since there can't be any heat flow along a surface of constant temperature. It's the presence of a temperature gradient that dictates the direction of heat transfer. This sets the stage for complex thermal analysis in engineering where the vector nature of heat flux must be accounted for in design and troubleshooting.

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Most popular questions from this chapter

Consider a steam pipe of length \(L\), inner radius \(r_{1}\), outer radius \(r_{2}\), and constant thermal conductivity \(k\). Steam flows inside the pipe at an average temperature of \(T_{i}\) with a convection heat transfer coefficient of \(h_{i}\). The outer surface of the pipe is exposed to convection to the surrounding air at a temperature of \(T_{0}\) with a heat transfer coefficient of \(h_{o^{*}}\) Assuming steady one-dimensional heat conduction through the pipe, \((a)\) express the differential equation and the boundary conditions for heat conduction through the pipe material, \((b)\) obtain a relation for the variation of temperature in the pipe material by solving the differential equation, and (c) obtain a relation for the temperature of the outer surface of the pipe.

Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the variation of temperature in the direction of heat transfer will be linear is (a) plane wall (b) cylindrical shell (c) spherical shell (d) all of them (e) none of them

A spherical container, with an inner radius \(r_{1}=1 \mathrm{~m}\) and an outer radius \(r_{2}=1.05 \mathrm{~m}\), has its inner surface subjected to a uniform heat flux of \(\dot{q}_{1}=7 \mathrm{~kW} / \mathrm{m}^{2}\). The outer surface of the container has a temperature \(T_{2}=25^{\circ} \mathrm{C}\), and the container wall thermal conductivity is \(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Show that the variation of temperature in the container wall can be expressed as \(T(r)=\left(\dot{q}_{1} r_{1}^{2} / k\right)\left(1 / r-1 / r_{2}\right)+T_{2}\) and determine the temperature of the inner surface of the container at \(r=r_{1}\).

A \(1200-W\) iron is left on the iron board with its base exposed to ambient air at \(26^{\circ} \mathrm{C}\). The base plate of the iron has a thickness of \(L=0.5 \mathrm{~cm}\), base area of \(A=150 \mathrm{~cm}^{2}\), and thermal conductivity of \(k=18 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. The outer surface of the base plate whose emissivity is \(\varepsilon=0.7\), loses heat by convection to ambient air with an average heat transfer coefficient of \(h=\) \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) as well as by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}=295 \mathrm{~K}\). Disregarding any heat loss through the upper part of the iron, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate, \((b)\) obtain a relation for the temperature of the outer surface of the plate by solving the differential equation, and (c) evaluate the outer surface temperature.

Consider a \(1.5\)-m-high and \(0.6-\mathrm{m}\)-wide plate whose thickness is \(0.15 \mathrm{~m}\). One side of the plate is maintained at a constant temperature of \(500 \mathrm{~K}\) while the other side is maintained at \(350 \mathrm{~K}\). The thermal conductivity of the plate can be assumed to vary linearly in that temperature range as \(k(T)=\) \(k_{0}(1+\beta T)\) where \(k_{0}=18 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\beta=8.7 \times 10^{-4} \mathrm{~K}^{-1}\). Disregarding the edge effects and assuming steady onedimensional heat transfer, determine the rate of heat conduction through the plate. Answer: \(22.2 \mathrm{~kW}\)

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