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EX2: Cylindrical Shell

4: Cylindrical Geometry (Example 2)

A hollow cylinder has circular inner and outer surfaces. The cross-section, shown in Figure 3, is a ring of inner radius and outer radius . We consider two cases of symmetric, steady state boundary conditions in which the temperature distribution depends on one space variable. As in , the conductivity is assumed to be constant.



Figure 2:





2a. FLOW ALONG THE CYLINDER

Suppose that the cylindrical tube has finite length , with plane end faces perpendicular to the axis of the tube, and that the lateral (curved) surfaces are insulated. Let each face be maintained at a constant temperature, one at and the other at . We define the x-axis to coincide with the tube's axis, with the ends of the tube lying in the planes and , such that at and at . (See Figure 4.)

Since the lateral surfaces are insulated there is no heat flow through these curved walls; in other words, the heat flux vector has a zero component normal to the walls. In fact, it can be shown, as in Example 1, that is always in the x-direction. Then an argument similar to that in Example 1 yields in this case the same expressions for the rate of heat flow and temperature distribution, given by equations (3.4) and (3.5).



Figure 3:





2b. RADIAL FLOW

Now suppose that the cylindrical tube, of inner radius and outer radius , has infinite length, and that the inner cylindrical surface is maintained at constant temperature , the outer at constant temperature . We wish to determine the temperature distribution and the direction and magnitude of the heat flow within the material.

Since has constant values on each of the cylindrical surfaces and , the temperature varies in the radial direction, depending only on , and heat flows in the radial direction.

The most convenient coordinates for this geometry are cylindrical coordinates, consisting of polar coordinates in a cross-sectional plane and a third (Cartesian) coordinate along the axis of the cylinder. In this problem we do not need the complete form of the gradient in cylindrical coordinates. Here the temperature and heat flow depend only on , in which case reduces to a single term:



Equation 17

The vector denotes a unit vector in the r (radial) direction.

Let us restrict our attention to a tube of finite (unit) length. In this tube consider a cylindrical surface of radius , which we call . On (that is, for fixed ) is constant and the heat flux vector (normal to the cylinder) has a constant magnitude . Since the area of is (the tube is of unit length), the rate of heat flow through (in watts) is . Heat flows outward if has positive sign.

Because this is a steady state process, the rate of heat flow through is the same for all , . In other words, (a constant which we determine below), or



Equation 18

Note that the heat flux varies inversely with ; in other words, is constant.

Using equations (4.1) and (4.2) to substitute for and in the heat conduction law, , we obtain the simple differential equation



Equation 19

A differential equation is an equation containing a derivative of an unknown function with respect to an independent variable. Equation (3.3), , is also a differential equation, even simpler than (4.3), for the right hand side of (3.3) is a constant.

 
 
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