Майнор / Книги / Chapter-2---One-dimensional--steady-state-heat-co_2021_Heat-Transfer-Enginee

2.5.1  Critical radius of insulation for cylinder

Consider a cylindrical pipe of radius r and length L in the direction perpendicular to the plane of the paper. It is necessary to add insulation with thermal conductivity of k to this pipe to reduce heat transfer. In the face of it, the proposal looks like a perfect engineering solution for the problem at hand. However, as r2 increases conduction becomes “more difficult” but because the area for convection given by 2πr2L increases, the increase in conduction resistance will be offset by a decrease in convection resistance. Hence, there is an interplay of these two competing phenomena, which also suggests that there would be a particular r2 at which the total resistance is an extremum (Fig. 2.7).

The heat transfer through the cylinder is given by

we need to differentiate the denominator (Rtotal) in Eq. (2.57) with respect to r2 and set it to zero to determine the value of r2 at which Rtotal becomes stationary.

In Eq. (2.60), rc is known as the critical radius of insulation. Second order conditions are required to check if Rtotal is maximum or minimum at r2 = rc

FIGURE 2.7

Schematic representation of conduction in a cylinder with insulation, and convection on the outside.

26 CHAPTER 2  One-dimensional, steady state heat conduction

FIGURE 2.8

Variation of thermal resistance and heat transfer with thickness of insulation.

Alternatively we can plot Q and Rtotal against r2 and examine what happens. A qualitative variation of thermal resistance and heat transfer with the thickness of

insulation is shown in Fig. 2.8.

From r1 to rc, the total resistance keeps decreasing. This is due to the dominance of the decreasing convection resistance compared to conduction resistance in

Eq. (2.57). Beyond rc1, Rtotal increases due to the dominance of the increased conduction resistance over the decreasing convection resistance in Eq. (2.57). The result is

counterintuitive in the sense that the total resistance is a minimum at r2 = rc, and so to have an insulating effect, r2 must be much greater than rc.

2.6  Steady state conduction in a spherical shell

A spherical shell or simply a sphere is an essential geometry in heat transfer engineering that is used in a variety of applications, such as nuclear reactor waste disposal, ball bearings (solid sphere), and rocket nozzles (nose cone can have a hemispherical shape) to name a few.

Consider a spherical shell of which a cross-sectional view is shown in Fig. 2.9, along with the geometrical details and the two boundary conditions.

FIGURE 2.9

Schematic representation of a cross-sectional view of a sphere, for studying one dimensional steady conduction.

28 CHAPTER 2  One-dimensional, steady state heat conduction

The heat transfer through the sphere can be calculated by using Fourier’s law as follows (Please recognize that A = f (r) and so we need to be cautious in evaluating Q)

Invoking the electrical analogy, we can obtain an expression for the conduction resistance across the shell, Rcond,sphere

It is instructive to mention that Q (At r = r1) = Q (At r = r2).

2.7  Steady state conduction in a composite wall, cylinder and sphere

2.7.1  Composite wall

A simple composite plane wall made of three materials with thermal conductivities ka, kb, with convection boundary conditions at the end wall kc and three thicknesses La, Lb, and Lc respectively is shown in Fig. 2.10, is undergoing one dimensional steady state conduction. A resistance circuit can be drawn, as shown in Fig. 2.11.

Under steady state, with no heat generation we have the following:

This may be alternatively written as (Refer to Fig. 2.7)

Using the Componendo-Divedendo rule, i.e., adding all the numerators and denominators respectively and equating the result to Q, we have

FIGURE 2.10

Schematic representation of a composite wall undergoing steady state conduction.

FIGURE 2.11

Schematic representation of a resistance network for the composite wall under consideration, under heat conduction.

In Eq. (2.76b), Rtotal can be calculated as

Eq. (2.77) is for a series connection and is much more powerful than what it appears to be. Without solving for conduction equation in three materials with two Robin conditions, we straight away solved the problem. Whenever two or more materials are kept side by side along the direction of heat transfer, their resistances are considered to be in series as shown in Fig. 2.11. The individual resistances in the composite wall are given by

FIGURE 2.13

Schematic representation of an electrical resistances network for solving a heat conduction problem with resistances in parallel.

the other, their thermal resistances are considered to be in parallel and are shown in Fig. 2.13. Let the boundary temperatures be T1 and T2 with T1 > T2, as before.

The total heat transfer through the slab can be calculated as

In Eq. (2.87), Rab is calculated as

2.7.1.2  Series-parallel connection

A series-parallel connection is shown in Fig. 2.14, and the corresponding resistance network is shown in Fig. 2.15. Let the temperatures at x = 0 and x = (L1 + L2 + L3) be T1 and T4 respectively with T1 > T4.

FIGURE 2.14

Schematic representation of a series parallel combination of thermal resistances.

FIGURE 2.16

Schematic representation of heat conduction in a chip that is under consideration in

Example 2.1. Please note that this figure is not to scale.

Thus the maximum temperature in the assembly is 80 °C at the interface between chip and joint.

2. When considering a finite thermal contact resistance