4.3. CONDUCTION THROUGH SOLIDS IN SERIES

4.3A. Plane Walls in Series

In the case where there is a multilayer wall of more than one material present, as shown in Fig. 4.3-1, we proceed as follows. The temperature profiles in the three materials A, B, and C are shown. Since the heat flow q must be the same in each layer, we can write Fourier's equation for each layer as

Equation 4.3-1


Figure 4.3-1. Heat flow through a multilayer wall.


Solving each equation for ΔT,

Equation 4.3-2


Adding the equations for T1T2, T2T3, and T3T4, the internal temperatures T2 and T3 drop out and the final rearranged equation is

Equation 4.3-3


where the resistance RA = ΔxA/kAA, and so on.

Hence, the final equation is in terms of the overall temperature drop, T1T4, and the total resistance, RA + RB + Rc.

EXAMPLE 4.3-1. Heat Flow Through an Insulated Wall of a Cold Room

A cold-storage room is constructed of an inner layer of 12.7 mm of pine, a middle layer of 101.6 mm of cork board, and an outer layer of 76.2 mm of concrete. The wall surface temperature is 255.4 K inside the cold room and 297.1 K at the outside surface of the concrete. Use conductivities from Appendix A.3 for pine, 0.151; for cork board, 0.0433; and for concrete, 0.762 W/m · K. Calculate the heat loss in W for 1 m2 and the temperature at the interface between the wood and cork board.

Solution: Calling T1 = 255.4, T4 = 297.1 K, pine as material A, cork as B, and concrete as C, a tabulation of the properties and dimensions is as follows:


The resistances for each material are, from Eq. (4.3-3), for an area of 1 m2,


Substituting into Eq. (4.3-3),


Since the answer is negative, heat flows in from the outside.

To calculate the temperature T2 at the interface between the pine wood and cork,


Substituting the known values and solving,


An alternative procedure for calculating T2 is to use the fact that the temperature drop is proportional to the resistance:

Equation 4.3-4


Substituting,


Hence, T2 = 256.79 K, as calculated before.


4.3B. Multilayer Cylinders

In the process industries, heat transfer often occurs through multilayers of cylinders, as for example when heat is being transferred through the walls of an insulated pipe. Figure 4.3-2 shows a pipe with two layers of insulation around it, that is, a total of three concentric hollow cylinders. The temperature drop is T1 − T2 across material A, T2 − T3 across B, and T3 − T4 across C.

Figure 4.3-2. Radial heat flow through multiple cylinders in series.


The heat-transfer rate q will, of course, be the same for each layer, since we are at steady state. Writing an equation similar to Eq. (4.2-9) for each concentric cylinder,

Equation 4.3-5


where

Equation 4.3-6


Using the same method of combining the equations to eliminate T2 and T3 as was done for the flat walls in series, the final equations are

Equation 4.3-7


Equation 4.3-8


Hence, the overall resistance is again the sum of the individual resistances in series.

EXAMPLE 4.3-2. Heat Loss from an Insulated Pipe

A thick-walled tube of stainless steel (A) having a k = 21.63 W/m · K with dimensions of 0.0254 m ID and 0.0508 m OD is covered with a 0.0254-m-thick layer of an insulation (B), k = 0.2423 W/m · K. The inside-wall temperature of the pipe is 811 K and the outside surface of the insulation is at 310.8 K. For a 0.305-m length of pipe, calculate the heat loss and also the temperature at the interface between the metal and the insulation.

Solution: Calling T1 = 811 K, T2 the interface, and T3 = 310.8 K, the dimensions are


The areas are as follows for L = 0.305 m:


From Eq. (4.3-6), the log mean areas for the stainless steel (A) and insulation (B) are


From Eq. (4.3-7) the resistances are


Hence, the heat-transfer rate is


To calculate the temperature T2,


Solving, 811 − T2 = 5.5 K and T2 = 805.5 K. Only a small temperature drop occurs across the metal wall because of its high thermal conductivity.


4.3C. Conduction Through Materials in Parallel

Suppose that two plane solids A and B are placed side by side in parallel, and the direction of heat flow is perpendicular to the plane of the exposed surface of each solid. Then the total heat flow is the sum of the heat flow through solid A plus that through B. Writing Fourier's equation for each solid and summing,

Equation 4.3-9


where qT is total heat flow, T1 and T2 are the front and rear surface temperatures for solid A, and T3 and T4 are those for solid B.

If we assume that T1 = T3 (front temperatures the same for A and B) and T2 = T4 (equal rear temperatures),

Equation 4.3-10


An example would be an insulated wall (A) of a brick oven where steel reinforcing members (B) are in parallel and penetrate the wall. Even though the area AB of the steel would be small compared to the insulated brick area AA, the higher conductivity of the metal (which could be several hundred times larger than that of the brick) could allow a large portion of the heat lost to be conducted by the steel.

Another example is a method of increasing heat conduction to accelerate the freeze-drying of meat. Spikes of metal in the frozen meat conduct heat more rapidly into the insides of the meat.

It should be mentioned that in some cases some two-dimensional heat flow can occur if the thermal conductivities of the materials in parallel differ markedly. Then the results using Eq. (4.3-10) would be affected somewhat.

4.3D. Combined Convection and Conduction and Overall Coefficients

In many practical situations the surface temperatures (or boundary conditions at the surface) are not known, but there is a fluid on both sides of the solid surfaces. Consider the plane wall in Fig. 4.3-3a, with a hot fluid at temperature T1 on the inside surface and a cold fluid at T4 on the outside surface. The convective coefficient on the outside is ho W/m2 · K and hi on the inside. (Methods for predicting the convective h will be given later, in Section 4.4 of this chapter.)

Figure 4.3-3. Heat flow with convective boundaries: (a) plane wall, (b) cylindrical wall.


The heat-transfer rate using Eqs. (4.1-12) and (4.3-1) is given as

Equation 4.3-11


Expressing 1/hiA, ΔxA/kAA, and 1/hoA as resistances and combining the equations as before,

Equation 4.3-12


The overall heat transfer by combined conduction and convection is often expressed in terms of an overall heat-transfer coefficient U defined by

Equation 4.3-13


where ΔToverall = T1T4 and U is

Equation 4.3-14


A more important application is heat transfer from a fluid outside a cylinder, through a metal wall, to a fluid inside the tube, as often occurs in heat exchangers. In Fig. 4.3-3b, such a case is shown.

Using the same procedure as before, the overall heat-transfer rate through the cylinder is

Equation 4.3-15


where Ai represents 2πLri, the inside area of the metal tube; AA lm the log mean area of the metal tube; and A0 the outside area.

The overall heat-transfer coefficient U for the cylinder may be based on the inside area Ai or the outside area A0 of the tube. Hence,

Equation 4.3-16


Equation 4.3-17


Equation 4.3-18


EXAMPLES 4.3-3. Heat Loss by Convection and Conduction and Overall U

Saturated steam at 267°F is flowing inside a -in. steel pipe having an ID of 0.824 in. and an OD of 1.050 in. The pipe is insulated with 1.5 in. of insulation on the outside. The convective coefficient for the inside steam surface of the pipe is estimated as hi = 1000 btu/h · ft2 · °F, and the convective coefficient on the outside of the lagging is estimated as ho = 2 btu/h · ft2 · °F. The mean thermal conductivity of the metal is 45 W/m · K or 26 btu/h · ft · °F and 0.064 W/m · K or 0.037 btu/h · ft · °F for the insulation.

  1. Calculate the heat loss for 1 ft of pipe using resistances if the surrounding air is at 80°F

  2. Repeat, using the overall Ui based on the inside area Ai.

Solution: Calling ri the inside radius of the steel pipe, r1 the outside radius of the pipe, and r0 the outside radius of the lagging, then


For 1 ft of pipe, the areas are as follows:


From Eq. (4.3-6), the log mean areas for the steel (A) pipe and lagging (B) are


From Eq. (4.3-15) the various resistances are


Using an equation similar to Eq. (4.3-15),

Equation 4.3-19


For part (b), the equation relating Ui to q is Eq. (4.3-16), which can be equated to Eq. (4.3-19):

Equation 4.3-20


Solving for Ui,

Equation 4.3-21


Substituting known values,


Then to calculate q,



4.3E. Conduction with Internal Heat Generation

In certain systems heat is generated inside the conducting medium; that is, a uniformly distributed heat source is present. Examples of this are electric-resistance heaters and nuclear fuel rods. Also, if a chemical reaction is occurring uniformly in a medium, a heat of reaction is given off. In the agricultural and sanitation fields, compost heaps and trash heaps in which biological activity is occurring will have heat given off.

Other important examples are in food processing, where the heat of respiration of fresh fruits and vegetables is present. These heats of generation can be as high as 0.3 to 0.6 W/kg or 0.5 to 1 btu/h · lbm.

1. Heat generation in plane wall

In Fig. 4.3-4 a plane wall is shown with internal heat generation. Heat is conducted only in the x direction. The other walls are assumed to be insulated. The temperature Tw in K at x = L and x = −L is held constant. The volumetric rate of heat generation is W/m3 and the thermal conductivity of the medium is k W/m · K.

Figure 4.3-4. Plane wall with internal heat generation at steady state.


To derive the equation for this case of heat generation at steady state, we start with Eq. (4.1-3) but drop the accumulation term:

Equation 4.3-22


where A is the cross-sectional area of the plate. Rearranging, dividing by Δx, and letting Δx approach zero,

Equation 4.3-23


Substituting Eq. (4.1-2) for qx,

Equation 4.3-24


Integration gives the following for constant:

Equation 4.3-25


where C1 and C2 are integration constants. The boundary conditions are at x = L or −L, T = Tw, and at x = 0, T = T0 (center temperature). Then, the temperature profile is

Equation 4.3-26


The center temperature is

Equation 4.3-27


The total heat lost from the two faces at steady state is equal to the total heat generated, , in W:

Equation 4.3-28


where A is the cross-sectional area (surface area at Tw) of the plate.

2. Heat generation in cylinder

In a similar manner an equation can be derived for a cylinder of radius R with uniformly distributed heat sources and constant thermal conductivity. The heat is assumed to flow only radially, that is, the ends are neglected or insulated. The final equation for the temperature profile is

Equation 4.3-29


where r is distance from the center. The center temperature T0 is

Equation 4.3-30


EXAMPLE 4.3-4. Heat Generation in a Cylinder

An electric current of 200 A is passed through a stainless-steel wire having a radius R of 0.001268 m. The wire is L = 0.91 m long and has a resistance R of 0.126 ohms. The outer surface temperature Tw is held at 422.1 K. The average thermal conductivity is k = 22.5 W/m · K. Calculate the center temperature.

Solution: First the value of must be calculated. Since power = I2R, where I is current in amps and R is resistance in ohms,

Equation 4.3-31


Substituting known values and solving,


Substituting into Eq. (4.3-30) and solving, T0 = 441.7 K.


4.3F. Critical Thickness of Insulation for a Cylinder

In Fig. 4.3-5 a layer of insulation is installed around the outside of a cylinder whose radius r1 is fixed and with a length L. The cylinder has a high thermal conductivity and the inner temperature T1 at point r1 outside the cylinder is fixed. An example is the case where the cylinder is a metal pipe with saturated steam inside. The outer surface of the insulation at T2 is exposed to an environment at T0 where convective heat transfer occurs. It is not obvious if adding more insulation with a thermal conductivity of k will decrease the heat-transfer rate.

Figure 4.3-5. Critical radius for insulation of cylinder or pipe.


At steady state the heat-transfer rate q through the cylinder and the insulation equals the rate of convection from the surface:

Equation 4.3-32


As insulation is added, the outside area, which is A = 2πr2 L, increases, but T2 decreases. However, it is not apparent whether q increases or decreases. To determine this, an equation similar to Eq. (4.3-15) with the resistance of the insulation represented by Eq. (4.2-11) is written using the two resistances:

Equation 4.3-33


To determine the effect of the thickness of insulation on q, we take the derivative of q with respect to r2, equate this result to zero, and obtain the following for maximum heat flow:

Equation 4.3-34


Solving,

Equation 4.3-35


where (r2)cr is the value of the critical radius when the heat-transfer rate is a maximum. Hence, if the outer radius r2 is less than the critical value, adding more insulation will actually increase the heat-transfer rate q. Also, if the outer radius is greater than the critical, adding more insulation will decrease the heat-transfer rate. Using values of k and ho typically encountered, the critical radius is only a few mm. As a result, adding insulation on small electrical wires could increase the heat loss. Adding insulation to large pipes decreases the heat-transfer rate.

EXAMPLE 4.3-5. Insulating an Electrical Wire and Critical Radius

An electric wire having a diameter of 1.5 mm and covered with a plastic insulation (thickness = 2.5 mm) is exposed to air at 300 K and h0 = 20 W/m2 · K. The insulation has a k of 0.4 W/m · K. It is assumed that the wire surface temperature is constant at 400 K and is not affected by the covering.

  1. Calculate the value of the critical radius.

  2. Calculate the heat loss per m of wire length with no insulation.

  3. Repeat (b) for insulation being present.

Solution: For part (a), using Eq. (4.3-35),


For part (b), L = 1.0 m, r2 = 1.5/(2 × 1000) = 0.75 × 10–3 m, A =r2L. Substituting into Eq. (4.3-32),


For part (c) with insulation, r1 = 1.5/(2 × 1000) = 0.75 × 10–3 m, r2 = (2.5 + 1.5/2)/1000 = 3.25 × 10–3 m. Substituting into Eq. (4.3-33),


Hence, adding insulation greatly increases the heat loss.


4.3G. Contact Resistance at an Interface

In the equations derived in this section for conduction through solids in series (see Fig. 4.3-1), it has been assumed that the adjacent touching surfaces are at the same temperature, that is, that completely perfect contact is made between the surfaces. For many engineering designs in industry, this assumption is reasonably accurate. However, in cases such as in nuclear power plants, where very high heat fluxes are present, a significant drop in temperature may be present at the interface. This interface resistance, called contact resistance, occurs when the two solids do not fit tightly together and a thin layer of stagnant fluid is trapped between the two surfaces. At some points the solids touch at peaks in the surfaces and at other points the fluid occupies the open space.

This interface resistance is a complex function of the roughness of the two surfaces, the pressure applied to hold the surfaces in contact, the interface temperature, and the interface fluid. Heat transfer takes place by conduction, radiation, and convection across the trapped fluid and also by conduction through the points of contact of the solids. No completely reliable empirical correlations or theories are available to predict contact resistances for all types of materials. See references (C7, R2) for detailed discussions.

The equation for the contact resistance is often given as follows:

Equation 4.3-36


where hc is the contact-resistance coefficient in W/m2 · K, ΔT the temperature drop across the contact resistance in K, and Rc the contact resistance. The contact resistance Rc can be added to the other resistances in Eq. (4.3-3) to include this effect for solids in series. For contact between two ground-metal surfaces, hc values on the order of magnitude of about 0.2 × 104 to 1 × 104 W/m2 · K have been obtained.

An approximation of the maximum contact resistance can be obtained if the maximum gap Δx between the surfaces can be estimated. Then, assuming that the heat transfer across the gap is by conduction only through the stagnant fluid, hc is estimated as

Equation 4.3-37


If any actual convection, radiation, or point-to-point contact is present, it will reduce this assumed resistance.

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