Chapter 3 Specifically Heat Transfer Rates For Cases And

PROBLEM 3.136 (Cont.)

(ii) For He, the ideal gas constant, specific heat at constant volume, and ratio of specific heats are:

The mean free path is

The plate separation, L, is

COMMENTS: The critical plate separation associated with helium is 7.1 × 10-3 m/ 24.7 × 10-6 m =

290 times greater than that for air. The thermal resistance associated with molecule–surface

PROBLEM 3.137

KNOWN: Thickness of of parallel aluminum plates and air layers. Wall surface temperatures.

FIND: The conduction heat flux through (a) aluminum wall, (b) air layer, (c) air layer contained

between two aluminum sheets and (d) composite wall consisting of 8 aluminum sheets and 7 air

layers.

SCHEMATIC:

ASSUMPTIONS: (1) Ideal gas behavior. (2) Nanoscale effects within the solid are not important.

PROPERTIES: Table A.4 (T = 300 K): Air; cp = 1007 J/kg⋅K, kAir = 0.0263 W/m∙K. Figure 2.8: Air;

M = 28.97 kg/kmol, d = 0.372 × 10-9 m. Table A.1 (T = 300 K): Pure Aluminum, kAl = 237 W/m∙K.

ANALYSIS:

(a) Case A: Aluminum Wall

PROBLEM 3.137 (Cont.)

(b) Case B: Air Layer

(c) Case C: Air Layer between two Aluminum Sheets

This case involves a resistance due to molecule–molecule interactions, as well as molecule–surface

collisions. For air, the ideal gas constant, specific heat at constant volume, and ratio of specific heats

From Equation 2.11 the mean free path of air is

In addition, the aluminum sheets pose a cumulative thermal resistance of

Hence, the conduction heat flux is

Continued…

PROBLEM 3.137 (Cont.)

The aluminum sheets pose a cumulative thermal resistance of

Hence, the conduction heat flux is

(d) Case D: Seven Air Layers between Eight Aluminum Sheets

In addition, the aluminum sheets pose a cumulative thermal resistance of

Problem 3.137 (Cont.)

Hence, the conduction heat flux is

In addition, the aluminum sheets pose a cumulative thermal resistance of

Hence, the conduction heat flux is

The predicted heat fluxes are summarized below.

Case Ltot = 600

µ

m Ltot = 600 nm

Continued…

PROBLEM 3.137 (Cont.)

COMMENTS: (1) For the Ltot = 600

µ

m cases, it is readily evident that the highest heat flux

corresponds to Case A in which conduction occurs exclusively through the high thermal conductivity

aluminum. The lowest heat flux is associated with conduction through the pure air layer (Case B). For

(2) For the Ltot = 600 nm cases, we again observe that the largest heat flux is associated with

conduction exclusively within the aluminum (Case A). However, consideration of the other three

cases reveals nanoscale behavior that would be unexpected from the macroscale point–of–view.

(3) Nanoscale effects could become important in the solid as the thickness of the solid approaches the

mean free path. See Table 2.1.

PROBLEM 3.138

KNOWN: Knudsen number, specific heat ratio and thermal accommodation coefficient for an ideal

gas and solid surface.

FIND: Expression for the the ratio of the thermal resistance due to molecule-surface collisions to the

thermal resistance associated with molecule-molecule collisions, Rt,m-s/Rt,m-m .

ASSUMPTIONS: (1) Ideal gas behavior.

ANALYSIS: The expressions for Rt,m-m and Rt,m-s are

Associating the critical Knudsen number, Kncrit, with Rt,m-s/Rt,m-m = 0.01, we may plot the value of the

critical Knudsen number for

γ

= 1.4 and 1.67 over the range 0.01 ≤

a

t ≤ 1 as shown below.

COMMENTS: (1) Relatively large Knudsen numbers are associated with more significant surface–

molecule collisions. (2) The critical Knudsen number is relatively insensitive to the specific heat ratio,

γ

.

PROBLEM 3.139

KNOWN: Thickness of alternating tungsten and aluminum oxide layers, interface thermal resistance,

thermal conductivities of tungsten and aluminum oxide thin films.

FIND: (a) Effective thermal conductivity of the nanolaminate. Comparison with bulk thermal

conductivities of aluminum oxide and tungsten, (b) Effective thermal conductivity of the nanolaminate

using bulk values of the thermal conductivity of aluminum oxide and tungsten.

SCHEMATIC:

Aluminum oxide

ASSUMPTIONS: (1) Steady–state, one–dimensional conditions, (2) Constant properties.

PROPERTIES: Table A.1, tungsten (300 K): kT = 174 W/m⋅K. Table A.2, aluminum oxide (300 K):

kA = 36 W/m⋅K.

ANALYSIS: (a) Consider a unit cell consisting of one layer of aluminum oxide, one layer of

tungsten, and two interfaces of unit cell thickness 2

d

= 1.0 nm as shown in the schematic. The sum of

the thermal resistances is

The effective thermal conductivity is

COMMENTS: (1) The effective thermal conductivity is dominated by the interface resistances and

is relatively insensitive to the thermal conductivity of the two materials. Although the interface

resistance is very small compared to typical contact resistance values (see Table 3.2), by using

extremely thin layer thicknesses, many such interfaces may be packed into the laminated structure,

resulting in very small values of the effective or bulk thermal conductivity. The material service

PROBLEM 3.140

KNOWN: Dimensions of and temperature difference applied across thin gold film.

FIND: (a) Energy conducted along the film, (b) Plot the thermal conductivity along and across

the thin dimension of the film, for film thicknesses 30 ≤ L ≤ 140 nm.

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x– and y-directions, (2) Steady-state

conditions, (3) Constant properties, (4) Thermal conductivity not affected by nanoscale effects

associated with 250 nm dimension.

PROPERTIES: Table A.1, gold (bulk, 300 K): k = 317 W/m⋅K.

ANALYSIS:

a) From Fourier’s law,

(b) The spanwise thermal conductivity may be found from Eq. 2.9a,

x

PROBLEM 3.140 (Cont.)

The plot is shown below.

COMMENT: Nanoscale effects become less significant as the thickness of the film is increased.

PROBLEM 3.141

KNOWN: Dimensions of an aluminum tube. Thickness and number of alternating aluminum and

aluminum oxide layers used as insulating coating. Aluminum mean free path.

FIND: Thermal resistance for unit tube length for wall of aluminum tube and nanocomposite coating.

Comment on effectiveness of coating as thermal insulator.

SCHEMATIC:

ASSUMPTIONS: (1) Steady–state, one–dimensional conditions, (2) Constant properties, (3) The

coating can be treated as a plane wall since it is thin in relation to the tube diameter.

PROPERTIES: Table 3.6, Aluminum/aluminum oxide:

,

′′

tb

R

= 95 × 10–10 m2⋅K/W. Table A.1, pure

aluminum (300 K): kA,bulk = 237 W/m⋅K; Table A.2, sapphire (300 K): kS,bulk = 46 W/m⋅K. Given,

l

mfp,A = 35 nm.

ANALYSIS: The thermal resistance of the tube wall (for a unit length) is:

Referring to Table 2.1, for aluminum oxide the layer thickness of 60 nm is greater than the critical

value Lcrit,x, therefore no correction is needed to its bulk thermal conductivity. For aluminum, however,

considering the discussion beneath Equations 2.9, with

d

/

l

mfp,A = 60/35 = 1.7 < 7, the correction is

Continued …

PROBLEM 3.141 (Cont.)

COMMENTS: The thermal resistance of the nanocomposite is dominated by the interface

resistances and is relatively insensitive to the thermal conductivity of the two materials. Although the

interface resistance is very small compared to typical contact resistance values (see Table 3.2), by

using extremely thin layer thicknesses, many such interfaces may be packed into the nanocomposite

structure, resulting in a large total thermal resistance.