Two-Dimensional Conduction:Flux Plots

and Shape Factors

Chapter 4Sections 4.1 and 4.3

General Considerations

General Considerations

xq

• Two-dimensional conduction:– Temperature distribution is characterized by two spatial coordinates, e.g., T (x,y).– Heat flux vector is characterized by two directional components, e.g., and .yq

• Heat transfer in a long, prismatic solid with two isothermal surfaces and two insulated surfaces:

Note the shapes of lines of constant temperature (isotherms) and heat flow lines (adiabats).

What is the relationship between isotherms and heat flow lines?

Solution Methods

The Heat Equation and Methods of Solution• Assuming steady-state, two-dimensional conduction in a rectangular domain with constant thermal conductivity and heat generation, the heat equation is:

2 2

2 2

,0

q x yT T

x y k

• Solution Methods:

– Exact/Analytical: Separation of Variables (Section 4.2) Limited to simple geometries and boundary conditions.

– Approximate/Graphical : Flux Plotting 0q

Of limited value for quantitative considerations but a quick aid to establishing physical insights.

– Approximate/Numerical: Finite-Difference, Finite Element or Boundary Element Method.

Most useful approach and adaptable to any level of complexity.

Flux Plots

Flux Plots

• Utility: Requires delineation of isotherms and heat flow lines. Provides a quick means of estimating the rate of heat flow.

• Procedure: Systematic construction of nearly perpendicular isotherms and heat flow lines to achieve a network of curvilinear squares.

• Rules:

– Sketch approximately uniformly spaced isotherms on the schematic, choosing a small to moderate number in accordance with the desired fineness of the network and rendering them approximately perpendicular to all adiabats at points of intersection.

– Draw heat flow lines in accordance with requirements for a network of curvilinear squares.

– On a schematic of the two-dimensional conduction domain, identify all lines of symmetry, which are equivalent to adiabats and hence heat flow lines.

Flux Plots (cont.)

Example: Square channel with isothermal inner and outer surfaces.

– Note simplification achieved by identifying lines of symmetry.– Requirements for curvilinear squares:

Intersection of isotherms and heat flow lines at right angles

Approximate equivalence of sums of opposite sides

– Determination of heat rate:

2 2

ab cd ac bdx y

(4.20)

iq Mq jTM k y

x

1 2

Mk T

N

1 2

Mq k T

N (4.24)

Shape Factor

The Conduction Shape Factor• Two-dimensional heat transfer in a medium bounded by two isothermal surfaces at T1 and T2 may be represented in terms of a conduction shape factor S.

1 2q Sk T T (4.25)

• For a flux plot,M

SN

(4.26)

• Exact and approximate results for common two-dimensional systems are provided in Table 4.1. For example,

Case 6. Long (L>>w) circularcylinder centered in squaresolid of equal length

• Two-dimensional conduction resistance:

2

1n 1.08 /

LS

w D

1

2cond DR Sk

(4.27)

Problem: Flux Plot

Problem 4.6: Heat transfer from a hot pipe embedded eccentrically in a solid rod.

Schematic

ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Length >> diametrical dimensions, (4) Constant thermal conductivity.

Flux Plot

ANALYSIS: For the symmetrical section and four temperature increments (N = 4), the flux plot is:

For the pipe the heat rate per unit length is

1 2q W

q kS T T 0.5 4.26 150 35 C 245 W/m.m K

COMMENTS: Because the curvilinear squares are irregular in the lower, right-hand quadrant of the flux plot a finer network would be needed to obtain a more accurate estimate of the shape factor. Determine the error associated with the flux plot by using a result from Table 4.1 to

compute the actual value of the shape factor.

Problem: Shape Factor

Problem 4.27: Attachment of a long aluminum pin fin (D=5mm) to abase material of aluminum or stainless steel. Determinethe fin heat rate and the junction temperature (a) withoutand (b) with a junction resistance.

The heat flow lines shown in the figure presume a fin effectiveness of 1f . How would the lines look

for 1f ?

Schematic:

Problem: Shape Factor (cont)

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Large base material, (4) Infinite fin.

PROPERTIES: Aluminum alloy, k = 240 W/mK, Stainless steel, k = 15 W/mK.

ANALYSIS: (a,b) From the thermal circuit with the junction resistance, the heat rate and junction temperature are

,

b bf

tot b t j f

T T T Tq

R R R R

(1)

,j f f t jT T q R R (2)

From Case 10 of Table 4.1, S=2D. Hence, from Eq. (4.27)

11 1 2 2 0.005b b b bR Sk Dk m k

c

25 2, ,

2With A = , the junction resistance is43 10 0.005 4 1.528t j t j c

D

R R A m K W m K W

1/ 2

1/ 21/ 2 32 2

With for an infinite fin (Table 3.4) and ,

50 0.005 240 4 16.4

f c b

f c

q hPkA P D

R hPkA W m K m W m K K W

Problem: Shape Factor (cont.)

Without t, jR With t, jR

Base Rb (K/W) qf (W) Tj (C) qf (W) Tj (C) Al alloy 0.417 4.46 98.2 4.09 92.1 St. steel 6.667 3.26 78.4 3.05 75.1

COMMENTS: (1) Why is the effect of the base material on the heat rate and the junction temperature substantial for the stainless steel and not for the aluminum?

(2) Why is the relative effect of the contact resistance on the heat rate and the junction temperature more pronounced for the aluminum alloy base than for the stainless steel?