heat transfeer conduction two three dimensions

8/12/2019 Heat Transfeer Conduction Two Three Dimensions

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CONDUCCION EN DOS Y TRES DIMENSIONES


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CONDICIONES DE CONTORNO


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BOUNDARY CONDICIONS CONSTANTS

Thus, the quantities Cn, sinh (nX/L) must be the coefficients of the Fourier

sine series for f(x) in the interval 0


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Principle of Superposition

The linearity of the Laplace equation requires that a linear combination of solutions is

itself a solution; this is the principle of superposition. A problem having more than

one non homogeneous boundary condition can be resolved into a set of simpler

problems each with the physical geometry of the original problem and each having

only one non homogeneous boundary condition. The solutions to the simpler

problems can be superposed (at the geometric point being considered) to yield the

solution to the original problem.

EXAMPLE


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NON HOMOGENOUS BOUDARY CONDITIONS

Determine the temperature at the (1/4, 1/4)


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CONDUCTIVE SHAPE FACTOR

The figure represents a heated pipe with a thick layer of insulation. The inner surface of

the insulation is at uniform temperature T1, the outer surface is at uniform temperature To,

and there is a resulting outward heat flux for Ti > To. Constructing uniformly spaced lines

perpendicular to the isotherms results in a group of heat flow lanes. In Fig. there are four

such lanes in the quadrant selected for study, and the other three quadrants would be similar

due to the problem symmetry. If we can determine the rate of heat transfer for a single

lane, then we can easily find the total. Note that there is no heat transfer across one of the

radial lines, such as line a-6, because there is no angular temperature gradient.


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Techniques for freehand plotting include:

1. Identify all known isotherms.

2. Apply symmetry (geometrical and thermal) to reduce the art work.

3. Begin, if possible, in a region where the adiabatics can be uniformly spaced.

4. Begin with a crude network sketch to find the approximate locations of isotherms

and adiabatics.

5. Continuously modify the network by maintaining adiabatic lines normal to

isothermal lines while forming curvilinear squares.

Freehand Plotting


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Electrical Analog

Electrical Analog

NUMERICAL ANALYSIS

Consider a general two-dimensional body as shown in Fig. 3-6. The body

has uniform thickness L in the z-direction and no temperature gradient in

that direction. Choosing an appropriate Ax and Ay, the body is dividedinto a network of rectangles, each containing a single nodal point at

its center. It is convenient to consider the heat transfer as occurring

between nodal points only, these being connected by fictitious rods

acting as conductors resistors for the heat flow. Thermal energy is

considered to be stored at the nodal points only. The horizontal and

vertical conductance are given by


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C t S l ti


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Computer Solution


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R l ti T h i


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Determine the steady state temperatures

at the four interior nodal points of Fig. 3-9.

The nodal equations, obtained with the aid

of (3.14),are

The nodal equations, obtained

Relaxation Technique


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which comprise a set of four linear algebraic equations containing the four

unknown nodal temperatures. The

relaxat ion method of solu t ion proc eeds as fo l low s:

1. Assume (guess) values for the four unknown temperatures. Goodinitial guesses help to minimize the ensuing work.

2. Since the initial guesses will usually be in error, the right side of each

nodal equation will differ from zero; a residual wil l exist due to

inaccu racies in the assumed values. Cons equent ly, we replace the

zeros in equations (I) thr ough (4)w ith RI,R2,R3,and R4,respectively :

3. Set up a unit change table such as Table 3-2, which shows the effectof a one-degree change of temperature at one node upon the residuals.

The fact that a block (overall) unit change has the same effect upon all

residuals is unusual, this being due to the overall problem symmetry.


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4. Calculate the initial residuals for the initially assumed temperatures using the

residual equations (5) through (8).

5. Set up a relaxat ion table such as Table 3-3. Begin with the in i t ia lly

assum ed temperatures and the result ing initial residuals. The left-hand column

records the changes from the initially assumed temperature values.

Notice that the procedure begins by relaxing the largest initial residual (or

perhaps by making a block change, a technique useful when all residuals are of

the same sign).


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the present problem, we should begin by reducing R2 or R3. Arbitrari ly choose R2

and proceed by over-relaxing slightly. At this point the convenience afforded by Table

3-2 becomes evident; this facilitates rapid calculation of the changes in the residuals

without recourse to the equations. Notice that the +20change in T, reduced the

residuals at nodes 1 and 2 but unfortunately increased R3.

The first row in Table 3-3 shows the new residuals and temperatures; the only

temperature changed is underlined. Proceeding, we next relax the largest resulting

residual, this being R3.Following a temperature change of +25 at node 3, we seethatR4=0. This does not necessari ly mean that we have obtained the correcttemperature at node 4,but rather that the set of as yet incorrect temperature

values happens to satisfy eq. (4) exactly. Proceeding, the largest residual is now

R2,which is reduced to 0 by a +5 change in T,. This also reduces all remaining

residuals to zero. A check is made by substituting the temperatures thus obtained

into eqs. (I) through (4);thi s ver i f ies the solution.

GaussSeidel Method


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GaussSeidel Method

To illustrate this very popular method of solution of a set of nodal equations,

return to the figure, and associated temperature nodal equations (1), (2),

(3),and (4).Namin g the s et of equations (3.26a) we have

Next, we rearrange these equations to have the unknown temperature at

each node to appear on one side of the equation alone and with the

coefficient +l. This is done by solving the first equation in (3.16a)for T,, the

second for T2,and so forth, yielding (3.16b).


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Proceeding to solve (3.166)with the Gauss-Seidel method, an initial set of

temperatures T1 throughT4 is assumed. Obviously, reasonablechoices h ere wil l faci l i tate the solu t ion, and the temperatures must

be bounded by 200 C and 500 C. Choosing Tl = 450 C, T2= 350, T3=

250, and T, = 350, all in "C,is a reasonable gu ess. We next in sert

these T2and T4values in the r ight s ide of the f irst equat ion

of (3.16b) and calculate a new estimate of Tl, viz.

Proceeding to calculate new estimates of T2, T3,and T4, always us ing

the newest value avai lable for each temperature, yields (3.I6d).


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At this point we have completed one iteration. To proceed, we next use

the newest temperature values from set (3.16d)together with the

solution set (3.I6b) to calculate the next iteration set, always using

the newest T value available. Thus, to calculate a new T,, defined in

set (3.I6b), we use the T2and T4 values from (3.26d),obtaining

Again, proceeding to calculate new estimates for T2, T,, and T4,

a lways using the newest value available for each temperature,

yields (3.16e).


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The next (third) iteration through the set of equations yields

and the next (fourth) iteration yields

All temperatures are in "C. At this point we question the value of additional

iterations. Usually, temperature values within 5 a few per cent accuracy are

suitable for engineering calculations. If we use m as a superscr ip t todenote the previous iteration and nz + 1 to identify the iteration just

completed, a computer program could use a test such as


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heat transfeer conduction two three dimensions

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