Radiative heat transfer problems

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RADIATIVE HEAT

TRANSFER (ME 672)

HOMEWORK 4

(Q 3.3 and Q 5.10)

Name: Eldwin Djajadiwinata

Student ID: 434107763

Lecturer: Prof. Dr. Abdulaziz Almujahid


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Question 3.3

Given:

Find:

Relevant ratio of absorbtance to emittance for solar incidence of off-normal and collectortemperature of 400 K.

Solution:

We will first find the total-hemispherical emittance as follows.

(1)

Based on the above equation, Eq. (1), we need to find in order to obtain the .

(2)

Since is not a function of , it can be taken out from the integration. Furthermore, theblack body intensity is diffuse (independent of direction). Thus, it can be completely taken

out of the integration and cancel out between the at the numerator and that at thedenominator.

(3)


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For

(4)

(5)

For

(6)Substitute Eq. (5) and (6) into Eq. (1)

[ ]

(7)

From the blackbody radiation function table (Table 12.1, Fundamentals of Heat and Mass

Transfer 6thedition, Incropera et al.) we get:

At Hence, the total-hemispherical emittance is:

(8)


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Now, let us find the absorptance. Since the sun irradiates the surface from only one direction,

but over the entire spectrum, we need to find the total, directional absorptance at angle of

off-normal. Accepting the fact that under any conditions

(p.763-764, Fundamentals of Heat and Mass Transfer 6thedition, Incropera et al.)

one can write

{

Let us find the total, directional absorptivity, at angle of off-normal

(9)

[ ]at suns temperatureof 5800 K

(10)

At Substitute into Eq. (10)


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(11)

Finally relevant ratio of absorptance to emittance is

(12)


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Question 5.10

Given:

Both pipe isothermal at Both pipes are diffusely emitting and reflecting with The duct wall is isothermal at The duct wall is diffusely emitting and reflecting with

Find:

Radiative heat loss from the pipes

Assumption:

The pipes and duct wall surfaces are gray surfaces and, thus, Solution:

We let both pipes denoted as surface 1 and 2 and the duct wall as surface 3. The flux

equations for net radiative exchange within a gray, diffuse enclosure is as follows.


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Equation for i = 1

( )

Realizing that (no external radiation); (surfaces 1 and 2have the same temperature); and ; the equation becomes

(13)

Substituting the given values into Eq.(13) we obtain

(14)

Equation for i = 2

( )

Realizing that (no external radiation); (surfaces 1 and 2have the same temperature); and ; the equation becomes

(15)Substituting the known values into Eq.(15) we obtain

(16)


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Equation for i = 3

( )

Again, realizing that (no external radiation) and the equationbecomes

(17)

Substituting the known values into Eq.(17) we obtain

(18)

Since , , and are given, it remains to find the view factors to solve for , , and from Eq.(14), (16),and(18).

Finding view factors

symmetric geometrical configuration. is found using equations given in Table 13.1 (two parallel cylinders configuration), p.816, Fundamentals of Heat and Mass Transfer, 6thedition, Incropera et al.


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From summation rule

where is the length of the pipes and duct wall.Due to symmetric configuration, one can conclude that

From summation rule again we get

Solving Eq.(14), (16),and(18)in conjunction with the obtained view factors above, we will get ,, and . We will solve these set of equations using EES software (attached at the end of this homework).

We obtain

The heat loss rate is


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To validate this result we will check the heat absorbed by the duct which must match the heat

loss from the pipe.

It is proven that . The very small difference between them (is dueto round off error. This means the result is valid.

Radiative heat transfer problems

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