RADIATIVE HEAT
RADIATIVE HEAT
TRANSFER (ME 672)
TRANSFER (ME 672)
HOMEWORK 4
HOMEWORK 4
(Q 3.3 and Q 5.10)
(Q 3.3 and Q 5.10)
Name: Eldwin Djajadiwinata
Name: Eldwin Djajadiwinata
Student ID: 434107763
Student ID: 434107763
Lecturer: Prof. Dr. Abdulaziz Almujahid
Lecturer: Prof. Dr. Abdulaziz Almujahid
Question 3.3
Given:Find:
Relevant ratio of absorbtance to emittance for solar incidence of
off-normal and collector temperature 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, the black 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 the denominator.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 6th edition, Incropera et al.) we get:
At
Hence, the total-hemispherical emittance is:
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 6th edition, Incropera et al.) one can write
{
Let us find the total, directional absorptivity,
at angle of
off-normal
∫
∫
∫
∫
∫
∫
∫
∫
∫
(9)
∫
∫
∫
[
]
at sun’s temperature of 5800 K(10)
At
(11) Finally relevant ratio of absorptance to emittance is
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.
Equation for i = 1
∑
∑
(
)
Realizing that
(no external radiation);
(surfaces 1 and 2 have 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 2 have the same temperature); and
; the equation becomes
(15) Substituting the known values into Eq. (15) we obtainEquation for i = 3
∑
∑
(
)
Again, realizing that
(no external radiation) and
the equation becomes
(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, 6th edition, Incropera et al.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
SolvingEq.(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 home work).We obtain
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