# ME 345 Heat Transfer ```ME 345 Heat Transfer
Lecture 14
Lecture Outline:
2) Analytical solutions
3) Approximate graphical solutions
You now have T(x,y)! The

T
(k
) term is back in the equation! The steady 2-d energy equation is:
y
y

T

T
(k
)
(k
)  q  0
x
x
y
y
IF: q  0 and k = constant, then this becomes:
2T
x 2

2T
y 2
0
which is also written as
 2 T  0 (this is just notation!)
How many boundary conditions do we need to fully solve this differential equation?
EXAMPLE: a “fat” fin…
T=20C
Tb=80C
There are three basic solution techniques:
I.
Analytical: (you usually aren’t this lucky…) these you can just look up in books
II. Approximate graphical Techniques: use of this depends on the geometry of the situation,
is particularly useful for first guesses
III. Numerical: the usual way in real problems, especially as computers improve.
Analytical:
The most commonly used technique is separation of variables. In the text they show you one example…
y
T2
2T
W
T(x,y)
T1
L
T1
x 2
T1

2T
y 2
0
x
TWO things to note:
1) You have all four boundary conditions, and
2) To solve, they NONDIMENSIONALIZE and NORMALIZE the temperature: i.e. they change the
equation variables to something with NO units, AND so that all of the values of  are between 0 and
1!

T  T1
T2  T1
One of the advantages of this is that if you get a value of  that is not 0-1, then you know you have done
something wrong!
The solution for (x,y) is given in Equation 4.19: p. 189:
( x, y) 
2

ny )
( 1) n 1  1
nx sinh(
L
sin
n

W
n
L
sinh(
n 1
L)


the results are also shown graphically in Figure 4.3 on p. 189.
NOTE: this solution is only good for this particular problem, with this particular boundary conditions!
Other solutions have been done, and are given in various references, but are not useful very often. We
will go on to other more useful techniques.
Approximate Graphical Techniques:
These are useful only for the case of steady state with NO internal generation, and constant k!
They are also most useful for cases in which the boundaries are adiabatic or isothermal.
FLUX PLOT: a graph showing the direction of travel of q” and the corresponding isotherms.
Example: 1-d heat transfer in a wall
The horizontal dotted lines are ADIABATICheat DOES NOT flow over them!!!!
70C
20C
The vertical solid lines are ISOTHERMSheat flows PERPENDICULAR to them!!!!
It is very important to note: isotherms are normal (perpendicular) to heat flow lines and adiabats!!!
BASIC TECHNIQUE FOR CONSTRUCTING A FLUX PLOT:
1) Draw the picture and identify if the heat transfer is 1-d or 2-d
2) Identify any physical symmetry: lines of symmetry are adiabatic heat flow lines!
3) Note that isothermal boundaries are normal to the flow of q”
4) Try to draw lines in each direction so that the area consists of “squares” (or at least as close as you
can get.)
If you do this correctly, the same amount of heat will be flowing in each “lane”. Then all you have to do
M
q
q
i
 Mq i
where M = number of heat flow lanes
Eq. 1
i 1
Tj
q i  kA i
and
x is the length of one square,  is the depth “into”: the paper, and Tj is the T across
one square.
x
 k( y  )
Tj
where
x
Also note that T should be equal for each set of squares!!!
N
This means that the total T 
 T
j
 NT j
Eq. 2
j1
Where N = the number of temperature increments
Combine eqs. 1 and 2 to get the total q, using the fact that since you have “squares”, x = y
and that Tj = T/N
then
q
Mk (  y  ) T
M
q
kT
 x
N
N
CONDUCTION SHAPE FACTOR:
This means that:
q = S k T
where S = M  /N
S is known as the “shape factor”
People have already done the work on this for common geometries.
To use the tables in the book, you must be able to assume:
2) 2-d
3) q  0 , and
4) k = constant.
Let’s try some examples (and be sure to go over Example 4.1, pp. 194-196.)
Group exercises:
1) Find q for the following shape: let  = 1.
T=0