Last Rev.: 23 JLY 08
Extended Fin / Mathcad Tutorial 2 : MIME 3470
Page 1
Grading Sheet
~~~~~~~~~~~~~~
MIME 3470—Thermal Science Laboratory
~~~~~~~~~~~~~~
Laboratory 2
EXTENDED FIN / MATHCAD TUTORIAL 2
Student’s Name / Section №
№ OF DEVIATIONS
(Minus 2 points per deviation)
THE DELIVERABLE OF THIS LAB IS AN EXACT DUPLICATION
OF PAGES 6 THROUGH 10. TWO (2) POINTS WILL BE
DEDUCTED FOR EACH DEVIATION FROM AN EXACT
DUPLICATION. A *.DOC FILE WITH CORRECT MARGINS,
HEADER, ETC. CAN BE DOWNLOADED FROM THE COURSE
WEB SITE WHERE THE *.PDF FILE WITH THE FULL REPORT
IS ALSO AVAILABLE.
COMMENTS
d
GRADER—
TOTAL
(100 PTS Max)
Last Rev.: 23 JLY 08
Extended Fin / Mathcad Tutorial 2 : MIME 3470
MIME 3470—Thermal Science Laboratory
~~~~~~~~~~~~~~
Laboratory №. 2
Experimental, Analytical, &
Numerical Solutions for an
EXTENDED FIN &
MATHCAD TUTORIAL 2
~~~~~~~~~~~~~~
PREPARED BY: STUDENT’S NAME SECTION: №
~~~~~~~~~~~~~~
Originally, this experiment was performed around the 10th week.
The lab write up took at least twice as much time as a normal write
up and there was much confusion as to what was required of the
students. It was too aggressive a lab for a mere two-hour lab
course. Thus, the lab was turned basically into a Mathcad tutorial
for the 2nd week and then actual data will be taken in the 10th week.
OBJECTIVE—Here the intention is to
1. Study the basic principles of conduction heat transfer in association with free convection cooling,
2. Demonstrate the utility of spike fins for heat transfer enhancement,
3. Illustrate the use of finite difference solvers in heat conduction, and
4. Further the student’s introduction to Mathcad.
THEORY—The laws that describe heat transfer in a fin are
Fourier’s Law of Heat Conduction and Newton’s Law of Cooling.
Both are based on observation of physical phenomena and are constitutive relations. Fourier’s Law is based upon the fact that the conductive heat, Qx, is directly proportional to the negative of the
temperature gradient. For one-dimensional heat conduction this is
T
Qx  
,
(1)
x
where
Qx  heat flowing in the positive x direction,
T  absolute temperature at x, K.
The negative sign is used because heat flows from higher to lower
temperatures. For heat flow in the +x direction, the term dT/dx is
negative. So for the inputted energy to be a positive value, the negative
sign is used. Besides the temperature gradient, it should be apparent
that the amount of heat let through is directly proportional to the cross
sectional area normal to the direction of heat flow and will differ from
one type of material to another. Introducing the thermal conductivity
as the constant of proportionality representing the material, Fourier’s
law is then expressed as
T
Qx  kAXS
,
(2)
x
where AXS  cross sectional area of fin, m2,
k  thermal conductivity of the fin material, W/m∙K.
Heat passes out of an object and into a fin by conduction; but, the
passage of that heat to the ambient air (or any other fluid) is defined by
Newton’s law of cooling. This is not a law (but a constitutive relation
defining the heat transfer coefficient, h) and is not about cooling, per
se; however, the name has held. This relation is expressed by
(3)
Qs  hAs Ts  Tamb  ,
where
s  subscript indicating the surface as opposed to the
cross section of the fin,
As  surface area of fin having a perimeter of length P, m2,
h  heat transfer coefficient, W/m2∙K,
Tamb  ambient temperature of surrounding fluid, K.
Analytical Solution—Consider the rod (spine) attached to a surface
maintained at a constant temperature shown in Figure 1.
Page 2
L
T  Tamb
T  Tb
Cross Sectional
dx
x
Area, AXS
Figure 1—Fin attached to a surface of constant base temperature, Tb
For the general, differential
element shown in Figure 2,
the steady-state energy
balance can be written
where all the heat in from
the left-hand side of the
element equals that passing
out to the right-hand side
plus the heat passed to the
fluid via Newton cooling. In
equation form this is

  kAXS dT

dx




x
T  Tamb
Heat conducted
out by convection
Heat
conducted
out by
conduction
at x + x
Heat
conducted
in by
conduction
at x
x
x
c.v.
Figure 2—Heat flow through
a small element

  kAXS dT

dx


  h As T  Tamb 


x  x 
Px
 dT
dT 


 dx x  x dx 
x   hP T  T

amb 

 kAXS
x

 


m2


Taking the limit as x  0 yields
d 2T
dx 2
2
 m
 T  Tamb  .
(4)
hP
kA XS
Note that the relation uses m2. Forgetting this has caused most
engineers grief at one time or another. Two types of boundary
conditions will be considered in this lab—these are:
1) Case A, Infinite Fin—The fin is infinitely long such that there is no
tip to consider. If there were a tip at infinity, it would never see the
heat introduced by the base in this century. Thus, at infinity, the
condition always exists that T = Tamb. With T = Tamb at the tip, no
difference in temperature exists to enter into Newton’s relation,
and thus, no heat transfer to the ambient occurs at infinity.
The above differential equation is more easily solved with a
change in variable. Use  = T Tamb. Then d2/dx2 = d2T/dx2 and
then Equation 4 becomes
d 2
(5)
 m2  0 .
dx2
Assuming a solution of the form erx and substituting into the
differential equation yields two roots
r 2erx  m2erx  0  r 2  m2  r  m, m .
So, the general solution is
  Ae mx  Be mx .
Apply the boundary conditions
at x = , T = Tamb = finite, thus B = 0;
at x = 0, T = Tb, thus A = Tb.
Thus, the solution is
  T  Tamb  Tbemx .
(6)
Note: Tb must be an absolute temperature.
2) Case B, Tip Convection—Here there is convective (Newton’s
law) heat transfer at the tip as well as the sides. Separation of
variables is used to obtain a solution. The only difference between
an undergraduate heat transfer course and a graduate conduction
Last Rev.: 23 JLY 08
Extended Fin / Mathcad Tutorial 2 : MIME 3470
course is the math. The undergraduate is simply told what
relation is the solution to a certain set of boundary conditions. In a
graduate course, the student gets to see how the relations are
derived. Such derivations are not difficult once the mathematical
underpinnings are taught. The solution to this case is
h
sinhmL  x 
coshmL  x  
mk
(7)
  T  Tamb 
h
sinhmL
coshmL 
mk
T = Tb= T0




x
i 1

i

i 1

Cell
Cell
Half
Half Cell
Cell
x
N 1
N



Cross Sectional
Area, A XS
Figure 3— Discretetized fin problem
Numerical Solution—An alternate way of solving Equation 4 is to
use a differenced form of the equation and solve for the temperature
distribution, usually, iteratively. In such a procedure, the spatial
domain (x distance along the fin) is discretetized into a number of
segments or cells as shown in Figure 3. Heat flows into and out of these
cells just as with the smaller differential element of Figure 2. The temperature throughout each cell is assumed to be constant—spatially
lumped temperature. The numerical solution for the cell cannot grasp
the concept that the segment has some x dimension—it only sees the
cell as having one uniform temperature. Thus, it is convenient for the
engineer to consider the entire cell as a point or node having the same
temperature as the cell. Unless otherwise prompted, this node is usually positioned at the center of the cell. For this solution, the nodes are
equally spaced with nodes at the tip and base. The objective of the
numerical solution is to determine the temperature at each node, Ti.
i
Ein cond

c.v.
Eout cond
Ti 1  2Ti  Ti 1
x 2

hP
Ti  T .
k rod AXS
The solution procedures below require a single coefficient for each
Tn. As Ti exists on the right-hand side of the equation, further
rearrangement is needed to yield
 2
1
hP 
1
hP
Ti 1
 Ti 


T (13)
  Ti 1
2
2
krod AXS 
AXS
x


x 2 k
 x 
rod









c
a
a
b
The terms a, b, and c are constants composed of dimensions and
physical properties that do not vary with temperature.
Now, consider the boundary
conditions; i.e., the base and
the tip. The equal spacing of
nodes causes the cells at the
base and tip to be half as
long as the midbody cells.
The temperature at the base
node (i = 0) is already
known, Ti–1 = T0 = Tb. So, for
the first FULL cell (i = 1), the
Equation 13 becomes
x / 2
Ein cond
iN
Eout conv
Eout conv
Figure 5—Half cell at tip






2
hP
1
hP
1  .(14)



T  Tb
  Ti 1
Ti 

2
k rod A XS 
k rod A XS
x 2
x 2
  x 
 
 











c
a
a  i 1
b

For the half cell at the tip, energy flows in by conduction. However,
this energy flows out by convection from both the sides and the tip
(see Figure 5). The differenced energy balance for this is
 k rod AXS
TN  TN 1
x 

 h AXS  P TN  T  .
x
2 

Again, grouping similar Tn terms to have a single coefficient for
each yields
Eout conv
Figure 4—Heat flow through a cell
With a knowledge of differencing, one could directly transform
Equation 4 onto the new format. As the student may be new to this
concept, the more basic concept of the energy balance will be
again used. For the cell shown in Figure 4 having Node i at its
center, the energy conducted in FROM Node (Cell) (i – 1) is:
Ein cond  krod AXS dT .
(8)
dx
In differenced form, this differential equation is
Ein cond  k rod AXS T  k rod AXS Ti  Ti 1 . (9)
x

x

Must be (  ). i.e.,
NOT Ti 1 Ti 
Similarly, energy conducted TO Node (i + 1) is:
Eout cond  krod AXS Ti 1  Ti .
(10)
x
For uniform cell temperature and perimeter P, the lateral convection
across the cell’s side area, As (=Px), is
Eout conv  hAs Ti  T  .
(11)
At steady state, the energy in equals that out; thus, one can write
Ein cond  Eout cond  Eout conv
Ti  Ti 1
T T
 krod AXS i 1 i  hPxTi  T  (12)
x
x
Rearranging this gives the final form of the differenced equation
 krod AXS
Page 3
k A
x  
x 
k A 


TN 1 rod XS   TN  rod XS  h AXS  P    T h AXS  P  .
x
x
2 
2 









 
d
f
e
(15)
What has been described by these differenced equations is a system
of N linear equations with N unknown temperatures. Such can be
solved using either of two methods:
Solution to Simultaneous Equations Using Matrices—With this
procedure the student should be familiar; but, a brief review is
presented. Given a system of N linear equations, each of the form
A1x1  A2 x2    AN xN  B , such equations can be arranged in
matrix form as
A1, 2  A1, N   x1   B1 
 A1,1

  

 A2,1 A2, 2  A2, N   x2    B2  .
 








 

  

 AN ,1 AN , 2  AN , N   xN   BN 

 
 
Α
X
B
Substituting Equations 13, 14, and 15, into Equation 16 gives
(16)
Last Rev.: 23 JLY 08
Extended Fin / Mathcad Tutorial 2 : MIME 3470
0 0 0 0   T1  c  Tba 
b a 0 0

  

a b a 0  0 0 0 0   T2   c 


0 a b a


0 0 0 0 T3
c 

  

0 0 0 0   T4   c  .
0 0 a b




      
0 0 0 0
b a 0 0  T8   c 

  

 0 0 0 0  a b a 0  T9   c 
0 0 0 0
0 a b a  T10   c 

  

0 0 d e  T11   f 
 0 0 0 0

 



 
Α
B
T
(17)
NN
N 1
 
 A 1 N  N BN 1
Identity Matrix, I

I N  N XN 1  XN 1  A 1 N  N BN 1




N 1
Equation 14, for Node 1, becomes,
c  Tb  0 a  T2a
,
T1 

b 

New
(21)
Value
(18)
NN
(20)
New
Value Use Temperatures From
Previous Iteration
To properly use the inverse matrix, this simple relation must be
maintained. The intent is to multiply both sides of the matrix
equation by the inverse matrix of A, i.e., A1. The only position
the A1can be placed and maintain the relation of Equation 18 is at
the beginning of each side of the equation; i.e.,
1
c  Ti 1a  Ti 1a
,
b 

Ti 

To solve for T, one cannot simply divide B by A as is done in
simple algebra. Instead the inverse of Matrix A must be used. The
reader will remember that there is a simple relation for the indices
of A, T, and B. This is
AA X
Page 4
(19)
N 1
This is the solution to X—or in the case of the fin, T.
Solution Using Successive Iterations—Instead of a matrix solution,
one can assume that all of the nodes of the rod are initially at room
temperature and the base node is suddenly brought up to its steady
state value, Tb. Then, in an iterative manner the nodes are heated as
energy flows from the base node to the next, and the next,(refer to
plots at the bottom of Page 9). This is time consuming but
instructive. It is not the same as the actual problem as the base node is,
in reality, slowly brought up to its steady-state temperature while
simultaneously passing on energy to other nodes. Even though there
is no real time increment between iterations, a similar process could
be performed using time steps. When real time steps are used, some
algorithms become unstable instead of converging to a solution. This
is because the time change (step) is too big for the spatial increment,
x. To obtain convergence, the time step is reduced—this process is
called relaxation. An alternate form of relaxation is when a
calculation tells one that the temperature at a node should be changed
to some value in the next iteration but the programmer only programs
in a percentage of the specified change.
So the reader asks: Why use this method when the simpler matrix
method provides an answer? The answer is that this lab course is
teaching techniques. That the simultaneous equations are all linear is
a fortuitous happening. Usually, one or more of the simultaneous
equations are not linear and the matrix method is inapplicable.
This method requires Equations 13, 14, and 15 to be rearranged to
express the unknown temperature of any node, Ti, in terms of the
temperatures of the adjacent nodes. Equation 13, for a general
midbody node, becomes,
Use Temperatures From
Previous Iteration
and, Equation 15, for the tip, becomes
f  TN 1d
TN 

e 

New
Value
.
(22)
Use Temperatures From
Previous Iteration
With the base node at temperature Tb and after initializing all of the
unknown temperature nodes (1N) to the ambient temperature,
Tamb, the algorithm is performed by sweeping from Node 1 to Node N
and at each node applying the appropriate equation. Thus, for the first
sweep, only the temperature of Node 1 can be altered. With successive
sweeps (iterations), the temperatures further out along the fin begin to
change. The solution is reached when the change in temperature of all
nodes is less than a user specified tolerance. A visual inspection of the
change in temperature value at each node could be performed in
stead of using a tolerance, per se. With no change in temperature
between iterations, this solution is the steady-state temperature
distribution of the fin as steady-state means no change.
Empirical, Natural Convection Correlations—Here are summarized appropriate empirical correlations that have been developed for
common immersed (external flow) geometries. The correlations are
suitable for most engineering calculations and are usually of the form
hL
NuL 
 CRa nL .
(23)
k
where,
h  average heat transfer coefficient, W/m2∙K. For turbulent flows, this is independent of L,
C  Coefficient which is a function of Ra L (Fig. 6),
L  characteristic length of the geometry, m,
k  thermal conductivity of fluid, W/m∙K,
n = 1/4 or 1/3 for laminar or turbulent flow, respecttively (function of Ra L , see Figure 6)
Ra L  Rayleigh number
gTs  T L3

 Grashof number
= GrL Pr 
GrL
(24)
gTs  T L3

 Prandtl number
C p 

=
k

 acceleration of gravity, m/s2,
 1/Tf , K1,
 Ts  Tamb  / 2 , K,
=
Pr
g

Tf
Ts  surface temperature of fin, K,
  thermal diffusivity of fin, m2/s, and
  kinematic viscosity (aka viscous diffusivity) of
fluid, m2/s.
Note: all properties are evaluated at the film temperature, Tf,
Last Rev.: 23 JLY 08
Extended Fin / Mathcad Tutorial 2 : MIME 3470
Note also: in the real fin of this lab, the surface temperature along the
fin, Ts, is not constant (not isothermal). However, glancing ahead to
Figure 8, this temperature varies little in terms of absolute
temperature. Thus, for this lab, an average value of surface
temperature will be used in computing film temperature, Tf.
Page 5
base of the rod will be at some lower (but constant) temperature. DO
NOT let the water in the container boil away.
Once steam can be observed coming out of the vent pipe, take
temperature readings generated by the eleven thermocouples
mounted in the fin. Wait, say, ten minutes and measure the
temperatures again. Wait another five or ten minutes and again
measure the temperatures. When the temperatures for each
thermocouple stop changing, steady state heat transfer has been
reached. At steady state, the heat loss from the rod will exactly equal
the heat added through its base and the temperature at any given point
along the rod will remain unchanged with time. For the calculations,
use only the last, steady-state values recorded.
The equations developed above are for free or natural convection,
students should avoid excessive talking or moving around as this
induces forced air currents which will increase the convection from
the fin and lower the temperatures being measured. To demonstrate
this, choose one thermocouple and simply blow air over it. It will be
observed that the temperature readout from that thermocouple
immediately drops. Note also: there are other air currents in the lab
which increase the heat flow from the fin—this will be considered.
Tb
360
Figure 6—Free convection Nusselt number for vertical plate
Natural Convection along a Vertical Plate—Equations of the form
given by Equation 23 have been developed for the vertical plate and
are plotted in Figure 6. The coefficient C and the exponent n depend
on the Rayleigh number range; for Rayleigh numbers less than 104,
the Nusselt number should be obtained directly from the figure.
More recently, the following correlation has been recommended that
may be applied over the entire range of RaL
2


0.387Ra1L/ 6


(25)
NuL  0.825 
 .
8
/
27


1  0.492/Pr 9 / 16


The foregoing results may be applied for constant heat flux (e.g., a
heating filament), as well as for constant surface temperature (isothermal). They maybe also applied to vertical cylinders of height L, if the
boundary layer thickness, , is much less than the cylinder diameter,
D. This condition is known to be satisfied when D / L ~
 35 / Gr1 / 4 .


L
Again, the real fin is not isothermal; but, an approximate procedure
for determining this variation is to use Nu L correlations obtained for
the isothermal plate but Ra L of Equation 24 (and thus Nu L ) should
be defined in terms of the average surface temperature along the fin.
Figure 7—Extended Fin Experimental Set Up (Photo: Greg Hill, BSME ‘03)
EXPERIMENTAL PROCEDURE—A rod penetrates the top of an
aluminum container as shown in Figure 7. The container is partially
filled with water and heated using an electric heating element. When
the water boils and steam can be seen coming out of the vent pipe, it
can be assumed that the portion of the rod inside the container, as well
as the inside surfaces of the container, are at the saturation
temperature of water. The outside surface of the container and the
Tamb
L
300
240
Texp 180
i
120
60
0
0
0.1
0.2
0.3
0.4
x
0.5
0.6
0.7
i
Figure 8— Experimentally measured, absolute temperatures, Texp, along
the length of the fin, L. These are bounded by the base
temperature, Tb, and the ambient temperature, Tamb. Note that
the change in ABSOLUTE temperatures is not great. This is the
basis of assuming an isothermal rod.
CALCULATION PROCEDURE
1. Experimental—Graph the experimental data and fit a curve to that
data. This curve will be used to compare other curves (see below).
2. Analytical—Using the base temperature from the steady-state
data, plot curves for Case A and Case B along with the experimental data. It should be obvious from Item 1 that the fin is not
isothermal. Yet, if the experimental temperature distribution is
plotted using absolute temperatures as shown in Figure 8, the
change in temperature over the length of the rod is small. Thus,
this experiment assumes an isothermal bar for its calculations.
The example of Appendix B may be helpful in these calculations.
3. As the actual data will not reflect pure, natural convection, but will
also reflect other air currents near the experiment, the temperatures of the experimental data will be lower than those predicted
by Cases A and B. These two cases used an average heat transfer
coefficient developed Equation 26 ( h  NuLkair / L ).Manually
change the average heat transfer coefficient until the curve for the
more realistic Case B lies atop the curve for the experimental data.
4. Numerical—Use this new value of heat transfer coefficient to perform numerical solutions for the temperature distribution using
the matrix method and method of successive iterations. Plot experimental, tip convection, matrix, and successive iteration temperatures on one graph. Explain any discrepancies in the results.
NOTE: Thermocouple spacing does not match the equally spaced
nodes used to develop the differenced equations. Do not try to match
the thermocouple spacing; instead use 10 evenly spaced nodes.
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ORDERED DATA, CALCULATIONS, AND RESULTS— Read “BEFORE STARTING this Mathcad tutorial” on Page 10.
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Page 10
BEFORE STARTING this Mathcad tutorial, note the following:
1. When using a Mathcad object within MS Word, occasionally, when one opens the object, all of their work from a previous session
just disappears. Although part of one’s engineering education is to learn to fiddle with things which do not work as advertised, the
student does not need this extra aggravation. Thus, it is recommended that the student prepare his/her Mathcad work using
Mathcad as a stand-alone program. Then, paste that work into a Mathcad object when done. So one asks: Why even bother to use
Mathcad objects? The answer is that the student is learning to produce seamless documents with text, equations, graphics, and
other objects which will impress one’s boss and clients—and be easy to grade by the TA.
2. To present a multipage Mathcad calculation in an MS Word document (such as the 5 objects just above), note that the individual
objects do not communicate with one another. Thus, the Mathcad programming shown in each object must be preceded by the
appropriate programming. So, after one prepares his/her work using just Mathcad as discussed in Item 1, paste that work INTO a
Mathcad object. Then for the second Mathcad page to be shown; copy the Mathcad object and paste it below the first one and then
open the second object and scroll down until “the work of the second page” is properly aligned in the object’s window. Close the
object and proceed to the next “page” if there is one.
3. Mathcad has a number of predefined variables and units; e.g., “g” is the acceleration of gravity and “A” is amperes. If the user
defines A as, say, area, Mathcad will place a green squiggle under the A to warn the user that that the predefined variable is being
redefined. This is good!; but, sometimes annoying. To turn off the squiggles, type [CTRL + SHIFT + r].
4. Depending on the version of Mathcad, the units of a variable may be reported differently; e.g., m/s as to m s-1. Ignore differences.
DISCUSSION OF RESULTS (For actual lab later in term)
CONCLUSIONS (For actual lab later in term)
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APPENDICES
Appendix A—PHYSICAL PROPERTIES (@300K)
Copper
Density, 
Specific heat, Cp
Thermal conductivity, k
Thermal diffusivity,  = k/ Cp
Air
8933 kg/m3
385 J/kgK
401 W/mK
117 m2/s
Appendix B—EXAMPLE PROBLEM—A glass-door firescreen, used
to reduce exhilaration of room air through a chimney, has a height of
0.71m and a width of 1.02m and reaches a temperature of 232C. If the
room temperature is 23C, estimate the convection heat rate from the
fireplace to the room.
Solution
Known: Glass screen situated in fireplace opening.
Find: Heat transfer by convection between screen and room air.
Schmetic:
Density, 
1.1614 kg/m3
Specific heat, Cp
Kinematic viscosity, 
Thermal conductivity, k
Thermal diffusivity,  = k/ Cp
Prandtl Number
1.007J/kgK
15.89106 m2/s
26.3103 W/mK
22.5106 m2/s
0.707
and from [another relation] it follows that transition to turbulence occurs on the panel. The appropriate correlation is given by Equation 25.

0.387Ra 1L/ 6

Nu L  0.825 

1  (0.492 / Pr) 9 / 16



and
where h may be obtained from knowledge of the Rayleigh number.
Using Equation 24,
 9.8 m  1 232  23C 0.71m 3


s  400 K 
gTs  T L3 
Ra L 



m 
m
 38.3  10 6
 26.4  10 6

2 
s 
s2 

 1.813  10 9
APPENDIX C—HISTORICAL SKETCHES
Baron Jean Baptiste Joseph Fourier (1768-1830)
This French Mathematician and
Physicist, famous for his pioneer
work on the representation of
functions by trigonometric series,
was born at Auxere, France on
March 21, 1768. He was the son of a
tailor who became a teacher of
mathematics at age sixteen at the
military school in Auxere. He later
joined the faculty at the Ecole
Normale at Paris in the year of its
founding (1795) when he was
twenty-seven. His teaching success
soon led to the offer of the Chair of Analysis at the Ecole Polytechnique
and in 1807, he was made a member of the Academy of Sciences.


Nu L  k

L
147 33.8  10 3
q  hATs  T 
Analysis:—The rate of free convection heat transfer from the panel to
the room is given by Newton’s law of cooling, q  h As Ts  T  ,
2
2
1/ 6


0.387 1.813  10 9


 0.825 
 147
8 / 27 
9
/
16


1  (0.492 / 0.690)
Hence, h 
Assumptions: 1. Screen is at a uniform temperature Ts.
2. Room air is quiescent.
Properties: From a properties table, Air (Tf = 400K):
k = 33.810-3 W/mK
 = 26.410-6 m2/s
 = 38.310-6 m2/s
Pr = 0.690
 = (1/Tf) = 0.0025K –1


8 / 27 


W
  7.00
2
m K



W 

mK 
0.71m
 7.00
W
m2  K

1.02m  0.71m 232  23C  1060W

Comments:
1. If h were computed from Equation 23, with C = 0.10 and n = 1/3,
we would obtain h  5.8W / m 2  K and the heat transfer predicttion would be approximately 20% lower than the foregoing result.
This difference is within the uncertainty normally associated with
using such correlations.
2. Radiation heat transfer effects are often significant relative to free
convection. Using Equation 1.6 and assuming  = 1.0 for the glass
surface and Tsur = 23C, the net rate of radiation heat transfer
between the glass and the surroundings is

4
q  As  Ts4  Tsur




W 
 232  2734  23  2734
 11.02  0.71m 2  5.67  10 8
2
m K4 

 2355W
Hence in this case radiation heat transfer exceeds free convection
heat transfer by more than a factor of 2.
Fourier's masterpiece was his mathematical theory of heat
conduction stated in Theorie Analytique de la Chaleur (1822). As one
of the most important books published in the 19th-century, it marked
an epoch both in the history of pure and applied mathematics. In it,
Fourier developed the theory of the series known by his name and
applied it to the solution of boundary-value problems in partial
differential equations. This work brought to a close a long controversy,
and henceforth it was generally agreed that almost any function of a
real variable can be represented by a series involving the sines and
cosines of integral multiples of the variable.
After a long and distinguished career, Fourier died in Paris on May 16,
1830 at age 62.
http://www.me.utexas.edu/~me339/Bios/fourier.html
See also
http://wwwgap.dcs.st-and.ac.uk/~history/Mathematicians/Fourier.html
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Sir Isaac Newton (1643-1727)
Re: what is the history behind Newton's law of cooling?
Date: Sat Nov 4 23:02:29 2000
Posted By: Benjamin Monreal,
Grad Student, Physics, MIT
Area of science: Physics
ID: 973198815.Ph
Message:
Re: [HM] Finite difference arrays
-----Message d'origine----De : M. Robert Showalter <showalte@vms2.macc.wisc.edu>
A : historia-matematica@chasque.apc.org
Date : mercredi 27 janvier 1999 16:25
Objet : [HM] Finite difference arrays
> ... I'm sure these finite difference arrays must have a history that
goes back farther. It seems likely that finite differencing arrays for
calculating polynomials must have been familiar objects for many
mathematicians in times past. How might I trace the matter back
farther? What resources might I use? Could someone please tell me
more about the history of these finite difference arrays?
Bonjour Anna,
You're (sort of) in luck... Newton
presented the Law of Cooling in a
paper [published anonymously]
entitled "Scala Graduum Caloris",
published (in Latin of course!) in
the Philosophical Transactions of
the Royal Society of London, in
1701. I found an English translation of this paper online at
http://web.bham.ac.uk/winterhs
/Newton.htm (Isn't the Web
amazing!)
However, reading 300-year-old
scientific papers is even harder
than reading modern ones. I'll
try to sum it up for you, which
should make it easier to follow.
Page 12
Anne Michel-Pajus (Annie.Michel-pajus@wanadoo.fr)
Thu, 28 Jan 1999 07:38:34 -0000
Some indications (not exhaustive): Finite differences are related to
the problem of interpolation, and construction of tables.
http://artzia.com/History/Biogra
phy/Newton/
Basically, Newton had to make up
his own temperature scale— you
http://astro.if.ufrgs.br/newton/
couldn't just buy a Celsius
thermometer back in 1701—so
the first part of the paper discusses how he did this. He thought up a
bunch of "reference temperatures"—the temperature at which water
boils, the temperature at which metals glow faintly, the temperature at
which bismuth melts, etc. He measured the temperatures of these
events by asking "How much does 10,000 units of volume of linseed oil
expand at this temperature"?
Imagine building a crude thermometer, filled with linseed oil instead of
mercury. You raise the temperature of the thermometer, the oil
expands and rises up the tube. Then, instead of writing numbers 0-100
on the tube and calling them "degrees Celcius", Newton put little marks
on the thermometer and labeled them "lead melts here; water boils
here; bismuth melts here" and so on. Newton then put his own
numerical scale on the tube, running from 0 (ice) to 192 (red-hot coals),
and called the units "degrees of heat".
Then, for the cooling law itself, he heated a "pretty thick piece of iron
red-hot". It was put in a cool place, and samples of the "temperature
flags" were placed on it—a bit of bismuth, a bit of lead, a drop of water,
etc. Newton observed how long it took before the bar cooled enough to
solidify the lead, solidify the bismuth, to stop glowing, etc., and
comparing these times to the marks on his imaginary linseed
thermometer, he deduced that the rate of heat loss was linear with
temperature; i.e. "The excess of degrees of heat [temperatures] of the
iron... were in geometrical progression, when the times were in
arithmetical progression."
I don't know whether he'd invented enough calculus yet to turn the idea
of "geometrical progression" into a differential equation, but that's how
the experiment was done.
Interesting stuff indeed! And it gives you an idea of some of the difficulties of doing science way back then... you had to invent your own
temperature scales, know all sorts of chemistry, do all of the math by
hand, etc. Newton was an alchemist, remember, so these sort of games
with bismuth and tin and red-hot iron was right up his alley.
Hope this helps! Thanks for asking!
-Ben Monreal
MadSci Network: Physics
http://www.madsci.org/posts/archives/nov2000/973522810.Ph.r.html
They are used by Indian and Arab mathematicians: among others,
Brahmagupta, in 665 (Khanda Khadyaka), and Al-Biruni (X-XI,
Qanun al-Mas'udi) for the calculation of sinus tables for astronomy.
Then by Briggs for the tables of logarithms (Aritmetica logarithmica,
1624), with methods maybe found by Harriot (around 1610). A
general formula is given by Gregory (letter to Collins 1670), but the
main work is done by Newton (from 1675, letter to John Smith,
then in Methodus differentialis, Letter to Oldenburg 1676,
Principia...) Leibniz (in Historia and Origo... 1715) tells that finite
differences are the source of his invention of Calculus.
Bibliography: History of algorithms, Chabert et alii, Springer-Verlag,
translation of Histoire d'Algorithmes, Belin.
Best wishes from Versailles, Anne Michel-Pajus
http://sunsite.utk.edu/math_archives/.http/hypermail/historia/jan99/0171.html
History of heat transfer
This doesn't pretend to be a complete history but I am slowly building
up a few things that have caught my eye.
Dividing heat transfer up into conduction, convection and radiation it
is odd that each of these has a famous name attached to it and that
the names in question are probably better known for something else.
The understanding of conduction certainly started with Fourier,
although he is probably better known for Fourier series.
Modern understanding of radiation is certainly due to Planck who
produced the theoretical law that gives the energy distribution with
wavelength, from which the other laws can be proved. He is better
known for inventing quantum mechanics (e.g. Planck's constant),
which first appeared in his radiation theory.
Much earlier Newton started the study of forced convection heat
transfer with Newton's law of cooling. Again he is more famous for
something else, though precisely what escapes me at the moment.
Anyway, all I have at the moment in my history is Newton's original
paper .
http://web.bham.ac.uk/winterhs/History.htm
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APPENDIX D—DATA SHEET FOR EXTENDED FIN EXPERIMENT
Time/Date
____________________________
Lab Partners
____________________________
____________________________
____________________________
____________________________
Data
Rod Diameter

Rod Length

Distance from Base to 2nd Thermocouple

Distance Between Thermocouples

Distance from Last Thermocouple to Tip

Rod Material

Ambient Temperature

.
3/8” 
.
23” 
.
3/4” 
.
2-1/2” 
.
3/4” 
. Copper
.
72°F


Sketch of Apparatus with
Appropriate Dimensions
Temperature Readings—When ALL the temperatures cease to change with time, steady-state has been achieved. Such will be the
last (not necessarily 5th) column of numbers recorded. The steady-state readings are the values to be used in the calculations.
Thermocouple
Number
Run 1
Run 2
Temperature Reading
Run 3
Run 4
Run 5
1
173°F
177
176
174

2
169
172
171
168

3
157
163
163
162

4
130
135
135
132

5
122
129
129
130

6
112
117
118
117

7
105
111
114
113

8
100
105
107
108

9
96
101
104
104

10
94
99
102
102

11
92
96
98
99

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