CONVECTION EXPERIMENT
LEADER: THOMAS SALERNO
JUNE 5, 2006
PARTNERS:
GREG ROTHSCHING
STEPHEN JOHNSON
JEN DIROCCO
i
ABSTRACT:
The convection experiment was separated into two parts. For the first part, we attached a flat
aluminum plate inside an air duct. We proceeded to supply a constant 20 watts to the plate while
we ran air over the duct at forced velocities of 0, 2.5, 5, and 10 meters per second. Measurements
were taken of the steady state temperature of air and of the surface of the plate. The second part
of the experiment was exactly like the first, except a finned plate was used for our heated surface.
This experiment had three main objectives. The first was to correlate the effect the forced air
velocity had on the heat transfer coefficient. The second was to analyze how well our predictive
equations were at matching the experimental heat transfer coefficients. Finally, we needed to
determine how the addition of fins to the surface aided the heat transfer rate.
It was determined from our experiment that the heat transfer coefficient increased linearly
with an increase in the square root of air velocity for both the flat and finned plates, over the
range of forced convection. From a jump in velocity of 2.5 to 10 m/s, the heat transfer coefficient
over the flat plate jumped from 39.4 to 78.1 W/m2 s, while that for the finned plate jumped from
32.1 to 55.8 W/m2 s. However, the limiting value to convection is not zero at zero forced air
velocity, but instead a finite value due to natural convection, though the predictive equations were
not able to model this effect quite accurately. Also, we saw that the predictive equations yield
excellent results in predicting the trend of the heat transfer coefficient, but not the absolute
numbers, yielding errors of 80% on the flat plate and 40% on the finned plate. Nevertheless, if
one point is taken then we can add a correction coefficient to the predictive equations to increase
error to an acceptable one of only 7%. Finally, we determined that the addition of fins increased
the heat transfer rate at all velocities, but that its effectiveness diminished as air velocity
increased. For the low velocity reading it was able to reduce the surface temperature by 400C, but
at the high velocity it only reduced the surface temperature by 180C.
The results of this experiment are of great importance for both process and design engineers.
This data shows that they can increase the heat transfer coefficient of their system by increasing
the flow rate of the heated fluid. Also that predictive equations can be used as long as their
geometry matches that of the assumptions for the equations. And if it is close, a simple fudge
factor can be added by reading only one data point. It makes the engineer aware that even at zero
velocity, a finite and substantial amount of convection will still take place. Finally, the addition of
fins will greatly help to increase the heat transfer rate at low fluid velocities, but will not provide
a substantial increase at higher fluid velocities.
ii
TABLE OF CONTENTS
ABSTRACT: ____________________________________________________________ ii
TABLE OF CONTENTS_____________________________________________________ iii
INTRODUCTION:_________________________________________________________ 1
THEORY: ______________________________________________________________ 2
Introduction to Convection: The Convection Problem ____________________________ 2
Figure 1: Convection over flat plate. ______________________________________________ 3
The Velocity Boundary Layer _________________________________________________ 5
Developing the Boundary Layer _____________________________________________________ 5
Figure 1: Flow of fluid over a flat plate. ___________________________________________ 5
Figure 2: Velocity boundary layer forming over a flat plate. __________________________ 6
Describing the Flow Conditions _____________________________________________________ 6
Momentum Equation of the Boundary Layer __________________________________________ 7
The Navier-Stokes Equation _______________________________________________________ 10
The Thermal Boundary Layer _______________________________________________ 11
Developing the Boundary Layer ____________________________________________________ 11
Figure 3: Flow of fluid over a flat plate. __________________________________________ 11
Energy Equation of the Boundary Layer _____________________________________________ 12
The Flat Plate in Parallel Flow _______________________________________________ 13
Solving the Momentum Boundary Layer Equations ____________________________________ 13
Figure 4: Velocity Profile in thermal boundary layer. _______________________________ 14
Solving the Thermal Boundary Layer Equations ______________________________________ 18
Figure 5: Temperature Profile in thermal boundary layer. __________________________ 20
Putting it all Together ____________________________________________________________ 23
Figure 6: Temperature Profile in thermal boundary layer. __________________________ 24
The Flat Plate in Free Convection ____________________________________________ 26
Physical Considerations __________________________________________________________ 26
Figure 7: Boundary Layer development on heated vertical plate. _____________________ 26
The Governing Equations _________________________________________________________ 27
Solving the Governing Equations ___________________________________________________ 29
Finned Surfaces ___________________________________________________________ 31
Physical Situation _______________________________________________________________ 31
Figure 8: Conduction and Convection through parallel fin. __________________________ 31
Establishing Steady State ___________________________________________________ 34
Figure 9: Front view of convection duct. __________________________________________ 35
Figure 10: Side view of convection duct. __________________________________________ 36
Figure 11: Flat Plate and air flow direction. _______________________________________ 36
Figure 12: Finned Plate and air flow direction. ____________________________________ 37
PROCEDURE: __________________________________________________________ 37
RESULTS: _____________________________________________________________ 38
Steady State ______________________________________________________________ 38
Table 1: Steady state temperature values. ________________________________________ 38
Flat Plate Heat Transfer Coefficient __________________________________________ 38
Table 2: Heat transfer coefficients along flat plate. ________________________________ 39
iii
Figure 13: Heat transfer coefficients versus velocity for flat plate. ____________________ 40
Table 3: Heat transfer coefficients along flat plate. ________________________________ 41
Figure 14: Heat transfer coefficients versus velocity for flat plate. ____________________ 41
Finned Plate Heat Transfer Coefficient ________________________________________ 42
Table 4: Heat transfer coefficients along finned plate. ______________________________ 42
Figure 15: Heat transfer coefficients versus velocity for finned plate. __________________ 43
Table 5: Heat transfer coefficients along finned plate. ______________________________ 44
Figure 16: Heat transfer coefficients versus velocity for finned plate. __________________ 44
Effectiveness of Fins ________________________________________________________ 44
Figure 17: Temperature Differences recorded for flat and finned plates. ______________ 45
Discussion of Results: __________________________________________________ 46
Effectiveness of Air Velocity on Heat Transfer Coefficient ________________________ 46
Comparison of Experimental results to theoretical ______________________________ 47
Forced Convection ______________________________________________________________ 47
Figure 18: Turbulence over flat plate. ___________________________________________ 48
Figure 19: Turbulence around fins. _____________________________________________ 50
Natural Convection ______________________________________________________________ 51
Effectiveness of Adding Fins to a Flat Plate ____________________________________ 52
Conclusions: __________________________________________________________ 53
Recommendations: _____________________________________________________ 55
References: ___________________________________________________________ 58
APPENDIX:____________________________________________________________ 59
Appendix 1: Sample Calculations ____________________________________________ 59
Run1: Flat Plate Low velocity _____________________________________________________ 59
Table 6: Experimental data obtained for flat plate, at 2.5 meters per second. ___________ 59
Figure 20: Steady state calculation for the first run.________________________________ 60
Table 7: Experimental data obtained for flat plate. ________________________________ 62
Run5: Finned Plate Low velocity ___________________________________________________ 62
Fin Effectiveness for Runs 1 and 5 _________________________________________________ 64
iv
INTRODUCTION:
In the field of chemical engineering and in every day life itself, convection is one of the
most important forms of heat transfer. From running 150,000 lb/hr of hot water through a
heat exchanger, to simply turning on a fan to cool down on a hot day, the principles and
uses of convection remain the same.
Essentially, forced convection is one of the most significant designs of heat exchangers.
In order to increase the temperature of their inputs, which allows a faster rate of reaction,
or decrease the temperature of waste, or possibly to recover heat that would otherwise be
wasted, a majority of processing plants have need of heat exchangers. For example, the
pulp and paper industry use heat exchangers to preheat their milling water before it is
sent to the separator and, also just as significant, to cool their waste water before it is sent
to water treatment. This last principle would permit a plant to recover the heat that would
be wasted and channel it for use in other areas of the plant. (CADDET) However, diverse
operations will require varying amounts of heat transfers. Thus, process engineers at
these plants need to calculate in the area needed for the heat transfer they desire. Such
information is only available as a result of convection.
Natural convection has many applications. Free convection, strongly, influences the heat
transfer from pipes and transmission lines, as well as from electric baseboard
refrigeration units to the surrounding air. It is, also, relevant to the environmental
sciences, where it manipulates oceanic and atmospheric motions. The most relevant use
of natural convection for the chemical engineer, however, is in the cooling of electronic
components. Both the performance reliability and life expectancy of electronic equipment
are inversely related to the component temperature of the equipment. The relationship
between the reliability and the operating temperature of a typical silicon semi-conductor
device demonstrates that a reduction in the temperature corresponds to an exponential
increase in the reliability and life expectancy of the device. Therefore, controlling the
temperature of the device by natural convection (which is a free resource) is of vital
importance. (Icoz and Jaluria) Computer engineers, by studying and mastering natural
convection, are better able to arrange the area, position, and location of heat sinks to cool
their electronic components.
1
To study the effects of convection, I selected to flow air at room temperature on both a
flat plate and a finned plate at an elevated temperature for a total of eight trials through a
convection chimney. In the first four trials, a flat vertical aluminum plate was placed near
the top of the chimney and heated with a constant 20 watts input. Then, air was flowed at
2.5, 5, and 10 meters per second over the plate. Temperature readings of the inlet air and
the surface of the plate at steady state were recorded. To find the effect of natural
convection, I repeated this setup for an air velocity of zero meters per second. For the
remainder of the final four trials, the experiment was then repeated with a finned plate in
place of the flat plate. With this data, I was able to correlate how the heat transfer
coefficient changed with the air velocity. I was, also, capable of comparing the heat
transfer rate of the finned plate to the flat plate at the same air velocity. As a final
outcome of the recorded data from my experimentation, I can test predictive equation
results with those obtained from the experiment.
This experiment had three main objectives. First, to study the effect the air velocity had
on the heat transfer coefficient; second, to examine the accuracy of current predictive
equations and finally, to find the effectiveness of the fins at different air velocities. With
this information, the principles of designing heat exchangers take form.
THEORY:
Introduction to Convection: The Convection Problem
Heat transfer due to convection involves the transfer of energy between a fluid at one
temperature moving over a solid surface at another temperature. Consider the situation
shown in Figure 1.
2
T8
A8
u8
Ts A s
L
Figure 1: Convection over flat plate.
Here a fluid, namely air, of velocity u (meters/sec) and temperature T∞ (°C) flows over
a flat aluminum surface of length L (meters) and area As (meters2). The temperature of
this surface is assumed to be uniform at Ts (°C) such that Ts > T∞. This will result in a
temperature gradient forming from the surface and extending into the fluid. We know
from Newton’s Law of Cooling that heat transfer at the surface will occur via a rate that
is proportional to the difference between the surface temperature Ts and the temperature
of the fluid T∞. Thus, we can write the local heat flux (heat transferred per unit area) as
(1)
q
 h(Ts  T )
A
Where q (Watts) is the heat transferred per unit time, A (meters2) is the surface area
available for heat transfer, and h (Watts/m C) is the constant of proportionality known as
the local heat transfer coefficient. Note that both q and h are referred to as local. This is
because flow conditions will vary from point to point along the plate causing both q and h
to vary with position x (meters). The total heat transfer rate Q may be obtained by
integrating the local heat flux over the entire surface. That is,
(2)
Q
q
 A dA
s
As
or from Equation 1 we can write,
3
(3)
Q
q
 A dA   h(T
s
As
s
 T )dAs (Ts  T )  hdAs
As
As
We can then define an average heat transfer coefficient h (W/m C) as,
 hdA
s
(4)
h
As
 dAs

1
As
 hdA .
s
As
As
Thus, the total heat transfer rate may also be expressed as
(5)
Q  hAs (Ts  T ) .
Note, that for our flat plate, we are considering flow in only one direction, namely the x
direction. Thus, we may rewrite Equation 4 as
L
(6)
h
1
hdx .
L 0
Equation 5 illustrates two very fundamental uses. First, with this equation, we have
available a method to experimentally measure our average heat transfer coefficient over a
flat plate. In short, if one has all the variables (measure of heat flux being convected from
the plate; the surface area of the plate, the temperature of the flowing air, and the steady
state temperature of the surface of the plate ) needed in the equation then, one is able to
obtain the average heat transfer coefficient for the flat plate.
Secondly, to find the total heat transferred from a heated plate to a flowing fluid, we only
need to know the average heat transfer coefficient, surface area of the plate and the
temperature of the surface and of the bulk fluid. However, in order to arrive at the
average heat transfer coefficient, we must, first, know how the local heat transfer
coefficient varies with x direction along the plate. This is dependent upon numerous fluid
properties such as viscosity, density, thermal conductivity, specific heat etcetera as well
4
as the flow conditions along the plate. This multitude of independent variables in
determining the average heat transfer coefficient has been termed the problem of
convection. One way to determine all of these variables is to consider the concept of
boundary layers.
The Velocity Boundary Layer
Developing the Boundary Layer
To develop the concept of boundary layers, consider, again, the flow situation
represented by Figure 1 below.
T8
A8
u8
Ts A s
L
Figure 1: Flow of fluid over a flat plate.
When fluid particles make direct contact with the surface, we assume their velocity is
reduced to zero due to viscous action between the surface and the fluid. These particles
will then act to retard the motion of particles in the adjoining fluid layer to a lower
velocity, which will then retard the motion of the next fluid layer and so on. The
retardation of fluid motion is due to shear stresses  (N/m2) acting between parallel fluid
layers. This retarding of motion will continue until we reach a distance where the velocity
of the fluid particles flowing over the plate are unaffected by the plate’s presence and
continue at a velocity of u . It is this vertical distance from the plate to the first particle
whose velocity is unaffected is  (meters) that is known as the boundary layer thickness.
Accordingly, we can develop a boundary layer velocity profile which shows the manner
in which u varies with the y direction throughout the boundary layer. This is represented
below in Figure 2.
5
u8
T8
A8
u8
TS
AS
L
Figure 2: Velocity boundary layer forming over a flat plate.
This boundary layer will form whenever a fluid flows over a flat surface. Its shape will
depend on how the shear stress varies with shear strain
u
in the fluid as well as the flow
y
conditions we are experiencing. Assuming, we have a Newtonian fluid, we can evaluate
the shear stress as,
(7)
 
u
y
where  (kg m/sec) is the viscosity of the fluid. As for the flow conditions, there are
typically three patterns seen.
Describing the Flow Conditions
First, in laminar flow, the motion of air particles is very orderly with all particles at a
position y moving in a straight line parallel to the pipe walls. Secondly, in turbulent flow,
the particles move around and have different velocity components in all directions.
Thirdly, a transition flow has mixed properties of the previous two. From observation, it
was hypothesized that fluid travels in a laminar motion when it is moving slowly and
converts to turbulent motion when it has a faster velocity. But this information was not
very useful in that it only gave us qualitative definitions of our flow. As a result, in the
1880’s, Osbourne Reynolds designed an experiment which involved inserting dye into
the center of a water stream which was flowing through a glass tube and where the
6
velocity was controlled by a valve. After many experiments, he developed the following
equation,
(8)
Re x 
 ux

where Re is called the Reynolds number, ρ is density (kg/m3), and x is the distance along
the plate in the x direction. He found that the dye made a straight line and stayed in the
center when Re < 5  105 (laminar), and that the dye formed fishhooks when Re > 107
(turbulent). Later this term was given a more scientific understanding, namely
(9)
Re 
inertial forces
.
viscous forces
Thus, when inertial forces such as pressure and mass flow overcome the viscous forces,
we have a large Re and our flow is very turbulent, however, if the viscous forces are
larger, than all of the particles stick together and stay in line and we arrive at laminar
flow.
Momentum Equation of the Boundary Layer
In order to study the motion of particles, we need a method to model the interaction and
motion of any fluid element in the boundary layer. This data can be obtained by
performing a momentum balance on the fluid particles and developing what is known as
an equation of motion for them.
Consider an arbitrary volume element. Momentum can be transported into and out of this
volume element by three different mechanisms. The first is through convection, the bulk
flow of fluid across the surface. The second is via forces that act on the surface of the
volume element and the third is by body forces which act on the entire volume of the
element in order to change its momentum. We are able to arrange these three mechanisms
to fit a verbal differential equation shown below.
Rate of
Net rate of
Net rate of
Net rate of
7
(10)
change of
momentum
=
momentum
+
momentum
convected into V
+
momentum
creation by
creation by
surface forces
body forces
Equation 10 is the verbal representation of the momentum balance on this fluid element.
The next step is to put this equation into mathematical terms.
The rate of momentum change caused by convection is determined by the rate at which
mass will enter and leave this fluid element and the velocity of this fluid. The local
volumetric flow rate of fluid across a surface element dA is the velocity of fluid
multiplied by the area, or in other words, u·dA. Thus, the rate of mass transfer in and out
of this element is given by ρ(u·dA). Multiplying this by the velocity determines our first
term of the equation. Note, that this equation is negative because it assumes the
momentum is being convected into the surface element. If momentum is in fact leaving
the surface element in a larger amount than the dot product, then, this will ensure that the
term turns out positive.
(11)
Rate of momentum convected   u  u  d A
The rate of momentum change created by surface forces is caused by molecular motion
and interactions within the fluid itself. This rate of molecular diffusion can be described
by the shear stress on the volume element multiplied by the area on which this stress acts.
These shear stresses are acting to compress this volume element and, thus, this term must
also be negative.
(12)
Rate of momentum created
   d A
by surface forces
Body forces acting on our volume element may be gravitational, surface, electrical, or
magnetic. However, in the present situation, the most common forces are only
gravitational and surface, and will be the only ones considered here. The gravitational
force is, simply, the acceleration due to gravity pulling down on the mass of the element,
where the surface force is the external pressure acting to compress our volume element.
8
(13)
Rate of momentum created
  Pd A   g dV
by body forces
Finally, we can write the rate of accumulation of momentum into and out of this volume
element as,
(14)
Rate of accumulation of momentum  

  u  dV
t
Thus, the momentum balance can now be written mathematically as:
(15)

  t   u  dV     uu  d A     d A   Pd A    gdV
C .V .
C .S .
C .S .
C .S .
C .V .
Mathematically, C.V. signifies an integral across the control volume of our fluid element
and C.S. represents an integral across the surface of our fluid element. However as it
stands, this equation cannot be solved easily. In order to simplify this equation, we would
need to organize our data into the same units and transform the integrals into differentials
which can be more easily applied to a given situation. Fortunately, we have from Gauss’s
theorem which is the ability to convert each integral over the control surface to an
integral over the control volume thereby standardizing the integrals. This is achieved by
using the gradient vector, as shown below:
(16)
  u  d A     udV
C .S .
C .V .
Thus, we can now write Equation 15 as:
(17)

  t   u  dV     uudV     dV  PdV   pgdV
C .V .
C .V .
C .V .
C .V .
C .V .
9
Now, Equation 17 illustrates that all terms have a triple integral over the control volume.
Provided that no math teachers are looking, we can cancel out all the integrals and arrive
at:
(18)

 dV          P  p g
t
Equation 18 is what is generally referred to as the equation of motion for a fluid. Just to
be certain nothing is lost in the math, we will define the physical significance of each
term. The term on the left side of the equal sign is the rate at which we accumulate
momentum in our fluid element. On the right side of the equation, the first term describes
the change in momentum due to convection via fluid flow. The second term accounts for
change in momentum due to molecular diffusion which is based off the viscosity of the
fluid and the shear stresses placed on it. The third term describes the change of
momentum that is caused by a pressure drop due to the friction experienced by flowing
along a plate. Finally, the last term on the right takes into account the gravitational force
on our fluid element.
The Navier-Stokes Equation
Inserting Equation 7 into Equation 18, we have developed what is known as the NavierStokes equation where we assume that we have a Newtonian fluid, a constant density,
and a constant viscosity. This equation is shown below for flow in the x direction in a
Cartesian coordinate system:
(19)

v x
P
2
 v  v x  
  v x  g x 
t
x
However, we could greatly simplify Equation 19 by factoring in a few more assumptions.
First, we will assume that flow is only in the x direction and that momentum is changing
only in the y direction. Secondly, we will again assume that the pressure drop for our
external flow is negligible. Applying these assumptions to Equation 19 and dividing
through by  and noting that  

is the molecular diffusity of the fluid, we arrive at

10
the working momentum equation of the boundary layer under laminar flow conditions,
namely,
(20)
 u
u 
 2u
.
u




 x
y 
y 2

The Thermal Boundary Layer
Developing the Boundary Layer
The development of the thermal boundary layer is similar to that of the velocity boundary
layer. Consider the flow situation represented by Figure 3, shown below.
T8
T8
A8
u8
TS
AS
L
Figure 3: Flow of fluid over a flat plate.
At the leading edge, the temperature profile of the fluid is uniform. However, once the
fluid makes contact with the plate the particles at the surface will achieve thermal
equilibrium at the plate’s surface temperature. These particles will, then, exchange energy
with the layer of fluid particles above them, who will exchange energy with the fluid
particles above them. This will form a temperature gradient in the fluid from the surface
at a temperature of Ts to a distance  t (meters) where the temperature is that of the bulk
fluid T∞. Here  t is known as the thermal boundary layer thickness and is the region of
the fluid in which the temperature gradients exist.
11
Energy Equation of the Boundary Layer
Just as we developed the momentum equation of the boundary layer, we will develop the
energy equation of the boundary layer by performing an energy balance on the fluid
particles inside the boundary layer. Energy is carried in and out of any arbitrary volume
element by three mechanisms. Convection is the first mechanism by the bulk flow of
fluid across the surface. The second is through conduction, the energy transfer from
molecular vibrations of particles adjacent to the control element. With the third being
viscous work where the body forces act on the entire volume in order to change its
momentum. By combining all three mechanisms, we arrive at a verbal energy equation
shown below in Equation 20.
(21)
Energy convected Energy convected Heat conducted Net viscous work




in at x
in at y
in at y
done on element
Energy convected Energy convected Heat conducted


out at x  dx
out at y  dy
out at y  dy
Note: that energy being convected into a control volume element is calculated by the
mass flow rate entering multiplied by the enthalpy of that mass. In mathematical terms
we transcribe it as,
(22)
Energy convected   C pu x  positionTx  position  dy
Also, the energy being carried in by conduction is, according to Fourier’s Law, equal to
the conduction heat transfer coefficient multiplied by the temperature gradient at this
position. Hence, in mathematical terms we can write.
(23)
 T 
Energy conducted  kdx  
 y  y  position
Finally, the viscous work being done on the element is equal to the viscous-shear force
acting on the element ( 
unit time (
u
dx ) multiplied by the distance through which it moves per
y
u
dy ). So, we may write this mathematically as,
y
12
2
(24)
 u 
Viscous work     dxdy
 y 
Therefore, we can plug Equations 22,23, and 24 into Equation 21 to yield the differential
energy balance,
 T
 u   
 u 
T
 2T
 C p u

T  
dxdy

k
dxdy     dxdy


2
y
y
 x y  
 y 
 x
2
(25)
 u  

Then using the continuity relation  
  0 and dividing through by  C p , we
 x y 

arrive at,
(26)
 T
T 
 2T
  u 
u x   y    y 2   C  y 



p 
2
which is the energy equation of the boundary layer under laminar flow conditions. In this
equation  
k
, is known as the thermal diffusity of the fluid. However, this can be
C p
further simplified by realizing that the second term on the right side of Equation 26 is
only of importance for highly viscous fluids and is therefore negligible for air in our
operating temperatures. This leaves us with our working energy equation for laminar
boundary layers, namely,
(27)
 T
T 
 2T
u




.
 x
y 
y 2

The Flat Plate in Parallel Flow
Solving the Momentum Boundary Layer Equations
Consider the flat plate situation shown below in Figure 4.
13
A
H
A
u8
dy
δ
dx
Figure 4: Velocity Profile in thermal boundary layer.
In order to solve for a velocity profile and the boundary layer thickness along the plate,
we must perform a momentum balance and a force balance on this arbitrary element.
Starting with the momentum balance we have,
(28)
Net Momentum Flow  Momentum carried out  Momentum carried in
Since we know that momentum in our situation is only carried in and away by mass, we
can make use of the equation Momentum  mass flow * velocity and write for the
momentum entering through the left side as
H
(29)
 u dy
2
0
and the momentum through the right side is
H

d 
2

u
dy


u
dy

 dx .
0
dx  0

H
(30)
2
14
The momentum flow through the bottom must be zero since there is a solid plate there,
and we are left with only the momentum through the top. This equation is complicated.
First, we need to know the mass flow entering the top. This can be found from the
difference between the mass exiting through the right and the mass entering through the
left as there is no creation or destruction of mass in this volume element. Thus, the mass
flow through the top is,
(31)
H
H
H
H


d 
d 

u
dy
dx


u
dy


u
dy



   u dy  dx .


dx  0
dx  0
0
0


This mass must carry with it a momentum equal to its mass flow rate times its velocity.
Since this area is above the boundary layer, we know that its velocity must be that of the
free stream. Therefore the momentum through the top is,
(32)
H

d 
u    u dy  dx .
dx  0

Combining Equations 32, 30, and 29 into Equation 28 yields us with our net momentum
flow out of our element as,
(33)
Net Momentum Flow 
H
H


d 
d 
2

u
dy
dx

u


   udy  dx .

dx  0
dx  0


Then using the product rule from calculus, we can rewrite the second term in this
equation as
(34)
u
H
H
H



du 
d 
d 
   udy  dx     uu dy  dx 
   udy  dx .
dx  0
dx  0
dx  0



Accordingly, we can rewrite Equation 33 in more useful terms as
15
Net Momentum Flow 
(36)
H
H
 d H


 
du 
d 
2

u
dy
dx


uu
dy
dx

 



   udy  dx 

dx  0
dx  0
 dx  0


 
H
 du  H

d 
     (u  u )udy  
   udy 
dx  0
 dx  0

Next, we need to do a force balance on this same volume element. The force at the left
and right sides are due to pressure force. The force at the left side is pH and that at the

 dp  
right side is  p    dx  H . Due to shear stress from the wall, the force at the bottom
 dx  

can be written as  w dx    dx
u
y
, which is the definition of viscosity. That leaves
y 0
only the force at the top. However since this section is outside our boundary layer, we
know there are now velocity gradients here and thus no force. Collecting terms, we arrive
at the net force balance as
Net Force  Force out  Force in
(37)

 dp  
Net Force  pH   p    dx  H   w dx .
 dx  

 dp 
Net Force   w dx  H   dx
 dx 
Then from Newton’s Law, the net force on an object is equal to its net increase in
momentum, we can set Equations 36 and 37 equal to each other to obtain.
(38)
H
 du  H

d 
 dp 
 w  H        (u  u )udy       udy 
dx  0
 dx 
 dx  0

For our system, we will assume that pressure across the plate is relatively constant
allowing us to cross out the second term on the left hand side of the equation. As a direct
result of the constant pressure assumption, we know the free stream velocity is not a
function of x, consequently, we can cross out the last term on the right hand side of the
16
equation. Finally from our definition of the boundary layer, we know that u=u∞ for y> δ,
thus we can change the limit on the integral to δ instead of H. This leaves us with,
(39)
u
w  
y
y 0


d 
    (u  u )udy 
dx  0

where we have brought in our definition of viscosity to rewrite the shear stress as a
function of the velocity gradient.
The next step is to find an equation for the velocity profile for use in Equation 39. Using
an approximate development, we can assume that this relation will be in the form of a
simple polynomial. To calculate approximately how many terms we need, let us list the
conditions that must be satisfied: (a) the velocity at the surface must be zero due to
viscous forces; and since our definition of our boundary layer is the only region in which
velocity gradients exist we arrive at: (b) the velocity at the top of the boundary layer
must be equal to the free stream velocity; and (c) the velocity gradient must be zero. For
our final condition, since we are at a constant pressure situation, we can solve Equation
20 with our other conditions to obtain (d) the velocity gradient must stop changing at the
plate surface. We can express these mathematically as,
(a)
u =0 at y=0
(b)
u = u at y= 
(c)
u
 0 at y= 
y
(d)
 2u
 0 at y=0.
y 2
The simplest polynomial that could satisfy all four conditions is,
(40)
u  C1  C2 y  C3 y 2  C4 y 3 .
17
Applying the four conditions (a) through (d), we arrive at an expression for the velocity
as a function of y position in the boundary layer,
(41)
u 3 y  1 y 
    
u 2    2   
3
for Re  5 1010 .
We can now plug Equation 40 into Equation 39, perform the integration and solve for

,
x
d  39
 3  u
 u2   

dx  280
 2 
140 
140 
 d 
dx 
dx
13  u
13 u
(42)
2
2

140  x
 Const
13 u
@ x  0,   0  Const  0 

x

4.64
Re
1
2
x
2
2

140  x
13 u
for Re  5 1010
which will then tell us the boundary layer thickness as a function of x along the plate.
Solving the Thermal Boundary Layer Equations
In order to find the temperature gradient in the boundary layer, we must solve the energy
equation of the thermal boundary layer, Equation (27). The conditions that this equation
must satisfy are: (a) the temperature at the surface of the plate must equal the surface
temperature because we have zero velocity at this point and the volume elements here
must be in thermal equilibrium with the plate; (b) the temperature gradient should be zero
at the top of the thermal boundary layer. This is a direct result of our definition that the
thermal boundary layer is the area in which we experience temperature gradients, thus the
temperature gradients must stop as we leave the boundary layer; (c) the temperature at
the top of the boundary layer must equal the free stream temperature, this is also a direct
result of our definition of the thermal boundary layer, since we are outside the area of
18
temperature gradients, we are outside the area which is affected by the plate, thus our
temperature must equal the free stream temperature we originally had; (d) finally the rate
of change of the temperature gradient must also be zero at the wall because the velocities
are zero at the wall thus there is no driving force to change the temperature gradient. We
can express these conditions mathematically as
(a) T=TS at y=0
(b)
T
=0 at y=δT
y
(c) T=T∞ at y= δT
(d)
 2T
=0 at y=0.
y 2
We now assume that the temperature profiles at various x positions must all have the
same functional dependence on the y position. Consequently, the simplest method to
solve Equation (27) with the four conditions (a) through (d) is to fit them to a third degree
polynomial with arbitrary constants, namely,
(43)
T  Ts
 C1  C2 y  C3 y 2  C4 y 3
T  Ts
Then after applying the four conditions, we are able to arrive at
(44)
3 y
T  T  
2  T
 1 y 
  
 2  T 
3
for Re  5  1010
Now, we have found an equation for the temperature profile of the thermal boundary
layer. However, we now need to find a relation for the thermal boundary layer thickness.
This can be achieved by an integral analysis of the energy equation of the boundary layer
shown again in Figure 5 below.
19
A
H
A
u8
T8
δT
δ
dx
dq w   kdx
dT
dy
w
Figure 5: Temperature Profile in thermal boundary layer.
Performing an energy balance yields
(45)
Energy convected in  Viscous work within  Heat transfer at wall  Energy convected out
H
The energy convected into the left side is, C p  uTdy ; the energy convected out the right
0
H
 d 
H

side is,  C p   uTdy     C p   uTdy   dx ; the energy due to mass flow through the


0
 dx 
0

top is, C pT
H

d 
    udy   dx ; the net viscous work done within this element is,
dx   0

 H  du 2 
T
. Combining all
     dy  dx ; and the heat transfer at the wall is dqw  kdx
y w
dy
 0  

of the above relations into Equation 45 yields,
(46)
2
H

d 
  H  du  
T
    dy   
   (T  T )udy   
dx   0
y

  C p  0  dy 
w
20
However, the net viscous work term is negligible unless we have a highly viscous fluid,
which air is not; therefore we will ignore this term and arrive at,
(47)
H

d 
T
   (T  T )udy    
dx   0
y

w
This is the energy boundary layer equation for our laminar flow over a flat plate. The
next step in calculating the thermal boundary layer thickness is to insert our temperature
profile Equation 44 and the velocity distribution Equation 41 into Equation 47, which
will then look like this,
(48)
H
 d  H

d 
   (T  T )udy       (   )udy  
dx   0
  dx   0

3
3
H
d   3  y  1  y    3  y  1  y   
  u
 1              dy 
dx  0  2  T  2   T    2    2     

 


T
y

w
3
2T
where we have defined  = T – TS. Allow us to assume δT is smaller than δ, which is the
case for air and most fluids, we can change the integrals to sum up to δ instead of H.
Performing the necessary integration we arrive at,
(49)
2
4
d   3  T 
3   T    3  
 u     
   
dx   20    280      2   T 
 
 

 
 
and because T <1 we can assume  T  is negligibly small compared to  T  and we

 
 
4
2
can write
21
(50)
2
3
d      3  
u    T   
20
dx      2  T 
 
 
performing the differentiation gives
(51)

  t
2d

 2 2   t    
1
u 
    dx
10 







3


      t   d  
    dx    


  






But from the development shown in Equation 42 we know,
2
2

140  x
so that we can
13 u
write,
  t
d
 t 
 t    
    4x   
 
   dx


3
(52)
2
  t
d
 t    
but noting that  
    dx


2

  13 

,
 14 



 1 d  3
 t

  we realize that Equation 52 is a first order
 3 dx   


linear differential equation whose solution is
3
13 
 t 
4

Cx

 
14 
 
3
(53)
For our situation, we will assume that the boundary layer starts forming at x=0 but that
the heat of the plate doesn’t start until x=x0. These will give us our two boundary
conditions, namely,
22
(a)
 t  0 @ x  x0
(b)
 t



  0 @ x  x0

We can solve these simultaneously with Equation 53 to yield
(54a)
 t


1/ 3
  x0 3/ 4 
1

1/ 3
  1.026 Pr 1   x   ,

   
or if the heat of the plate starts immediately at x=0,

(54b)  t

1

1/ 3
  1.026 Pr

for Re  5 1010 .
In these equations we noticed a ratio that has been in numerous equations throughout this
development. This ratio is
 CP 

. Looking more deeply into the properties of this

k
ratio we see it is a ratio of the kinematic viscosity. A ratio of this kind of viscosity
conveys information about the rate at which momentum diffuses through a material, and
the thermal diffusivity, which conveys how heat diffuses through a material, each by
molecular vibrations. It is through this ratio that we discover the relative magnitudes of
momentum and thermal diffusivity in the material, which is the information needed to
determine the relative sizes of the momentum and thermal boundary layers. This link
between the velocity field and the temperature field is so important that it was named, as
is shown in Equations 54, Pr, the Prandtl number.
Putting it all Together
Visualize, again, the situation where we have a fluid flowing over a hot plate, shown
below in Figure 6.
23
T8
T8
A8
u8
TS
AS
L
Figure 6: Temperature Profile in thermal boundary layer.
The temperature of the surface is at Ts, the temperature of the fluid is at T∞, and the
boundary layer thickness is at a height of δT. At the wall we have zero velocity due to
viscous forces, thus the heat transfer from the wall into the boundary layer must be done
through conduction. Therefore, our heat flux into the boundary layer must be
(55)
Q
T
 ks
As
y
surface
We can now combine this with Equation 5 and Equation 6 to obtain,
k
(56)
h
T
y
wall
Ts  T

3 k
.
2 t
We can now plug in Equations 54 and Equation 42 to yield,
(57)
hx  .332k Pr
1/ 3
Re x1/ 2
x
  x0 3/ 4 
1    
  x  
1/ 3
However, from our development, we realize that we can use this equation for all similar
geometries. Hence, we wish to nondimensionalize this equation such that it will be
24
applicable to all flat plates that we will encounter. We will divide both sides by x / k
which will give us,
(58)
  x0 3/ 4 
hx x
1/ 3
1/ 2
Nu x 
 .332 Pr Re x 1    
k
  x  
1/ 3
It is at this point that we have defined another useful dimensionless parameter known as
the Nusselt number. As is stated in the above paragraph this makes for an important
parameter. Through the development of a flat plate, we expect that we can create a single
correlation for a given similar geometry and then use dimensionless parameters such that
this correlation can be used for all fluids and all absolute dimensions of the relative
geometry. The Nu, Pr, and Re all provide us with this ability. In fact, from this
development, we can now state with a fair amount of certainty that h is a function of Nu
which is a function of Re and Pr. Thus for all other geometries, we can run numerous
experiments with different fluids and create correlations for h based on different
manipulations of the these three dimensionless parameters. Since most of the geometries
are very difficult to solve analytically, this is the exact procedure we use to obtain the
correlations for a number of turbulent flows in different geometries.
Back to the analysis, we want to extend our relation for the heat transfer coefficient for
the flat plate in forced flow to include one more step. We want to have the average heat
transfer coefficient over the entire plate such that we can use an average equation for the
entire plate instead of integral analysis every time.
From the definition of average, we can calculate
L
 h dx
x
(59)
h
0
L
 dx
 2hL
0
Thus we can plug Equation 59 into Equation 58 and obtain the working relationship for
forced laminar flow over a flat plate as,
25
(60a)
Nu 
hL
 .664 Pr1/ 3 Re x1/ 2
k
1/ 2
L
.332 Pr1/ 3 k  u x 
(60b) h 
0   
L
for Re  5 1010
3/ 4
 1    x0  
1

    
 x    x  
1/ 3
dx for Re  5 1010
The analysis to arrive at this equation assumed that all physical properties were constant.
If this is not the case, it is suggested by many textbooks to use properties at the film
temperature,
(61)
Tf 
Tw  T
.
2
The Flat Plate in Free Convection
Physical Considerations
In our previous discussion, we examined the situation of a flat plate and a fluid with a
forced velocity flowing over it causing forced convection. However, convection can also
result when there are density gradients caused by temperature gradients producing the net
effect of buoyancy forces on the molecules present due to the gravitational field.
Consider the situation represented in Figure 7
ρ = fucn(y)
T8
ρ8
U = func(y)
Ts
Figure 7: Boundary Layer development on heated vertical plate.
26
In this situation, a heated vertical plate at Ts is surrounded by air at T∞. The fluid close to
the plate is at a higher temperature and less dense than fluid far removed. Therefore,
buoyancy forces induce a free convection boundary layer in which the fluid at a higher
temperature rises vertically, entraining fluid from the fluid far removed. The resulting
velocity distribution, shown in Figure 7, is unlike that we have seen before and thus
requires a new development.
The Governing Equations
In order to analyze this situation, we again need the momentum and energy equations that
we derived from the related conservation principles. Thus, for momentum, we return to
Equation 19,
(19)

v x
P
2
 v  v x  
  v x  g x 
t
x
We can, again, simplify this equation by assuming that flow is only in the x direction and
that momentum is changing only in the y direction. However, we cannot ignore the
pressure drop term, as its value is no longer negligible nor can we ignore the body force
term that gravity imparts on our system. Applying these to Equation 19 and dividing
through by  , noting that  
(62)

is the molecular diffusity of the fluid, we arrive at,

u
u
1 p
 2u
u 

 g  2 .
x
y
 x
y
We can rewrite this equation by noting that the x pressure gradient in the boundary layer
must be equal to the pressure gradient in the surrounding fluid, but in the surrounding
fluid u=0. Applying this we can solve Equation 62 for the surrounding fluid to find,
(63)
p
   g
x
Substituting Equation 63 into Equation 62 yields,
27
(64)
    
u
u
 2u
u 
 g
  2 .
x
y
y
  
Note: the new term in this equation represents the buoyancy force present in our system
due to the density gradient. We can make this equation more functional if we assume that
density variations are caused solely by temperature variations, in which case we may take
advantage of the fluid property known as the volumetric thermal expansion coefficient.
This property measures the change in density with a change in temperature at constant
pressure.
(65)
 
1   
  T  p
However, we will further simplify Equation 65 with the Boussinesq approximation,
which allows us to approximate the derivative so that we arrive at,
(66)
1   
1   
  

  T 
 T  T
which can be solved for    and then directly substituted into Equation 64 to yield
the working momentum equation for free laminar flow along a vertical plate,
(67)
u
u
u
 2u

 g  (T  T )  2 .
x
y
y
Again note, that the buoyancy force as we are representing it only has an effect on our
momentum equations, thus we can rewrite our working expressions for the mass and
energy equations exactly how they were for forced convection. This provides us with our
three working equations from which we can solve our free laminar boundary layer,
namely,
28
(68)
u 

0
x y
(69)
u
(70)
 T
T 
 2T
u x   y    y 2 .


u
u
 2u

 g  (T  T )  2
x
y
y
Solving the Governing Equations
If we nondimensionalize the equations, we realize that we can obtain much more
practical expressions for our correlations. Hence, we define the following variables
x 
T  T
x  y  u  
, y  , u  ,  , T  
, where L is a characteristic length and u0
L
L
u0
u0
Ts  T
is an arbitrary reference velocity. Applying these substitutions into Equations 69 and 70
yield the following nondimensionalized free laminar boundary layer equations
(71)

g  (Ts  T ) L 
u 
1  2u 
 u
u


T 
x
y
u0 2
Re L y2
(72)
u


T 
1  2T 
 T



x
y  Re L Pr y 2
where we have a new dimensionless parameter as a result of the buoyancy force,
g  (Ts  T ) L
. However, in its present form it is problematic because it requires some
u0 2
2
u L
reference velocity. Thus, we will multiply this parameter by Re L 2   o  and obtain
  
what is known as the Grashof number,
g  (Ts  T ) L  uo L 
g  (Ts  T ) L3
.
GrL 
   
u0 2
2


2
(73)
29
The Grashof number is the free convection equivalent to the Reynolds number in forced
convection. Where Reynolds was the ratio of inertial to viscous forces, Grashof is the
ratio of the buoyancy force to the viscous force acting on the fluid.
In order to solve Equations 71 and 72 for our situation, we can take two directions. The
first involves us creating differential elements and balancing momentum and forces,
deriving expressions for the velocity and temperature distributions, and finally solving for
a functional relationship of the boundary layer thickness. Such a route is quite tedious
and provides below satisfactory results for its “exact” correlation. The second method we
can take is to recognize that in forced convection, the Nusselt number was a function of
the dimensionless parameters present in forced convection, Pr and Re. In consequence,
we can assume that once again we are going to experience the Nusselt number to be a
function of the dimensionless parameters present and write,
(74)
Nu L  function(Pr,GrL ) .
Then, we simply need to run a multitude of experiments to determine the correct
correlation for our system. This is the method utilized by Churchill and Chu, according to
Welty, Wicks, Wilson, and Rorrer. In their process, they defined a new dimensionless
parameter which is simply a manipulation of our current dimensionless parameters. This
new parameter is called the Rayleigh number and is defined as,
(75)
RaL  GrL Pr 
g  (Ts  T ) L3

.
They then anticipated the functional relationship of Equation 74 would be
(76)
Nu L 
hL
 CRa L n .
k
Churchill and Chu proceeded to run multiple experiments in the range of RaL  109 and
found the working relationship for laminar flow to be,
30
(77)
Nu L  .68 
.670 Ra1/L 4
  .492 9 /16 
1  
 
  Pr  
4/9
.
Finned Surfaces
Physical Situation
It is a frequent practice to add extended surfaces to a flat plate to increase the rate of heat
transfer away from the surface. As shown in Figure 8 below, we have added rectangular
fins to the plate.
dqconvection = hf *P dx * (Tfin @ x - T8 )
P
t
z
dx
L
Figure 8: Conduction and Convection through parallel fin.
Heat generated in the surface is being conducted through the fin and then convected away
to the surrounding fluid. The temperature at the base of the fin is TS and the temperature
of the surrounding fluid is T∞. To begin our analysis of this situation, we must first
conduct an energy balance around our fin.as calculated below
31
Energy conducted in  Energy conducted out  Energy lost by convection
kA
(78)
dT
dT
 kA
dx
dx
 hPdx(T  T )
x  dx
 dT d 2T 
dT
 kA 
 2 dx   hPdx(T  T )
dx
 dx dx

2
d T hP
  0
dx 2 kA
kA
where   T  T . In order to solve the above differential equation, we need two
boundary conditions. To simplify the math, we will assume the tip of the fin is insulated,
which basically means that there is no driving force for heat transfer at the tip which is
mostly the case. Thus, we can now label our two conditions as follows: (a) the
temperature of the fin at its base is equal to the temperature of the surface; and (b) there is
no temperature gradient at the tip (insulated). Mathematically we can write these
conditions as,
(a)
  0 @ x  0
(b)
d
0 @
dx
xL
Solving the above second order linear differential Equation 78 with the two boundary
conditions a and b, we find the temperature distribution along the fin as,
(79)
 cosh(m( L  x))

0
cosh( mL)
where we have defined a new constant, m 2 
hP
, to make the math easierr to follow.
kA
Noting that all of the heat lost by the fin via convection must be conducted into the base
of the fin by conduction, we may write an expression for the total heat lost by the fin as,
32
q  kA
(80)
dT
dx
x 0
1
1 

q   kA 0 m 


2 mL
1  e 2 mL 
 1 e
q  hPkA 0 tanh(mL)
To make this relation more applicable, we would like to combine the heat convected from
the plate and the heat convected from the fin into one equation. In order to do this, we
define the fin efficiency as follows,
(81)
Fin efficiency 
actual heat transferred
f
heat which would be transferred
if entire fin area were
at base temperature
For our case ,we can determine this mathematically as
(82)
f 
hPkA 0 tanh(mL) tanh(mL)

.
hPL 0
mL
For this reason, we can describe convection from a flat plate with fins as,
(83)

q  h0 A0  h f Af  f
 T  T 
s

where h0 A0 are for the flat plate without fins and h f Af  f from our equations will be the
heat convected from the fins. Consequently, all we need to do is measure the areas of the
flat plate and the fins and the temperature of the surface and the surrounding air, and the
total heat produced within the plate to find the heat transfer coefficients of the plates and
of the fins.
Of particular note here, we may also find an estimated heat transfer coefficient of the fins
by using Equation 60, which we developed for flat plates. This is a good approximation
33
because the fins are in essence flat plates but with less uniform temperatures. However,
because the conductivity of the plate is very high, we can assume that the temperature is
fairly uniform.
Establishing Steady State
There were a total of eight runs used for this experiment. Six of these were for forced
convection with the other two for natural convection. As for the latter, steady state values
were obtained by allowing the system to run for an hour and a half. Measurements were
then taken for each variable. Because of the extended run time, these readings were
assumed to be steady state values.
For the six forced convection runs, measurements were taken every 5 minutes for an
hour, or until the readings started leveling out. This data was then inserted into Newton’s
Law of Cooling model to determine the steady state values. From Newton’s Law of
Cooling, we know the rate of change in temperature is proportional to the driving force.
In this case, a constant times the difference between the steady state temperature and the
current temperature, or in mathematical terms,
(84)
dT
 k T  T 
dt
Solving this separable differential equation we arrive at
(85)
T  T  exp  Kt  , K  Constant
Thus, we can graph Temperature versus and guess values of K to give us a good
correlation. The y-intercept of this graph will be the steady state temperature the system
is trying to attain.
34
EQUIPMENT:
For these experiments, the medium of heat transfer was air, which was flowed through a
duct with the use of a pump. The duct (chimney) has a cross sectional area of 120mm by
70mm and stood about 1.5m tall. Air velocity was controlled by the variac controller on
the pump and could be further adjusted by a slide cover in front of the chimney.
Measurements of air velocity were made by a portable analog anemometer mounted on
the duct. Readings were taken in meters per second. A 20 watt power control circuit
manufactured by Armfield Technical Education Co., HTG-B Serial Number 3681-3,
which had a direct reading digital watt meter imbedded in it, heated the surface.
Temperature measurements were taken to a resolution of .10C through the use of
thermistor probes and a digital temperature indicator which read in 0C which is located on
the control circuit. All equipment is shown below in Figures 9 and 10.
Chimney
Viewing Window
Anemo meter
Figure 9: Front view of convection duct.
35
Boundary Layer Profile Measurement
Heated Surface
Inlet Air Measurement
Temperature Probe
20 watts
Pump and Slide
Cover
Power Supply
Figure 10: Side view of convection duct.
There are two surfaces that will be used in this experiment both are made from an
aluminum alloy with an estimated conductivity of 166.5
Watts
. The first surface is a flat
m 0C
plate, with a length of 100mm and a width of 110mm. Subsequently, the second surface
is a finned plate, which has 9 fins that are 4mm thick at the base, 2mm thick at the top,
100mm long (in direction of air flow) and 68mm in depth. These 9 fins have a horizontal
spacing of 12-13mm across a flat plate identical to the first surface. Pictures of these
surfaces are shown below in Figures 11 and 12.
T8
A8
u8
Figure 11: Flat Plate and air flow direction.
36
T8
A8
u8
Figure 12: Finned Plate and air flow direction.
PROCEDURE:
1.
Insert the flat plate in the proper position in the chimney. Plug in the
power supply to the flat plate. Attach the temperature sensor into the flat
plate.
2.
Insert the Temperature probe into the air inlet hole.
3.
Turn on the anemometer. Turn on the pump and adjust the variac
controller and the slide cover to achieve an air velocity of 2.5m/sec.
4.
Turn on the power source. Adjust power output to 20 Watts.
5.
Start taking readings every 5 minutes until the system reaches steady state,
or after an hour of measurement time, which ever comesfirst.
6.
Move the temperature probe into each of the three profile holes near the
middle of the plate. Allow the probe 5 to 10 minutes to reach steady state
and record the reading.
7.
Document all data.
8.
Repeat steps 2 through 7 for air velocities of 5 and 10 m/sec. Then, repeat
steps 2 through 7 once more with the pump turned off to record a
measurement for natural convection. Note this last run may take up to two
hours to reach steady state. Readings do not have to be taken for every 5
minutes as long as you leave enough time for the system to reach steady
state.
9.
Turn the power source off. Unplug the flat plate power source and
temperature sensor. Remove the flat plate. Insert the parallel fins. Plug the
power source and the temperature sensor into the parallel fins.
37
10.
Repeat steps 2 through 8 for the parallel fin setup.
11.
Turn the power supply off.
12.
Clean up the lab.
RESULTS:
Steady State
For the six forced convection runs, we arranged the data to a graph of temperature versus
exp(k  t ) . The data for this can be seen in Appendix 2. The two natural convection runs
had actual steady state measurements taken during their analysis, therefore no
manipulation of data was required to obtain their steady state values. The steady state
values of all runs are shown in Table 1 below.
Table 1: Steady State Temperatures
Flat Plate:
Air Velocity(m/sec)
Air Inlet Temp (°C)
Plate Surface Temp (°C)
0.0
25.5
93.8
2.5
25.7
71.8
5.0
24.9
58.8
10.0
25.0
47.1
Finned Plate:
Air Velocity(m/sec)
Air Inlet Temp (°C)
Plate Surface Temp (°C)
0.0
25.0
47.1
2.5
25.0
31.1
5.0
25.0
29.9
10.0
25.0
28.9
Table 1: Steady state temperature values.
Flat Plate Heat Transfer Coefficient
During this experiment, I determined the heat transfer coefficient across a flat plate at
four different velocities each by three different methods. The first was to measure it
38
experimentally. The second method involved using the theoretical analysis presented in
the theory section of this report to come up with a predicted estimate of the heat transfer
coefficient. Estimating the heat transfer coefficient with the final method is a direct result
of comparisons of the experimental and predicted results. It involves a minor adjustment
to the theoretical prediction equation to account for turbulent mixing.
I will begin to explain my results for the flat plate by comparing the first two methods of
estimating the heat transfer coefficient, experimental and predicted.
In order to determine the experimental heat transfer coefficient, I measured the power
output, the area of the plate, the steady state temperature of the air, and calculated the
steady state temperature of the plate, during the lab. From this data, I used Equation 5 to
determine the average heat transfer coefficient over the plate for all four experiments.
The second calculation involved using a theoretical analysis assuming laminar flow of the
air over the flat plate. During the lab trials, I recorded the velocity of the air, and
calculated the film temperature of the air. With this data, I was able to use Equation 60 to
calculate the average heat transfer coefficient over the plate for the three forced
convection runs and using Equation 77 for the natural convection prediction. Table 2
shows the results of these calculations.
Table 2: Heat transfer coefficients along flat plate (W/sqm deg C)
Flat Plate:
Air Velocity(m/sec)
Experimental
Experimental Error
Theory
Percent Error in Predictions
0.0
26.6
12.1%
6.8
292.8%
2.5
39.4
7.7%
22.0
79.6%
5.0
53.6
6.4%
30.9
73.7%
10.0
78.1
1.4%
44.4
75.9%
Table 2: Heat transfer coefficients along flat plate.
If one were to examine the table , they would be able to note that both experimental and
predicted results follow the same trend of increasing with increasing velocity. In order to
more clearly see the effect that velocity has on the heat transfer coefficient, it is
convenient to graph the data in Table 2. This is shown in Figure 13 below.
39
Heat transfer coefficients versus velocity for laminar flow over flat plate
90
h (Watts per meter squared degrees Celsius)
80
70
60
50
Predicted
Experimental
Linear (Predicted)
40
30
20
10
0
0
0.5
1
1.5
2
2.5
3
3.5
(Velocity (meters per second))^.5
Figure 13: Heat transfer coefficients versus velocity for flat plate.
From this graph, it is clear to see that both the experimental and predicted heat transfer
coefficients increase almost linearly with the root of air velocity. Also, from Figure 13, it
is obvious that the theoretical predicted equation is not providing good results for our
experimental conditions. This conclusion led to the third method of prediction, turbulent
mix predictions. This prediction method is exactly the same as the original theoretical
prediction method except that I use a different coefficient in Equation 60. The new
predicting equation for this method is shown below,
(86)
Nu 
hL
 1.169 Pr1/ 3 Re x1/ 2 .
k
Using this new method for my predicted heat transfer coefficients and comparing it to my
experimental measurements are shown below in Table 3.
40
Table 3: Heat transfer coefficients along flat plate (W/sqm deg C)
Flat Plate:
Air Velocity(m/sec)
Experimental
Experimental Error
Theory
Percent Error in Predictions
0.0
26.6
12.1%
0.0
#DIV/0!
2.5
39.4
7.7%
38.7
2.0%
5.0
53.6
6.4%
54.4
-1.4%
10.0
78.1
1.4%
78.2
-0.1%
Table 3: Heat transfer coefficients along flat plate.
The most important thing to note from this table is the large decrease in percent error in
the predictions. With this new equation, we have obtained an average error of only 1%. In
order to find the functional relationship of the heat transfer coefficient and the air
velocity, it is more suitable to see the data from Table 3 in a graph. This is shown in
Figure 14 below.
Heat transfer coefficients versus velocity for laminar flow over flat plate
90
80
h (Watts per meter squared degrees Celsius)
70
60
50
Turbulent Mix Predicted
Experimental
Linear (Turbulent Mix Predicted)
40
30
20
10
0
0
0.5
1
1.5
2
2.5
3
3.5
-10
(Velocity (meters per second))^.5
Figure 14: Heat transfer coefficients versus velocity for flat plate.
Figure 14 above, illustrates that the new predictive equation also follows the trend of heat
transfer coefficients increasing linearly with the root of air velocity. Also from this graph,
41
one is able to visualize that our experimental data lines up well with the new predicted
data.
Finned Plate Heat Transfer Coefficient
The analysis of the finned plate is similar to that of the flat plate. Once again, we will
compare three determinations of the heat transfer coefficient. The first will be via
experiment. We will measure the temperature of the air, the steady state temperature of
the plate, the area of the plate, and the power input to the plate. This data however is not
enough to solve for the heat transfer coefficient for the fins. According to Equation 83,
we also need the heat transfer coefficient of the plate. I am going to presuppose that this
value is equal to the value found for the flat plate at the same velocity in the previous set
of experiments. Thus with all of this data, we are able to solve for our experimental
value. Next, we will calculate the value of the heat transfer coefficient as predicted from
theory for laminar flow over the plate. This calculation is exactly the same as it was for
the flat plate, with the use of Equation 60. The data for this part of the experiment is
shown below in Table 4
Table 4: Heat transfer coefficients along finned plate (W/sqm deg C)
Finned Plate:
Air Velocity(m/sec)
Experimental
Experimental Error
Theory Percent Error in Predictions
0.0
5.1
20%
6.6
-22.9%
2.5
32.1
13.2%
22.0
45.7%
5.0
41.5
11.8%
31.1
33.2%
10.0
55.8
2.1%
44.0
26.6%
Table 4: Heat transfer coefficients along finned plate.
It is important once again to note the trend and the percent error from this table. As
before both the experimental and predicted heat transfer coefficients increase with
increasing air velocity. In addition, a large error is seen in our predictions from theory
and our experimental results. In order to determine the effect of air velocity on the heat
transfer coefficient, it will be more convenient to see the data in Table 4 in graphical
form. This is presented below in Figure 15.
42
Heat Transfer Coefficients for flow over Finned Plate
h (watts per meter squared degree celcius)
70
60
50
40
Predicted
Experimental
30
Linear (Predicted)
20
10
0
0
0.5
1
1.5
2
2.5
3
3.5
(Velocity (m eters per second))^.5
Figure 15: Heat transfer coefficients versus velocity for finned plate.
Two things become apparent from Figure 15. First, both the experimental data and
predictive data increase with increase velocity and second, our predictions from Equation
60 once again do not provide satisfactory results. Consequently, I tried to change the
coefficient from Equation 60 and create a new turbulent mix predicting equation shown
below.
(87)
Nu 
hL
 .90 Pr1/ 3 Re x1/ 2
k
Note the newly developed predicting correlations, Equations 86 and 87 are not the same.
The next step was to compare the results obtained experimentally with those obtained
from Equation 87. The results of these calculations are shown below in Table 5.
43
Table 5: Heat transfer coefficients along finned plate (W/sqm deg C)
Finned Plate:
Air Velocity(m/sec)
Experimental
Experimental Error
Theory Percent Error in Predictions
0.0
5.1
20%
0.0
#DIV/0!
2.5
32.1
13.2%
29.9
7.4%
5.0
41.5
11.8%
42.2
-1.8%
10.0
55.8
2.1%
59.7
-6.7%
Table 5: Heat transfer coefficients along finned plate.
From this table, it should be noted that our error in predictions have dropped
substantially, as we now only have an average error of 6%. We would like to again graph
this data such that we can more easily see the relationship between the heat transfer
coefficient and the air velocity. This is shown below in Figure 16.
Heat Transfer Coefficients for flow over Finned Plate
70
h (watts per meter squared degree celcius)
60
50
40
Turbulent Mix Predicted
Experimental
Linear (Turbulent Mix Predicted)
30
20
10
0
0
0.5
1
1.5
2
2.5
3
3.5
(Velocity (meters per second))^.5
Figure 16: Heat transfer coefficients versus velocity for finned plate.
From this figure, it should be noted that our newly predicted heat transfer coefficients
follow the same trend of increasing linearly with the root of air velocity. And again, our
experimental data line up quite well with our newly predicted data.
Effectiveness of Fins
44
To satisfy my last objective, I need to compare how much the addition of fins helped with
the heat transfer from a plate. The easiest way to accomplish this is to run the same
amount of power into the flat plate and the finned plate, then compare the difference
between the temperature of the surface with the temperature of the inlet air for each. In
order to analyze this data, it should be noted that at a constant q,
h flat Aflat T flat  h fin Afin T fin . Thus a smaller
T would indicate that better heat transfer
is being established because more of the heat being pumped in must be being carried
away.
Thus in Figure 17 below, I have graphed my T ’s for each run from the fin and the
plate.
Overal Heat Transfet Com parison (a)
80
70
Delta T (degrees Celsius)
60
34 deg
50
Flat Plate
40
Finned Plate
30
40 deg
20
29 deg
18 deg
10
0
0
2
4
6
8
10
12
Velocity (m eters per second)
Figure 17: Temperature Differences recorded for flat and finned plates.
Figure 17 reveals that in all cases the fin has increased the heat transfer rate. However,
another important aspect of Figure 17 is the decreasing difference of T fin with T flat as
45
velocity increases. This implies that the fin is becoming less effective at increasing the
heat transfer rate at higher velocities.
Discussion of Results:
Effectiveness of Air Velocity on Heat Transfer Coefficient
The first objective of my experiment was to find the effect that increasing air velocity
would have on the heat transfer coefficient, both for a flat plate and a finned plate.
Figures 13 and 15 reveal the answer to this correlation. The heat transfer coefficient
increases linearly with the square root of air velocity. This relationship agrees with the
theoretical correlation produced by Equation 60.
This result is logical. An increase in the free stream velocity, will increase the velocity of
the air in the boundary layer. Let us imagine we are an air molecule near the surface of
the plate in the boundary layer. At the beginning of the boundary layer, we are floating by
at a temperature close to that of the free stream temperature. As we float by at a lower
temperature than the plate, we cause heat transfer from the plate to us. This will result in
our temperature increasing, which means that we as we move down the plate, our
temperature gradient becomes less and we are causing less transfer. Thus at 10cm down
the plate, we are causing significantly less heat transfer. However, if we increase our
velocity, then we will be effecting the same temperature gradient in the beginning but not
sit at the same point long enough to absorb as much heat. Instead, we will be replaced
more quickly by the air molecules behind us at the same temperature. Thus, our
temperature gradient 10 cm down the plate will not be as diminished as it was at the
lower velocity. The net result of this will increase the heat transfer coefficient.
This result was in agreement with both the experimental and the theoretical predictions.
The significance of these results is most evident in the design of heat exchangers. If we
are operating a heat exchanger to heat up fluid A by using fluid B, but it is noted that we
need to increase the outlet temperature of fluid A by more than its current value, then we
46
must increase the amount of heat transferred. From the results above, we know that if we
increase the velocity of fluid B, then the average heat transfer coefficient will also rise,
allowing us to transfer more heat to fluid A and thus raising its outlet temperature.
Comparison of Experimental results to theoretical
Forced Convection
The second objective of my experiment is to compare my results obtained from these
experiments with those predicted by theory. From Tables 2 and 4, we see that the
predicted results are not in agreement with the experimental results. For the flat plate,
ignoring natural convection, our average prediction error was around 80%, while for the
finned plate our average error did drop, but still remained relatively high at a value of
35%. Comparing this to the average experimental error which stayed relatively low for
both setups at an average of 6%, it would appear that the predictive equations do not give
excellent results.
However, this may not necessarily be the case. Looking back at Tables 2 and 4, or by
looking at Figures 13 and 15, we can see a couple of important characteristics. First, the
error in the prediction seems to diminish as we increase the velocity. I rationalized this
information to mean that as the flow becomes more turbulent from a Reynolds number
standpoint, though still laminar, then the predicting equation produced better results.
Next, we see a large decrease in experimental error when we move from the flat plate to
the finned plate, however, the error is still substantial. I interpreted this to be the fact that
the boundary layer was forming over a plate that was separated from the surface of the
chimney, extending away from it.
As a result of combining the above observations, I came up with the following
conclusion. Because the flat plate is along the level of the chimney, though a bit raised,
then we must be starting a boundary layer along the chimney before even reaching the
flat plate. Of course ,only the flat plate is heated thus we would have to use Equation 60b
to find the average heat transfer coefficient. However, without doing the math, we can
see that this cannot fully explain what is happening. If this were truly the case ,then from
Equation 42, we know the boundary layer will be thicker over the plate then it would
47
have been from our original assumption of the boundary layer starting at the leading edge
of the plate. But a thicker boundary layer would push the free stream temperature farther
away from the plate, making it harder to transfer heat. Therefore this result would only
lower the predicted heat transfer coefficient and thus increase our error. Something else
must be happening.
It is my hypothesis that the boundary layer is starting along the chimney, however,
because the flat plate is a bit raised, it is causing turbulence in the boundary layer when it
reaches the flat plate. This may cause the mixing in the boundary layer to be turbulent
even though it is still at a low Reynolds number as shown below in Figure 18.
Turbulence begins because the
boundary layer formed over the
duct wall hits the edge of the
flat plate which is slightly
raised, thus disturbing the molecules in the boundary layer to
form a turbulent mix
u8
u8
T8
A8
u8
Duct Wall
Figure 18: Turbulence over flat plate.
This is not a far stretch to explain my results.. As people have run the Reynold’s
experiment described earlier and found turbulence at low Reynolds numbers by adding
factors to increase mixing. Thus if we experience better mixing in the boundary layer, we
will increase the heat transfer coefficient because we will ensure that molecules farther
up in the boundary layer at a lower temperature are being forced closer to the plate. This
would increase the average temperature gradient close to the plate and result in higher
experimental heat transfer coefficient as compared to if the flow was perfectly laminar.
48
To test this theory with my data I had to make a conjecture. Because the Reynolds
probably remains at the laminar level, assuming the velocity of molecules in the turbulent
mixing boundary layer haven’t changed much, than the correlation of Equation 60 that
the Nusselt number is a function of Reynolds to the one half power and Prandtl to the one
third power probably remains true. However, due to the turbulent mixing I feel that the
coefficient of this equation should increase, because the increase of the average
temperature gradient near the plate should proportionally raise the heat transfer
coefficient. I fitted the data using
(88)
Nu 
hL
 function(Pr1/ 3 Re L1/ 2 )  Constant  Pr1/ 3 Re x1/ 2 .
k
From the analysis of the flat plate I obtained,
(86)
Nu 
hL
 1.169 Pr1/ 3 Re x1/ 2 .
k
I then extended this analysis to the finned plate. For this situation, the boundary layer
could not have started early because the fins are protruding into the air with no surface
before them. However, as a result of their blunt, non aerodynamic shape, they might be
causing the incoming air to also form a turbulent mix boundary layer. When the air hits
the fins, as shown in Figure 19 below, it will cause the stream lines to form eddy circles
which result in a turbulent mix at the low Reynolds number.
49
Turbulence begins because the
incoming air will h it the blunt
side of the fin, causing the
mo lecules to be disturbed in
many different directions. This
causes the boundary layer to
have a slight turbulent mix.
u8
T8
A8
u8
Figure 19: Turbulence around fins.
Thus, I performed the same analysis as before using Equation (88) for the finned plates I
obtained,
(87)
Nu 
hL
 .90 Pr1/ 3 Re x1/ 2
k
Returning to the comparison of the experimental results with the new predictive
equations using turbulent mixing, we have Figures 14 and 16 as well as Tables 3 and 5.
Our new predictive equations yield us excellent results when compared with those
obtained from experiment.
With so few data points, it is hard for me to tell whether these equations are truly
accurate. However, a couple of features of these equations do give me confidence in
declaring these equations as a more accurate model to what is happening in this lab. First,
both of the constant increased. This agrees with my theory that mixing is turbulent and is
increasing heat transfer proportionately. Second, the constant for the flat plate is larger
than the constant for the small plate. This is logical because they are both experiencing a
different type of situation when they become a turbulent mix. The flat plate has a built up
50
boundary layer which is being distorted allowing it to already be built up on the plate as a
turbulent mix layer. However, the fins boundary layer only forms once it hits the fins’
leading edge, causing it to alleviate some of the turbulence into the free stream air above
the plate instead of retaining it all in the boundary layer. Subsequently, the fins boundary
layer will be less of a turbulent mix and thus it has a lower coefficient.
In summary, the theoretically derived predictive equations, do not provide excellent
results. However, the defining equations used to develop these equations, Equation 88,
which led to Equations 86 and 87, does provide excellent results. Therefore, although we
probably cannot increase accuracy above 30% without building the exact situation, we
can quickly develop equations which can achieve an accuracy of about 15% by running a
simple convection experiment.
Natural Convection
Again, looking back at Tables 2 and 4, or by looking at Figures 13 and 15, we see that
our predictive equations for natural convections, also, do not provide satisfactory results.
The error over the flat plate is about 300%, while the error for the finned plate was close
to an acceptable value of -22%. In all probability, these percent errors are not giving the
true accuracy of the equation because of only one data point for each measurement. But
from the large error over the flat plate, it is my conclusion that the predictive equations
for natural convection will not provide good results.
I don’t have an answer as to why the predictive results do not match the experimental
results. My only guess would be the assumptions used to create the natural convection
models probably aren’t applicable to all situations. Therefore if an engineer wishes to use
natural convection in his process, he should try to determine the heat transfer coefficients
from small scale experiments of his situations rather than use the predictive equations.
However, these conclusions are only a result of one data point for each situation and thus
I would need more data to make a more solid judgment.
On the other hand, this doesn’t mean that the results from this experiment are useless.
Although, we weren’t able to accurately predict the values of the heat transfer coefficient
at natural convection, we were able to prove that natural convection is present. Our
51
experiment has concluded that the limiting value to convection is not zero at zero velocity
as forced convection theory predicts. but that heat transfer still takes place, and at a
sufficient level, when there is a zero velocity atmosphere.
The significance of the above results is to enlighten engineers about natural convection. If
they are armed with this information they would be prepared and can account for its
presence in their design if it calls for a piece of equipment at a certain temperature. Or
more importantly, if their design only requires a small amount of heat transfer to keep an
electronic device at a low temperature, then they can save money by not installing a
pump to create forced convection, but instead let the density gradient due the work for
free in creating the necessary heat transfer rate.
Effectiveness of Adding Fins to a Flat Plate
My final objective was to determine how much the fins helped in the heat transfer from
the flat plate at the same velocity. From Figure 17, we see the temperature differences for
both the finned plate and the flat plate. Since the heat being convected away from each is
a constant, the lower T , signifies a greater heat transfer rate has been achieved. From
the first two data points, it appears that the fins become more effective moving from
natural convection to forced convection. With the data from this figure, we see that for
every run the fins provide a faster heat transfer rate than the flat plate. However, this is
fairly obvious because we are adding more area for heat transfer.
Another, and more important issue, is the trend we see in the effectiveness of the fins. For
all of our runs at constant q, we have, h flat Aflat T flat  h fin Afin T fin . For the fins, we have
increased area to increase the heat transfer rate. This provided us with a lower T .
However, as our air velocity increased, the increase in both h flat and h fin , to obtain higher
heat transfer rates overshadowed the increased area provided by the fins. Thus, the effect
of the extra area provided by the fins dropped. Accordingly as the air velocity increased,
the heat transfer rate of the flat plate greatly increases, however the heat transfer rate of
the fins does increase, but not as dramatically. Therefore, the effectiveness of adding the
fins drops as we increase the air velocity.
52
If an engineer is trying to decide whether he should add the extra cost of installing fins to
his surface, the variable he should look at is his operating fluid velocity. If the velocity is
high, the gain in heat transfer rate will probably not be worth the cost. However, if he is
operating at a fairly low fluid velocity, it will probably be worth the cost to add the fins,
because the increase in heat transfer will be substantial.
Conclusions:
This experiment provided four results that are very important in the life of a process or a
design engineer. First, this experiment proved that the heat transfer coefficient increased
linearly with an increase in the square root of air velocity. This is shown in Figures 13
and 15. The significance of this result is most obvious in the design of heat exchangers. If
the engineer has an old exchanger that he would like to use and performs all of the
calculations at a given flow rate but finds that he is not providing enough fluid to heat up
the cold product, he now knows that he only needs to increase the flow rate of the heating
fluid to increase the heat transfer between them. However, I should note that an increase
in velocity usually also means an increase in pressure drop, the engineer should be aware
of this in making his design.
Secondly, this experiment has proven the theoretically derived equations for convection
but have large errors associated with them in trying to predict the absolute values of the
heat transfer coefficient if not applied to the exact geometry described with their
assumptions as shown in Tables 2 and 4. Nonetheless, the equations are useful for
predicting the trend. It is recommended to first run small scale experiments to obtain a
constant fudge factor which can be applied to the correlation stated in the predictive
equations. This was shown to reduce error to an acceptable level as shown in Tables 3
and 5. The significance of this result is it allows engineers to use correlations of heat
transfer coefficients for the design of heat exchangers. It would be too expensive for the
engineer to try multiple heat exchangers, run tests and pick the right one. Thus, the
engineer must run through the equations derived for convection heat transfer in heat
exchangers to predict the type, length, and area of his heat exchanger. This data proves
that such correlations will yield acceptable results.
53
The third major result of this experiment is the effectiveness of fins. At all velocities, this
extended surface provided a great heat transfer rate. Therefore, if the engineer wishes to
achieve a large heat transfer rate, but is limited in his length/area requirement, he will be
able to simply add fins to the surface. However, he should note that the effectiveness of
the added fins decreases as the velocity increases, as is shown in Figure 17. The
significance of this result comes into play when an engineer is designing a heat exchanger
which must dissipate a certain amount of heat to the surroundings. Before he chooses his
design, the engineer should check the velocity of the flowing fluid. If the velocity is high,
the added fins will probably not be worth the cost. If however, the velocity is low, the
cost of adding the fins will be overshadowed by the large increase in the heat transfer
rate.
Finally, the last result is that the heat transfer coefficient is a function of geometry only
not Area. This was first noted in the theory section when Area dropped out and all of the
correlations were made for similar geometries. This was, again, proven in the experiment
when the same predictive equation, Equation 60 was used for the flat plate, at an area of
.011 m 2 and for the fins of the finned plate, each at an area of .0068 m 2 . Both of the
adjusted equations based from Equation 60 provided great results, thus the geometries of
the flat plate and the fins (flat surface) allowed us to use the same equation, the difference
in their areas had no effect.
54
Recommendations:
In my experiment I had trouble gaining too much confidence in my conclusions because
of the lack of data points that I was able to attain. Thus for the next time this experiment
is run I have a few suggestions for the experimenter. First he might want to concentrate
on getting all 8 runs on the flat plate to attain enough data to ensure himself of the
correlations. Or he might want to keep the same 4 runs on a flat plate, 4 runs on an
extended surface, but instead try to make both the fins and the flat plate, more
aerodynamic thus to ensure a fully laminar boundary layer will develop.
55
Nomenclature:
A=Area=(m 2 )
 J 
Cp  heat capacity= 

0
 kg C 
f =Fin
Gr=Grassof Number
m
g=Acceleration due to gravity=  2 
s 
 W 
h=heat transfer coefficient=  2 0 
m C
 W 
k=Conductive heat transfer coefficient=  0 
m C
L=Length=(m)
Nu=Nusselt number
 N 
P=pressure=  2 
m 
Pr=Prandtl number
q=rate of heat transfer=(W)
Ra=Rayleigh number
Re=Reynolds number
s
=surface
T=Temperature=  0 C 
t=time=(s)
t
=thermal
m
u=velocity=  
s 
V=Volume=  m 3 
w
=wall
x=vertical position=(m)
y=horizontal position=(m)
56
 m2 
α=thermal diffusivity= 

 s 
β=volumetric thermal expansion coefficient=  K -1 
δ=boundary layer thickness=(m)
η=fin efficiency
 kg 
μ=viscosity= 

ms
 m2 
ν=momentum diffusivity= 

 sec 
θ=T-T =  0 C 
 kg 
ρ=density=  3 
m 
 N 
2 
m 
  shear stress= 
 m3 
υ=specific volume= 

 kg 
Δ=Change in a variable=final-initial

=Property at free stream, far removed from system
 Average Property
57
References:
1.
Centre for the Analysis and Dissemination of Demonstrated Energy Technologies.
Energy Conservation in the Pulp and Paper Industry. CADDET Analyses Series.
March 28, 2001.
2.
Holman J.P. Heat Transfer 9th Edition. McGraw Hill, Inc. (2002)
3.
Icoz and Jaluria. Rutger University. Design of Cooling Systems for Electronic
Equipment Using Both Experimental and Numerical Inputs. Journal of
Electronics Packaging. May 31, 2004.
4.
McCabe, W. L. and Smith, J. C. Unit Operations of Chemical Engineering, 3rd
Edition, McGraw-Hill Book Company, 1976.
5.
Principles of Flow. Ed. Alicat ScientificTM. 5 July 2005.
http://www.alicatscientific.com/flow_principles.php
6.
Welty, J.R., et al. Fundamentals of Momentum, Heat, and Mass Transfer – 4th Ed.
John Wiley & Sons, Inc. New York. (2001)
58
APPENDIX:
Appendix 1: Sample Calculations
Run1: Flat Plate Low velocity
Given Variables:
Area  .011 m 2  measured 
Length  .1 m  measured 
W
 tabulated 
m 0C
q  20 W  measured 
k plate  166.5
u  2.5
m
 measured 
s
Experimental Data
Data Points Time (min) Air InletTemperature (°C) Plate Surface Temperature (°C)
exp(-kt) K
1
0
25.7
1.00
0.05
2
5
25.7
35.5
0.76 R Squared
3
10
25.7
45.7
0.58
4
15
25.7
51.2
0.44
5
20
25.7
56
0.33
6
25
25.7
59.5
0.25
7
30
25.7
62.3
0.19
8
35
25.7
64.6
0.15
9
40
25.7
66.3
0.11
10
45
25.7
67.8
0.08
11
50
25.7
69.1
0.06
12
55
25.7
70
0.05
1.00
Table 6: Experimental data obtained for flat plate, at 2.5 meters per second.
Calculating Steady State Temperature:
The time, air inlet temperature, and the plate surface temperature were measured
during the lab. Then a K value was assumed to be one. Then data was calculated
of the form exp  K * t  . Then the Temperature of the Surface was graphed
versus exp  K * t  and then sent through a linear regression through excel. Then
59
K was goalseeked until the correlation coefficient was closest to unity. The graph
below shows the final result of this calculation.
Temperature of Surface
(deg C)
Establishing Steady State
80
70
60
50
40
30
20
10
0
0.00
Series1
Linear (Series1)
y = -47.016x + 71.8
R2 = 0.9981
0.20
0.40
0.60
0.80
exp(-K*t)
Figure 20: Steady state calculation for the first run.
From Equation85, we calculate,
Ts , steady state  71.80 C
Calculating Film Temperature
T film 
Ts , steady state  T
2

71.8  25.7
 48.75
2
Calculating Properties at Film Temperature:
W
 tabulated 
m0C
W
kair  .031 0  tabulated 
mC
kg
air  1.94 105
 tabulated 
ms
kg
 air  1.102 3  tabulated 
m
Pr f  .703  tabulated 
k plate  166.5
Calculating Experimental h:
From Equation 1 we have,
60
q
 h(Ts  T )
A
q
h
A(Ts  T )
h
20W
.011m 2  71.8  25.7  C
0
hexp  39.44
W
m 0C
Calculating Predictive h:
From Equation 8 we have
Re L 
 uL


kg
m
 2.5  .1m
3
m
s
 14394
5 kg
1.94 10
ms
1.102
And from Equation 60
hL
 .664 Pr1/ 3 Re L1/ 2
k
.664 Pr1/ 3 Re L1/ 2 k
h
L
Nu 
.664  .703
1/ 3
 14394 
h
hlaminar  21.96
for Re  5 1010 x0  0
1/ 2

W 
  .031 0 
m C

.1m
W
m 2 0C
Calculating Predictive Error:
hexp  hlaminar
100%
h laminar
39.44  21.96
Prediction Error 
100%
21.96
Prediction Error  79.6%
Prediction Error 
Calculating Experimental Error:
The error from q is derived from the fluctuations of the value read during the
experiment, the error from A was determined based on the accuracy of reading a
ruler, and finally the error in temperatures was given from the manufacturers of
the thermisters.
61
q  hA Ts  T 
h
q
A Ts  T 
h 
q A Ts T



q
A
Ts
T
1.4 .0005
.1
.1



20 .011 71.8 25.7
 h  .1207  12.1%
h 
Calculating the turbulent mix coefficient for Nusselt number:
Based on the larger error associated with using Equation 60 as my prediction
equation, I moved to calculate my own equation based on turbulent mixing over
the flat plate. This equation uses the same correlation as equation 60 but has an
adjusted coefficient to account for the turbulent mixing. In order to calculate this
coefficient, I used the percent error from all three forced convections runs and
summed them. Then I sought to minimize this error by guessing different values
of the coefficient, using goalseek in Excel. The data from this calculation appears
below.
Forced Convection Run #
New coefficient Re
Pr
h est
h exp
error
1
1.168197768 14393.94
0.703 38.63252 39.4399527 0.807435
2
1.168197768 29362.59
0.705 54.33877 53.633682 0.705084
3
1.168197768 60632.94
0.706 78.12178 78.1217878 1.14E-05
1.512531
Table 7: Experimental data obtained for flat plate.
Run5: Finned Plate Low velocity
Given Variables:
62
Area plate  .011 m 2  measured 
Length  .1 m  measured 
Length fins  .1m (measured )
Depth fins  .068m (measured )
Perimeterfins  .206m (measured )
W
 tabulated 
m 0C
q  21 W  measured 
k plate  166.5
m
 measured 
s
W
 39.44 2 0  calculated 
m C
u  2.5
hplate
Calculating heat transfer coefficient and efficiency
From our previous development we have, Equations 82 and 83,

q  h0 A0  h f A f  f
f 
m2 
 T  T 
s

hPkA 0 tanh(mL) tanh(mL)

hPL 0
mL
hf P
kA
where A0 is the free are of the flat plate. This area is the total area of the flat plate
minus the area of the 9 fins sitting on the flat plate. Thus we have,
A0  Aplate  Af  .011  9*.0004  .0074
The only unknown in the above equations is hf, however, these equations are too
complex to try and solve for hf explicitly. The calculation of hf is therefore a trial
and error procedure solving all three of these equations simultaneously. Thus we
have,
63


W
0
21W   39.44 2 0  .0074m 2  h f .1224m 2 f   31.06  25  C
m C


h f  .206m
 .1m)
W
2
166.5 0  .1224m
mC
h f  .206m
 .1m
W
2
166.5 0  .1224m
mC
tanh(
f 
And from Excel we obtain,
h f  32.08
W
m 2 0C
 f  .836
Fin Effectiveness for Runs 1 and 5
Given Variables:
Ts , fin  31.10 C
T , fin  250 C
Ts , flat  71.80 C
T , flat  25.10 C
Effectiveness of adding fins:
T fin  Ts , fin  T , fin
T fin  31.10 C  250 C  6.10 C
T flat  Ts , flat  T , flat
T flat  71.80 C  25.10 C  46.10 C
Effectiveness compares T flat  T fin  46.10 C  6.10 C  400 C
Reviewed and Validated by:
Thomas Salerno
____________________________
Jennifer DiRocco
____________________________
Greg Rothsching
____________________________
Stephen Johnson
____________________________
64
Appendix 2: Raw Data
Run 1: Flat Plate Low Velocity
Air Speed (m/sec)
Power Input (Watts)
Area (m*m)
2.5k plate
20L plate (m)
0.011Air Viscosity
Steady State Values
166.5Air Density
0.1Pr
0.000019423Tf
1.102
0.703
48.75
Plate Surface
Temperature
Time (min) Air InletTemperature (°C) (°C)
Fin T1 (°C) Fin T2 (°C)
25.7
71.8
+-0.5
Data Points
Experimental Data
Data Points
Plate Surface
Temperature
Time (min) Air InletTemperature (°C) (°C)
1
0
25.7
exp(time) K
1.00
0.05
2
3
5
10
25.7
25.7
35.5
45.7
0.76 R Squared
0.58
1.00
4
5
6
7
8
9
10
11
12
15
20
25
30
35
40
45
50
55
25.7
25.7
25.7
25.7
25.7
25.7
25.7
25.7
25.7
51.2
56
59.5
62.3
64.6
66.3
67.8
69.1
70
0.44
0.33
0.25
0.19
0.15
0.11
0.08
0.06
0.05
Temperature of Surface
(deg C)
Establishing Steady State
80
70
60
50
40
30
20
10
0
0.00
Series1
Linear (Series1)
y = -47.016x + 71.8
R2 = 0.9981
0.20
0.40
0.60
0.80
exp(-K*t)
65
Run 2: Flat Plate Med Velocity
Air Speed (m/sec)
Power Input (Watts)
Area (m*m)
Data Points
5k plate
20L plate (m)
0.011Air Viscosity
Steady State Values
166.5Air Density
0.1Pr
0.00001914Tf
1.124
0.705
41.85
Plate Surface
Time (min) Air InletTemperature (°C) Temperature (°C)
Fin T1 (°C) Fin T2 (°C)
24.9
58.8
Experimental Data
Data Points
Plate Surface
Time (min) Air InletTemperature (°C) Temperature (°C)
exp(time)
1
0
25
69.6
1
0.0581
6
5
24.9
66.3 0.747888
11
10
63.9 0.559337 0.995365
16
15
63.1 0.418322
21
20
25.1
61.9 0.312858
26
25
61.2 0.233983
31
30
60.6 0.174993
36
35
60.1 0.130875
41
40
59.7 0.09788
46
45
24.8
59.5 0.073203
51
50
59.3 0.054748
56
55
59.2 0.040945
67
66
65
y = 9.821x + 58.805
R2 = 0.9954
64
63
Series1
Linear (Series1)
62
61
60
59
58
0
0.2
0.4
0.6
0.8
66
Run 3: Flat Plate High Velocity
Air Speed (m/sec)
Power Input (Watts)
Area (m*m)
Data Points
Data Points
10k plate
166.5Air Density
1.14475
19L plate (m)
0.1Pr
0.706
0.011Air Viscosity
0.00001888Tf
36.055
Steady State Values
Air
InletTemperature Plate Surface
Time (min)
(°C)
Temperature (°C)
Fin T1 (°C) Fin T2 (°C)
25
47.11
Time (min)
1
6
11
16
21
26
31
36
41
46
51
56
61
Experimental Data
Air
InletTemperature Plate Surface
(°C)
Temperature (°C)
0
24.9
5
10
15
25.1
20
25
30
25.2
35
40
45
50
55
60
exp(time)
58.5
53.1
50.7
49.2
48.2
47.8
47.5
47.3
47.2
1 0.110469
0.575598
0.331313 0.999026
0.190703
0.109768
0.063182
0.036368
0.020933
0.012049
0.006935
0.003992
0.002298
54
53
y = 10.53x + 47.111
R2 = 0.999
52
51
Series1
50
Linear (Series1)
49
48
47
46
0
0.2
0.4
0.6
0.8
Run 4: Flat Plate Natural Convection
Air Speed (m/sec)
Power Input (Watts)
0k plate
20L plate (m)
166.5Air Density
0.1Pr
1.06025
0.701
67
Area (m*m)
0.011Air Viscosity
Steady State Values
0.00001996Tf
Plate Surface
Time (min)Air InletTemperature (°C)Temperature (°C) Fin T1 (°C)
25.5
93.8
Data Points
59.65
Fin T2 (°C)
Experimental Data
Plate Surface
Time (min)Air InletTemperature (°C)Temperature (°C) exp(time)
1
0
25.2
47.3
1 0.035255
6
5
55.6 0.838388494
11
10
60.6 0.702895267 0.999246
16
15
21
20
26
25
31
30
36
35
41
40
46
45
51
50
56
55
61
60
Data Points
87
25.5
91.75 0.046553076
37.1
Run 1: Fin Plate Low Velocity
Air Speed (m/sec)
Power Input
(Watts)
Area
Data Points
Data Points
2.5Efficiency Estimated
Te
0.236445618Po fin
0.208Ao fin
20Efficiency Calculate
0.836044514Pl fin
0.204Al fin
0.011Area fin
0.1224k fin
166.5L fin
Steady State Values
Air InletTemperature
Plate Surface
Fin T2
Time (min)(°C)
Temperature (°C)
Fin T1 (°C) (°C)
Fin T3 (°C)
25
31.062
26.6
26.5
26.2
Experimental Data
Air InletTemperature
Plate Surface
Fin T2
Time (min)(°C)
Temperature (°C)
Fin T1 (°C) (°C)
Fin T3 (°C)
1
0
6
5
25.5
28.5
11
10
30.1
16
15
30.7
21
20
30.9
26
25
31
31
30
26
31
26.6
26.5
26.2
36
35
41
40
46
45
51
50
68
56
61
55
60
31.5
31
30.5
30
Series1
y = -6.4873x + 31.062
R2 = 0.9992
29.5
Linear (Series1)
29
28.5
28
0
0.1
0.2
0.3
0.4
0.5
Run 2: Fin Plate Med
Velocity
Air Speed
(m/sec)
Power Input
(Watts)
Area
Data Points
Efficiency
5Estimated
0.350340136Po fin
0.208Ao fin
20Efficiency Calculate
0.799479113Pl fin
0.204Al fin
0.011Area fin
0.1224k fin
166.5L fin
Steady State Values
Air
Time InletTemperature Plate Surface Temperature Fin T1
Fin T2
Fin T3
(min) (°C)
(°C)
(°C)
(°C)
(°C)
25
29.9
26.8
26.75
26.6
Experimental Data
Air
Time InletTemperature Plate Surface Temperature Fin T1
Fin T2
Fin T3
Data Points (min) (°C)
(°C)
(°C)
(°C)
(°C)
1
0
31
6
5
29.9
11
10
29.9
16
15
26.4
29.9
21
20
29.9
26.8
26.8
26.6
69
26
31
36
41
46
51
56
61
25
30
35
40
45
50
55
60
Run 3: Fin Plate High
Velocity
Air Speed
(m/sec)
Power Input
(Watts)
Area
Data Points
Data Points
10Efficiency Estimated
0.418492739Po fin
0.208Ao fin
20Efficiency Calculate
0.750989434Pl fin
0.204Al fin
0.011Area fin
0.1224k fin
166.5L fin
Steady State Values
Air
Time
InletTemperature Plate Surface
Fin T1 Fin T2 Fin T3
(min)
(°C)
Temperature (°C)
(°C)
(°C)
(°C)
25
28.879
26.7
26.6 26.57
Time
(min)
1
0
6
5
11
10
16
15
Experimental Data
Air
InletTemperature Plate Surface
(°C)
Temperature (°C)
26.4
Fin T1
(°C)
27.2
28.6
28.8
28.9
26.7
Fin T2
(°C)
26.6
Fin T3
(°C)
26.57
70
29
28.8
28.6
28.4
y = -1.6795x + 28.879
R2 = 0.999
28.2
Series1
Linear (Series1)
28
27.8
27.6
27.4
27.2
27
0
0.2
0.4
0.6
0.8
1
1.2
Run 4: Fin Plate Natural Convection
Air Speed
(m/sec)
Power Input
(Watts)
Area
Efficiency
Estimated
0.925287356Po fin
0.208Ao fin
Efficiency
Calculate
0.969085768Pl fin
0.204Al fin
0.011Area fin
0.1224k fin
166.5L fin
Steady State Values
Air
InletTemperature Plate Surface
Data Points Time (min)(°C)
Temperature (°C) Fin T1 (°C) Fin T2 (°C) Fin T3 (°C)
25
59.8
58.5
58
55.1
Experimental Data
Air
InletTemperature Plate Surface
Data Points Time (min)(°C)
Temperature (°C) Fin T1 (°C) Fin T2 (°C) Fin T3 (°C)
1
0
29.4
6
5
34.5
11
10
38.6
16
15
21
20
26
25
31
30
36
35
41
40
46
45
51
50
71
56
61
55
60
26.4
5
8
.
59.85
58
55.1
72