Analysis and Comparison of the Performance of Concurrent and
Countercurrent Flow Double Pipe Heat Exchangers
by
David Onarheim
An Engineering Project Submitted to the Graduate
Faculty of Rensselaer Polytechnic Institute
in Partial Fulfillment of the
Requirements for the degree of
MASTER OF ENGINEERING IN MECHANICAL ENGINEERING
Approved:
_________________________________________
Ernesto Gutierrez-Miravete, Engineering Project Adviser
Rensselaer Polytechnic Institute
Hartford, CT
MAY 2012
(For Graduation May 2012)
.
© Copyright 2012
by
David Onarheim
All Rights Reserved
ii
CONTENTS
Analysis and Comparison of the Performance of Concurrent and Countercurrent Flow
Double Pipe Heat Exchangers ...................................................................................... i
LIST OF TABLES ............................................................................................................ vi
LIST OF FIGURES ......................................................................................................... vii
LIST OF SYMBOLS ......................................................................................................... x
ACKNOWLEDGMENT ................................................................................................. xii
ABSTRACT ................................................................................................................... xiii
1. Introduction and Background ...................................................................................... 1
1.1
Heat Exchanger Analysis Theory....................................................................... 1
1.1.1
Log Mean Temperature Difference ........................................................ 3
1.1.2
Heat Exchanger Effectiveness (ε) .......................................................... 3
1.1.3
NTU Method .......................................................................................... 3
1.1.4
Thermal Entrance Length in Pipe Flow ................................................. 4
1.2
Description and History of Previous Graetz Problem Solutions........................ 6
1.3
Finite Element Analysis Theory ........................................................................ 7
2. Problem Description and Methodology ....................................................................... 9
2.1
Defining Material Properties .............................................................................. 9
2.2
Methodology and Approach ............................................................................... 9
2.2.1
Finite Element Analysis Modeling ........................................................ 9
2.2.2
Defining Variable Temperature and Velocity ...................................... 10
3. Results and Discussion .............................................................................................. 11
3.1
The Modified Graetz Problem Results ............................................................. 11
3.1.1
The Modified Graetz Problem COMSOL Model ................................ 11
3.1.2
The Modified Graetz Problem COMSOL Mesh .................................. 14
3.1.3
The Modified Graetz Problem Study Results ...................................... 18
3.1.4
The Modified Graetz Problem Calculations ........................................ 19
iii
3.1.5
3.2
3.3
3.4
3.5
Turbulent Flow with Constant Wall Temperature ............................... 21
Flow in a Pipe with Wall Conduction .............................................................. 25
3.2.1
Laminar Flow with a Pipe Wall COMSOL Model .............................. 25
3.2.2
Laminar Flow with a Pipe Wall Problem Calculations ........................ 26
3.2.3
Turbulent Flow in a Pipe with Wall Conduction ................................. 28
Flow in a Concurrent Flow Heat Exchanger .................................................... 31
3.3.1
Laminar Flow in a Concurrent Heat Exchanger COMSOL Model ..... 31
3.3.2
Turbulent Flow in a Concurrent Heat Exchanger ................................ 34
3.3.3
Laminar Flow in a Concurrent Heat Exchanger Problem Calculations 37
Flow in a Counter-Current Heat Exchanger..................................................... 44
3.4.1
Laminar Flow in a Counter-current Heat Exchanger COMSOL Model
.............................................................................................................. 44
3.4.2
Turbulent Flow in a Counter-Current Heat Exchanger ........................ 47
3.4.3
Laminar Flow in a Counter-current Heat Exchanger Problem
Calculations .......................................................................................... 50
Fouling on Heat Transfer Surfaces .................................................................. 54
3.5.1
Flow in a Concurrent Flow Heat Exchanger with Fouling .................. 55
3.5.2
Laminar Flow in a Concurrent Heat Exchanger with Fouling COMSOL
Model ................................................................................................... 55
3.5.3
Laminar Flow in a Concurrent Heat Exchanger with Fouling Problem
Calculations .......................................................................................... 56
3.5.4
Laminar Flow in a Counter-current Heat Exchanger with Fouling
COMSOL Model .................................................................................. 61
3.5.5
Laminar Flow in a Counter-current Heat Exchanger with Fouling
Problem Calculations ........................................................................... 62
4. Conclusion ................................................................................................................. 65
5. References.................................................................................................................. 66
6. APPENDIX................................................................................................................ 67
6.1
Laminar Concurrent Flow Heat Exchanger Data ............................................. 67
6.2
Laminar Counter-Current Flow Heat Exchanger Data .................................... 69
iv
6.3
Laminar Concurrent Flow Heat Exchanger with Fouling Data ....................... 71
6.4
Laminar Counter-Current Flow Heat Exchanger with Fouling Data-Fouling
Layer .001 m .................................................................................................... 73
6.5
Laminar Counter-Current Flow Heat Exchanger with Fouling Data-Fouling
Layer .004 m .................................................................................................... 75
v
LIST OF TABLES
Table 1: COMSOL Model Initial Conditions .................................................................. 11
Table 2: User Defined Material Properties ..................................................................... 12
Table 3: Modified Graetz Problem COMSOL Equations .............................................. 13
Table 4: Mesh Extension Study Results ......................................................................... 17
Table 5: Modified Graetz Problem Comparison ............................................................ 20
Table 6: Results from Mesh Refinement ........................................................................ 21
Table 7: Change in Temperature Along the Channel of Water ...................................... 27
Table 8: Fluid Properties ................................................................................................ 38
vi
LIST OF FIGURES
Figure 1: Basic Heat Exchanger Design [5] ..................................................................... 2
Figure 2: Graetz Problem Temperature Profile [12]......................................................... 4
Figure 3: Nusselt Number for Various Pr Numbers [12] ................................................. 6
Figure 4: Modified Graetz Problem Geometry ............................................................... 12
Figure 5: Modified Graetz Problem Velocity Profile ..................................................... 14
Figure 6: Modified Graetz Problem Temperature Profile .............................................. 14
Figure 7: User Defined Mesh ......................................................................................... 16
Figure 8: Centerline Temp vs. Mesh Element Number .................................................. 17
Figure 9: Initial Value Variance ..................................................................................... 18
Figure 10: Modified Graetz Problem Centerline Temperature....................................... 19
Figure 11: Laminar Flow Velocity Profile ..................................................................... 22
Figure 12: Turbulent Flow Velocity Profile ................................................................... 22
Figure 13: Turbulent Flow Centerline Temperature ....................................................... 23
Figure 14: Turbulent Model for the Modified Graetz Problem ...................................... 24
Figure 15: Velocity Profile for Flow Through a Pipe ..................................................... 25
Figure 16: Temperature Profile of Flow Through a Pipe ............................................... 26
Figure 17: Outflow Temperature Distribution for the Modified Graetz Problem .......... 28
Figure 18: Outflow Temperature Distribution for the Modified Graetz Problem with a
Pipe Wall ......................................................................................................................... 28
Figure 19: Velocity Profile for Laminar Flow in a Pipe ................................................. 29
Figure 20: Velocity Profile for Turbulent Flow in a Pipe .............................................. 29
Figure 21: Temperature Profile of Turbulent Flow Through a Pipe .............................. 30
Figure 22: Velocity Profile for Concurrent Heat Exchanger .......................................... 31
Figure 23: Temperature Profile for Concurrent Heat Exchanger ................................... 32
Figure 24: Concurrent Flow Heat Exchanger Temperature Change .............................. 33
Figure 25: Temperature Change Across the Outlet Flow ............................................... 34
Figure 26: Laminar Flow Developing Velocity Profile .................................................. 34
Figure 27: Velocity Profile for a Turbulent Concurrent Flow Heat Exchanger ............. 35
Figure 28: Temperature Profile for a Turbulent Concurrent Flow Heat Exchanger ...... 36
Figure 29: Turbulent Concurrent Flow Heat Exchanger Temperature Change ............. 36
vii
Figure 30: Cooling Water Flow Rate Effect on Oil Outlet Temperature ....................... 41
Figure 31: Cooling Water Flow Rate Effect on the Change in Oil Temperature ........... 42
Figure 32: Temperature Change in the Fluids vs. the Difference in Inlet Temperatures 43
Figure 33: Velocity Profile for Countercurrent Heat Exchanger .................................... 44
Figure 34: Temperature Profile for Countercurrent Heat Exchanger ............................. 45
Figure 35: Outlet of the Inner Pipe, Inlet of the Outer Pipe ........................................... 45
Figure 36: Inlet of the Inner Pipe, Outlet of the Outer Pipe ........................................... 46
Figure 37: Counter-current Flow Heat Exchanger Temperature Change ....................... 47
Figure 38: Turbulent Flow Arrow Velocity Profile ........................................................ 48
Figure 39: Velocity Profile for a Turbulent Counter-current Flow Heat Exchanger ..... 48
Figure 40: Temperature Profile for a Turbulent Counter-current Flow Heat Exchanger 49
Figure 41: Turbulent Counter-current Flow Heat Exchanger Temperature Change ...... 49
Figure 42: Cooling Water Flow Rate Effect on Oil Temperature for Counter-Current
Flow ................................................................................................................................. 52
Figure 43: Cooling Water Flow Rate Effect on the Change in Oil Temperature for
Counter-Current Flow ...................................................................................................... 53
Figure 44: Temperature Change in the Fluids vs. the Difference in Inlet Temperatures
for Counter-Current Flow ................................................................................................ 54
Figure 45: Fouling Layer Thermal Conductivity Expression ......................................... 55
Figure 46: Cooling Water Flow Rate Effect on Oil Temperature for Concurrent Flow
with Fouling ..................................................................................................................... 58
Figure 47: Fouled and Non-fouled Concurrent Flow Heat Exchanger Comparison ...... 59
Figure 48: Cooling Water Flow Rate Effect on the Change in Oil Temperature for
Concurrent Flow with Fouling......................................................................................... 60
Figure 49: Temperature Change in the Fluids vs. the Difference in Inlet Temperatures
for Concurrent Flow with Fouling ................................................................................... 61
Figure 50: Fouled and Non-fouled Counter-current Flow Heat Exchanger Comparison
......................................................................................................................................... 63
Figure 51: Concurrent and Counter-current Flow Heat Exchangers with and without
Fouling ............................................................................................................................. 64
viii
ix
LIST OF SYMBOLS
A= Area (𝑚2 )
𝐶𝐶 = Heat capacity rate for the cold fluid, 𝑚̇𝑐 × 𝐶𝑝,𝑐 , (W/K)
𝐶ℎ = Heat capacity rate for the hot fluid, 𝑚̇ℎ × 𝐶𝑝,ℎ , (W/K)
𝐶𝑝 = Specific Heat at Constant Pressure (J/ kg K)
𝐶𝑟 = Heat capacity ratio (
𝐶𝑚𝑖𝑛
⁄𝐶
)
𝑚𝑎𝑥
𝐶𝑚𝑖𝑛 = Minimum of 𝐶𝐶 and 𝐶ℎ (W/K)
𝐶𝑚𝑎𝑥 = Maximum of 𝐶𝐶 and 𝐶ℎ (W/K)
D= Diameter of a circular tube (m)
h= Heat transfer coefficient (W/𝑚2 K)
k= Thermal conductivity (W/m K)
L= Flow length of a tube (m)
𝐿∗ = Dimensionless length,
𝐿
𝐷𝑅𝑒𝑃𝑟
𝑚̇= Mass flow rate (Kg/s)
Nu= Nusselt number, hD/k
P= Pressure (N/m^2)
Pe= Peclet Number, Re Pr
Pr= Prandtl Number,
𝜇𝐶𝑝
𝑘
q= Heat (J)
𝑞̇ = Heat transfer rate (W or J/s)
r= Radial distance of a circular tube (m)
Re= Reynolds Number,
𝜌𝑉𝐷
𝜇
𝑅𝑓 = Fouling factor ( 𝑚2 k/W)
T= Temperature (C, K)
𝑇𝑚∗ (𝐿)=Dimensionless Temperature
𝑇𝑚 (𝐿)= Outlet Temperature (C, K)
𝑇𝑤 = Wall Temperature (C, K)
𝑇𝑜 = Initial temperature of fluid flow (C, K)
U= Overall heat transfer coefficient (W/𝑚2 k)
x
V= Velocity (m/s)
𝑉𝑖 = Inlet Velocity (m/s)
𝑉𝑖 = Inlet Velocity (m/s)
µ= Dynamic viscosity (Pa s)
ε= Heat exchanger effectiveness
Subscripts c and h denote cold and hot fluid flow
Subscripts i and o denote inlet and outlet fluid flow, or inner and outer pipe
xi
ACKNOWLEDGMENT
I’d like to thank my family (Ken, Marj, and Dan Onarheim), friends, girlfriend (Jessica
Baker), and advisor (Professor Ernesto Gutierrez-Miravete) for supporting me during
work on this project and my master’s degree.
xii
ABSTRACT
Concentric tube heat exchangers utilize forced convection to lower the
temperature of a working fluid while raising the temperature of the cooling medium. The
purpose of this project was to use a finite element analysis program and hand
calculations to analyze the temperature drops as a function of both inlet velocity and
inlet temperature and how each varies with the other. These results were compared
between concurrent and countercurrent flow and between concurrent and countercurrent
flow with fouled piping. To determine the best heat transfer rate, both laminar and
turbulent flow was analyzed.
xiii
1. Introduction and Background
There are many uses for heat exchangers from car radiators, to air conditioners,
to large condensers in power plants. Just in submarines alone, heat exchangers are used
for: hydraulic cooling, air conditioning and ventilation, electrical device cooling, cooling
of different types of coolant systems, in purification means, and in the nuclear reactor
and steam generators themselves to provide the means of propulsion. But for all
applications the effectiveness of these heat exchangers are dependent on many factors.
Not only does the viscosity and density of the fluids affect the heat transfer due to being
a factor of the Reynolds number and therefore Nusselt number, but the inlet velocity
(mass flow rate) and temperatures of the fluids are proportional to the heat transfer rate.
𝑞̇ = 𝑚̇ × 𝐶𝑝 × (𝑇𝐻 − 𝑇𝑐 )
[1]
This project looks at the heat exchange between fluids in concentric tube heat
exchangers. In this type of heat exchanger, forced convection is caused by fluid flow of
different temperatures passing parallel to each other separated by a boundary, pipe wall.
Basically, one fluid flows through a pipe while the second fluid flows through the
annulus between the inner pipe and outer pipe hence making the pipe walls of the inner
tube the heat transfer surfaces. Several assumptions will have to be made to make it
easier to focus on the inlet velocity and temperature dependence on heat exchanger
temperature drop. Not only will the viscosity and density remain constant for the hand
calculations, but specific heat and overall heat transfer coefficients will be assumed
constant. Any effects from potential and kinetic energy are assumed negligible.
Examining the marketplace for applications for concentric tube heat exchangers
or double pipe heat exchangers, one finds that they are used in areas where extreme
temperature crosses are needed, there are high pressure and temperature demands, and
there are low to medium surface area requirements for the job.
1.1 Heat Exchanger Analysis Theory
Two types of analysis for parallel flow heat exchangers to determine temperature
drops are the log mean temperature difference and the effectiveness-NTU method. Each
method is dependent upon the conditions provided or being solved for. The equation for
heat transfer using the log mean temperature difference becomes:
1
q  UATlm  UA 
T2  T1

ln  T2


T
1

[2]
where the only change for parallel and countercurrent flow is how the ΔT’s are defined.
The NTU (number of transfer units) method uses the effectiveness number of the type of
heat exchanger to determine the amount of heat transfer.
𝑞̇ = 𝜀 × 𝐶𝑚𝑖𝑛 × (𝑇ℎ,𝑖 − 𝑇𝑐,𝑖
[3]
The effectiveness of the types of heat exchangers is as follows:

Parallel Flow:
Counter Flow:  
1  exp[  NTU (1  C r )]
1  Cr
[4]
1  exp[  NTU (1  Cr )]
for 𝐶𝑟 < 1
1  Cr exp[  NTU (1  Cr )
[5]
In general the heat flux is comprised of three factors: the temperature difference,
the characteristic area, and an overall heat transfer coefficient. An approximate value for
the transfer coefficient U (W/𝑚2 k) is 110-350 for water to oil. In the case where fouling
is present on the heat exchanger tubes, the following can be used in the case of tubular
heat exchangers:
UA 
R f ,i
1


hi Ai
Ai
ln(
1
Do
[6]
)
R f ,o
Di 1

2kL ho Ao
Ao
𝑅𝑓 is defined as the fouling factor with units of 𝑚2 K/W. An approximate value of .0009
is used for fuel oil, while .0001 - .0002 is used for seawater and treated boiler feedwater.
Figure 1: Basic Heat Exchanger Design [5]
2
1.1.1
Log Mean Temperature Difference
In order to determine the amount of heat to be transferred in a heat exchanger or
the intensity at which the heat from fluid flow will be transferred, the log mean
temperature difference is calculated. As the name suggests, it is the logarithmic average
of the hot and cold fluid channels of a heat exchanger at both the inlet and outlets. The
log mean temperature difference is defined in terms of Δ𝑇1 and Δ𝑇2 which are defined
depending on whether flow is concurrent or counter current. The larger the temperature
difference, the larger the value of heat that is transferred. The basic equation is:
∆𝑇𝐿𝑀 =
∆𝑇2 −∆𝑇1
∆𝑇
𝐿𝑁( 2⁄∆𝑇 )
1
[7]
For concurrent flow: ∆𝑇2 = 𝑇ℎ,𝑜 − 𝑇𝑐,𝑜 𝑎𝑛𝑑 ∆𝑇1 = 𝑇ℎ,𝑖 − 𝑇𝑐,𝑖
For counter-current flow: ∆𝑇2 = 𝑇ℎ,𝑜 − 𝑇𝑐,𝑖 𝑎𝑛𝑑 ∆𝑇1 = 𝑇ℎ,𝑖 − 𝑇𝑐,𝑜
1.1.2
Heat Exchanger Effectiveness (ε)
The effectiveness ε is the ratio of the actual heat transfer rate to the maximum
possible heat transfer rate:

qactual
,0    1
q max
[8]
The effectiveness equation is usually defined by the type of heat exchanger. The
equations for effectiveness include the value of NTU (number of transfer units) and 𝐶𝑟
(ratio of heat capacities). These values are arranged into different equations depending
upon the type of heat exchanger (equations 4 and 5).
1.1.3
NTU Method
This is another method in determining the heat transfer rate and is based on the
“number of transfer units.” For any heat exchanger, the effectiveness can be found to be
a function of the NTU and ratio of heat capacity rates. By definition NTU is:
𝑈𝐴
𝑁𝑇𝑈 = 𝐶
𝑚𝑖𝑛
[9]
As shown above, the effectiveness of a double pipe heat exchanger, whether it be
concurrent or countercurrent, can be solved based on the NTU number and the ratio of
heat capacity rates of the fluids, 𝐶𝑟 . This method is typically used when some of the inlet
3
or outlet temperature data is not available or needs to be solved for. Using this method,
the amount of heat transferred can be determined by the following equation:
𝑞 = 𝜀 × 𝐶𝑚𝑖𝑛 × (𝑇ℎ,𝑖 − 𝑇𝑐,𝑖 )
[10]
Therefore the outlet temperatures for the hot and cold fluids can be calculated as
follows:
𝑞
𝑇𝑐,𝑜 = 𝑇𝑐,𝑖 + ⁄𝐶
𝑚𝑎𝑥
𝑞
𝑇ℎ,𝑜 = 𝑇ℎ,𝑖 − ⁄𝐶
𝑚𝑖𝑛
[11]
[12]
To determine the heat capacity rate for each fluid, the mass flow rate for the fluid is
multiplied by the specific heat of the fluid. The smaller value of these is labeled
𝐶𝑚𝑖𝑛 while 𝐶𝑚𝑎𝑥 is denoted as the larger value.
1.1.4
Thermal Entrance Length in Pipe Flow
Figure 2: Graetz Problem Temperature Profile [12]
The development of fluid flow and temperature profiles of a fluid after
undergoing a sudden change in wall temperature is dependent on the fluid properties as
well as the temperature of the wall. This thermal entrance problem is well known as the
Graetz Problem. From reference [2] for incompressible Newtonian fluid flow with
constant ρ and k, the equation of energy becomes:
𝜕𝑇
𝜕𝑇
𝜌𝐶𝑝 ( 𝜕𝑡 + 𝑉𝑟 𝜕𝑟 +
𝑉𝜃 𝜕𝑇
𝑟
𝜕𝑇
1 𝜕
𝜕𝑇
1 𝜕2 𝑇
𝜕2 𝑇
+ 𝑉𝑧 𝜕𝑧 ) = 𝑘 [𝑟 𝜕𝑟 (𝑟 𝜕𝑟 ) + 𝑟 2 𝜕𝜃2 + 𝜕𝑧 2 ] + 𝜇𝛷𝑣
𝜕𝜃
[13]
The term 𝜇𝛷𝑣 represents the dissipation function which is negligible. The velocity
field of the flow in the tube is assumed as Poiseuille flow where 𝑉𝑟 = 𝑉𝜃 = 0. The
velocity is given in the form of the following equation [10]:
𝑟2
𝑉𝑧 = 𝑉0 (1 − 𝑅2 )
4
[14]
𝑉0 is defined as the maximum velocity at the center of flow. The velocity definitions
then simplify the general energy equation in cylindrical coordinates to the following:
𝜕𝑇
1 𝜕
𝜕2 𝑇
𝜕𝑇
𝜌𝐶𝑝 𝑉𝑧 𝜕𝑧 = 𝑘[𝑟 𝜕𝑟 (𝑟 𝜕𝑟 ) + 𝜕𝑧 2 ]
[15]
Equation 15 is also expressed as equation 16 since the thermal diffusivity of the fluid is
defined as 𝛼 = 𝑘⁄𝜌𝐶 .
𝑝
𝜕𝑇
1 𝜕
𝜕𝑇
𝜕2 𝑇
𝑉𝑧 𝜕𝑧 = 𝛼[𝑟 𝜕𝑟 (𝑟 𝜕𝑟 ) + 𝜕𝑧 2 ]
[16]
Neglecting dissipation and any conduction axially, equation 16 reduces to the following:
𝜕𝑇
𝑉𝑧 𝜕𝑧 =
𝛼 𝜕
𝜕𝑇
𝑟 𝜕𝑟
(𝑟 𝜕𝑟 )
[18]
The velocity distribution is assumed to be known when using this equation and can be
several different types of flow. For low Prandtl number materials such as liquid metals
the temperature profile (T) will develop faster than the velocity profile (u) and u will be
constant. For high Prandtl number materials such as oils or when the thermal entrance
(sudden change in wall temperature) is fairly far down the entrance of the duct/ tubing
the velocity is expressed as:
𝑢 = 2𝑢̅(1 −
𝑟2
𝑟02
)
[19]
The velocity profile can also be developing and can be used for any Prandtl number
material assuming the velocity and temperature profiles are starting at the same point.
For the purposes of this paper as previously mentioned Poiseuille flow is assumed and
the equation 19 is used to describe the velocity field of the fluid flowing through the
constant wall temperature tubing. Analyzing the paper from Sellars [9] where he extends
the Graetz problem, our above equations from White [12] match the equations used in
the study.
𝜇𝜌𝐶𝑝
𝜕𝑇 𝑘 𝜕
𝜕𝑇
𝑟 2
=
(𝑟 ) 𝑎𝑛𝑑 𝑢 = 2𝑢𝑚 [1 − ( ) ]
𝜕𝑥 𝑟 𝜕𝑟 𝜕𝑟
𝑟𝑜
There have been numerous analytical solutions developed for the Graetz problem
with different types of flow. For laminar flow with a developing velocity profile, the
mean Nusselt number can be approximated based on the relationship illustrated below
between the log mean Nusselt number and the Graetz number for various Prandtl
numbers.
5
Figure 3: Nusselt Number for Various Pr Numbers [12]
An approximation for the mean Nusselt number was given by Hausen (1943) for fluid
with their Prandtl number >1 (especially for use with oils). This is given by equation 20
below.
𝑁𝑢𝑚 = 3.66 +
.075⁄ ∗
𝐿
.05
1+ 2
[20]
⁄
(𝐿∗ 3 )
𝐿
Where 𝐿∗ = 𝐷 𝑅𝑒 𝑃𝑟
1.2 Description and History of Previous Graetz Problem Solutions
The classic Graetz problem which continues to provide background for the
development and understanding of compact heat exchangers has been refined and
expanded upon since initially introduced in 1883. The original problem has a fluid with a
fully developed velocity profile and uniform temperature enter a tubing or duct that is
maintained at a constant temperature. This could be heating or cooling the flowing fluid
just as long as it was different from the initial value of the fluid flow. This classic
problem neglected any viscous dissipation, axial heat conduction, or heat generation by
the fluid. The purpose of the solution to this problem was to determine the temperature
distribution and any connection between the wall temperature and the heat flux to the
6
fluid. Using a separation of variables technique, Graetz found a solution in the form of
an infinite series in which the eigenvalues and functions satisfied the Sturm-Louiville
system. While Graetz himself only determined the first two terms, Sellars, Tribus, and
Klein were able to extend the problem and determine the first ten eigenvalues in 1956.
Even though this further developed the original solution, at the entrance of the tubing the
series solution had extremely poor convergence where up to 121 terms would not make
the series converge.
Schmidt and Zeldin in 1970 extended the Graetz problem to include axial heat
conduction and found that for very high Peclet numbers (Reynolds number multiplied by
the Prandtl number) the problem solution is essentially the original Graetz problem.
Similar to the original problem which showed poor convergence near the ducting
entrance, they discovered up to a 25% deviation in the local Nusselt number which made
the results in this region questionable.
The purpose of this paper is to not redo the various numerical solutions presented by
multiple groups over the past century as there doesn’t appear to be a definitive solution
that has proven convergence everywhere. A modified Graetz problem will be introduced
in a finite element program with certain dimensions, fluid properties, and tubing
temperature in order to analyze the velocity and temperature changes as a building block
to eventually analyzing a compact heat exchanger for the same conditions. Both a
developing velocity and developing temperature profile will be investigated.
1.3 Finite Element Analysis Theory
By definition, finite element analysis is the term applied to the numerical
technique which is used to solve partial differential equations and/ or integral equations.
For problems involving complex geometries or regions/ bodies with irregularities, it
becomes difficult to arrive at a numerical solution for the problem and only approximate
values as specific points can be found. The finite element method will divide the
geometric region of concern into elements or sub-regions in which mathematical
functions can be derived to represent the geometric body in its entirety. The COMSOL
computer program used in this project is a finite element program. A typical finite
element program consists of: a pre-processor, a mesh generator, a processor or solver,
7
and a post-processor. The pre-processor part of the program consists of building a model
of the item to be analyzed and the application of boundary conditions. The boundary
conditions consists of any constraints or loads being applied in the statics/ dynamics
region or any velocity or temperature conditions for the fluid dynamics and heat transfer
aspects. In additions to boundary condition definition, the properties of the materials
involved are also defined, and many programs have a library in which the properties of
common materials are stored and able to be used for definition.
The mesh generator breaks up the model into elements which are geometric bodies
which produce the stiffness or material properties of part of the structure. The element
geometry is defined by nodal locations or conductivity. These elements can be modified
to be smaller or larger or coarser or more refined. The mesh created from the model
applies the geometric and boundary conditions as well as the material properties to the
entire structure. The processor portion of the finite element program has the equations of
heat transfer, fluid flow, as well as material property equations in order to solve the
defined model. In the COMSOL program there are 3 different types of non-linear solver
which can be used for this purpose. How the solver develops a solution can also be
modified by increasing or decreasing the tolerance of convergence that is required for a
solution to be obtained, or by changing the order in which the solver solves the
equations. The post-processor portion of the program allows examination of the results
in the form of 1D, 2D, and 3D plots of velocity and temperature profiles as well as
arrow, surface, and contour plots. It is this portion of the program that allows the finite
element analysis to be used in whatever fashion is needed.
8
2. Problem Description and Methodology
For this project, developing laminar and turbulent incompressible fluid flow was
analyzed in three heat exchanger cases: parallel flow, countercurrent flow, and flow in a
fouled heat exchanger. The resulting temperature difference was compared and
determined as a function of the inlet velocity and inlet temperatures. The overall
objective for this project was to determine the max temperature difference in these cases
for both laminar and turbulent flow for a variety of flow rates and inlet temperatures. To
simplify the number of variables, water and oil were chosen as the fluids to maintain
viscosities and densities of the fluids constant. The type of heat exchanger used was of
concentric tube design. Water was the cooling medium and oil the working fluid.
2.1 Defining Material Properties
Water was used as the base fluid flowing through tube or pipe. Its material
properties were derived from tables based on the temperature which was being
calculated in the model. The material was defined in COMSOL using its material
browser. For the modified Graetz problem model certain properties were defined by the
user prior to computing the model, these properties were: thermal conductivity, density,
heat capacity at constant pressure, ratio of specific heats, and dynamic viscosity. For the
modified Graetz problem with axial conductions as well as for the heat exchanger
models the material library properties in COMSOL were used.
2.2 Methodology and Approach
2.2.1
Finite Element Analysis Modeling
A finite element analysis was done using COMSOL for the fluid flow and
convective heat transfer. A 2D axisymmetric model was chosen to depict the tubing the
fluid was flowing through. The type of physics to be applied was then added. For the
baseline model (the modified Graetz problem) the physics used was laminar fluid flow
and then non-isothermal flow was chosen. This allowed for definition of not only the
fluid parameters but also the heat transfer of the constant wall temperature to the fluid.
The second model added a pipe wall to the modified Graetz problem while the third
model introduced a second pipe concentric to the first and was analyzed for fluid flow in
9
the same direction. The fourth model reversed the fluid flow for the cooling medium,
which was chosen as water. The material library was used for definition of properties for
oil and water. The fifth and sixth models added on to the second and third models a layer
of fouling for both types of flow and determined the effect on not only the flow but the
resultant temperature differences. These models were repeated using turbulent flow
which added complexity to each model. Post-processing plots developed in COMSOL
were used for analysis. In addition to this, the COMSOL information was exported to
excel to better compare and analyze the data. Hand calculations for the temperature
differences were also done to verify results.
2.2.2
Defining Variable Temperature and Velocity
In the COMSOL computer application, temperature, velocity, and various fluid
parameters are easily defined and changed by the left-hand tab. For the modified Graetz
problem, non-isothermal flow was used to define the fluid flow parameters and
temperature distribution, but in the later models, conjugate heat transfer equations were
added. This allowed for laminar flow parameters as well as heat transfer equations to be
added. For the fluid flowing both an inlet and outlet point was chosen. Under these the
velocity field incoming is defined as well as if there is any viscous stress at the outlet.
Now that the velocity is defined, the heat transfer in solids is added when conjugate heat
transfer is used for models with pipe walls, or heat transfer in non-isothermal flow is
used. Under this tab (right clicking on the flow tab) these are many applications that can
be defined from heat flux, heat conduction, cooling, insulation, to temperature definition
and outflow. For the purposes of the models in this paper, temperature is defined in this
method both for incoming fluid as well as the constant wall temperature as defined in the
beginning models. Now that temperature and velocity of the fluid and/ or tubing or pipe
wall is defined, the models can be meshed and solved. The parameters are easily
changed and many iterations with various values can be performed.
10
3. Results and Discussion
3.1 The Modified Graetz Problem Results
To begin the COMSOL analysis of temperature difference in fluid flow the base
condition must first be analyzed. The first condition is that of fluid passing through a
tube with a constant wall temperature, as described before this is known as the Graetz
problem. However, since the fluid flowing in the model will be developing as it flows
and is not already fully developed as it’s assumed in the classical Graetz problem, this
will be referred to as a modified Graetz problem. A base model was run in COMSOL
and the analysis was compared to hand calculations to verify. The input data for the
problem were as follows:
Table 1: COMSOL Model Initial Conditions
Flow Parameters
L=
1.0 m
D=
.1 m
k=
0.64
0.000547 Pa s
µ=
ρ=
988 kg/𝑚3
𝐶𝑝 =
4181 J/Kg k
3.1.1
The Modified Graetz Problem COMSOL Model
As previously described, the physics used for modeling was non-isothermal
laminar flow. The water was selected to be flowing through a tube or pipe of length 1m
with a diameter of .1m. The inlet flow of the water was set initially at .0001 m/s and
varied for two other cases: .01 m/s and .001 m/s. The temperature of the water flowing
into the tubing was set at 50℃ or 323.15 K while the wall temperature remained constant
at 30℃ or 303.15 K. This temperature difference was also varied for two other cases.
Figure 4 shows the geometry of the model in COMSOL.
11
Figure 4: Modified Graetz Problem Geometry
The material properties of the fluid were then defined. Water at 50 C was used
and the properties used for temperature determination were user defined. These values
were entered into the material browser and are shown below in table 2.
Table 2: User Defined Material Properties
The physics used was for non-isothermal flow and laminar flow and heat transfer
physics equations were applied to define the fluid flow as well as the heat transferred
from the constant wall temperature to the water. For fluid flow the inlet and outlet points
of flow were defined with the water velocity defined at the inlet point. For heat transfer,
12
the temperature of the water flowing at the inlet was defined as well as the temperature
of the wall. The outlet of fluid flow was also defined as outflow in terms of the heat
transfer physics. The equations used by COMSOL for the non-isothermal flow are
summarized in table 3 below and are from the fluid tab under non-isothermal flow in the
COMSOL model.
Table 3: Modified Graetz Problem COMSOL Equations
After initializing a mesh of the model, results were obtained for not only the
velocity profile but also the temperature profile. Figures 5 and 6 show the velocity and
temperature profiles computed with the model, respectively. It can easily be seen that the
thermal entrance length to a fully developed temperature profile is much longer than the
entrance length for the velocity profile. In fact by approximately 15% of the length of
the tube, the velocity profile is already fully developed and the velocity nearly constant,
while by the end of the length of tubing the temperature profile was still not at a constant
value. In the original Graetz problem, the velocity profile would have been constant
throughout the length of the tubing since it enters already fully developed.
13
Figure 5: Modified Graetz Problem Velocity Profile
Figure 6: Modified Graetz Problem Temperature Profile
3.1.2
The Modified Graetz Problem COMSOL Mesh
Initially the physics controlled mesh was used in COMSOL but looking at the
study results it was discovered that the results were dependent upon the refinement of
14
the mesh and the initial values tab of the COMSOL model. The initial values are defined
to only be an initial guess for the final solution derived by the non-linear solver in
COMSOL. However, it was found that varying the temperature in this initial values tab
would vary the centerline outlet temperature even though the temperature of inlet flow
and surface temperature were previously defined. It was also discovered that the initial
tolerance of 10−3 as defined by COMSOL allowed for a very large variance in the outlet
temperature just by changing the refinement of the model. Ideally refining the model
should change the value slightly as the model becomes more refined since more
elements are added to the mesh; the temperature being solved for should become closer
and closer to the desired value. However by starting at the extremely coarse and going to
the fine mesh, the outlet temperature changed by almost 10 degrees and the change was
not linear.
To improve the results and take out the uncertainty that was being created by
changing the mesh refinement, the tolerance of the solver was changed to 10−4 and a
different type of mesh was created. Instead of using the triangular type elemental mesh
or unstructured mesh which COMSOL automatically defines when the physics
controlled mesh is selected, the user controlled mesh option was used and a free quad
mesh was defined. This allowed for more of a rectangle shape to the mesh elements or a
structured mesh along the length of the tubing toward the middle of the flow. Along the
wall of the tubing boundary layer meshing was added which refined the mesh elements
and added extra elements along the wall where the temperature and velocity profiles are
developing and there is more change in the flow at these points. This allows for
COMSOL to have the solver focus more on the boundary that has complicated change to
it than on the steady flow in the middle of the tubing. Figure 7 shows an example of this
mesh with the additional layers applied around the wall of the tubing.
15
Figure 7: User Defined Mesh
It took several iterations of attempting to find the best mesh to yield the best
result. Ultimately as the number of elements increases the outlet temperature on the
centerline should level out and gradually approach a certain value instead of varying
higher and lower around several values. By changing the number and thickness of the
boundary layers a more accurate mesh was able to be obtained. The maximum size of
the elements in the mesh were changed while the number of boundary layers kept
constant to increase and decrease the number of elements in the model (lowering the
maximum elements size increased the total number of mesh elements in the model).
Table 4 below shows the results from increasing the mesh elements on centerline
temperature for the case of 𝑉𝑖 =.0001m/s. The variance in centerline temperature was
from 306.0347 K to 305.2428 K for a difference of .7919 K instead of 10 K. The number
of boundary layers was 40 with the stretching factor at 1.2 and the thickness adjustment
factor at 15.
16
Table 4: Mesh Extension Study Results
Mesh Effectiveness
Number of Mesh Elements
2150
2279
2408
2948
3388
3696
4095
4500
5875
8183
10, 376
Centerline Outlet Temp (K)
306.0347
305.9992
306.0265
305.9083
305.8629
305.8664
305.6118
305.6001
305.2428
305.7506
305.6016
Plotting these numbers on a scatter plot shows that as the element size increases
the outlet temperature gradually gets closer to a constant centerline temperature. Figure 8
shows this relationship. An exponential trend-line was added to illustrate the
temperatures gradual approach to a constant value.
306.2
Temperature (K)
306
305.8
305.6
305.4
y = 308.55x-0.001
305.2
305
2000
3000
4000
5000
6000
7000
8000
9000
10000
11000
Number of Elements in the Mesh
Centerline Outlet Temperature (K)
Power (Centerline Outlet Temperature (K))
Figure 8: Centerline Temp vs. Mesh Element Number
17
Since the initial value for the temperature of the modified Graetz problem was
causing an unexpected variance in the results, its effect on this new mesh was also
documented. Using the most refined mesh (element number of 10, 376) the initial value
of temperature was varied from 283.15 K to 323.15 K and the resulting centerline
Temperature (F)
temperature was fairly constant as shown in figure 9.
310
309
308
307
306
305
304
303
302
301
300
Centerline Outlet Temp (F)
280
290
300
310
320
330
Initial Value Temp (F)
Figure 9: Initial Value Variance
This study proved the change in initial values and mesh refinement only affected
the results by a fraction of a percent vice several percent when boundary layer elements
were used in the mesh. To further refine the mesh and provide more accurate results, the
element size near the center of the fluid flow was enlarged and made more rectangular
by changing the size of the quad elements. This mesh was then proven accurate like the
previous study by verifying that changing the number of elements and initial values
didn’t vary the outcome by more than a percent of a fraction. This type of element array
now proven was applied to the following models which added on to this modified Graetz
problem model.
3.1.3
The Modified Graetz Problem Study Results
Using COMSOL’s post-processing capabilities, a 1D line graph was plotted
along the center of the tubing to track the temperature as it changes along the center of
the tubing. Figure 10 shows the temperature trend as the fluid cools from its inlet
temperature to near the constant wall temperature.
18
Figure 10: Modified Graetz Problem Centerline Temperature
To determine the outlet temperature of the center of flow a point evaluation was
done under the derived values tab of the post-processor results of the model. This
yielded 305.3221 K. In order to verify the results, the velocity was changed at the inlet
of the tube and compared to hand calculations for both .001 m/s and .01 m/s inlet
velocity, in addition to the initial case of .0001 m/s inlet velocity
3.1.4
The Modified Graetz Problem Calculations
The outlet temperature of the fluid is determined by using the mean Nusselt
number of the fluid flow. The Nusselt number approximation initially used was equation
20 from White’s Viscous Fluid Flow and proposed by Hausen (1943) for PR>1. First the
Reynolds number is calculated for the initial conditions. For the purpose of analysis the
flow is considered incompressible Newtonian flow.
𝑅𝑒 =
𝜌𝑉𝐷
𝜇
=
(988)(.0001)(.1)
(5.47𝑥10−4 )
= 18.062
[21]
The Prandtl number is calculated using the material properties of water at the inlet
temperature.
𝑃𝑟 =
𝐶𝑝 𝜇
𝑘
=
(4181)(5.47𝑥10−4 )
19
.64
= 3.57
[22]
The dimensionless length value is defined as
𝐿
(1)
𝐿∗ = 𝐷𝑅𝑒𝑃𝑟 = (.1)(18.062)(3.57) = .15508
[23]
The outlet temperature is defined as
𝑇𝑚 (𝐿) = 𝑇𝑤 − (𝑇𝑤 − 𝑇𝑜 ) ∗ 𝑇𝑚∗ (𝐿)
[24]
Since there is a relationship between 𝑇𝑚∗ (𝐿) and the mean nusselt number, if the
Nusselt number is obtained from the approximation equation, the outlet temperature can
then be determined. Using equation 20, the Nusselt number is calculated.
𝑁𝑢𝑚 = 3.66 +
.075⁄ ∗
𝐿
1+.05⁄ 2
𝐿∗3
= 3.66 +
−1
.075⁄
.15508
1+.05⁄
.155082/3
= 4.0722
∗
𝑁𝑢𝑚 = 4𝐿∗ 𝑙𝑛𝑇𝑚∗ (𝐿) ∴ 𝑇𝑚∗ (𝐿) = 𝑒 (−4𝐿 𝑁𝑢𝑚) = .07997
[25]
[26]
𝑇𝑚 (𝐿) = 𝑇𝑤 − (𝑇𝑤 − 𝑇𝑜 ) ∗ 𝑇𝑚∗ (𝐿) = 30 − (30 − 50) ∗ (. 07997)
𝑇𝑚 (𝐿) = 31.5994 C = 304.7494 K
[27]
This was then compared to the centerline temperature of the fluid at the end of
the tubing (at z-=1.0 m) and a percent error was calculated between the expected and
actual (COMSOL value). Table 5 shows this particular case as well as two other cases.
The inlet velocity was varied to .001 and .01 m/s and the centerline temperature obtained
both by hand and by COMSOL. Overall the derived values of the outlet temperature are
all near the values of the COMSOL model with less than 2% error. The Hausen equation
is noted to have an approximation error of 5%. There should also be some error expected
due to the fact that this analysis was done for a modified Graetz problem which included
a developing velocity profile as well as a developing temperature profile. However, the
percent error for this would be small due to the fact that the velocity is nearly fully
developed as soon as it enters the constant wall temperature tubing.
Table 5: Modified Graetz Problem Comparison
Inlet V
(m/s)
0.0001
0.001
0.01
Inlet Temp
(C)
50
50
50
Wall Temp
(C)
30
30
30
Expected Value Calc (K)
304.7494
316.646
317.644
20
COMSOL Value (K)
305.3221
321.9023
323.0481
% Error
0.187572
1.632887
1.672847
Several other numerical methods were attempted in order to create the best
numerical solution for the modified Graetz problem underneath these initial conditions.
As described previously the mesh was changed in the original model to a mesh which
showed little to no variation in centerline temperature when the number of elements or
initial values changed. Table 6 below shows the comparison to the previous hand
calculations and how they compare to the previous models results.
Table 6: Results from Mesh Refinement
Inlet V
(m/s)
0.0001
0.001
0.01
Exp. Value Calc (K)
304.7494
316.646
317.644
COMSOL Value (K)
305.3221
321.9023
323.0481
% Error
0.187572403
1.632886749
1.672846861
COMSOL Refined (K)
305.93648
322.812
323.15
While the percent error is slightly higher with the refined mesh which included
boundary layer mesh elements, the consistency of the results were far superior to the
original mesh. Originally the results were highly dependent on the initial temperature
value the non-linear solver was using even though they should be mutually exclusive as
well as dependent on element size and amount as described.
3.1.5
Turbulent Flow with Constant Wall Temperature
Originally the modified Graetz problem COMSOL models were modeled using
laminar flow. To analyze and determine the difference flow types have on the velocity
and temperature profiles, turbulence was added to the model. Figure 11 below shows the
developing velocity profile of laminar flow.
21
% Error
0.388015
1.91009
1.703853
Figure 11: Laminar Flow Velocity Profile
Figure 12 below shows the velocity profile for turbulent flow.
Figure 12: Turbulent Flow Velocity Profile
As opposed to the laminar flow, the turbulent flow has a more even flow and
distributed as it enters, flows, and then exits the tubing. Under the non-isothermal tab,
the RANS turbulence model was turned on essentially talking the same laminar flow
modified Graetz problem model but changing the flow from laminar to turbulent. The kε turbulence type model was used with Kays-Crawford heat transport. The same quad
22
element mesh with boundary layer elements added was used with an accuracy tolerance
of 10−4. The centerline temperature along the length of the tubing had more of a linear
relationship while the laminar flow was more gradual in lowering towards the outlet
temperature as described previously. Figure 13 shows the centerline temperature for
turbulent flow. The main difference between the laminar and turbulent flows was that for
both .0001 m/s and .001 m/s the outlet centerline temperature was approximately 322 K
whereas in laminar flow the lowest velocity flow actually lowered the centerline
temperature down to approximately 305 K due to the fluid flowing slower and having
more time to undergo heat transfer with the wall. In the turbulent flow, the fluid is mixed
and temperature more evenly distributed so that for the slowest of velocities the
centerline temperature doesn’t lower nearly as much.
Figure 13: Turbulent Flow Centerline Temperature
For a velocity of .0001 m/s the centerline temperature was 322.25503 K and for a
velocity of .001 m/s the centerline temperature was 322.86295 K. For the largest of the
velocities that have been used (.01 m/s) and the same geometry, the type of solver being
used had to be modified. The same mesh and boundary layer elements were used as
previously described, but with this velocity, geometry, and turbulent model defined the
stationary solver would not converge and determine a solution. Due to this, the solver
was changed from a fully coupled to segregated in order for the non-linear solver to
23
divide up the solution process into sub-steps. Also the type of solver was changed from
MUMPS to SPOOLES. Once these changes were made, the same mesh and parameters
were solved and a solution obtained. For a velocity of .01 m/s the centerline temperature
was 323.14911 K. Figure 14 shows not only the type of solver used, but the temperature
profile for the turbulent model used for a velocity of .01 m/s.
Figure 14: Turbulent Model for the Modified Graetz Problem
24
3.2 Flow in a Pipe with Wall Conduction
To add to the modified Graetz problem a pipe wall was added and the heat transfer
to the fluid flow from the pipe wall was analyzed. In addition to this the heat conduction
through the pipe wall was taken into account. A pipe wall of .02 m was added to the
original model and the same 3 velocity profiles were analyzed. An accuracy tolerance of
10−4 was used as before as well as the previously defined quad element mesh with
boundary layers applied.
3.2.1
Laminar Flow with a Pipe Wall COMSOL Model
The development of the velocity and temperature profiles of flow through a steel
pipe is shown below in Figures 15 and 16. Just as in the modified Graetz problem
example, the velocity profile develops extremely fast while the temperature profile takes
the entire length of the pipe to come close to an approximately constant temperature.
Figure 15: Velocity Profile for Flow Through a Pipe
25
Figure 16: Temperature Profile of Flow Through a Pipe
For a velocity of .0001 m/s the centerline temperature was 305.93998 K. For a
velocity of .001 m/s the centerline temperature was 322.83099 K. For a velocity of .01
m/s the centerline temperature was 323.14999 K. All 3 temperatures are approximately
the same or larger than the results from the modified Graetz problem. This could be
attributed to some of the cooling effect being lost due to the thickness of the pipe wall,
which depending on the thermal conductivity would act like a thermal insulation
boundary.
3.2.2
Laminar Flow with a Pipe Wall Problem Calculations
To analyze the flow a lumped parameter model was used and the temperature
change determined at various points along the length of the pipe. Heat transferred from
the wall will be equal to the heat transferred to the water.
̇
𝑄𝑤𝑎𝑙𝑙 𝑜𝑢𝑡𝑙𝑒𝑡 = 𝑞̇ × ∆𝐴 = 𝑞2𝜋𝑅∆𝐿
[28]
𝑄𝑤𝑎𝑡𝑒𝑟 = 𝜌𝐶𝑝 ∆𝑇∆𝑉 = 𝜌𝐶𝑝 ∆𝑇𝜋𝑅 2 ∆𝐿
[29]
𝑞̇ 2𝜋𝑅∆𝐿
∆𝑇 = 𝜌𝐶
𝑝
𝜋𝑅 2 ∆𝐿
2𝑞̇
= 𝜌𝐶
𝑝𝑅
[30]
To determine a basic change in temperature and therefore outlet temperature, the
heat flux along the length of the flow was graphed using the original laminar flow
modified Graetz problem model and determined at various points along the flow path.
26
Using an excel spreadsheet and the above equations, the outlet temperature was
determined and could be used as comparison to the COMSOL value of laminar flow
with a pipe wall. Table 7 below shows the outlet temperature based on the heat flux
along the length of the fluid channel.
Table 7: Change in Temperature Along the Channel of Water
Length (m)
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Heat Flux at the Wall (W/𝑚2 )
45.71
58.05
73.29
92.19
115.87
145.86
185.05
240.52
304.94
32844.5
ΔT (K)
-0.00044
-0.00056
-0.00071
-0.00089
-0.00112
-0.00141
-0.00179
-0.00233
-0.00295
-0.31804
Outlet Temp (K)
323.1495574
323.1494379
323.1492903
323.1491073
323.148878
323.1485876
323.1482081
323.147671
323.1470472
322.8319572
While the resultant temperature at the outlet of the pipe has a very close
temperature to that of the .001 m/s velocity for flow through a pipe, these heat flux
values were from the modified Graetz problem model of velocity .0001 m/s so there
does appear to be a small error in the calculations. Taking a line average of the outlet of
the flow from the centerline to the inner wall in the model with pipe wall heat
conduction and from the centerline to the tubing in the modified Graetz problem verified
that the averages were very similar. The modified Graetz problem resulted in an average
of 304.78102 K while the modified Graetz problem with a pipe wall resulted in an
average of 304.32395 K. Also when a line graph was created of the temperature at the
outlet of the flow for each model the curves shapes were identical with the exception of
the model with the pipe wall which had a small slanted horizontal line to the right where
a small amount of temperature rise was seen across the pipe wall going to outside to
inside. Figures 17 and 18 show the similarity in outlet temperature distribution between
the Graetz problem and pipe wall models.
27
Figure 17: Outflow Temperature Distribution for the Modified Graetz Problem
Figure 18: Outflow Temperature Distribution for the Modified Graetz Problem with a Pipe Wall
3.2.3
Turbulent Flow in a Pipe with Wall Conduction
The turbulence model was added to flow in the pipe with a wall and the same
dimensions, velocities, and temperatures were used. In order to demonstrate that the
laminar flow had a developing velocity profile while the turbulent model had velocity
which was already developed with very little change, the velocity field z component was
plotted for both models, and the velocity graph of the centerline velocity across the
28
length of the pipe was shown. This velocity relationship is shown in figures 19 and 20
for both models below.
Figure 19: Velocity Profile for Laminar Flow in a Pipe
Figure 20: Velocity Profile for Turbulent Flow in a Pipe
This shows that for laminar flow, the velocity was changing from approximately
10 X 10−5 m/s to approximately 20 X 10−5 m/s, while for turbulent flow the step
change between the entrance and exit was only from 10 to 10.2. This large variation in
29
laminar flow velocity indicates a developing flow, while the small change in turbulent
flow velocity indicates steady flow already exists.
The centerline temperatures from this model were extremely similar to the
turbulent flow through the tubing with no pipe wall (modified Graetz problem) and the
difference between the laminar and turbulent flow in a pipe wall was similar to the
differences between laminar and turbulent flow in the modified Graetz problem. Figure
21 shows the temperature profile for the turbulent model with velocity at .0001 m/s.
Figure 21: Temperature Profile of Turbulent Flow Through a Pipe
For a velocity of .0001 m/s the centerline temperature was 322.25321 K. For a
velocity of .001 m/s the centerline temperature was 322.83265 K. For a velocity of .01
m/s the centerline temperature was 323.1494 K. Just as in the modified Graetz problem
when flow was changed from laminar to turbulent, the centerline temperature does not
drop as much at the lowest velocity due to the better mixing and more evenly distributed
flow. The approximate 322 K was similar between both turbulent models as expected.
This was also the main difference between the laminar and turbulent model for flow
through a pipe with wall heat conduction.
30
3.3 Flow in a Concurrent Flow Heat Exchanger
3.3.1
Laminar Flow in a Concurrent Heat Exchanger COMSOL Model
Adding onto the COMSOL model of flow through a pipe with a pipe wall, a
second pipe and pipe wall were added. Flow was defined to be flowing in the same
direction with the outer flow at a lower temperature cooling the inner fluid. For the
purposes of simplifying the model for development, the same type of pipe was used as in
the previous model, the same fluid, water, was used for both sides of the fluid flow, and
the same dimensions and temperatures were used. Once the model was made and
analyzed the velocity, temperatures, and materials could be changed for further
investigation. Figures 22 and 23 shows how the velocity and temperature profiles are
affected by adding a channel of a 2nd fluid flowing around the original pipe.
Figure 22: Velocity Profile for Concurrent Heat Exchanger
31
Figure 23: Temperature Profile for Concurrent Heat Exchanger
For concurrent flow heat exchangers the hotter fluid will lower in temperature as
it loses heat to the cooler fluid which will then rise in temperature due to the heat
transfer. A 1D plot was made to determine this temperature development. First a line
graph of the temperature distribution along the centerline (the hotter fluid) was made.
Then a second curve was created of the temperature along the length of the pipe in the
middle of the flow in the outer tube. Figure 24 shows this gradual temperature change in
both flow paths. This is the correct curve form already proven for concurrent flow heat
exchangers.
32
Figure 24: Concurrent Flow Heat Exchanger Temperature Change
Looking at the end of the 1 m heat exchanger, the flow closest to the centerline
was the hottest for the inner fluid and the flow closest to the outside of the inner pipe
was the hottest for the outer fluid. This is due to the flow closest to the inside wall of the
inside pipe experiences more of the heat transfer to the colder fluid of the outer pipe. The
flow closest to the outside wall of the inner pipe receives more of the heat energy and
therefore has a higher temperature nearest the inner pipe for the colder outer flow. This
leads to a downwards sloping curve from the 0.0 m to the .05m mark for the inner flow
as well as a downward slope from .07 m to 1.2 m when temperature is graphed along the
radius at the outlet of the heat exchanger. In addition to this, figure 25 also shows the
slight heat conduction in the steel pipe. It’s very slight, but does show that a portion of
the heat energy is transferred to the pipe wall and not the flow parallel to one another.
33
Figure 25: Temperature Change Across the Outlet Flow
3.3.2
Turbulent Flow in a Concurrent Heat Exchanger
In the laminar flow model, an arrow surface plot of the flow shows the
developing velocity profile of the inner and outer flow, and the typical parabolic shape
of the velocity as shown in figure 26.
Figure 26: Laminar Flow Developing Velocity Profile
34
The velocity profile for the turbulent model of the same concurrent flow heat
exchanger shows that there is very little change in the velocity of either fluid since the
velocity profile as previously discussed is already developed. The temperature profile
and resultant graphs of the centerline of both fluid flows shows very little change in
either the inner or outer fluid’s temperature. In the laminar case, the concurrent flow heat
exchanger yielded a gradually lowering hot fluid temperature with a similar gradually
increase in the cold fluid temperature, but with turbulence applied to the model, the
temperature of both fluids with the .0001 m/s velocity shows little to no change in either
fluid. Figures 27 and 28 show the effect the turbulent flow has on a concurrent flow heat
exchanger.
Figure 27: Velocity Profile for a Turbulent Concurrent Flow Heat Exchanger
35
Figure 28: Temperature Profile for a Turbulent Concurrent Flow Heat Exchanger
Figure 29 is the same as the plot in figure 24 (concurrent flow heat exchanger
temperature change), except that due to the little to no change in temperature because of
the turbulent flow, the hot and cold fluid variation with respect to the arc length looks
like a constant temperature is being maintained.
Figure 29: Turbulent Concurrent Flow Heat Exchanger Temperature Change
36
3.3.3
Laminar Flow in a Concurrent Heat Exchanger Problem Calculations
In order to analyze the concurrent flow heat exchanger better, an example heat
exchanger was designed in COMSOL using the existing model and an excel spreadsheet
made to document the hand calculated results. In the cases studied, engine oil was
assumed to be flowing through the inner pipe which was made of copper and cooled by
the outer concentric pipe in which water was flowing. Material properties such as
dynamic viscosity, density, Prandtl number, and thermal conductivity were obtained
from reference [6]. It was noted at this time that in the mesh that was previously used, no
boundary layer elements were added to the outside of the inner pipe where the cooling
water of the outer pipe was flowing across. For the oil and water heat exchanger design,
an additional boundary layer mesh was added to this surface. Comparing results for the
first case (.0001 m/s oil velocity with varying water velocity) with and without this
boundary layer showed only a small change in the outlet temperatures. The largest
difference was approximately .5K.
For comparison to the COMSOL model results, the outlet temperatures for the oil
and water were determined using a NTU-effectiveness method. An excel spreadsheet
was used so that during the differing cases which changed the fluid velocities and
temperatures, only these parameters had to be changed in the spreadsheet and the hand
calculated version of the outlet temperatures would automatically update. An example of
these calculations is as follows for the first case analyzed, oil velocity at .0001 m/s and
water velocity .0001 m/s. The hot inner fluid (oil) is flowing through 1 copper pipe 1
meter in length.
𝐷𝑜𝑢𝑡𝑒𝑟 𝑡𝑢𝑏𝑒 = 𝐷𝑜 = .14 𝑚,
𝐴𝑜𝑢𝑡𝑒𝑟 𝑡𝑢𝑏𝑒 = 𝐴𝑜 = 𝜋𝐷𝐿 = 𝜋(. 14 𝑚)(1 𝑚) = .4398 𝑚2
𝐷𝑖𝑛𝑛𝑒𝑟 𝑡𝑢𝑏𝑒 = 𝐷𝑖 = .10 𝑚,
𝐴𝑖𝑛𝑛𝑒𝑟 𝑡𝑢𝑏𝑒 = 𝐴𝑖 = 𝜋𝐷𝐿 = 𝜋(. 10 𝑚)(1 𝑚) = .3142 𝑚2
The cross sectional area of each fluid flow is:
𝐴𝑜𝑖𝑙 = 𝜋𝑟 2 = 𝜋(. 05 𝑚2 ) = .007854 𝑚2
𝐴𝑤𝑎𝑡𝑒𝑟 = 𝜋(𝑟𝑜 2 − 𝑟𝑖 2 ) = 𝜋((. 12 𝑚)2 − (. 07 𝑚)2 ) = .0298454 𝑚2
The inlet temperature of each fluid and its corresponding properties due to that
temperature is shown in table 8:
37
Table 8: Fluid Properties
Fluid Parameters for Oil
T=
125 C
T=
398.15 K
k=
0.134 w/m K
0.00915 Pa s
µ=
826 kg/m^3
ρ=
Pr=
159
Cp=
2328 J/Kg K
Fluid Parameters for Water
T=
20 C
T=
293.15 K
k=
0.600 w/m K
0.001003 Pa s
µ=
998.2 kg/m^3
ρ=
Pr=
6.99
Cp=
4182 J/Kg K
The mass flow rates are then calculated and used to determine the heat capacity rates.
826 𝑘𝑔
𝑚
𝑚̇𝑜𝑖𝑙 = 𝜌𝐴𝑣 = (
) (. 007854 𝑚2 ) (. 0001 ) = .0006487 𝑘𝑔/𝑠
3
𝑚
𝑠
998.2 𝑘𝑔
𝑚
2)
(.
𝑚̇𝑤𝑎𝑡𝑒𝑟 = 𝜌𝐴𝑣 = (
)
029845
𝑚
(.
0001
) = .002979 𝑘𝑔/𝑠
𝑚3
𝑠
𝐽
. 0006487 𝐾𝑔
𝐶𝑜𝑖𝑙 = 𝐶𝑝,𝑜𝑖𝑙 × 𝑚̇𝑜𝑖𝑙 = (2328
)(
) = 1.5102 𝑊/𝐾
𝐾𝑔𝐾
𝑠
𝐶𝑤𝑎𝑡𝑒𝑟 = 𝐶𝑝,𝑤𝑎𝑡𝑒𝑟 × 𝑚̇𝑤𝑎𝑡𝑒𝑟 = (4182
𝐽
. 002979 𝐾𝑔
)(
) = 12.4588 𝑊/𝐾
𝐾𝑔𝐾
𝑠
From this it can be defined for the analysis purposes that 𝐶𝑚𝑖𝑛 is 𝐶𝑜𝑖𝑙 and 𝐶𝑚𝑎𝑥 is
𝐶𝑤𝑎𝑡𝑒𝑟 . This yields our ratio of heat capacity rates to be:
𝐶𝑟 =
𝐶𝑚𝑖𝑛
𝐶𝑜𝑖𝑙
1.5102
=
=
= .12122
𝐶𝑚𝑎𝑥 𝐶𝑤𝑎𝑡𝑒𝑟 12.4588
The Reynolds number for the oil flow and then the Nusselt number for the heat transfer
from the oil to the water are as follows:
𝑅𝑒 =
𝑁𝑢 = 3.66 +
𝜌𝑣𝐷
4𝑚̇𝑜𝑖𝑙
4 × .0006487𝑘𝑔/𝑠
=
=
= .9027
𝜇
𝜋𝐷𝑖 𝜇𝑜𝑖𝑙 𝜋 × .10𝑚 × .00915 𝑃𝑎 𝑠
. 0668𝐴
𝐷
𝑤ℎ𝑒𝑟𝑒 𝐴 = 𝑅𝑒𝑃𝑟 ( 𝑖⁄𝐿) = (. 9027)(159)(. 10) = 14.35
.667
1 + .04𝐴
𝑁𝑢 = 4.435
The heat transfer coefficient of the inner pipe wall is expressed as follows:
𝑊
𝑘𝑜𝑖𝑙 × 𝑁𝑢 . 134 𝑚𝐾 × 4.435
ℎ𝑖 =
=
= 5.9435 𝑊⁄𝑚2 𝐾
𝐷𝑖
. 10 𝑚
The overall heat transfer coefficient is expressed in terms of UA. For this overall
coefficient, the heat transfer coefficient of the outer wall of the inner pipe is required.
38
For the purposes of the analysis it is assumed to be approximately half of the value for
the heat transfer coefficient of the inner wall. This overall coefficient is defined as
𝑊
follows with the thermal conductivity of copper being 393.11 𝑚 𝐾:
𝑈𝐴 =
=
1
𝐷
ln( 𝑜⁄𝐷 )
1
1
𝑖
+
+
ℎ𝑖 𝐴𝑖 ℎ𝑜 𝐴𝑜 2𝜋𝐿𝐾𝑐𝑜𝑝𝑝𝑒𝑟
1
ln(. 14𝑚⁄. 10𝑚)
1
1
+
+
(5.9435𝑊/𝑚2 𝐾)(.3142𝑚2 ) (. 5 ∗ 5.9435𝑊/𝑚2 𝐾)(.4398𝑚2 ) 2𝜋(1𝑚)(393.11𝑊/𝑚𝐾)
𝑈𝐴 = 0.768769 𝑊/𝐾
The value for the number of heat transfer units is:
𝑁𝑇𝑈 =
𝑈𝐴
. 768769 𝑊/𝐾
=
= .50903
𝐶𝑚𝑖𝑛
1.5102 𝑊/𝐾
Now that the heat capacity ratio and NTU values are determined for this concurrent
concentric tube heat exchanger the effectiveness value is calculated as follows:
𝜀=
1 − 𝑒 [−𝑁𝑇𝑈(1+𝐶𝑟 )] 1 − 𝑒 [−.50903(1+.12122)]
=
= .387872
1 + 𝐶𝑟
1 + .12122
The equation for heat transferred in the NTU-effectiveness method is in terms of this
effectiveness value as well as the minimum heat capacity.
1.5102 𝑊
𝑞 = 𝜀 × 𝐶𝑚𝑖𝑛 × (𝑇ℎ,𝑖 − 𝑇𝑐,𝑖 ) = (. 387872) (
) (398.15 𝐾 − 293.15 𝐾)
𝐾
= 61.50782 𝑊
From equations 11 and 12 we know the overall energy balance gives the outlet
temperatures of the fluids by subtracting or adding the value of the heat transferred
divided by the heat capacity of the fluid to the inlet temperature of that fluid. For this
case:
𝑞
𝑇𝑐,𝑜 = 𝑇𝑐,𝑖 + ⁄𝐶
= 293.15 𝐾 + 61.50782 𝑊⁄12.4588 𝑊/𝐾 = 298.0869 𝐾
𝑚𝑎𝑥
𝑞
𝑇ℎ,𝑜 = 𝑇ℎ,𝑖 − ⁄𝐶
= 398.15 𝐾 − 61.50782 𝑊⁄1.5102 𝑊/𝐾 = 357.4235 𝐾
𝑚𝑖𝑛
39
As a double check for this calculation, the log mean temperature difference was
determined using the outlet temperatures calculated and then compared to the log mean
temperature difference determined by equation 2.
𝑞 = 𝑈𝐴∆𝑇𝐿𝑀 ∴ ∆𝑇𝐿𝑀 =
∆𝑇𝐿𝑀 =
𝑞
61.50782𝑊
=
= 80.017𝐾
𝑈𝐴 . 768769𝑊/𝐾
(357.4235𝐾 − 298.0869𝐾) − (398.15𝐾 − 293.15𝐾)
∆𝑇2 − ∆𝑇1
=
=
∆𝑇
𝐿𝑁 ( 2⁄∆𝑇 ) 𝐿𝑁 (357.4235𝐾 − 298.0869𝐾)⁄(398.15𝐾
− 293.15𝐾)
1
∆𝑇𝐿𝑀 = 80.017𝐾 ∴ 𝑇ℎ𝑒 𝑐ℎ𝑒𝑐𝑘 𝑖𝑠 𝑆𝐴𝑇
After completing the model generation in COMSOL, the study of the heat
exchanger consisted of running the model with the same oil velocity of .0001 m/s but the
cooling flow velocity was increased from .0001 m/s to .001 m/s and then .01 m/s.
Maintaining the same fluid velocities for both, the inlet temperature of the cooling flow
was then increased thus lowering the temperature difference between the fluids. Figure
30 shows that as the cooling water flow increases the outlet temperature of the oil
lowers. For each increase of velocity (each increment was ten times the previous), the
outlet temperature of the hot fluid lowered by approximately 2K. So therefore as the
velocity increases for the colder fluid, the heat capacity rate for the cooling fluid will
increase which will decrease the ratio between the capacity rates and therefore change
the effectiveness of the heat exchanger. In the case of this concurrent flow heat
exchanger, the effectiveness increases which therefore increases the amount of heat
transferred, allowing the temperature of the oil to drop more and the temperature of the
water to raise more.
40
370
369
Oil Flow Oulet Temperature (K)
368
367.05901
367
366
365.1556
365
363.7862
364
Th,o (oil)
363
362
361
360
0
0.002
0.004
0.006
0.008
0.01
0.012
Cooling Water Velocity (m/s)
Figure 30: Cooling Water Flow Rate Effect on Oil Outlet Temperature
Figure 31 shows that as the cooling water flow increases the temperature change
of the hotter fluid increases. This is due to the fact that oil temperature is the lowest for
the larger the cooling flow.
41
35
34.3638
34.5
Change in Oil Temperature (K)
34
33.5
32.9944
33
32.5
Δ in Oil Temp
32
31.5
31.09099
31
30.5
0
0.002
0.004
0.006
0.008
0.01
Cooling Water Velocity (m/s)
Figure 31: Cooling Water Flow Rate Effect on the Change in Oil Temperature
As mentioned above, the velocity of the oil and water was held constant at .0001
m/s and the inlet temperature of the water was increased from 293.15 K to 303.15 K and
to 313.15K. Figure 32 depicts the temperature changes in both the hot and cold fluids as
the temperature drop between the fluids increase. For the smaller difference between the
inlet temperature of the oil and water, 85F, the change between the inlet and outlet for
both the cold and hot fluids is the smallest. But as the temperature difference increase to
95F and 105F, the temperature change between both the cold and hot fluids increases
linearly.
42
Change in Fluid Temperature Between Inlet and Outlet (K)
35
31.09099
28.83715
30
26.30021
25
20
16.62035
15.05942
Th,o-Th,i (Oil)
13.50046
15
Tc,o-Tc,i (Water)
10
5
0
75
80
85
90
95
100
105
110
Difference in Inlet Temperatures Between Fluids (K)
Figure 32: Temperature Change in the Fluids vs. the Difference in Inlet Temperatures
For each case the results were compared to the COMSOL values and the percent
difference calculated. Most of the results were in the range of 2-3% different. These
results are part of the results spreadsheet located in the appendix section. There are a
couple possible reasons for the difference between the actual (COMSOL) and calculated
values. First, the heat transfer coefficient for the outer portion of the inner pipe was
estimated in the hand calculations, and the COMSOL model used the previously
determined value from the material library. For better results, if this coefficient could be
user defined in the finite element program or the value the program uses recorded for use
in the hand calculations, a more accurate solution might have been obtained. This
affected the overall heat transfer coefficient and therefore the NTU value and the
effectiveness of the heat exchanger. Secondly, the material property values used in the
calculations were based on the inlet temperatures of the oil and water. To create a better
representation of the actual case, these should have been based off the average
temperature of the fluids. If a more in depth study could have been performed, the outlet
43
temperature should have initially been guessed and several iterations of the calculations
performed until the value of the outlet temperature settles out to a near constant value. In
this method the specific heat values, Prandtl numbers, thermal conductivity numbers,
viscosity, and densities would be based off the average temperature of the fluids (inlet
temperature plus outlet temperature divided by 2).
3.4 Flow in a Counter-Current Heat Exchanger
3.4.1
Laminar Flow in a Counter-current Heat Exchanger COMSOL Model
Adding onto the COMSOL model of flow through a pipe with a pipe wall, a
second pipe and pipe wall were added. Flow for each fluid was defined to be flowing in
opposite directions with the outer flow at a lower temperature cooling the inner fluid.
For the purposes of simplifying the model for development, the same type of pipe was
used as in the previous model, the same fluid, water, was used for both sides of the fluid
flows, and the same dimensions and temperatures were used. Once the model was made
and analyzed the velocity, temperatures, and materials could be changed for further
investigation. Figures 33 and 34 show how the velocity and temperature profiles are
affected by changing the direction of the outer, cooling fluid.
Figure 33: Velocity Profile for Countercurrent Heat Exchanger
44
Figure 34: Temperature Profile for Countercurrent Heat Exchanger
Figure 35 shows approximately the same drop in temperature from centerline to
the wall for the inner fluid as was shown in figures 17 and 18 for the Graetz problem and
for laminar flow in a pipe wall. Temperature is the same from .07m to .12m since this
represents the inlet temperature of the cooling water flow now that its direction is
reversed for the counter-current flow heat exchanger.
Figure 35: Outlet of the Inner Pipe, Inlet of the Outer Pipe
45
Figure 36 is the opposite temperature distribution curve where the cross section
of the heat exchanger was taken at the entrance of the inner, hotter fluid and the exit of
the outer, cooler fluid. The figure shows a constant temperature from 0m to .05m since
this is the inlet temperature of the oil. There is then a drop in temperature across the pipe
wall between the oil and water and then a large drop in temperature across the cooler
outer fluid, water. This shows that for the cooling outer fluid, the flow nearest the pipe
wall is hotter than the flow nearest the outer wall due to the heat transfer from the oil.
The flow nearest the outer wall remains approximately that of the temperature at the
inlet of the heat exchanger.
Figure 36: Inlet of the Inner Pipe, Outlet of the Outer Pipe
Looking at the difference in temperature profile of the inner fluid versus the
temperature profile of the outer fluid, the proven results of a counter-current heat
exchanger are obtained. The hotter fluid gradually lowers in temperature as the colder
fluid gradually rises in temperature to meet it. Figure 37 represents this relationship with
respect to the center of flow for both the oil and water. Figure 36 already showed that
even though the colder fluid gets hotter as expected when the hotter fluid gets colder,
that for the cooling flow there is a temperature drop across the flow from closest to the
inner pipe outside wall to the closest to the outer pipe inside wall.
46
Figure 37: Counter-current Flow Heat Exchanger Temperature Change
3.4.2
Turbulent Flow in a Counter-Current Heat Exchanger
For turbulent flow in the counter-current heat exchanger, the velocity profile was
almost exactly that of turbulent flow in a singular pipe. The extent of the velocity
distribution was between 9.8-10.2 X 10^-5 m/s which as discussed is due to the
developed flow already entering both pipes due to the turbulence being applied to the
model. Figure 38 shows that both velocity profiles are developed prior to heat transfer,
and figure 39 shows the extent of velocity distribution throughout the model.
47
Figure 38: Turbulent Flow Arrow Velocity Profile
Figure 39: Velocity Profile for a Turbulent Counter-current Flow Heat Exchanger
Although the velocity profiles were different between the concurrent and counter
current flow heat exchangers with turbulence applied, the temperature profiles between
the 2 types of heat exchangers were almost identical, and very little change is seen
between the hot and cold fluid along the length of the center of each fluid. Figures 40
and 41 show the turbulent temperature profiles in the counter-current type heat
exchanger.
48
Figure 40: Temperature Profile for a Turbulent Counter-current Flow Heat Exchanger
Figure 41: Turbulent Counter-current Flow Heat Exchanger Temperature Change
Figure 41 is the same as the plot of figure 24 (concurrent flow heat exchanger
temperature change), except that due to the little to no change in temperature because of
the turbulent flow, the hot and cold fluid variation with respect to the arc length looks
like a constant temperature is being maintained. It is also the same as figure 29 except
figure 41 is for the reversed flow of the cooling water (counter-current flow).
49
3.4.3
Laminar Flow in a Counter-current Heat Exchanger Problem Calculations
The only equation that is different between the two heat exchangers and heat
transfer performances is the effectiveness of the heat exchanger. However, this number
is directly proportional to the amount of heat transferred and therefore proportional to
the change in outlet temperature as well. Where the effectiveness equation for the
concurrent flow concentric tube heat exchanger is fairly simple, the counter-current flow
equation adds more terms (see equation 5). The same conditions discussed and analyzed
in section 3.3.3 for laminar flow in a concurrent heat exchanger were also used in a
counter-current heat exchanger. Again oil was flowing in the inner pipe as the hotter
fluid and water was used to cool it by flowing in the outer pipe. Using the conditions for
the first case of counter-current flow where oil velocity is .0001 m/s and cooling water
flow is .0001 m/s, the counter-current calculations were performed. The heat capacity
ratio and NTU values determined for the concurrent concentric tube heat exchanger are
the same for the counter-current heat exchanger, but the effectiveness value is calculated
as follows:
1 − 𝑒 [−𝑁𝑇𝑈(1−𝐶𝑟 )]
1 − 𝑒 [−.50903(1−.12122)]
𝜀=
=
= .390964
1 − 𝐶𝑟 𝑒 [−𝑁𝑇𝑈(1−𝐶𝑟 )] 1 − (.12122 ∗ 𝑒 [−.50903(1−.12122)] )
The equation for heat transferred in the NTU-effectiveness method is in terms of this
effectiveness value as well as the minimum heat capacity.
1.5102𝑊
𝑞 = 𝜀 × 𝐶𝑚𝑖𝑛 × (𝑇ℎ,𝑖 − 𝑇𝑐,𝑖 ) = (. 390964) (
) (398.15 𝐾 − 293.15 𝐾)
𝐾
= 61.99814 𝑊
From equations 11 and 12 we know the overall energy balance gives the outlet
temperatures of the fluids by subtracting or adding the value of the heat transferred
divided by the heat capacity of the fluid to the inlet temperature of that fluid. For this
case:
𝑞
𝑇𝑐,𝑜 = 𝑇𝑐,𝑖 + ⁄𝐶
= 293.15 𝐾 + 61.99814 𝑊⁄12.4588 𝑊/𝐾 = 298.1263 𝐾
𝑚𝑎𝑥
𝑞
𝑇ℎ,𝑜 = 𝑇ℎ,𝑖 − ⁄𝐶
= 398.15 𝐾 − 61.99814 𝑊⁄1.5102 𝑊/𝐾 = 357.0988 𝐾
𝑚𝑖𝑛
50
As a double check for this calculation, the log mean temperature difference was
determined using the outlet temperatures calculated and then compared to the log mean
temperature difference determined by equation 2.
𝑞 = 𝑈𝐴∆𝑇𝐿𝑀 ∴ ∆𝑇𝐿𝑀 =
∆𝑇𝐿𝑀 =
𝑞
61.99814 𝑊
=
= 80.64597 𝐾
𝑈𝐴 . 768769 𝑊/𝐾
(357.0988 𝐾 − 293.15 𝐾) − (398.15 𝐾 − 298.1263 𝐾)
∆𝑇2 − ∆𝑇1
=
=
∆𝑇
𝐿𝑁 ( 2⁄∆𝑇 ) 𝐿𝑁 (357.0988 𝐾 − 293.15 𝐾)⁄(398.15
𝐾 − 298.1263 𝐾)
1
∆𝑇𝐿𝑀 = 80.64597 𝐾 ∴ 𝑇ℎ𝑒 𝑐ℎ𝑒𝑐𝑘 𝑖𝑠 𝑆𝐴𝑇
Just as in the concurrent heat exchanger model, the cooling water velocity was
increased while the oil velocity remained constant. Figure 42 shows that like the
concurrent heat exchanger, as the cooling flow increases, the hotter fluid’s outlet
temperature will decrease. As explained previously, the velocity changes the heat
capacity rate which in turn affects the ratio of capacity rates and then the heat exchanger
effectiveness. As shown, the counter-current flow causes a larger drop between velocity
increases. It was seen in the concurrent flow model that for the velocity increase by a
multiple of ten, about a 2K drop in oil outlet temperature was seen. In the case of
counter-current flow, the velocity increase causes a drop in oil outlet temperature of
approximately 4K, twice that of its concurrent flow counterpart. While this shows a
better heat transfer in the counter-current type flow, it was anticipated that there be a
larger temperature drop for each velocity increase for the counter-current flow than
concurrent. However, the concurrent flow heat exchanger actually caused the oil outlet
temperature to be lower at each velocity increment.
It should be noted that for these examples of the counter-current heat exchanger
the percent difference for velocity variation was less than 5% like the concurrent flow
heat exchanger, except that the first iteration in case 1 (oil and water velocity of .0001
m/s) the difference was 8.1%. Even though the differences were similar to the concurrent
flow heat exchanger, they were consistently higher in the iterations of case 1. This
higher percent difference compiled with the fact that the material properties were based
on the inlet temperature and not average, and that the heat transfer coefficient of the
outer wall of the inner pipe was estimated, could have caused enough error to make the
temperature drop in the counter-current heat exchanger model not as much as it should
51
have been. It could also have been due to the low velocities and temperatures chosen for
the model. As it were, the hand calculations showed a consistently lower oil outlet
temperature for the counter-current flow than concurrent flow as cooling flow increased
as well as a consistently rising cooling water outlet temperature. However, the difference
in outlet temperatures between models was very small. For example, in the first iteration
of case 1, the counter current flow showed a lowering oil outlet temperature of
357.0988K while the concurrent flow lowered the oil temperature to 357.4235K. The
results for the counter-current analysis can be found in the appendix section.
376
375.54683
Oil Flow Oulet Temperature (K)
375
374
373
372
371.25821
371
Th,o (oil)
370
369
367.44359
368
367
0
0.002
0.004
0.006
0.008
0.01
Cooling Water Velocity (m/s)
Figure 42: Cooling Water Flow Rate Effect on Oil Temperature for Counter-Current Flow
Figure 43 shows the temperature change of the oil from inlet to outlet as the
cooling water flow increased. As in the concurrent model examples, the inlet
temperatures remained the same, and just cooling velocity changed.
52
32
30.70641
Change in Oil Temperature (K)
30
28
26.89179
26
Δ in Oil Temp
24
22.60317
22
20
0
0.002
0.004
0.006
0.008
0.01
Cooling Water Velocity (m/s)
Figure 43: Cooling Water Flow Rate Effect on the Change in Oil Temperature for Counter-Current
Flow
Figure 44 shows how varying the temperature difference between the two fluids
at the inlet of flow (by changing the water’s inlet temperature) affects the overall change
in temperature for each fluid. As seen in figure 32 for the concurrent flow model, the
relationship is linear. As the difference of the fluids gets larger and larger (the cooling
water gets colder and colder) the change in each fluid’s temperature increases verifying
the proportionality between temperature difference and heat transferred (equation 1).
What’s interesting to note and not expected, was that for concurrent flow, the oil
temperature between inlet and outlet changed more than the cooling water. There was an
approximate 2K change in inlet and outlet oil temperature per 10K delta temperature as
well as for the cooling water. But the Oil changed from 26.30021K to 28.83715K to
31.09099K while water changed from 13.50046K to 15.05942K to 16.62035K which is
almost reverse to the counter-current model. In the counter-current model, the oil
53
changed the least, from 19.16949K to 20.96748K to 22.60317K while the water changed
the most, 25.842385K to 28.62676K to 31.33479K.
Change in Fluid Temperature Between Inlet and Outlet (K)
35
31.33479
28.62676
30
25.842385
25
22.60317
20.96748
19.16949
20
Tc,o-Tc,I (Water)
15
Th,i-Th,o (Oil)
10
5
0
75
80
85
90
95
100
105
110
Difference in Inlet Temperatures Between Fluids (K)
Figure 44: Temperature Change in the Fluids vs. the Difference in Inlet Temperatures for CounterCurrent Flow
3.5
Fouling on Heat Transfer Surfaces
Throughout the life of the heat exchanger the heat transfer surfaces, which are the
inside and outside of the concentric pipes in a double pipe heat exchanger, there will be
some corrosion, deterioration, or wear on the surface that will ultimately lower the
effectiveness of the heat exchanger performance. A form of this “fouling” is marine
growth. For systems that utilize seawater as a means of cooling and have it running
through a portion of the heat exchanger, sea-life in the form of mussels, barnacles, and
other organisms will attach to the sides of the pipe, especially in the warmer waters. This
not only significantly decreases the ability for heat transfer to occur since the thermal
conductivity of the solid material drops, but the surface roughness increases which in
turn increases the amount of drag or resistance in the pipe to flow, therefore affecting
54
flow velocity. Marine growth not only makes inspection and maintenance of systems
more difficult, but increases the corrosion of the metals used in the heat exchanger
design. In submarine uses, these seawater heat exchanger systems use chlorination and
filters to protect the system from sea growth.
3.5.1
Flow in a Concurrent Flow Heat Exchanger with Fouling
3.5.2
Laminar Flow in a Concurrent Heat Exchanger with Fouling COMSOL
Model
To approximate a layer of fouling on the outer wall of the inner pipe, a piecewise
expression for the thermal conductivity was defined from 0.050 m to .069 m of the inner
pipe to be 393.11
𝑊
𝑚𝐾
which is the thermal conductivity of copper used in the
calculations based off of the inlet temperature of the oil. Then the thermal conductivity
was defined to be 2.7
𝑊
𝑚𝐾
from 0.069 m to 0.070 m. 2.7 was chosen as the thermal
conductivity of the marine growth since the thermal conductivity would be low and this
is the approximate value for calcium carbonate conductivity which is a component of
some types of sea life. A plot of this piecewise expression for thermal conductivity is
show below in figure 45.
Figure 45: Fouling Layer Thermal Conductivity Expression
55
3.5.3
Laminar Flow in a Concurrent Heat Exchanger with Fouling Problem
Calculations
Fouling affects the heat transfer of the heat exchanger by adding a term onto the
overall heat transfer coefficient equation. Depending upon whether the fouling is on the
inside wall, outside wall, or both will determine how many terms are in the heat transfer
coefficient equation. This equation uses a fouling factor which is dependent on the type
of fouling. For marine growth and sea water as discussed previously, the representative
fouling factor is .0001-.0002 𝑚2 𝐾 ⁄𝑊 . In equation 6 a fouling factor was added to the
outer wall of the inner pipe where the cooling water is flowing over transferring heat
from the oil. This makes the overall heat transfer coefficient equations as follows:
𝑈𝐴 =
𝑈𝐴 =
=
1
𝐷
ln( 𝑜⁄𝐷 )
𝑅
1
1
𝑖 + 𝑓,𝑜
+
+
ℎ𝑖 𝐴𝑖 ℎ𝑜 𝐴𝑜 2𝜋𝐿𝐾𝑐𝑜𝑝𝑝𝑒𝑟 𝐴𝑜
[31]
1
𝐷
ln( 𝑜⁄𝐷 )
𝑅𝑓,𝑜
1
1
𝑖
+
+
+
ℎ𝑖 𝐴𝑖 ℎ𝑜 𝐴𝑜 2𝜋𝐿𝐾𝑐𝑜𝑝𝑝𝑒𝑟
𝐴𝑜
1
2
ln(. 14𝑚⁄. 10𝑚)
(.0002 𝑚 𝐾⁄𝑊 )
1
1
+
+
+
(5.9435𝑊/𝑚2 𝐾)(.3142𝑚2 ) (. 5 ∗ 5.9435𝑊/𝑚2 𝐾)(.4398𝑚2 ) 2𝜋(1𝑚)(393.11𝑊/𝑚𝐾)
(.4398𝑚2 )
𝑈𝐴 = 0.7685006 𝑊/𝐾
The value for the number of heat transfer units is:
𝑁𝑇𝑈 =
𝑈𝐴
. 7685006 𝑊/𝐾
=
= .50885
𝐶𝑚𝑖𝑛
1.5102 𝑊/𝐾
Now that the heat capacity ratio and NTU values are determined for this concurrent
concentric tube heat exchanger the effectiveness value is calculated as follows:
𝜀=
1 − 𝑒 [−𝑁𝑇𝑈(1+𝐶𝑟 )] 1 − 𝑒 [−.50885(1+.12122)]
=
= .387771
1 + 𝐶𝑟
1 + .12122
The equation for heat transferred in the NTU-effectiveness method is in terms of this
effectiveness value as well as the minimum heat capacity.
1.5102𝑊
𝑞 = 𝜀 × 𝐶𝑚𝑖𝑛 × (𝑇ℎ,𝑖 − 𝑇𝑐,𝑖 ) = (. 387771) (
) (398.15𝐾 − 293.15𝐾)
𝐾
= 61.49188 𝑊
From equations 11 and 12 we know the overall energy balance gives the outlet
temperatures of the fluids by subtracting or adding the value of the heat transferred
56
divided by the heat capacity of the fluid to the inlet temperature of that fluid. For this
case:
𝑞
𝑇𝑐,𝑜 = 𝑇𝑐,𝑖 + ⁄𝐶
= 293.15𝐾 + 61.49188 𝑊⁄12.4588 𝑊/𝐾 = 298.08563 𝐾
𝑚𝑎𝑥
𝑞
𝑇ℎ,𝑜 = 𝑇ℎ,𝑖 − ⁄𝐶
= 398.15𝐾 − 61.49188 𝑊⁄1.5102 𝑊/𝐾 = 357.43402 𝐾
𝑚𝑖𝑛
As a double check for this calculation, the log mean temperature difference was
determined using the outlet temperatures calculated and then compared to the log mean
temperature difference determined by equation 2.
𝑞 = 𝑈𝐴∆𝑇𝐿𝑀 ∴ ∆𝑇𝐿𝑀 =
∆𝑇𝐿𝑀 =
𝑞
61.49188 𝑊
=
= 80.015 𝐾
𝑈𝐴 . 7685006 𝑊/𝐾
(357.43402 𝐾 − 298.08563 𝐾) − (398.15 𝐾 − 293.15 𝐾)
∆𝑇2 − ∆𝑇1
=
∆𝑇
𝐿𝑁 ( 2⁄∆𝑇 ) 𝐿𝑁 (357.43402 𝐾 − 298.08563 𝐾)⁄(398.15
𝐾 − 293.15 𝐾)
1
=
∆𝑇𝐿𝑀 = 80.015𝐾 ∴ 𝑇ℎ𝑒 𝑐ℎ𝑒𝑐𝑘 𝑖𝑠 𝑆𝐴𝑇
Since the overall heat transfer coefficient is inversely proportional to the fouling
factor, by adding this term into the denominator of the expression, UA will be lower for
a fouled heat exchanger. This causes the NTU for the heat exchanger to be lower and
then in turn the effectiveness since they are proportional. With the effectiveness
lowering due to the fouling, so does the amount of heat transferred since effectiveness is
proportional to the heat transferred. The fouling caused the heat to drop from 61.50782
W to 61.49188 W, for the first case of the oil and water heat exchanger. Chapter 6 of the
paper, the appendix has the excel spreadsheets used for the analysis. Similar to the nonfouled laminar concurrent flow heat exchanger calculations, the percent difference
between the COMSOL values and the calculated values ranged from 0.03% to
approximately 3.6% except for the cooling water outlet temperature for the first case
(both velocities .0001 m/s, oil temperature 398.15 K, and water temperature 293.15 K).
The reason discrepancies in the results can be attributed to the fact that an approximate
value was used for the marine growth fouling factor and the piecewise expression used
by COMSOL was a crude estimate to develop a fouling layer to the model. The value of
thermal conductivity used for the marine growth could also have caused some error. For
57
a more in depth study, a geometric region could be created and meshed into the model
after an analysis was done on what was the best mesh to apply.
Comparing the two types of concurrent heat exchangers, it can be shown that for
the heat exchanger with fouling, the amount of heat transfer is not as much as that of a
non-fouled heat exchanger. Figure 46 shows that while an increase of cooling water flow
will cool the hotter fluid in approximately the same manner, the amount of which the oil
outlet temperature drops is lower for the fouled heat exchanger.
374
Oil Flow Oulet Temperature (K)
372
370.74566
370
367.5886
368
Th,o (oil)
365.63366
366
364
362
360
0
0.002
0.004
0.006
0.008
0.01
0.012
Cooling Water Velocity (m/s)
Figure 46: Cooling Water Flow Rate Effect on Oil Temperature for Concurrent Flow with Fouling
From the non-fouled to the fouled heat exchanger we see approximately a 2 K
rise in temperature, while for both heat exchangers the increase in temperature by ten
times will show an approximate 2 K drop in temperature. Figure 47 shows both fouled
and non-fouled concurrent heat exchangers and how the cooling water flow rate affects
both.
58
374
Oil Flow Oulet Temperature (K)
372
370.74566
370
367.5886
368
Th, o (oil) With Fouling
367.05901
366
365.63366
365.1556
Th, o (oil) No Fouling
363.7862
364
362
360
0
0.002
0.004
0.006
0.008
0.01
0.012
Cooling Water Velocity (m/s)
Figure 47: Fouled and Non-fouled Concurrent Flow Heat Exchanger Comparison
Figure 48 is the plot of the change in oil temperature as cooling water flow rate
increases. As expected the temperature drop increases proportionately with the amount
of cooling. Comparing this plot to figure 31, which showed the same relationship for the
non-fouled concurrent flow heat exchanger, provides the fact that the curves are almost
identical so the same relationship exists, but that for the fouled heat exchanger the oil
temperature does not drop as much. This is expected since the amount of heat transfer is
less due to the lower thermal conductivity of the heat transfer surfaces for the fouled heat
exchanger. Figure 48 shows that there is an approximate 3.5-2 K difference in the
temperature drops from .0001 to .001 to .01 m/s for the cooling water velocity.
59
33
32.51634
Change in Oil Temperature (K)
32
31
30.5614
30
Δ in Oil Temp
29
28
27.40434
27
0
0.002
0.004
0.006
0.008
0.01
Cooling Water Velocity (m/s)
Figure 48: Cooling Water Flow Rate Effect on the Change in Oil Temperature for Concurrent Flow
with Fouling
Similar to the non-fouled heat exchanger in figure 32, figure 49 shows that in the fouled
heat exchanger there is still a proportional relationship between the changes in fluid
temperature as the amount of temperature difference between the two fluids increases.
Since the temperature difference between the inlet flows is proportional to the heat to be
transferred (equation 3), it was expected that the temperature change for each fluid
would increase as the difference in inlet temperatures increased. Comparing the fouled
and non-fouled cases showed that there was a consistent drop at each increment for the
temperature difference of the oil of about 3.5 K between the non-fouled and fouled
cases. This proves that in the fouled heat exchanger, the lower heat transfer caused the
oil temperature to not drop as much as in the original non-fouled case. It was interesting
to note that in the non-fouled case the temperature drop of the oil was higher than the
water for each increment, but in the fouled heat exchanger, the temperature change
between the inlet and outlet of the water was larger at each increment than the oil.
60
Change in Fluid Temperature Between Inlet and Outlet (K)
35
28.78301
30
26.16604
27.40434
23.52009
25
25.1718
22.73502
20
Th,i-Th,o (Oil)
15
Tc,o-Tc,i (Water)
10
5
0
75
80
85
90
95
100
105
110
Difference in Inlet Temperatures Between Fluids (K)
Figure 49: Temperature Change in the Fluids vs. the Difference in Inlet Temperatures for
Concurrent Flow with Fouling
3.5.4
Laminar Flow in a Counter-current Heat Exchanger with Fouling
COMSOL Model
To analyze how fouling affects counter-current flow, the same oil and water heat
exchanger for counter-current flow analysis was used and the same cases analyzed but
with an additional fouling layer added to the outer wall of the inner pipe. As in the
concurrent flow fouling analysis, a piecewise expression was defined for the inner wall
which defined the inner wall to have a thermal conductivity of the copper to a point and
then the thermal conductivity of the marine growth from that point to the outside of the
pipe where the cooling water is flowing over it. Like the concurrent flow model, the
fouling layer initially was chosen to be from .069 m to .070 m for the inner wall. The
results which will be discussed were found to actually lower the oil outlet temperature
slightly more than the non-fouled example. Further analysis gave the same results if the
fouling layer was anywhere from .067 m to .070m to .069 m to .070 m. Once a fouling
61
layer of .004 m was created, the heat transfer was lower than the non-fouled example
which is what was expected.
3.5.5
Laminar Flow in a Counter-current Heat Exchanger with Fouling Problem
Calculations
As discussed in chapter 3.4.3 (laminar flow in a counter-current heat exchanger),
the difference between the concurrent and counter-current heat exchangers is the
effectiveness of each type of heat exchanger. The calculations for this have already been
discussed. As discussed in chapter 3.5.3 (laminar flow in a concurrent heat exchanger
with fouling problem calculations), the layer of marine growth fouling affects the
expression for the overall heat transfer coefficient (UA) which then in turn affects the
NTU value and the effectiveness of the heat exchanger which is proportional to the heat
transferred which directly affects what the outlet temperature becomes. Since an excel
spreadsheet was already made up for the counter-current heat exchanger for every case
analyzed, the only expression that had to be changed was that a fouling factor term was
added to the equation for the overall heat transfer coefficient (equation 31). These
spreadsheets can be found in the appendix chapter of the paper.
Like the concurrent flow example, initially a fouling layer of .001 m was used in
the piecewise fouling expression, but the COMSOL values for the outlet temperatures of
the oil were slightly lower than those for the non-fouled heat exchanger when it’s
expected that the outlet temperature of the oil should be higher in the non-fouled case
since there is less heat transfer due to the fouling layer. In the first case with
temperatures constant at 398.15 K for the oil and 293.15 K for the water and the cooling
water flow rate increased, the oil outlet was 375.5468 K for the non-fouled case and
375.20507 K for the fouled example with cooling flow at .0001 m/s. For cooling flow at
.001 m/s the non-fouled case was 371.2582 K while the fouled heat exchanger was
370.81375 K. The difference between each example was very slight and could be
attributed to the fact that the fouling layer was crudely estimated by the piecewise
expression and the fouling layer estimate of .001 m. The COMSOL model of the
counter-current heat exchanger with fouling was run several times using different layers
of fouling. It was found that for a fouling layer of .004 m the outlet temperatures for the
fluid flow were higher for the fouled heat exchanger as expected.
62
Figure 50 depicts the relationship of cooling water flow rate against outlet
temperature of the oil for both the non-fouled and fouled heat exchangers. It can be seen
that for each increment of cooling water flow rate increase, the oil outlet temperature is
higher for the fouled case, since the layer of lower thermal conductivity prohibits the
same amount of heat transfer as in the case of the non-fouled heat exchanger.
378
376.40116
Oil Flow Oulet Temperature (K)
376
375.20507
374
372
371.25583
Th, o (oil) Fouling
Th, o (oil) No Fouling
370
370.81375
367.94207
368
367.11019
366
0
0.002
0.004
0.006
0.008
0.01
Cooling Water Velocity (m/s)
Figure 50: Fouled and Non-fouled Counter-current Flow Heat Exchanger Comparison
Combining the fouled and non-fouled cases for both the concurrent and countercurrent
flow heat exchangers finds that as expected the temperature drop increases as cooling
flow increases and the amount of temperature drop decreases as fouling is applied to the
walls of the heat exchanger piping. However, not expected was that fact that the
concurrent flow heat exchanger actually caused a lower oil outlet temperature (cooled
the working fluid more) than the counter-current flow heat exchanger did. Figure 51
63
combines
as
four
cases
of
this
relationship.
378
Oil Flow Oulet Temperature (K)
376
374
372
Th, o (oil) Fouling Counter
Current
370
Th, o (oil) No Fouling
Counter Current
368
Th, o (oil) No Fouling
Concurrent
366
Th, o (oil) Fouling
Concurrent
364
362
0
0.002
0.004
0.006
0.008
0.01
Cooling Water Velocity (m/s)
Figure 51: Concurrent and Counter-current Flow Heat Exchangers with and without Fouling
64
4. Conclusion
-Talk about findings of velocity and inlet temp on outlet temp and the difference
between concurrent and countercurrent HX
-Talk about whether laminar or turbulent flow is better and the difference between the 2
-How did fouling affect the heat transfer?
-Talk about findings during the process of the project that the initial values and mesh
refinement can change the ending results unless a proper mesh is produced and verified
to give consistent results. This may involves changing mesh conditions (boundary
layers), changing the tolerance of solution convergence, or changing the type of nonlinear solver. If not done, the results could be inaccurate analysis and results that in the
industrial and business application could lead to developing and marketing the wrong or
improperly designed heat exchanger that not only could cause damage but could be a
personnel hazard in the industrial workplace.
65
5. References
[1] Beek, W.J., K.M.K. Muttzall, and J.W. van Heuven. Transport Phenomena. 2nd ed.
New York: John Wiley & Sons, Ltd., 1999.
[2] Bird, Byron R., Warren E. Stewart, and Edwin N. Lightfoot. Transport Phenomena.
Revised 2nd ed. New York: John Wiley & Sons, Inc., 2007.
[3] Blackwell, B.F. “Numerical Results for the Solution of the Graetz Problem for a
Bingham Plastic in Laminar Tube Flow with Constant Wall Temperature.”
Sandia Report. Aug. 1984.
[4] Conley, Nancy, Adeniyi Lawal, and Arun B. Mujumdar. “An Assessment of the
Accuracy of Numerical Solutions to the Graetz Problem.” Int. Comm. Heat Mass
Transfer. Vol.12. Pergamon Press Ltd. 1985.
[5] http://www.britannica.com/EBchecked/topic/130908/concentric-tube-heat-exchanger
Encyclopedia Britannica, 2006.
[6] Kays, William, Michael Crawford, and Bernhard Weigand. Convective Heat and
Mass Transfer. 4th ed. New York: The McGraw-Hill Companies, Inc.,
2005.
[7] Lemcoff, Norberto. “Heat Exchanger Design.” Groton. 10 July 2008.
[8] Lemcoff, Norberto. “Project: Heat Exchanger Design.” Groton. 17 July 2008.
[9] Sellars J., M. Tribus, and J. Klein. “Heat Transfer to Laminar Flow in a Round Tube
or Flat Conduit—The Graetz Problem Extended.” The American Society of
Mechanical Engineers. New York. 1955.
[10] Subramanian, Shankar R. “The Graetz Problem.”
[11] Valko, Peter P. “Solution of the Graetz-Brinkman Problem with the Laplace
Transform Galerkin Method.” International Journal of Heat and Mass Transfer
48. 2005.
[12] White, Frank. Viscous Fluid Flow. 3rd ed. New York:
Companies, Inc., 2006.
The McGraw-Hill
[14] W.M Kays and H.C. Perkins, in W.M. Rohsenow and J.P Harnett, Eds., Handbook
of Heat Transfer, Chap. 7, McGraw-Hill, New York, 1972.
66
6. APPENDIX
6.1 Laminar Concurrent Flow Heat Exchanger Data
CASE 1:
Iteration No.
Velocity of oil= .0001 m/s
1
2
3
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Th,I (Celsius)
Vh, I (m/s)
A oil flow
A water flow
ρ Oil (kg/m^3)
ρ Water (kg/m^3)
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
0.439823
0.314159
20
0.0001
125
0.0001
0.007854
0.029845
826
998.2
0.002979
0.000649
4182
2328
0.439823
0.314159
20
0.001
125
0.0001
0.007854
0.029845
826
998.2
0.029791
0.000649
4182
2328
0.439823
0.314159
20
0.01
125
0.0001
0.007854
0.029845
826
998.2
0.297914
0.000649
4182
2328
Cc (w/k)
Ch (w/k)
Cmin/Cmax
12.45877
1.510264
0.121221
124.5877
1.510264
0.012122
1245.877
1.510264
0.001212
μ (Pa s)
Pr
Re
k oil (w/m*k)
Nusselt Number
hi (w/m^2*k)
k Copper (w/m*k)
UA (w/k)
NTU
ε
q (w)
Tc,o (Celsius)
Tc,o (Kelvin)
Tc,o (COMSOL)
Tc,o Percent Diff
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
Th,o Percent Diff
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768769
0.50903
0.387872
61.50782
24.93691
298.0869
309.7703
3.771614
84.27347
357.4235
367.059
2.625067
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768769
0.50903
0.397797
63.08173
20.50632
293.5063
295.9225
0.816473
83.23133
356.2313
365.1556
2.443964
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768769
0.50903
0.398809
63.2422
20.05076
293.0508
293.1495
0.033675
83.12507
356.1251
363.7862
2.105942
67
CASE 2:
Iteration No.
Velocity of water and oil= .0001 m/s
1
2
3
4
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Th,I (Celsius)
Vh, I (m/s)
A oil flow
A water flow
ρ Oil (kg/m^3)
ρ Water (kg/m^3)
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
0.43982297
0.31415927
20
0.0001
125
0.0001
0.00785398
0.02984513
826
998.2
0.00297914
0.00064874
4182
2328
0.439822972
0.314159265
30
0.0001
125
0.0001
0.007853982
0.02984513
826
995.6
0.002971381
0.000648739
4179
2328
0.439823
0.314159
40
0.0001
125
0.0001
0.007854
0.029845
826
992.2
0.002961
0.000649
4179
2328
0.439822972
0.314159265
20
0.0001
150
0.0001
0.007853982
0.02984513
811
998.2
0.002979141
0.000636958
4182
2440
Cc (w/k)
Ch (w/k)
Cmin/Cmax
12.4587672
1.51026412
0.12122099
12.41740188
1.51026412
0.121624808
12.375
1.510264
0.122042
12.45876723
1.554177302
0.124745673
μ (Pa s)
Pr
Re
k oil (w/m*k)
Nusselt Number
hi (w/m^2*k)
k Copper (w/m*k)
UA (w/k)
NTU
ε
q (w)
Tc,o (Celsius)
Tc,o (Kelvin)
Tc,o (COMSOL)
Tc,o Percent Diff
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
Th,o Percent Diff
0.00915
159
0.90273224
0.134
4.43545035
5.94350347
393.111
0.76876929
0.5090297
0.38787175
61.5078227
24.9369108
297.936911
309.77025
3.82003733
84.2734662
357.273466
367.05901
2.66593206
0.00915
159
0.90273224
0.134
4.435450351
5.94350347
393.111
0.768769293
0.509029701
0.38783566
55.64475678
34.48119158
307.4811916
318.20942
3.371436464
88.15561228
361.1556123
369.31285
2.20876087
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768769
0.50903
0.387798
49.78263
44.02284
317.0228
326.6505
2.947377
92.03713
365.0371
371.8498
1.832099
0.00564
104
1.437943262
0.132
4.46366455
5.892037207
391.3795
0.762112671
0.490364047
0.376916352
76.15332906
26.11242891
299.1124289
313.80913
4.683324888
101.0008742
374.0008742
386.5914
3.25680441
68
6.2 Laminar Counter-Current Flow Heat Exchanger Data
CASE 1:
Iteration No.
Velocity of oil= .0001 m/s
1
2
3
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Th,I (Celsius)
Vh, I (m/s)
A oil flow
A water flow
ρ Oil (kg/m^3)
ρ Water (kg/m^3)
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
0.439823
0.314159
20
0.0001
125
0.0001
0.007854
0.029845
826
998.2
0.002979
0.000649
4182
2328
0.439823
0.314159
20
0.001
125
0.0001
0.007854
0.029845
826
998.2
0.029791
0.000649
4182
2328
0.439823
0.314159
20
0.01
125
0.0001
0.007854
0.029845
826
998.2
0.297914
0.000649
4182
2328
Cc (w/k)
Ch (w/k)
Cmin/Cmax
12.45877
1.510264
0.121221
124.5877
1.510264
0.012122
1245.877
1.510264
0.001212
μ (Pa s)
Pr
Re
k oil (w/m*k)
Nusselt Number
hi (w/m^2*k)
k Copper (w/m*k)
UA (w/k)
NTU
ε
q (w)
Tc,o (Celsius)
Tc,o (Kelvin)
Tc,o (COMSOL)
Tc,o Percent Diff
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
Th,o Percent Diff
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768769
0.50903
0.390964
61.99814
24.97627
298.1263
324.4848
8.123192
83.94881
357.0988
375.5468
4.912309
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768769
0.50903
0.39812
63.13294
20.50674
293.5067
294.0418
0.181983
83.19742
356.1974
371.2582
4.05669
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768769
0.50903
0.398841
63.24734
20.05077
293.0508
293.15
0.033855
83.12167
356.1217
367.4436
3.081268
69
CASE 2:
Iteration No.
Velocity of water and oil= .0001 m/s
1
2
3
4
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Th,I (Celsius)
Vh, I (m/s)
A oil flow
A water flow
ρ Oil (kg/m^3)
ρ Water (kg/m^3)
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
0.43982297
0.31415927
20
0.0001
125
0.0001
0.00785398
0.02984513
826
998.2
0.00297914
0.00064874
4182
2328
0.439822972
0.314159265
30
0.0001
125
0.0001
0.007853982
0.02984513
826
995.6
0.002971381
0.000648739
4179
2328
0.439823
0.314159
40
0.0001
125
0.0001
0.007854
0.029845
826
992.2
0.002961
0.000649
4179
2328
0.439822972
0.314159265
20
0.0001
150
0.0001
0.007853982
0.02984513
811
998.2
0.002979141
0.000636958
4182
2440
Cc (w/k)
Ch (w/k)
Cmin/Cmax
12.4587672
1.51026412
0.12122099
12.41740188
1.51026412
0.121624808
12.375
1.510264
0.122042
12.45876723
1.554177302
0.124745673
μ (Pa s)
Pr
Re
k oil (w/m*k)
Nusselt Number
hi (w/m^2*k)
k Copper (w/m*k)
UA (w/k)
NTU
ε
q (w)
Tc,o (Celsius)
Tc,o (Kelvin)
Tc,o (COMSOL)
Tc,o Percent Diff
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
Th,o Percent Diff
0.00915
159
0.90273224
0.134
4.43545035
5.94350347
393.111
0.76876929
0.5090297
0.39096374
61.9981432
24.9762663
297.976266
324.48479
8.1694195
83.9488074
356.948807
375.54683
4.95225125
0.00915
159
0.90273224
0.134
4.435450351
5.94350347
393.111
0.768769293
0.509029701
0.390937449
56.08978617
34.51703075
307.5170308
331.77676
7.312064066
87.86094237
360.8609424
377.18252
4.327235957
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768769
0.50903
0.39091
50.18212
44.05512
317.0551
338.9924
6.471313
91.77262
364.7726
378.9805
3.748976
0.00564
104
1.437943262
0.132
4.46366455
5.892037207
391.3795
0.762112671
0.490364047
0.379811816
76.7383374
26.15938447
299.1593845
332.1002
9.918938781
100.6244639
373.6244639
396.93667
5.87302908
70
6.3 Laminar Concurrent Flow Heat Exchanger with Fouling Data
CASE 1:
Iteration No.
Velocity of oil= .0001 m/s
1
2
3
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Th,I (Celsius)
Vh, I (m/s)
A oil flow
A water flow
ρ Oil (kg/m^3)
ρ Water (kg/m^3)
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
0.439823
0.3141593
20
0.0001
125
0.0001
0.007854
0.0298451
826
998.2
0.0029791
0.0006487
4182
2328
0.439823
0.314159
20
0.001
125
0.0001
0.007854
0.029845
826
998.2
0.029791
0.000649
4182
2328
0.439823
0.314159
20
0.01
125
0.0001
0.007854
0.029845
826
998.2
0.297914
0.000649
4182
2328
Cc (w/k)
Ch (w/k)
Cmin/Cmax
12.458767
1.5102641
0.121221
124.5877
1.510264
0.012122
1245.877
1.510264
0.001212
μ (Pa s)
Pr
Re
k oil (w/m*k)
Nusselt Number
hi (w/m^2*k)
k Copper (w/m*k)
UA (w/k)
NTU
ε
q (w)
Tc,o (Celsius)
Tc,o (Kelvin)
Tc,o (COMSOL)
Tc,o Percent Diff
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
Th,o Percent Diff
0.00915
159
0.9027322
0.134
4.4354504
5.9435035
393.111
0.7685006
0.5088518
0.3877712
61.49188
24.935631
298.08563
321.93301
7.4075594
84.284022
357.43402
370.74566
3.590504
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768501
0.508852
0.397691
63.06488
20.50619
293.5062
296.7322
1.087163
83.24249
356.2425
367.5886
3.086634
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768501
0.508852
0.398702
63.22525
20.05075
293.0507
293.1495
0.033687
83.13629
356.1363
365.6337
2.597508
71
CASE 2:
Iteration No.
Velocity of water and oil= .0001 m/s
1
2
3
4
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Th,I (Celsius)
Vh, I (m/s)
A oil flow
A water flow
ρ Oil (kg/m^3)
ρ Water (kg/m^3)
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
0.439823
0.3141593
20
0.0001
125
0.0001
0.007854
0.0298451
826
998.2
0.0029791
0.0006487
4182
2328
0.439822972
0.314159265
30
0.0001
125
0.0001
0.007853982
0.02984513
826
995.6
0.002971381
0.000648739
4179
2328
0.439823
0.314159
40
0.0001
125
0.0001
0.007854
0.029845
826
992.2
0.002961
0.000649
4179
2328
0.439822972
0.314159265
20
0.0001
150
0.0001
0.007853982
0.02984513
811
998.2
0.002979141
0.000636958
4182
2440
Cc (w/k)
Ch (w/k)
Cmin/Cmax
12.458767
1.5102641
0.121221
12.41740188
1.51026412
0.121624808
12.375
1.510264
0.122042
12.45876723
1.554177302
0.124745673
μ (Pa s)
Pr
Re
k oil (w/m*k)
Nusselt Number
hi (w/m^2*k)
k Copper (w/m*k)
UA (w/k)
NTU
ε
q (w)
Tc,o (Celsius)
Tc,o (Kelvin)
Tc,o (COMSOL)
Tc,o Percent Diff
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
Th,o Percent Diff
0.00915
159
0.9027322
0.134
4.4354504
5.9435035
393.111
0.7685006
0.5088518
0.3877712
61.49188
24.935631
298.08563
321.93301
7.4075594
84.284022
357.28402
370.74566
3.630963
0.00915
159
0.90273224
0.134
4.435450351
5.94350347
393.111
0.76850064
0.508851816
0.387735146
55.63033552
34.48003021
307.6300302
329.31604
6.585166575
88.16516111
361.1651611
372.9782
3.167219663
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768501
0.508852
0.387698
49.76973
44.0218
317.1718
336.6701
5.791513
92.04568
365.0457
375.415
2.762091
0.00564
104
1.437943262
0.132
4.46366455
5.892037207
391.3795
0.761848649
0.490194168
0.376818481
76.13355498
26.11084175
299.2608418
329.06374
9.05687702
101.0135974
374.0135974
391.08175
4.364343921
72
6.4 Laminar Counter-Current Flow Heat Exchanger with Fouling
Data-Fouling Layer .001 m
CASE 1:
Iteration No.
Velocity of oil= .0001 m/s
1
2
3
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Th,I (Celsius)
Vh, I (m/s)
A oil flow
A water flow
ρ Oil (kg/m^3)
ρ Water (kg/m^3)
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
0.439823
0.314159
20
0.0001
125
0.0001
0.007854
0.029845
826
998.2
0.002979
0.000649
4182
2328
0.439823
0.314159
20
0.001
125
0.0001
0.007854
0.029845
826
998.2
0.029791
0.000649
4182
2328
0.439823
0.314159
20
0.01
125
0.0001
0.007854
0.029845
826
998.2
0.297914
0.000649
4182
2328
Cc (w/k)
Ch (w/k)
Cmin/Cmax
12.45877
1.510264
0.121221
124.5877
1.510264
0.012122
1245.877
1.510264
0.001212
μ (Pa s)
Pr
Re
k oil (w/m*k)
Nusselt Number
hi (w/m^2*k)
k Copper (w/m*k)
UA (w/k)
NTU
ε
q (w)
Tc,o (Celsius)
Tc,o (Kelvin)
Tc,o (COMSOL)
Tc,o Percent Diff
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
Th,o Percent Diff
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768501
0.508852
0.390861
61.98178
24.97495
298.125
327.8878
9.077138
83.95964
357.1096
375.2051
4.822809
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768501
0.508852
0.398013
63.11604
20.5066
293.5066
293.9691
0.15734
83.20861
356.2086
370.8138
3.938674
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768501
0.508852
0.398734
63.23039
20.05075
293.0508
293.15
0.033859
83.13289
356.1329
367.1102
2.990192
73
CASE 2:
Iteration No.
Velocity of water and oil= .0001 m/s
1
2
3
4
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Th,I (Celsius)
Vh, I (m/s)
A oil flow
A water flow
ρ Oil (kg/m^3)
ρ Water (kg/m^3)
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
0.439823
0.3141593
20
0.0001
125
0.0001
0.007854
0.0298451
826
998.2
0.0029791
0.0006487
4182
2328
0.439822972
0.314159265
30
0.0001
125
0.0001
0.007853982
0.02984513
826
995.6
0.002971381
0.000648739
4179
2328
0.439823
0.314159
40
0.0001
125
0.0001
0.007854
0.029845
826
992.2
0.002961
0.000649
4179
2328
0.439822972
0.314159265
20
0.0001
150
0.0001
0.007853982
0.02984513
811
998.2
0.002979141
0.000636958
4182
2440
Cc (w/k)
Ch (w/k)
Cmin/Cmax
12.458767
1.5102641
0.121221
12.41740188
1.51026412
0.121624808
12.375
1.510264
0.122042
12.45876723
1.554177302
0.124745673
μ (Pa s)
Pr
Re
k oil (w/m*k)
Nusselt Number
hi (w/m^2*k)
k Copper (w/m*k)
UA (w/k)
NTU
ε
q (w)
Tc,o (Celsius)
Tc,o (Kelvin)
Tc,o (COMSOL)
Tc,o Percent Diff
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
Th,o Percent Diff
0.00915
159
0.9027322
0.134
4.4354504
5.9435035
393.111
0.7685006
0.5088518
0.3908605
61.981776
24.974953
298.12495
327.88778
9.0771384
83.959645
356.95964
375.20507
4.8627875
0.00915
159
0.90273224
0.134
4.435450351
5.94350347
393.111
0.76850064
0.508851816
0.390834248
56.07497939
34.51583833
307.6658383
334.89334
8.130200998
87.87074647
360.8707465
376.84412
4.23872171
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768501
0.508852
0.390807
50.16887
44.05405
317.2041
341.8226
7.202134
91.78139
364.7814
378.653
3.663396
0.00564
104
1.437943262
0.132
4.46366455
5.892037207
391.3795
0.761848649
0.490194168
0.379711442
76.71805756
26.15775671
299.3077567
336.44947
11.03931395
100.6375125
373.6375125
396.54266
5.776212692
74
6.5 Laminar Counter-Current Flow Heat Exchanger with Fouling
Data-Fouling Layer .004 m
CASE 1:
Iteration No.
Velocity of oil= .0001 m/s
1
2
3
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Th,I (Celsius)
Vh, I (m/s)
A oil flow
A water flow
ρ Oil (kg/m^3)
ρ Water (kg/m^3)
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
0.439823
0.314159
20
0.0001
125
0.0001
0.007854
0.029845
826
998.2
0.002979
0.000649
4182
2328
0.439823
0.314159
20
0.001
125
0.0001
0.007854
0.029845
826
998.2
0.029791
0.000649
4182
2328
0.439823
0.314159
20
0.01
125
0.0001
0.007854
0.029845
826
998.2
0.297914
0.000649
4182
2328
Cc (w/k)
Ch (w/k)
Cmin/Cmax
12.45877
1.510264
0.121221
124.5877
1.510264
0.012122
1245.877
1.510264
0.001212
μ (Pa s)
Pr
Re
k oil (w/m*k)
Nusselt Number
hi (w/m^2*k)
k Copper (w/m*k)
UA (w/k)
NTU
ε
q (w)
Tc,o (Celsius)
Tc,o (Kelvin)
Tc,o (COMSOL)
Tc,o Percent Diff
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
Th,o Percent Diff
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768501
0.508852
0.390861
61.98178
24.97495
298.125
330.0611
9.675823
83.95964
357.1096
376.4012
5.125254
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768501
0.508852
0.398013
63.11604
20.5066
293.5066
294.1269
0.210899
83.20861
356.3586
371.2558
4.012657
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768501
0.508852
0.398734
63.23039
20.05075
293.0508
293.15
0.033859
83.13289
356.2829
367.9421
3.168754
75
CASE 2:
Iteration No.
Velocity of water and oil= .0001 m/s
1
2
3
4
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Th,I (Celsius)
Vh, I (m/s)
A oil flow
A water flow
ρ Oil (kg/m^3)
ρ Water (kg/m^3)
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
0.439823
0.3141593
20
0.0001
125
0.0001
0.007854
0.0298451
826
998.2
0.0029791
0.0006487
4182
2328
0.439822972
0.314159265
30
0.0001
125
0.0001
0.007853982
0.02984513
826
995.6
0.002971381
0.000648739
4179
2328
0.439823
0.314159
40
0.0001
125
0.0001
0.007854
0.029845
826
992.2
0.002961
0.000649
4179
2328
0.439822972
0.314159265
20
0.0001
150
0.0001
0.007853982
0.02984513
811
998.2
0.002979141
0.000636958
4182
2440
Cc (w/k)
Ch (w/k)
Cmin/Cmax
12.458767
1.5102641
0.121221
12.41740188
1.51026412
0.121624808
12.375
1.510264
0.122042
12.45876723
1.554177302
0.124745673
μ (Pa s)
Pr
Re
k oil (w/m*k)
Nusselt Number
hi (w/m^2*k)
k Copper (w/m*k)
UA (w/k)
NTU
ε
q (w)
Tc,o (Celsius)
Tc,o (Kelvin)
Tc,o (COMSOL)
Tc,o Percent Diff
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
Th,o Percent Diff
0.00915
159
0.9027322
0.134
4.4354504
5.9435035
393.111
0.7685006
0.5088518
0.3908605
61.981776
24.974953
298.12495
330.06108
9.6758235
83.959645
357.10964
376.40116
5.1252539
0.00915
159
0.90273224
0.134
4.435450351
5.94350347
393.111
0.76850064
0.508851816
0.390834248
56.07497939
34.51583833
307.6658383
336.85101
8.664118795
87.87074647
361.0207465
378.06195
4.507516169
0.00915
159
0.902732
0.134
4.43545
5.943503
393.111
0.768501
0.508852
0.390807
50.16887
44.05405
317.2041
343.5672
7.67336
91.78139
364.9314
379.8715
3.932947
0.00564
104
1.437943262
0.132
4.46366455
5.892037207
391.3795
0.761848649
0.490194168
0.379711442
76.71805756
26.15775671
299.3077567
339.06417
11.72533603
100.6375125
373.7875125
397.85497
6.049304212
76