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

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 ........................................................................................................ ix
ACKNOWLEDGMENT .................................................................................................. xi
ABSTRACT .................................................................................................................... xii
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 ........................ 7
1.3
Finite Element Analysis Theory ........................................................................ 8
2.
Problem Description and Methodology ..................................................................... 10
2.1
Defining Material Properties ............................................................................ 10
2.2
Methodology and Approach ............................................................................. 10
2.2.1
Finite Element Analysis Modeling ...................................................... 10
2.2.2
Defining Variable Temperature and Velocity ...................................... 11
3.
Results and Discussion .............................................................................................. 12
3.1
The Modified Graetz Problem Results ............................................................. 12
3.1.1
The Modified Graetz Problem COMSOL Model ................................ 12
3.1.2
The Modified Graetz Problem COMSOL Mesh .................................. 15
3.1.3
The Modified Graetz Problem Study Results ...................................... 19
3.1.4
The Modified Graetz Problem Calculations ........................................ 20
iii

3.1.5
Turbulent Flow with Constant Wall Temperature ............................... 22
3.2
Flow in a Pipe with Wall Conduction .............................................................. 26
3.2.1
Laminar Flow with a Pipe Wall COMSOL Model .............................. 26
3.2.2
Laminar Flow with a Pipe Wall Problem Calculations ........................ 27
3.2.3
Turbulent Flow in a Pipe with Wall Conduction ................................. 29
3.3
Flow in a Concurrent Flow Heat Exchanger .................................................... 32
3.3.1
Laminar Flow in a Concurrent Heat Exchanger COMSOL Model ..... 32
3.3.2
Turbulent Flow in a Concurrent Heat Exchanger ................................ 35
3.3.3
Laminar Flow in a Concurrent Heat Exchanger Problem Calculations 38
3.4
Flow in a Counter-Current Heat Exchanger..................................................... 45
3.4.1
Laminar Flow in a Counter-current Heat Exchanger COMSOL Model
.............................................................................................................. 45
3.4.2
Turbulent Flow in a Counter-Current Heat Exchanger ........................ 48
3.4.3
Laminar Flow in a Counter-current Heat Exchanger Problem
Calculations .......................................................................................... 51
3.5
Fouling on Heat Transfer Surfaces .................................................................. 55
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 ................................................................................................... 56
3.5.3
Laminar Flow in a Concurrent Heat Exchanger with Fouling Problem
Calculations .......................................................................................... 57
3.5.4
Laminar Flow in a Counter-current Heat Exchanger with Fouling
COMSOL Model .................................................................................. 62
3.5.5
Laminar Flow in a Counter-current Heat Exchanger with Fouling
Problem Calculations ........................................................................... 63
4.
Conclusion ................................................................................................................. 66
5.
References .................................................................................................................. 69
6.
APPENDIX................................................................................................................ 70
6.1
Laminar Concurrent Flow Heat Exchanger Data ............................................. 70
6.2
Laminar Counter-Current Flow Heat Exchanger Data .................................... 72
iv

6.3
Laminar Concurrent Flow Heat Exchanger with Fouling Data ....................... 74
6.4
Laminar Counter-Current Flow Heat Exchanger with Fouling Data- Fouling
Layer .001 m .................................................................................................... 76
6.5
Laminar Counter-Current Flow Heat Exchanger with Fouling Data- Fouling
Layer .004 m .................................................................................................... 78
v

Table 1: COMSOL Model Initial Input Data .................................................................. 12
Table 2: User Defined Material Properties ..................................................................... 13
Table 3: Modified Graetz Problem COMSOL Equations .............................................. 14
Table 4: Mesh Extension Study Results ......................................................................... 18
Table 5: Modified Graetz Problem Comparison ............................................................ 22
Table 6: Results from Mesh Refinement ........................................................................ 22
Table 7: Change in Temperature Along the Channel of Water ...................................... 28
Table 8: Fluid Properties ................................................................................................ 39
vi

Figure 1: Basic Heat Exchanger Design [4] ..................................................................... 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 ............................................................... 13
Figure 5: Modified Graetz Problem Velocity Profile ..................................................... 15
Figure 6: Modified Graetz Problem Temperature Profile .............................................. 15
Figure 7: User Defined Mesh ......................................................................................... 17
Figure 8: Centerline Temp vs. Mesh Element Number .................................................. 18
Figure 9: Initial Value Variance ..................................................................................... 19
Figure 10: Modified Graetz Problem Centerline Temperature ....................................... 20
Figure 11: Laminar Flow Velocity Profile ..................................................................... 23
Figure 12: Turbulent Flow Velocity Profile ................................................................... 23
Figure 13: Turbulent Flow Centerline Temperature ....................................................... 24
Figure 14: Turbulent Model for the Modified Graetz Problem ...................................... 25
Figure 15: Velocity Profile for Flow Through a Pipe ..................................................... 26
Figure 16: Temperature Profile of Flow Through a Pipe ............................................... 27
Figure 17: Outflow Temperature Distribution for the Modified Graetz Problem .......... 29
Figure 18: Outflow Temperature Distribution for the Modified Graetz Problem with a
Pipe Wall ......................................................................................................................... 29
Figure 19: Axial Centerline Velocity for Laminar Flow in a Pipe ................................. 30
Figure 20: Axial Centerline Velocity for Turbulent Flow in a Pipe ............................... 30
Figure 21: Temperature Profile of Turbulent Flow Through a Pipe .............................. 31
Figure 22: Velocity Profile for Concurrent Heat Exchanger .......................................... 32
Figure 23: Temperature Profile for Concurrent Heat Exchanger ................................... 33
Figure 24: Concurrent Flow Heat Exchanger Temperature Change .............................. 34
Figure 25: Temperature Change Across the Outlet Flow ............................................... 35
Figure 26: Laminar Flow Developing Velocity Profile .................................................. 35
Figure 27: Velocity Profile for a Turbulent Concurrent Flow Heat Exchanger ............. 36
Figure 28: Temperature Profile for a Turbulent Concurrent Flow Heat Exchanger ...... 37
Figure 29: Turbulent Concurrent Flow Heat Exchanger Temperature Change ............. 37
vii

Figure 30: Cooling Water Flow Rate Effect on Oil Outlet Temperature ....................... 42
Figure 31: Cooling Water Flow Rate Effect on the Change in Oil Temperature ........... 43
Figure 32: Temperature Change in the Fluids vs. the Difference in Inlet Temperatures 44
Figure 33: Velocity Profile for Countercurrent Heat Exchanger .................................... 45
Figure 34: Temperature Profile for Countercurrent Heat Exchanger ............................. 46
Figure 35: Outlet of the Inner Pipe, Inlet of the Outer Pipe ........................................... 46
Figure 36: Inlet of the Inner Pipe, Outlet of the Outer Pipe ........................................... 47
Figure 37: Counter-current Flow Heat Exchanger Temperature Change ....................... 48
Figure 38: Turbulent Flow Arrow Velocity Profile ........................................................ 49
Figure 39: Velocity Profile for a Turbulent Counter-current Flow Heat Exchanger ..... 49
Figure 40: Temperature Profile for a Turbulent Counter-current Flow Heat Exchanger 50
Figure 41: Turbulent Counter-current Flow Heat Exchanger Temperature Change ...... 50
Figure 42: Cooling Water Flow Rate Effect on Oil Temperature for Counter-Current
Flow ................................................................................................................................. 53
Figure 43: Cooling Water Flow Rate Effect on the Change in Oil Temperature for
Counter-Current Flow ...................................................................................................... 54
Figure 44: Temperature Change in the Fluids vs. the Difference in Inlet Temperatures for Counter-Current Flow ................................................................................................ 55
Figure 45: Fouling Layer Thermal Conductivity Expression ......................................... 56
Figure 46: Cooling Water Flow Rate Effect on Oil Temperature for Concurrent Flow with Fouling ..................................................................................................................... 59
Figure 47: Fouled and Non-fouled Concurrent Flow Heat Exchanger Comparison ...... 60
Figure 48: Cooling Water Flow Rate Effect on the Change in Oil Temperature for
Concurrent Flow with Fouling......................................................................................... 61
Figure 49: Temperature Change in the Fluids vs. the Difference in Inlet Temperatures for Concurrent Flow with Fouling ................................................................................... 62
Figure 50: Fouled and Non-fouled Counter-current Flow Heat Exchanger Comparison
......................................................................................................................................... 64
Figure 51: Concurrent and Counter-current Flow Heat Exchangers with and without
Fouling ............................................................................................................................. 65
viii

A= Area ( 𝑚 2
)
𝐶
𝐶
= Heat capacity rate for the cold fluid, 𝑚̇ 𝑐
× 𝐶 𝑝,𝑐
, (W/K)
𝐶 ℎ
= Heat capacity rate for the hot fluid, 𝑚̇ ℎ
× 𝐶 𝑝,ℎ
, (W/K)
𝐶 𝑝
= Specific Heat at Constant Pressure (J/ kg K)
𝐶 𝑟
= Heat capacity ratio (
𝐶 𝑚𝑖𝑛 ⁄
𝐶 𝑚𝑎𝑥
)
𝐶 𝑚𝑖𝑛
= Minimum of 𝐶
𝐶
and 𝐶 ℎ
(W/K)
𝐶 𝑚𝑎𝑥
= Maximum of 𝐶
𝐶
and 𝐶 ℎ
(W/K)
D= Diameter of a circular tube (m) h= Heat transfer coefficient (W/ 𝑚 2
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, 𝜇𝐶 𝑝 𝑘
𝑊
𝑄̇ = Heat flux ( 𝑚 2
) q= Heat (J) 𝑞̇ = Heat transfer rate (W or J/s) r= Radial distance of a circular tube (m) 𝑟
0
, 𝑅 = Radius of the tubing in the original Graetz problem
Re= Reynolds Number, 𝜌𝑉𝐷 𝜇
𝑅 𝑓
= Fouling factor ( 𝑚 2
k/W)
T= Temperature (C, K)
𝑇 ∗ 𝑚
(𝐿) =Dimensionless Temperature
𝑇 𝑚
(𝐿) = Outlet Temperature (C, K) ix

𝑇 𝑤
= Wall Temperature (C, K)
𝑇 𝑜
= Initial temperature of fluid flow (C, K)
∆𝑇
𝐿𝑀
= Log Mean Temperature Difference (K)
U= Overall heat transfer coefficient (W/ 𝑚 2
k) u= Velocity (m/s) 𝑢̅ = Mean Velocity (m/s)
V= Velocity (m/s)
𝑉 𝑖
= Inlet Velocity (m/s)
ε= Heat exchanger effectiveness
µ= Dynamic viscosity (Pa s) 𝜌 = Density (kg/ 𝑚 3
)
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 x

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. xi

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. Each heat exchanger model was built in steps and analyzed along the way until both concurrent flow and countercurrent flow heat exchanger models were developed. The results were compared between each model and between concurrent and countercurrent flow with fouled piping. Turbulent flow was also analyzed during the development of the heat exchangers to determine its effect on heat transfer. It was found that an increase in cooling flow lowered the hotter fluid and that the larger the inlet temperature differences between the fluids the larger the heat transfer. While as expected the fouled heat exchanger had a lower performance and therefore cooled the working fluid less, the performance of the countercurrent heat exchanger unexpectedly was very similar to slightly worse to that of the concurrent heat exchanger. The percent error between the hand calculations and finite element analysis can be attributed to several factors, most importantly the slight difference in temperature dependent material properties, the assumption of the outer wall heat transfer coefficient, and the nature of fouling.
xii

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 the Nusselt number, but the inlet velocity
(mass flow rate) and temperatures of the fluids are proportional to the heat transfer rate. 𝑞̇ = 𝑚̇ × 𝐶 𝑝
× (𝑇
𝐻
− 𝑇 𝑐
) [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.
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

𝐿𝑀
∆𝑇
2
−∆𝑇
1
𝐿𝑁(
∆𝑇
2
⁄
∆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.
𝑚𝑖𝑛
ℎ,𝑖
𝑐,𝑖
[3]
The effectiveness of the types of heat exchangers is as follows:
Parallel Flow:
 
1
 exp[

NTU ( 1

C r
1

C r
)]
[4]
Counter Flow:
 
1
1


C exp[ r
 exp[
NTU

( 1
NTU

( 1
C r

)]
C r
)
for 𝐶 𝑟
< 1 [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/ 𝑚 2
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

1 h i
A i

R f , i
A i
1
 ln(
D o
D i
2
 kL
)
1 h o
A o

R f , o
A o
[6]
𝑅 𝑓
is defined as the fouling factor with units of 𝑚 2
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 [4]
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 Δ 𝑇
1
and Δ 𝑇
2
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:
∆𝑇
𝐿𝑀
=
∆𝑇
2
−∆𝑇
1
𝐿𝑁(
∆𝑇
2 ⁄
∆𝑇
1
)
For concurrent flow: ∆𝑇
2
= 𝑇 ℎ,𝑜
− 𝑇 𝑐,𝑜
𝑎𝑛𝑑 ∆𝑇
1
= 𝑇 ℎ,𝑖
− 𝑇 𝑐,𝑖
For counter-current flow: ∆𝑇
2
= 𝑇 ℎ,𝑜
− 𝑇 𝑐,𝑖
𝑎𝑛𝑑 ∆𝑇
1
= 𝑇 ℎ,𝑖
− 𝑇 𝑐,𝑜
[7]
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:
  q actual
, 0
 q max
 
1 [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 𝐶 𝑟
(ratio of heat capacities). These values are arranged into different equations depending upon the type of heat exchanger (equations 4 and 5). As shown in equation 3, the effectiveness of a heat exchanger is directly proportional to the amount of heat transferred.
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:
𝑁𝑇𝑈 =
𝑈𝐴
𝐶 𝑚𝑖𝑛
[9]
3

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, 𝐶 𝑟
.
This method is typically used when some of the inlet or outlet temperature data is not available or needs to be solved for which would not allow for the use of the log mean temperature difference method. Using this method, the amount of heat transferred can be determined by the following equation: 𝑞̇ = 𝜀 × 𝐶 𝑚𝑖𝑛
× (𝑇 ℎ,𝑖
− 𝑇 𝑐,𝑖
) [10]
Therefore the outlet temperatures for the hot and cold fluids can be calculated as follows:
𝑇 𝑐,𝑜
= 𝑇 𝑐,𝑖
+ 𝑞̇
⁄
𝐶 𝑚𝑎𝑥
𝑇 ℎ,𝑜
= 𝑇 ℎ,𝑖
− 𝑞̇
⁄
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
𝐶 𝑚𝑖𝑛 while 𝐶 𝑚𝑎𝑥
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: 𝜌𝐶 𝑝
(
𝜕𝑇
𝜕𝑡
+ 𝑉 𝑟
𝜕𝑇
+
𝜕𝑟
𝑉 𝜃
𝜕𝑇 𝑟 𝜕𝜃
+ 𝑉 𝑧
𝜕𝑇
) = 𝑘 [
𝜕𝑧
1 𝑟
𝜕
𝜕𝑟
(𝑟
𝜕𝑇
) +
𝜕𝑟
1 𝑟 2
𝜕
2
𝑇
𝜕𝜃 2
+
𝜕
2
𝜕𝑧
𝑇
2
] + 𝜇𝛷 𝑣
[13]
4

The term 𝜇𝛷 𝑣 represents the dissipation function which is negligible. The velocity field of the flow in the tube is assumed as Poiseuille flow where 𝑉 𝑟
= 𝑉 𝜃
= 0 . The velocity is given in the form of the following equation [10]:
𝑉 𝑧
= 𝑉
0
(1 − 𝑟
2
𝑅 2
) [14]
𝑉
0
is defined as the maximum velocity at the center of flow. The velocity definitions and the fact that the temperature field is assumed steady and axisymmetric
𝜕𝑇
𝜕𝜃
= 0 , then simplify the general energy equation in cylindrical coordinates to the following: 𝜌𝐶 𝑝
𝑉 𝑧
𝜕𝑇
𝜕𝑧
= 𝑘[
1 𝑟
𝜕
𝜕𝑟
(𝑟
𝜕𝑇
𝜕𝑟
) +
𝜕
2
𝜕𝑧
𝑇
2
] [15]
Equation 15 is also expressed as equation 16 since the thermal diffusivity of the fluid is defined as ⁄ 𝑝
.
𝑉 𝑧
𝜕𝑇
𝜕𝑧
= 𝛼[
1 𝑟
𝜕
𝜕𝑟
(𝑟
𝜕𝑇
𝜕𝑟
) +
𝜕
2
𝜕𝑧
𝑇
2
] [16]
Neglecting dissipation and any conduction axially, equation 16 reduces to the following:
𝑉 𝑧
𝜕𝑇
𝜕𝑧
= 𝛼 𝑟
𝜕
𝜕𝑟
(𝑟
𝜕𝑇
)
𝜕𝑟
[17]
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 will develop faster than the velocity profile and 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 − 𝑟
2 𝑟
2
0
)
The velocity profile can also be developing and can be used for any Prandtl
[18] number material assuming the velocity and temperature profiles are starting at the same point [12]. For the original Graetz problem, Poiseuille flow was assumed and equation
18 was used to describe the fully developed velocity field of the fluid flowing through the constant wall temperature tubing. Analyzing the paper from Sellars [9] where he extends the Graetz problem, this equation for velocity is also used. For the purposes of this paper and the use of the finite element program, a constant value for the inlet velocity was used. This means a modified Graetz problem was introduced and analyzed.
5

Instead of just a developing temperature profile in a fully developed flow, in the case of this modified Graetz problem, both the velocity and temperature profiles are developing.
This means while the velocity in the original Graetz problem already exhibits the velocity described in equation 18 at the inlet, the velocity in the modified Graetz problem of this paper develops to this velocity profile during flow in the tubing.
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.
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 19 below.
𝑚
.075
⁄
𝐿 ∗
.05
1+
(𝐿∗
2
⁄
3
)
[19]
Where 𝐿 ∗ =
𝐿
𝐷 𝑅𝑒 𝑃𝑟
6

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 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 [9] 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.
7

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 at 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, 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 consist 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
8

element analysis to be used in whatever fashion is needed. Post-processor results allow for the analysis of temperature change across the center of flow, derive the outlet temperature of flow, plot the relationship of temperature or velocity profiles along the arc length, as well as other calculations.
9

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 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 for the heat exchangers and the viscosities and densities of the fluids were maintained constant. The type of heat exchanger used was of concentric tube design. Water was the cooling medium and oil the working fluid.
Water was used as the base fluid flowing through tubing or piping. 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 pipe wall conduction as well as for the heat exchanger models the material library properties in COMSOL were used.
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
10

model introduced a second pipe and its wall concentric to the first and was analyzed for fluid flow in 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 third and fourth 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 performed and compared against the COMSOL 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. The point at which the fluid outflows is also defined for the heat transfer. 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.
11

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 Input Data
Flow Parameters
L=
D=
1.0 m
.1 m k= 0.64 W/m K
µ= 0.000547 Pa s
ρ=
988 kg/ 𝑚
3
𝐶 𝑝
= 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 used for all three cases. Figure 4 shows the geometry of the model in COMSOL.
12

Figure 4: Modified Graetz Problem Geometry
The material properties of the fluid were then defined. Water at 50 ℃ was used and the properties that were temperature dependent 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 physics equations for laminar flow and heat transfer were applied to the model in order 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
13

point. For heat transfer, 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 nonisothermal flow are summarized in table 3 below and are from the fluid tab under nonisothermal 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.
14

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
15

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 K 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.
16

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 mesh that produced more accurate results was 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 shows the results from the effect that increasing the mesh elements had on centerline temperature for the case of 𝑉 𝑖
=.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.
17

Table 4: Mesh Extension Study Results
Mesh Extension
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
306
305,8
305,6
305,4
305,2 y = 308,55x -0,001
305
2000 3000 4000 5000 6000 7000 8000
Number of Elements in the Mesh
Centerline Outlet Temperature (K)
9000 10000 11000
Степенная (Centerline Outlet Temperature (K))
Figure 8: Centerline Temp vs. Mesh Element Number
18

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 was fairly constant as shown in figure 9.
310
309
308
307
306
305
304
303
302
301
300
280 290 300 310
Initial Value Temp (K)
320 330
Centerline Outlet Temp (K)
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 or structured 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. The curve illustrates the developing
19

temperature profile and shows that even at the exit of the tube, neither a constant value or the temperature of the wall is reached.
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.
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
19 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
= 𝑘𝑔
(988 𝑚3
)(.0001 𝑚/𝑠)(.1 𝑚)
(5.47𝑥10 −4 𝑃𝑎 𝑠)
= 18.062
[20]
20

The Prandtl number is calculated using the material properties of water at the inlet temperature.
𝑃𝑟 =
𝐶 𝑝 𝜇
= 𝑘
𝐽
(4181 𝑘𝑔 𝐾
)(5.47𝑥10
−4
𝑃𝑎 𝑠)
𝑊
.64 𝑚 𝐾
= 3.57
The dimensionless length value is defined as
𝐿 ∗
𝐿
=
𝐷𝑅𝑒𝑃𝑟
(1 𝑚)
=
(.1 𝑚)(18.062)(3.57)
= .15508
[21]
[22]
The outlet temperature is defined as
𝑇 𝑚
(𝐿) = 𝑇 𝑤
− (𝑇 𝑤
− 𝑇 𝑜
) ∗ 𝑇 ∗ 𝑚
(𝐿) [23]
Since there is a relationship between 𝑇 ∗ 𝑚
(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 19, the Nusselt number is calculated.
𝑁𝑢 𝑚
= 3.66 +
.075
⁄
𝐿 ∗
1+.05
⁄ 2
𝐿 ∗ 3
= 3.66 +
.075
⁄
.15508
1+.05
⁄
.15508
2/3
= 4.0722
𝑁𝑢 𝑚
=
−1
4𝐿 ∗ 𝑙𝑛𝑇 ∗ 𝑚
(𝐿) ∴ 𝑇 ∗ 𝑚
(𝐿) = 𝑒 (−4𝐿 ∗ 𝑁𝑢 𝑚
) = .07997
𝑇 𝑚
(𝐿) = 𝑇 𝑤
− (𝑇 𝑤
− 𝑇 𝑜
) ∗ 𝑇 ∗ 𝑚
(𝐿) = 30℃ − (30℃ − 50℃) ∗ (. 07997)
𝑇 𝑚
(𝐿) = 31.5994℃ = 304.7494 K
[24]
[25]
[26]
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.
21

Table 5: Modified Graetz Problem Comparison
Inlet V
(m/s)
0.0001
0.001
0.01
Inlet Temp
( ℃ )
50
50
50
Wall Temp
( ℃ )
30
30
30
Expected Value Calc (K) COMSOL Value (K) % Error
304.7494
316.646
317.644
305.3221
321.9023
323.0481
0.187572
1.632887
1.672847
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. The COMSOL refined column shows the new results of the COMSOL model with a refined mesh.
Table 6: Results from Mesh Refinement
Inlet V
(m/s)
0.0001
0.001
0.01
Exp. Value Calc (K) COMSOL Value (K)
304.7494
316.646
317.644
305.3221
321.9023
323.0481
% Error
0.187572403
1.632886749
1.672846861
COMSOL Refined (K) % Error
305.93648
322.812
323.15
0.388015
1.91009
1.703853
While the percent error is slightly higher (still less than 2%) 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.
22

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 is distributed as it enters, flows, and then exits the tubing. Under the non-isothermal tab, the RANS turbulence model was turned on essentially taking the same laminar flow modified Graetz problem model but changing the flow from laminar to turbulent. The k-
ε turbulence type model was used with Kays-Crawford heat transport. The same quad
23

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 not having the turbulence effect to enhance the momentum and heat transport. 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 was 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 fully coupled to segregated in order for the non-linear solver to divide
24

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
25

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 three velocity profiles were analyzed. An accuracy tolerance of 10 −4
was used as before as well as the previously defined quad element mesh (structured 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
26

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 three 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. Therefore, the change in temperature can be defined in terms of the equations for the amount of heat to be transferred. 𝑞̇ 𝑤𝑎𝑙𝑙 𝑜𝑢𝑡𝑙𝑒𝑡
= 𝑄̇ × ∆𝐴 = 𝑄̇2𝜋𝑅∆𝐿 𝑞̇ 𝑤𝑎𝑡𝑒𝑟
= 𝜌𝐶 𝑝
∆𝑇∆𝑉 = 𝜌𝐶 𝑝
∆𝑇𝜋𝑅 2 ∆𝐿
[27]
[28]
∆𝑇 =
𝑄̇2𝜋𝑅∆𝐿 𝜌𝐶 𝑝 𝜋𝑅 2 ∆𝐿
=
2𝑄̇ 𝜌𝐶 𝑝
𝑅
[29]
27

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.
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/ 𝑚
2
)
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 hand calculation for temperature at the outlet of the pipe has a very close temperature to that of the COMSOL model for .001 m/s velocity flow through a pipe (322.83099 K), 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. The only exception being in 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 from outside to inside since the inner fluid is hotter than
28

the constant wall temperature. Figures 17 and 18 show the similarity in outlet temperature distribution between the modified Graetz problem and pipe wall models.
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. Figures 19 and 20 show the computed values of the axial centerline velocity for laminar flow (figure 19) and
29

turbulent flow (figure 20) conditions, respectively. In both cases, the flow develops through approximately the same entrance length, as expected. However, the centerline velocity for fully developed flow has a much higher value. This is also to be expected since turbulence enhances momentum transport and causes the velocity profile across the tube radius to become much more uniform (see also figures 11 and 12).
Figure 19: Axial Centerline Velocity for Laminar Flow in a Pipe
Figure 20: Axial Centerline Velocity for Turbulent Flow in a Pipe
30

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 laminar flow velocity indicates a developing flow. The small change in turbulent flow velocity is indicative of a more uniform velocity profile in the center of flow due to the enhanced momentum transfer from eddy motion, random quick fluctuations of fluid flow, which is a part of turbulent flow.
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 uniform fluid flow. The approximate 322 K was similar between both turbulent models as expected.
31

This was also the main difference between the laminar and turbulent model for flow through a pipe with wall heat conduction.
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 2 nd fluid flowing around the original pipe.
Figure 22: Velocity Profile for Concurrent Heat Exchanger
32

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.
33

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.
34

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
35

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 more developed and evenly distributed than the laminar flow profile. 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
36

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
37

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 .5 K.
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.
𝐷 𝑜𝑢𝑡𝑒𝑟 𝑡𝑢𝑏𝑒
= 𝐷 𝑜
= .14 𝑚,
𝐷 𝑖𝑛𝑛𝑒𝑟 𝑡𝑢𝑏𝑒
= 𝐷 𝑖
= .10 𝑚,
𝐴 𝑜𝑢𝑡𝑒𝑟 𝑡𝑢𝑏𝑒
= 𝐴 𝑜
= 𝜋𝐷𝐿 = 𝜋(. 14 𝑚)(1 𝑚) = .4398 𝑚 2
𝐴 𝑖𝑛𝑛𝑒𝑟 𝑡𝑢𝑏𝑒
= 𝐴 𝑖
= 𝜋𝐷𝐿 = 𝜋(. 10 𝑚)(1 𝑚) = .3142 𝑚 2
The cross sectional area of each fluid flow is:
𝐴 𝑜𝑖𝑙
= 𝜋𝑟 2 = 𝜋(. 05 𝑚 2 ) = .007854 𝑚 2
𝐴 𝑤𝑎𝑡𝑒𝑟
= 𝜋(𝑟 𝑜
2
− 𝑟 𝑖
2 ) = 𝜋((. 12 𝑚) 2 − (. 07 𝑚) 2 ) = .0298454 𝑚 2
The inlet temperature of each fluid and its corresponding properties due to that temperature is shown in table 8:
38

Table 8: Fluid Properties
T=
T= k=
µ=
ρ=
Pr=
𝐶 𝑝
=
Fluid Parameters for Oil
125 ℃
398.15 K
0.134 W/m K
0.00915 Pa s
826 kg/ 𝑚
3
159
2328 J/Kg K
Fluid Parameters for Water
T=
T= k=
µ=
ρ=
Pr=
𝐶 𝑝
=
20 ℃
293.15 K
0.600 W/m K
0.001003 Pa s
998.2 kg/ 𝑚
3
6.99
4182 J/Kg K
The mass flow rates are then calculated and used to determine the heat capacity rates. 𝑚̇ 𝑜𝑖𝑙
= 𝜌𝐴𝑣 = (
826 𝑘𝑔 𝑚 3
) (. 007854 𝑚 2 ) (. 0001 𝑚
) = .0006487 𝑘𝑔/𝑠 𝑠 𝑚̇ 𝑤𝑎𝑡𝑒𝑟
= 𝜌𝐴𝑣 = (
998.2 𝑘𝑔 𝑚 3
) (. 029845 𝑚 2 ) (. 0001 𝑚
) = .002979 𝑘𝑔/𝑠 𝑠
𝐶 𝑜𝑖𝑙
= 𝐶 𝑝,𝑜𝑖𝑙
× 𝑚̇ 𝑜𝑖𝑙
= (2328
𝐽
𝐾𝑔𝐾
) (
. 0006487 𝐾𝑔 𝑠
) = 1.5102 𝑊/𝐾
𝐶 𝑤𝑎𝑡𝑒𝑟
= 𝐶 𝑝,𝑤𝑎𝑡𝑒𝑟
× 𝑚̇ 𝑤𝑎𝑡𝑒𝑟
= (4182
𝐽
𝐾𝑔𝐾
) (
. 002979 𝐾𝑔 𝑠
) = 12.4588 𝑊/𝐾
From this it can be defined for the analysis purposes that 𝐶 𝑚𝑖𝑛
is 𝐶 𝑜𝑖𝑙
and 𝐶 𝑚𝑎𝑥
is
𝐶 𝑤𝑎𝑡𝑒𝑟
. This yields our ratio of heat capacity rates to be:
𝐶 𝑟
=
𝐶 𝑚𝑖𝑛
𝐶 𝑚𝑎𝑥
=
𝐶 𝑜𝑖𝑙
𝐶 𝑤𝑎𝑡𝑒𝑟
=
1.5102
12.4588
= .12122
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 = 𝜌𝑣𝐷 𝜇
=
4𝑚̇ 𝑜𝑖𝑙 𝜋𝐷 𝑖 𝜇 𝑜𝑖𝑙
4 × .0006487𝑘𝑔/𝑠
= 𝜋 × .10 𝑚 × .00915 𝑃𝑎 𝑠
= .9027
𝑁𝑢 = 3.66 +
. 0668𝐴
1 + .04𝐴 .667
𝑤ℎ𝑒𝑟𝑒 𝐴 = 𝑅𝑒𝑃𝑟 (
𝐷 𝑖
𝐿
𝑁𝑢 = 4.435
The heat transfer coefficient of the inner pipe wall is expressed as follows: ℎ 𝑖
= 𝑘 𝑜𝑖𝑙
× 𝑁𝑢
𝐷 𝑖
=
. 134
𝑊 𝑚 𝐾 × 4.435
. 10 𝑚
= 5.9435 𝑊 𝑚
2 𝐾
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.
39

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 ℎ 𝑖
𝐴 𝑖
+
1
1 ℎ 𝑜
𝐴 𝑜
+ ln (
𝐷 𝑜 ⁄ ) 𝑖
2𝜋𝐿𝐾 𝑐𝑜𝑝𝑝𝑒𝑟
=
1
1
(5.9435 𝑊/𝑚 2 𝐾)(.3142𝑚 2 )
+
1
(. 5 ∗ 5.9435 𝑊/𝑚 2 𝐾)(.4398𝑚 2 )
+ ln (. 14𝑚 . 10𝑚 )
2𝜋(1𝑚)(393.11 𝑊/𝑚 𝐾)
𝑈𝐴 = 0.768769 𝑊/𝐾
The value for the number of heat transfer units is:
𝑈𝐴
𝑁𝑇𝑈 =
𝐶𝑚𝑖𝑛
=
. 768769 𝑊/𝐾
1.5102 𝑊/𝐾
= .50903
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+𝐶 𝑟
)]
1 + 𝐶 𝑟
=
1 − 𝑒 [−.50903(1+.12122)]
1 + .12122
= .387872
The equation for heat transferred in the NTU-effectiveness method is in terms of this effectiveness value as well as the minimum heat capacity. 𝑞̇ = 𝜀 × 𝐶 𝑚𝑖𝑛
× (𝑇 ℎ,𝑖
− 𝑇 𝑐,𝑖
) = (. 387872) (
1.5102 𝑊
𝐾
) (398.15 𝐾 − 293.15 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:
𝑇 𝑐,𝑜
= 𝑇 𝑐,𝑖
+ 𝑞̇
⁄
𝐶 𝑚𝑎𝑥
𝑇 ℎ,𝑜
= 𝑇 ℎ,𝑖
− 𝑞̇
⁄
𝐶 𝑚𝑖𝑛
= 298.0869 𝐾
= 357.4235 𝐾
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.
40

𝑞̇ = 𝑈𝐴∆𝑇
𝐿𝑀
∴ ∆𝑇
𝐿𝑀
= 𝑞̇
𝑈𝐴
=
61.50782 𝑊
. 768769 𝑊/𝐾
= 80.017 𝐾
∆𝑇
𝐿𝑀
=
∆𝑇
2
− ∆𝑇
1
𝐿𝑁 (
∆𝑇
2 ⁄
∆𝑇
1
)
=
(357.4235 𝐾 − 298.0869 𝐾) − (398.15 𝐾 − 293.15 𝐾)
𝐿𝑁
(357.4235 𝐾 − 298.0869 𝐾)
⁄
(398.15 𝐾 − 293.15 𝐾)
=
∆𝑇
𝐿𝑀
= 80.017 𝐾 ∴ 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.
41

370
369
368
367
366
365
364
363
362
361
367,05901
365,1556
363,7862 Th,o (oil)
360
0 0,002 0,004 0,006 0,008
Cooling Water Velocity (m/s)
0,01 0,012
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. Therefore there is a directly proportional relationship with cooling water flow rate and the temperature drop in the working fluid.
42

35
34,5
34
33,5
34,3638
32,9944
33
32,5
32
31,5
Δ in Oil Temp
31,09099
31
30,5
0 0,002 0,004 0,006
Cooling Water Velocity (m/s)
0,008 0,01
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. This illustrates the intensity of the heat transfer is directly proportional to the difference in temperature between the two fluid in which the transfer is occurring.
43

35
30
25
20
15
10
5
26,30021
13,50046
28,83715
15,05942
31,09099
16,62035
Th,o-Th,i (Oil)
Tc,o-Tc,i (Water)
0
75 80 85 90 95 100 105
Difference in Inlet Temperatures Between Fluids (K)
110
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
44

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) which could be calculated since the outlet temperature was estimated.
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
45

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 modified 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
46

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 cooling water 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, 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. Note in figure 37 that the temperature profiles are started from the oil inlet of z= 0 m, so that this is actually where the water exits and explains the lowering in temperature for the cooling water flow as the
47

cross section of piping goes from 0 m to 1.0 m in length. Had the curve for the water been from 1.0 m to 0 m, the opposite shape of the oil temperature profile would be obtained as in the concurrent flow heat exchanger plot.
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 more evenly developed flow already entering both pipes due to the turbulence being applied to the model. Figure 38 shows both evenly distributed velocity profiles prior to heat transfer, and figure 39 shows the extent of velocity distribution throughout the model.
48

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 two 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.
49

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).
50

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 chapter 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−𝐶 𝑟
)] 𝜀 =
1 − 𝐶 𝑟 𝑒 [−𝑁𝑇𝑈(1−𝐶 𝑟
)]
=
1 − 𝑒 [−.50903(1−.12122)]
1 − (.12122 ∗ 𝑒 [−.50903(1−.12122)] )
= .390964
The equation for heat transferred in the NTU-effectiveness method is in terms of this effectiveness value as well as the minimum heat capacity. 𝑞̇ = 𝜀 × 𝐶 𝑚𝑖𝑛
× (𝑇 ℎ,𝑖
− 𝑇 𝑐,𝑖
) = (. 390964) (
1.5102 𝑊
𝐾
) (398.15 𝐾 − 293.15 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:
𝑇 𝑐,𝑜
= 𝑇 𝑐,𝑖
+ 𝑞̇
⁄
𝐶 𝑚𝑎𝑥
𝑇 ℎ,𝑜
= 𝑇 ℎ,𝑖
− 𝑞̇
⁄
𝐶 𝑚𝑖𝑛
= 298.1263 𝐾
= 357.0988 𝐾
51

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. 𝑞̇ = 𝑈𝐴∆𝑇
𝐿𝑀
∴ ∆𝑇
𝐿𝑀
= 𝑞̇
𝑈𝐴
=
61.99814 𝑊
. 768769 𝑊/𝐾
= 80.64597 𝐾
∆𝑇
𝐿𝑀
=
∆𝑇
2
− ∆𝑇
1
𝐿𝑁 (
∆𝑇
2 ⁄
∆𝑇
1
)
=
(357.0988 𝐾 − 293.15 𝐾) − (398.15 𝐾 − 298.1263 𝐾)
𝐿𝑁
(357.0988 𝐾 − 293.15 𝐾)
⁄
(398.15 𝐾 − 298.1263 𝐾)
=
∆𝑇
𝐿𝑀
= 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 2 K 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 4 K, 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
52

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 357.0988
K while the concurrent flow lowered the oil temperature to 357.4235 K. The results for the counter-current analysis can be found in the appendix section.
371
370
369
368
376
375,54683
375
374
373
372
371,25821
367,44359
Th,o (oil)
367
0 0,002 0,004 0,006
Cooling Water Velocity (m/s)
0,008 0,01
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, as the cooling water velocity increased and the inlet oil temperature remained constant, the oil temperature drop increased as well.
53

32
30
28
26,89179
26
24
22
22,60317
30,70641
Δ in Oil Temp
20
0 0,002 0,004 0,006
Cooling Water Velocity (m/s)
0,008 0,01
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 2 K change in inlet and outlet oil temperature per 10 K ∆𝑇 as well as for the cooling water. But the oil changed from 26.30021 K to 28.83715 K to 31.09099 K while water changed from 13.50046 K to 15.05942 K to 16.62035 K which is almost reverse to the counter-current model. In the counter-current model, the oil changed the least, from
54

19.16949 K to 20.96748 K to 22.60317 K while the water changed the most, 25.842385
K to 28.62676 K to 31.33479 K.
35
31,33479
30
28,62676
25,842385
25
22,60317
20,96748
19,16949
20
15
Tc,o-Tc,I (Water)
Th,i-Th,o (Oil)
10
5
0
75 80 85 90 95 100 105
Difference in Inlet Temperatures Between Fluids (K)
110
Figure 44: Temperature Change in the Fluids vs. the Difference in Inlet Temperatures for Counter-
Current Flow
3.5.1
Flow in a Concurrent Flow Heat Exchanger with Fouling
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
55

turn increases the amount of drag or resistance in the pipe to flow, therefore affecting 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.
3.5.2
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 in
COMSOL, 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 𝑚 𝐾
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 𝑚 𝐾
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
56

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 𝑚 2 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:
1
1 ℎ𝑖𝐴𝑖
+
1 ℎ𝑜𝐴𝑜
+ ln (
𝐷𝑜
⁄
𝐷𝑖
)
2𝜋𝐿𝐾𝑐𝑜𝑝𝑝𝑒𝑟
+
𝑅𝑓,𝑜
𝐴𝑜
[30]
𝑈𝐴 =
1 ℎ 𝑖
1
𝐴 𝑖
+ ℎ 𝑜
1
𝐴 𝑜
+ ln (
𝐷 𝑜
𝐷 𝑖
2𝜋𝐿𝐾 𝑐𝑜𝑝𝑝𝑒𝑟
+
𝑅 𝑓,𝑜
𝐴 𝑜
=
1
1
(5.9435𝑊/𝑚 2 𝐾)(.3142𝑚 2 )
+
1
(. 5 ∗ 5.9435𝑊/𝑚 2 𝐾)(.4398𝑚 2 )
+ ln (. 14𝑚 . 10𝑚 )
2𝜋(1𝑚)(393.11𝑊/𝑚𝐾)
+
(.0002 𝑚
2 𝐾
𝑊
(.4398𝑚 2 )
𝑈𝐴 = 0.7685006 𝑊/𝐾
The value for the number of heat transfer units is:
𝑈𝐴
𝑁𝑇𝑈 =
𝐶 𝑚𝑖𝑛
=
. 7685006 𝑊/𝐾
= .50885
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+𝐶 𝑟
)]
=
1 + 𝐶 𝑟
1 − 𝑒 [−.50885(1+.12122)]
= .387771
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. 𝑞̇ = 𝜀 × 𝐶 𝑚𝑖𝑛
× (𝑇 ℎ,𝑖
− 𝑇 𝑐,𝑖
) = (. 387771) (
1.5102𝑊
𝐾
) (398.15𝐾 − 293.15𝐾)
= 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
57

divided by the heat capacity of the fluid to the inlet temperature of that fluid. For this case:
𝑇 𝑐,𝑜
= 𝑇 𝑐,𝑖
+ 𝑞̇
⁄
𝐶 𝑚𝑎𝑥
= 298.08563 𝐾
𝑇 ℎ,𝑜
= 𝑇 ℎ,𝑖
− 𝑞̇
⁄
𝐶 𝑚𝑖𝑛
= 357.43402 𝐾
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. 𝑞̇ = 𝑈𝐴∆𝑇
𝐿𝑀
∴ ∆𝑇
𝐿𝑀
= 𝑞̇
𝑈𝐴
=
61.49188 𝑊
. 7685006 𝑊/𝐾
= 80.015 𝐾
∆𝑇
𝐿𝑀
=
∆𝑇
2
− ∆𝑇
1
𝐿𝑁 (
∆𝑇
2 ⁄
∆𝑇
1
)
=
(357.43402 𝐾 − 298.08563 𝐾) − (398.15 𝐾 − 293.15 𝐾)
𝐿𝑁
(357.43402 𝐾 − 298.08563 𝐾)
⁄
(398.15 𝐾 − 293.15 𝐾)
=
∆𝑇
𝐿𝑀
= 80.015 𝐾 ∴ 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 transfer rate 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 non-fouled 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) which was about 7.4%. The reason for 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 a more in depth study, a geometric region could
58

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
372
370,74566
370
368
367,5886
366
364
365,63366 Th,o (oil)
362
360
0 0,002 0,004 0,006 0,008
Cooling Water Velocity (m/s)
0,01 0,012
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.
59

374
372
370,74566
370
368
366
367,5886
367,05901
365,1556
364
362
365,63366
363,7862
Th, o (oil) With Fouling
Th, o (oil) No Fouling
360
0 0,002 0,004 0,006 0,008 0,01
Cooling Water Velocity (m/s)
0,012
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 3.5-2.0 K difference in the temperature drops from .0001 to .001 to .01 m/s for the cooling water velocity.
60

33
32
31
30,5614
30
29
32,51634
Δ in Oil Temp
28
27,40434
27
0 0,002 0,004 0,006
Cooling Water Velocity (m/s)
0,008 0,01
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
61

change between the inlet and outlet of the water was larger at each increment than the oil.
35
30
28,78301
25
23,52009
22,73502
26,16604
25,1718
27,40434
20
15
Th,i-Th,o (Oil)
Tc,o-Tc,i (Water)
10
5
0
75 80 85 90 95 100 105
Difference in Inlet Temperatures Between Fluids (K)
110
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 outer wall of the inside pipe which defined the pipe 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
62

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 layer of .004 m was created, the heat transfer was lower than the nonfouled 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 30). 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
63

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.
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
376,40116
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
Cooling Water Velocity (m/s)
0,008 0,01
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 combines all four cases of this relationship. While this was unexpected, it should be
64

noted that the hand calculations showed that the counter-current heat exchanger did lower the oil outlet temperature slightly more than the concurrent heat exchanger. The small fraction of better heat transfer was most likely masked by the percent error in the hand calculations and the COMSOL model applications.
378
376
374
372
370
368
366
364
362
0
Th, o (oil) Fouling Counter
Current
Th, o (oil) No Fouling
Counter Current
Th, o (oil) No Fouling
Concurrent
Th, o (oil) Fouling
Concurrent
0,002 0,004 0,006 0,008
Cooling Water Velocity (m/s)
0,01
Figure 51: Concurrent and Counter-current Flow Heat Exchangers with and without Fouling
65

The main objective of this project was to analyze the fluid flow in double pipe heat exchangers and the subsequent performance of these heat exchangers. To facilitate this analysis, the COMSOL finite element analysis program was used to perform the modeling and calculations. In order to verify the development of each model, the models were built in stages and each stage analyzed and verified. The first stage was a modified
Graetz problem model where velocity and temperature profiles were analyzed for fluid flow in a tube of 1.0 m in length. The profiles showed that the temperature profile took the entire length of the flow to approach an almost constant value while the velocity profile developed extremely fast. This basic model was verified by hand calculations and the percent difference was seen to be less than 2%. While this was a basic model the analysis did bring to light several factors with COMSOL that needed addressing prior to proceeding.
During the first part of the project the initial values and mesh refinement of the
COMSOL model were found to change the ending results unless a proper mesh is produced and verified to give consistent results. This involved changing mesh conditions
(boundary layers), changing the tolerance of solution convergence, and changing the type of non-linear solver. If not done, the results were found to be inaccurate and inconsistent. If these models were being used in an industrial or business application it could have led 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. Once the appropriate mesh was found, it was one less variable to be concerned with and didn’t need to be changed in the subsequent more complex models.
The next stage added a pipe wall to the flow and essentially half of a double pipe heat exchanger was created. The outlet temperatures for the flow were, as expected, larger than for the modified Graetz problem which had no pipe wall to restrict its heat transfer. Even though the outlet temperature was only higher by a fraction, it did verify the concern that for the best heat transfer in a heat exchanger, the pipe walls needed to be of a material that had a high thermal conductivity.
66

These first two models also introduced turbulent flow to the models to experiment with how this affected the cooling of the flow from the constant wall temperature. It was discovered that even though there is a more evenly distributed velocity flow with the turbulent model, the laminar flow had approximately the same amount of heat transfer as the turbulent models. It should be noted, that for the lower velocity of .0001 m/s, the laminar model actually cooled the outlet flow to 305.93648 K in the modified Gratez problem while the turbulent model only cooled the flow to 322.25503 K. It was also discovered that the centerline temperature from start to finish (0 to 1 m) had a more parabolic, gradual slope lowering from the inlet temperature of 323.15 K to 305.93648 K while the turbulent flow had an almost linear relationship with the centerline temperature against the arc length.
The concurrent and counter current models first were verified to be providing the same cooling relationship as expected from these types of heat exchangers. All models with and without fouling showed that as the cooling water flow increased, so did the amount of heat transfer by a decreasing oil outlet temperature and increasing oil temperature difference. All models also demonstrated that the difference in temperature between the fluids was a driving force in the amount of heat transfer and that there was a linear relationship. The COMSOL results were found to be fairly consistent with hand calculations with most of the values within 5% of each other. However, several instances were shown to be around 5-8% different. Several things could have caused this difference. First of all, the material properties which were temperature dependent were considered constant based on the inlet temperature of the fluid flow for the hand calculations instead of using the average temperature whereas the COMSOL values were from the material library. Secondly, the heat transfer coefficient for the outside of the inner pipe wall had to be estimated and there was no relationship between the estimate and the value the COMSOL model had used. If a more in depth study could have been performed, this coefficient should have been better estimated by a piecewise function like the fouling layer or in another way.
As expected, when a fouling layer was added to each type of heat exchanger the amount of heat transfer lowered and the outlet temperature was affected by not lowering as much. Therefore no matter what type of heat exchanger flow is used, if there is some
67

deterioration to the heat transfer surface, the performance of the subject heat exchanger will suffer. In fact the difference in outlet temperature between the concurrent and countercurrent heat exchangers was so low in the hand calculations (fraction of a Kelvin) that the percent error of the model made the performance of the concurrent model slightly better than the countercurrent model. However, the fouled heat exchangers performance was much lower than the non-fouled heat exchanger. There was an approximate 2 K rise in outlet temperature for a fouled concurrent heat exchanger and an approximate 1 K rise in outlet temperature for a fouled countercurrent heat exchanger
. vice a fraction of a Kelvin change between the concurrent and countercurrent heat exchangers. This finding proves it is more important for engineers and developers to focus on the method of preventing damage to the heat transfer surfaces and the type of material chosen than it is to focus on the type of flow. The more time that is spent researching how a material will perform over time and the type of corrosion that occurs with the material or ways to prevent corrosion and deterioration in the system will be more efficient.
68

[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.”
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[4] Concentric tube heat exchanger: operating principle with parallel flow. Art.
Encyclopædia Britannica Online. Web. 12 Apr. 2012.
< http://www.britannica.com/EBchecked/media/5861/Operating-principle-of-aparallel-flow-heat-exchanger >.
[5] 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.
[6] Kays, William, Michael Crawford, and Bernhard Weigand. Convective Heat and
Mass Transfer. 4 th
ed. New York: The McGraw-Hill Companies, Inc.,
2005.
[7] Lemcoff, Norberto. “Heat Exchanger Design.” Lecture notes from Mechanical
Engineering Foundations 2. Groton. 10 July 2008.
[8] Lemcoff, Norberto. “Project: Heat Exchanger Design.” Lecture notes from
Mechanical Engineering Foundations 2. 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. Paper No. 55-SA-66 AD-A280 848. New York. 1955.
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< http://web2.clarkson.edu/projects/subramanian/ch560/notes/Graetz%20Problem
.pdf
>
[11] Valko, Peter P. “Solution of the Graetz-Brinkman Problem with the Laplace
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[12] White, Frank. Viscous Fluid Flow. 3 rd
ed. New York: The McGraw-Hill
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[13] 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.
69

CASE 1:
Iteration No.
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
1
0.439823
0.314159
20
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)
0.0001
125
0.0001
0.007854
0.029845
826
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
998.2
0.002979
0.000649
4182
2328
Cc (w/k)
Ch (w/k)
Cmin/Cmax
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)
12.45877
1.510264
0.121221
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
Tc,o Percent Diff 3.771614
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
84.27347
357.4235
367.059
Th,o Percent Diff 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
Velocity of oil= .0001 m/s
2
0.439823
0.314159
20
3
0.439823
0.314159
20
0.001
125
0.0001
0.007854
0.029845
826
0.01
125
0.0001
0.007854
0.029845
826
998.2
0.029791
0.000649
4182
2328
124.5877
1.510264
0.012122
998.2
0.297914
0.000649
4182
2328
1245.877
1.510264
0.001212
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
70

CASE 2:
Iteration No.
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Velocity of water & oil= .0001 m/s
1 2 3 4
0.439823 0.439823 0.439823 0.439823
0.314159 0.314159 0.314159 0.314159
20 30 40 20
0.0001 0.0001 0.0001 0.0001
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)
Cc (w/k)
Ch (w/k)
Cmin/Cmax
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)
125
0.0001
125
0.0001
125
0.0001
150
0.0001
0.007854 0.007854 0.007854 0.007854
0.029845 0.029845 0.029845 0.029845
826 826 826 811
998.2 995.6 992.2 998.2
0.002979 0.002971 0.002961 0.002979
0.000649 0.000649 0.000649 0.000637
4182 4179 4179 4182
2328 2328 2328 2440
12.45877 12.4174 12.375 12.45877
1.510264 1.510264 1.510264 1.554177
0.121221 0.121625 0.122042 0.124746
0.00915 0.00915 0.00915 0.00564
159 159 159 104
0.902732 0.902732 0.902732 1.437943
0.134 0.134 0.134 0.132
4.43545 4.43545 4.43545 4.463665
5.943503 5.943503 5.943503 5.892037
393.111 393.111 393.111 391.3795
0.768769 0.768769 0.768769 0.762113
0.50903 0.50903 0.50903 0.490364
0.387872 0.387836 0.387798 0.376916
61.50782 55.64476 49.78263 76.15333
24.93691 34.48119 44.02284 26.11243
Tc,o (Kelvin)
Tc,o (COMSOL)
297.9369 307.4812 317.0228 299.1124
309.7703 318.2094 326.6505 313.8091
Tc,o Percent Diff 3.820037 3.371436 2.947377 4.683325
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
84.27347 88.15561 92.03713 101.0009
357.2735 361.1556 365.0371 374.0009
367.059 369.3129 371.8498 386.5914
Th,o Percent Diff 2.665932 2.208761 1.832099 3.256804
71

CASE 1:
Iteration No.
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
1
0.439823
0.314159
20
0.0001
Th,I (Celsius)
Vh, I (m/s)
A oil flow
A water flow
ρ Oil (kg/m^3)
ρ Water (kg/m^3)
125
0.0001
0.007854
0.029845
826
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
998.2
0.002979
0.000649
4182
2328
Cc (w/k)
Ch (w/k)
Cmin/Cmax
μ (Pa s)
Pr
12.45877
1.510264
0.121221
0.00915
159
Re k oil (w/m*k)
Nusselt Number hi (w/m^2*k) k Copper (w/m*k)
UA (w/k)
NTU
ε
0.902732
0.134
4.43545
5.943503
393.111
0.768769
0.50903
0.390964 q (w)
Tc,o (Celsius)
61.99814
24.97627
Tc,o (Kelvin) 298.1263
Tc,o (COMSOL) 324.4848
Tc,o Percent Diff 8.123192
Th,o (Celsius) 83.94881
Th,o (Kelvin) 357.0988
Th,o (COMSOL) 375.5468
Th,o Percent Diff 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
Velocity of oil= .0001 m/s
2
0.439823
0.314159
20
0.001
3
0.439823
0.314159
20
0.01
125
0.0001
0.007854
0.029845
826
998.2
0.029791
0.000649
4182
2328
124.5877
1.510264
0.012122
125
0.0001
0.007854
0.029845
826
998.2
0.297914
0.000649
4182
2328
1245.877
1.510264
0.001212
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
72

CASE 2:
Iteration No.
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Velocity of water & oil= .0001 m/s
1 2 3 4
0.439823 0.439823 0.439823 0.439823
0.314159 0.314159 0.314159 0.314159
20 30 40 20
0.0001 0.0001 0.0001 0.0001
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)
Cc (w/k)
Ch (w/k)
Cmin/Cmax
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
125
0.0001
125
0.0001
125
0.0001
150
0.0001
0.007854 0.007854 0.007854 0.007854
0.029845 0.029845 0.029845 0.029845
826 826 826 811
998.2 995.6 992.2 998.2
0.002979 0.002971 0.002961 0.002979
0.000649 0.000649 0.000649 0.000637
4182 4179 4179 4182
2328 2328 2328 2440
12.45877 12.4174 12.375 12.45877
1.510264 1.510264 1.510264 1.554177
0.121221 0.121625 0.122042 0.124746
0.00915 0.00915 0.00915 0.00564
159 159 159 104
0.902732 0.902732 0.902732 1.437943
0.134 0.134 0.134 0.132
4.43545 4.43545 4.43545 4.463665
5.943503 5.943503 5.943503 5.892037
393.111 393.111 393.111 391.3795
0.768769 0.768769 0.768769 0.762113
0.50903 0.50903 0.50903 0.490364
0.390964 0.390937 0.39091 0.379812
61.99814 56.08979 50.18212 76.73834
24.97627 34.51703 44.05512 26.15938
297.9763 307.517 317.0551 299.1594
324.4848 331.7768 338.9924 332.1002
8.16942 7.312064 6.471313 9.918939
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
83.94881 87.86094 91.77262 100.6245
356.9488 360.8609 364.7726 373.6245
375.5468 377.1825 378.9805 396.9367
Th,o Percent Diff 4.952251 4.327236 3.748976 5.873029
73

CASE 1:
Iteration No.
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
1
0.439823
0.3141593
20
0.0001
Th,I (Celsius)
Vh, I (m/s)
A oil flow
A water flow
ρ Oil (kg/m^3)
ρ Water (kg/m^3)
125
0.0001
0.007854
0.0298451
826
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
998.2
0.0029791
0.0006487
4182
2328
Cc (w/k)
Ch (w/k)
Cmin/Cmax
μ (Pa s)
Pr
12.458767
1.5102641
0.121221
0.00915
159
Re k oil (w/m*k)
Nusselt Number hi (w/m^2*k) k Copper (w/m*k)
UA (w/k)
NTU
ε
0.9027322
0.134
4.4354504
5.9435035
393.111
0.7685006
0.5088518
0.3877712 q (w)
Tc,o (Celsius)
61.49188
24.935631
Tc,o (Kelvin) 298.08563
Tc,o (COMSOL) 321.93301
Tc,o Percent Diff 7.4075594
Th,o (Celsius) 84.284022
Th,o (Kelvin) 357.43402
Th,o (COMSOL) 370.74566
Th,o Percent Diff 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
Velocity of oil= .0001 m/s
2
0.439823
0.314159
20
0.001
3
0.439823
0.314159
20
0.01
125
0.0001
0.007854
0.029845
826
998.2
0.029791
0.000649
4182
2328
124.5877
1.510264
0.012122
125
0.0001
0.007854
0.029845
826
998.2
0.297914
0.000649
4182
2328
1245.877
1.510264
0.001212
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
74

CASE 2:
Iteration No.
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Th,I (Celsius)
Vh, I (m/s)
A oil flow
Velocity of water & oil= .0001 m/s
1 2 3 4
0.439823 0.439823 0.439823 0.439823
0.314159 0.314159 0.314159 0.314159
20
0.0001
30
0.0001
40
0.0001
20
0.0001
125
0.0001
125
0.0001
125
0.0001
150
0.0001
0.007854 0.007854 0.007854 0.007854
A water flow
ρ Oil (kg/m^3)
ρ Water (kg/m^3)
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
0.029845 0.029845 0.029845 0.029845
826 826 826 811
998.2 995.6 992.2 998.2
0.002979 0.002971 0.002961 0.002979
0.000649 0.000649 0.000649 0.000637
4182 4179 4179 4182
Cph (j/kg*k)
Cc (w/k)
Ch (w/k)
Cmin/Cmax
μ (Pa s)
Pr
Re k oil (w/m*k)
Nusselt Number hi (w/m^2*k) k Copper (w/m*k)
2328 2328 2328 2440
12.45877 12.4174 12.375 12.45877
1.510264 1.510264 1.510264 1.554177
0.121221 0.121625 0.122042 0.124746
0.00915 0.00915 0.00915 0.00564
159 159 159 104
0.902732 0.902732 0.902732 1.437943
0.134 0.134 0.134 0.132
4.43545 4.43545 4.43545 4.463665
5.943503 5.943503 5.943503 5.892037
393.111 393.111 393.111 391.3795
UA (w/k)
NTU
ε q (w)
Tc,o (Celsius)
Tc,o (Kelvin)
Tc,o (COMSOL)
0.768501 0.768501 0.768501 0.761849
0.508852 0.508852 0.508852 0.490194
0.387771 0.387735 0.387698 0.376818
61.49188 55.63034 49.76973 76.13355
24.93563 34.48003 44.0218 26.11084
298.0856 307.63 317.1718 299.2608
321.933 329.316 336.6701 329.0637
Tc,o Percent Diff 7.407559 6.585167 5.791513 9.056877
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
84.28402 88.16516 92.04568 101.0136
357.284 361.1652 365.0457 374.0136
370.7457 372.9782 375.415 391.0818
Th,o Percent Diff 3.630963 3.16722 2.762091 4.364344
75

CASE 1:
Iteration No.
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
1
0.439823
0.314159
20
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)
0.0001
125
0.0001
0.007854
0.029845
826
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
998.2
0.002979
0.000649
4182
2328
Cc (w/k)
Ch (w/k)
Cmin/Cmax
μ (Pa s)
12.45877
1.510264
0.121221
0.00915
Pr
Re k oil (w/m*k)
Nusselt Number
159
0.902732
0.134
4.43545 hi (w/m^2*k) k Copper (w/m*k)
5.943503
393.111
UA (w/k)
NTU
ε q (w)
Tc,o (Celsius)
Tc,o (Kelvin)
0.768501
0.508852
0.390861
61.98178
24.97495
298.125
Tc,o (COMSOL) 327.8878
Tc,o Percent Diff 9.077138
Th,o (Celsius) 83.95964
Th,o (Kelvin) 357.1096
Th,o (COMSOL) 375.2051
Th,o Percent Diff 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
Velocity of oil= .0001 m/s
2
0.439823
0.314159
20
3
0.439823
0.314159
20
0.001
125
0.0001
0.007854
0.029845
826
0.01
125
0.0001
0.007854
0.029845
826
998.2
0.029791
0.000649
4182
2328
124.5877
1.510264
0.012122
998.2
0.297914
0.000649
4182
2328
1245.877
1.510264
0.001212
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
76

CASE 2:
Iteration No.
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
Vc, I (m/s)
Velocity of water & oil= .0001 m/s
1 2 3 4
0.439823 0.439823 0.439823 0.439823
0.314159 0.314159 0.314159 0.314159
20 30 40 20
0.0001 0.0001 0.0001 0.0001
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)
Cc (w/k)
Ch (w/k)
Cmin/Cmax
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)
125
0.0001
125
0.0001
125
0.0001
150
0.0001
0.007854 0.007854 0.007854 0.007854
0.029845 0.029845 0.029845 0.029845
826 826 826 811
998.2 995.6 992.2 998.2
0.002979 0.002971 0.002961 0.002979
0.000649 0.000649 0.000649 0.000637
4182 4179 4179 4182
2328
0.00915
159
2328
0.00915
159
2328
0.00915
159
12.45877 12.4174 12.375 12.45877
1.510264 1.510264 1.510264 1.554177
0.121221 0.121625 0.122042 0.124746
2440
0.00564
104
0.902732 0.902732 0.902732 1.437943
0.134 0.134 0.134 0.132
4.43545 4.43545 4.43545 4.463665
5.943503 5.943503 5.943503 5.892037
393.111 393.111 393.111 391.3795
0.768501 0.768501 0.768501 0.761849
0.508852 0.508852 0.508852 0.490194
0.390861 0.390834 0.390807 0.379711
61.98178 56.07498 50.16887 76.71806
24.97495 34.51584 44.05405 26.15776
Tc,o (Kelvin)
Tc,o (COMSOL)
298.125 307.6658 317.2041 299.3078
327.8878 334.8933 341.8226 336.4495
Tc,o Percent Diff 9.077138 8.130201 7.202134 11.03931
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
83.95964 87.87075 91.78139 100.6375
356.9596 360.8707 364.7814 373.6375
375.2051 376.8441 378.653 396.5427
Th,o Percent Diff 4.862787 4.238722 3.663396 5.776213
77

CASE 1:
Iteration No.
Ao (m^2)
Ai (m^2)
Tc,I (Celsius)
1
0.439823
0.314159
20
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)
0.0001
125
0.0001
0.007854
0.029845
826
Mc (kg/s)
Mh (kg/s)
Cpc (j/kg*k)
Cph (j/kg*k)
998.2
0.002979
0.000649
4182
2328
Cc (w/k)
Ch (w/k)
Cmin/Cmax
μ (Pa s)
12.45877
1.510264
0.121221
0.00915
Pr
Re k oil (w/m*k)
Nusselt Number
159
0.902732
0.134
4.43545 hi (w/m^2*k) k Copper (w/m*k)
5.943503
393.111
UA (w/k)
NTU
ε q (w)
Tc,o (Celsius)
Tc,o (Kelvin)
0.768501
0.508852
0.390861
61.98178
24.97495
298.125
Tc,o (COMSOL) 330.0611
Tc,o Percent Diff 9.675823
Th,o (Celsius) 83.95964
Th,o (Kelvin) 357.1096
Th,o (COMSOL) 376.4012
Th,o Percent Diff 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
Velocity of oil= .0001 m/s
2
0.439823
0.314159
20
3
0.439823
0.314159
20
0.001
125
0.0001
0.007854
0.029845
826
0.01
125
0.0001
0.007854
0.029845
826
998.2
0.029791
0.000649
4182
2328
124.5877
1.510264
0.012122
998.2
0.297914
0.000649
4182
2328
1245.877
1.510264
0.001212
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
78

CASE 2:
Iteration No.
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)
1 2 3 4
0.439823 0.439823 0.439823 0.439823
0.314159 0.314159 0.314159 0.314159
20
0.0001
Velocity of water & oil= .0001 m/s
30
0.0001
40
0.0001
20
0.0001
125
0.0001
125
0.0001
125
0.0001
150
0.0001
0.007854 0.007854 0.007854 0.007854
0.029845 0.029845 0.029845 0.029845
826 826 826 811
998.2 995.6 992.2 998.2
0.002979 0.002971 0.002961 0.002979
0.000649 0.000649 0.000649 0.000637
4182 4179 4179 4182
Cph (j/kg*k)
Cc (w/k)
Ch (w/k)
Cmin/Cmax
μ (Pa s)
Pr
2328 2328 2328 2440
12.45877 12.4174 12.375 12.45877
1.510264 1.510264 1.510264 1.554177
0.121221 0.121625 0.122042 0.124746
0.00915 0.00915 0.00915 0.00564
159 159 159 104
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)
0.902732 0.902732 0.902732 1.437943
0.134 0.134 0.134 0.132
4.43545 4.43545 4.43545 4.463665
5.943503 5.943503 5.943503 5.892037
393.111 393.111 393.111 391.3795
0.768501 0.768501 0.768501 0.761849
0.508852 0.508852 0.508852 0.490194
0.390861 0.390834 0.390807 0.379711
61.98178 56.07498 50.16887 76.71806
24.97495 34.51584 44.05405 26.15776
298.125 307.6658 317.2041 299.3078
330.0611 336.851 343.5672 339.0642
Tc,o Percent Diff 9.675823 8.664119 7.67336 11.72534
Th,o (Celsius)
Th,o (Kelvin)
Th,o (COMSOL)
83.95964 87.87075 91.78139 100.6375
357.1096 361.0207 364.9314 373.7875
376.4012 378.062 379.8715 397.855
Th,o Percent Diff 5.125254 4.507516 3.932947 6.049304
79