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This is to certify that the
thesis entitled

Optimization of a Variable
Core Geometry Radiator

presented by

Liaquat Ali Khan

has been accepted towards fulfillment
of the requirements for

Master's degree ”Mechanical Engineering

'WM

Major professor

Date—MLZLJQL

0-7639 MS U is an Affirmative Action/Equal Opportunity Institution

 

 

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TO AVOID FINES Mum on or More dd. duo.

DATE DUE DATE DUE DATE DUE

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

MSU In An Nflrmdlvo Action/Equal Opponunlty Institulon
Wm:

 

OPTIMIZATION OF A.VARIABLE
CORE GEOMETRY RADIATOR

BY

Liaquat Ali Khan

A THESIS

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

MASTER OF SCIENCE

Department of Mechanical Engineering

1995

ABSTRACT

OPTIMIZATION OF A VARIABLE
CORE GEOMETRY RADIATOR

BY

piaquat Ali Khan

An investigation on the cost of a heat exchanger by
using different geometries, specifically annular fins, has
been conducted..The effect of different parameters on the
cost was investigated, such as material cost, labor cost,
fuel cost which is used for operating the pump and fan for
the whole life of the radiator and also the effect of

interest rate of the money used for operating cost. Analysis

showed that the cost of the radiator is a step function,
which limits the available techniques of the optimization.
Calculations are performed by using a spread sheet (Micro
Soft Excel) for the different parameters of the geometries
and results showed that there is no improvement in the cost
by arranging different geometries in combination rather than
using only one type of geometry. By this analysis we are

able to find which geometry gives the minimum cost.

to my parents, Hafiz Muhammad Ashraf and Bano for

their loving devotion

ACKNOLODGEMENTS

I am grateful for the opportunity to work with my major
professor, C.W. Somerton; for his guidance and encouragement
conducting this research and the experience gained from the
discussion on the topic. Also, his open, frank discussions
concerning professional approach was enlightening and
appreciated.

The commitment of the remaining committee, J.V.Beck and
Abraham Engada also recognized. Their comments in the defence
were instructive.

The Department of Mechanical Engineering is acknowledged
for the continued generous support. Special thanks are
extended to C.W.Somerton for his cooperation in my' M.S.
degree program.

Finally, I would like to thank my family, who have

endured through my education and given love and

encouragement My parentsare appreciated for their endless
unassuming support,without which this point could have never

reached.

vi

TABLE OF CONTENTS

List of Tables ................................ xvii
List of Figures .............................. xviii
Nomencl a ture ................................. xxi i 1

vii

Chapter

1 Introduction ........................................ 1
1.0 Heat Exchanger Analysis ........................... 5
1.0.1 Basic Analysis ............................. 5

1.0.2 Dimensionless Analysis .................. 11

1.1 Literature Review ....................... 15

2 Method of Solution ...........................18

2.1 Sizing Process ................................... 18
2.2 Rating Process ................................... 19
2.3 Heat Exchanger Analysis .......................... 20

viii

2.3.1 Sizing Problem ........................... 20

2.4 Concerning Equations ............................. 21
2.4.1 Frontal Area ............................. 21
2.4.2 Fixed Parameters for a Geometry .......... 22
2.4.4 Friction Factor Equation ................. 24

2.5 Calculated Values for a Geometry ................ 25
2.5.1 Fin Length ................................. 25
2.5.2 Fin Perimeter ............................ 26
2.5.3 Fin Cross-Sectional Area ................. 27
2.5.4 Inside Surf. Area/Outside Surf. Area ..... 28
2.5.5 Volume of Cell Solid/Volume of Cell ...... 29

2.6 Method to Find the Outlet Temperatures .......... 29

of Coolant (Ethylene Glycol)

ix

.10

.11

.12

.13

.14

.15

.16

.17

Finding of Specific Heats ....................... 30

Calculation of Pradtle Number ................... 30
Effectiveness ................................... 3O
Calculation of Ntu .............................. 31
Outside Reynold's Number ........................ 32
Outside Heat Transfer Coefficient..§ ............ 32
Inside Heat Transfer Coefficient ................ 33
Overall Surface Efficiency(no) .................. 34
Overall Heat Transfer Coefficient ............... 35
2.15.1 Fouling .................................. 35
Depth and Number of Tubes Required .............. 37
Power Requirements .............................. 39

X

2.17.1 Outside Pressure Change .................. 39

2.17.2 Inside Pressure Change ................... 40
2.17.3 Outside Power Loss ....................... 41
2.17.4 Inside Power Loss ........................ 42
2.17.5 Total Operating Power Required ........... 42

2.18 Rating Problem .................................. 42
2.19 Cost Function ................................... 46
2.19.1 Material Cost ............................ 46
2.19.1.1 Tubes Cost ....................... 47

2.19.2.2 Fins Cost ........................ 48

2.19.2 Fabrication Cost ......................... 48
2.19.3 Fan and Pump Cost ...................... 49

xi

2.19.4 Pump and Fan Operation Cost ............ 49

2.20 Microsoft Excel Spread Sheet ................... 51
2.20.1 Spread Sheet For Sizing Problem ......... 51
2.20.1.1 Part "A" Exchanger Fixed ....... 57

Parameter for a Design

2.20.1.1 Part "8" Heat Exchanger Core...57

Configuration Parameters

2.20.2 Spread Sheet for Rating Problems ........ 66
2.21 Optimal Heat Exchanger ...................... 67
2.21.1 Optimization Using One Type ............ 67

of Geometry

2.21.2 Optimization Using Mixing .............. 68
of Geometry

xii

3 Results and Discussion ................................70

3.0 Source of Given Data ........................... 70
3.0.1 Hot Fluid (Ethylene Glycol) .............. 70
3.0.2 Cold Fluid (Air) ....................... 71

3.0.3 Some Engine Properties ................. 71

3.0.4 Thickness of the Copper Tubes .......... 71

3.1 Cost of Radiator by Using One Type ............. 72

of Geometry by Sizing Method

.1 Cost of Radiator Using Geometry .......... 72
"All only

.2 Cost of Radiator Using Geometry .......... 73
"B" only

.3 Cost of Radiator Using Geometry .......... 75
II C" only

xiii

3.1.4 Cost of Radiator Using Geometry .......... 76

"D" only
3.1.5 Cost of Radiator Using Geometry .......... 78

"E" Only
3.1.6 Cost of Radiator Using Geometry .......... 79

n F u only
3.2 Comparison of Cost for Different Geometries ..... 81
3.3 Comparison of Minimum Cost for Different ........ 83

Geometries

3.4 Comparison of Ntu at Minimum Cost ............... 84
3.5 Conclusion for the One Type of .................. 85

Geometries Used

3.5.1 Breakdown of Minimum Cost .................. 85
3.6 Cost of Radiator by Using Combination of ........ 86
Geoametries

xiv

4 Conclusions and Recommendations...............89

for Future WOrk...

List of References..............................114

Appendices

Appendix A-1:

Appendix A-2:

Appendix A-3:

Appendix 8-1:

Appendix C-2:

Appendix C-3:

Derivation of s-Ntu Relationship ..... 91

Derivation of M or Fin Efficiency....96

Pressure Loss in the Heat Exchanger..101

Comparison of Mass Flow, Cost, ....... 103

Ntu,8 & # of Tubes for Geometry 'A'

Comparison of Mass Flow, Cost, ...... 104

Ntu,8 & # of Tubes for Geometry '8'

Comparison of Mass Flow,Cost, ..... 105

Nt& # of Tubes for Geometry 'C'

XV

Appendix C-4: Comparison of Mass Flow,Cost, ...... 106-

Ntu,8 & # of Tubes for Geometry 'D'

Appendix C—5: Comparison of Mass Flow,Cost, ...... 107

Ntu,8 & # of Tubes for Geometry 'E'

Appendix C-6: Comparison of Mass Flow,Cost, ...... 108

Ntu & # of Tubes for Geometry

xvi

LIST OF TABLES

TABLE 2.1: Showing the different important ............. 19

parameters of the six geometries.

TABLE 3.1: Show that the cost is minimum in ............. 83

case of geometry 'B'

TABLE 3.2: Showing the cost of different ............... 87

combinations

xvii

FICHHRE

FICKHUB

FICHIRE

FICHIRE

FIINIRE

FICHIRE

LIST OF FIGURES

Heat flow through a cross flow heat ...... 2

exchanger

Annular fins, with liquid(hot fluid) is..3
flowing inside the tubes and Air(gas)

flowing outside
Generic heat exchanger, nonspecific ...... 7
design exchanging energy between the

two fluids

Thermal circuit for a heat exchanger ..... 9

wall neglecting any fouling

Graphic relationship between Colburn....23

Factor and Reynold's number

Showing the thickness of tubes and ...... 26

fin diameter

xviii

FIGURE

FIGURE

FIGURE

FIGURE

FIGURE

FIGURE 3.1:

FIGURE 3.1a:

Showing the arrangement of tubes in ..... 28

radiator to get width and height

Thermal resistance circuit showing all..36
resistances and also fouling resistances

of cold side and hot side fluids.

Showing the sketch of radiator for ...... 38

length, width and height

Showing the internal and outer .......... 47

diameter of the tubes

Showing tube and fin cross .............. 48

sectional areas

Showing the cost of geometry ‘A’ Vs ...... 72

different mass flow rates.

Showing the cost Vs Ntu and ............. 73
effectiveness at different costs for

geometry ‘A'.

xhc

FIGURE 3.2: Showing the cost Vs different

FIGURE

FIGURE

FIGURE

FIGURE

FIGURE

FIGURE

FIGURE

.2a: Showing cost Vs Ntu and effectiveness..

.3a: Showing cost Vs Ntu and effectiveness..

.4a: Showing cost Vs Ntu and effectiveness.

.Sa: Showing cost Vs Ntu and effectiveness..

mass flow rate for geometry ‘B’.

at different costs for geometry ‘8’.

.3: Showing the cost Vs mass flow for ........

geometry ‘C’.

at different costs for geometry ‘C’

.4: Showing cost Vs mass flow rate for ......

geometry ‘D’.

at different costs for geometry ‘D’.

.5: Showing cost Vs mass flow rate for ......

geometry ‘E’.

at different costs for geometry ‘E’.

74

.74

75

.76

77

..77

78

.79

FIGURE 3.6:

FIGURE

FIGURE

FIGURE

FIGURE

FIGURE

FIGURE

FIGURE

A-l.

1:

Showing cost Vs mass flow rate for ..... 80

geometry ‘F’.

Showing cost Vs Ntu and effectiveness..81

at different costs for geometry ‘F’.

Showing the comparison of minimum ...... 82

mass flow for the six geometries.

Showing the comparison of minimum ...... 83

mass flow for the given six geometries.

Showing the comparison of Ntu for ...... 84

the six geometries at minimum cost.

Showing the distribution of cost of....86
geometry ‘B’ which is the minimum

cost of all the six.

Showing cost for the combination of....88

different geometries (A,B,C,D,E,& F).

Showing the cross flow heat ........... 91

exchanger phenomenon

xxi

FIGURE A-2.1:

FIGURE A-2.2:

FIGURE A-3.l:

Heat flow in small segment dx ......... 96

Showing the dotted circle to .......... 99

diameter dm

Heat exchanger core model for ........ 101

pressure drop analysis is based on the

minimum free flow area in the core

xxii

Arabic

>

NOMENCLATURE

Area[m?]
specific heat [J/kg K]
heat capacity [W/K]

mass flow rate [kg/s]

enthalpy [J]

specific enthalpy [J/kg]

convective heat transfer coefficient
[W/mzK]

xxiii

Temperature [C9 of K]

heat [W]

thermal conductance [W/mzld
thermal resistance [K/W]
number of transfer units.
thermal conductivity [W/mfi
tube thickness [m]

height of the radiator [m]
width of the radiator [m]
slope of the straight line.
fin pitch [# fins/meter]
diameter [m]

y-intercept of straight line
reynold's number

coulburn factor

constant

length [m]

perimeter [m]

xxiv

Pr

Nu

KC

Ke

CF

Ml

CO

volume [n3]

prandtle number

nusselt number

friction factor

constant

Cartesian coordinates [m]
total number of terms
acceleration due to gravity
entrance loss coefficient
exit loss coefficient
cost of fuel [$1
miles/year

cost [S]

interest rate [%]

principle amount [$]

common ratio

Power [W]

J objective function

Cs cost component related to heat exchanger
sizes annual equivalent.

CP cost component related to the pumping
power annual equivalent.

CH Cost component related to the supply of
heating or cooling effect to the
radiator. (heat exchange)

Greek

8 effectiveness

_ change

v specific volume [m3/kg]

n efficiency

p density [kg/mfi

u dynamic viscosity

6 free flow area/frontal area

a heat transfer area/total volume

N”

Subscripts

in

out

min

max

tot

correction factor

inlet

outlet

cold fluid wall side
hot fluid wall side
minimum

maximum

wall

total

ratio

inside

outside

mean

xxvii

avg

act

COI'IV

cond

surf

fu

average
fin side
hydraulic
solid
actual
convection
conduction
surface

\

fan

pump

fouling

:mwdii

Chapter 1

Introduction

Heat exchangers provide for the transfer of heat between
two moving fluids. These devices are used in power
generation, chemical and food processing, heating, air
conditioning and motor vehicles. It is the most recognizable
heat transfer devices and one of the most widely used. Heat
exchangers are classified based on flow arrangement and type
of construction. There are many types of heat exchanger
designs, each type with its own characteristics that make it
suitable for a particular application. In motor vehicles the
heat exchangers are called radiator. In a radiator the hot
fluid is moving in the tubes and atmospheric air is the
cooling fluid. It is a cross flow heat exchanger and these are

the most successful type of heat exchangers.

 

E
E
E

 

 

 

 

 

 

 

 

 

 

——> a / / 53%

K L} D

C K r“ 7

_ \\ \ r
H (J -3

C h \ mam...
/’ / ma Ench-
k \

 

 

 

Fig’
.1-1: Heat flow through a Cross Flow heat exchanger

The design of heat exchangers, in general, requires
consideration of the heat transfer occurring between the two
fluids in addition to the mechanical energy needed to overcome
the frictional forces to move the fluids through the heat
exchanger. These two design criteria can. be generally
classified as heat transfer and, pressure drop. It is
typically desired to achieve large heat transfer yet maintain
a small pressure drop. Large heat transfer rates can he
obtained by either having a large heat transfer area or having
a large heat transfer convection coefficient. Unfortunately,
both of these conditions cause an increase in the pressure
drop, since a larger area gives more frictional area resulting

in an increase in the pressure drop and the larger flow rate

3

to increase the convection coefficient would likewise increase
the pressure drop. There is a trade-off in these two design
criteria; a beneficial gain in one criteria is usually at
expense of the other criteria and a compromise must be

established.

Focusing on the heat transfer, consider a simplistic heat
exchanger (radiator) that has two fluid streams separated by
a thin wall. One fluid is moving inside the tubes and the
cold air (atmospheric air) is moving across the tubes. Heat is

transferred from the tube fluid to the air.

\

 

 

 

 

 

Fig 1-2: Annular fins, with liquid(hot fluid) is
flowing inside the tubes and Air(gas) outside.

An effective way to increase surface area density is to

4

make use of secondary surfaces, or fins, on one or both fluid
sides of the surface. Fig. 1.2 illustrates a finned circular
tube surface in which circular fins have been attached to the
outside of circular tubes (annular fins). Such an arrangement
is frequently used in gas-to-liquid heat exchangers where
optimum design demands a maximum of surface area on the gas
side. Therefore, the annular fin type heat exchanger has been

chosen to study for the car radiator.

Axial conduction of heat is typically neglected in the
analysis of heat transfer in a heat exchanger. This
assumption may be approximate for most cases, but under
certain circumstances the effect of axial conduction can

become important.

Cost optimization is the most important feature of the
heat exchanger. Considering the effect of cooling fluid mass
flow rate and different annular fins geometries to optimize

the cost of the radiator will be the focus of this thesis.

Once the type of geometry used, fluid used, inlet and
outlet temperatures are selected, the problem is to get the
proper type of tube geometry or combination of different tube
geometries with optimum cooling fluid mass flow in order to

minimize cost of the radiator. The inlet temperature of air is

5
taken 5032. This temperature is being achieved in some parts

of asia.

The use of a variable core geometry is an attempt to make
heat transfer uniform through the heat exchanger by increasing

the surface area as the temperature difference decreases.

The remaining sections of this chapter will review the
analysis of a heat exchanger cost optimization of heat
exchangers and will conclude with a literature review. The
method of solution is presented next. Heat exchanger analysis,
spread sheet, optimization of cost and cost function will be
addressed in chapter two, chapter three will present the
results of the investigation with corresponding discussion.
A summary and the resulting conclusions will be given in

chapter four in addition to recommendations for future work.

1.0 Heat Exchanger Analysis:

1.0.1 Basic Analysis:

Figure 1.3 shows a heat exchanger that transfer energy

between two moving fluids through a wall; the geometry and

-

6
construction of this heat exchanger may be considered
arbitrary. A general thermal analysis of this heat exchanger
will be performed. First, taking the heat exchanger as a
control volume and applying an overall energy balance,
assuming no interactions (work or heat) with the surroundings

and steady state, results in an enthalpy balance (H).
in out ...... (1.1)

This can be written in terms of the inlet, outlet and existing

conditions using specific enthalpy and mass flow rate as
Ii‘lflhl-II in+mChCl in=thI out+mChCI out

and rearranged to group the fluids

)=m (13 | -h

mH(hH| in-hHl out

Assuming the fluid behaves as an incompressible liquid or
an ideal gas and a negligible pressure change, the change in

the enthalpy can be expressed as

dh=C dT
P

....... (1.4)

 

 

HUHWMU

 

THKmt<$-———- IWH -—-~E- THM?

 

 

V

TC,out (_ __ mC —— 4 TC,in

 

 

CdUFWMi

 

 

 

 

 

 

Fig.1-3: Generic heat exchanger, nonspecific design exchanging
energy between the two fluids.

This can be approximated by differences (dh.=h2 -hl) using
the approximate form of equation (1.4) in equation (1.3), the

results for the overall energy balance are;

q=mc (T -T )=n'1C (T -T ,) ..... (1.5)

H p,H H.121 H, out C p.C C.out C.1n

Introducing the new term as the heat capacity

...... (1.6)

CIhC
p

8

equation (1.5) can be rewritten as

q=CH(T -T )=C (T

[Lin H. out C coat-TC, in

)
....... (1.7)

In general, the heat transfer between the tmm> fluid

streams is a function of the following six parameters.

q=f(mH’ mCTH,in’ T

C,in’TC.out’TH,out) °°°°°° (1'8)

where the first four are typically given design parameters
and the last two are. desired results. Thus, in order to
obtain a solution more information is needed. The additional
information will come from the heat transfer analysis of the

wall separating the two fluids.

A circuit describing the thermal communication of the two
fluids is shown in Figure 1.4. This circuit shows the
resistance that impedes the heat flow, neglecting any fouling
on the heat exchanger wall. The total resistance of the

series circuit is the sum of the individual resistances.

_ ....... (1.9)
R tot _R.‘1'+Rw+RC

 

..J-N.AAA.IW%LJVVLJVV\_
RC amt Ru (um MI

 

 

 

Fig.1-4: Thermal circuit for a heat exchanger wall
neglecting any fouling.

Also, the overall conductance of the heat exchanger wall,
\
that will give the total amount of heat transferred if the

temperature difference is known, is

 

COC

These terms are analogous to an electrical circuit, with
the temperature difference as the voltage potential and the
heat flow as the current. The heat transfer over a different

element of length dx is

...... (1.11)

dq=UP[TH(x) -T x) ]dx

C(

The heat transfer over the entire length of the heat
exchanger is the sum of these differentials elements over the

total length

10

L L
giqu:UPf[7h(X)—T}(X)]dx ...... (1.12)
0 0

The product UP was assumed constant and thus could be taken

out of the integral, but (TH - T depends on x and cannot be

C )
removed from the integral. Because this integral is not known

in general, we will define the mean temperature difference as

L
l
(AT)m=—I—l{[\TH(X)-TC(X)ldX ....... (1.13)

By substituting equation (1.13) into equation (1.12), the
total heat transfer can be written as

q=UA(ATLn
...... (1.14)

For geometries that are more complicated, evaluating
equation (1.13) is difficult if not impossible. This suggests
that another approach may be necessary to remain a general

analysis.

11

1.0.2 Dimensionless Analysis:

To introduce another approach, consider the parameters

that the mean temperature difference is a function of

(AThfif(T

H, in’

T

C, in’

CC'CH'UA) ...... Cl.15)

It depends on the inlet temperatures and heat capacities
of the fluids and the wall conduction. The number of
independent parameters can be reduced by considering a second

function times the temperature difference at the inlet.

(AT)m=g(CC,CH,UA)(T -T .) ...... (1.16)

HI in C: 1n

Substituting equation (1.16) into equation (1.14) the

functional dependence for the heat transfer is given by

q=UAg(CC,CH,UA) (T -T . )

H, in C, 1n

....... (1.17)

As typical in heat transfer, scaling will be introduced to
jprovide dimensionless parameters. Defining the maximum

possible heat transfer as:

12

qmax_Cmin H,in TC, in

where C

min

equation (1.18)

= min(Cu ,Cc)

q _ UA

 

 

g(CC, CH, UA)

qmax min

Introducing another function that

independent variable gives

 

 

 

 

UA UA
q = (CR, >
qmax Cmin Cmin
Where
C : Cmin
R C
max
Finally, noting that equation (1.20)

final function as

 

 

and dividing equation

is terms of

...... (1.18)

(1.17) by

...... (1.19)

scaled

...... (1.20)

...... (1.21)

can be written for a

...... (1.22)

13
This demonstrates that the performance of a heat exchanger
can be expressed in terms of three dimensionless variables.
The first was given previously in equation (1.21) as the ratio

of the heat capacities. The second is the effectiveness

urn—Tom ...... (1.23)

Effectiveness (e) is the ratio of actual heat transfer
given by (1.7) to the maximum possible heat transfer. The

number of transfer units (NTU) is the last dimensionless

parameter v

UA
Cm... ...... (1.24)

 

NTU=

(NTU) is the ratio of the heat exchanger ability to transfer

energy to the minimum fluids ability to retain energy.

Previously mentioned demensionless parameters at least for
the moderately simple heat exchanger geometry, can result
during the analytical analysis of the heat transfer occurring
in a heat exchanger. For example, consider a cross flow one
fluid mixed and one unmixed heat exchanger. The correlation

describing the performance of this heat exchanger.

14

(i) for Cmn1(unmixed)

cmax (mixed)

€=(€%)(l-epr-C}[l-exp(-NTU)])
R

..... (1.25)

(11) for Cm-

in (mixed)

Cmax (unmixed)

\

ezl—exp(—C;1[1-exp(<NTU)l)

15

1.1 Literature Review:

Previously the optimization of cost is done to optimize
the heat transfer in radiative heat exchangers. The idea is
that ii? we optimize the heat transfer then the cost also
optimize.

Kuprys[1] used the concentric heating element with an
additional central heat exchanger in the form of field's tube
is analyzed. Experiment showed that effective thermal power
developed by such heat exchangers depends basically on the
flowrates, distribution and initial pressure of the cooling
gas. A number of design and operating parameters (heating-
element emissivity, cross-sectional area of the field tube,
etc) affect the effective thermal power developed by the heat
exchanger under the specified constraints on the temperature
of its elements. He used a computer package to find the

optimum heat flux in a radiator.

Another way adopted by Kovarik [2] was to get a maximum of

the objective function” Basically this is the aim. of

16
optimization. An objective function, J, is defined as the

ratio of performance to cost.

 

 

J'= ‘H (1 27)
Cs + Cp + Ch ...... .
From equation (1.27)
J'= l
(Cs*Cp)/H+a4 ..... (1.28)

Clearly, the position of the maximum of J coincides with
the minimum of the first term in the denominator of the right

hand side of equation (1.28), and is independent of the value

of the energy cost factor a4 . Therefore an optimal heat
exchanger is optimal for any cost of energy. Hence by
applying the cmmindzation techniques we can found the

optimum of J.

Evans [3] used the fundamental principles of differential
second law analysis to find the optimum number of transfer

units.

NTU : (l/b) ln(l+an) --------- (1.29)

17
Equation (1.29) gives the optimum number of transfer units for
all elementary flow arrangements (i.e, counterflow, parallel
flow and elementary cross flow) and for all possible values of

the heat capacity rate ratios.

The optimization done in this thesis differ from previous

work. in the formulation and solution of the problem.
Previous techniques of optimization involves the
optimization of the heat flux [1], optimization of objective

(performance to cost) function by getting the maxima [2], and
the~ optimization. of ‘transfer ‘units [3]. Although, the
previous techniques of optimization are better to some
extent, these are not the direct approaches to develop the
appropriate cost function. and. the use the optimization.
Therefore, the tasks are to develop an appropriate cost
function, modify the existing spread sheet program, and

perform the iterative calculation to minimize cost function.

Chapter 2

Method of Solution

The method employed to get the optimum design is the
Microsoft Excel Ver-4. Two different spread sheets, one for

sizing and one for rating problems, were employed.

Operating data used was for of General Motor Cavelier
3.1L. The radiator is of cross flow, annular fin type. Air is

cooling fluid and ethylene glycol is the working(hot) fluid.

2.1 Sizing Process:

In first step, a sizing analysis was done for uniform
cores using the 6 (six) different geometries A,B,C,D,E, and F
one ‘by one.For‘ one type: of geometry sizing' is done by
iterating with the mass flow of air to get the optimum mass
flow where the cost is minimum. This is done by changing the
outlet temperature of air with a particular mass flow of air
such that we achieved the required outlet temperature of hot

fluid (ethylene hlycol).

By getting the cost for different mass flow rates we get

the minimum cost required for geometries A,B,C,D,E, & F at a

18

19
particular mass flow rate. From these observations we can then
get the geometry and air mass flow rate for which the cost is

minimized.

 

 

 

 

 

 

 

 

S.No Geometry Fin Pitch Ctr-Ctr Dis Ctr—Ctr Dis.
1 A 289/m 0.0248 m 0.0203 m
2 B 343/m 0.0247 m 0.0203 m
3. C 343/m 0.0247 m 0.0203 m
4 D 276/m 0.0313 m 0.0343 m
5 E 343/m 0.0313 m 0.0343 m
6 F 343/m 0.0469 m 0.0343 m

 

 

 

 

 

 

Table 2.1: Showing the important features of the six
geometries used

2.2 Rating Process:

For the variable geometry core, that is a core consisting
of two or more geometries,it is useful to perform a rating
problem.By adding one layer of a particular type of geometry
and using the mass flow rate that gives the minimum cost in
the sizing problem for that particular geometry the outlet
temperatures are calculated. If outlet temperature of hot
fluid is equal of slightly less than the required one then the
process stops, otherwise, another layer of same type of
geometry or of other type is added. This process is continued

until we get the required temperature. This process is

20
iterated with all possible combination

to get the minimum possible cost.

2.3 Heat Exchanger Analysis:

In heat exchanger analysis there are two types of

problems:

1— Sizing Problem

2- Rating Problem

2.3.1 Sizing Problem:

In the sizing problem we know the inlet and outlet
temperatures of the fluids, mass flow rates, frontal area of
the exchanger the geometry used and its parameters. Hence all
the operating conditions are known or can be calculated from
the first law of thermodynamics. We can calculate
effectiveness (e) from the given informations but we have to
know the e-Ntu relationship according to the geometry
specification (annular fin one fluid mixed (air) and one
unmixed (ethylene glycol)). Once the Ntu is calculated, the

heat transfer surface area can be determined.

2.4 Concerning Equations:

(cooling and hot fluid) the problem is set up in the following

After deciding the type of heat exchanger, working fluids

manner.

Cold Fluid

(Air)

1- Inlet temp. of air TC,in
2- Outlet temp- of air TC,out
3— Average temp. TH,avg
4- Cp'c
5- fit (Air)
5- cc :Cmax or Cmin

Other inputs include the material of the tubes and the fins.

Tube Material

Fin Material

Copper(Cu)

2.4.1 Frontal Area:

We know that height

Aluminum(Al)

(H)

21

and width

Hot Fluid

(Ethylene Glycol)

Inlet temp. T

H,in

Outlet temp. T

H.0ut

Average temp. Tmaw

C H

D.

fih(Ethylene Glycol)

CH =Cmax or C

min

(W)

of the radiator.

22
which

is fixed for a particular design then we can calculate the

frontal area

Frontal Area = A = W x H -------- (2.1)

2.4.2 Fixed Parameters for a geometry:

I used the informations about he geometry (annular fin) form

[1]. The following'jparameter‘ are fixed. parameters for' a

particular geometry and these are read out directly from the

geometry specifications (Appendix “D")

(a) Free flow area/frontal area, 0
(b) Heat transfer area/Total area, a
(c) Flow passage hydraulic diameter, 4r =(4xtube dia)
(d) Tube outer diameter = DO
(e) Inside tube diameter = Di
(f) Fin diameter = Df
(g) Fin Pitch (Fins/meter) = P
(h) Fin thickness = t
(i) Fin Area/Total Area = Af/At

(j) Exch. Height: Ctr. to Ctr. Tube Distance = h

II
D;

(k) Exch. Depth: Ctr. to Ctr. Tube Distance

23

The spread sheet program requires specification of the
constant and exponent associated with the colburn factor
jH=CRem. The six types of annular fin geometries used have,
correlating relations that are both straight lines curves. All
have been approximated as a straight line, and it appear

accuracy is not very much affected.

The standard equation for the straight line is

y=mx+b ------- (2.1)

Where, m = slope of the line

b = y intercept

 

lnjH

 

 

InRe

 

 

 

Fig.2-1: Graphic relationship between Colburn factor
and Reynold's number.

24

lnjH = m lnRe + b ..... (2.2)

Slope m is as follow:

 

 

(y'iy ) (lnj -ln' )
m: 2 1 = ”'2 3’“ ..... (2.3)
(xz-xl) ln(Re2)-ln(Re1)
and the y—intercept
bzlnjH'Z—mlnRez ...... (2.4)
lnjH = lnRem + b
lnjH = lnRem + lneb
lnjH = lnRem eb
j”: eb Rem
...... (2.5)
m- lnjH'Z-lnjH’1
ln(Re2)-ln(Re1) ...... (2.6)
Hence, C = eb ...... (2.6a)
jH = C Rem ...... (2.7)

Equation (2.5) gives the relation for the Colburn Factor.

2.4.4 Friction Factor Equation:

Similar to Colburn's factor, we approximate the friction
factor relation to be a power law and find the equation in a

similar way.

25

 

f = C Rem ...... (2.8)

C = eb ...... (2.9)

b=lnf2-mlnRe2 """ (2'10)
m: lan-lnf1

lnIRez)‘1n<Rel> ..... (2.11)

2.5 Calculated Values For a Geometry:
\ .

2.5.1 Fin Length:

Since the focus has been on annular fin type heat

exchangers the fins may be modeled as shown bel

cur.“

26

 

 

 

 

 

 

 

Df

 

 

 

 

 

and fin diameter.

tubes

of

Fig.2-2: Showing the thickness

....(2.12)

 

erimeter:

2.5.2 Fin P

 

27

Pf=II (Df+Do)

2.5.3 Fin Cross-Sectional Area:

...... (2.14)

28

2.5.4 Inside Surf. Area/Outside Surf. Area(U.C):

 

 

 

 

 

Fig.2-3: Showing the arrangement of tubes in radiator to get
width and height.

A- Inside Surface Area of Tubes

1 _

:4: — (Surf. Area of Fins)x2 + Nacked surf. area of tubes

 

 

AC nDiL
Ao LP(ED§—EDZ)2+[nD L-LPnD t]
4 4 O O O
...... (2.15)
A D.
_C= 1
A (02_ 2

29

 

 

 

 

AC D1.
-—= ...... 2.16
A0 0 . SP(D§-Dj) +DO-PDot ( )
2.5.5 VOlume of cell Solid/Volume of Cell:
[ Vs 1 = Volume of Solid material
11
Van C9 Total VOlume
H 2_ 2 H 2_ 2
v Zwo Di)L+PLZ(Df Do)t
(-—3—) 11:
Vtot ce HWLI
( VS ) :n‘(D:-Di2)+Pn(D§-D02)t
ll
Vcot C9 4HW
..... (2.17)

2.6 Method to find the outlet Temperature of Coolant
(Ethylene Glycol):

Outlet temperature of working fluid can be found by the

use of first law of thermodynamics.

Heat lost by working fluid(Ethy. Gly.) = Heat gain by air
(Ith) C(TC, out—TCin) = (me) H( TH, in-TH, out) ..... (2 . 18)

which specifies all of the temperatures.

30

2.7 Finding of Specific heats (Cp '3):
The values of Cp '5 are found at their respective mean
temperature. The existing spread sheet incorporate a curve

fit to data to represent Cplas a function of temperature.

2.8 Calculation of Prandtle Number:

The Prandtl number is

 

./*U
n

The values of u, Cp and K are evaluated at the mean

temperature .

2.9 Effectiveness (e):

The effectiveness now can be calculated by the following
formula. It is the ratio of the actual heat transfer to the

theoretical heat transfe

CHIT T )_ CC(T

H. in H. out:

T )

qact _

 

= _ C. out. C, in
q Cmin ( TH. ill-TC. in) Cmin ( TH. in-TC, in)

The spread sheet uses an if statement to determine Cm”.

31

2.10 Calculation of Ntu:

There are different relationships between 8 and Ntu for
different conditions. The relationship for the cross flow heat
exchanger with one fluid mixed (air) and one unmixed (ethylene
glycol) is being used in this analysis.
then relation

If air is (C and Ethylene Glycol in (C

max) min)

 

is;
~ 1
Ntu=-ln[1+(27)ln(l-ecgl
R ..... (2.21)
and if air is (Cmfi) and Ethylene Glycol (me)
Ntu=-(—£0]1HI?1n(l-e)+l]
CR R ...... (2.22)
Where
Cmin
C1:

max

32

2.11 Outside Reynold's number:

We can find the Reynold's number by using the following

 

formula.
GD
Re=——3
u
Where,
:mHCH
oA

fr

2.12 Outside Heat Transfer Coefficient:

Outside heat transfer coefficient is extracted

colburn factor

 

. GCP ....
hO—Jfiprzn

Where,
jH = C Rem

from the

.(2.24)

Value of m, C and b can be found by using equations

(2.11),(2.9) and (2.10) respectively. Prandtl number of the

outside fluid has been evaluated by the equation (2.18)

33

2.13 Inside Heat Transfer Coefficient:

If inside Reynolds number is less than 2300 flow is laminar

and

 

and Nusselt Number = Nu = 4

But if flow is turbulent then

 

and

(

f
E) (RED-1000) Pr

Nu:

 

l
2

f 2
1+12.7(-§) (Pr 3-1) [6]

where f is the friction factor and it is

f=(0.79lnReD-1.64)'2
[6]

..... (2.28)

34

This relation is valid for O.5<Pr<2000 and 2300<Re <5 x 105

2.14 Overall Surface Efficiency (no ):

The quantity no is termed the overall surface efficiency
or temperature effectiveness of a finned surface. It is
defined such that, for the hot or cold surface, the heat

transfer rate is

‘q=nohA<Tb-T..> ...... (2.29)

It can be introduced easily into the expression for the
overall heat transfer coefficient,The overall surface fin

efficiency is related to the fin effectiveness as

A

no=l- (1-nf)

_£
A ..... (2.30)

For simplicity , it is assumed that a straight or pin fin of
length "L" can be used to model the annular fin. Assuming an

adiabatic tip

tanh(MLf)
MLf ..... (2.31)

 

n0

35

h P

o surf

A.K
..... (2.32)

The derivation of can be found in Appendix (A-2)

2.15 Overall Heat Transfer Coefficient:

Now, the overall heat transfer heat transfer coefficient
can be found. An essential , and often the most uncertain,
part of any heat exchanger analysis is determination of the
overall heat transfer coefficient. Overall coefficient is
defined in terms of the total thermal resistance to heat
transfer between two fluids. The coefficients can be
determined by accounting for conduction and convection
resistances between fluids separated by composite

plane and cylindrical walls respectively.

2.15.1 Fouling:

During normal heat exchanger operation surfaces are often
subject to fouling by fluid impurities, rust formation, or
other reaction between the fluid and the wall material. The
deposition of a film or scale on the surface can greatly

increase the resistance to heat transfer between the fluids.

h“

36
This effect can be treated by introducing in additional
thermal resistance, termed the fouling factor, Rf . Its value

depends (Hi the operating temperatures, fluid velocity, and

length of service of heat exchanger.

 

_NVLNV\_AN\_NV\_AM_
RC (RDC Ru (RDH RH

 

 

 

Fig.2-4: Thermal resistance circuit showing all
resistances and also fouling resistances
of cold side and hot side fluids.

1 1 Rf, C IR“) 1
_= + +Rw+ +
UA nOhAC (noAlc “0A,, nohA

 

 

I:

..... (2.33)

Where, Rhcenxiihklare the fouling factors. Because ethylene
glycol which is corrosion resistance and also anti-freeze,
fouling is reduced to its minimum value. So, in further
analysis fouling will be ignored. Rw depends on the geometry

used. For cylindrical

37

tube using the annular fin ln(Der)/(2nKLLAJ is the tube wall

 

 

resistance.
D
(..__O
D.
=[ IA +ln. 1L +hl ]-1
hH—H ZHK— Cm“ ..... (2.34)
C

0

Hence we need to know the five parameter hH, }%/AC,.ILO , h

L/AQ_Adl of these can be calculated through equations (2.28),

(2.15) .(2.25), & (2.30).

\

2.16 Depth and Number of Tubes Required:

The sizing problem really comes down to determining the
depth of the heat exchanger core and the number of tubes
required. First of all we find the total volume

A
=—5 ...... (2.35)
a

V

tot

Where A0 is the required heat transfer surface area on the
fin side and has been determined form the E-Ntu analysis

NtuC'min
AC=——————— ..... (2.36)
UC

38

The depth of the heat exchanger can then be calculate

 

Depth=d=vt°t
A

...... (2.37)
fr

 

 

 

 

 

 

 

 

 

7T \
\\____.._...._
>|

 

 

 

Fig.2-5: Showing the sketch of radiator for
length, width and height.

The number of tubes for a particular geometry for one pass

can be calculated by dividing depth with the center to

center distance of tubes.

Frontal Area.(A£Q

Exch. Depth Ctr-Ctr. Tube Dist.

Number of Tubes in one pass

39

Similarly the number of passes (number columns in the depth

Depth of Exchanger(d)

umber of Passes Depth Columns = ‘
Exch.Deptthr-Ctr. Tube.Dist

 

direction) can be calculated by Total Number of Tubes = (#

of tubes in one pass) x
(# of passes in depth columns)

..... (2.37)

2.17 Power Requirements:

Power requirements for heat exchangers required

Outside pressure change

Inside pressure change

Outside power loss

Inside power loss

Total operating power

2.17.1 Outside Pressure Change:

According to equation (2.8) friction factor can be found

f = C Recm

There two conditions for the outside pressure loss.

40
If,

2 v0 food/Mm
(l+o)(—-l)+(——.—)<0 ...... (2.38)
v. OAfVI

if this condition satisfied than the change in pressure is

zero

AP

II
CD

If, the above condition do not satisfy than the pressure drop

 

is
v0 deVv vimi2
AP.=[(1+02) (——1)+(———.’”
1 v. oAfv1 202.112
1 f
...... (2.39)
2.17.2 Inside Pressure Change:
For the tube side flow,
If Re <2300, then the friction factor is
=2
Re ..... (2.40)

But, if the Reynolds number is larger than 2300 then the

w”

41

equation (2.40) no longer valid. The equation for this case

is as follow.

1
(0.791nRe-1.64)2 ..... (2.41)

f:

 

Inside pressure loss can be found by the following

relationship.

P 8m§wnfi
A :
o H2(D.)szH2 ..... (2.42)

l

 

Details of this development can be found in Appendix A-3

2.17.3 Outside Power Loss:

Power required to maintain the mass flow of air to get the
required temperature. of the hot side fluid. This is actually

the fan power required.

nu APo

Outside Power required = P '= ——————— ..... (2.43)

0

pl

‘C'.

42

2.17.4 Inside Power Loss:

Power required to maintain the mass flow of hot Fluid
(ethylene glycol) to get the required outlet temperature, so

this is the pumping power required.

nu APl

Inside Power required = P.:= -——————- ..... (2.44)

1

p2

2.17.5 Total Operating Power required:

Total operating power is the sum of Fan Power and Pumping

Power (outside power loss + inside power loss).

2.18 Rating Problems:

For the rating problem we know the core geometry, the mass
flow rates and the emmering fluid temperature.If the heat
transfer rate and exchanger effectiveness are predicted rate
the resulting outlet temperatures can be determined. This is
like using a heat exchanger given to us off the shelf, where
we know everything about it except for its operating

conditions.

a.“

43

The rating problem follows the solution methodology outlined

below
(i) First of all find the CE“
temperature,

if both of the temperatures are not

values at the inlet temperatures.

(ii) Calculate the Cmin and Cmax (me)

CR: Cmin/Cmax

(iii) Find the ho,hi,nb, Ai/AO , as in

(iv)

and CD.C at the mean
known then we find the

and then calculate

the sizing problem.

hi can be found out by first checking the Re number to

ensure the type of flow whether it is laminar or turbulent.

 

UD
Re:___[3
V
NuK
__ H
hi- D

If flow is laminar Nu=4

If flow is turbulent

(-§)(ReD-1000)Pr

 

Nu=
f .1. .2.
1+12.7(§) 2 (Pr 3—1)

..... (2.46)

..... (2.47)

....... (2.48)

44
and f = (0.79lnRe - 1.64)‘2

This correlation is valid for O.5<Pr<2000 and 2300<Pr<5x10S

(v) Now we calculate the h3(cmter heat transfer coefficient)

by the same equation(2.25) as in the sizing problem.
(vi) The Ntu can now be calculated.

(vii) Next we find the effectiveness by using the appropriate

effectiveness Ntu equation.

Since we are using an annular fin heat exchanger with one
fluid mixed(air) and one fluid unmixed(ethylene glycol), then
if C

: Air, and Cmin = Ethylene Glycol

max

€=(€%)(l-exp(-C;1[l-9Xp(-Ntu))1
R
...... (2.49)

and if

C = mixed and Cmin = unmixed

max

e=1-exp(C;1(l-eXpl-C§(Ntu))1)

45

 

...... (2.50)
(vi) Next the Ntu can be found by
Ntu: HA
min ...... (2.51)

With the effectiveness known, the actual heat transfer rate

is determined from

Where,

..... (2.52)

(vii) Now by applying the let law of thermodynamics we can

calculate the exit temperatures of hot and cold fluid.

__ qact
TH,ouc—TH,in- C ..... (2.53)

H

T T" qact

 

..... (2.54)

The operating power required will be calculated in the same

manner as in the sizing’ problem in sectio

46
2.17.5,12.l7.6,2.17.3

2.19 Cost Function:

The cost is defined, for this purpose, as the annual
equivalent of all the present value of expenditure necessarily
incurred in acquiring and operating any equipment over its
useful lifetime.

The cost function. associated. with. the cost of ‘heat
exchanger is as follow:

CostE C = Material Cost
+ Fabrication Cost
+ Fan Capital Cost
+ Pump Capital Cost
+ Operating Cost

(Pump & Fan)
..... (2.55)

2.19.1 Material Cost:

Material cost includes the cost of tubes and fins.

Material Cost = Tube Material Cost + Fin Material Cost

..... (2.56)

47

2.19.1.1 Tubes Cost:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig.2-6: Showing the internal and outer diameter of the
tubes.

Tube Cost = (n/4)(Dj — Df) (width of Radiator)
x (# of tubes in each column)
x (# of columns of tubes)
x (Cost per kg of tube material<Cu))

..... (2.57)

2.19.1.2 Fins Cost:

48

 

 

   
  
  
  
  
 

//
\\\\\\\\\

 

459k ,Azzgzay
\gzpd'llllllllll'72Z§$§\
V " s

 
 
   

////

"’///// /

 
   
 

 

 

 

 

Di

 

 

Fig.2-7: Showing tube and fin cross sectional areas

Fin Cost 2

X

2.19.2 Fabrication Cost:

(n/4) (Df - Df)t

(Fin Pitch)

(# tubes)

(width of Radiator)

(cost of fin material per kg(Al))
....(2.58)

The cost associated to manufacture the radiator is the

fabrication cost.

It is

"n' times of material cost. The value

49
of "n" depends on the experience and the process used. I am

using the value of n=2 im this analysi

Fabrication Cost = n x (Material Cost)

....(2.59)

2.19.3 Fan and Pump Cost:

The capital costs of fan and pump is fixed, and there is no

need to include in the total cost of the radiator for

optimization.

2.19.4 Pump and Fan Operating Cost:

Cost associated with the operating power required for fan

and pump.

 

By checking the units

=[(kW)x($/kw.hr)(# or hours of operating/year)

50

= S/year (U.S. Dollars)

Since these operating costs are represented as annual
costs and we wish to compare with capital costs, the value
of money must be accounted for. We are taking the interest
rate of 7% (because recently in United States of America the
interest rate on cars sold on installments is this) but this

rate is flexible.

Life of the radiator is also flexible. I have taken the
life of the radiator 10 years.
The series for the 7% compound interest is the Geometric

Progression (G.P.).

Cpp +(CPF x I + C...) + [(Cp. x I + opp)x1+ch]+---

Upto 11th term.

 

Cpp ..... (2.61)

Now for n=10 the N=10+l=11.

Hence after 10 years the term will be 11th term.

: P(r”-1)
” (r-l)

51

..... (2.62)

This will gives the total cost associated with the

operating power of fan and pump.

Hence, the Cost Function is

Cost 2 Fin Cost + Tube Cost + F(fabrication Cost)
+ operating Cost associated with fan and pump

+ Interest cost associated with fan and pump

operating cost.

\

2.20 Microsoft Excel Spread Sheet:

To solve the problem the existing spread sheet made by

Edith Brown [7] was modified into two spread sheets now

(1) For Sizing Problems

(2) For Rating Problems

2.20.1 Spread Sheet For Sizing Problem:

In the spread sheet some cells are the inputs and the some

cells are the calculated results. Red border cells are the

user's input and the other one's are of calculated results.

52

 

 

 

1. Free Flow Area/Frontal Area:

 

Surface Area / Volume:
Colburn Factor Re

 

. Colburn Factor Re constant:

Friction Factor Re

 

 

Friction Factor Re constant:

Diameter
.toCtr. Tube dist:
.-to-Ctr. Tube dist:

 

53

1Jnside Tube Diameter
2. Fin Diameter:
3.Ftn Pitch
Area/1' otal Area:
thickness

 

 

 

Image-9159!. I

1 070693697

 

6004.439525

0.935833879

0.

010137135

14.

At Tave

Tin IT

9.3950] 1.0] [+00 0 . 97640248

T of Tube

Flow:
I

Turbulent Flow

 

Friction factor

Friction factor

Power
Power

7-Total Operating Power

0041190551
0. 14484
106.091734

1395

151 .4708504

 

'C'1 I

of Tube Material

of Fin material I

of tube material

of fin material |

Cost Factor

of Fuel in

Miles/Year of car

Miles

Total horse

of the

O-Lite ot the Radiator

1-lnterest Rate/Year

I

"c.2'9

Total

13.3! .

55

 

 

Cost for the entire life of radiator

Total Cost of Radiator

"D”

Correlations:

_ Outside (Fin Side) Fluid:

 

1012.04 .

2060 87

2592 62

NA

 

Vlec.

g 2 03505

4 44E-02

_ 4 735-03

NA.

T. Cond.
2.9lE-02
I 405-01
20012-01

N A.
6 5315-01

T Out

9.787477181

7.1

35.78411213

1498680417

0.987019616

1.07

Prandtl

7.00EvOI

6 52E+02

4 76E+0|

N A. NA.

9 825402

 

Inside (Tube Side) Fluid:

I014 29

2l85.85

2700.76

NA.

 

Vise.

2 l6E-05

2.l2E-02

2 4815-03

NA

3 03E-04

T Coed.

31315-02

l 38E-Ol

2 635-0]

NA.

6 76E-Ol

 

 

 

 

9 BlE-Ol

8 45E+02

l 07E+03

NA.

Prandtl

6 96E~0|

3.37E+02

2 55E+0|

NA

I88E+00

 

57

2.20.1.1 Part "A" Exchanger Fixed Parameter for a Design:

Red border cells shows the user input or the input by some
feed-data. It has five parameter. Four are user input and the one

is calculated.

1- Fins material thermal conductivity Al(237 w/m°k)
2- Tubes material thermal conductivity Cu(401 w/m°k)
3- Height of Exchanger (in m)
4- Width of exchanger (in m)
5- Frontal area in (m2)
Frontal area is calculated by multiplying parameter 3 and 4.
This is done by specifying equation at that cell which has the

multiplication of the cell number of 3 and 4.

For example in the existing spread sheet.

at E11 = $E$9 + $E$10

2.20.1.2 Part "B" Heat Exchanger Core Configuration

Parameter 3

This part deals with the fixed and calculated parameters of
heat exchanger, for a particular geometry of a heat exchanger core

configuration.

58

Part "B-1":

I am using the six different geometries of a annular fin, and
I have restrict to only these six geometries, therefore, I have
made the spread sheet user friendly. For doing this I have used
the code of different geometries. For example for Geometry-A
configuration if we put "1" then all the fixed data relating the

geometry "A" will automatically entered at the required cells.

So I have made the cell "C27" is the input cell for a geometry
code. Making reference of(this cell and using "IF" statement one

can easily entere the 15 fixed parameters for a geometry.

These fifteen parameters are as follow;

1— Free flow Area/Frontal Area

2- Surface Area/Volume

3— Colburn Factor, Re exponent

4- Colburn Factor, Re constant

5- Friction Factor, Re exponent

6- Friction Factor, Re constant

7- Hydraulic Diameter.

8— Exch. Height ctr. to ctr. Tube dist.
9- Exch. Depth ctr. to ctr. Tube dist.

10- Outside tube diameter

59
11- Inside tube diameter
12- Fin diameter
13- Fin pitch
14— Fin Area/Total Area

15- Fin Thickness.

For example for parameter 1 (o) for

Geom.-A : 0.538
Geom.—B = 0.524
Geom.-C = 0.494
Geom.—D = 0.449
Geom.-E = 0.443

Geom.-F = 0.628

By using IF statement at cell F31

=IF($C$27=1,0.538,IF($C$27=2,0.524,IF($C$27=3,0.494,IF($C$27=4,0.

449,IF($C$27=5,0.443,IF($C527=6,0.628,"ERROR"))))))

It compares with IF statement and replace the value of parameter

for the respective geometry.

In this part five parameters are calculated which are as follow;

60

1- Fin Length

2- Fin Perimeter
3— Fin Cross-sectional area
4- Inside surface Area/outside surf. Area (U.C.)

5- Volume of Cell solid/Volume of Cell

These values are calculated by using the equations (2.12)

,(2.13), (2.14), (2.16) and (2.17) respectively.

The only difference in inputting the formula is not to enter
variable directly, insted.specify the cell numbers to enter the

formula

Part "B-2 " :

This part gives the opportunity to the user to select the tube
fluid and the outside (cooling) fluid. In this part we listed only

the codes for different fluids.

Fluid Code
Air 1
Engine Oil 2

Ethylene Glycol 3

61
Superheated water vapor 4
Water 5

Part "B-3": Fluid Properties:

In this section we evaluate the fluid properties for hot and
cold fluid, once selecting the two; by entering the appropriate

code in cell ”B65" and ”F65"

In this five (5) quantities are input quantities.

1- Mass flow rate of cooling fluid in (kg/s)

2— Mass flow rate of hot fluid in (kg/s)
3- Timc of Cold fluid in (K)
4— TOW,C of Cold fluid in (K)
5- 'T of Hot fluid in (K)

in,H

All other parameters are either calculated or replaced by
comparing two conditions.

Calculated values are of T of both

of hot fluid and Tav

out e

fluid (hot and cold)

Specific heats of hot and cold (Cp) fluid
Dynamic viscosity of hot and cold (u) fluids

Thermal conductivity of hot and cold fluids

62
Density of hot and cold fluids.

Prandtl number of hot and cold fluids.

These quantities are replaced by comparing the appropriate
condition through the macros in Part-D at the end of spread
sheet. The value of CD is calculated at the mean temperature and
finalized value is taken when the constraint codition is satisfied.

This give the correct values for the properties of the fluid.

Part "B-4" Calculated parameters for heat Exchanger Design:

This part include all the calculated parameters relating the
final design of the heat exchanger. In this part we get all the

desired values regarding the design of any heat exchanger. There

are sixteen parameters.

1— Effectiveness e by using Equation(2.20)

2 " Cmin/ Cmax

3- Ntu by using equations (2.21) & (2.22)

4- Outside Reynolds number through equations(2.23)

5— Colburn factor by using equation(2.7)

6- Outside heat transfer coefficient by
equation(2.25)

7- Inside Reynold number through(

8- Inside heat transfer coefficient by

equation(2.28)

63

9- no through equation(2.30)
10- nf fin efficiency by using equation(2.3l)
11- Outside(fin side) overall coefficient

equation(2.34)
12- Total volume needed by using equation(2.35)
13- Depth of exchanger by using equation(2.36)
14- Number of tubes (Height)
15- Number of passes, depth columns.

16- Total number of tubes by using equation(2.37)

From these quantities we can find how many tubes
are required and hOW' many passes are there in a the heat
exchanger.We also get the performance parameters such as
effectiveness(e), Ntu (net transfer units), and fin efficiencey

(nf) I etc.
Part-B-S Exchanger Power Requirements:
In this section we have three parameters which will be taken by

comparing the statements. All of them are related to the outside

fluid (air)

HG
ll

Specific volume at Tin

NC.
ll

Specific volume at Tout

11 2 Specific volume at T (mean temperature)

me an

'C'

64

There are seven calculated parameters regarding the power

required to run fan pump. These are

Outside friction factor Q) w equation(2.8)
Inside friction factor fit»!
equation(2.38)or(2.39)

Outside pressure change by equation(2.40)
Inside pressure change by equation(2.4l)
Outside power loss (Fan) by equation(2.42)
Inside power loss (pump) by equation(2.43)
Total power loss through equation(2.44)

‘.

Hence we get the operating power for fan and pump.

Part C:

This part is concerned with the cost of the heat exchanger.

Part C-l:

This part include the parameters which are user defined and

supplied the user, therefore, these are red bordered cells. These

are as follow.

1- Density of tube material.

2- Density of fin material.

65

3- Cost per kg of tube material.

4- Cost per kg of fin material.

5- Fabrication cost factor.

6- Cost of fuel in (S/gallon).

7- Miles/year of a car(approx. car running).
8- MPG (miles per gallon).

9- Total horse power of the engine.
10- Life of the radiator.

11- Interest rate/year.

Part C-2:

This part include all the calculated parameters regarding the
cost of heat exchanger. If we are able to calculate eight (8)
quantities then we obtain the total cost of the heat exchanger.

These are as follow:

1- Tube material used.

2- Fin material used.

3- Tube material cost.

4- Fin material cost.

5- Fabrication cost.

6- Operating cost (Fan & Pump) per year.

7- Multiplication factor for each year.

8- Total operating cost for the entire life of

radiator.

66

9- Total cost.

This will complete the design and will calculate the cost of

heat exchanger.

Part-D:

This is a small macro and gives the value of the properties of
fluids used for cooling and the working fluid.

These properties are:

Specific heat

Dynamic Viscosity. ‘.
Thermal conductivity
Density and

Specific volume of outside fluid (Air). at T. T

1n'

0“ , and at

the mean temperature (T

mean ) '

2.20.2 Spread sheet for Rating Problems:

There are no major differences between the rating and sizing

spread sheets. The differences are in part "B-3" and "B-4".

In "B-3" now only two temperature are known.

T. of cold fluid (air)

1n

Tin of Hot fluid (Ethylene Glycol)

and the other two temperature are calculated.

67
Tout of cold fluid by Equation (2.51)

Tout of hot fluid by Equation (2.52)

2nd difference is in part "B-4". In this the depth of the
exchanger is a user input and the formulae used for
effectiveness and Ntu are little bit different those are
equation(2.49) or (2.50) for effectiveness and (2.50.b) for Ntu. In
this the only is inlet temperatures of the hot and cold fluids and
the depth of exchanger and we get the required outlet temperatures
of hot and cold fluids. Keep changing the depth of exchanger until

we get the required outlet temperature of hot fluid.

2.21 Optimal Heat Exchanger:

An optimal heat exchanger is defined as one that, while

satisfying imposed constraints, achieves the required task at the

lowest possible cost.

2.21.1 Optimization with uniform geometry:

In this step the sizing problem spread sheet is use. By using
the one type of geometry A,B,C,D,E, and F and iterating with the
mass flow rate of air and the outlet temperature of air we get the

required outlet temperature of ethylene glycol.

Then taking the mass flow rate of air then we change the

68
outlet temperature of air (cold fluid) such that we achieve the
required outlet temperature of ethylene glycol. If the temperature
of ethylene glycol is more than the required, we increase the
outlet temperature of the air, and if it is less then we decrease

the temperature of air.

This process continues for different mass flow rates of
ethylene glycol for all the geometries. By the observations we can
able to decide which geometry and at what mass flow rate of air we

get the minimum cost (optimal solution).

2.21.2 Optimization with variable geometries:

After determining the optimum cost using the unifrm geometry
core,we now want to further reduce the cost by using different
geometries together. Single type of geometry optimization will

show that which geometry gives the minimum cost at what mass flow.

The spread sheet used in this case is the rating problem spread
sheet, in which we calculate the outlet temperatures of air and
ethylene glycol. The depth of the exchanger is now the input. By
using one layer of tubes of particular geometry which gives the
minimum cost and also the proper mass flow of air, we get the
outlet temperature of air and ethylene glycol. A 2nd layer is then
added of same geometry or of other geometry keeping mass flow

constant. When we are adding the 2nd layer the outlet temperatures

69
achieved by adding the first layer will be the inlet temperatures
of air and ethylene glycol. This process continue with different

combinations of geometries, until we achieve the minimum cost.

Chapter 3

RESULTS AND DISCUSSION

3.0 Source of given data:

I have taken the data of General Motor Car Cavalier-uplevel.

Engine Capacity is 3.1L.

Height of Exchanger H 0.36 meters

Width of Exchanger “W = 0.6 meters
Thermal Conductivity of Tubes(Cu) = 401 w/m k
Thermal Conductivity of Fins(Al) = 237 w/m k

Working Fluid (Hot Fluid) Ethylene Glycol

Cooling Fluid is ambient air.

3.0.1 Hot Fluid (Ethylene Glycol)

1- Mass flow rate of hot fluid (m°H ) = 0.87 kg/s

2- Temperature leaving engine entering radiator TH in = 370 K

- 360.89 K

3- Temperature leaving radiator entering engine TH out-

70

3.0.2

71

Cold Fluid (Air)

Mass flow rate (m.°C ) (Found through iteration the optimize

mass flow)

Temperature entering radiator T = 323 K

C,in

Temperature leaving radiator (Calculated through formula)
Some Engine Properties:

Engine capacity 2 3.1 L

160 HP.

Horse Power of the engine

Miles per gallon (MPG) 27 miles/gallon

Thickness of the Copper tubes:

Thickness of the copper tubes is taken 0.0025m from the hand book

[6].

72
3.1 Cost of Radiator by using one Type of Geometry by Sizing
method.
In this we use geometry A,B,C,D,E and F one by one. By taking one
type of geometry and iterating through mass flow rate of air to get

the minimum cost of the radiator.

3.1.1 Cost of Radiator using Geometry 'A' only:

By iterating through the mass flow rate of air starting from 0.95
kg/s to 0.70 kg/s. Table 3.1 in appendix C-l shows the mass flow
rate of air, Cost, Ntu and Effectiveness. Figures (3.1)and (3.1a)
also tells the same story. Minimum cost is $345.48 incurred at 0.81

kg/s.

 

 

Graph Blw Cost & Mass Flow

= I 343.82

Cast in 3

 

 

 

 

Graph 3.1: Show: the cost of Geometry 'A' with
dillerent man flow and the optimum mu: How
is 0.81 kg/I where can in $345.49.

 

 

 

Fig.3.1: Graph cost vs mass flow for the geometry "A"

73

 

 

 

 

 

 

 

 

 

Gnufiflwwffif,NhH&Cbst
g 1+,
.. ‘53: ,2
3 . __--...----~*"“"’M Seriss1
0.34“...
g 061’ . ...... - .
g ‘ (m- ~- * SenesZ
- [141'
B
0.24»
.3; 0‘
m +H-H—H—0—H-4—9-‘l—O-‘I-H-H-O-O-O-O-H
<9 0 "2‘9 0 "s e a?!“
a 3;5? 3 $ 3325 8 8
g m m m (‘7 (‘7 m
Costint

 

 

 

Graph 3.1.1: Showing the cost vs Ntu and Bflectiveneu
at. different costs.

 

 

 

Fig.3.1a: Graph effectiveness & Ntu vs cost for geometry ”A"

‘.

3.1.2 Cost of Radiator using Geometry 'B' only:

Table 3.2 in appendix C-2 shows the mass flow rate of air, cost,
Nut and Effectiveness for the geometry 'B'. Minimum cost is $240.38
incurred at 0.81 kg/s of air. Graphs (3.2) and (3.2a) also tells

the same story.

74

 

 

Graph blw Cosi & Mass Flow

300
250
100
50
0

Cast in S
G;
D

0.95
092
089
[188
083
077
074
071
088

 

Mass Flow In Kqu

 

 

Graph 3.:2 Showing the cost 7: different mass flow
te. it showing that the minimum cost
ii at 0.81:3/3 which it $240.38

 

 

 

Fig.3.2: Graph cost vs mass: flow for geometry ”B"

 

 

Graph bM Eff.. Ntu 8. Cost

 

 

Series1

Senssz

 

 

 

Efiediveness & Ntu

 

 

 

 

Graph 3.2.1: Seriea 1 showing the Ntu and Series 2

showing the effectiveness vs different
cotte.

 

 

 

Fig.3.2a: Graph effectiveness & Ntu vs cost for geometry ”B"

75

3.1.3 Cost of Radiator using Geometry 'C' only:

Table 3.3 in appendix (C—3) showing the mass flow of air, cost, Ntu
and effectiveness for geometry 'C'. Minimum cost is $324.86
incurred at 0.84 kg/s. Graphs (3.3) and (3.3a) also shows the same

data in the graphical form.

 

 

Graph blw Cost (Miss Flow

45!]
400
350
300
250
200
150
100

5|]

Cost in 8

r:
0.95 —<—T—-—<';—~

0.93 .

r-cnrxmm—mh-mm
m «(scam arsr:~:5
QDODQQQODD

0.71

 

 

Mass Flow in Kg]:

 

Graph 3.3: Showing the cost vs mass flow for
geometry 'C’. At mats flow .84kg/a
the cost is $323.87.

 

 

 

Fig.3.3: Graph cost vs mass flow for the geometry ”C"

76

 

 

 

 

 

 

 

 

 

Grapli 5N E”.. "TU & COSt
1.4 v
1.2 J) M .r

g ,3

08 1 a-..» ~~~~~~~

I: ....~<'—’ .

5 08 ‘1.- ........ --*""’ Senesl
0.6 .. ..- . . ' SeriesZ
0.4 1

I” 02 o

0 'L-H-Q-O-t-H-O-t-H—O-O-H-O—t-O-O—O-O-O-fl—H
N m a) ("I N s- s- (O s- to '- fx (*3
u (D to m' v 11" «i N N .— .—‘ c m
N N N N N N f‘ m m N Ix n N
m m m ('1 m m m m m m ("I m '

Cosfin 8

 

 

 

Graph 3.3.1: Series 1 is showing the Ntu and Series 2
is showing effectiveness vs different costs.

 

 

Fig.3.3a: Graph showing tahe effectiveness & Ntu vs cost.

‘

3.1.4 Cost of Radiator using Geometry 'D' only:

Table 3.4 in the appendix C-4 shows the same four quantities as in
table 3.3 for geometry 'D'. Minimum cost at 0.86 kg/s is $442.38.
Graphs (3.4) and (3.4a) are also the graphical representation of

the data in table 3.4.

77

 

 

Cost Vs Mass Flow (D)
800
500
«400
.5 |-
8200'
100 "j
0|
s:;ssss;2:22:
aoocoooodoodo
Muss Flmv in Kgls

 

 

 

Graph-4: Showing Cost vs Mass Flow Rate for
Geometry "D". Minimum Cost
is $442.38 at 0.86 Kg/s

 

 

 

Fig.3.4: Graph cost vs mass flow for geometry "D"

 

 

 

 

 

 

 

 

Eff, NtuVs Cost
1.4
£12 .,»”
.5 1 _,-. J -1. 2'
£03 M_.,-_~«»~-"“‘”"" Seriss1
:
20-5 “SeriesZ
"B
a 0.4
a:
L” 0.2
vavrvrvvv . . .1rrvvvj
mvmwaenqwg—Qm
WanNNv—GG’WMNI—
33333333383333
Cosiin!

 

 

 

Graph 3.4.1: Series 1 is showing Ntu and Seriese 2
is showing Effectiveness for geometry 'D'

 

 

 

Fig.3.4a: Graph effectiveness vs Ntu for geometry ”D”

78
3.1.5 Cost of Radiator using Geometry 'E' only:
Table 3.5 in appendix C—5 shows the four quantities as in the other
tables like 3.3 and 3.4 for geometry 'E'. Minimum cost is $408.29
at mass flow of air 0.92 kg/s. Graphs (3.5) and (3.5a) also tells

the same story of table 3.5 in graphical form.

 

 

 

CostVs Mass Flow
son
500
-400
E
8200
mu
0 . .
mm—mnmm—mnmm—
mmmgmwmqfifififih
DDDDGDDODOGDQ
Mass Flow in Kgls

 

 

Graph 3.5: Showing cost vs mass flow. It is showing
that cost is minimum at 0.93kg/s which
is 409.30

 

 

 

Fig.3.5: Graph cost vs mass flow for the geometry "E"

79

 

 

Eff" Ntu Vs Cost

 

 

Ssriesl

 

SeriesZ

 

 

 

nnvnNfiNnmNoiNu?

mm‘e‘g‘g—ommmmmm

c3 mmmmoooo

v vvvvmmmm
Cosine.

 

 

 

Graph-5a: Showing HTU. Bffec. vs Cost for
Geometry ”E".

 

 

 

Fig.3.5a: Graph effectiveness & Ntu vs cost for geomerty “E"

‘
\

3.1.6 Cost of Radiator using Geometry 'F' only:

Table 3.6 in appendix (C-6) shows the mass flow rate of air,
cost, Ntu and effectiveness for the geometry 'F'. At the optimum
mass flow of 0.92 kg/s the minimum cost is $344.23. Graphs (3.6)

and (3.6a) on the next two pages also tells the same story.

 

80

 

 

 

CmflVstthw
450
400
350
a 300
.E 250 ‘
15D * .
100
50
0 .
3858338517232;
D D O D o C) D D D D D D 0
Mass Flow in Kg]:

 

Graph 3.6: Showing the cost vs mass flow for Geom. ’P'.
It shows at 0.92kg/s the min. cost is $344.23

 

 

 

Fig.3.6: Graph cost vs mass flow for the goemtry "E"

81

 

 

EfiNNhFWSOMflUW

-‘
Lb

 

 

 

 

 

 

 

1.2 J- r
2 w"
1 .. ,,...-"‘
d ’._J_._,-' '
o Fer—w" “mm .
g 0.8 «gym-r“ Senes‘l
6
g 0.6 .. SsrissZ
o 0.4 “
3
Lu 02 ..
0 d—H-i—o-++H—H—+—H—H—+-:-+—o—r-o—+-H-1
m ‘0 .—_ ‘0. ". “3. N "t N. (D. V. 0'! in.
3!, v m co (.0 in mi v N .— .— a: 00
v IN N N N h N r— v— v— V' V
0') m 0') m f") m m V V Q’ V V
Gnu S

 

Graph 3.6.1: Series 1 is showing the Ntu and Series 2 is
showing Effectiveness of Geometry ’l”.

 

 

Fig.3.6a: Graph effectivehess & Ntu vs cost for geometry ”F”

3.2 Comparison of cost for Different geometries:

When we compare the cost of all the six geometries for mass flow
between 0.95 to 0.70. This is better way to present the results.
Table 3.7 in the appendix C-7 shows the numeric comparison and

the graph (3.7) on the next page shows the story in graphical

form.

 

82

 

 

Cost Vs Geom.

450
400
350
300
250
200
150
100

50

Cast in t

 

A B C D E F
Geom- Nnms

 

 

 

Graph 3.10: Showing the comparision of minimum cost
for the six geometries.

 

 

 

Fig.3.7: Graph showing the comparision of minimum cost vs the
geometries

83

3.3 Comparison of minimum cost for different geometries:

 

 

 

 

 

 

 

Geom. Mass Flow Cost ($)
A 0.81 345.48
B 0.8 240.38
C 0.84 324.86
D 0.86 442.38
E 0.93 408.29
F 0.92 344.23

 

 

 

 

 

(Table 3.7 shows that the cost is minimum.in case of
geometry 'B')

Graph (3.8) shows the minimum cost Vs optimum mass flow of air

for the six geometries using individually.

 

 

Chatlflu4n1lflaa5ns FZHDUJ

0.94
0.92
0.9
0.33
0.05
0.84
0-82
0.8 .
0.70 .3.
r+++

0.7a .

s9++

0-74 : :23:
072 '

Kgm

§+é+

Mass Flow in

 

GD E3 0: [:1 E5 fig

Graph 3.7: Showing the comparision of minimum mass
flow for the given six geometries.

 

 

 

 

 

 

Fig.3.8: Graph showing the comparison of optimum.mass flow vs
the geometries.

84

3.4 Comparison of Ntu at minimum cost:

Similarly if we draw the graph for the Ntu of the different

geometries graph (3.9) obtained.

 

 

Opt- Ntu Vs Geom.

 

 

Nu:

 

 

 

 

 

 

 

 

(Seoaninnn:

 

 

 

Graph 3.9: Showing the comparision of Ntu for the six
geometries at the minimum cost.

 

 

 

Fig.3.9: Graph showing the comparison on Ntu vs geometries at the
their respctive minimum cost.

85
3.5 Conclusion for the one type of geometries used:

All the calculations shows that by using the geometry "B" with

air mass flow rate 0.80 kg/s gives the minimum cost of all the

six geometries used.

Hence the minimum cost of the radiator by using geometry 'B'
with mass flow 0.80 kg/s is $240.38. And this is the optimum cost

for a annular fin radiator core with one type of geometry.

3.5.1 Breakdown of minimum cost:

1- Cost of Tube Material (Cu) = $ 39.15
2— Cost of Fin Material (A1) = $ 35.78
3- Fabrication Cost : $149.87
4- Operating Cost = $ 15.58

Graph (3.10) represent the same story in pie graph on the

next page.

86

 

 

Distr. of min. Cost

6%

16%

 

I Tube.Cosi
E] Fm Cast
I Feb. Cost
El Opsr. Cost

 

 

 

 

 

 

 

Graph 3.11: Showing the distribution of cost of geometry
’ .which is the minimum cost of all the six.

 

 

 

Fig.3.10: Graph showing the distribution of cost of geometry "B”

3.6 Cost of Radiator by using combination of geometries:

To further minimize the cost we used the different
combinations of the given six geometries to further minimize the
cost. For this purpose we use 2nd spread sheet in Appendix B—2
for rating problems of radiator. Results for using a particular
geometry at a time shows that further cost can improve by

combining geometries A,B,C & F.

87

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Comb. Geom. Mass Flow Cost (S)
4B-2C 0.80 296.70
3B-3C 0.80 301.44
ZB-4C 0.84 307.73
lB-5C 0.84 312.37
4C-2B 0.80 307.64
3C-3B 0.80 302.96
2C-4B 0.80 296.70
1C-SB 0.80 291.95

Alt B-C 0.80 301.84
4B-2A 0.80 289.91
3B-3A 0.80 300.10
2B-4A 0.81 342.19
1B-5A 0.81 294.11
4B-2F 0.80 265.89
3B-4F 0.91 292.47
ZB-SF 0.91 285.73

1B-7F 0.91 312.31

 

 

 

 

 

(Table 3.8 Showing the cost of different combinations)

Graph (3.11) below also shows the same situation. Minimum cost is
for the case using 4B-2F with mass flow rate 0.80 kg/s is

$265.89.

 

88

 

 

 

 

 

Cost Vs Combination Diff- Geom.
350
300
2m]‘
0
.5 200 «-
8 150 -.
o
100-)
50 ..
U::::%;:;;fi‘::..:a
Combination of Geom.

 

 

Graph 3.12: Showing cost for the combination of different

geometries (A,B,C,D,E & I“)

 

Fig.3.11: Graph showing cost for the combinaion of different

geometries (A,B,C,D,E & F}

 

Chapter #4

Conclusions and Recommendations for future work

By using one type of annular fin geometry and iterating
with air mass flow rate the minimum cost is achieved by using
Geometry "B" configuration with an air massflow is 0.80 kg/s

This 0.80 kg/s mass flow is the optimum mass flow at which
the cost is minimum. This is done using sizing spread sheet in

Appendix—"B-l".

Using the combinations of different geometries A,B,C,D,E
& F utilizing the use rating spread sheet (see appendix "B-
2") there is no further improvement in the cost. When a
combination like 4B—2C used the mass optimum mass flow of B is
used and if 4C—ZB is used the optimum mass flow rate of C is

used and similarly for other combinations.

Analysis shows that for a uniform core, using the sizing
spread sheet the cost is minimum for geometry ”B" at the mass
flow of 0.8 kg/s of air ($240.38). The effeciveness, Ntu of
geometry ”B" are also more than all the other geometries used.
Variable geometry cores are not as cost effective as uniform

89

90

core of geometry ”B".

Cost can be reduced further with the combination of
different geometries by iterating through mass flow rate of
air. Hence we get the optimum mass flow where the cost is

minimum.

For the future work, we have both the spread sheets for
sizing (appendix A-1) and rating (appendix A-2) problems of
the heat exchanger. By using these spread sheets especially
rating spread sheet, taking the combinations of different
geometries and iterate with air mass flow to get the minimum
cost. This mass f10W' is the optimum. mass flow for a

particular combination. This is quiet long iterative process

which is not inclueded in this thesis.

However, it can be accomplised by using these spread
sheets given in appendix "B—l", and "B-2" then the cost of

radiator can be further reduced.

Appendix-A—l

Derivation of e—Ntu Relation

Cross flow arrangement and the idealized temperature conditions are pictured in the

following figure.

 

 

 

 

 

 

 

 

 

Catlin-stun
Z—r r <‘
: 2 I: go
._y. .1
"' (
.,~ _+ 2.,-
_.. UO
- E

 

 

 

 

 

 

 

COLD FLUID FLOW LENGTH

 

 

Fig.A—1.1: Showing the cross flow heat exchanger phenomenon

91

92

Uniform flow distribution for the cold stream will be specified.

So that with respect to the frontal area Af, .

 

 

= ............. (Al-1)

Cc = C“mm (mixed) and CH = Cmin (unmixed)

A uniform distribution to the heat transfer area A will be assumed so that:

dA

A
= _ .................. Al-2
dc; cc ( )

In both cased chc is the differential capacity rate of the cold fluid associated with the cold

fluid frontal area dA and the heat transfer surface dA.

Figure shows that for temperature conditions as a function of hot fluid flow length
suggested that a condenser type of effectiveness depression is applicable.

Now we define F

 

F = c .......... (Al-3)

C,in

93

For the parallel flow [4] the e-Ntu relation is

e = [1 - exp(-Ntu(1+ CR ))]

 

1 + CR ............ (A1 -4)
Hence here the condenser or evaporator condition is applicable
for
CR = O

F = 1- exp (-Ntu) ............... (Al-5)
From equation (Al-2)

F = 1- exp (-Ntu) = Constant ............... (Al-6)
An energy balance for the hot fluid flowing in the entire tubes

dq = 'CH (11“ ............... (A1‘7)

dq can be written as follow

dq = F (ti-i ' tC,in) dCc ............... (Al-8)

94

By the combination of equation (Al-7) and equation (a1-8) we get

 

 

 

 

 

 

dt dC c M
" =-F ‘-.-.-F—‘ ’ ................ (Al-9)
th — ’an C h Ch A I,
CC, CH, Af, are the total magnitudes and are not variables.
Integration yields
I out _tc in C
”—'—+ = exp(—F( C )) ............... (Al-10)
tHJn - C,in CH
For the case CR = Cmin, it follows from the equation of e
f . -t
e = M .................. (Al-11)
tHJn _ [C,in
So the equation (Al—10) become,
(”W 7 I” 7 "J” “Hi" = e'F‘m ................. (Al-12)
(”JR—[C,in
— 0”” 4”,...) + (“m -t..-,.) = e'HC") .............. (Al-13)
(tI-lJn-tcjn) (’HJn‘tcjn)

95

By putting the value of F

e = (CLXI — exp(—CR [1 — exp(Ntu)]))

R

For the case

Cc = Cmin (mixed)

CH = Cmam (unmixedO

‘

and taking the definition of effectiveness

_ CH(tH.in _tH.om)
Cc(tH.m _tC‘J‘n)

 

by using the similar approach we get

a = 1- exp {-CR"(1-exp[-CR(NtU)D)

............. (Al-14)

............ (Al-15)

.............. (Al-16)

........ (Al-17)

96

Appendix-A—Z

Derivation of M or Fin Efficiency

 

 

 

 

 

i / f
qcond I : qoond
at xx : : x+dx\
/ , I ,r
I |
l I
1. I |
x x+dx

 

 

 

FigB- 1: flea! flair in SIM/l saga/11611! 0'!

For “’M’ an eigen value from the fin equation

By using the energy balance

Ac[qcond.x+At— qcondx] = _ qwm Psurf ............. (AZ-1)

97

A And: AX . Pan-f
PM = Perimeter

Divide both sides with Ax.

Ac[qcond x+Ar _ qcond.x]

Ax _

 

_-qconv Psurf

Lim Ax 0

1

d .
Ac ;(qcond) : " qconvpsurf

Equation (AZ-4) is by fourier series.

According to Newton’s law of cooling transforms.

q... = h.(T— T...)

Put the values in equation (AZ-3)

d dT
A—-K— =—h T—T .P

........... (AZ-2)

.............. (AZ-3)

.................. (AZ-4)

................ (AZ-5)

98

 

dzT
- —hc(T- £0).me

-KAC er __

dzT
(#2

 

KAC -hc(T- £0).ij = o

 

(127' he?“
(1x,— KA " (T—T.)=o

C

dzT
6&2

 

—M2(T—7;,)=0 .............. (AZ-6)

M2 = thM

 

 

M: c “"f ....... (AZ-7)

Calculation of PC and AW, at some mid point:

Now we calculate the PM and Ac at some mid point of the annular fin.

 

 

, = _ = . .................. (AZ-8)

99

 

 

 

 

Fig.A—2.2: 520nm? M6 dot/ed circle to dame/6r flip.

‘

By putting the values of Ac and Psurf from equations (AZ-7) and (AZ-8) in equation (AZ-6)

we get

 

 

A. = ((21: 927:3 .............. (AZ-9)
2 2
A MD. _ a.)
P = ””1 _ 4 4
”'” L L
f f

T!
P = — ............. A2-1 0
surf 2 L] ( )

 

100

By putting the values of Ac and Psurf in equation (AZ-7) we get the following value of M.

 

 

h _ .
Mz‘] .(D. D') ............ (AZ-11)
’1 K1 L1

Appendix A—3:

Pressure Loss in the Heat Exchanger

 

 

 

 

 

 

.1 . b .2
——:——+ 1 '—:—9
_§_> /////////////////22 a

: W E
___';_;. . .__.:_7\.
m: a a! a

'2 a

 

 

Fig A3 — 1: flea! ere/Eager core model for pressure drop aria/1573
6' 13' based or] Me [ma/mm free flair area 121 Me eere.

By the definition of the enterance and exit loss coefficients Kc Ke . and by the integration

of the momentum equation (for the integration of momentum equation see [1]) through

101

102

exchanger cores is:

AP G2 v v. Av v.
= __0_ K +1-02 +2 _'__1+ -——-l— 1— 2—K —' ......... A .-l
P; ZECRKC ) (v )fAv ( 6 JV] (3)

 

0 C 0 0

However for flow normal to tube banks or through wire matrix surfaces, as might be
employed in periodic-flow-type exchanger, enterance and exit loss effects are accounted

for in the fiiction factor, and the equation becomes (with KC and kc = 0)

AP G2 v " v. Av
—=——°— 1+ 2 ——'—-l + ———"i ........... A3-2
R chRR 6 )(Vo )fAcvo] ( )
(1 + c 2 )(i - 1) Flow acceleration
v0
f A h Flow friction
AC v0

For multipass arrangements, losses in the return headers must be accounted for

separately, as must any losses in inlet and exit headers and associated ducting.

Showing the comaprison of the , mass flow, cost, Ntu,

Appendix B-l

effeciveness, and number of tubes for geometry ’A'.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1_====.
S.No. Mass Flow Cost Ntu Eff. # of Tubes
1 0.95 348.82 0.7449 0.4738 102
2 0.94 348.58 0.7579 0.4794 102
3 0.93 348.30 0.7688 0.4840 102
4 0.92 348.05 0.7815 0.4894 102
5 0.91 347.80 0.7943 0.4947 102
6 0.90 347.56 0.8089 0.5006 102
7 0.89 347.31 0.8220 0.5059 102
8 0.88 347.06 0.8364 0.5117 102
9 0.87 346.82 0.8510 0.5174 102
10 0.86 346.61 0.8675 0.5238 102
11 0.85 346.36 0.8825 0.5296 102
12 0.84 “346.15 0.9001 0.5362 102
13 0.83 345.92 0.9167 0.5423 102
14 0.82 345.70 0.9347 0.5489 102
15 0.81 345.49 0.9538 0.5557 102
16 0.80 394.65 0.9738 0.5627 117
17 0.79 394.46 0.9956 0.5702 117
18 0.78 394.25 1.0162 0.5772 117
19 0.77 394.05 1.0387 0.5847 117
20 0.76 393.87 1.0631 0.5926 117
21 0.75 393.67 1.0865 0.6000 117
22 0.74 393.50 1.1142 0.6085 117
23 0.73 393.32 1.1425 0.6170 117
24 0.72 393.15 1.1715 0.6255 117
25 0.71 392.97 1.2003 0.6338 117
26 0.70 438.91 1.2350 0.6434 131

 

 

 

 

 

 

 

103

Appendix B-2

Showing the comaprison of the , mass flow, cost, Ntu,

effeciveness, and number of tubes for geometry ’8'.
“a

 

  

 

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

S.No Mass Flow Cost Ntu Eff. # of Tubes
1 0.95 243.52 0.7449 0 4738 73
2 0.94 243.30 0.7574 0 4791 73
3 0.93 243.06 0.7688 0.4840 73
4 0.92 242.82 0.7821 0.4896 73
5 0.91 242.61 0.7943 0.4947 73
6 0.90 242.40 0.8084 0.5004 73
7 0.89 242.17 0.8215 0.5057 73
8 0.88 241.96 0.8364 0.5117 73
9 0.87 241.76 0.8520 0.5178 73 i
10 0.86 241.54 0.8663 0.5234 73 1|
11 0.85 241.34 0.8831 0.5298 73
ll 12 0.84 \24l.15 0.9001 0.5362 73
13 0.83 240.50 0.9173 0.5426 73 u
14 0.82 240.76 0.9354 0.5491 73
ll 15 0.81 240.57 0.9545 0.5559 73
I[ 16 0.80 240.38 0.9738 0.5628 73
'1 17 0.79 287.81 0.9956 0.5701 88 H
18 0.78 287.63 1.0162 0.5772 88
19 0.77 287.45 1.0387 0.5847 88
20 0.76 287.28 1.0624 0.5923 88 n
21 0.75 287.12 1.0887 0.6006 88 “
lrzz 0.74 286.96 1.1142 0.6085 88 H
II 23 0.73 286.80 1.1425 0.6170 88
|L24 0.72 286.65 1.1715 0.6255 88
25 0.71 286.49 1.2011 0.6340 88
26 0.70 286.35 1.2350 0.6434 88
27 0.69 286.20 1.2690 0.6256 88

 

 

 

 

 

 

 

104

Appendix B-3

Showing the comaprison of the , mass flow, cost, Ntu,
effeciveness, and number of tubes for geometry 'C'.

 

Cost Ntu

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

S.No. ‘Mass Flow Eff. # of Tubes
1 0.95 327.16 0.7449 0 4738 88
2 0.94 326.85 0.7574 0.4791 88
i 3 0.93 326.54 0.7699 0.4845 88 1|
I 4 0.92 326.20 0.7815 0.4894 88
S 0.91 325.89 0.7943 0.4947 88
6 0.90 325.60 0.8084 0.5004 88
7 0.89 325.28 0.8215 0.5057 88
8 0.88 325.00 0.8364 0.5117 88
9 0.87 324.70 0.8510 0.5174 88
10 0.86 324.41 0.8663 0.5234 88
11 0.85 324.14 0.8831 0.5298 88
12 0.84 ‘323.87 0.9001 0.5362 88
13 0.83 373.10 0.9173 0.5426 102
14 0.82 372.83 0.9347 0.5489 102
15 0.81 372.57 0.9538 0.5557 102
16 0.80 372.31 0.9738 0.5628 102
17 0.79 372.08 0.9955 0.5702 102
18 0.78 371.81 1.0156 0.5770 102
19 0.77 371.57 1.0380 0.5845 102
20 0.76 371.20 1.0624 0.5923 102
21 0.75 371.10 1.0864 0.6000 102
22 0.74 370.89 1.1142 0.6085 102
23 0.73 370.68 1.1424 0.6170 102
24 0.72 423.50 1.1715 0.6255 117
25 0.71 423.28 1.2012 0.6340 117
26 0.70 423.08 1.2342 0.6432

 

 

 

 

105

 

Appendix B-4

Showing the comaprison of the , mass flow, cost, Ntu,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

effeciveness, and number of tubes for geometry ’D'.
S.No. Mass Flow Cost Ntu Eff. # of Tubes
I 1 0.95 446.30 0.7449 0.4738 93
{2 0.94 445.84 0.7568 0.4789 93
3 0.93 445.38 0.7688 0.4840 93 n
[4 0.92 444.94 0.7815 0.4894 93
[5 0.91 444.40 0.7943 0.4947 93
6 0.90 444.06 0.8078 0.5002 93
[ 7 0.89 443.64 0.8220 0.5059 93
[8 0.88 443.23 0.8370 0.5119 93
9 0.87 442.80 0.8516 0.5176 93
10 0.86 442.38 0.8663 0.5234 93
11 0.85 492.07 0.8835 0.5296 104
|r12 0.84 '491.70 0.9001 0.5362 104 H
II 13 0.83 491.31 0 9173 0.5425 104 “
r714 0.82 490.92 0.9347 0.5489 104 H
15 0.81 490.56 0.9545 0.5559 104
16 0.80 490.19 0.9738 0.5628 104 N
17 0.79 489.55 0.9956 0.5702 104
18 0.78 489.49 1.0162 0.5772 104
19 0.77 489.13 1.0380 0.5845 104
20 0.76 488.82 1.0631 0.5926 104
21 0.75 543.11 1.0865 0.6000 116
22 0.74 542.81 1.1141 0.6085 116
23 0.73 542.51 1.1424 0.6170 116
24 0.72 542.20 1.1715 0.6255 116 J
| 25 0.71 541.88 1.2003 0.6338 116
u 26 0.70 541.62 1.2350 0.6434 116 u

106

Appendix 8-5

Showing the comaprison of the , mass flow, cost, Ntu,
effeciveness, and number of tubes for geometry ’E’.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

S.No. Mass F10W’ Cost Ntu Eff. # of Tubes
II 1 0.95 409.33 0.7449 0.4738 81
n 2 0.94 408.82 0.7574 0.4791 81
I 3 0.93 408.30 0.7699 0.4845 81
ll 4 0.92 464.50 0.7815 0.4894 93
r 5 0.91 463.98 0.7943 0.4947 93
6 0.90 463.50 0.8084 0.5004 93
7 0.89 462.97 0.8215 0.5057 93
8 0.88 462.50 0.8364 0.5117 93
9 0.87 462.01 0.8510 0.5174 93
10 0.86 461.56 0.8676 0.5238 93
11 0.85 461.08 0.8831 0.5298 93
12 0.84 K“460.63 0.9001 0.5362 93
13 0.83 460.18 0.9173 0.5426 93 u
14 0.82 459.72 0.9348 0.5489 93
15 0.81 459.32 0.9552 0.5562 93 “
16 0.80 458.87 0.9738 0.5628 93
17 0.79 458.48 0.9956 0.5702 93 1
18 0.78 510.08 1.0162 0.5772 104
19 0.77 509.67 1.0380 0.5845 104
20 0.76 509.28 1.0616 0.5921 104 7T
21 0.75 508.90 1.0864 0 6000 104
22 0.74 508.56 1.1142 0.6085 104
23 0.73 508.21 1.1425 0.6170 104
24 0.72 507.85 1.1715 0.6255 104 i
25 0.71 507.50 1.2012 0.6340 104 H
26 0.70 563.92 1.2342 0.6432 116 ‘“

 

107

Showing the comaprison of the , mass flow,

Appendix B-6

cost, Ntu,

effeciveness, and number of tubes for geometry 'F'.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

S.No. Mass Flow Cost Ntu Eff. # of Tubes
1 0.95 345.03 0.7449 0.4738 70 H
2 0.94 344.77 0.7574 0.4791 70 1
3 0.93 344.49 0.7688 0.4840 70 n
#4 0.92 344.23 0.7815 0.4894 70
1 5 0.91 377.10 0.7943 0 4947 77 i
If 6 0.90 376.86 0.8083 0.5004 77 n
F7 7 0.89 376.61 0 8220 0.5057 77 1|
8 0.88 376.36 0.8364 0.5117 77
9 0.87 376.11 0.8510 0.5174 77
10 0.86 375.87 0.8663 0.5234 77
11 0.85 375.65 0.8831 0.5298 77
I 12 0.84 “375.41 0.9001 0.5362 77
13 0.83 375.19 0 9173 0.5475 77 l
14 0.82 374.95 0.9348 0.5489 77
15 0.81 374.74 0.9545 0.5559 77
16 0.80 412.39 0.9738 0.5627 85
17 0.79 412.19 0.9956 0.5702 85
18 0.78 411.97 1.0162 0 5772 85
19 0.77 411.77 1.0387 0.5847 85 u
20 0.76 411.57 1.0616 0.5921 85 i
r—21 0.75 411.37 1.0865 0.6000 85
22 0.74 411.18 1.1142 0.6085 85
23 0.73 448.88 1.1424 0.6170 93
24 0.72 448.70 1.1715 0.6255 93
25 0.71 448.52 1.2012 0 6340 93
26 0.70 448.56 1.2350 0.6434 93 “

 

108

Appendix 01

For Geometry “A”

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

u u no 4 J12! .

Tube outside diameter = 0.381n = 7.62x10e-3m
Fin pitch = 343 per m

Flow passage hydraulic dia = 3.929x10e-3m
Fin thicknesslaverage] = 4.6x10e-4m

Free flow areaflrontal area = 0.524

Heat transfer areaflotal area =835 s.m!c.m
Fin arealtotal area=0.91

Ref: LondonlS]

109

Appendix 02

For geometry “B”

Rs ab"

 

Tuhe outside diameter I 0.30ln I 0.05 x toe-3m

Fin pltch = 7.34 per ln = 206 per m

Flow passage hydraulle dlameter. =- 0.0154 ft = 4.75 x 100-3 m
Fln thlclmess [average] t = 0.010 In. aluminum 8 0.46 x10e-3 In
Free flow areaflrontal area = 0.5300

Heat transfer arealtotal volume. =140 s.lvc.tt =450 s.mlc.m

Fln arealtatal area = 0.092

Note: Experimental uncertainty for heat transfer results possibly somewhat

greater than the nominal O!- 5,‘ quoted for the other surfaces because
of the necessity of estimatlng a contsnct reslstsnce In the hlmetal tubes.

110

Appendix 03

For geometry “C”

' ?

 

' ' Reno" hW'

D8

Tube outside diameter=0.42in=l.06'6x1 0e-2m
Fin pitch = 343 per rn

Flow passage hydraulic diameter = 4.425x10e-3m
Fin thicknesslaverage] t= 4.8x10e-4m

Free flow arealfrontal area = 0.494

Heat transfer araflotal volume 446 s.m!c.m

Fin areaftotal ara=0.8?6

lll

Appendix C-4

For geometry “D”

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

mum—M

Tube outside diameter = 0.845ln=16.30x10e-3m

Fin pftch=7.0per 1n=276 per m

Flow passage hydraulic diameter = 0.0219 ft = 8.60x10e-3 m
Fin thickness =0.010 in = 0.25x10e-3 m

Free-flow areaflrontal area =0.449

Heat transfer arealtatal volume = 269x10e-3 rn
Fin arealtotal area = 0.830

Note: Minimum free-flow sea is in spaces transverse to flow.

112

Appendix C-5

For geometries “E” & “F”

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 
 
   
  

 
 

 
 
   

 

 

Note: Minimum free-flow area is ln'spaces transverse to flow.

Ref: LondonlS]

113

cm .
0.000 “
ooao -
a 1
‘B' K ‘
om I mum mu
tlnml mu
case 1 111111 ii iii
‘ . ' Wu;
3 x‘ \ to sun ran V
m 1» c “be. '
. D d _ — .
secs 1 In - ~ “It“ I
—_°_J. 1 1 1 tut"
nabs * . he: no" 0 our
to ' so so as so so no
Tube outside dlameter=0.646 ln=16.30x10e-3 m
Fln pitch=6.7 per In =343 per m
Fin thickness =0.010 in =0.25 x10e-3 m
Fin arealtotal area=0.062
A 8
Flow passage hydraulic diameter 0.01797 0.0303
Free flow arealfrontal area 0.443 0.620
Heat transfer arealtotla volume 00.7 65.7 s.ftfc.ft

LIST OF REFERENCES

AJ. KUPRAYS, V.V.LAPPO, M.M.TAMONIS and O.L.TUTLYTE,
“Optimization of Heat Transfer in Radioactive Heat Exchangers”, Heat

Transfer Soviet Research, Vol.21, no.3, pp 299- 307, May-June 1989.

M.KOVARIK, “Optimal Heat Exchangers:, Journal of Heat Transfer
Transactions ASME, Vol 111, no.2, pp287-293, May 1989.
RB. EVANS, “Two principles of differential second law analysis for heat

exchanger for heat exchanger design”, Approaches to design and Optimization

of thermal systems, AES~Vol.7 pp 1- 12, Nov 27-Dec 2 1988.

Standard Handbook for Mechanical Engineers. M.37, ed-7.

W.M. Kays, A.L. London, “Mc.Graw-Hill Book Company”.

Frank P. IncrOpera, David P. Dewitt. Introduction to Heat Transfer, second ed,

Brown, Edith, ME490 Report, Summer 1994.

R.K. Shah, A.L. London, Advances in Heat Transfer, Laminar Flow forced

convection in ducts.

R.K. Shah, A.D. Kraus, Compact Heat Exchangers.

114

115

10 W.M. Kays , A.L.London, Compact Heat Exchangers, Basic Heat Transfer and

Flow Friction Design Data.

"‘0101111101“