16fin+and+tube+heatexchanger

Determination of heat transfer correlations for plate-fin-and-tube heat exchangers

Dawid Taler

Abstract This paper presents a numerical method for determining heat transfer coef?cients in cross-?ow heat exchangers with extended heat exchange surfaces.Coef?-cients in the correlations de?ning heat transfer on the li-quid-and air-side were determined using a non-linear regression method.Correlation coef?cients were deter-mined from the condition that the sum of squared?uid temperature differences at the heat exchanger outlet,ob-tained by measurements and those calculated,achieved minimum.Minimum of the sum of the squares was found using the Levenberg-Marquardt method.The outlet tem-perature of the?uid leaving the heat exchanger was cal-culated using the mathematical model describing the heat transfer in the heat exchanger.Since the conditions at the liquid-side and those at the air-side are identi?ed simul-taneously,the derived correlations are valid in a wide range of?ow rate changes of the air and liquid.This is especially important for partial loads of the exchanger, when the heat transfer rate is lower than the nominal load. The correlation for the average heat transfer coef?cient on the air-side based on the experimental data was compared with the correlation obtained from numerical simulation of3D?uid and heat?ow,performed by means of the commercially available CFD code.The numerical predic-tions are in good agreement with the experimental data. List of Symbols

A area,m2

A f?n surface area,m2

A in,A oval inside and outside cross section area of the

oval tube,m2

A mf tube outside surface between?ns,m2

A min minimum free?ow frontal area on the air

side,m2

A rs smooth tube outside surface area,m2

c speci?c heat,J/kg K

"c mean speci?c heat,J/kg K

C covariance matrix d h hydraulic diameter of air?ow passages,m

d z hydraulic diameter on th

e liquid side,4A in/

P in,m

D diagonal matrix with positive diagonal

elements

f c￠,f c￠￠measured temperature of coolin

g liquid at the

inlet and outlet of heat exchanger,°C

f am￠measured air inlet temperature,°C

h convection heat transfer coef?cient,W/m2K

h o weighted heat transfer coef?cient,W/m2K

I identity matrix

J Jacobian

k thermal conductivity,W/mK

(k)iteration number

L heat exchanger?ow length,m

L ch length of automotive radiator,m

m_mass?ow rate,kg/s

m_c mass?ow rate of cooling liquid?owing inside

the tubes,kg/s

m_a air mass?ow rate,kg/s

m_Ic mass?ow rate of cooling liquid?owing

through the?rst row of tubes at the?rst pass,

kg/s

N1,N2number of transfer units for the liquid and

gas side,respectively

Nu c liquid-side Nusselt number,h c d z/k c

Nu a air-side Nusselt number,h a d h/k a

n r total number of tubes in the?rst row of heat

exchanger,n r=n g+n d

n g,n d number of tubes in the?rst row in the?rst

(upper)and the second(lower)pass of heat

exchanger,respectively

p1pitch of tubes in plane perpendicular to?ow

(height of?n),m

p2pitch of tubes in direction of?ow(width of

?n),m

P con?dence interval of the estimated

parameters,%

P in,P o inside and outside perimeter of the oval tube,

respectively,m

Pr Prandtl number,l c p/k

chi

total heat transfer rate in exchanger between

hot liquid and cold air,W

r in inside radius of?n,m

r o outside radius of?n,m

r residual vector

Re c liquid-side Reynolds number,w c d z/m c

Re a air-side Reynolds number,w max d h/m a

s?n pitch(center-to-center spacing),m

Heat and Mass Transfer40(2004)809–822

DOI10.1007/s00231-003-0466-4

809

Received:18December2002

Published online:8November2003

óSpringer-Verlag2003

D.Taler

Department of Power Installations,

Academy of Mining and Metallurgy,

Al.Mickiewicza30,Paw.B-3,30-059Cracow,Poland E-mail:taler@https://www.360docs.net/doc/9312762270.html,.pl

Tel.:(0-48)126173078

Fax:(0-48)126173113

S sum of temperature difference squares,K2

T temperature,°C

T bf?n base temperature,°C

T c￠,T c￠￠inlet and outlet temperature of cooling liquid, respectively,°C

T cm outlet temperature of cooling liquid after the ?rst pass,°C

T d￠,T d￠￠, T d￠￠￠air temperature at inlet and after the?rst and second row of tubes at the second(lower) pass,respectively,°C

T g￠,T g￠￠, T g￠￠￠air temperature at inlet,after the?rst,and second row of tubes at the?rst(upper)pass, respectively,°C

T g,i￠,T g,i￠￠air temperature at inlet and outlet of the

control volume,°C

t a=2nàm the(1–a/2)th quantile of the Student?s t-dis-tribution with n–m degrees of freedom

T a air temperature,°C

T am￠,T am￠￠mean inlet and outlet temperature of air,°C

"T f mean temperature of the?n,°C

U A overall heat transfer coef?cient,W/m2K

_V c volume?ow rate,m3/s

w0average frontal?ow velocity,m/s

w max mean axial velocity in the minimum free?ow area,m/s

x vector of unknown parameters

x,y,z Cartesian coordinates

Greek symbols

d f?n thickness,m

d r tub

e wall thickness,m

g f?n ef?ciency

l?uid dynamic viscosity,Pa?s

m kinematic viscosity,m2/s

q?uid density,kg/m3

Introduction

Plate-?n exchangers are used in heating engineering, refrigeration,and automotive industry(engine radiators and air-conditioning systems).In order to enhance heat transfer on the air-side the plate?ns form a series of channels that create narrow passages for the external air ?ow and act as extended surfaces on the smooth tubes.

An essential,and,often most uncertain part of any two-?uid heat exchanger analysis is the determination of heat transfer coef?cients on both?uid sides.Heat transfer coef?cients for exchanger surfaces are usually obtained experimentally using the modi?ed Wilson plot technique [1–3]for the test data analysis.Fluid side heat transfer coef?cients are determined from an overall heat transfer coef?cient,which is known from the energy balance of a heat exchanger.The original and modi?ed Wilson plot techniques are based on the linear regression and their applications are limited to the power-law relationship between Nusselt(Nu),Reynolds(Re)and Prandtl(Pr) https://www.360docs.net/doc/9312762270.html,ing the Wilson method for determining the correlation for the heat transfer coef?cient would imply the necessity to carry out numerous time-consuming iterations,as the?n ef?ciency is a function of heat transfer coef?cient,that should be determined.

The correlations for the average air-side heat transfer coef?cient are often determined using the method described by Kays and London[4–6].They carried out experiments in an open-loop steam-to-air test apparatus. The tube-side heat transfer coef?cient was assumed to be very high and was implicitly included in the air-side heat transfer coef?cient.

Experimental data in the Kays and London method[4] are approximated by the following function:

j St Pr2=3?feReT;e1Twhere St=Nu/(Re Pr)is the Stanton number.

Substituting St in equation(1),yields the formula for the Nusselt number Nu

Nu?Re feReTPr1=3:e2TThe experimental results f(Re i)are approximated by the various functions.

The next cause of inadequate accuracy of the corre-lation formulas determining the heat transfer coef?cients on the air side,is assuming an in?nite heat transfer coef?cient on the?uid side in the tubes[4]or deter-mining that coef?cient from known correlations[1,5–8]. In the work[4],the heating medium in the tubes was saturated water vapor,which has a very high heat transfer coef?cient when condensing.The condensate, however,?owing on the internal surface of the pipe, reduces considerably the heat transfer coef?cient.The assumption of an in?nite heat transfer coef?cient inside tubes is therefore incorrect.

In many papers dealing with plate-?n and?nned heat exchangers the problem of accurate determining the heat transfer coef?cient inside tubes is ignored,assuming that it is very high,compared with the heat transfer coef?cient on the air side.However,the?ow velocities of liquid inside tubes are often small,and the water-side resistance can be noticeable.Therefore,neglecting the tube-side heat transfer coef?cient or its inaccurate calculation has a sig-ni?cant in?uence on the experimental determination of the air-side heat transfer coef?cient,when the?uid mass ?ow rate inside the tube is small.

In papers[3,5]the water-side heat transfer coef?cient h w was determined using the Dittus-Boelter formula,and in paper[2]by the Sider-Tate formula.The Dittus-Boelter formula in form

Nu?0:023Re0:8Pr n;e3Twith n=0.4for water heating and n=0.3for water cooling,is being recommended for Reynolds numbers Re>10,000.In the Reynolds number range of2,300

More accurate values of the heat transfer coef?cient can be found using the Gnielinski formula[9]or from the the Pietukhov formula[10].

810

It yields a higher accuracy,compared to the Dittus and Boelter equation(3).

Gnielinski?s correlation approximates90%of the pub-lished results with20%accuracy in the range of0.5

It follows from the above discussion that the heat transfer coef?cients both on the air and liquid side should be determined simultaneously.

Another source of errors in the published results on ?nned or plate-?n heat exchangers is calculating the ef?-ciency of hexagonal or rectangular?ns.Such form follows from geometric division of the plate?n.In order to determine the ef?ciency of such?ns the Schmidt formula [13]or the sector method[14]is being applied.

The Schmidt formula replaces a hexagonal or rectan-gular?n by a circular one of the same surface area,and then the effectiveness of that circular?n is being calcu-lated.

In the sector method,the hexagonal or rectangular?n is being divided into n circular sectors with an inner radius equal to the outer radius of the tube on which the ?n is set.The outer radius of the sector results from the shape of the analysed?n.The ef?ciency of a sector with an outer radius of r o and inner radius of r in is to be deter-mined in the same way as for a circular?n of the same ratio r o/r in.Then the weighted average of the ef?ciencies of all n sectors composing the?n is calculated.

This way of determining the ef?ciency of the?n is inaccurate,as it takes no account of the two-dimensional character of the temperature?eld in hexagonal or rect-angular?ns.

Although until now the accuracy of both the Schmidt and the sector method was not assessed,they are used in calculating the effectiveness of plate-?n heat exchangers, as there are no other suitable methods.The majority of cited papers[5,8]applies the Schmidt method.In the papers[1,7]the sector method is used.One should note that only circular tubes were analysed.

The heat transfer coef?cients on the air-side were also determined using the naphthalene sublimation method [15],based on the heat and mass transfer analogy.

Using naphthalene sublimation,heat transfer is mod-eled on a surface at constant temperature,whereas actually the temperature of plate-?ns and tubes differ from each other and are non-uniform.Furthermore,the Schmidt number of naphthalene is Sc=2.5,and the Prandtl number for air is Pr=0.71.Therefore,the mass transfer coef?cient is not numerically equal to the heat transfer coef?cient.The local heat transfer coef?cients are obtained by transforming the local mass transfer coef?cients,mea-sured by the naphthalene sublimation technique into local heat transfer coef?cients by employing the heat and mass analogy:Nu?ShePr=ScTn;e4T

where n can be selected as1/3or0.4.The symbol Sh in formula(4)denotes the Sherwood number,and Sc the Schmidt number.

In recent years,some researchers[16]attempt model-

ling heat transfer in plate-?n exchangers by solving the equations for continuity,momentum and energy on the air

side using numerical methods.The governing equations

were solved numerically assuming that the surface tem-perature of tubes and plate?ns is constant.Thus,heat conduction in the?n and tube wall was neglected.The

actual air-side heat transfer does not occur under constant

wall temperature and the usage of numerical modeling

results for the design of plate-?n exchangers is limited.

Notwithstanding the considerable interest on the side of industry and great efforts of researchers,heat exchangers

with extended surface are still insuf?ciently investigated

and the results of their study are questionable.Heat exchangers consisting of round tubes in staggered arrangements[16–17]were usually investigated.There are

no experimental data on plate-?n exchangers with tubes of

oval or elliptical cross-section.Such heat exchangers,

having an in-line pipe arrangement,are used when the

height of the exchanger and pressure drop on the air-side

must be small,as is the case in automotive radiators.

The objective of this paper is to present a numerical method for obtaining correlations for both?uid sides.

Primary measurements consist of the?ow rates of each

?uid stream,its inlet and outlet temperatures and pressure drops in both streams between the inlet and outlet of the

heat exchanger.The problem of determining non-dimen-sional average heat transfer coef?cients is formulated as a parameter estimation problem by selecting the functional

form for the Nusselt number Nu=f(Re,Pr).There are n parameters,x=(x1,…,x n)T,to be determined such that computed outlet tube side?uid temperatures T c,i￠￠agree in

the least-squares sense with the experimentally obtained temperatures f c,i￠￠.The non-linear,least-squares problem

will be solved using the Levenberg-Marquardt method [18].In order to calculate the outlet?uid temperature T c,i￠￠

as the function of searched parameters,a mathematical

model of the heat exchanger,will be developed.The?nite element method will be used to calculate the?n ef?ciency.

The results of the experimental analysis of the plate-?n-

tube automotive radiator are presented in this paper.The tested two-pass radiator consists of two inline rows of oval

tubes with smooth plate?ns.The air and coolant mass

?ows,the total pressure drops through the radiator and

inlet and outlet temperatures were measured in eighteen measurement points.The new correlations for the heat transfer coef?cients on the air and coolant sides will be developed using the presented technique.Uncertainties in

the estimated parameters are determined by calculating

the95%con?dence intervals for temperature measure-

ments with unknown standard deviations.The main advantage of the presented method is that it does not

require any knowledge of,or solution for,the complex

?uid?ow?eld.It can be used for determining heat

transfer characteristics in different types of heat exchangers.

811

The local and average heat transfer coef?cients will also be calculated numerically.The results of the computa-tional studies of3D laminar?ow and heat transfer in the tested automotive radiator generally con?rm well the experimental measurements.

Mathematical model of a heat exchanger

Cross-?ow heat exchangers have usually a complex con-struction,characterized by several tube rows in several passes.After each tube row,the air temperature distribu-tion is varying.This makes it necessary to determine the air temperature in that row separately.Designing a mathematical model of the heat exchanger calls for taking account of the way in which the individual tube rows are interconnected at the?uid entering side.

At?rst,a general procedure for building a mathemat-ical model describing a heat transfer in the cross-?ow exchangers will be presented.Subsequently,a model of the two-pass double row automotive radiator will be devel-oped.Figure1depicts a diagram of a single-pass cross-?ow heat exchanger with a single row of tubes.

The tubes are fed from one manifold,thus the tem-perature of the cooling liquid on the tube inlet is uniform and amounts to T1￠.The cooling?uid temperature is a function of the x coordinate only,as for any speci?c value of x,the temperature is equal in all tubes and does not change in the direction of the z axis.The temperature of the coolant at the outlet is equal in all tubes and is T1￠￠.The air(gas)stream,having a uniform temperature T2￠,?ows across the row of tubes.The air temperature T2is a function of the x and y coordinates,i.e.it changes on the width of the exchanger and along the?ow way in the y direction.Along the height of the exchanger,the air tem-perature is constant for a given value of the coordinates x and y.The mass?ow of the cooling?uid is m_1and of air m_2.Energy balance equations,which de?ne temperature changes of the?uid and gas,in most cases being the air, have the following form[19–20]:

dT1

?àN1T1extTàT m2extT

? e5T

@T2ext;ytT

@yt

?N2T1extTàT2ext;ytT

? e6T

where x,y denote Cartesian coordinates(Fig.1),x+=x/L x, y+=y/L y–dimensionless coordinates,L x and L y–heat exchanger dimension along the x and y axes,and T1and T2 denote temperature of the?uid inside the tubes and gas temperature.Numbers of transfer units N1and N2are de?ned by

N1?

U A A

m1c p1

;N2?

U A A

m2c p2

;e7T

where A is heat transfer area,usually a smooth tube outer surface area,U A–coef?cient of the heat transfer related to A area,m_1and m_2–mass?ow rates of the?uid and gas, and c p1and c p2–speci?c heat of the?uid and gas under constant pressure.

The quantity T m2(x+)is the mean temperature of air on the length L y de?ned as

T m2extT?

yt?0

T2ext;ytTdyt:e8T

Since the equations(5–6)are?rst order in both x and y, two initial conditions are needed for each of the coordi-nates.They are

T2j yt?0?T02;e9TT1j xt?0?T01:e10TOverall heat transfer coef?cient U A is calculated using the following formula:

U A

h c

d r

h o

e11T

where the weighted heat transfer coef?cient h o is de?ned by

h o?h a

A mf

A rs

A f

A rs

ág feh aT

;e12T

where h c is a heat transfer coef?cient at the inner surface, d r–a tube wall thickness,k–a thermal conductivity of the tube wall,h o–a weighted heat transfer coef?cient from the gas(air)side related to outer surface A rs of the smooth tube,h a–a heat transfer coef?cient on the air side,A mf–an outer surface of the smooth tube between?ns,A rs–an outer surface of the smooth tube,A f–a?n surface area,g f–an ef?ciency of the?n,de?ned as the ratio of the heat?ow transferred by the physical?n to the heat?ow,transferred by an isothermal?n of the base temperature T bf.

A smooth plate?n is divided into equivalent?ns,usu-ally rectangular in shape.Ef?ciency of the equivalent?ns was calculated by means of Finite Element Method

(FEM).

Fig.1.A single row cross-?ow heat exchanger

812

The solutions to equations (5–6)with initial condi-tions (9–10)are,respectively,

T 2ex t;y tT?T 1ex tTàT 1ex tTàT 02??

exp eàN 2y tT;e13T

T 1ex tT?T 02tT 01àT 02àáexp àN 1

N 2

?1àexp eàN 2T x t&'

e14T

Equation (13)and (14)allow determining the air and cooling liquid temperature along their ?ow way.Substi-tuting (13)into (8)yields the expression for the mean air temperature on a thickness L y of the row T m 2ex t

R 10

T 2ex t;y tTdy t

?T 1ex tTà12

?T 1ex tTàT 02 ?1àexp eàN 2T :e15T

The air outlet temperature in the ?rst tube row obtained

from equation (13)is given by

T 002ex tT?T 2ex t;y tTj y t?1

?T 02teT 01àT 02Texp àN 1

N 2?1àexp eàN 2T x t

??1àexp eàN 2T :

e16T

In order to evaluate the heat rate of the tube row,i.e.the heat rate ?owing from the hot liquid to the cooling air stream,it is necessary to ?nd the mean air temperature on the L x width of the exchanger according to T 002m ?Z 10T 002

ex tTdx t?Z 1

T 0

2teT 01àT 02Tè?exp àN 1

N 2e1àe àN 2Tx t !

?1àexp eàN 2T 'dx t;T 00

T 0

tN 2N 1

T 01àT 0

2àá1àexp eàB T? ;e17T

where

B ?N 1

N 2

1àexp eàN 2T? :

e18T

All the heat transfer equations presented so far apply to heat exchangers with one row of tubes.Then,a mathematical model of a heat exchanger with two tube rows will be developed.The analyzed automotive radiator is a two-pass heat exchanger with two tube rows (Fig.2a).We will ?rst

analyze a single pass exchanger consisting of two tube

rows (Fig.2b),which are fed in parallel from one mani-fold.The tube number and sizes in the ?rst and second

row are identical.The mass ?ows of the cooling liquid ?owing through the ?rst and second tube row are also

equal.The same air mass stream ?ows crosswise through the ?rst and second row.The system of ?rst-order

differential equations describing the heat exchange in the

second tube row (Fig.2b)is as follows

@T II

2@y t?N 2?T 3ex tTàT II 2ex t;y tT ;e19TdT 3dx

?àN 1?T 3ex tTàT II m 2ex t

T ;e20Twhere

T II m 2?

Z 1

T II 2ex t;y tTdy t:

e21T

The temperature T II 2x t;y t

eTis the air temperature in the second tube row.Since the air temperature changes

continuously,then the outlet temperature in the ?rst row is the inlet temperature in the second row of tubes.Thus,the initial condition for the air in the second tube row is T II 2y t?0 ?T 002ex t

e22Twhere T 2￠￠(x +)the air temperature distribution at the outlet of the ?rst tube row is de?ned by equation (16).The

?uid temperature in the second row is speci?ed at its inlet T 3ex tTx t?0j ?T 03;e23Twhere T 3￠=T 1￠.Equation (19)is integrated ?rst using the

method of variable separation.Equation (20)is a

linear Fig.2.Automotive radiator with two tube rows and two passes.

(a)Automotive radiator,(b)Single-pass,cross-?ow heat exchanger with two tube rows 813

non-homogenous equation of the ?rst order,which will be solved using the variation of parameters method.Solution

of the equation system (19–20)with the initial conditions takes the following forms:–air T II 2ex t;y tT?T 3ex tTà?T 3ex tTàT 002ex tT exp eàN 2y t

T:e24TThe mean air temperature in the second tube row T m 2II

is determined from

T II

m 2?R 10

T II 2ex t;y tTdy t?T 3ex tTy tt1N 2exp eàN 2y tTh i 1

0àT 002ex tT?1N 2exp eàN 2y tT 1

0:Evaluating the integral it follows that

T II m 2?T 3ex tT1àB N 1

tT 002ex tTB N 1

:e25T–cooling liquid T 3ex tT?T 02t?Cx tteT 03àT 02T exp eàBx tT;e26Twhere C is given by

C ?B T 01àT 0

2àá1àexp eàN 2T? :e27TThe air temperature at the outlet of the second tube row is evaluated from equation (24)T 0002?T II 2ex t;y tTj y t?1

?T 3ex tT?1àexp eàN 2T tT 002ex t

Texp eàN 2T:e28TSubstituting for T 3(x +)from equation (26)and for T 2￠￠(x +)from equation (16)yields T 0002ex tT?T 02tCx ttT 03àT 02

àá??exp eàBx tTèé?1àexp eàN 2T? t?T 02tT 01àT 02àá?1àexp eàN 2T? exp eàBx tT exp eàN 2T:e29T

To determine the heat transfer rate from the tubes to the air,the mean air temperature at the outlet of the sec-ond tube row is needed.Integrating T 2￠￠￠(x +)with respect to x +yields

T 0002m ?Z 10

T 0002ex tTdx t?1àexp eàN 2T? ?Z 1

T 02tCx texp eàBx tTtT 03àT 0

2àáexp eàBx tT??dx tte àN 2

Z 1

T 02tT 01àT 0

2àáexp eàBx tT1àexp eàN 2T? è

édx t

e30Tfrom which it follows that

T 002m ?T 02teT 01àT 02

T&?1àexp eàN 2T 2

1àexp eàB T

B àexp eàB T !t?1àexp eà2N 2T

1àexp eàB T

:e31TA mathematical model of a two-pass internal-combustion motor radiator will be now developed,applying formulas

for the ?uid temperature distribution in the heat ex-changer consisting of two tube rows.

The radiator is a double-pass heat exchanger (Fig.2a),consisting of two tube rows.The layout of the radiator is shown in Fig.3.The cooling liquid ?ows in parallel through both tube rows.The tube outlets in the upper pass converge into one manifold where the coolant from the

?rst and second tube row are mixed.Upon mixing the coolant with the temperature T 1￠￠from the ?rst tube row and the coolant with the temperature T 2￠￠,the feeding water temperature of the second lower pass is T cm .In second lower pass the total mass ?ow rate splits into two

?ow rates m _c .At the outlet from the ?rst tube row in the bottom pass the coolant temperature is T 3￠￠,and from the

second row is T 4￠￠.Upon mixing the cooling liquids from the ?rst and second row,the ?nal coolant temperature is T c ￠￠.This T c ￠￠temperature is thus the radiator exit tem-perature.The air stream m _a ?ows crosswise through both

tube rows.Assuming that the air inlet velocity w 0is the

same in the upper and lower pass,the mass ?ow rate of air through the upper pass is m _g =m _a n g /n r ,where n g is the

number of tubes in the ?rst row of the upper pass and n r is the total number of tubes in the ?rst row of the upper and

lower pass.The mass ?ow rate across the tubes in the

lower pass is m _d =m _a n d /n r ,where n d is the number of

tubes in the ?rst row of the bottom pass.The coolant temperature at the exit of the ?rst pass results from the

energy balance equation _m c "c c eT cm TT cm ?_m c 2"c c T 001àáT 001t_m c 2

"c c T 002àáT 002;

e32Twhere "c c eT Tis the mean speci?c heat over the temperature

range from the reference temperature T ref =0°C to T (°C).The speci?c heat of the coolant was assumed to be a linear function of the temperature:c c =a +bT ,where a and b are

constants.Thus,the temperature T cm determined from equation (32)is

T cm ?

àa t?????????????????????????????????????a 2t2b "c c eT cm TT cm p :e33TThe temperature T c ￠￠(Fig.3)of the coolant leaving the

radiator may be calculated in the similar fashion

T 00c ?àa t??????????????????????????a 2t2b "c c eT 00c p Tb

:e34TEquations (33)and (34)allow to determine the tem-peratures T cm and T c ￠￠after a few iterations.The air

814

temperature after the radiator is calculated according to energy balance _m g i 000gm t_m d i 000dm ?_m a i 00am :

e35T

Assuming that speci?c heat of air c pm is constant,yields

n g n r _m a c pm T 000gm tn d n r _m a c pm T 000dm

?_m a c pm T 00am ;

T 00am ?n g n r T 000gm tn d n r T 000

dm :

e36T

The mean air temperature T am ￠￠given by equation (36)is necessary to calculate the heat transfer rate in the radiator.The rate of heat transferred in the automotive radiator from the hot coolant to the cold air may be calculated as follows

_Q ?_m c "c c j T 0c T 00c T 0c

àT 00c àá?_m a c pm T 00am àT 0am àá:e37Twhere "c c j T

0c T 00c is the mean speci?c heat of the coolant over the temperature interval from T c ￠￠to T c ￠.Until now,the direct problem has been considered in that both heat transfer

coef?cients on the coolant and air-side are known.Then,the inverse problem will be solved.3

Identification of heat transfer coefficients on the coolant and air sides

The issue of determining heat transfer coef?cients of the air-and coolant-sides is the inverse heat transfer problem.The following values are known from the measure-ments (Fig.3):–coolant mass ?ow rate m _c ,–air ?ow rate m _a ,

–inlet ?uid temperature T c ￠,–inlet air temperature T am ￠,–

outlet ?uid temperature T c ￠￠.

Furthermore,construction of the heat exchanger is

known,as well as the materials it has been made of.

Correlations for the heat transfer coef?cients on the side of the coolant,h c ,as well as the air,h a ,are determined

simultaneously.

Initially,a speci?c form of correlation equations is as-sumed for non-dimensional heat transfer coef?cients at the side of the air

Nu a h a d h =k a ?Nu a eRe a ;Pr a ;x 1;...;x na Te38T

and at the side of the coolant

Nu c h c d z =k c ?Nu c eRe c ;Pr c ;x na t1;...;x m T;where na \m :

e39TSymbol m stands for the total number of coef?cients

present in the correlation formulas,whereas na denotes the number of unknown coef?cients in the expression that de?nes the Nusselt number on the side of the air.The Reynolds and Nusselt numbers were determined based on

the hydraulic diameters.Equivalent hydraulic diameters on the side of the air,d h ,and the ?uid,d z ,are speci?ed by

the following expressions:d h ?4A min L

A 0f tA 0mf

;e40Td z ?

4A in

P in

;e41T

where the ?n surface of a single passage A f ￠and the tube outside surface between two ?ns A mf ￠￠are given by A 0f ?2á2ep 1p 2àA oval T?4ep 1p 2àA oval T;A 0mf ?2A mf ?2P o es àd f T:

The minimum cross-section area for transversal air ?ow through the tube array,related to one tube pitch p 1,is

A min ?es àd f Tep 1àd min T:

The air-side Reynolds number Re a =w max d h /m a for the foregoing correlation (38)is based on the maximum

?uid

https://www.360docs.net/doc/9312762270.html,yout of a two-pass plate-?n and

tube heat exchanger with two inline

tube rows,1–?rst tube row in upper pass,2–second tube row in upper pass,3–?rst tube row in lower pass,4–second tube row in lower pass

815

velocity w max occurring within the tube row,and is de?ned by w max ?s áp 1es àd f Tep 1àd min T"T

am t273T 0

am t273w 0;e42Twhere w 0is the air velocity before the radiator.As the tubes in the radiator are set in line,w max is the air velocity in the passage between two tubes.The thermo-physical properties of the coolant were determined at the mean

temperature "T c ?eT 0c tT 00c T=2,where T c ￠and T c ￠￠denote the inlet and outlet temperatures.All properties appearing in the equation (39)for the air are also evaluated at the mean temperature "T am ?T 0am tT 00am àá=2.Outlet temperature of the liquid T c ￠￠is a function of all the coef?cients:x 1,x 2,…,x m .We must adjust coef?cients x 1,x 2,…,x m to minimise the sum of squares of the tem-perature differences S ?X n i ?1r 2i ex 1;...;x m T?X

n i ?1f 00c ;i àT 00

c ;i ex 1;...;x m Th i 2?min :e43THence,we see that the solution for the non-linear least

squares problem needs to be found.

In order to solve the least squares problem,a Leven-berg-Marquardt method [18]will be applied.In the

Levenberg-Marquardt method,it is necessary to determine vector x ?,for which the sum

S ex T?

X n i ?1

?r i ex T 2e44Tattains the minimum,where x ?ex 1;x 2;...;x m TT

?x 1x 2...x m

8>>><>>>:9>>>=

>>>;;where m 6n :e45TThe symbol n denotes the number of measurement points,whereas m denotes the number of searched coef?cients.According to the Levenberg-Marquardt method,coef?-cients x k +1are calculated iterative,according to the for-mula

x ek t1T?x ek Ttd ek T;

e46T

where the direction of d ek Tand Jacobian J are speci?ed

through the following formulas

d ek T?àJ ek Th i T J ek Ttl ek TD ek T&'à1J ek Th i

r ek T;e47TJ ?

@r i

@x j

;i ?1;...;n ;j ?1;...;m :

Components of the vector r =(r 1,…,r n )T

are the values

of measured and calculated temperature differences.It is usually assumed that the matrix D (k )=I m ,i.e.matrix D (k )is a unit matrix of the dimensions m ·m .If l (k )

approaches zero,then the Levenberg-Marquardt method is transformed into the Gauss-Newton method.When l (k )

becomes very large,then the Levenberg-Marquardt method becomes a steepest-descent method.As it is seen,the Levenberg-Marquardt method is made up of a com-bination of the Gauss-Newton method and the steepest-descent method.

Initially,the value of the coef?cient l (k )was chosen so that the sum of squares S (x (k +1))was minimum.It re-quired,however,the search of the minimum of the func-tion in each step,which has become highly cost-ineffective.In 1963,Marquardt published a modi?ed Levenberg algorithm,for which the initial value of the l (1)parameter a small positive value was assumed,such as l (1)

=0.01.If,at the k -th iteration,the step d ek Tof (46)reduces S (x )(44),he sets x ek t1T?x ek Ttd ek Tand divides l by a constant factor,e.g.l (k +1)=l (k )/10,to push the algorithm closer to Gauss-Newton method,which near the minimum

is quickly convergent and precise.If after substituting Equation (46)to (44)the value of the sum S (x )does not

decrease,then l (k )is raised several times,e.g.setting l (k )=

10l (k )

and each time Equation (44)is recalculated.The

process is repeated as long as the value S (x )is decreased.In newer algorithms value l (k )is determined by searching for the minimum of the function S (x (k +1))=S (l (k ))using simple line search algorithms.The Levenberg-Marquardt method provides conver-gence of the iteration process even if the initial

approximation x (1)is far from x ?,in which the function S (x ?)reaches minimum,i.e.remains feature speci?c for the steepest-descent method.If x (k +1)

is close to the x ?

slow convergence can occur when the steepest-descent method is used.Therefore,in this case,due to the

presented selection of l (k ),the Levenberg-Marquardt

method is not quite different from the Gauss-Newton method,providing fast convergence and high precision of search for x ?,if x (k +1)is close to x ?.The Levenberg-Marquardt method has been proved to be a good gen-eral-purpose algorithm for least-squares problems.Gen-erally it is robust and works well.Simultaneously,coef?cients and exponents in the formulas,which de?ne heat transfer coef?cients on the side of air and cooling liquid,are determined.

The equation de?ning heat transfer coef?cient on the side of the air was taken in the form

Nu a ?x 1Re x 2

a Pr 1=3a ;

e48T

where Re a =w max d h /m a .

The Nusselt number Nu c for coolant is approximated using the expression

Nu c ?n

8eRe c àx 3TPr c

1tx 4n àá1=2Pr 2=3

c à1

àá1td z l 2=3"#;e49Twhere Re c =w c d z /m c ,d z –equivalent tube diameter,l –tube length.

If x 3=1000and x 4=12.7are assumed then the expression (49)becomes the correlation given by Gni-elinski [9].The thermo-physical properties of the ?uid

needed for the calculation of numbers Nusselt Nu c ,816

Reynolds Re c and Prandtl Pr c are calculated at the mean temperature,de?ned as "T c ?0:5T 0c tT 00c àá.4Air-and coolant-side heat transfer correlations for the automotive radiator The correlations for the heat transfer coef?cients in an automotive radiator were determined according to the method described in previous section.The tested auto-motive radiator is used for cooling the spark ignition en-gine of a cubic capacity of 1580cm 3.The cooling ?uid for the engine is a 35%water solution of mono-ethylene gly-col.The cooling liquid,warmed up by the engine is sub-sequently cooled down by air in the radiator.The radiator

consists of 38tubes of the oval cross-section,out of which 20are located in the upper pass with 10tubes per each row (Fig.2and 4).In the lower pass there are 18tubes with 9tubes per each row.The radiator is 520mm wide,359mm high and 34mm thick.The outer diameters of the oval tube are:d min =6.35mm,d max =11.82mm.The thickness of the tube wall is d r =0.4mm.The number of plate ?ns equals 520.The dimensions of the single plate tube are as follows:length –359mm,height –34mm and thickness –d f =0.08mm.The plate ?ns and the tubes are made of aluminum,and the radiator cross-section in horizontal plane is presented in Figure 4.The path of the coolant ?ow is U-shaped.The cross-section of the inlet chamber and the reverse chamber is depicted in Fig.4.The chambers are made of plastic.The pitches of the tube arrangement are as follows:perpendicular to the gas ?ow direction p 1=18.5mm and longitudinal p 2=17mm.A diagram of the equivalent ?n,into which the plate ?ns were partitioned,is presented in the Figure 5.

The temperature distribution in the ?n was determined for different heat transfer coef?cients h f by FEM using ANSYS [21].The ?nite element mesh and an example of the calculated temperature distribution are shown in Fig.6and Fig.7,respectively.Subsequently,the ef?ciency of the ?n was determined for various heat transfer coef?cients

using the formula g f ?"T

f àT a àá=T bf àT a àá,where T bf –?n base temperature,T a –air temperature,"T

f –mean temperature of the ?n surface.Then,the obtained values of the ?n ef?ciency were approximated usin

g the least-squares method [22]by the function

g f ?a tch f 1tbh f

;

e50T

where a =0.99999962,b =0.0024757218,c =

0.00036516734.Changes of the ?n ef?ciency g f with

respect to the heat transfer coef?cient h f are shown in Fig.8.

In order to determine heat transfer correlations the

measured data obtained by a manufacturer from the standard tests for the automotive radiators were used.The measured data are shown in Table 1.Next,by minimising the sum of squares (43),the parameter values x 1,x 2,x 3and x 4were obtained using the Levenberg-Marquardt method.In order to calculate the outlet temperature of the

cooling ?uid T c ￠￠,a mathematical model of the radiator,described in section 3,is used as well as the experimental data presented in Table 1.Parameter values x 1…,x 4,minimising the sum (44)are:x 1?0:069633x 2?0:60372

x 3?427:5321

x 4

?22:21919:

e51

TFig.4.Horizontal cross-section of the ?rst pass in the tested

automotive radiator;1–?rst row of tubes,2–second row of tubes,3–inlet header of the radiator,4–outlet header of the ?rst

pass

Fig.5.Diagram of the rectangular equivalent ?n on the oval tube;

d f =0.08mm,A f =513.64?10–6m

Fig.6.Dividing of the plate ?n into ?nite elements 817

The expression (48)for the Nusselt number on the side of the cooling ?uid,as well as the expression (49)de?ning the Nusselt number on the side of air,have the form

Nu a ?0:06963Re 0:6037a

Pr 1=3

a ;2006Re a 61500;e52T

Nu c ?n

8Re c à427:53eTPr c

1t22:22

??n 8

q Pr 2=3

à1àá1td z l 2=3"#;30006Re c 621000:

e53T

The exponent x 2=0.60372present in the Reynolds num-ber of the formula (52)is close to the exponent 0.625in the correlation for in-line,high-?n tube arrays derived by Schmidt [23].Obtained values of the Nusselt Nu c number for selected values of the Reynolds number,Re c ,are compared in Table 2.

Inspection of the results shown in Table 2implies that the values of the Nusselt number based on the correlation (53)are smaller than those derived from the Gnielinski equation.The differences are noticeable for larger Rey-nolds numbers.A reason for these differences may be the oval shape of the cross-section of the tubes utilized in the automotive radiator and the application of ethylene glycol

as the coolant.

A very good ‘‘?t’’of the calculated outlet temperatures T c ,i ￠￠to measured temperatures f c ,i ￠￠,i =1,…,18is achieved (Table 3).The value of the minimum sum of squares,S ,

de?ned by the formula (43)for the determined value of the

parameters,is extremely small and equals S =0.324K 2

.This proves that the accuracy of the mathematical model

of the radiator,described in section 3,is very high.Additionally,it con?rms the appropriateness of the form of equations (48)and (49)which approximate Nusselt numbers for both the coolant-and air-sides as well as the high accuracy of experimental data.5

Confidence intervals of the determined parameters in the correlations for the heat transfer coefficients at the sides of the air and the coolant

The real values of the determined parameters ~x 1;...;~x m are found with the probability of P =(1–a )?100[%]in the following intervals [18]

x i àt a =2n àm s t ?????c ii p 6~x i 6x i tt a =2

n àm

s t ?????c ii p e54

TFig.8.Ef?ciency g f of the equivalent ?n versus the heat transfer coef?cient h f

Table 1.Thermal measurement results for the automotive radiator No.w 0[m/s]V

_c ?103[m 3/s]f am ￠[°C]f c ￠[°C]f c ￠￠[°C]1. 1.51 1.390019.594.792.22. 2.50 1.391419.394.791.03. 4.00 1.389419.594.789.8

4. 5.50 1.388919.794.788.8

5.7.00 1.390020.494.788.2

6.8.51 1.391120.494.78

7.5

1.510.833320.194.590.58.

2.510.835019.594.688.79. 4.010.835619.494.686.810. 5.500.832219.394.785.711.7.000.834720.394.784.812.8.510.831120.194.78

3.813. 1.500.280319.893.983.51

4. 2.510.277820.294.080.11

5. 4.000.276720.494.177.71

6. 5.500.275319.794.075.11

7.7.010.278920.494.174.41

8.52

0.2789

20.3

94.0

72.9

Table https://www.360docs.net/doc/9312762270.html,parison of the Nusselt Nu c number at the inner

surface of the automotive radiator tube calculated according to Gnielinski correlation and equation (53)No.

Re c

Gnielinski formula [9]

Equation (53)1.300021.1618.322.600045.5034.573.900067.0649.344.1200087.1863.295.15000106.3776.696.18000124.8989.697.21000142.87

102.36

Fig.7.Temperature distribution T f (°C)in the ?n determined by

FEM;h f =50W/m 2K,T bf =100°C,T a =0°C

818

where the x i –parameter value determined using the least squares method,t n –m a /2–quantile of the t-Student distri-bution for the con?dence level 1–a and n –m degrees of freedom.

Value s t 2is an estimate of the variance calculated according to

s 2t ?P h i ?1

T 00

c ;i ex 1;...;x m Tàf 00c ;i

h i 2n àm ?S ex 1;...;x m Tn àm :e55Twhere n –denotes the number of measurement points,and

m –denotes the number of searched parameters.

Values c ii are the diagonal elements c ii of the covariance matrix de?ned by the following formula

C es Tx ?eJ es TTT J

es Th i à1

,where (s )denotes the number of the

last iteration while J is the Jacobian.

In this case,the following values are under consider-ation:n =18(Table 1),and m =4.Quantile t n –m a /2for 95%con?dence level (a =0.05)is:t 140.025=2.1448.Having solved the non-linear least squares problem,the temper-ature differences of the calculated and measured outlet temperatures were obtained.Sum of the squared temper-ature differences is S =0.32367[K 2].Limits of the 95%con?dence interval,calculated according to (54)are:0:0666056~x 160:072661;0:5967736~x 260:610667;427:18906~x 36427:8752;21:858876~x 4622:57951:

e56T

The con?dence intervals of the coef?cients x 1,…,x 4are small,which results from good accuracy of the developed mathematical model of the radiator and small measure-ment errors.

Numerical simulation

In order to determine local and average heat transfer coef?cients on the ?ns and tubes the ?uid ?ow and heat transfer was studied numerically.Pioneering experiments on the air-side transfer coef?cients were carried out by Saboya and Sparrow [15].The heat-mass transfer analogy,in conjunction with the naphtalene sublimation technique,was used to investigate local and average heat transfer coef?cients in two-row plate ?n and tube heat exchanger with circular tubes [15].The local mass transfer coef?-cients on this geometry have also been measured by

Kru

¨ckels and Kottke [24].They used a chemical method based on absorption,chemical reaction and coupled col-our reaction.The solid surface is coated with a wet ?lter paper and ammonia to be transferred is added as a short gas pulse.The locally transferred mass is visible as colour density distribution and the colour intensity corresponds to the local mass (heat)?ow.The measurements revealed high values of the heat transfer coef?cient on the leading edge of the ?n due to the presence of developing boundary layers.

In this paper,the air and heat ?ow in the tested two-row automotive radiator was simulated numerically by using the CFD program FLUENT [25].The three-dimen-sional (3D)?ow is treated as laminar since the air-side Reynolds Re a number is less than 1550.Owing to the complicated construction of the radiator,the numerical study of the whole radiator is very dif?cult.Due to the symmetry,the 3D ?ow through the single narrow passage between ?ns was simulated.The temperature distribution in the adjacent plate ?ns and tube walls was also calcu-lated.In this way,the effect of non-uniform heat transfer coef?cient on the tube and ?n surfaces,as well as,the effect of the tube-to-tube heat conduction through the ?ns on the heat transferred from the cooling liquid to the air is also accounted for.The liquid ?ow and heat transfer inside the tubes was not modeled.A schematic diagram of the physical problem under study is presented in Fig.9.The ?ns are made of aluminum with thermal conductivity of k =207W/mK.The computations were conducted for the data set No.15from the Table 1.The uniform frontal velocity of w 0=4m/s and uniform temperature of T am ￠=20.4°C before the radiator were assumed.At the symmetry planes normal gradients are set to zero.The boundary condition of the third kind (convection surface condition)is speci?ed at the inside surface S of the oval tubes

àk @T r @n

S ?h c T r j S

àT c àá:e57TThe tube-side heat transfer coef?cient h c =1286W/m 2K was obtained from Equation (53).The liquid temperatures T 1￠￠=84.3°C and T 2￠￠=85.8°C at the outlet of the ?rst pass (Fig.3)were calculated using the developed analytical model of the radiator.The air-side Reynolds number based on the maximum mean axial velocity in the mini-mum free ?ow area was Re a =601.Figure 10shows the grid used for the 3D computations in the air and solid (tube and ?ns)regions of the automotive radiator.The ?ow passage between the ?ns is divided into ?ve layers of control volumes,while on the half of the ?n thickness is only one layer (Fig.10).The distribution of the heat transfer coef?cients on the ?n surface obtained from the numerical simulation is shown in Figure 11.Examination of the Figure 11shows that the heat transfer coef?cient is noticeably higher on the forward part of the ?n due to the developing boundary layers.These boundary layers grow in the ?ow direction and cause a corresponding reduction in the heat transfer coef?cients.Stagnation ?ow on the front of the tube in the ?rst row produces high heat

transfer near the base of the ?n and at the forward part of the tube circumference.Behind the tubes,the low-velocity

Table 3.Differences between the calculated and measured

temperatures at the radiator outlet No.(T c,i ￠￠–f c,i ￠￠)[K]No.(T c,i ￠￠–f c,i ￠￠)[K]1.–0.1620510.0.000502.–0.0142111.–0.020423.–0.0188412.0.013464.–0.0424813.–0.114945.–0.0730114.0.286566.–0.0396615.–0.155317.–0.02221216.0.172028.0.0514417.–0.220099.

0.18526

18.

0.06129

819

wake regions exist.Then,in the regions downstream of the tubes very low heat transfer coef?cients can be observed.Relatively low heat transfer coef?cients are encountered on the portions of the ?n that lie downstream of the minimum ?ow cross sections.The heat transfer rates are especially low in the dead spaces in the regions behind the tubes.It was found that the second-row ?n and tube are less effective as heat transfer surfaces than those in the ?rst

row.Thus,the average heat transfer coef?cient "h

over the surface may be calculated as

"h ?R

h dA R dA ;e58T

where dA is in?nitesimal area of the air-side surface.The average heat transfer coef?cients are:on the plate

?n surface,"h

f ?59:6W/m 2K,on the tube surface in the ?rst row,"h

t 1?59:3W/m 2K,and on the tube surface in the second row,"h

t 2?18:4W/m 2K.The average value for the whole passage,calculated for the ?n surface and

for the tube surfaces between the ?ns is "h

a ?58:6W/m 2K.The average heat transfer coef?cient over the second row tube surface is especially low due to the presence of the wakes in front of the tube and in the region behind the tube.

The key to heat transfer enhancement on the ?n asso-ciated with the ?rst tube row is a major boundary layer contribution,while the portion of the ?n adjacent to the second tube row has no boundary layer contribution and only a small vortex contribution.The regions behind the tubes contribute very little to the performance of the heat exchanger.The numerical simulations using Fluent were conducted for all the data points shown in Table 1.Next,

the average Nusselt numbers Nu a ?"h

a d h =k a and the Colburn factors j ?Nu a =eRe a Pr 1=3Twere calculated.The least squares method was used to correlate the

data

Fig.9.Schematic of two-row

plate ?n and tube heat

exchanger con?guration;the dimensions are given in

millimeters

Fig.10.Mesh of ?nite volumes for the automotive radiator

820

j n ;i ?j n eRe a ;i T,i =1,…,18.The following correlation was obtained

j n ?0:08207Re à0:4132

:e59TThe correlation j =f (Re a )obtained from equation (52)is

j ?0:06963Re à0:3963

:e60TThe j factors obtained from the numerical simulation show good agreement with the correlation (60)based on the experimental results (Fig.12).The relative error e =100(j n –j )/j does not exceed 10percents.Examination of Figure 12indicates that at high Reynolds numbers numerical and experimental j factors are closer than at lower Reynolds numbers.The lower values of the air-side heat transfer coef?cients may results from the contact resistance between the ?ns and the tubes.The numerical

simulation does not account for the thermal contact con-ductance,because the homogenous tube and ?n material is assumed.On the other hand,in the developed data

reduction method contact resistance between tube and ?n has implicitly been included in the air-side heat transfer coef?cient.7

Conclusions

A general numerical method for determining heat transfer correlations in cross-?ow heat exchangers with extended surfaces has been presented.The inverse problem of determining non-dimensional average heat transfer coef-?cients is formulated as a parameter estimation problem by selecting the functional form for the Nusselt number Nu =f (Re,Pr).There are m parameters x =(x 1,…,x m )T to be determined such that computed outlet tube side ?uid temperatures T c,i ￠￠agree in the least-squares sense with the experimentally acquired temperatures f c,i ￠￠.The non-linear least-squares problem was solved using the Levenberg-Marquardt method.Primary measurements consist of the ?ow rates of each ?uid streams and their inlet and outlet tempera-tures.In order to calculate the outlet ?uid temperature,T c,i ￠￠,as the function of searched parameters a mathe-matical model of the heat exchanger was developed.The temperature distributions on the ?uid and air sides can easily be calculated using the proposed model.The

results of the experimental investigation of the plate-?n-tube automotive radiator are presented.The tested,two-pass radiator consists of two inline rows of oval tubes with smooth plate ?ns.The air and coolant mass ?ows,the air temperature in front of the radiator,and the inlet and outlet temperatures of the cooling liquid were measured in eighteen measurement points.The new correlations for the heat transfer coef?cients on the air and coolant sides were developed using the presented technique.The correlation for the average heat transfer coef?cient on the air-side based on the experimental data was compared with the correlation obtained from numerical simulation of 3D ?uid and heat ?ow,per-formed by means of the commercially available CFD code.The numerical predictions are in good agreement with the experimental data.

The main advantage of the presented method is that it

does not require any knowledge,or solution of the com-plex ?uid ?ow ?eld.It can be used for determining heat transfer characteristics of different type of heat exchang-ers.The developed analytical mathematical model of the

plate-?n-and-tube heat exchangers can be used in the

performance and design calculations of automotive radi-ators.It can be useful in the design of heating,ventilating,

air conditioning and refrigeration equipment.The new correlations for the friction factors and the air side heat transfer coef?cient for inline oval tube geometries with plain ?ns can be used in the design of air-cooled heat exchangers.The proposed methods for identi?cation of heat transfer,both on ?uid and air sides,are effective mathematical tools in determining new heat exchanger characteristics.References

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Fig.12.Air-side Colburn j factor as a function of Reynolds

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