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    LAB 4

    PLATE HEAT EXCHANGER

    1.0OBJECTIVE

    1.1 Todemonstrate the working principles of a plate heat exchanger operating under

    parallel flow conditions and counter flow conditions.

    1.2 To determine the overall heat transfer coefficient for both parallel and counter flowsystem.

    2.0 INTRODUCTION

    The SOLTEQHeat Tra!"er Ser#$%e U$t &P'ate Heat E(%)a*er+ &,-e'/HE104P

    HE104PA+has been designed specifically to demonstrate the working principles ofindustrial heat exchangers in the most convenient way possible in the laboratoryclassroom. The equipment consists of a plate heat exchanger mounted on a supportframe. The external surface of the piping is insulated. Two temperature measuringdevices are installed in both the inside and outside tubes to measure the fluidtemperatures accurately. The flow rates are measured using independent flow metersinstalled in each line.

    .0 THEOR3Plate heat exchangers are used extensively in the food and beverage industries due tothe fact that they are easily taken apart for cleaning and inspection.

    The general equation for heat transfer across a surface is:

    Q = U A Tm 1!

    "here#$ % heat transfer per unit time# "& % the overall heat transfer coefficient# "'m2()* % heat transfer area# m2.

    +Tm % the mean temperature difference# the temperature driving force#()

    The mean temperature difference is normally expressed in terms of log-meantemperature difference#

    ,or counter-current flow:

    - 1 -

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    ( ) ( )( )

    ( )12

    21

    1221

    lntT

    tT

    tTtTT

    lm

    =

    2!

    ,or co-current' parallel flow:

    ( ) ( )( )

    ( )22

    11

    2211

    lntT

    tT

    tTtTT

    lm

    =

    !

    +Tlm % log mean temperature differenceT1 % inlet hot water temperatureT2 % outlet hot water temperature

    t1 % inlet cold water temperaturet2 % outlet cold water temperature

    ,rom the energy balance principle:

    Power /mitted % Power *bsorb 0 Power oss

    "here# Power /mitted# , ,( )E H H H H in H outW Q Cp T T =

    Power *bsorbed# , ,( )A C C C C out c inW Q Cp T T =

    /fficiency for the system can be calculated by applying the following equation:

    Power Absorbed100%

    Power Emitted=

    verall heat transfer coefficient#m

    Power Absorbed

    t AreaU=

    "here#

    *rea % surface of contact area

    % width x length! x number of plates 3 1!

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    *n example of schematic diagram for the flat plate heat exchanger# which is described indetail# is shown as below:

    Figure 1:)ounter )urrent ,low

    Figure 2: )o- current 'Parallel

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    4.0 EQUIP,ENT AND SPECIICATIONS

    4.1 E56$78et

    Figure 3:4ear view of the 5eat Transfer 6ervice &nit

    1. Pump2. )irculation 7alve# 72. Pump 8nlet 7alve# 719. "ater Tank. 5eater

    ;. evel 6witch. "ater Tank )over

    - 4 -

    21

    4

    5

    3

    78

    6

    8

    9

    8

    7

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    Figure 4:,ront view of the 5eat Transfer 6ervice &nit

    ?. Temperature )ontroller

    1@. Temperature 6elector11. Aain 6witch12. "ater utlet1. "ater 8nlet19. Temperature 8ndicator1. ,low Aeter #,811;. ,low Aeter# ,821. ,low Aeter )ontrol 7alve# 7>

    1?. )ontrol 7alve2@. Plate 5eat /xchanger21. Temperature 6ensor# T122. Temperature 6ensor# T22. Temperature 6ensor# T29. Temperature 6ensor# T92. Temperature 6ensor# T

    - 5 -

    9

    1

    2

    51

    3

    1

    8

    1

    9

    1

    4

    1

    0

    1

    6

    1

    1

    2

    5

    2

    0

    1

    7

    1

    5

    2

    4 2

    2

    2

    3

    2

    1

    4.2 S7e%$"$%at$-!

    Area = surface of contact area

    = (widt ! "en#t) ! (number of $"ates 1)= (0&12 ! 0&23) ! (' 1)= 0&14 m2

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    9.0 EXPERI,ENTAL PROCEDURES

    5.2 Experiment 1: Parallel Flow Arrangement

    .2.1 6et hot water inlet temperature on the temperature controller at ;@ )..2.2 Position of the valves for parallel flow.

    Position Valve

    $en 4, 5

    *"ose 3, '

    .2. The cold water flow rate#$)is set constant at PA..2.9 6et hot water flow rate $5at PA..2. 4ecord the hot and cold water temperatures at inlet and outlet onceconditions have stabiliBed.

    5.3 Experiment 2: Counter Current Flow Arrangement

    ..1 6et hot water inlet temperature on the temperature controller at ;@ )...2 Position of the valves for counter current flow.

    Position Valve

    $en 3, '

    *"ose 4, 5

    .. The cold water flow rate# $)is set constant at PA...9 6et hot water flow rate $5at PA... 4ecord the hot and cold water temperatures at inlet and outlet once

    conditions have stabiliBed.

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    :.0 RESULT AND CALCULATION

    1. Table of data for exp 1 )-)&44/CT' P*4*/!

    8nitial values used:)ontrolled hot water temperature % ;@ ()5ot water flow rate# $5 % .@ PA)old water flow rate# $) % .@ PA

    Power Power Power Efficiency tm U

    Emitte !"sor"e #ost $ %& '(m) %&

    ' ' '

    )*)&*T8C

    1& outHinHHHH TTCpQWEmittedPower ,,, =

    2& inCoutCCCC TTCpQWAbsorbedPower ,,, =

    3& AbsorbedPowerEmittedPowerWLostPower =,

    4& %100, xEmittedPower

    AbsorbedPowerEfficiencySystem =

    5&

    ( ) ( )

    ( )

    ( )

    , , , ,

    , ,

    , ,

    ,

    "n

    H in C in H out C out

    m

    H in C in

    H out C out

    T T T T LMTD t

    T T

    T T

    =

    '&Areaxt

    AbsorbedPowerUtCoefficienTransferHeatOerall

    m

    =,

    2. Table of data for exp 2 )&CT/4 )&44/CT!

    - + -

    *+,'!,E- 3, t(in) (.*) 4/&0

    4, t(out) (.*) 42&10

    t,ae (.*)

    ot(#mD)

    *$,ot (6#&6)

    &+#.

    '!,E- 2, t*(out) (.*) 40&30

    1, t*(in) (.*) 30&00

    tc,ae (.*)

    co"d(#mD)

    *$,co"d(6#&6)

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    8C8T8* 7*&/6 &6/=:)ontrolled hot water temperature % ;@ ()5ot water flow rate# $5 % .@ PA)old water flow rate# $) % .@ PA

    *+,'!,E-

    3, t(in) (.*) 4&004, t(out) (.*) 3/&20

    t,ae (.*)

    ot(#mD)

    *$,ot (6#&6)

    &+#.'!,E- 1, t*(out) (.*) 4'&/0

    2, t*(in) (.*) 2/&+0

    tc,ae (.*)

    co"d(#mD)

    *$,co"d(6#&6)

    Power Power Power Efficiency tm U

    Emitte !"sor"e #ost $ %& '(m) %&

    ' ' '

    . =etermine the system efficiency of parallel and counter current flow arrangement.

    6how your calculations.

    &!#&U#!,/+

    1&Power emitted = 7*$ (in- out)2&Power absorbed = 7***$* (*out *in)

    3&Power "ost = $ower emitted - $ower absorbed

    4&Efficienc8 9 = %100emittedpower

    absorbedpower

    5&:o# mean tem$erature difference

    ( ) ( )

    ( )( )

    , , , ,

    , ,

    , ,

    ,

    "n

    H in C in H out C out

    m

    H in C in

    H out C out

    T T T T LMTD t

    T T

    T T

    =

    - / -

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    '&era"" eat transfer coefficient, ; = areat

    absorbedpower

    m

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