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    Faculty of Engineering and TechnologyDepartment of Chemical Engineering Chemical Engineering laboratory (2 )

    Title of experiment: " r h l E t h ehlT dna l hS " Number of experiment: " 2" Done by : Ramzy Kilani 0115131

    Amal suliman 0100463Mayson Qawooq 0115281Doaa bdarneh 0124657

    Date of performing experiment: 3/4/2134 Date of submitting report: 21/4/2134 Type of report: "Short Report"

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    Rsults:Table (1) Results of the experiment

    trialinletwater T

    outletwater T

    inletwater P

    outletwater P

    inletsteam T

    outletsteam T Avg water Avg staem

    1 292.15 304.15 0.4 0.7 378.15 338.15 298.15 338.152 293.15 298.15 0.49 0.69 378.15 339.15 295.65 336.93 294.15 303.15 0.8 0.5 381.15 339.15 298.65 339.94 295.15 308.15 1 0.49 382.15 339.15 301.65 341.95 295.15 313.15 1.15 0.41 381.15 341.15 304.15 342.65

    Table (2) Results of the experiment

    inletsteam P

    steam

    steam

    Cpsteam Hfg(kJ) Tlm

    Vol.Flowsteam

    mass.Flowsteam

    Vol.Flowwater

    mass.Flowwater

    1 0.50.0000

    1965.25

    42.004

    82257.63

    658.894834

    642.03333

    E-050.0196268

    310.0002916

    670.2873558

    33

    2 0.50.0000

    1964.19

    31.999

    22257.63

    630.273372

    37 0.000021.92839E-

    050.0002666

    670.2633733

    33

    3 0.50.0000

    1963.21

    41.993

    72257.63

    659.994945

    491.83333

    E-050.0176589

    230.0002333

    330.2309439

    77

    4 0.50.0000

    1962.54

    81.993

    22257.63

    657.706128

    392.08333

    E-050.0200530

    830.0002166

    670.2144938

    03

    5 0.50.0000

    1962.25

    4 1.9922257.63

    656.285229

    792.31667

    E-050.0222922

    180.00017833

    30.1766194

    07

    Calculations of the fifth trial : Calculation of overall heat transfer coefficient:

    trial Re Nu hi ho Ud Uc

    1 1648.9117 7.3377 394.7635 8500.0000 325.7890 412.2553

    2 1399.3470 7.2452 386.8937 8500.0000 326.8577 411.07503 1123.0575 7.0718 374.6236 8500.0000 324.4589 410.1236

    4 1043.0620 6.9617 368.4479 8500.0000 321.4520 409.2558

    5 844.5679 6.6571 351.6692 8500.0000 320.6369 409.0120

    Rd Qw Qs QL QL%0.0006 14419.6307 33475.1580 1780.2540 0.05320.0006 5505.0294 32650.1455 1897.6750 0.05810.0006 8687.0732 32500.2315 1941.1530 0.05970.0007 11654.1991 31152.2350 1980.3450 0.0610

    0.0007 13286.9367 31015.2324 1996.1566 0.0640

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    23035.3)37)(030.2)(014(. m LN D A Oo

    LM T F U q

    Fig .2. Parallel flow heat exchanger.

    14.19105

    1931

    33.3193165105

    56

    191053165

    ln

    191053165

    ln

    0

    1

    2

    12

    P

    R

    C

    T T

    T T T LM

    Fig.3. correction factor

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    F=.99

    hhhh T cpmmq is the latent heat of condensation.

    From steam tables (at T sat =105C0=378K) = 2225 KJ.Kg -1

    1.2.482.1855699.3035.3

    33970

    97.33

    ))65105(7.41813102.14()3102.142225000(

    /3102.143.22.985)1.(42.14

    1.1.1817.4

    3580852

    65105

    2

    K mW TLM F A

    qU

    KW q

    q

    T cpmmq

    skg mkg sml Qm

    K Kg KJ CP

    K C T T

    T

    oo

    hhhh

    hh

    h

    hout hinh

    Calculation of the heat transfer coefficientQ= 4 = 17.5 L.min -1

    3

    1

    23

    4

    2

    4

    22.985

    252

    31192

    6234.2)1012(

    4

    10967.2

    4

    10967.2

    m Kg

    C TcTc

    T

    ms Di A

    QV

    OOUT inC

    TURBULANT

    VDi 55

    10200310.6105

    012.6234.222.985Re

    Fully developed flow 10Di =.12 m

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    12

    12

    4.8.4.8.

    ..8500

    ..854930012.

    64559.15891

    15891261.3620310023.Pr Re023.

    K mW h

    K mW h

    K

    Dh NU

    o

    i

    ii

    Calculation of the fouling resistance

    .30351.33703.2014..83158.23703.2012.

    111

    111

    2

    2

    m LN D Am LN D A

    A K R R

    A R

    A R

    Ah Ah AU

    A K R R

    A R

    A R

    Ah Ah

    AU

    oo

    ii

    i gla ss

    iO

    o

    fo

    i

    fi

    ooiiOO

    i gla ss

    iO

    o

    fo

    i

    fi

    ooii

    OO

    R fo =zero. (Steam moves in the shell so it assumed to be clean)

    12

    4573

    4

    11

    003145505.

    10851083854.483158.2

    10561285.31013086.4106320.1

    10851083854.483158.23035.38500

    183158.2854930

    13035.3482.185

    1

    ..728001681.)(75.662

    5.90432

    W K m R

    R

    R

    k mW GLASS K C ThTc

    T

    fi

    fi

    fi

    O

    Calculation of the heat losses

    KW q

    q

    H H m H H mq hihohCiCoc

    91.968

    ))26950002335000(01126(.))142400217700(0667(.

    ))(())((

    ..

    ..

    ....

    H m H mqinlet outlet

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    Objectives:

    -To determine the overall heat transfer corfficient using steam on the shellside and cooling water on the tube side.

    -To investigate the effect of flow pattern (co-current or counter current) onthe heat transfer coefficient.

    -To establish energy balance in the heat exchanger.

    General DescriptionShell and Tube Heat Exchangers are one of the most popular types ofexchanger due to the flexibility the designer has to allow for a widerange of pressures and temperatures. There are two main categoriesof Shell and Tube exchanger:1. those that are used in the petrochemical industry which tend to be

    covered by standards from TEMA, Tubular Exchanger ManufacturersAssociation

    2. those that are used in the power industry such as feedwater heatersand power plant condensers.

    Regardless of the type of industry the exchanger is to be used in there area number of common featuresA shell and tube exchanger consists of a number of tubes mountedinside a cylindrical shell. Figure 1 illustrates a typical unit that may befound in a petrochemical plant. Two fluids can exchange heat, one fluidflows over the outside of the tubes while the second fluid flows throughthe tubes. The fluids can be single or two phase and can flow in aparallel or a cross/counter flow arrangement.

    http://www.thermopedia.com/content/1121/#SHELL_AND_TUBE_HEAT_EXCHANGERS_FIG1http://www.thermopedia.com/content/1121/#SHELL_AND_TUBE_HEAT_EXCHANGERS_FIG1http://www.thermopedia.com/content/1121/#SHELL_AND_TUBE_HEAT_EXCHANGERS_FIG1http://www.thermopedia.com/content/1121/#SHELL_AND_TUBE_HEAT_EXCHANGERS_FIG1
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    Figure 1. Shell and tube exchanger.

    The shell and tube exchanger consists of four major parts: Front Header this is where the fluid enters the tubeside of the

    exchanger. It is sometimes referred to as the Stationary Header. Rear Header this is where the tubeside fluid leaves the exchanger

    or where it is returned to the front header in exchangers with multipletubeside passes.

    Tube bundle this comprises of the tubes, tube sheets , baffles andtie rods etc. to hold the bundle together.

    Shell this contains the tube bundle .

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    DISSCUSSION: RELIABILITY OF DATA:

    1. Heat transfer is thermal energy transit due to temperature difference from high temperature

    region to low temperature region.The data shows increase in cooled water (tube fluid) temperature and decrease in steams (hellfluid) temperature then the data is accurate and logic.

    2. The pressure drop is necessary for moving water in the tubes and steam in the shellduring the experiment.

    3. While increasing the flow rate of cold water in the tubes the heat transfer from steam tocooled water will be greater so the condensate flow rate will increase.

    4. changing the flow rate of cold water cause the system to be under transient state so weshall wait about (10-15 ) minutes until the variables in the heat transfer process returnconstant with time again.

    5. The pressure drop of steam in the shell is greater than of the pressure drop of water in

    tubes so we put the steam in the shell because it is stronger than tubes.

    COMPARISION OF RESULTS WITH LITERRATURE FINDINGS:Table (4) Comparison between experimental and theoretical resultsVariables Experimental results Representative results

    Overall heat transfercoefficient

    223.147 W/m .K 200-700 W/m .K

    Heat transfercoefficient

    3744.927 W/m .K 2500-100000 W/m .K

    Fouling resistance .00209955644 m .K/W .0001 m .K/W

    Heat transfer 27.09 KW 968.91 KW

    1. Overall heat transfer coefficient is less than the representative one because of the heat lossfrom the device to the surrounding.2. Fouling resistance is high because maintenance is annually and the fluid is dirty.3. Heat loss to the surrounding is very huge because the steam is in the shell side and the

    boundary is glass and it is not a good insulator.4. Changing the water flow rate causes the system to be in transient state the variables will

    change with time so the system need time to return to steady state.5. Heat transfer coefficient inversely proportional to water flow water, because the thermalenergy transfer from steam to water will be decreasing.

    QWATER , q , hi

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    CONCLUSIONS AND RECOMMENDATIONS :CONCLUSIONS:1. the steam usually is in the tube side and the cold water in the shell side to minimize the

    heat loss to surrounding but in this experiment its the opposite because the pressure dropof steam is very high and it cause cracking in the tubes and the mechanical cleaning ismuch more practical for tubes than for the shells so the steam pass through it.

    2. Controlling the pressure of steam at the main supply before it enters the shell byincreasing it gradually.

    3. Counter current flow has more efficient heat transfer coefficient than parallel flow.4. Importance of steam trap to get red of condensate in the steams stream and to prevent

    builds up of pressure on the whole system.5. Some valves stay open to the atmosphere to keep the fluids moving by establishing

    pressure drop.6. the diameter of tubes are very small and large number to get larger effective area as

    possible perpendicular to the heat transfer direction7. Segmental baffles vertically cut by 25% fitted along the shell to direct the flow aroundthe tubes, increase velocity, promote cross flow, and help supporting the tubes.

    8. log mean temperature difference method should be adjusted by a correction factor because the driving force (T) will vary along the length of the exchanger and the flow patterns is a mixture of cocurrent, counter current, and cross flow.

    RECOMMENDATIONS:1. Instead of doing five trials on parallel flow onlyTwo trials on parallel flow and three trials on counter flow.

    2. Doing maintenance every six months

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    References:1. FRANK P.INCROPERA and DAVID P.D EWITT, Fundamentals of

    Heat and Mass Transfer, 4 th edition,(1996)

    2.

    TEMA Seventh Edition. (1988) Tubular Exchanger ManufacturersAssociation.3. Saunders, E. A. D. (1988) Heat Exchangers Selection, Design and

    Construction , Longman Scientific and Technical.4. Hewitt , G. F, Shires, G. L., and Bott, T. R. (1994) Process Heat

    Transfer, CRC Press .5. Boiler and Pressure Vessel code, ASME (American Society of

    Mechanical Engineers).

    6.

    British Master Pressure Vessel Standard, BS 5500.

    http://www.thermopedia.com/authors/1/http://www.thermopedia.com/authors/1/http://www.thermopedia.com/authors/1/
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