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    Alexandria University
    Faculty of Engineering
    Chemical Engineering Department
    Chemical Engineering Diploma






    DESIGN OF HEAT TRANSFER EQUIPMENT






    CASE STUDY ON

    VERTICAL TYPE OIL COOLER
    HEAT EXCHANGER











    CASE STUDY ON

    VERTICAL TYPE OIL COOLER
    HEAT EXCHANGER



    Submitted To

    Prof. Dr. Hassan A. Farag,
    Professor of Chemical Engineer,
    Chemical Engineering Department,
    Alexandria University.

    By: Islam M. M. El-Attar,
    & Mohamed A. El-Barsi.
    Students of Chemical Engineering Diploma
    Chemical Engineering Department,
    Alexandria University.



    Jan. 2009



    Table of *******:


    Summary………………………………………..4
    Introduction…………………………………………..5
    Problem statement ……………………………………..6
    Design calculations…………………………………….6
    Specification Sheet…………………………………….10
    Acknowledgment ……………………………………..11
    Table of nomenclature ………………………….12
    References………………………………………13
    Appendix………………………………………..14
































    Summary
    Intelligent selection of heat-transfer equipment requires an understanding of the basic theories of heat transfer and the methods for design calculation. In addition, the problems connected with mechanical design, fabrication, and operation must not be overlooked.
    Determination of appropriate coefficients of heat transfer is required for design calculations on heat-transfer operations. These coefficients can some times be estimated on the basis of past experience, or they can be calculated from empirical or theoretical equations developed by other workers in the field. Many semiempirical equations for the evaluation of heat-transfer coefficients have been published.























    Introduction

    As with all types of design, calculations should begin by identifying the necessary assumptions and segregating the important from unimportant variables.

    At the top of the list is the function. Next come requirements for the plant site. Finally, at the bottom of the list, come the variables affecting design. For the case under study, these data are:
    Process requirements: oil temperature, flow rate, properties of oil, allowable pressure drop, and calculated duty.
    Site data: temperature of water, density of water and specific heat of the water, width limitations to the plant, and horsepower required to deliver the fluid.
    Design variables: water outlet temperature, or mass rate of water flow, heat-transfer coefficient, water pressure drop, required surface, and tube arrangement.

    The procedure is as follows:
    1. Identify all process and site data.
    2. Assume the layout of the tube bundle, and water temperature rise or mass flow rate.
    3. For the assumed values, calculate film coefficients and overall heat-transfer coefficient, effective temperature difference, and surface; check this surface against the assumed layout.
    4. When the required surface fits the assumed layout, calculate the outside pressure drop and check this against the allowable pressure drop.
    5. When surface and tubeside pressure drop are verified, calculate the water side pressure drop and pump horse power.



    Problem Statement

    A vertical shell-and-tube heat exchanger with two tube passes and one shell pass is being used to cool 172,000 Ibm/h of lubricating oil from 154˚F to 122 ˚F at 8 atm. The oil passes through the shell side, and water at 95˚F is used to cool the oil. The tubes are copper with an OD of 3/4 in. and 0.064 in. thickness, and the tubes are in line. The exchanger contains a total of 440 tubes (220 tubes per pass). Ten segmental baffles with 25 percent cut are used on the shell side, and the baffles are spaced equally 1 ft. apart. The inside diameter of the shell is 20 in. the clearance between tubes is 0.25 in., the tube length is 12 ft. long, and the flow rate of water in the tubes is 187,500 Ibm/h. the shell can be considered to be full-packed. The water enters at 95 ˚F and leave at 110˚F at 10 atm.

    Design calculations

    From the Appendix, [2]
    k for copper =234 Btu/(h)(ft2)( ˚F/ft).
    at (95+110)/2=103˚F,
    µ for water = 0.71 centipose,
    k for water = 0.362 Btu/(h) )(ft2)( ˚F/ft),
    cp for water = 0.999 Btu/ (Ibm)(˚F),
    ρ for water = 62 Ibm/ft3,
    average oil film temperature at [(154-110)+(122-
    95)]/2=36˚F,
    µ for oil = 135 centipose,
    k for oil = 0.1 Btu/(h)(ft2)( ˚F/ft),
    cp for oil = 1.1 Btu/ (Ibm)(˚F),
    ρ for oil = 54 Ibm/ft3,
    at (154+122)/2=138 ˚F
    µ for oil = 17 centipose






    For the tube side
    Flow area per tube = (3.14)(0.6)2/4 = 0.283 in2.

    Mass velocity = G = (187,500)(144)/(0.283)(220)
    = 434,000 Ibm/(h)(ft2)

    NRe = DG/ µ = (0.6)(434,000)/(12)(0.71)(2.42) = 13,000
    cp µ/k = (0.999)(0.71)(2.42)/0.362 = 4.7
    µw at 36 ˚F =1.6 centipose

    µ/ µw= 0.71/1.6 = 0.44

    from eqn 26 [1]
    hi= (k/D)(0.023)(DG/ µ)0.8(cp µ/k)1/3(µ/ µw)0.14
    hi= 486 Btu/(h)(ft2)( ˚F)

    for the shell side
    no. of tube rows = 20/(0.75+0.25) = 20
    free area between baffles = (20)(0.25)(1)/12 = 0.42 ft2

    Gs= 180,500/0.42 = 430,000 Ib/(h)(ft2)

    DoGs/ µf = (0.75)(430,000)/(12)(135)(2.42) = 100
    (cp µ/k)f = (1.1)(135)(2.42)/0.1 = 3594

    From eqn. 29 [1]
    ho = (kf/Do)(ao/Fs)( DoGs/ µf)0.6(cp µ/k)f1/3

    ao = 0.26 Fs=1.6
    ho =[(0.1)(12)/.75][0.26/1.6][100]0.6[3594]1/3

    ho = 63 Btu/(h)(ft2)( ˚F)


    A fouling coefficient of 2000 Btu/(h)(ft2)( ˚F) is adequate for the water, and of 1000 Btu/(h)(ft2)( ˚F) is adequate for the oil.

    Basing Ud on the inside tube area
    1/Ud = A/ h'A'f + A/ h"Af"+ Axw/kAw+ A/ h'd A'f + A/hd"Af"

    = 1/486 + 0.6/(63)(0.75) +
    (0.6)(0.064)/(234)(0.664)(12) + 1/2000 + (0.6)/(1000)(0.75)
    = 0.00206 + 0.013 + 0.0000206 + 0.0005 + 0.0008
    = 0.0164
    Ud = 61 Btu/(h)(ft2 of inside area)( ˚F)

    q = m. cp ∆t
    = (172,000)(1.1)(154-122) = 6*106 Btu/h

    A = q / Ud ∆toa,m = 6*106/(65)(36) = 2564 ft2 of inside tube area
    Length per tube = L = 2564 / (0.1529)(440) = 35 ft
    But the available length is 20 ft.
    Calculations of pressure drop
    For tube side
    From eqn 30 [1] - ∆Pi= Bi2fiGi2Lnp/gcρiDiΦi

    Bi = 1 Φi = 1.02 (µi/ µw)0.14

    Φi = 1.02(0.44)0.14 = 0.9

    fi from fig. [2] = 0.006 np = 2

    - ∆Pi =(1)(2)(0.006)(434,000)2(20)(2)/(3600)2(32.17)(62)(0.6/12)(0.9)
    =77 psf





    For the shell side
    From eqn 31 [1]
    - ∆Po= Bo2f 'NrGs2/gcρo
    f ' = [0.044 + 0.08xl/(xT-1)0.43+1.13/xl]( DoGs/ µf)-0.15
    = [0.044+ 0.08(1/0.75)/(1/0.75 – 1 )0.43+1.1/(1/0.75)(100)-0.15
    =0.234

    Bo = 11 Nr = (20)(3/4) = 15
    - ∆Po = (11)(2)(0.234)(15)(430,000)2/(32.17)(3600)2(54)
    = 640 psf = 4.5 psi




















    HEAT EXCHANGER
    Date May 1997
    By Siemens Identification: Item cooler
    Item no. 60899925/01
    No. required II
    Function: Cooler for lubricating oil from the Steam Turbine
    Operation: continuous
    Type: Vertical
    Fixed tube sheet
    Expansion ring in shell
    Duty 6,000,000 Btu/h outside area 1800 sq.ft.
    Tubes 3/4 in. diam.
    With 0.064 in. thickness
    1" Centers in line Pattern
    440 tubes each 24 ft long
    2 passes
    Tube material copper Tube side:
    Fluid Handled cooling water
    Flow rate 187,500 Ib/h
    Pressure 10 atm.
    Temperature 95˚F to 110˚F
    Head Material copper
    Shell : 20 in. diam. I pass
    (Transverse baffles Tube support req'd)
    (Longitudinal baffles 0 req'd)
    Shell material carbon steel lined Shell side:
    Fluid handled lube' oil
    Flow rate 172,000 Ib/h
    Pressure 8 atm.
    Temperature 154 ˚F to
    (Constant temp.)
    Utilities: Untreated cooling water
    Controls: cooling water rare controlled by outlet oil temperature
    Insulation: not required
    Tolerances: Tubular Exchangers Manufacturers Siemens
    Comments and drawings: Location and sizes of inlets and outlets are
    shown on drawing


    Specification sheet for oil cooler heat exchanger






    اهداء

    اهداء الى كل من ساهم فى هذا البحث. كما اود ان اتقدم بخالص الشكر والتقدير الى

    السيد الاستاذ الدكتور/
    حسن عبد المنعم فرج
    استاذ الهندسة الكيميائية
    كلية الهندسة جامعة الاسكندرية
    على جهدة المتواصل وعطائة المستمر فى مجال الهندسة الكيميائية واشرافة على اتمام هذا البحث.
    واخص بالشكر ك/ محمد اسماعيل على حسن تعاونة

    ........... وشكرا

    اسلام مصطفى محمود










    Table of nomenclature

    h film coefficient of heat transfer, Btu/(h)(ft2)( ˚F)
    D diameter, ft
    k thermal conductivity, Btu/(h)(ft2)( ˚F/ft).
    G mass velocity, Ib/(h)(ft2)
    µ absolute viscosity, Ib/(h)(ft)
    cp heat capacity, Btu/ (Ibm)(˚F),
    L heated length of straight tube, ft
    Bi correction factor to account friction, dimensionless
    fi fanning friction factor, dimensionless
    np number of tube passes, dimensionless
    gc conversion factor in Newton's law of motion, 32.17
    (ft) (Ibm)/(s2) (Ibf)
    ρ density, Ibm/ft3
    Φi correction factor for nonisothermal flow, dimensionless
    f ' special friction factor, dimensionless
    Nr number of tubes in exchanger, dimensionless
    ao constant for evaluating outside film coefficient, dimensionless
    A area of heat transfer, ft2
    xT ratio of pitch transvers to flow to tube diameter, dimensionless
    xl ratio of pitch parallel to flow to tube diameter, dimensionless
    Fs safety factor to account bypassing on shell side exchanger









    References

    [1]Max S. Peters, & Klaus D. Timmerehaus, " Plant Design and Economics for Chemical Engineers ", 4th edition, p.579, McGraw Hill book company, New York, 1980.

    [2]Robert H. Perry, Don W. Green, & James O. Maloney, "Perry's Chemical Engineers' Handbook", 7th edition, P.2-327&334, McGraw Hill book company, New York, 1997.

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