Thermal Design of Tube and Shell Heat Exchanger and Verification by HTRI Software

DOI : 10.17577/IJERTV9IS120224

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Thermal Design of Tube and Shell Heat Exchanger and Verification by HTRI Software

Ranjeet Handibag

Mechanical Engineering Department Pimpri Chinchwad College of Engineering Pune, India

Dr. Umesh Potdar

Associate Professor Mechanical Engineering Department

Pimpri Chinchwad College of Engineering Pune, India

Ajinkya Jadhav

Mechanical Engineering Department Pimpri Chinchwad College of Engineering Pune, India

Abstract Heat exchangers are the devices used to transfer heat from one fluid to another. Fluids can be either gases or liquids. plate heat exchanger, Double pipe heat exchanger, shell and tube heat exchanger, condensers, evaporators and boilers are the most common types of heat exchangers. They are widely used in petroleum refineries, sewage treatment, air conditioning, power station, space heating and petrochemical plants. Tube and shell tube and shell heat exchanger is type of heat exchanger in which two fluids which are at different temperatures are separated by solid wall. this article includes thermal design calculation and verification of all the process parameters which are required for proper functioning of the compressor system as a whole. the design calculations of tube and shell heat exchanger are verified by HTRI(Heat Transfer Research Inc.) software. This is the software for the rating, design, and/or simulation of a wide variety of heat transfer equipment, including shell-and-tube and non-tubular exchangers, air coolers and economizers, and fired heaters.

Keywordstube and shell heat exchanger, HTRI, temparature, compressor.


    Industries often use compressed gases such as air, hydrogen, acetylene, oxygen, methane, etc. The volume required of these compressed gases is very high so it is required to compress these gases efficiently. For this reason intercoolers are used. Intercoolers are placed after compression or between two stages of compression. Intercoolers can be placed horizontally or vertically but horizontal intercoolers are more effective thus horizontal intercoolers generally used.

    Intercoolers are generally placed in between two stages of compression as it cools down the hot fluid before sending it to second compression. Thus volume of fluid decreases which increases the volumetric capacity and decreases the work done by motor.

    Fig 1: 2 stage compressor block diagram [6]

    Fig 2: Tube and shell heat exchanger [12]

    Following are the functions of an intercooler used in compressor:

    • Atmospheric air contains moisture, and furthermore, the air may pick up oil vapor as it passes through some parts of the compressors. Cooling air down to or below its initial temperature will remove moisture down to the dew point, improving the quality of the air.

    • Intercoolers improve the efficiency of compressor. As the volume of hot air reduces on cooling volumetric efficiency increases.

    • Every 4 °C rise in inlet air temperature results in higher energy consumption by 1 percent to achieve equivalent output. Thus lower the temperature of intake air, more is the energy efficiency of a compressor.

  2. METHODOLOGY Problem identification

    Selection of a heat exchanger

    Thermal design

    Verification of design


    To cool down the air from Thermal design of heat exchanger based on the input parameters which are given in the specification sheet and verification of design i.e. verification of the overall heat transfer coefficient also, finding out the acceptability of the pressure drop.


    Thermal design of the heat exchanger (intercooler) is according to the Kern's method. It is simple to apply, accurate enough for preliminary calculations. The selection criteria for the heat exchanger, design procedure and its calculations are discussed in the subsequent sections Selection Criteria for Shell and Tube Heat Exchanger

    • Materials of construction

    • Operating pressure and temperature

    • Flow rates

    • Flow arrangements

    • Fouling tendencies

    • Maintenance, inspection, cleaning, extension

    • Overall efficiency

      1. Shell:

        Shell is the cylindrical vessel container through which fluid flows and the tube bundle is placed inside the shell. Shells are usually casted from standard steel pipe with satisfactory corrosion allowance.

        Fig 3: Shell Design [7]

        BEW type shell type is selected. It is simple in construction and easy to clean as it has floating rare end head thus tube sheet can be easily removed for cleaning.

      2. Tube and tube layout

        The enough tube thickness ensures that it will bear the internal pressure along with the adequate corrosion allowance. Longer tube less is the shell diameter at the expense of higher shell pressure drop. Stainless steel, copper, bronze and alloys of copper- nickel are the commonly used tube materials.

        The shortest center to center distance between the adjacent tubes is Pitch. The general pattern of tubes is square or triangular pattern. The tube count is the number of tubes that can be accommodated in a given shell ID.

        Fig 4: Tube Layout [1]

        For this application triangular layout is selected as it is more effective and easy to clean

      3. No of passes

        To obtain greater heat transfer co-efficient and also to reduce scale formation, number of passes is chosen to get the required tube side fluid velocity. The tube passes vary from 1 to 16. A single tube pass is selected for the design.

      4. Baffle

      Liquid is maintained in a state of turbulence ensures the higher heat transfer coefficients. The turbulence is induced outside the tubes which cause the liquid to flow at right angles to the axes of the tubes. This causes considerable turbulence even when a small amount of liquid flows through the shell.

      One from the various types of baffles is used to increase the fluid velocity by diverting the flow across the tube bundle to obtain higher transfer co-efficient. The distance

      Between adjacent baffles is called baffle-spacing. The segmental baffle used in design shown in the figure 4..

      The % cut for segmental baffle refers to the cut away height from its diameter.

      Fig 5: Segmental baffle [3]



    Gas Handled


    Capacity of compressor



    Suction Temperature



    Relative humidity (Min / Max)



    Suct. Pres. at compr. flange

    Bar (g)


    Cooling Water Temp: In / O

    ° C



    Quantity required


    1 No.

    Code of construction

    TEMA C






    Tube side


    Shell side


    Corrosion allowance


    Tube side


    Shell side



    Bar (g)

    Tube side


    Shell side


    Design Pressure

    Kg/cm² (g)

    Tube side


    Shell side


    Hydro Test Pressure

    Kg/cm² (g)

    Tube side


    Shell side


    Design Temperature


    Tube side


    Shell side


    Max.permissible Pr .Drop

    Kg/cm² (g)

    (By Vendor)

    Tube side


    Shell side


    Water consumption



    Air (Gas) Temperature






    No. of passes per Shell



    Material of construction



    Tube Sheets




    Baffle Plates


    Tie Rods




    Fouling factor


    Tube side


    Shell side


    Tube OD size



    Tube thickness



    Tube length



    Tube pitch



    • Outlet Condition Of Air (At Outlet Of intercooler) Pressure P2= 2.8806 kg/cm2

      Temperature T2= 40.0 RH = 1.00

      Saturated pressure of vapour PV2 = 0.0739 kg/cm2

      Weight of air per kg of dry air

      = (0.622*PV2*RH2)/(P2- PV2*RH2)

      = (0.622*0.0739*1)/( 2.8806- 0.0739*1)

      = 0.01637 kgs

    • Temperature of air at the inlet of the intercooler (adiabatic compression)

      T2/T1 = (P2/P1)(-1)/ T2/38 = (2.8806/1.0269)(1.4-1)/1.4

      T2 = 144.56

      T1, T2 = temperature at the i/p and o/t of L.P. cylinder resp. P1, P2 = Pressure at the i/p and o/t of L.P. cylinder resp.

    • Weight of dry air;

      = P*V/R*T kgs/hr

      = ((1.026-0.7*0.06689)*736.2*104)/29.271*311.0

      = 792.60 kgs/hr

      Weight of vapour in = 0.0294*792.60 = 23.30 kg/hr

    • Heat duty = (flow rate * specific heat * temperature rise)

    Dry air = 792.6*0.245(144.56-40) = 20304.19 kcal/hr Vapour = 23.30*0.45(144.56-40) = 1096.311 kcal/hr

    Total = 21400.5 kcal/hr

    Hot fluid

    Cold fluid


    High Temperature

    Del t5=144.56


    Del t2=104.56

    Low temperature


    Del t6=32.0

    Del ta=8.0



    Del t3=104.56

    Del t4=8.0

    (Note- Temperatures in above table are in )

    1. Thermo-Physical Properties Of Hot And Cold Fluids

      • Inlet Condition Of Air (To L.P. Cylinder) Pressure P1= 1.0269 kg/cm2

        Temperature T1= 38.0 . RH = 0.7

        Saturated pressure of vapour PV1= 0.06689 kg/cm2

        • LMTD

        = (Del t2-Delt1)/ ln(Del t2(Del t1)

        = (104.56-8.0)/ ln(104.568.0)

        = 37.56

        Weight of air per kg of dry air:

        = (0.622*PV1*RH1)/(P1- PV1*RH1) (1)

        = (0.622*0.06689*0.7)/( 1.0269- 0.06689*0.7)

        = 0.0294 kg

    2. Shell Side Calculations

      • Flow Area

        As = (Ds*C*Bp)/(144*Pt)

        = (6*0.105*3.590)/(144*0.4803)

        = 0.0327 ft2

        Ds = Diameter of shell = 6 inch Pt = Pitch of tubes = 0.480 inch

        Do = outer diameter of tube = 0.375 inch Bp = baffle pitch = 3.59 inch

        C = Pt-do = 0.480-0.375 = 0.105

      • Mass Velocity

        Gs = W/As

        = 11025/0.0327

        = 337155.96 lbs/hr.ft2

        W = weight of water = 5000*2.205 = 11025 lbs/hr

      • Reynolds No.

        Res = Des*Gs/MYU

        = 0.0248*337155.96/1.808

        = 4624.70

        Des = Equivalent Diameter



        = 4*(Pt/2 * 0.86Pt 1/2 * d2 /4)1/2 * do

        = 0.2977 inch = 0.0248 ft MYU = viscosity of water at 35

        = 0.7191 cp = 1.808 lb/

      • Shell Side Heat Transfer Coefficient Ho:

        Ho = 0.36 * (K/Des) * (Des*Gs/MYU)0.55 * (C*MYU/K)0.33 * (MYU/MYUW)0.14

        =0.36 * (0.3630/0.0248) * (4624.70)0.55 *

        (1.0*1.808/0.3630)0.33 * 10.14

        = 928.16 Btu/hr.ft2.

      • Shell side thermal resistance Ros = 1/Ho = 1/928.16 = 1.077*10-3

        Shell side fouling factor, Rdo = 0.002 Thermal resistance Ro:

        Ro = Ros + Rdo

        = 1.077*10-3 + 0.002

        = 3.077*10-3 hr..ft2

    3. Tube Side Calculations

      • Flow area

        At = (No of tubes * flow area per tube)/(No of passes


        = (94 * 0.0730)/(1*144)


        M= weight of air = weight of dry air + weight of vapour

        = (792.6+23.3) * 2.205 = 1799.3 lbs/hr

      • Reynolds No :

        Res = di * Gt/MYU

        = 0.02541 * 37794.1/0.0505

        = 19016.79

        MYU = viscosity of air

        = 0.0208 cp = 0.0505 lb/fthr

      • Tube Side Heat Transfer Coefficient Hi :

        Hi = JH * (K/di) * (C*MYU/K)0.33 * (MYU/MYUW)

        = 70 * (0.0183/0.02541) * (0.245*0.0505/0.0183)0.33 * 10.14

        = 44.30 Btu/hr.ft2.

        Referring to graph in Kerns book fig.24 page 834, JH corresponding to the Reynolds no 19016.79 is 70

      • Tube side thermal resistance Tube side fouling factor rdi = 0.001

        ri = 1/ Hi

        = 1/44.30 = 0.0225

        Ri = ri + rdi

        = 0.001 + 0.0225 = 0.0235

        Tube wall resistance = rw

        rw = d/(24*K) ln(d/(d-2*t))

        = 0.375/(24*117) ln(0.375/(0.375-2*0.035))

        = 2.75*10-5

      • Total resistance Rtotal :

        = Ri * (Ao/Ai) + Ro + rw

        = 0.0235 * (0.375/0.305) + 3.07*10-3 + 2.75*10-5

        = 0.03199

    4. Overall heat transfer coefficient U

      U = 1/Total resistance

      = 1/0.03199

      =31.258 Btu/hr.ft2 = 152.61 kcal/hrm²°C

    5. Pressure drop

      • Pressure drop shell side Ps

        Refer the graph figure 29 of kern: corresponding to Reynolds No. friction factor F = 0.0025

        Ps = F * G2s * Ds * (N+1)/(5.22 * 1010 * s * De * PHIs)

      • Mass velocity :

        = 0.0476 ft

        Gt = M/At

        = 1799.3/0.0476

        = 0.0025 * 337155.962 * 0.5 * 13/(5.22 * 1010 * 1* 0.0248 *1 )

        = 1.45 psi = 0.09983 kg/cm2 F = friction factor

        N = no of baffles

        = 37794.1 lbs/hr.ft2

        s = sp. Gravity

      • Pressure drop tube side Pt

    Refer the graph figure 26 of kern: corresponding to Reynolds No. friction factor F = 0.0002

    Pt = F * Gt 2 * L * N/(5.22 * 1010 * di * s * PHIt)

    = 0.0002 * 37794.12 * 5.25 *1/(5.22 *1010 * 0.025*


    = 0.4 psi



    Calculated value

    HTRI value


    Overall heat transfer coefficient





    Less than 10% of variation

    Heat duty

    21400.5 kcal/hr

    21896 kcal/hr

    Less than 10% of variation

    Shell side pressure drop

    0.09983 kg/cm2

    0.049 kg/cm2

    < 0.5 kg/cm2

    Tube inside pressure drop

    0.044 kg/cm2

    0.053 kg/cm2

    <0.1 kg/cm2


    Calculated value

    HTRI value


    Overall heat transfer coefficient





    Less than 10% of variation

    Heat duty

    21400.5 kcal/hr

    21896 kcal/hr

    Less than 10% of variation

    Shell side pressure drop

    0.09983 kg/cm2

    0.049 kg/cm2

    < 0.5 kg/cm2

    Tube inside pressure drop

    0.044 kg/cm2

    0.053 kg/cm2

    <0.1 kg/cm2

    L = tube length per pass in ft N = no of baffles

    PHIt = MYU/MYUW = 1

    s = sp. Gravity

    Pr = 4 * (N/s) * V2/(2 * g)

    = 4 * 1/0.0027 * 0.162/(2 * 32.2 ) = 0.25 psi

    PTt = Pt + Pr

    = 0.4 + 0. 25

    = 0.65 psi = 0.044 kg/cm2


    The software results look like figure no 6. The main parameter which will be verifies is the overall heat transfer coefficient, heat duty and pressure drop.

    The difference between calculated value and HTRI value of overall heat transfer coefficient and Heat duty (heat exchanged) is less than 10% which is acceptable. Calculated value and HTRI value of shell side presure drop and tube inside presure drop are less than allowable pressure drop hence the values of the calculated parameters are varified.


At first thermal design is done by using kerns method. The input parameters are taken from specification sheet. The values of Overall heat transfer coefficient, Heat duty, pressure drop, are obtained.

HTRI software is used to verify analytical thermal design. It gaves close results as that of obtain from thermal design. The comparision between calculated values and software obtained values concludes the varification of the parameters.

The proven theoretical methods are in good agreement with the software results.

Fig 6: HTRI Result [13]


HTRI incorporates a user-friendly interface that reduces time and increases efficiency. It can load a case from CD or a read only directory. If a user loads a case marked as read only, HTRI opens and runs the case. The results are given by the software are in less time as compared to manual calculations.

HTRI can estimate the shell weight. It has an approximation procedure for estimating the weight. It does not rigorously determine the material needed to construct the shell but rather divides the diameter into suitable portions. The values obtained are reasonable but not exact

Thus after all the input is put in the software, it generates the results which can further be used for analyzing them with required output. If found satisfactory user may move ahead with the design or he shall again change the input values and obtain the required results.


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  12. Image of tube and shell heat exchanger

  13. Image genarated by Heat Transfer Research, Inc. , HTRI Xchanger Suite, [Online]. Available: [Accessed Jan 2019].

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