DESIGNING STEPS OF SHELL AND TUBE HEAT EXCHANGER

 

DESIGNING STEPS OF SHELL AND TUBE HEAT EXCHANGER

Shell and tube heat exchanger can be designed in three types.

U-Tube heat exchanger

Straight tube (one pass tube side)

Straight tube (two pass tube side)

HEAT TRANSFER DEPENDS UPON

The heat transfer in a heat exchanger depends upon:

  • Area
  • Thickness
  • Time in contact
  • Mean temperature difference between the liquids
  • Material through which heat transfer takes place
  • Flow Patterns

  • Parallel FlowCounter Current FlowShell and Tube with bafflesCross Flow
  • Physical Properties needed for design

  • Density: 912 Kg/m3Specific Heat = 2.14 KJ/Kg.oCViscosity: 6*10-3Ns/m2Thermal Conductivity: 0.2758 W/m.oC
  • Mass flow rate of the process stream                 m = 2035.8 Kg/hrMass flow rate of the Water                                       m = .81904 Kg/hrHeat duty: Q=m Cp ∆T

    = 2035.8*2.14*85/3600

    = 103.34kW

  • LMTD =  52.810 OCAssumed Calculat  Value of UD AssumedUD   = 300 W/m 2C

     

    (rc vol6 table 12.1)

  • R      = (T1-T2)/ (t2-t1) = 2.833      = (t2-t1)/ (T1- t= 0.35R & S intersect at fig 12.19 RC VOL 6 so from the value of Ft is 0.83
  • Heat Transfer Area

  • Heat Transfer AreaA=Q/Ud     t=103.3 /43.83*300=7.859m2
  • Total Length of Tubes

  • TUBE DIMENSIONS:O.D. = 20 mmI.D. = 16 mmLength of tube = Lt = 4.88 m 

    (Ref:rc vol.6 table 12.3)

  • Number of Tubes:

  • Area of One TubeAssume L = 8ft (From RC VoLAt     =0.218m2No. of tubes  = Ao/AtNt    = 36
  • Tube Bundle Diameter (Db).

  • Db = O.D* (Nt/K)1/nFor square pitch patternn=2.291K=0.156

    (table 12.4 RC vol.6)

     

    Db = 0.216 m

  • Shell Diameter (Ds):

  • Ds = Db+BDCSplit ring floating head is selected. (From Figure 12.10 ,RC volume 6)BDC = 52mm=0.052mDs  = 0.268 m
  • Shell Side Heat Transfer Coefficient:

  • Baffle spacing, lb = DS/5= 0.268/5= 0.0536 mPitch = 1.25 * O.D of TubesPt = 0.020 m

    Cross Flow Area

  • As =0.002886 m2
  • Mass VelocityGs = M/AsGs = 195.945 Kg/s.m2Reynold’s No.

    NRe = Gs.do/µ

    NRe =4.70 *102

    Jh  Factor

    jh = 1.6×10-2         (Fig. 12.31, Vol. 6)

  • Pr = Cpµ/k= 4.67hoc = (jH/do).k.Re.Pr1/3283.936  = W/m .o C
  • Fn, Tube row correction factor

  • (BC) baffle cut is 25% of shell dia     (RC vol.6)Baffle cut height(Hc) = 0.25xDs= 67.16m mHeight between baffle tips (Ds-2Hc)= 0.135 m

    Ncv = 0.135/0.025= 5.4

    From Fig 12.32, RCVol. 6

    Fn =0.98

    From fig12.32 of R.C volume 6 Fn=0.98

  • Fw, Window factor

  • Hb = (Db/2)-Ds(0.5-0.25)= 41.154m mBundle cut, Bb = Hb/Db=0.207Ra’ = 0.16 (Fig 12.41, RC Vol. 6)Tubes in one window, Nw = NtxRa‘

    = 36×0.16 = 6

    Rw =Npx 2Nw/Nt = 0.33

    Fw  = 1.05 (Fig. 12.33, Vol. 6)

  • Fb, Bypass correction factor

  • Ab = lb(Ds-Db) = 0.002782 m2α = 1.35 (for turbulent flow fromRC Vol. 6)Ns/Ncv=1/5(try one slip for each five vertical rows)Fb = exp[-α.(Ab/As)(1-2(Ns/Ncv)1/3]

    = 0.71006(from Fig. 12.34, Vol. 6)

  • Fl, Leakage correction factor

  • AL/AS  =0.5167FL=1-βL(Atb + 2Asb/AL)βL=0.57 (fig12.35, R.C volume 6)Atb=.001490

    Asb =.002887

    FL=0.7069

  • Total shell side coefficient (hs),

  • hs = hoc x Fn x Fb x Fw x Flhs =207.314W/m2 .oC
  • Shell side Pressure Drop

  • ¢Cross flow zone∆Pc = ∆Pi F’b F’LRe = 654.5, for square pitchjf = 8 x 10-2 (Fig. 12.36, Vol. 6)

    us = .2153 m/s

    ∆Pi, ideal tube bank pressure drop

  • ∆Pi = 1616.288 N/m2
  • F’b, bypass correction factor for pressure drop

  • (α=4)  for turbulent flowF’b= exp[-α.(Ab/As)(1-2(Ns/Ncv)1/3]F’b =0.3373
  • F’L, leakage factor for pressure drop

  • β’l = 0.34( Fig. 12.38, Vol.6)F’l = 1-β’l(Atb+2Asb/Al)F’l = 0.7069∆Pc = ∆Pi x F’b x F’l

    ∆Pc = 17.4kPa

  • Window Zone Pressure Drop

  • ∆Pw = F’l(2+0.6Nwu)ρuz2/2For baffle cut 0.25, Ra = 0.16 (Fig. 12.41, Vol. 6)Aw = (πDs2/4 x Ra) – (Nwπdo2/4)= 0.0433m2

    uw = 0.076 m/s

    uz = 0.1279 m/s

    Nwu = Hb/Pt’= 41.154/25=

    ∆Pw = F’l(2+0.6Nwu)ρuz2/2= 15.734 N/m2

    End Zone Pressure Drop

  • Pe=∆Pi[(Nwu+Ncv/Ncv)]Fb= 34.693 N/m2
  • Total Shell Side Pressure Drop(∆Ps )

  • Number of baffles, Nb = 2.44/0.053 – 1Nb = 45∆Ps = 2∆Pe + ∆Pc(Nb – 1) + Nb x ∆Pw=8278.83  N/m2 or 1.2006 Psi
  • Tube side Co-efficient

  • Mean temperature = (20+70)/2 = 45oCTube cross-sectional area = π/4 x d02= 201 mm2As 4 passes are used so number of tubes per pass = 206/4 = 53Total flow area = 53*201*10^-6= 0.01065 m2

    Mass velocity = Flow Rate of Water/ Total Flow Area

    = 354.436 kg/s.m2

  • ρ = 995 kg/m3
  • Water linear velocity,

    ut = water mass velocity/ density of water

    ut = 0.4095 m/s

    hi = 4200(1.35+0.02t)ut0.8/di0.2  (From Rc Vol.6)

    hi = 2491.79 W/m2.oC

  • Overall Heat Transfer Coefficient

  • kw = 45 (stainless steel)hod = 5000 W/m2.o  Chid = 3000  W/m2.o  C              (Table 12.2, 12.6, Vol.6) 

    Uo = 323.084 W/m2. oC

  • Tube side Pressure drop

  • Re =  8149.05 (for tube side)jf =  4 x 10-3 ( Fig. 12.2Vol. 6)Np, Number of passes = 2∆Pt= Np[8jf(L/di)+2.5]ρut2/2

    Pt = 2060 N/m2 or 0.298Psi

  • SPECIFICATION SHEET

  • Heat  Exchanger  Type Shell & Tube
    Equipment  Id E-5(b)
    Pass 1-2  Shell  & Tube
    No. Of  Tubes 36
    Outer Diameter of  Tube 0.025 m
    Inner Diameter of  Tube o.o16 m
    Pitch Type Square Pitch
    Pitch 0.025 m
    Shell Diameter 0.268 m
    Baffle Spacing 0.0536 m
  • Baffle Cut 25 %
    Bundle Cut 0.22
    No. of  Baffles 36
    Shell  Side  Coefficient 207.31 W/m2  0C
    TUBE  Side Coefficient 3241.607 W/m2  0C
    Over all Coefficient Supposed 300 W/m2  0C
    Over all coefficient Supposed 336.08 W/m2  0C
    Shell Side Pressure Drop 1.2006Psi
    Tube Side Pressure Drop 0.298 Psi

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