# 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
Updated: September 1, 2019 — 6:14 am