CONDENSER DESIGN

CALCULATE HEAT DUTY – Q

Shell Side: (cold)                                             Tube Side: (hot)

Water and MEG                                                     Water

Inlet temperature of the process (condensing)stream

= T₁ =T =161.6 ^oF =72 ^oC

Outlet temperature of the process stream

= T₁ =T=161.6 ^oF =72 ^oC

Inlet temperature of the water                = t₁  = 77 ^oF=25 ^oC

Outlet temperature of the water = t₂  = 113^oF=45 ^oC

Mass flow rate of the process stream = m  = 33322 lb/hr= 15128 kg/hr

 of Process Stream    = 1001 B.T.U / lb = 2327.65553 K J / Kg

Q = m * = 15128 * 2327.65553 = 3.52127 *10⁷  KJ/ hr

                  = 1001 * 33322=3.337 *10⁷Btu /hr

C_p of cooling water=1 B.T.U/ lb. hr = 4.187 KJ/kg k

Mass flow rate of cooling water

 = (3.52127 *10⁷ ) /(4.18738 * (45-25))       m= 420310.55 Kg /hr

  =(3.337*10⁷)/(1*(113-77))         m =926944.44 lb / hr

LMTD = ((161.6-77)-(161.6-113)) / Ln ((161.6-77)/(161.6-113))

LMTD = 64.9  ^oF

LMTD = ((72-25)-(72-45)) / Ln ((72-25)/(72-45))

 LMTD =36 ^oC

Caloric Temperature

Tc and tc              Tc=T=161.6 ^oF =72 ^oC

tc=(t1+t2)/2           t₁=77 ^oF  =  25 ^oC                  t₂ =113^oF=45 ^oC

tc=(25+45)/2= 35 ^oC

Overall Design heat transfer Coefficient

Assumed  Ud=  513 BTU/hr/sq.feet/F = 2914.4597W /m² K

ESTIMATED AREA

=Q/(Ud*LMTD)

= 3.52127 *10⁷ / ( 2914.459 * 36 )*1000 / 3600

A = 93.2 m²

TUBE SIDE CALCULATIONS

Tubes: ¾ in. dia  . 14 BWG

Effective Length of the tubes L=5m

O.D of the tubes=0.75 in  = 0.0625   ft   = 0.01905 m

I.D of the tubes =0.584 in = 0.04867 ft = 0.01484 m

SURFACE AREA PER LINEAR FT OF TUBE

 =0.1963 ft2/ft= 2.3556 in2/in =0.0598m²/m

Surface area =at =  5*0.0598 = 0.292 m²

 = 3.1408 ft²

NO OF TUBES

nt=A/at =93.2/0.292 =  318.752

Density of the water ρ=60.2 lb/ft³=964 kg / m³

VOLUMETRIC FLOW RATE OF WATER

V=w/ ρ    V= 420310.55  /(964*3600)   V=0.121m³/sec

VELOCITY OF WATER

  v = 4  ft/sec =1.2 m /sec

TOTAL CROSS SECTIONAL AREA

Ax=V/v     Ax  =0.121/1.2      Ax = 0.10083 m²

Cross Sectional Area of a tube   x=0.546 in² =0.00035226 m² 

TUBES PER PASS

 tp=Ax/x                 tp = 0.10083 / 0.00035226 = 286.2

No of tube pass

np=nt / tp                 np= 318.752/286.2 = 1.1137

Corrected  number of tube pass

 np =1

SHELL AND TUBE SPECIFICATION

Corrected value of number of tubes

Correct nt=361

Corresponding Shell Dia

 D=21.25 in =0.53975 m

Corrected A

A=nt*a t= 361*0.292     A  = 105.412 m²

CORRECTED Ud 

Ud   = Q/(A_cor*LMTD)

Ud  = 3.52127 *10⁷ /(105.412*36)*1000/36000

       =  2570.8   W/m² K=452.96 BTU/ hr ft^{2 o}F

TUBE   CROSS SECTIONAL AREA

a_c=0.268 in²=0.0001729 m²

Total Cross sectional Area

Act=at*nt /np

At=0.0001729*361 / 1    At=0.0624 m²

MASS VELOCITY

Gt=w/At     Gt=420310.55 /0.0624    Gt=6733810.058  kg/ hr m²

VELOCITY

v=Gt/Rhoc

v=6733810.058 /(964*3600)

v=1.940 m/sec  (assumption is true)

CALORIC TEMPERATURE

tc= 35 ^oC

INSIDE TUBE HEAT TRANSFER COEFFICIENT BASED ON THE OUTER DIAMETER                

hi=   1440 BTU/hr/sq.feet/F =8180.928   W/m² K          

hio  = hi*id/OD

hio  =1440*0.584 / 0.75

hio  =1121.3  BTU/hr/sq.feet/F  = 6370.32 W/m² K          

SHELL SIDE CALCULATIONS

Shell Diameter= 21.25 in = 0.53975 m

Pitch =Pt=15/16 in =0.0238 m

B=ID=0.5397 m

Number of baffles  = nb=5/0.5397 =9.26

CORRECTED BAFFLE NUMBER

 Nb =10   O.D of the tubes=   0.75 in  =  0.0625  ft   = 0.01905 m

 I.D of the tubes =   0.584 in =0.04866 ft   = 0.0148 m   

       

CLEARANCE   

C=pt-OD=0.0238-0.01905  = 0.00475 m= 0.1875 in

B=L/nb*12

B= 5/10= 0.5 m =19.68 in

SHELL CROSS SECTIONAL AREA

as=ID*c*B/(pt)

as = 0.53975 * 0.00475*0.625/(0.0238)

as=0.0657968 m²  =0.7083 ft²

Mass Flow rate =wf= 15128 kg/hr

MASS VELOCITY

Gs= wf / as;

Gs = 15128 /0.0657968=229919.996

Gs =229919.996 kg/m²hr

at= 0.292m²  np= 1

CONDENSING LOAD

Gl=w/(L*nt^2/3)

Gl=15128/(5*361^2/3)

Gl=59 kg / m hr =39 lb / hr ft

Dirt Factor

Rd= 0.0005/5.6812    = 0.000088 m² K/W

Overall Clean Coefficient    Uc=Ud/(1-Ud*Rd)     Uc=Ud/(1-Ud*Rd)

Uc=452.96 /(1-452.96 *0.0005)

Uc=585 BTU/hr/sq.feet/F   = 3323.5 W/m² K

hio = 1121.3  BTU/hr/sq.feet/F=  6770.329 W/m² K   = 22933.186 KJ /m² hr

ho=Uc*hio/(hio-Uc)  ho= 1225.7 BTU/hr/sq.feet/F

ho =6963.44 W/m² K

WALL TEMPERATURE

 tw=tc+hio/(hio+ho)*(Tc-tc)

tw=35+24195.06 /(24195.06 +3852.72)*(72-35)     tw = 66

tw = 54.3 ^oC= 129.7818 ^oF    tf=(Tc+tw)/2

tf = (72 + 54.3)/2 = 63.15 ^oC= 145 .6909 ^oC

kf at  tf' = 0.3860 Btu ft / hr /sq.ft / F

sg at tf= 0.9750

μf at tf=0.45 Cp       ho=1800 Btu/hr ft^{2 o}F               

ho =  10119.24 W/m² K

OVER ALL CLEAN COEFFICIENT

Uc=hio*ho/(hio+ho) =326 Btu / hr ft  F =1852.073863 W/m² K

OVER ALL DESING COEFFICIENT

Ud=Uc/(1+Uc*Rd) = 280 Btu ft / hr ft  F= 1590.73 W/m² K

 =5723 KJ /hr m²

PRESSURE DROP (SHELL SIDE)

At   T = 72 ^oC

m_ap =0.00042 kg /m sec =1.512 kg / m hr

De = 0.01397 m   

Re_s= D_eG_s/m= 0.01397 ´ 229919.996 /0.00042 *3600

                    = 2124.32

For   Res  =  2124.32

Friction factor for shell side f = 0.4032 ft²/ ft²

No. of crosses, N + 1  = 11

SHELL SIDE PRESSURE DROP

                   f G_s ² D_s  (N + 1)

Δ P_s =  ½(                                     )

                  2.54x 10¹¹  D_e  S F_s

R_e.s       = 2124.32    =0.0028                                      

D_e           = 0.01396m.        s       = 0.002169

G_s           =229919.996 kg / hr m²

                 0.028*(229919.996) ² x0.53975 x 11

Δ P_s =                                                                  

                 2  x 2.54 x 10¹¹ x 0.01396x 0.002169

      = 571.kg/m²=0.812 psi

PRESSURE DROP (TUBE SIDE)

Reynolds No.

Re_t       = DG/m  = 0.014846733810/(0.000543600 )  = 50469.57

for Re_t = 50469.57

Friction factor for tube side f = 0.02448

            = 

            = 1527 kg / m² =2.17psi

    = (4´n / s)´(v²/2g) =

     =4*1/0.964*0.21=0.87  psi

     =4*1/0.964*147.79 =613.2kg / m²

 =2140 kg /m²     = 3.041 psi