HEAT EXCHANGER DESIGNING

HEAT EXCHANGER

FUNCTION :

           

To cool the dissolving solution from 65 °C to 35 °C before crystallization.

Design :

        Two heat exchangers connected parallel having same specifications will work on half of the total flow rates.

To cool fluid from 63 °C to 35 °C with water being heated from 25 °C to 40 °C.

           

T1 = 63 °C = 145.4 °F                        ,           t1 = 25 °C = 77 °F

T2 = 35 °C = 95 °F                 ,           t2 = 40°C = 104 °F

Flo 0w rate of Fluid                                           

Flow rate of water

Hot		Cold	Difference
145.4 95	High Temp. Low Temp.	104 77	41.4°F =  ΔT2 18 °F =    ΔT1

Heat Balance:

Heat to be removed      =          2891802.84

63 °C                          35 °C

40 °C                           25 °C       (Eq. 5.14 By Kern P / 89)

LMTD Correction :

                                        0.395              From fig. 18 by Kern P / 828)

Ft for 1 ─   2 exchanger = 0.525          (Very Low)

(From fig. 19 by Kern P / 829)

Ft for 2 ─   4 exchanger = 0.92

Corrected LMTD         =       LMTD x  Ft

=       28.04 x  0.92

=       25.85° F

Let Tubes available for service are  &1" Δ Pitch &13BWG

(From Kern, Page 128)

Inner diameter  of tubes           =          0.56 in.

(From Kern, Page 843, Table10)

(Page 843 by Kern)

Assuming U D from Kern, Page840

U D                  =          40

Q                     =          U DA (LMTD)

(From Kern, page 89, eq.5.13)

                                                                                     2

Total length of tubes                              =         

                                                            =                                  14247.14 ft

Let length of each tube                          =                                  16 ft

(From Richardson & Coulson, Vol 6, Page 645)

Number of tubes                       =                                

`                                                                       =                                  890.45 tubes

                                                                                                (From Page 842 by Kern)

For  “OD and 1" Δ Pitch for 4 passes.

Nearest No. of tubes                                        =                      882

(Table 9, Page 842 by Kern)

And internal diameter of Shell is 35 in.

(From Kern, page 842, Table 9)

A         =          NtdoL

=          8820.196316  

=          2770.185 ft2

Corrected UD                       40.38

This is nearly equal to assumed value of U D = 40

In our case the flow rate of cold fluid H2O is less than hot fluid (slurry) , also slurry is corrosive therefore slurry is taken in tubes and water is taken in shell.

Let,

Baffle spacing               =          B          =          12 in.

            (25 % cut segmented baffles)

(By Kern Page 156)

Clearance       =      Cl    =      PT    -    OD

                                                            =          1.0     -    0.75 in

                                                            =          0.25 in

Shell Side: (water, cold fluid)

At,

=          90.5 °F

For Water,

µ          =          0.8cp   =          0.8×2.42

                                   

(Page 822 by Kern)

(By Kern, Table 10 Page800)

(Page 804 by Kern)   

                                                De        =          0.73 in.

(From fig 28, Page 838 By Kern)

=          10434.68

                                    JH    =      58

(From fig 28, Page 838 By Kern)

(Eq. 6.15b by Kern)

Let,

(By Kern, page111)

                                                          =          1

=          1.83

Tube side (Hot Slurry Fluid):

(Table 10, Page 843 by Kern)

(Eq. 7.48 By Kern)

Eq. 7.2 By Kern, Page 138)

=          0.0466ft

=                 120.2 °F

µ              =      4Cp

                                                            =          4 2.42

                                                                        =          1332.22

=          48.75

K         =          0.9 K water

                                                                                                (Kern page 161)

=          0.9  0.37

=          0.333

=          343.35

JH     =       3.7

(From Kern Fig.24, Page 834)

(Eq. 6.15, By Kern, Page 111)

Where,

(From Kern, Page 111)

=          2.51

So,

(From Kern, Page105, and Eq. 6.5)

(Eq. 6.71, Page 106 by Kern)

(Eq. 6.10, Page 107 by Kern)

=          0.0026

Pressure drop:

Shell side:

N Re    =          10434.68

                                                f           =          0.0021

Fig 29, Page 839 by Kern)

                                                 S         =          1

                                                Ds        =         

                                                            =          2.917 ft

No. of crosses                                      =          N + 1

                                        =          16

For 2 – Passes

16 * 2         =       32

(Eq. 7.44, Page 147 by Kern)

=6.82 Psi

6.82 Psi < 10 Psi  (Allowable pressure drop)

Tube side (Slurry, hot fluid)

For,

N Re    =          1332.22

f           =          0.00048

(From Kern, Fig26, Page 836)

(From Pak Arab NP Manual)

(Eq. 7.45, Page 148 by Kern)

=       1.5 Psi

Exit and Entrance Losses:

                            

For,

(Fig. 27, Page 837 by Kern)

(Eq. 7.46, Page 148 by Kern)

=       0.105 Psi

(Eq. 7.47, Page 148 by Kern)

=          1.5 + 0.105

=          1.605

1.605 < 10 Psi (Allowable Pressure Drop)

As,         

 And  Both less than 10 Psi

So,

ASSUMPTIONS FOR DESIGN ARE VALID