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CHAPTER 6

SAMPLE CALCULATIONS FOR SHELL-AND-TUBE HEAT

EXCHANGERS AND GASKETED-PLATE HEAT EXCHANGERS

6.1 ANALYSIS OF SHELL-AND-TUBE HEAT EXCHANGER

6.1.1 Objective-Process

Distilled water with a mass flow rate of 50 kg/s enters a baffled shell-and-

tube heat exchanger at 32ºC and leaves at 25ºC. The heat will be transferred to

150 kg/s of raw water coming from a supply at 20ºC. It is required to design the

heat exchanger for this purpose. A single-shell and single-tube pass is preferable.

The tube diameter is ¾ in. (19-mm O.D. with 16-mm I.D.) and tubes are laid out

on 1-in. square pitch. The maximum length of the heat exchanger is 8 m because

of space limitations. The tube material is 0.5 Cr alloy (k=42.3 W/m.K). Assume

total fouling resistance of 0.000176 m2.K/W. Note that the surface over design

should not exceed 30%. The maximum flow velocity through the tube is also

suggested to be 2 m/s to prevent erosion. Assume baffle spacing of 0.5 m.

Perform thermal and hydraulic analysis of the heat exchanger.

145



6.1.2 Process Specifications

Tube-Side Data

Cold Fluid (raw water)

Inlet temperature (ºC) 20

Mass flow rate (kg/s) 150

Density (kg/m3) 998.20

Thermal conductivity (W/m.K) 0.6044

Dynamic viscosity (N.s/m2) 9.832 x 10-4

Specific heat (J/kg.K) 4181.6

Prandtl number 6.808

Shell-Side Data

Hot Fluid (distilled water)


Outlet temperature (ºC) 25

Average temperature (ºC) 28.5

Mass flow rate (kg/s) 50






6.1.3 Design of Shell-and-Tube Heat Exchanger

6.1.3.1 Sample Calculation by Kern Method

As given in section 6.1.1, a single-shell and single-tube pass, shell-and-

tube heat exchanger, where tubes are laid out on 1-inch square pitch, is to be

designed. The hot fluid (distilled water) is in the shell and the cold fluid (raw

water) is flowing in the tubes.

146



Properties of hot water (shell-side):

Water properties are calculated at ( ) 5.2822532 =+ ºC and are given in section

6.1.2.

Properties of cold water (tube-side):Water properties are calculated at 21 ºC and are given in section 6.1.2.

Tube:

31016 −×=id m tube inside diameter

31019 −×=od m tube outside diameter

1= p N number of tube passes

The tube material is 0.5 Cr alloy:

3.42=t k W/m.K tube thermal conductivity

Tube layout:

Tubes are laid out on 1-inch square pitch:

21054.2 −×=T P m

Total fouling:

000176.0= ft R m2.K/W

TUBE-SIDE CALCULATIONS:Number of tubes:

From Eq. (4.2),

692.373016.022.998

11504422=

×××

××==

π π ρ icc

pc

t d u

N m N

Number of tubes is rounded off as 374.

374=t N tubes

Tube-side Reynolds number:

From Eq. (4.3),

4

410249.3

10832.9

016.022.998Re ×=

×

××==

−c

icct

d u

µ

ρ

Since , the flow is turbulent.10000Re >t

Tube-side Nusselt number:

The Nusselt number can be calculated from Eq. (4.4) as

147



−

+

=

1Pr 2

7.1207.1

Pr Re2

3

22

1

c

ct

t

f

f

Nu

where can be calculated from Eq. (4.5) as f

( ) 3210797.528.3Reln58.1 −−

×=−= t f

Therefore,

552.225

1808.6

2

10797.57.1207.1

808.610249.32

10797.5

3

22

13

43

=

−

×+

×××

×

=−

−

t Nu

Tube-side heat transfer coefficient:

From Eq. (4.7),

2.8520016.0

6044.0552.225 =×==

i

ct i

d

k Nuh W/m2.K

SHELL-SIDE CALCULATIONS:

The total flow area through the tubes is:

0752.0374016.044

22 =××== π π t it N d A m

2

Shell inside diameter:

The tube pitch ratio is given by Eq. (4.11) as

337.1019.0

0254.0===

o

T

d

P PR

From section 4.2.2.1, the tube count calculation constant for one-tube pass is

93.0=CTP

and the tube layout constant for square pitch is

0.1=CL

Then the shell diameter is estimated from Eq. (4.12) as

( ) ( )2

122

2

122 374337.1019.0

93.0

0.1637.0637.0 ×××××== π π t o s N PRd

CTP

CL D

575.0= s D m

Shell inside diameter is rounded off as 580 mm.

148



580.0= s D m

Shell equivalent diameter:

From Eq. (4.15), the shell equivalent diameter for square pitch is:

0242.0019.0

4019.00254.04

44

22

22

=×

×−

=

−

=π

π

π

π

o

oT

ed

d P

D m

The baffle spacing is assumed to be

5.0= B m

From Eq. (4.19), the tube clearance is:

3104.6019.00254.0 −×=−=−= oT d P C m

The bundle cross flow area is calculated from Eq. (4.18) as

0731.05.0104.60254.0

580.0 3 =×××== − BC P

D A

T

s s m2

Shell-side Reynolds number:

Shell-side Reynolds number is calculated from Eq. (4.21) as

4

410999.1

10292.8

0242.0

0731.0

50Re ×=

××=

=

−h

e

s

h s

D

A

m

µ

Therefore, the flow is turbulent.Shell-side heat transfer coefficient:

Shell-side heat transfer coefficient is calculated from Eq. (4.22) as

( ) 3

155.043

1

55.0 641.510999.10242.0

6144.036.0Pr Re

36.0×××

×== h s

e

ho

D

k h

9.3769=oh W/m2.K

Overall heat transfer coefficient:

Clean overall heat transfer coefficient based on the outer surface area is calculated

from Eq. (4.23) as

11

3.42

016.0

019.0ln

2

019.0

2.8520

1

016.0

019.0

9.3769

1ln

2

11

−−

+×+=

++=t

i

o

o

ii

o

o

ck

d

d

d

hd

d

hU

2.2256=c

U W/m2.K

149



Fouled overall heat transfer coefficient based on the outer surface area is

calculated from Eq. (4.24) as

1

ln

211

−

+++=t

i

o

o ft

ii

o

o

f k

d

d

d Rhd

d h

U

1

4

3.42

016.0

019.0ln

2

019.01076.1

2.8520

1

016.0

019.0

9.3769

1

−

−

+×+×+= f U

9.1614= f U W/m2.K

Outlet temperature calculation:

The cold water outlet temperature is calculated from Eq. (4.25) as

( ) ( )33.295293

6.4181150

2983057.4178501

21

2 =+×

−××=+

−= c

pcc

hh phh

c T cm

T T cmT

K

LMTD calculations:

For counter flow:

67.933.295305211 =−=−=∆ ch T T T K

5293298122 =−=−=∆ ch T T T K

From Eq. (4.27), the log-mean temperature difference is:

08.7

5

67.9ln

567.9

ln2

1

21 =

−=

∆

∆

∆−∆=

T

T

T T LMTD K

From Eq. (3.25),

19.0293305

29333.295

11

12 =−

−=

−

−=

ch

cc

T T

T T P

From Eq. (3.26),

00.329333.295

298305

12

21 =−

−=

−

−=

cc

hh

T T

T T R

Therefore, the correction factor is assumed to be as LMTD

94.0= F

150



The mean temperature difference is therefore

66.608.794.0)( =×==∆ LMTD F T m K

Heat balance for fluids:

The heat duty of the heat exchanger is:

( )( ) ( )( ) 6

21 10463.12983057.417850 ×=−×=−= hh phh T T cmQ W

From Eq. (4.26a), the surface area of the heat exchanger for the fouled condition

is:

09.13666.69.1614

10463.1 6

=×

×=

∆=

m f

f T U

Q A m2

Similarly, from Eq. (4.26b), the surface area of the heat exchanger for the clean

condition is:

41.9766.62.2256

10463.1 6

=×

×=

∆=

mc

cT U

Q A m2

The over surface design is:

40.1==c

f

A

AOS (40%)

The over surface design may not be more than 30% for economical reasons.

Assume 20% over surface design; then cleaning scheduling must be arranged

accordingly:

1.188020.1

2.2256

20.1=== c

f

U U W/m2.K

From Eq. (4.24), the corresponding total resistance will be:

510865.82.2256

1

1.1880

111 −×=−=−=c f

ft U U

R m2.K/W

For 20% over surface design, the surface area of the heat exchanger becomes:89.11641.9720.120.1 =×== c f A A m2

Length of the heat exchanger:

From Eq. (4.30), the length of the heat exchanger is:

24.5019.0374

89.116=

××==

π π ot

f

d N

A L m

Length of the heat exchanger is rounded off as 5.5 m.

151



5.5= L m

Shell diameter:

Shell diameter can be calculated from Eq. (4.31) as

5.02

5.02

5.5019.034.189.116

93.00.1637.0637.0

×××=

=

Ld PR A

CTP CL D

o f

s

56.0= s D m

Shell diameter is rounded off as 0.60 m.

62.23m60.0 == s D in

Tube-side pressure drop:

The tube-side pressure drop can be calculated from Eq. (4.32):

24

4 2

cc p

i

pt

t

u N

d

LN f p ρ

+=∆

where ( )( ) ( )( ) 324210797.528.310249.3ln58.128.3Reln58.1 −−−

×=−××=−= t t f

and 1= p N

423

1039.22

22.99814

016.0

15.510797.54×=××

×+

××××=∆

−

t p Pa

9.23=∆ t p kPa

Shell-side pressure drop:

The shell-side pressure drop can be calculated from Eq. (4.33):

( )

seh

sb s s s

D

D N G f p

φ ρ 2

12 +=∆

where the friction factor is calculated from Eq. (4.34) as s f

( )( ) ( )( ) 271.010999.1ln19.0576.0expReln19.0576.0exp 4 =××−=−= s s f

The bundle cross flow area is calculated from Eq. (4.18) as

0756.05.0104.60254.0

6.0 3 =×××== − BC P

D A

T

s s m

2

The shell-side mass velocity is:

46.6610756.0

50===

s

h s

A

mG

kg/m2.s

The wall temperature is:

152



83.2972

298305

2

33.295293

2

1

222

1 2121 =

++

+×=

++

+= hhcc

w

T T T T T K

The dynamic viscosity at the wall temperature is:

410001.9 −×=w

µ kg/m.s

989.010001.9

10292.814.0

4

414.0

=

×

×=

=

−

−

w

h s

µ

µ φ

The number of baffles is:

1015.0

5.51 =−=−=

B

L N b

Hence, the pressure drop in the shell-side is:

( ) 4

2

1064.1989.00242.041.9962

60.011046.661271.0 ×=×××

×+××=∆ s p Pa

4.16=∆ s p kPa

Tube-side pumping power:

Assume pump efficiency as 0.80.

80.0= pη

Tube-side pumping power is:

341049.4

80.02.998

1039.2150 ×=×

××=∆= pc

t ct

pm P η ρ

W

49.4=t P kW

Shell-side pumping power:


80.0= pη

Shell-side pumping power is:

34

1003.180.041.996

1064.150×=

×

××=

∆=

ph

sh s

pm P

η ρ

W

03.1= s P kW

The selected shell-and-tube heat exchanger for this purpose has the following

parameters which will be rated by the Bell-Delaware method:

153



Table 6.1 Results of Kern Method

Shell diameter (m) 0.600

Number of tubes 374

Length of the heat exchanger (m) 5.5

Tube outside diameter (m) 0.019

Tube inside diameter (m) 0.016

Baffle spacing (baffle cut 25%) (m) 0.5

Tube pitch (m) 0.0254

Number of passes 1

Tube-side heat transfer coefficient (W/m2.K) 8520

Shell-side heat transfer coefficient (W/m2.K) 3770

Clean overall heat transfer coefficient (W/m2.K) 2256

Fouled overall heat transfer coefficient (W/m2.K) 1880

Tube-side pressure drop (kPa) 23.9

Shell-side pressure drop (kPa) 16.4

Tube-side pumping power (kW) 4.5

Shell-side pumping power (kW) 1.0

According to the given specifications, the size of the heat exchanger was

estimated. Such a heat exchanger may also be available. Then this heat exchanger

can be rated for design by the use of Bell-Delaware method which will be given

next.

6.1.3.2 Rating Problem by Bell-Delaware Method

Input data:

Heat exchanger to be rated by Bell-Delaware method is given in the following

table:

154



Table 6.2 Input data in Bell-Delaware method

Tube inside diameter (m) 0.016 Specific heat of the shell-side

fluid (J/kg.K)

4178.7

Tube outside diameter (m) 0.019 Dynamic viscosity of the shell-

side fluid (kg/m.s)

410292.8

−×

Tube pitch (m) 0.0254 Dynamic viscosity of the shell-

side fluid evaluated at wall

surface temperature (kg/m.s)

410001.9

−×

Shell inside diameter (m) 0.600 Thermal conductivity of the

shell-side fluid (W/m.K)

0.6144

Length of the heat exchanger

(m)

5.5 Prandtl number of the shell-side

fluid

5.641

Baffle spacing (m) 0.5 Density of the shell-side fluid

(kg/m³

)

996.41

25% baffle cut as percent of

the shell inside diameter

25 Specific heat of the tube-side

fluid (J/kg.K)

4181.6

Clearance between two

adjacent tubes (m)

3104.6

−×

Dynamic viscosity of the tube-

side fluid (kg/m.s)

410832.9

−×

Inside shell diameter to tube

bundle bypass clearance, for

fixed tube sheet (m) [3]

31015

−×

Thermal conductivity of the

tube-side fluid (W/m.K)

0.6044

Central baffle spacing (m) 0.5 Thermal conductivity of the

tube material (W/m.K)

42.3

Inlet baffle spacing (m) 0.5 Prandtl number of the tube-side

fluid

6.808

Outlet baffle spacing (m) 0.5 Density of the tube-side fluid

(kg/m³)

998.20

Number of passes 1 Velocity of the tube-side fluid

(m/s)

2

Number of tubes 374 Inlet temperature of the hot fluid

(K)

305

Over surface design 0.20 Outlet temperature of the hot

fluid (K)

298

Mass flow rate of the shell-

side fluid (kg/s)

50 Inlet temperature of the cold

fluid (K)

293

Mass flow rate of the tube-

side fluid (kg/s)

150 Outlet temperature of the cold

fluid (K)

295.3

155



Shell-side Reynolds number:

From Eq. (4.52), the diameter of the circle through the centers of the tube located

within the outermost tubes is:

( ) ( ) 566.0019.0015.06.0 =+−=+−=obb sctl

d L D D m

From section 4.3.4 (1), for 90º layout,

0254.0, == T eff T P P m

From Eq. (4.53), the cross flow area at the shell centerline with one baffle spacing

is:

( ) ( )

2

,

m079.0

019.00254.00254.0

566.0015.05.0

=

−+×=

−+=

m

oT

eff T

ctl bbm

S

d P P

D L BS

The maximum shell-side cross flow mass velocity is:

46.634079.0

50===

m

s s

S

m M

kg/m2.s

Then the Reynolds number on the shell-side is:

4

41045.1

10292.8

46.634019.0Re ×=

×

×==

− s

so s

M d

µ

Ideal shell-side heat transfer coefficient:

From Table (4.2),

370.01 =a 395.02 −=a 187.13 =a 370.04 =a

203.0Re14.01 4

3 =+

=a

s

aa

The Colburn j-factor for an ideal tube bank from Eq. (4.101) is:

( ) 3395.04

203.0

1 10386.81045.1

019.0

0254.0

33.1370.0Re33.1 2 −−×=××

×=

= a

s

a

o

T

i

d

P a j

The cross flow area at the centerline of the shell for one cross flow between two

baffles is, from Eq. (4.18):

0756.05.0104.60254.0

6.0 3 =×××== − BC P

D A

T

s s m2

From Eq. (4.100), the ideal tube bank-based coefficient is:

156



14.03

2

=

w

s

s ps

s

s

s psiid

c

k

A

mc jh

µ

µ

µ

14.0

4

43

2

43

10001.910292.8

10292.87.41786144.0

0756.0507.417810386.8

×

××

×××

×××= −

−

−−id h

4.7232=id h W/m2.K

Segmental baffle window correction factor,c

J :

From Eq. (4.61), the angle intersecting the diameter of the circle through the

centers of the outermost tubes is:

1161002521

566.06.0cos2

10021cos2 11

=

×−××=

−= −− c

ctl

sctl B

D Dθ deg

From Eq. (4.64), the fraction of number of tubes in one baffle window is:

( ) ( )179.0

2

116sin

360

0.116

2

sin

360=

×−=−=

π π

θ θ ctl ctl w F

From Eq. (4.65), the fraction of number of tubes in pure cross flow between the

baffle cut tips is:

642.0179.02121 =×−=−= wc F F

From Eq. (4.79), the segmental baffle window correction factor is:

012.1642.072.055.072.055.0 =×+=+= cc F J

Baffle leakage effect correction factor,l

J :

From Eq. (4.60), the centri-angle of the baffle cut intersection with the inside shell

wall is:

120

100

2521cos2

100

21cos2 11 =

×−×=

−= −− c

ds

Bθ deg

From Eq. (4.50), the inside shell-to-baffle clearance (diametral) is:

333 105.56.0004.0101.3004.0101.3 −−− ×=×+×=+×= s sb D L m

From Eq. (4.76), the shell-to-baffle leakage area is:

( ) ( )120360105.56.000436.036000436.0 3 −××××=−= −ds sb s sb L DS θ

310453.3 −×= sbS m2

From Fig. (4.14), the diametral tube-to-baffle hole clearance for the maximum

157



unsupported length greater than 36 in (the maximum unsupported length is greater

than 36 in, from Eq. (4.45) or from Fig. (4.8)) is:

64

1=tb L in = m4

10969.3 −×

From Eq. (4.77), the tube-to-baffle hole leakage area for one baffle is:

( )[ ] ( )

−−+= wt otbotb F N d Ld S 14

22π

( )[ ] ( ) 32 10675.3179.01374019. −×=

−××24 010969.3019.0

4

−

−×+×=π

tbS m2

From Eq. (4.80), the ratio of both leakage areas to the cross flow area is:

090.0

079.0

10675.310453.3 33

=×+×

=+

=−−

m

tb sblm

S

S S r

From Eq. (4.81), the ratio of the shell-to-baffle leakage area to the sum of both

leakage areas is:

485.010675.310453.3

10453.333

3

=×+×

×=

+=

−−

−

tb sb

sb s

S S

S r

From Eq. (4.82a), the baffle leakage heat transfer correction factor is:

( ) ( )[ ] ( )lm s sl r r r J 2.2exp144.01144.0 −−−+−=

( ) ( )[ ] ( ) 861.0090.02.2exp485.0144.01485.0144.0 =×−×−×−+−×=l J

Bundle bypass effect correction factor, :b

J

The effect of the tube lane partition bypass width is taken as the half of the tube

outside diameter:

3105.92

019.0

2

−×=== o pl

d L m

From Eq. (4.74), the bypass area within one baffle is:

( ) pl otl sb L D D BS +−=

where 585.0019.0566.0 =+=+= octl otl d D D m

( ) 0123.0105.9585.06.05.0 3 =×+−×= −bS m2

From Eq. (4.75), the fraction of the bypass area to the overall cross flow area is:

155.00788.0

0123.0===

m

b sbp

S

S F

From Table (4.1), the length for square pitch layout is: p P

158



0254.0== T p P P m

From Eq. (4.70), the number of effective rows crossed in one cross flow section,

that is, between the baffle tips, is:

811.111002521

0254.06.0

10021 =

×−×=

−= c

p

stcc

B P D N

ss N is the number of sealing strips (pairs) in one baffle and it is assumed to be 1:

1= ss N

From Eq. (4.83), the ratio of the number of sealing strips (pairs) in one baffle to

the number of tube rows crossed between baffle tips in one baffle section, is:

0847.0811.11

1===

tcc

ss

ss N

N r

From Eq. (4.84), the correction factor for bundle bypass is:

)3 21exp ss sbpbhb r F C J −−=

where for turbulent flow. Therefore,25.1=bhC

) 917.00847.021155.025.1exp 3 =×−××−=b J

Correction factor for variable baffle spacing at the inlet and/or outlet

sections, :s J

The inlet baffle spacing and the outlet baffle spacing are equal to the central baffle

spacing:

oi B B B ==

From Eqs. (4.92) and (4.93), the dimensionless length ratios are:

B

B L i

i = and B

Boo = L


1015.0

5.51 =−=−=

B

L N b

From Eq. (4.95), the correction factor for unequal baffle spacing at inlet and/or

outlet, is:

( ) ( ) ( )

( ) oib

n

o

n

ib s

L L N

L L N J

++−

++−=

−−

1

1 11

where 6.0=n for turbulent flow. Therefore,

159



( ) ( ) ( )

( )0.1

11110

11110 6.016.01

=++−

++−=

−−

s J

Correction factor for adverse temperature gradient in laminar flow,r

J :

From Eq. (4.72), the effective number of tube rows crossed is:

189.42

566.06.0

100

256.0

0254.0

8.0

2100

8.0=

−−

××=

−−

= ctl sc

s

p

tcw

D D B D

P N

From Eq. (4.87), the total number of tube rows crossed in the entire heat

exchanger is:

( )( ) ( ) ( ) 176110189.4811.111 =+×+=++= btcwtccc N N N N

Since , the correction factor for adverse temperature gradient in laminar

flow, is:

100Re > s

0.1=r J

Shell-side heat transfer coefficient (Bell-Delaware method):

From Eq. (4.99), the actual shell-side heat transfer coefficient is:

( ) ( ) 57750.10.1917.0861.0012.14.7232 =×××××== r sbl cid o J J J J J hh W/m2.K

Overall heat transfer coefficient:

Clean overall heat transfer coefficient based on the outer surface area is:

11

3.42

016.0

019.0ln

2

019.0

2.8520

1

016.0

019.0

5775

1ln

2

11

−−

×+×+=

++=k

d

d

d

hd

d

hU

i

o

o

ii

o

o

c

Table 6.3 Correction factors in Bell-Delaware method

Correction factors Values

c J 1.012

l J 0.861

b J 0.917

s J 1.0

r J 1.0

160



9.2847=cU W/m2.K

1

+=d d

hU

o

f

5775

1

= f U

1.1897= f

U

10463.1 ×=Q

∆=

m f

f T U

Q A

∆=

mc

cT U

Q A

==c

f

A

AOS

2.1== c

f

U U

510023.79.2847

1

3.2373

111 −×=−=−=c f

ft U U

R

Fouled overall heat transfer coefficient based on the outer surface area is:

1

ln

21

−

++k d

d

d Rh

i

o

o ft

ii

o

1

4

3.42

016.0

019.0ln

2

019.01076.1

2.8520

1

016.0

019.0

−

−

×+×+×+

W/m2.K

The heat duty of the heat exchanger and the mean temperature difference were

already calculated as:

6 W and 66.6=∆ mT K

Therefore, the surface area of the heat exchanger for the fouled condition is:

85.11566.61.1897

10463.1 6

=×

×= m2

The surface area of the heat exchanger for the clean condition is:

17.7766.69.2847

10463.1 6

=×

×= m2

The over surface design is:

50.117.77

85.115= (50%)

The over surface design should not be more than 30% for economical reasons.

Assume 20% over surface design; then cleaning scheduling must be arranged

accordingly:

3.23732.1

9.2847= W/m2.K

The corresponding total resistance will be:

m2.K/W

For 20% over surface design, the surface area of the heat exchanger becomes:

161



60.9217.7720.120.1 =×== c f A A m2

Length of the heat exchanger:

15.4

019.0374

60.92=

××

==

π π ot

f

d N

A L m

Length of the heat exchanger is rounded off as 4.5 m.

5.4= L m

Tube-side pressure drop:

24

4 2

cc p

i

pt

t

u N

d

LN f p ρ

+=∆

where ( )( ) ( )( ) 324210797.528.310249.3ln58.128.3Reln58.1 −−−

×=−××=−= t t f

Therefore, the tube-side pressure drop is:

423

1010.22

22.99814

016.0

15.410797.54×=××

×+

××××=∆

−

t p Pa

0.21=∆ t p kPa

Shell-side pressure drop (Bell-Delaware method):

From Table (4.2),

391.01 =b 148.02 −=b 30.63 =b 378.04 =b

( )009.1

10454.114.01

30.6

Re14.01378.04

3

4

=××+

=+

=b

s

bb

From Eq. (4.102),

( ) ( ) 0942.010454.1

019.0

0254.0

33.1391.0Re

33.1 148.04

009.1

12 =××

×=

=

−b

s

b

o

T

i

d

P b f

From Eq. (4.104), the pressure drop in an equivalent ideal tube bank in one baffle

compartment of central baffle spacing, is:

811.1110292.8

10001.9

41.9962

46.6340942.04

24

14.0

4

4214.0

2

×

×

××

×××=

=∆

−

−

tcc

s

w

s

sibi N M

f pµ

µ

ρ

98.908=∆ bi p Pa kPa91.0=

From Eq. (4.82b), the correction factor for baffle leakage effects for pressurel R

162



drop, is:

( )[ ] p

lm sl r r R +−= 133.1exp

where ( )[ ] ( )[ ] 577.08.0485.0115.08.0115.0 =++×−=++−= sr p

Therefore,

( ) 611.00904.0485.0133.1exp 577.0 =×+×−=l R

From Eq. (4.85), the correction factor for bundle bypass effects for pressure

drop, is:

b R

)3 21exp ss sbpbpb r F C R −−=

where for turbulent flow. Therefore,7.3=bpC

) 773.00847.021155.07.3exp 3=×−××−=b R


815.0

5.41 =−=−=

B

L N b

From Eq. (4.103), the combined pressure drop of all the interior cross flow

sections (baffle tip to baffle tip), is:

( ) ( ) 31001.3773.0611.01898.9081 ×=××−×=−∆=∆ bl bbic R R N p p Pa

01.3=∆ c p kPa

From Eq. (4.63), the gross window flow area, that is, without tubes in the window,

is:

( ) ( )0553.0

2

120sin

360

1206.0

42

sin

3604

22 =

×−××=

−=

π

π

π

θ θ π dsds swg DS m2

From Eq. (4.66), the segmental baffle window area occupied by the tubes, is:

019.0019.04

179.03744

22 =

×××=

=

π π owt wt d F N S m2

From Eq. (4.68), the net cross flow area through one baffle window, is:

0363.0019.00553.0 =−=−= wt wg w S S S m2

From Eq. (4.105),

04.9350363.0079.0

50=

×==

wm

sw

S S

mm

kg/m2.s

From Eq. (4.106), the combined pressure drop in all the windows for turbulent

163



flow, is:

( ) ( ) 611.041.9962

04.935189.46.028

26.02

22

×

×××+×=

+=∆ l

s

wtcwbw R

m N N p

ρ

31068.9 ×=∆ w p 68.9=Pa kPa

From Eq. (4.98), the pressure drop correction factor for unequal baffle spacing

at inlet and/or outlet, is:

s R

n

i

n

o

s B

B

B

B R

−−

+

=

22

where 2=n for turbulent flow. Therefore,

25.0

5.0

5.0

5.02222

=

+

=

−−

s R

Since all the baffle spacings are equal, 2= s R .

From Eq. (4.108), the pressure drop in the entrance and exit sections (in the two

end zones), is:

31090.12773.0811.11

189.4198.9081 ×=××

+×=

+∆=∆ sb

tcc

tcwbie R R

N

N p p Pa

90.1=∆ e p kPa

From Eq. (4.109), the total shell-side pressure drop is:

59.1490.168.901.3 =++=∆+∆+∆=∆ ewc s p p p p kPa

Tube-side pumping power:


80.0= pη

Tube-side pumping power is:

3

4

1094.380.02.998101.2150 ×=

×××=∆=

pc

t ct

pm P η ρ

W

94.3=t P kW

Shell-side pumping power:


80.0= pη

Shell-side pumping power is:

164



8.91480.041.996

10459.150 4

=×

××=

∆=

ph

sh s

pm P

η ρ

W

91.0= s P kW

Results of the rating analysis of the shell-and-tube heat exchanger that was

predicted by the Kern method (or available) are given in the following table:

Table 6.4 Results of the Bell-Delaware Method

Shell diameter (m) 0.600

Number of tubes 374


Tube outside diameter (m) 0.019

Tube inside diameter (m) 0.016

Baffle spacing (baffle cut 25%) (m) 0.5

Tube pitch (m) 0.0254

Number of pass 1

Surface area of the heat exchanger (m2) 92.6

Tube-side heat transfer coefficient (W/m2.K) 8520

Shell-side heat transfer coefficient (W/m2.K) 5775



Tube-side pressure drop (kPa) 21.0

Shell-side pressure drop (kPa) 14.6

Tube-side pumping power (kW) 3.9

Shell-side pumping power (kW) 0.9

There are two kinds of rating program. In the first case, the inlet and the

outlet temperatures (3 temperatures) and the mass flow rates are given. Therefore,

the heat transfer (heat load or heat duty) is fixed. Then the length of the heat

exchanger and the pressure drops for both streams are calculated, using log-mean

165



temperature difference method, as done in this thesis. In the second case, the heat

duty is not known because only inlet temperatures are given while the outlet

temperatures are not. On the other hand, the heat exchanger length is fixed and

outlet temperatures and the pressure drops are to be calculated. In this case, outlet

temperatures can be calculated by using ε-NTU method. One of the constraints in

this problem may be a required outlet temperature and the pressure drop ∆P. For

instance, if it is a water heater, outlet temperature may be less than 40ºC. The

outlet temperatures must then satisfy this process requirement. Note that in both

cases, the pressure drop requirement must be satisfied. Do not forget that Heat

Exchanger is a part of the existing system, part of a piping circuit!

6.2 ANALYSIS OF GASKETED-PLATE HEAT EXCHANGER

6.2.1 Objective-Process

Distilled water with a mass flow rate of 50 kg/s enters a gasketed-plate

heat exchanger at 32ºC and leaves at 25ºC. The heat will be transferred to 150

kg/s of raw water coming from a supply at 20ºC. The maximum permissible

pressure drop for each stream is 70 kPa. It is required to design the heat exchanger

for this purpose.

6.2.2 Process Specifications

Cold Fluid Data

Cold Fluid (raw water)


Flow rate (kg/s) 150



Dynamic viscosity (N.s/m2) 9.832 x 10

-4



166



Hot Fluid Data

Hot Fluid (distilled water)


Outlet temperature (ºC) 25Average temperature (ºC) 28.5

Flow rate (kg/s) 50






Constructional Data of the Proposed Plate Heat Exchanger

Plate material SS316

Plate thickness (mm) 0.6

Chevron angle (degrees) 45

Total number of plates 105

Enlargement factor 1.25

Number of passes single pass

Overall heat transfer coefficient (clean/fouled) (W/m2.K) 8000/4500

Total effective area (m2)

Fouling resistance (hot fluid/cold fluid) (m2.K/W)

110

0.00005/0

All port diameters (mm) 200

Compressed plate pack length, Lc, (m) 0.38

Vertical port distance, Lv, (m) 1.55

Horizontal port distance, Lh, (m) 0.43

Effective channel width, Lw, (m) 0.63

Thermal conductivity of the plate material (SS304) (W/m.K) 17.5

After giving the process requirements, this proposed heat exchanger may

be given by the manufacturer; or this heat exchanger may be available. So this

proposed one must be rated for the final design.

167



6.2.3 Rating/Design of a Gasketed-Plate Heat Exchanger

The properties of hot water and cold water are given in section 6.2.2.

Since the inlet temperature and the outlet temperature of the hot fluid are known,

from Eq. (5.32), the required heat duty of the gasketed-plate heat exchanger is:

h phhrh T cmQ ∆=

where 729830521 =−=−=∆ hhh T T T K

Therefore,

610463.177.417850 ×=××=rhQ W

The outlet temperature of the cold fluid is therefore:

33.2956.4181150

10463.1293

6

12 =×

×+=+=

pcc

rhcc

cm

QT T

K

From Eq. (5.28), for counter flow arrangement the logarithmic mean temperature

difference is:

∆

∆

∆−∆=∆

2

1

21,

ln

T

T

T T T cf lm

where 5293298121 =−=−=∆ ch T T T K

and 67.933.295305212 =−=−=∆ ch T T T K

Therefore,

08.7

67.9

5ln

67.95, =

−=∆ cf lmT K

The effective number of plates is:

10321052 =−=−= t e N N

Effective flow length between the vertical ports is:

55.1== veff L L m

From Eq. (5.7), the plate pitch is:

31062.3

105

38.0 −×===t

c

N

L p m

168



From Eq. (5.6), the mean channel flow gap is:

3331002.3106.01062.3 −−− ×=×−×=−= t pb m

From Eq. (5.8), one channel-flow area is:

33

1090.163.01002.3−−

×=××==wch bL A m

2

The single plate heat transfer area is:

068.1103

1101 ===

e

e

N

A A m

2

From Eq. (5.4),

35.12.055.1 =−=−= pv p D L L m

From Eq. (5.3), the projected plate area is:

851.063.035.11 =×== w p p L L A m2

Enlargement factor has been specified by the manufacturer, but it can be verified

from Eq. (5.2):

256.1851.0

068.1

1

1 === p A

Aφ

From Eq. (5.11), the channel equivalent diameter is:

33

1081.4256.1

1002.322 −−

×=××

==φ

b

De m

From Eq. (5.22), the number of channels per pass is:

5212

1105

2

1=

×

−=

−=

p

t cp

N

N N

Heat transfer analysis:

The mass flow rate per channel is:

96.052

50===

cp

h

ch N

mm

kg/s

From Eq. (5.21), the mass velocity is:

54.5051090.1

96.03=

×==

−ch

chch

A

mG

kg/m2.s

From Eq. (5.20), the hot fluid Reynolds number is:

7.293110292.8

1081.454.505Re

4

3

=×

××==

−

−

h

echh

DG

µ

169



Similarly, the cold fluid Reynolds number is:

5.247210832.9

1081.454.505Re

4

3

=×

××==

−

−

c

echc

DG

µ

Therefore, both fluids are in turbulent flow.

From Eq. (5.18), the hot fluid Nusselt number is:

17.0

3

1

Pr Re

=

w

bh

y

hhh C Nuµ

µ

where from Table (5.2), 3.0=hC and 663.0= y . Assuming wb = ,

( ) 24.1061641.57.29313.017.03

1

663.0 =×××=h Nu

From Eq. (5.19), the hot fluid heat transfer coefficient is:

135741081.4

6144.024.1063

=×

×==

−e

hhh

D

k Nuh W/m

2.K

Similarly, the cold fluid Nusselt number is:

( ) 03.1011808.65.24723.0Pr Re17.03

1

663.0

17.0

3

1

=×××=

=

w

bc

y

chc C Nuµ

µ

The cold fluid heat transfer coefficient is:

126981081.4

6044.003.101 3 =×

×== −

e

ccc

Dk Nuh W/m2.K

Let’s calculate the heat transfer coefficients from Eq. (5.14) just to see and

compare the results with the above ones.

[ ] ( )[ ]14.0

3

17.390/2sin0543.0728.025 Pr Re10244.7006967.02668.0

×+−=

++−

w

b

hhh Nuµ

µ β β

πβ

Again assuming wb = ,

[ ]( )[ ] ( ) 14.03

1

7.390/785.02sin0543.0728.0

25

1641.52931.7

785.010244.7785.0006967.02668.0

××

×××+×−=

+×××+

−

π

h Nu

18.121=h Nu

The hot fluid heat transfer coefficient is:

154831081.4

6144.018.1213

=×

×==

−e

hhh

D

k Nuh W/m2.K

Similarly, the cold fluid Nusselt number is:

170



[ ] ( )[ ]14.0

3

17.390/2sin0543.0728.025 Pr Re10244.7006967.02668.0

×+−=

++−

w

b

ccc Nuµ

µ β β

πβ

Again assuming wb = ,

( )[ ] ( ) 14.07.390/785.02sin0543.0728.0

25

1641.52472.5

785.010244.7785.0006967.02668.0

××

×××+×−=+×××+

−

π

c Nu

58.114=c Nu

The cold fluid heat transfer coefficient is:

144011081.4

6044.058.1143

=×

×==

−e

ccc

D

k Nuh W/m2.K

The hot fluid and the cold fluid heat transfer coefficients calculated from two

different correlations differ from each other by %14 and %13, respectively (the

second ones are %14 and %13 higher). The first results will be used here.

From Eq. (5.30), the clean overall heat transfer coefficient is:

5356

5.17

106

13574

1

12698

1

1

11

14

=

×++

=

++

=−

platehc

c

k

t

hh

U W/m2.K

From Eq. (5.31), the fouled (service) overall heat transfer coefficient is:

4225

01055356

11

1

1

5

=

+×+

=

++

=−

fc fh

c

f

R RU

U W/m2.K

From Eq. (5.31), the corresponding cleanliness factor is:

79.05356

4225===

c

f

U

U CF

From Eq. (5.33), the actual heat duty for fouled surface is:

6

, 1029.308.71104225 ×=××=∆= cf lme f f T AU Q W

Similarly, the actual heat duty for clean surface is:

6

, 1017.408.71105356 ×=××=∆= cf lmecc T AU Q W

From Eq. (5.34), the safety factor is:

25.21046.1

1029.36

6

=×

×==

r

f

sQ

QC

By replacing U in Eq. (4.28) by U from Eq. (4.29), the percentage over surface f f

171



design is:

ft c RU OS 100=

where m55 1050105 −− ×=+×=+= fc fh ft R R R2.K/W

Therefore,

78.261055356100 5 =×××= −OS

The surface area of the gasketed-plate heat exchanger is:

9.4808.74225

1046.1 6

,

=×

×=

∆=

cf lm f

r

T U

Q A m2

Pressure drop analysis:

From Table (5.2), the fluid friction coefficients are:

441.1= p K and 206.0=m

From Eq. (5.24), the hot fluid friction coefficient is:

278.07.2931

441.1

Re 206.0===

m

h

p

h

K f

Similarly, the cold fluid friction coefficient is:

288.05.2472

441.1

Re 206.0===

m

c

p

c

K f

Let’s calculate the fluid friction coefficients from Eq. (5.15) just to see and

compare the results with the above ones.

[ ] [ ]{ }1.290/2sin0577.02.023 Re10016.21277.0917.2++−−×+−=

πβ β β hh f

[ ] [ ]{ }1.290/2sin0577.02.023 7.2931785.010016.2785.01277.0917.2 +×××+−− ×××+×−= β π

h f

389.0=h f

Similarly, the cold fluid friction coefficient is:[ ] [ ]{ }1.290/2sin0577.02.023 Re10016.21277.0917.2

++−−×+−=πβ

β β cc f

[ ] [ ]{ }1.290/2sin0577.02.023 5.2472785.010016.2785.01277.0917.2 +×××+−− ×××+×−= β π

c f

406.0=c f

The hot fluid and the cold fluid friction coefficients calculated from two different

correlations differ from each other by %40 and %41, respectively (the second

172



ones are %40 and %41 higher). The first results will be used here.

From Eq. (5.23), the frictional pressure drop from hot and cold streams are

respectively:

( ) 17.03

217.02

141.99621081.4

54.505155.1278.042

4 −−

−

××××

××××=

=∆

w

b

he

ch peff hh

DG N L f p

µ µ

ρ

41060.4 ×=∆ h p Pa kPa0.46=

( ) 17.0

3

217.02

12.99821081.4

54.505155.1288.04

24

−

−

−

××××

××××=

=∆

w

b

ce

ch peff

cc D

G N L f p

µ

µ

ρ

41076.4 ×=∆ c p Pa kPa6.47=

From Eq. (5.26), the mass velocities of the hot fluid and cold fluid in the port are

respectively:

1592

4

2.0

50

4

22=

×

=

=

π π p

h ph

D

mG

kg/m2.s

4775

4

2.0

150

4

22=

×

=

=

π π p

c pc

D

mG

kg/m2.s

From Eq. (5.25), the port pressure drop for the hot fluid is:

322

108.141.9962

159214.1

24.1 ×=

×××==∆

h

ph

p ph

G N p

ρ Pa

8.1=∆ ph p kPa

Similarly, the port pressure drop for the cold fluid is:

422

1060.12.9982

477514.1

24.1 ×=

×××==∆

c

pc

p pc

G N p

ρ Pa

0.16=∆ pc p kPa

From Eq. (5.27), the total pressure drop for the hot fluid is:

8.478.10.46 =+=∆+∆=∆ phhth p p p kPa

Similarly, the total pressure drop for the cold fluid is:

6.630.166.47 =+=∆+∆=∆ pcctc p p p kPa

173



Pumping powers:


80.0= pη

The pumping power for the hot fluid is:

33

100.380.041.996

108.4750×=

×

××=

∆=

ph

thhh

pm P

η ρ

W

0.3=t P kW

The pumping power for the cold fluid is:

43

1019.180.02.998

106.63150×=

×

××=

∆=

pc

tccc

pm P

η ρ

W

9.11=t P kW

Results of the design of the gasketed-plate heat exchanger are given in the

following table:

Table 6.5 Results of the rating analysis of the proposed gasketed-plate heat

exchanger

Heat transfer area (m2

) 110


Total number of plates 105

Plate thickness (mm) 0.6

Number of pass 1/1

Hot fluid heat transfer coefficient (W/m2.K) 13574

Cold fluid heat transfer coefficient (W/m2.K) 12698



Over surface design (%) 26.78

Hot fluid pressure drop (kPa) 47.8

Cold fluid pressure drop (kPa) 63.6

Hot fluid pumping power (kW) 3.0

Cold fluid pumping power (kW) 11.9

174



175

The over surface design value shown in Table 6.5 is acceptable. The

calculations show that the proposed unit satisfies the process required and

pressure drop constraint, but a smaller heat exchanger can also be used. There are

2 alternatives to do this: (1) one can fix the size of the Chevron plate, then the

number of the plates will be reduced. (2) the number of the plates can be fixed as

105, but the plate size will be different, then one can select a smaller Chevron

plate.

chapter 6 total

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