Heat Exchanger Design
In partial fulfillment of the requirements in ChE 192 – Chemical Process Equipment Design
Submitte b!
"e S# $lbarico ChE 192 %
Submitte to
Sir &ino $pollo &uerrero
'arch 19( 2)1*
1
Design Problem+
Design a shell an tube heat exchanger to be use in cooling ,erosene using light crue oil# $ssume the -or,ing pressure for ,erosene to be * bars an light crue oil to be .#* bars# $lso( assume fouling allo-ance for ,erosene to be 2 x 1)/0 3m245C6/1 an for crue oil to be 7 x 1)/0 3m245C6/1# 8ther information is as follo-s+ 'ass flo- rate of ,erosene 1* ))) ,g3hr • :erosene inlet temperature 2))5C • :erosene outlet temperature 9)5C • 'ass flo- rate crue oil ;) ))) ,g3hr •
&i=en+
Light crude oil Tout = ?
"eat #$changer Kerosene
Flowrate = 0 000 kg/hr Tin = !5°C Tout = 90°C
Flowrate = 15 000 kg/hr Tin = 200°C
>equire+ Design Specifications of the Heat Exchanger
Solution+ 1. Duty of the heat exchanger ? Choose light crue oil in the tube since it is a irt! flui( ,erosene in the shell# @ube sie ata+
Shell sie ata+ :erosene Alo-rate 1* ))) ,g3hr @in 2)) 5C @out 9) 5C
2
2. Fluid physical properties
? $ssume properties =ar! -ith temperature linearl!# @able 1# Ph!sical properties of ,erosene# Property
Temperature 5C6 Specific heat ,"3,g5C6
Values
Linear Eqn vs T
2))
10*
9)
/
2#;2
2#0;
2#2.
0#111x1)/7@ 1#;.;9;
Thermal conductivity 3m5C6
)#17)
)#172
)#17*
/0#*0*0*0x1)/*@ )#17920
Density ,g3m76
.9)
;7)
;;)
/)#;2;2;2@ 7*#0*0*0*
Viscosity ms3m 26
)#22
)#07
)#)
/*#2;2;2;x1)/7@ 1#20;;9
Source: Towler 2!1". @able 2# Ph!sical properties of light crue oil 70 5$PI6# Property
Temperature 5C6 Specific heat ,"3,g5C6
Values
Linear Eqn vs T
;
*9
0)
/
2#)9
2#)*
2#)1
2#1)*2.7x1)/7@ 1#92*;9
Thermal conductivity 3m5C6
)#177
)#170
)#17*
/*#2.71*x1)/*@ )#17;1)*
Density ,g3m76
))
2)
0)
/2#)*2.72@ 2#1)*2.7
Viscosity ms3m 26
2#0
7#2
0#7
/)#)*@ .#2*
Source: Towler 2!1".
". Type of heat exchanger ? %se 2/. shell an tube heat exchanger#
#. Trial value for overall coefficient $ % Fase on Aigure 1( choose 2)) 3m25C#
Aigure 1# @!pical o=erall coefficients# &'tained from: Towler 2!1".
(. )ean temperature difference *T lm Qshell =−Qtube
mCpdT ´ ¿ crude mCpdT ´ ¿ kerosene =−¿
¿
(
kg 15000 hr
90
) ∫ (4.18 x 10
−3
(
T + 1.88 ) dT =−
200
70000 kg
hr
X
− ( 2.11 x 10 T + 1.93 ) dT ∫ ) 3
45
X =73.546013 ° C
∆ T lm =
∆ T 2− ∆ T 1 ln
( ) ∆ T 2 ∆ T 1
∆ T 2=( 200−73.55 ) ° C =120.45 ° C ∆ T 1=( 90 −45 ) ° C = 45 ° C
∴ ∆ T lm
=
(120.45 −45 ) °C ln (
120.45 °C 45 ° C
)
∆ T lm=78.835390 ° C ? It must be note that the greatest temperature ifference 12)#0*5C6 is at least 7)5A /1#115C6 an the minimum temperature ifference 0*5C6 is at least 1)5A /12#225C6( conforming to the stanars in temperature ifference# @able 7# Ph!sical properties of ,erosene at inlet( mean an outlet temperature# Property Temperature 5C6 Specific heat ,"3,g5C6 Thermal conductivity 3m5C6 Density ,g3m76 Viscosity ms3m26
Values 9)
10*
2))
2#2*7
2#07
2#;17
)#17*
)#172
)#17)
;;)
;7)
.9)
)#;;7
)#07
)#197
@able 0# Ph!sical properties of light crue oil 70 5$PI6 at inlet( mean an outlet temperature# Property Temperature 5C6 Specific heat ,"3,g5C6 Thermal conductivity 3m5C6 Density ,g3m76 Viscosity ms3m26
Values 0*
*9#2;7
;7#*0.
2#)21
2#)*1
2#)1
)#17*
)#170
)#177
70#;7;
19#;17
)0#.
0#)))
7#2.
2#*;7
Temperature correction factor:
T 1−T 2 200 ℃− 90 ℃ R = = =3.853429 * >+ t 2−t 1 73.546 ℃− 45 ℃
? S+
S=
t 2−t 1
= 73.546 ℃− 45 ℃ =0.184168 T −t 200 ℃− 45 ℃ 1
1
Aigure 2# @emperature correction factor# &'tained from: Towler 2!1".
? Correction factor+ Fase from Aigure 2(
Y ≈ 0.94
+. ,eat transfer area
Q=UAY ∆ T lm 90
A =
Q = UY ∆ T lm
A =76.79579 m
15000 kg
/ hr ∫ (4.18 x 10− T + 1.88) dT 3
200
( 200 W / m ℃ )( 0.94 )( 78.84 ℃ )
(
1 hr 3600 s
)(
1000 1 k
)
2
-. ,eat exchanger layout % %se split ring floating hea for efficienc! an ease of cleaning# ? Plain carbon steel can be use for the shell an tube since neither flui is corrosi=e# ? Crue oil is irtier than ,erosene( so put the crue through the tubes an the ,erosene in the shell# ? Aor tubes+ choose 2* mm )#)2* m6 outer iameter( 21 mm )#)21 m6 insie iameter( length 12 ft 7#.*;* m6( triangular pitch 1#2*Do( baffle cut 2*G# Number of tubes +
( )( ) $rea of 1 tube+ A 1 tube= ! d o " =! 0.025 m 3.6575 m = 0.287267 m
2
of tubes+
¿ tubes =
A
A 1 tube
=
76.80 m
2
0.287 m
2
=267.332248 ≈ 268 tubes
tubes3pass+
¿ tubes / p#ss= 268 / 6 ≈ 45
Tube-side velocity: 2
Cross section area of 1 tube+ A tube =
Atube $rea3pass+ p#ss
! d$ 4
=
! ( 0.021 m ) 4
2
=3.4636 x 10− m 4
=¿ tubes ∗ A tube =43∗3.46 x 10− m = 0.014894 m 4
2
2
2
p#ss
/ hr ∗1 hr 819.713 kg / m 70000 kg
olumetric flo-+
%´ =
& = '
3
3600 s
3
=0.023721 m / s
3
0.024 m / s %´ % t = = = 1.592711 m / s 2 Atube 0.0150 m p#ss
? @he tube/sie =elocit! is satisfactor! since =elocit! is bet-een 1 an 2 m3s#
Bundle and shell diameter:
Aigure 7# Constants for bunle iameter# &'tained from: Towler 2!1". 1 / n1
( )
) t (b= d o Funle iameter+ * 1
(b=( 0.025 m )
J from Aigure 7( at . pass( : 1)#);07( n12#099
(
256 0.0743
)
1 2.499
=0.650739 m
Aigure 0# Shell/bunle clearance# &'tained from: Towler 2!1".
Shell Diameter+
(S =( (S − (+ ) + (+
J from Aigure 0(
( (¿ ¿ S− ( + ) )#).7 m
¿
(S =( 0.063 + 0.6507 ) m =0.713739 m " 3.6576 m = =5.124564 <3Ds+ ( S 0.071 m ? Since <3DS is bet-een * an 1)( DS is the optimum shell iameter#
hell thic!ness:
Aigure *# Shell thic,ness# &'tained from: Towler 2!1".
? Shell to be use is rolle from plate -ith a minimum thic,ness of ;#9 mm# . /ndividual heat exchanger coefficients Tube-side heat transfer coefficient:
' d $ % t ( 819.713 kg / m 3)( 0.021 m)( 1.59 m / s ) =8342.65215 2 −3 >e!nols umber+ ) ℜ= , = 3.286 x 10 )s / m 2
Prantl umber+ <3Di+
Cp, ( 2.051 k / kgC )( 3.286 m)s /m ) ) -r = = =50.2958209 k 0.134 W / mC
" / ($=3.6576 m / 0.021 m =174.1714
Aigure .# >e!nols umber =s tube heat transfer factor# &'tained from: Towler 2!1". −3
Heat transfer factor+ Fase on Aigure .( . / =3.0 x 10 Heat transfer coefficient+ ht =
0.134 W / mC 0.021 m
( )
k , 0.33 ht = ) ℜ ) -r . / d$ , 0
( 8342.65 ) ( 50.30 )
0.33
0.14
( 3.0 x 10− ) ( 1 )=555.99689 W / m 3
2
° C
? Since ht is much greater than the assume %o 19)3m2C6( the heat transfer coefficient is satisfactor!# hell-side heat transfer coefficient
$rea of crossflo-+
A s =
( p t − d o ) ( s l + pt
( 0.0312 m−0.025 m ) ( 0.071 m ) =
5
0.0312 m
A s $rea of crossflo- per pass+
(
0.071 m
p#ss
=
As
¿ shell p#ss
=
)=
0.020377
0.020 m 2
2
=0.010188 m
2
W s 15000 kg / hr us = = =0.560218 m / s ' A s ( 730 kg / m3 )( 0.010 m2 )
Equi=alent iameter+
de=
1.10
do
d e=
( ' −0.917 d ) 2
2
t
o
1.10
( ( 0.031 m) −0.917 ( 0.025 m) ) =0.018270 m 0.025 m 2
2
3
(730 kg / m )( 0.018 m)( 0.56 m/ s ) ) = = =15458.8703 ℜ − >e!nols umber+ , 0.483 x 10 )s / m ' d e % s
3
2
2
Prantl umber+
Cp, ( 2.483 k / kgC )( 0.483 m)s / m ) ) -r = = =9.070131 k 0.132 W / mC
Aigure ;# >e!nols number =s shell heat transfer factor# &'tained from: Towler 2!1".
−3
Heat transfer factor+ Fase on Aigure ;(
( )
k , 0.33 h s= ) ℜ ) -r . / de , 0
Heat transfer coefficient+ 0.132 W / mC
h s=
. / =1.8 x 10
0.018 m
( 15458.87 ) ( 9.07 )
0.33
0.14
( 1.8 x 10− ) ( 1 )= 403.51363 W / m 3
2
°C
0. &verall coefficient 1
U o
1
= +
d o ln
1
+
ho hod 1
?
1
?
hod
d$
2 k 0
556.00 W / m ℃
d o ln
( )= do d$
2
?
1
?
h$
=0.0003 m
= 1
U o
1
) 2
℃ / W
℃ / W
(
0.025 m 0.021 m
2 ( 55 W / m ℃ ) 2
h$d
+
=0.0024782 m
0.025 m ln
2 k 0 1
(
1
d $ h $d h$
2
=0.0002 m
?
do
1
=
ho
( )+ do
)=
3.96 x 10
−5
2
m ℃ / W
℃ / W
1 2
403.513 W / m ℃
=0.001799 m
2
℃ / W
=0.002537 + 0.0002 + 3.96 x 10− + 5
0.025 m
( 0.0003 +0.001778 )
0.021 m 2
U o= 191.71206 W / m ° C ? @he o=erall heat transfer coefficient is satisfactor! since the assume heat transfer coefficient 2)) 3m25C6 is near 0#72G ifference6# 1!. ,eat exchanger pressure drops Tube-side pressure drop
? $ssumption of Aran,Ks 19;6 correction ue to pressure losses
[ (
−m
( )( )
" , ∆ -= ) p 8 . 1 d $ ,0
+ 2.5
)]
2
'u t 2
Aigure # @ube/sie friction factor# &'tained from: Towler 2!1".
? Ariction factor+ Fase from Aigure (
.1 =0.0032 1.52 m
¿
2
/s¿
3
( 819.713 kg / m )¿ ∆ -=
[(
∗
4 8 0.0032
(
3.6576 m 0.021 m
) ( ) + )] ¿ 1
2.5
∆ - = 26 821.1839 -# ? Pressure rop is satisfactor! since LP is belo- -or,ing pressure of crue oil .#* bars .* ))) Pa6# Correction factor for tube pressure rop follo-e Aran,Ks metho# hell-side pressure drop
( )( ) ( ) 2
( s " us , ∆ -= 8 .1 ' d e l+ 2 , 0
−0.14
Aigure 9# Shell/sie friction factor# &'tained from: Towler 2!1"
. 1 =0.031
? Ariction factor+ Fase from Aigure 9(
∆ - = 8∗0.032
(
0.73 m 0.018 m
)(
3.6576 m 0.15 m
)∗
3
730
kg / m ∗( 0.54 m / s ) 2
2
(1 )
∆ -= 26596.467 -# ? Pressure rop is satisfactor! since LP is belo- -or,ing pressure of ,erosene *) ))) Pa6# Baffle Thic!ness
T =
T =
√ √
&2 S 2
( 0.025 m ) ( 1.25 )
26821.18
2
2067857.12
T =0.0017795 m
11. &ptimiation
* bars
? $Musting the assume heat transfer coefficient( %( results to changes in the actual %o# Still( the conitions presente are the optimiNe parameters for the construction of a shell an tube heat exchanger#
12. Summary
Split ring( floating hea( 2 shell pass( . tube passes#
2. carbon steel tubes( 7#.*;. m long( 2* mm o##( 21 mm i##( @riangular pitch( pitch 71#2* mm#
Heat transfer area ;.#;9*;9 m2 base on outsie iameter6#
Shell i## ;2*#; mm( rolle from plate -ith thic,ness ;9 mm( baffle spacing 10*#1. mm( 2*G cut#
@ube/sie coefficient **. 3m25C( clean#
Shell/sie coefficient 0)7#* 3m25C( clean#
8=erall coefficient estimate 2)) 3m25C( irt!#
8=erall coefficient require 191#;1 3m25C( irt!#
Dirt3Aouling factors+ @ube/sie crue oil6 )#)))7 3m25C6/1 Shell/sie ,erosene6 )#)))2 3m25C6/1
o o
Pressure rops o o
1". Design
@ube sie( estimate 2#. barJ specifie .#* bar o=erall# Shell sie( estimate 2#.. barJ specifie * bar o=erall#
Aigure 1)# Cross/section of heat exchanger#
Aigure 11# Cross/section of heat exchanger sho-ing pipe arrangement#
Aigure 12#
1#. eferences •
@o-ler( &#( O ># Sinnot# 2)176# Chemical Engineering Design+ Principles( Practice an Economics of Plant an Process Design# 2 n E# Else=ier+ %:#
