Shell Diameter : Bundle diameter: Db = d0 {Nt/K}1/n Where, d o = tube diameter, Nt = no. of tubes=6 !or sin"le #ass, K=0.1$ and n=%.1&% !rom 'hi(h, Db = 1$.0){6/0.1$}1/%.1&% Db = %%&.6%& or %%)mm. !or (learan(e if 'e add ) mm *hen shell diameter: %60 mm
Tube side Co-efcient : +ean 'ater tem#erature= {%&&}/% = %$ 0 *ube *ube (rossse(tion area= /&Di% Where Di is inside diameter of tube = 1).2mm=0.01)2m 3rea = 0.0&10 m % +ean 4elo(it5 = 4elo(it5 of 'ater/area of tube = %.$$2$/0.0&10 = 2 K"/m %.s
7inear 4elo(it5 of 'ater= mean 4elo(it5/densit5 of 'ater = 2/$$) = 0.02& m/s *his 4elo(it5 is too lo' *herefore (hoose &#asses *ube *ube (rossse(tion area = {Nt/n#} {/&}D i% Where n#= no. of #asses !rom 'hi(h tube (ross se(tion area= 0.000)0 m% 3nd 'ater mean 4elo(it5= %.$$2/0.000)0 = $%.$& 8"/m %.s 3nd 7inear 'ater 4elo(it5= $%.$&/$$) = 0.$2 m/s
hi=
4200 [ 1.35 + 0.02 t ] u
d
0.8
0.2
Where, t = mean 'ater tem#erature= %$0u= linear 'ater 4elo(it5 = 0.$2 m/s d= i.d. of tube = 1).2 mm
from 'hi(h hi = &6%2.$)$ W/m %. 0'ith no. of #asses (han"ed, ne' 9hell diameter = Db = d0 {Nt/K}1/n 'here,
K= 0.12) and n=%.%)
'ith these 4alues, ne' shell diameter =%&0 mm, for shell (learan(e of %0mm ne' shell diameter = %60 mm.
Shell side Co-efcient: v
¿
l − ρ
¿
¿
ρ¿ g l ( ¿¿ µ
¿
l τh ]
❑1/ 3 N −1 /6
ρ¿
¿
hco =0.926 K l ¿
-al(ulation of mean tem#erature of (ondensate lm: 7et t' = tube 'all tem#erature 3ssume, h(o = 1200 W/m % 0-, 3t stead5 state, ;eat transfer rate throu"h (old lm= <4er all heat transfer h(o 3( >t(t'?
=
@( 3( >t(ta4"?
3( on both sides (an(els and substitutin" the 4alues of h(o = 1200 W/m % 0-
and @( =)%) W/m% 0-
and ta4"= >%.&)%?/% , t (= 26.62 0'e "et t' = 6%.6)) therefore, mean tem#erature of (ondensate lm = >t(t'?/% = >26.626%.6))?/% = 6$.6) 03ll the #h5si(al #ro#erties are (al(ulated at this tem#erature K l = liAuid thermal (ondu(ti4it5 of miture !or Butanol it is 0.1 W/m.K 3nd for +CK it is 0.1) W/m.K *herefore, 34era"e thermal (ondu(ti4it5= K me8WmK butanolWbutanol !rom 'hi(h, K l = 0.16 W/m. K l = densit5 of liAuid (ondensate = 06 K"/m = densit5 of 4a#or = %.$2 8"/m El = a4". 4is(osit5 of (ondensate liAuid 34era"e 4is(osit5 of (ondensate liAuid is (al(ulated as follo's:
1
μ
=
W of mek W of Butanol + μ of mek μ of butanol
Where, W is 'ei"ht fra(tion of +CK= 0.$2 and W of butanol = 0.10 E of +CK and Butanol are 0.16 and 1.% mNs/m % 9ubstitutin" the 4alues = 0.12)6$10 8"/m.s τh
E = 0.12)6$ mNs/m %
= horiFontal tube loadin" of (ondensate #er
unit len"th of tube >8"/m.s? W c
=
L N t
Where W( = (ondensate Go' K"/s =6&0/600 = 0.122 8"/s 7 = len"th of tube= 1.m and N t= no. of tubes = 6 9ubstitutin" τh
= 0.001) K"/m.s
Nr = a4era"e no. of tubes in 4erti(al tube ro' = >%/?NrH = >%/? >Db/It? Where, Db= diameter of bundle = %&0 mm and It = #it(h = %.1%) 9ubstitutin" the 4alues 'e "et
Nr = 6.21$ 9ubstitutin" the 4alues in the eAuation v
¿
l − ρ
¿
¿
ρ¿ g l ( ¿¿ µ
¿
l τh ]
❑1/ 3 N −1 /6
ρ¿
¿
hco =0.926 K l ¿
h (o = %&%.) W/m %. 0-. for sub(oolin", standard 4alue from 8ern, hosub= %.22 W/m % 0-
Overall Heat transer Co-Efcient:
<4erall heat transfer (oeJ(ient (an be (al(ulated from follo'in" formula:
hi + ¿
d0 1 d i hid ¿
d d0 ln 0 di d0 1 1 1 + + + hoc hod di ¿ 2 k w U oc =
1 ¿
enerall5 for or"ani( 4a#ors, hod = 10000 W/m % 0-
and
for (oolin" 'ater hid = &000 W/m % 0-
as 990&7 is sele(ted: K ' = 16. W/m 09ubstitutin" the 4alues 1
U oc =
1 2342.5
+
1 10000
0.01905ln +
19.05 15.748
2 x 16.3
+
19.05 15.748
x
1 4627.959
+
19.05 15.748
= % W/m % 0;en(e reAuired area for (ondensation: U oc∗∆T ¿
3(o =
109075.654
mc
=
Qcond
832 45.725
¿
m% LeAuired area for sub(oolin": hi+ ¿
d0 1 d i hid ¿
1
+
1
ho!ub hod
d 0 di d0 1 + di ¿ 2 k w
d 0 ln +
U o!ub =
1 ¿
9ubstitutin" the 4alues, hosub= %.22 and rest 4alues as same U o!ub
= %%.20 W/m% 0-
*herefore,
= %.6
x
1 4000
3rea reAuired for sub(oolin"= U o!ub∗∆ T
3sub =
16443.88
mc
¿
Q¿
¿
=
232.7 29.20
= %.&% m%
*otal area = %.6%.&% ).% m% ;ere,
" a!!umed " calculated
= 6.2$/).% = 1.%)$
i.e. %.)$M etra area is #ro4ided 'hi(h is more than 10M than reAuired in desi"n. *herefore 'e (an "o 'ith the desi"ned area of 6.2$ m%. 3rea (al(ulated for (ondensation : &.)& m % 3rea (al(ulated for 9ub(oolin" m%
: %.%)%
3rea #ro4ided for (ondensation in e(ess: = %.6%M of %.6 = .660 m % 3nd for sub(oolin", = %.&%1.% =.0$26 m% ;en(e (om#ared to total area of 6.2$ m % &).6M of total area should be #ro4ided for sub (oolin".
3ssumin" that tubes are uniforml5 distributed in the (ross se(tion of shell, 0.&)6= { 5 D %i}/>/&? D%i !rom 'hi(h 5 = 0.&)6 5 = 0.) for this 4alue of O5H, h/D i = 0.6 hen(e, .6M of shell must be submer"ed in the #ool of (ondensate to fa(ilitate sub(oolin". *his (an be a(hie4ed b5 #ro4idin" @loo# seal in the drain line of (ondensate. *herefore, .6M of %60 mm = 100 mm hei"ht.
Shell side Pressure drop:
9hell side Go' area: 3s=
( # t − d o ) B ! $ ! #t
%
x
Where, H = 1>h/di? =10.6=0.61 #t = tube #it(h = %.1%) do = tube <.D.=1$.0) mm Bs = BaPe s#a(in" = 0.%60 Ds= 9hell Q.D. = 0.%60
9ubstitutin" the 4alues, 3s = 0.00)) m % 9hell side 4elo(it5 = s/4 4 = %.2$ K"/m s =m/3s = >6)/600?/0.00)) = %1. K"/m %.s *herefore, 9hell side 4elo(it5 = %1./%.2$ = 2.& m/s CAui4alent diameter for trian"ular #it(h, de=
1.1
d0
( # − 0.907 d ) 2
2
t
o
!rom 'hi(h, de = 1.26 mm & ! de
Le5nolds number Le =
μ
9ubstitutin" 4alue E = 0.01 Le= $2$0 !or %)M baPe (ut se"ment and "i4en Le5nolds no. of $2$0 Rf =0.0
( )( )( )( )
$! ∆ #= 0.5∗8∗ ' f de
2
L ρ v u ! μ 2 B! μw
9ubstitutin" the 4alues
0.14
∆#
= .$2 KIa.
*his is the (al(ulated #ressure dro# on shell side
Tube side Pressure Drop :
*ube side #ressure dro# (an be (al(ulated for −m
( )( )
L μ ∆ #t = N ( [ 8 ' h d !i μ w
+ 2.5 ]
2
ρ u t 2
Rh (al(ulation: Le = >d4?/E Where, = $$) 8"/m, 4 = 0.$2 m/s, d=0.01)2&m 3nd E= 0.10 9ubstitutin", *herefore,
Le = 1$2 Rh= &10
7= %000 mm, di= 1).2&mm, ut = 0.$2 m/s, N# = & 9ubstitutin" in #ressure dro# formula, ∆ #t = 1%.60 KIa
F!"# HE"T E$CH"!%E& D"T": HEAT EXCHANGER SPECIFICATION SHEET
EQUIPMENT NAME AND TYPE: MEK CONDENSER AND SHELL AND TUBE (FIXED-TUBE SHEET) SERVICE: CONDENSING AND SUB-COOLING OF MEK AND BUTANOL VAPORS GENERATED FROM REACTOR IN PRESENCE OF HYDROGEN AS A NON-CONDENSIBLES
FLUID PROPERTIES DATA: FLUID STATE TEMPERATURE IN ( OC)
SHELLSIDE
TUBESIDE
VAPOR
LIQUID
1575 !C
"#!C
TEMPERATURE OUT ( OC)
$"!C
$#!C
DENSITY (KG%M $)
"&7
&&5
VISCOSITY (K'%)
!175*&+1! -$
SENSIBLE HEAT (,%KG% OK) LATENT HEAT (K,%KG) PROCESS DATA: HEAT DUTY (K.) FLO. RATE (KG%HR) PRESSURE DROP (KPA)
1&#
!!!1
#1*
#$"7 SHELLSIDE 1"551& *5 $&7
TUBE VELOCITY (M%S) CONSTRUCTION DATA: HEAT TRANSFER AREA (M ")
*7&
NUMBER OF TUBES%SHELL
*$
TUBE OUTSIDE DIAMETER (MM)
1&!5
TUBE B.G NUMBER
1*
TUBESIDE 1!7&"7" 1"*!
TUBE LENGTH (METER)
1$
TUBE PITCH AND ORIENTATION
TRIANGULAR
INSIDE SHELL DIAMETER (MM)
"*!
SHELL MATERIAL
CARBON STEEL/
TUBE MATERIAL
STAINLESS STEEL/
BAFFLE SPACING (MM)
"*!
'ECH"!C"# DES%!:
9hell thi(8ness: 9hell material is (hosen as -arbon steel. 3nd o#eratin" #ressure is 1.06&10 ) N/m% *herefore desi"n #ressure is 10M more than o#eratin" #ressure and that (omes to 1.1110 ) N/m% *sh =
# d $ 2 f) + #
;ere, Id= 0.11, D = %60, f = $) N/m% > for (arbon steel? and R= 0.) >Soint eJ(ien(5? 9ubstitutin" the 4alues, *sh = 0.120 mm *a8in" minimum thi(8ness as 6 mm and (orrosion allo'an(e of % mm total thi(8ness (omes as mm *a8in" standard thi(8ness, 9hell thi(8ness: 10 mm. Head : (Torrispherical head)
#¿ * ¿c T h= W 2 f' 1
Where, =
W = ⌊ 3 + + 4
* c ⌋ *ic
L(= (ro'n radius = %60 mm Li( = 6M of (ro'n radius=1).6 mm With these 4alues W = 1.22 3nd substitutin" the 4alues of W and I of 0.11 N/m% !=$) and R = 1 We "et thi(8ness of head = * h = 0.%) mm 9o, same thi(8ness of shell i.e. 10 mm of head is sele(ted. *hi(8ness of baPes is sele(ted as a 6 mm standard.
B3!!7C9 9I3-QN: Segment al bafflesar eused. Bafflespac i ng==260/ 5=52mm Thi ck nessof baffles=6mm Ti er odsandspaces Di ame t eroft i er od=10mm Numberoft i er ods=6
Desi"n of !lan"es: Shel l t hi c knes s=go =10mm
Fl an ge mat er i al –I S:2004–1962c l as s2 Gas k et mat er i al –as bes t osc ompos i t i on Bol t i ngst eel =5% CrMos t eel Al l owabl es t r es soffl angemat er i al –100MN/m 2 Al l owabl es t r es sofbol t i ngmat er i al ,Sg–138MN/ m2 Ou t s i d ed i a me t e r=B=26 0+( 2x1 0 ) =280mm Gasketwi dt h do/di [ ( y -pm) /( y -p{ m+1} ) ]0.5 =
m –gas k etf ac t or–2. 75 y–mi ndes i gns eat i ngs t r es s–25. 5MN/m 2 l e t ’ sas s umeGas k e tt hi c k nes s=1. 6mm Thus, do/di =1. 002 Le tdi oft hegas k e tequal 290mm [ 10mm gr eat ert hans hel l di a] do =0 . 2 90x1 . 0 02 . =0 . 2 90 . 5 8m= 29 0. 5 8mm
Meang as k e twi dt h=( 0. 290. 58–0. 290)/ 2 =0 . 2 9x1 0-4 T a k i n gg as k e twi d t ho f1 2mm, do =0 . 2 91m Bas i cgas kets eat i ngwi dt h,bo=5mm Di amet erofl oc at i onofgas ketl oadr eac t i oni s , G =di +N =0 . 2 90+0 . 0 12 =0 . 3 02 m Est i mat i onofbol tl oads Lo add uet od es i g npr es s u r e: H=πG2P/4 =( 3. 14x0. 302x0. 11)/4 =0. 0260MN Loadt ok eepj oi ntt i ghtunderoper at i on: Hp=πG( 2b ) mp 3
=3 . 1 4x 0 . 3 02x2x5x10 x2. 75x0. 11 =0. 00286MN
T ot al l oad =0. 0260+0. 00286 =0. 0288MN Loadt os eatgas k etunderbol t i ngupc ondi t i on Wg
=
πGby
=3. 14x0. 302x0. 005x25. 5 = 0. 1209MN Cont r ol l i ngl oad
=
0. 1589MN
Mi ni mum bol t i ngar ea = Am =
0. 12 09 / 1 38
=
8 . 7 6x1 0-4 m2
=
W / gS g
L e t st a k eM 18x2 4
2
Rootar eaofbol t=1. 54x10 m Ther ef or eno.bol t sr equi r ed 4
4 / =8 . 7 6x1 0 1. 54x10 =5 . 6 89
8no.ofbol t sar esel ect ed. R g1 go
= =
0. 027m / go 0. 707
=
10mm
=
1. 415g0 f orwel dl eg
Bol tc i r c l edi a me t e r , C
=
C
I . D+2( g1+R)
=
0 . 2 60+2( 1 . 4 15x0 . 0 10+0 . 0 27 )
=
0 . 3 42 3m
Cal c ul at i onofout s i dedi amet eroffl ange: A=C+Bo l td i a me t e r+0 . 0 2( mi n i mu m) =0 . 3 42 3+0 . 0 18 +. 0 2 =0 . 3 64 1m Ch ec kg as k e twi dt h : =( Ag *Sb) / πGN 4
= 8 x 1. 5 1x 1 0 x 138/ 3. 14x0. 30 2x0 . 012 =1 4. 6 49 6 Whi c hi sl es st han2y( 2x 25. 5) Ther ef or ec ondi t i oni ss at i s fied. Fl angemomentcomput at i on: a)Forope r a t i ngc ondi t i on: Wo W 1=
= ,B 4
2
#
W1+W2+W3
0.280
¿ ¿ ¿2 3.14 ¿
W 1=¿
=0 . 0 06 76 W2
=
HW1 2
2
Wher eH=( π/ 4) G =0. 78 5x0 . 3 02 =0 . 0 71 61 W2= = W3
0 . 0 71 63–0 . 0 06 76 0. 0648MN
= =
WoH
=
Hp( gasket l oad)
0. 00286MN
Tot al flangemoment ,
Mo=W 1a1 +W2a2 +W3a3 a1=( CB)/2
=( 0. 34230. 280) / 2=0. 03115
a3 =( CB)/2=( 0. 34230. 302) /2=0. 02015 a3=( a1+a2)/ 2=0. 02565 M0 =2 . 8 38x1 0-4MJ b)Forbol t i ngupcondi t i on( noi nt er nalpr essur e) : Mg = W.a3 W =( Am +Ab) / ( 2) .Sg
Am = ar eao fbol t = 8x1 . 5 6x1 0-4 =1 . 2 48x1 0-3 m 2 Am =Mi n i mu mb ol ta r e a.=1 . 3 8x1 0-3 m 2 Sg=138N/ mm 2 Th er e f o r e ,W =0 . 1 81 3MN a3 =0 . 0 20 15 Mg=0 . 1 81 3x0 . 0 20 15 3
Mg =3 . 6 5x1 0 MJ MG i sgr eat ert hanM0.Ther ef or eMgi sc ont r ol l i ng. Fl a nget hi c kne ss: - . / / t 2 = B 0 T
K=A/ B Wh er e ,A=fl an geou t s i d ed i a me t e r=0. 3 64 1m B=fl a ng ei n s i d ed i a me t er=0 . 28 0 K=( 0. 3641/ 0. 280) =1. 3 00
Fr om fi gur egi v e ni n7. 6 AtKv a l ueof1. 300,Y=10 3
As sume,Cf=1 andM =3. 65x10 MJ St=Al l o wa bl es t r e s s=1 00 MN/m 2 2
3
t =(3. 65x10 x10)/( 0. 280x100) =0 . 0 36 1m t=0. 074m Nozz l ef orcondenser : T ot al 5no.ofnoz z l esmus tbepr o vi dedf orc ondens er .Twonoz z l eson t ubes i deeac hc ool i ngwat ers uppl yi nandout .4No.noz z l esar epr ov i ded Ons hel l s i de
;eat e(han"er Ii#in": 9hell side Guid: Ta#or inlet #i#e siFin":
Ta#or Go' rate: 6) K"/hr.=0.1%2 8"/s " = %.2) K"/m 4olumetri( Go' rate: 6)/%.2) = %&.2 m hr = 0.0626$ m /s. <#timum #i#e siFe (an be (al(ulated from eAuation: !or stainless steel,
d o(t =293 &
0.53
−0.37
ρ
Where, Do#t= o#timum #i#e siFe = !lo' rate, K"/s=0.1%2 8"/s U= densit5 of 4a#or =%.2) 8"/m 9ubstitutin" the 4alues, Do#t= 1.) mm 9tandard of 0 NB >9-;.10? (an be used. @n(ondensed 4a#or: !lo' rate =11%.& 8"/hr = 0.01% 8"/s 3nd for same densit5 of %.2) 8"/m , !rom the abo4e formula, *he #i#e siFe of %) NB >9-;.10? (an be ed. -ondensate drain line: !lo' rate: )&).66 8"/hr=0.1)1)2 K"/s Densit5 of (ondensate : 02 8"/m 9ubstitutin" the 4alues in the same eAuation, *he #i#e siFe of 1) NB >9-;.10? is ed. -oolin" 'ater lines: !lo' rate: %.$$ 8"/s Densit5: $) 8"/m 9ubstitutin" the 4alues in same eAuation:
)0 NB >9-;.10? line siFe is ed *he same siFe of return line (an be used.
3s the nal #ro(essdesi"n sta"e is -om#lete, it be(omes #ossible to ma8e a((urate (ost estimation be(ause detailed eAui#ment s#e(i(ation and denite #lant fa(ilit5 information are a4ailable. Dire(t #ri(e Auotation based on detailed s#e(i(ation (an then be obtained from 4arious manufa(turers. ;o'e4er desi"n #roSe(t should #ro(eed to the nal sta"es before (osts are (onsidered and (ost estimate should be made throu"h out all the earl5 sta"es of the desi"n 'hen (om#lete s#e(i(ations are not a4ailable. C4aluation of (osts in the #reliminar5 desi"n is said #re desi"n (ost estimation. 9u(h estimation should be (a#able of #ro4idin" a basis for (om#an5 mana"ement to de(ide if further (a#ital should be in4ested in the #roSe(t. C4aluation of (osts in the #reliminar5 desi"n #hase is some time (alled "uesses
estimations. 3 #lant desi"n ob4iousl5 must #resent a #ro(ess that is (a#able of o#eratin" under (ondition 'hi(h 'ill 5ield a #rot. 3s the nal #ro(essdesi"n sta"e is -om#lete, it be(omes #ossible to ma8e a((urate (ost estimation be(ause detailed eAui#ment s#e(i(ation and denite #lant fa(ilit5 information are a4ailable. Dire(t #ri(e Auotation based on detailed s#e(i(ation (an then be obtained from 4arious manufa(turers. ;o'e4er desi"n #roSe(t should #ro(eed to the nal sta"es before (osts are (onsidered and (ost estimate should be made throu"h out all the earl5 sta"es of the desi"n 'hen (om#lete s#e(i(ations are not a4ailable. C4aluation of (osts in the #reliminar5 desi"n is said #re desi"n (ost estimation. 9u(h estimation should be (a#able of #ro4idin" a basis for (om#an5 mana"ement to de(ide if further (a#ital should be in4ested in the #roSe(t. C4aluation of (osts in the #reliminar5 desi"n #hase is some time (alled "uesses estimations. 3 #lant desi"n ob4iousl5 must #resent a #ro(ess that is (a#able of o#eratin" under (ondition 'hi(h 'ill 5ield a #rot.
