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Properties of isopropanol
`T = 51.5°C
m = 1.028 cp
Cp = 3096.17 J/kg K
k = 0.129W/mK
r = 762 kg/m2

Properties of water :
`T = 71
m = 0.407 cp
Cp = 4189.5 J/kg K
k = 0.664W/mK
r = 977.8 kg/m2

Flow rate of isopopanol = 65 000 kg/ h
mi = 18.06 kg/s
Q given = Qtaken
= mi*Cp*DT
=18.06*3096.17*( 72- 31)
=2292590 J/s
Q given =m* Cp*DT
2292590 = mw*4189.5*(97-45)
mw = 10.52 kg / s
Assumed overall heat transfer coefficient : 400 W/mK

97

72 DTLM= (97-72)-(45-31) = 18.97° C
45 In (97-72)
(45-31)


Q = A*Uİ*DTLM*F
2292590 = A*400*18.97*0.9
A= 335.7
From appendix A ; F= 0.9

Selected type of tubes ; ¾ inch 14 BWG
Xw = 0.0021 m
Dti= 0.0148 m
Dto=0.01905 m
Si ;= 1.73 10-4 m2
Assumed length of tubes : L = 20 ft (6.1m)
Area of the one tube : Atube = p*Dti*L
= p*0.0148*6.1
=0.28 m2
Number of the tubes ; Ntubes = Heat transfer area = 335.7 = 1199 tubes
Area of one tube 0.28

Number of tubes at the shell diameter ; NDtube = 4* Ntubes = 40
p

Diameter of the shell side : Ds = (NDtube –1) *2.54 10-2 + Dto
= (40-1)*2.54.10-2 + 0.01905
= 1 m

TUBE SIDE
m = u*r*S*Nt / 4
10.52= uwater*977.8*1.73.10-4*1199/ 4
uwater = 0.2 m/ s

Re = D*u*r = 0.0148*0.2*977.8 = 5689 Turbulent flow
0.407.10-3


From SEIDER EQUATION ismi doğrumu?

h i D ti = 0.023 ( Re)0.8 (Pr)0.33( m /mw)0.14
k

Pr = Cp*m = 4189.5*0.407.10-3 = 2.57
k 0.664

( m /mw)0.14 = (1.028/ 0.83)0.14 = 1.03

h i D ti = 0.023 ( 5689 )0.8 (2.57)0.33(1.23)0.14
k
hi = 1464.4Wm /K

SHELL SIDE
h0 = 0.36 k . De*G 0.55 Cp m 0.33 m . 0.14
De m k mw

De = 4* rh = 4* Cross sectional area
Wetted perimeter

Tubes were laid out triangular pattern.


Y = 2.54.10-2


Cross sectional area ; Yh/2 - pD2to/ 8
=2.54.10-4* (Ö3)*2.54. 10-4/ 4 - p(0.01905)2 / 8
=3.335.10 -5m m2


BERRİİN BU HESAPLAMALARA BAK..DE 0.0046 CIKACAK SEKİLDE DÜZELT.
Wetted perimeter ; pDto/2
=p*0.01905/ 2
=0.029m
De = 4 rh = 4* 1.369 = 0.0046 m
0.029

Baffle spacing ; Ib =0.2
Area of shell side ;As = Ds*l b* C’/Y
As = 1*0.2*(2.54 10-2 –0.01905)/ 2.54 10-4
= 0.05 m2

G = ms .= 18.06 = 361.2kg/m2s
As 0.05

h0 = 0.36 k . De*G 0.55 Cp m 0.33 m . 0.14
De m k mw

h0 = 0.36* 0.129 * 0.0046*361.2 0.55 3096.17*1.028 10-3 0.33 1.028 . 0.14
0.0046 1.028. 10-3 0.129 0.83

h0 =1742 Wm2/K
Ui = 1
Fouling factor hdi = 1700 (water)
hdo = 1020 (isopropanol)
Xw Di . = 0.0021*0.0148 = 4.907 10-6
K DL 377* 0.0168
DL = 0.01905-0.0148 . = 0.0168m
In (0.01905/0.0148)
1 / hdi =1 /1700 = 5.8 10-4
1 / hi =1 /1464.4 =6.8 10-4
Di / D0h0 = 0.0148 / 0.01905*1742 = 4.46 10-4
Di / D0hd0 = 0.0148 /0.01905*1020 =7.6 10-4
Ui = (1/ 5.8 10-4 +6.8 10-4 + 4.46 10-4 + 7.6 10-4 + 4.907 10-6)
= 404.7 Wm2 /K
Berrin birimleri kontrol et.
Pressure drop in the tube side;
DPt = Np ( 8 Jh( L/ di) (m /mw)-0.14 + 2.5 ) r u 2t/2
ut =0.2 m/s
Re =0.01905*0.2*977.8 /0.407 10-3
=9153
Jh =5.8 10-3 From appendix ????
DPt = 4 ( 8*5.8 10-3*( 6.1/ 0.0148 )*0.87-0.14 + 2.5 ) 977.8 (0.2)2 / 2
= 0.017 atm
Pressure drop in shell side
DPs = 8 Jf ( Ds / de ) ( L / Ib ) r us2 / m ( m /mw)-0.14
Gs = us*r
.us = 361.2/ 762=0.47 m/s
Re = 1*0.47*762/ 1.028 10-3
=348385
Jf = 6.5 10-3 From appendix ?????
DPs =8*6.5 10-3 ( 1/ 0.0046 )(0.86/0.0046) (6.1/ 0.2) (762 * 0.42 /2 ) (1.028/0.83)-0.14
=0.29 atm




Working of pump:
P1 + gz1 + u12 + hWp = P2 + gz2 + u22 + åhf
r 2 r 2
Point 1:before the pump
Point 2:exit of tube side in the heat exchanger
Assumptions: z1 = z2
åhf is negligible
u1 = u2
r =rw (water flow in tube side )
h = %70
Result : hWp = DP / rw
Wp = 0.017 105 / 977.8 * 0.7
= 2.48 W
We assumed that our system is worked while 330 days ( 7920 hour )
To find cost of electricity =0.12 $ / kWh * [ 2.48 10-3 *7920 kWh]
=2.2 $
To find cost of water = 1.8$ / 1000 gal *[ 264.2 gal/ 1 m3 ]* 0.011 m3 /s
= 5.12$
bu kısmı ekledim. Burayı ister annual cost ta yaz istersen burda kalsın..altttaki Purchased cost ın altına appendix yazmalısın ok!!!
Q = U*A*F*DTLM
2292590 = 412*A*0.9*18.97
A =325.9 m2
A =3507.9 ft2
Purchased Cost ;25 000$ * (1092 / 904 ) =30 200 $
DIRECT COST (DC)
Purchased equipment cost ;( % 100 PEC) = 30 200$
Instillation; ( % 25 PEC ) = 7 550$
Instrumentation and control ( % 5 PEC ) = 1 510$
Piping ( % 10 PEC ) = 3 020$
INDIRECT COST (IDC)
Engineering & supervising ( %5 DC ) = 2 114 $
Constructor’s fee ( % DC ) = 4 228 $
Contingency ( % FCI ) = 2 559 $
FIXED CAPITAL INVESMENT (FCI) = DC + IDC
= 42 280 + 6342 +0.05 FCI
= 51 181 $
MANUFACTURING COSTS
Direct production costs
Utilities cost (UC) = Electricity + water
Maintenance-repair & Operating labour costs ( % 3 FCI ) = 1 535$
Fixed Charges
Depreciation ( % 7 FCI ) = 3 583 $
Insurance ( %0.4 FCI ) = 205 $



(Arkadaslar Direk Orijinalini Sundum İngilizce Yerleri Kendiniz Cevirirsiniz )

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