condenser design pb

1. Condenser Design
Hot Fluid Cold Fluid
= Methanol = Water
T1= 73 C0 T1=20 C0
T2= 45 C0 T1= 55 C0
m = 2548.31kg/min
Cp = 4.3306 kj/kg Co
Heat Duty
For Methanol
m = 2548.31/60 = 42.472kg/s
Q= mcpΔT = 42.472×4.3306×103 ×(73-45)
Q = 5150.02 kW
Mass transfer of Water
Q/ cpΔT=m= 5150.02/(4.18) ×(55-20)
m= 35.202kg/s
LMTD
ΔTlm=(73-55)-(45-20)/ln(73-55/45-20)
ΔTlm=21.309C0
R= T1-T2/t2-t1
R= 73-45/55-20=0.8
S= t2-t1/T1-t1
= 55-20/73-20 = 0.66

2. ΔTlm= 19.6 Co
Ft = 0.92
Diameter
I.D = 14 mm O.D = 16 mm length = 4.88 m
Water tubes
Tube side Heat transfer coefficient
Surface area
A = π do L = π (16 × 103) 4.88 = 0.2453 m2
Estimated (predicted) heat transfer coefficient
U = 800 w/m2C0
Estimated area
A = Q/ UΔTlm
= 5150.02 × 103/ 800 × 19.6 = 328.445 m2
Number of tubes
Nt = 328.445/ 0.2453 = 1336
A = π.di
2/4 = π ( 14 × 10-3)2/4 = 1.5394× 10-4 m2
Foreight passes Nt = 167
Velocity ( water )
vt = 35.202 / 993.18 × 167 ×1.5394×10-4
= 1.379 m/s
Heat transfer coefficient
hi = 4200( 1.35+ 0.02t) ut
0.8/di
0.2
hi = 4200( 1.35+ 0.02×37.5) (1.379)0.8/(14)0.2
= 6728.1302 w/m2C0

3. Shell side heat transfer coefficient
Bundle diameter
Db = d0 ( Nt/k1)1/n
1
Db = 16 ( 1336/0.0331)1/2.643
= 884.766 mm
Clearance = 68 mm
Ds = 952.766 mm
Pt = 1.25 do = 1.25 (16) = 20
Number of tubes in center row
Db/Pt = 884.766/20 = 44
Nr = 2 × 44 /3 = 29
ρL = 753.6 kg/m3
µc = 0.3567 × 10-3 Pa.s
kL = 0.20155 kg/mC0
ρv = 0.8688 kg/m3
Իn = Wc/L×Nt = 42.472/4.88 ×1336 = 6.5144×10-3 kg/m.s
(hc)b = 0.95kL [ρL( ρL– ρv)g/ µc Իn ]1/3 Nr-1/6
(hc)b = 0.95× 0.20155[753.6( 753.6 – 0.8688)9.8/ 0.3567×10-3×6.5144 ×10-3 ]1/3
×(29)1/6
= 1461 W/m2C0
Overall heat transfer coefficient
hod = 0.0002 hid = 0.0017
1/vo = 1/(hc)b + hod + d0.ln(do/di)/2kw + do ( 1/hi + hid)
1/vo = 1/1461 + 0.002+(16×10-3)ln(16/14)/2×16 + 16( 1/6718.1302 + 0.00017 )/14
vo =760.24 w/m2.C0