Design of condenser

1. •Introduction
•A condenser is a type of heat exchanger in which
vapors are transferred into liquid state by removing
the latent heat with the help of a coolant such as
water.
•Condensers may be classified into two main types:
1.Those in which the coolant and condensing vapor
are brought into direct contact.
2.Those in which the coolant and condensate stream
are separated by a solid surface, usually a tube
wall

2. Different types of the Condenser
1.Double pipe and multiple pipe
2.Plate Condensers
3.Air-Cooled Condensers
4.Compact Condensers
5.Shell & tube type

4. DESIGN CALCULATIONS FOR CONDENSER
•Inlet temperature of the process stream ‘T
1
’ = 45
o
C
•Outlet temperature of the process stream ‘T
2
’ = 45
o
C
•Inlet temperature of the water ‘t
1
’ = 25
0
C
•Outlet temperature of the water ‘t
2
’ = 40
o
C
•Mass flow rate of the process stream ‘m’ = 8060 Kg/hr
•Enthalpy of Vapors of Process Stream = 1940 KJ/Kg
Removed ‘λ
1
’
T
2
= 45
o
C
t
1
= 25
o
C t
2
= 40
o
C
T
1
= 45
o
C
Condenser

5. Heat Load:
Q = m (λ
1
)
Q = 4343 KW
Mass flow rate of cooling water
ΔtC
Q
m
p
=


= 68.9 Kg/sec

Log Mean Temperature Difference
LMTD = (Dt2-Dt1)/ log (Dt2/Dt1)
LMTD = 14.4
o
C
C
p
= 4.2 KJ / Kg.K

6. Assumed Calculations
Assumed Value of Overall Coefficient ‘U
D
’ = 1000 W/m
2
C
True Mean Temperature Difference
Dimensionless Temperature Ratios
R
R = (T1-T2) / (t2-t1)
= (45-45) / (40-25)
= 0
S
S = (t2-t1) / (T1-t1)
= (40-25)/ (45-25)
= 0.75

7. From Literature the value of F
t


is 1

Dt
m
= F
t
x LMTD
= 1 x 14.4
= 14.4
o
C
Heat Transfer Area


= 301 m
2

Surface Area of single tube = 3.14 x 19 x 4.88 / 1000
= 0.292 m
2

No. of tubes = 301/.292
= 1030
Pitch ‘P
t
’ = 1.25 ´ 19.05= 23.8 mm

ΔtU
Q
A
D
=

8. Tube Bundle Diameter
Db = d
0
(Nt/K
1
)
1/n1
= 19 (1030/0.158)
1/2.263
= 920 mm
No. of tubes in centre row
Nr = Db / Pt
= 920 / 23.8
= 39
Shell Side Calculations
Estimate tube wall temperature T
w
Assume condensing coefficient of 4250 W/m
2
C (from literature)
Mean Temperature
Shell side =( 45+45) / 2 = 45
o
C
Tube side = (25+40) / 2 = 32.5
o
C
(45-T
w
) x 4250 = (45-25) x 1000
T
w
= 40.3
o
C

9. Physical Properties
Viscosity of the liquid ‘µ
L
’ = 0.8 mNs/m
2
Density of liquid ‘ρ
L
’ = 993 Kg/m
3
Thermal conductivity ‘k
L
’ = 0.571 W/m C
Average M. Wt. of Vapors = 42.8
Density of vapor = 29 x 273 x 1/(22.4 x 1 x (273+42))
= 1.12 Kg / m3

Condensate loading on a horizontal tube ’Ѓh’ = m / L x Nt
= 8060 / 3600 x (4.88 x 1030)
= 4.45 x 10
-3
Kg/m s
# of tubes in the vertical row ’Nr’ = 2/3 x 39 = 26 mm
Heat transfer coefficient in condensation
‘h
0
’ = 0.95 x k
L
( ρ
L
x (ρ
L
– ρ
v
) g / (µ
L
x Ѓh)
1/3
x Nr
-1/6
= 4396.0 W/m
2о
C
•As our assumed value is correct so no need to correct the
wall temperature

10. Tube Side Calculations

Tube cross sectional area = 3.14 / 4 x (19 x 10
-3
)
2 x
1030 / 4
= 0.073 m
2
Density of water at 30
0
C = 993 kg/m3
Tube velocity = m / (ρ
H2O
x A
t
)
= 68.9 / (993 x 0.073)
= 0.95m/s
Film heat transfer coefficient inside a tube
‘h
i
’ = 4200(1.35+0.02 x t) Vt
0.8
/ d
i
. 0 2
= 4809.67 W/m
2 0
C
From Literature take fouling factor as 6000 W/m
2 0
C
Thermal Conductivity of the tube wall material
‘K
w
’ = 50 W/m
0
C

11. Overall Coefficient
1/U
0
= 1/h
o
+ 1/h
od
+ (d
0
ln(d
0
/di))/2k
w
+d
0
/di x 1/h
id
+d
0
/d
i
x 1/h
i
= 0.001
U
0
= 1100.29 W/m
2 0
C
So assumed value is correct

12. Shell Side Pressure Drop
For pull through floating head with 45% cut baffles
From literature clearance = 88 mm
Shell internal diameter ‘Ds’ = Db+88
= 1008 mm
Cross flow area ‘A
s
’ = m
2

A= 0.205 m
2
Mass Velocity
G
t
= m / A
s
= 8060 / (3600 x 0.205)
= 10.92 Kg/s m
2
Equivalent diameter ‘de’ = 1.27 (Pt
2

-0.785d
0
2
) / d
0
= 19 mm
Viscosity of vapors ‘m’ = 0.009 mNs/m
2
Reynold’s No.
Re = d
e
G
t
/ m
= 19 x 10
-3
x 10.92 /0.009 x 10
-3

Re = 23053

13. From literature
jf = 0.029
By neglecting the viscosity correction factor



Where
Ds = dia of shell
L = Length of tubes
l
B
= baffle spacing
So
= 765 N/m2
= 0.765 Kpa
= 0.109 Psi

14. Tube side pressure drop

Viscosity of water ‘µ’ = 0.9 x 10
-3
Ns/m
2
Re = Vt ρ di /µ
= 0.95 x 993 x 16.56 x 10
-3
/ 0.6 x 10
-3
= 26036
From literature
jf = 0.0039
Where
Np = No. of tube passes
So
∆P
t
= 4119.8 N/m
2
= 4.119Kpa
= 0.59 psi
Acceptable

15. h
io
= hi ´I.D/O.D
h
io
= 4165.2 W/m
2 0
C

Clean Overall Coefficient:


= 2138.7 W/m
2 0
C

Design Overall Coefficient Calculated

dirt factor Rd = 0.0005

D
D
UU
UU
R
C
C
d
-
=

oio
oio
C
hh
hh
U
+
=

16. SPECIFICATION SHEET CONDENSER
Identification: Item condenser
No. Required = 8
Function: Condense vapors by removing the latent heat of vaporization

Operation: Continuous
Type: 1-4 Horizontal Condenser
Shell side condensation
Heat Duty = 4343 KW
Tube Side:
Fluid handled: Cold Water
Flow rate = 68.9 Kg/sec
Pressure = 14.7 psia
Temperature = 25
o
C to 40
o
C
Tubes: 0.75 in. Dia.
1030 tubes each 16 ft long
4 passes
23.8 mm triangular pitch
Pressure Drop = 0.59 psi
Shell Side:
Fluid handled = Steam
Flow rate = 8060 Kg/hr
Pressure = 10 KPa
Temperature = 45
o
C to 45
o
C
Shell: 39 in. dia. 1 passes
Baffles spacing = 3.5 in.
Pressure drop =0.109 psi
Utilities: Cold water
Ud assumed = 1000 W/m
2
C Ud calculated =1100.97 W/m
2
C
Rd = 0.0005

17. References
•Chemical Engineering Design
Volume 6 by Coulson &v Richardson’s
•Process Heat Transfer
by D.Q. Kern
•Plant Design & Economics for Chemical Engineers
5
th
Edition by Max S. Peters, Klaus D. Timmerhaus,
Ronald E. West
•Perry’s Chemical Engineers’ Handbook
by Robert H. Perry, Don. W. Green