Excerpt
1
Content
Title
page
1. Introduction ...2
2. Pre-heater design ...4
2.1 shell side coefficient ... 5
2.2 pressure drop ...6
3. Pump selection ...7
3.1 pipe selection ...7
3.2 cavitation... ....................................................................9
3.3 pump specification ...9
4. Distillation column design ...10
4.1 equilibrium data ...12
4.2 column design ...13
4.3 diameter design ... ....13
4.4 perforation area ... ...14
4.5 weir height ...15
5. Storage tank design ...16
6. Condenser design ...18
6.1 shell side coefficient ...20
6.2 tube side coefficient ...21
6.3 shell side pressure drop ...23
6.4 tube side pressure drop ...24
6.5 summary of proposed design ...24
7. Pipe insulation ...25
8. Notations ...26
9. Acknowledgement ...27
10. Refference ...28
2
1
Introduction
As it was shown in the design set up, the design project contains
some parts of typical plant unit operations that could be designed by the given
data's. In this paper the design analysis begins from the distillation column, by
collecting data's such as the vapor liquid equilibrium of the mixture and the
physical properties of the mixture from chemical engineering handbooks
[Perry].The McCabe Thiele method is used to determine the number of
theoretical stages by plotting the operating and stripping line .The column
diameter, perforations and tray spacing also designed.
Next, the heat exchanger design was done (U-tube) for heating
purpose. By using a steam as a heating media and the type of material of
construction, tube side and shell side pressure drops, generally thermal and
mechanical designs is done. Similar procedure was followed for heat exchanger
two (condenser).
Pump selection is performed by using Bernoulli's equation, in order to
calculate the power needed by the pump by taking into consideration the head
losses due to pipe roughness, fitting loss and heat exchanger shell side
pressure losses. Finally the net positive suction head available (NPSHA) was
checked to be above the net positive suction head required (NPSHR) which is
specified by the manufacturer.
A cylindrical liquid storage tank made of stainless steel was chosen
and the minimum surface area needed for the construction is calculated and
finally it is suggested that an insulating material with low thermal conductivity
is selected for the pipes after the heat exchanger one, for the distillation
column and for the pipe parts before and after the condenser.
Finally the design analysis does not consider the economic cost.
3
2. Pre-heater design
Cold
stream
(alcohol)
hot
stream
(steam)
m
c
= 4 Kg/s
m
h
=
T
ci
= 25
0
C
T
hi
= 170
0
C
T
co
= 81.8
0
C
T
ho
= 110
0
C
T
mean
= 25+81.8/2 = 54
0
C T
mean
= 170+110 /2 = 140
0
C
Physical properties of steam
Physical properties of alcohol
@T
mean
= 140
@
Tmean
= 140
C
p
= 2043 J/Kgk
C
p
= 3462 J/Kgk
µ = 0.0002838 Pa.s
µ = 0.000578 Pa.s
K = 0.0244 W/m.k
K = 0.42 W/m.k
= 0.579 Kg/m
3
mix
= 868 Kg/m
3
Heat load of the pre-heater is given by
Q = m
c
C
p
T
= 4 * 3462 * (81.8 - 25) = 786623.2 Watt
The heating steam flow rate would be
m
h
= Q / C
p *
T
m
h
= 786623.2 /2043 * (170 -110) = 6.42 Kg/s
T
lm
= (170-81.8) (110-25)/ ln (170-81.8)/(110-25) = 86.6
0
C
The correction factors for a one shell and two tube passes is found from a chart
by
R = 170 -110 /81.8 -25 = 1.056
And P = 81.8 -25 /170-25 = 0.4
Therefore Ft = 0.94
And
T
m
= T
lm
* Ft = 86.6*0.95 = 81.4
0
C
Assuming an overall heat transfer coefficient of Uo = 800 W/m
20
C
4
A = Q /Uo * T
m
= 786623.2 /800*81.4 = 12.08m
2
Selecting a 20mm o.d, 16mm i.d and 4.88m length pipe, square pitch tube
arrangement made from stainless steel.
Area of one pipe = *do *L = 3.1415*0.02*4.88 = 0.31 m
2
Number of tubes (Nt) = 12.08/0.31 = 39
Tube side mass flow rate is
Gt = m
t
/A = 6.424/0.31 = 20.7Kg/ m
2
s
Tube side velocity is
Vt = Gt/ = 20.7/0.579 = 35.75 m/s
Tube side convective heat transfer coefficient for water is given by
hi = 4200*(1.35 + 0.02*T)Vt
0.8
/d.i
0.2
hi = 4200*(1.35 + 0.02*140)35.75^0.8 /16^0.2 =
2.1 Shell side coefficient
The bundle diameter (Db) is
Db = do *(Nt/k)
1/n
= 20(78/0.156)
1/2.291
= 301.4mm
For a square pitch the value of k=0.156 and n=2.291
And from chart the bundle clearance is 12mm
then shell diameter is Ds= 12mm + 301.4mm = 313.4mm
The baffle space is given by
l
b
= Ds/5 = 313.4/5 = 62.7mm
Since it's a square pitch Pt = 1.25do = 1.25*20 = 25mm
The shell side cross sectional area is
As = (Pt-do)*D
s
*l
b
/Pt = (5/25)*313.4*62.7*10^-6 = 0.00393 m
2
Gs = m
s
/A
s
= 4Kg/s / 0.00393 m
2
= 1018Kg/ m
2
s
The Reynolds number is
Re = G
mix
*de /µ
mix
= 1018*0.0142 / 0.000578 = 25010
Prandts number is Pr = Cp*µ*/k =4.8
j
h
factor for 25% BC from chart is found to be jh = 3.8*10
-3
5
Therefore shell side convective heat transfer coefficient is
h
o
= (Kf/de)* jh*Re*Pr
1/3
= 4741W/ m
2 0
C
Finally, the overall heat transfer coefficient (U) is
1/U = 1/ho + 1/h
od
+ [do ln (do/d
i
) / 2kw] + (do/di)*1/hid + (do/di)*1/hi
U =784.5 W/m
20
C
Uassumed Ucaculated /Uassumed <20% ... It's acceptable
2.2 Heat exchanger pressure drops
1. Tube side pressure drop:
p = Np*8jf (L/di)Vt
2
/2(2.5 + (µ/µ
m
)
-m
Where m = 0.25 for laminar flow, Re <2100
= 0.14 for turbulent flow, Re>2100
P = Np[8jf(L/di)(µ/µ
w
)
-0.14
+ 2.5]* v
2
/2
Where N
p
is number of tube passes
J
f
is found from a chart
2 8x3.6x10 x
.
.
2.5
.
.
P = 4039926. 124 pa= 40.4 bar
2.3 Shell side pressure drop
8
2
.
8 3.8 10
.
.
.
.
= 29532.6 Pa = 0.3 bar
6
3. Pump selection
Given data's
m = 4 kg/s.
Volumetric flow rate = m/ = 4kg/s /798kg/m
3
= 5.01*10^-3 m
3
/s
T= 25
0
C
Selecting a centrifugal pump of 60% efficiency
3.1 Pipe selection
The optimum pipe diameter is calculated by
d
optimum
= 260G
0.52
-0.37
= 260*4^0.52 * 798^-0.37 40mm
So, using a commercial steel pipe with Nominal pipe size of 1
, schedule
40, 40mm i.d, and 48 mm o.d is selected.
Pipe roughness of 0.046 mm
From the given data's
We have 3 control valves (one gate valves and two globe valves) with loss,
f= 21 pipe diameter
= 2 Elbows (60 pipe diameters)
= Heat exchanger pressure drop = 30 bar = 3.83m
. .
30,000
798
9.81
3.83
- 3o meter vertical pipe & 10m horizontal pipe.
Solution
- Flow area =
.
1.256 10
- Flow mean velocity
.
.
3.98 /
- At T= 298
0
K,
- µ
mix
= 168*10^-3 Pa.s
-
mix
= 798kg/ m
3
by applying Bernoulli's equation
7
2
2
V
A
= is negligible because of longer tank diameter compared to the pipe.
h
A
= 0, h
B
= 30 m
pumpwork
h
H
Using Darcy equation
4
Re
798 0.04 3.98
1.168x10
108768and d
0.046
40
0.0015
From moody chart of R
e
V
s
,
0.00275
Therefore 4
4 0.00275
.
.
.
.
H
f
= 21 m
Total pipe length = 30 m+10M + (21 x 0.04) + (60x0.04) +3.83 m
L
tot
= 47.07 m
Pump work = (3.98
2
/ 2*9.81) +30m+21m=51.8m = 51.8m
Power used by the pump = gQh
= 798 x9.81 x 5.01 x10
-3
x51.8m
Power= 2.03 kW
Power to be supplied = P
out
/ efficiency = 2.03 / 0.6 = 3.38Kw = 4.5Hp
From pump characteristic curve- the pump will be a 5 inch impeller,
centrifugal pump of 5hp.
Excerpt out of 27 pages
- Quote paper
- Yasabie Abatneh (Author), 2013, Design Of Different Equipment, Munich, GRIN Verlag, https://www.grin.com/document/212472
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