 Open Access
 Total Downloads : 453
 Authors : Reetika Saxena, Sanjay Yadav
 Paper ID : IJERTV2IS90384
 Volume & Issue : Volume 02, Issue 09 (September 2013)
 Published (First Online): 14092013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Designing Steps for a Heat Exchanger
Reetika Saxena 1
M.Tech. Student in I.F.T.M. University, Moradabad
Sanjay Yadav 2
Asst. Prof. in I.F.T.M. University, Moradabad
Distillation is a common method for removing dissolved solids and to obtain pure water for drinking and for the purpose like battery water, electroplating etc. Distillation of water have certain problems and operational issues and too like as it is an energy consuming process.
Multiple effect distillation is a distillation process generally used for sea water distillation. It consists of stages (effect). In first stage feed water is heated by steam in tubes. Some of the water evaporates and this fresh steam flows into the tube of the next stage. But for better heat transfer, it is necessary to design a heat exchanger which fulfills the requirements of MED unit.
The most common problems in heat exchanger design are rating and sizing. The rating problem is concerned with the determination of the heat transfer rate, fluid outlet temperature, inlet temperature, heat transfer area and the sizing problem involves determination of the dimension of the heat exchanger. An heat exchanger (shell and tube type) is being designed here with proper dimension for 45 Kg/hr steam. This method is also used for other temperature range and increased mass flow rate of steam. Using this design procedure of heat exchanger we can also increases the boiler efficiency which produces steam for multi effect distillation unit.
Related input data for the design of heat exchanger, which is used for multiple distillation unit is given below
Input steam pressure = 4 bar gauge Standard atmospheric pressure = 1.013 bar
Absolute pressure = Atmospheric pressure + gauge Pressure Therefore Input steam pressure = 1.013 + 4 = 5.013 bar
Inlet Temperature of Steam, Thi = 152.7 C at 5.013 bar Outlet Temperature of Steam, Tho = 117 C
Inlet Temperature of Water, Tci = 110 C Specific Heat for Steam, Cph = 2.5 kJ/kg K Specific Heat for Water, Cpc = 4.18 kJ/kg K Mass flow rate of Steam, mh = 45 kg/h
= 45/3600
= 0.0125 kg/s Mass flow rate of Water, mc = 250 kg/h
= 250/3600
= 0.0694 kg/s

Energy Balance Eq.,
mc * Cph * ( Tco Tci ) = mh * Cph * (Thi Tho )
0.0694 * 4.18 * ( Tco 110 ) = 0.0125 * 2.5 * (152.7 117 )
0.290092 * Tco 31.91012 = 1.1156
0.290092 * Tco = 33.025745 Tco = 113.85 C
Outlet Temperature of Water, Tco = 113.85 C

Heat Transfer Rate,
Q = mc*Cph* ( Tco Tci )
= 0.0694 * 4.18 * (113.85 110 )
= 1.17 kW

Log Mean Temperature Difference, LMTD for Parallel Flow,
Tho Tco
T1
=
Thi – Tci
Thi
=
152.7 110
=
42.7C
T2
=
=
Tho – Tco
117 113.85
=
3.15C
Tci
T1
=
Thi – Tci
Thi
=
152.7 110
=
42.7C
T2
=
=
Tho – Tco
117 113.85
=
3.15C
Tci
LMTD, Tm = T1 T2
ln (T1 /T2)
= 42.73.15
ln (42.7)
ln (42.7)
3.15
Tm = 39.55 / 2.6067 = 15.17C

Temperature Correction Factor,
The selection of parameters R and P should be such that the value of correction factor, Ft is more than 0.75
Capacity Ratio, R = Tci Tco
Tho Thi
= (110 113.85) / (117 152.7)
= 0.12
Temperature Ratio, P = Tho Thi Tci Thi
= (117 152.7) / (110 152.7)
= 0.84
The value may taken from the chart:
Ft =
2+1
1
Fig.1: Relation between R, P and Ft
1
1
2 +1 2+1
2 +1+ 2+1
Ft =
0.122+1
0.121
10.84
10.840.12
20.84 0.12+1 0.122+1
20.84 0.12+1+ 0.122+1
Ft = 0.904
Both result are same, therefore Ft = 0.904

Mean Temperature Difference, DTm = Ft * LMTD
= 0.904 * 15.17
DTm = 13.71C

Overall Heat Transfer Coefficient,
Study based on, Steps for design of Heat Exchanger by Dr. Reyad Shewabkeh, Dept. of Chemical Engineering, King Fahd University of Petroleum & Minerals, The range of overall heat transfer coefficient for water is 800 1500 w/m2 C.
U = 945 w/m2 C
Table 1: Overall heat transfer coefficient for different combination

Provisional Area,
A = Q
U T
= 1170 94513.71
= 0.090 m2 = 900 cm2

Tube Outer Diameter,
Case I Number of Tubes, Nt = 7
Length of Tubes, L = 100 cm do = 900 / (3.14*7*100) = 0.41 cm
= 4.1 mm
Case II Number of Tubes, Nt = 5
Length of Tubes, L = 100 cm do = 900 / (3.14*5*100) = 0.57 cm
= 5.7mm
Case III Number of Tubes, Nt = 7
Length of Tubes, L = 50 cm do = 900 / (3.14*7*50) = 0.82 cm
= 8.2 mm
Case IV Number of Tubes, Nt = 5
Length of Tubes, L = 50 cm do = 900 / (3.14*5*50) = 1.15 cm
= 11.5 mm
Above design calculation are not feasible because calculations are based on LMTD 13.71C. MED Unit will provide best result when condensation of steam will take place with minimum temperature difference.
Study based on paper Porteous,A. (1975), Saline water distillation Process (1st Ed) Longman UK, London, 150p, it is clear that condensation can take place with a temperature difference of 2C.
Therefore our design calculation will be based on 4C. DTm = 4C
A = Q
U DT m
= 1170 9454
= 0.309523 m2
= 3095.23 cm2
Case I
Number of Tubes, Nt
Length of Tubes, L
=
=
7
100 cm
do
= 3095.23 / (3.14*7*100)
=
=
1.408 cm
14.08 mm
Case II
Number of Tubes, Nt
Length of Tubes, L
=
=
5
100 cm
do
= 3095.23 / (3.14*5*100)
=
=
1.8207 cm
18.21 mm
Case III
Number of Tubes, Nt
Length of Tubes, L
=
=
7
50 cm
do
= 3095.25 / (3.14*7*50)
=
=
2.816 cm
28.16 mm
Case IV
Number of Tubes, Nt
Length of Tubes, L
=
=
5
50 cm
do
= 3095.25 / (3.14*5*50)
=
=
3.942 cm
39.42 mm
Case V
Number of Tubes, Nt
Length of Tubes, L
=
=
5
75 cm
do
= 3095.25 / (3.14*5*75)
=
=
2.628 cm
26.28 mm
Case IV
Number of Tubes, Nt Length of Tubes, L
=
=
7
75 cm
do
= 3095.25 / (3.14*7*75)
=
=
1.877 cm
18.77 mm
Sr. No.
NUMBER OF TUBES,
Nt
LENGTH OF TUBES,
L (mm)
OUTER DIAMETER,
do (mm)
1.
7
1000
14.08
2.
5
1000
18.21
3.
7
500
28.16
4.
5
500
39.42
5.
5
750
26.28
6.
7
750
18.77
Table 2: Different configuration for tubes
We get best result in case III, therefore outer diameter of tube is 28.16 mm, And the configuration of the tube is
do
=
=
28.16 mm
0.02816 m
Nt
=
7
L
=
50 mm

Tube Pitch,
Pt = 1.25 * do
= 1.25 * 28.16
= 35.2 mm

Bundle Diameter,
Db = do [ Nt / K1 ]1/n
Table 3: Relation between constant K1 and n1
For Square pitch,
Pt = 1.25 * do
K1 = 0.0366
N = 2.63
Db = 28.16 * [ 7 / 0.0366 ]1/2.63
= 207.57 mm
= 0.207 m

Bundle diameter clearance,
For fixed floating head, BDC = 10 mm
Fig. 2: Bundle diameter clearance

Shell Diameter,
Ds = Db + BDC = 207.57 + 10
= 217.57 mm

Baffle Spacing,
Bs = 0.4 * Ds
= 0.4 * 217.57
= 87.03 mm

Area for cross flow,
As = (Pt do ) Ds Bs
Pt
= 35.228.6 217.5787.03
35.2
= 3550.33 mm2
= 3.55 * 103 m2

Shell side mass velocity,
Gs = Shell Side flow rate [kg /s]
As
Shell side flow rate = 0.0694 kg/s
3
3
Gs = 0.0694
3.55 10
= 19.55 kg/m2 s
= 1.955 * 105 kg/mm2s

Shell equivalent diameter for a square pitch arrangement,
1.27 [ P2 0.785 d2 ]
de =
t o
do
= 1.27 [ 35.2 2 0.785 28.162 ]
28.16
= 27.81 mm = 0.02781 m

Shell side Reynolds number,
Properties of water at 110C temp.
Density ,
=
951 kg/m3
Kinematic viscosity,
=
0.273 * 106 m2/s
Fluid thermal Conductivity, kf
=
0.62 W/mK
Specific heat, Cp
=
4233 J/kgK
e
e
R = Gs de
= Gs de
19.55 0.02781
=
=
9510.273106
= 2094.1
Re > 2000 therefore flow inside shell side is Transition and Turbulent.

Prandtle number,
r
r
P = Âµ Cp
kf
= Cp
kf
= 9510.273 106 4233
0.62
= 1.77

Nusselt number,
Nu
=
0.023 * (Re)0.8 * (Pr)n
n
=
0.4 for heating
=
0.3 for cooling
Nu
=
=
0.023 * (Re)0.8 * (Pr)0.4
0.023 * (2094.1)0.8 * (1.77)0.4
=
13.11

Heat transfer coefficient,
o
o
h = Nu kf
de

Tube inside diameter, di = do – t
= 13.11 0.62
0.02781
= 292.35 W/m2 K
Thickness of tube metal = 6 mm
di = 28.16 6
= 22.16 mm
= 0.02216 m

Tube side Reynolds number,
Properties of steam at 152.7C temp.
Dynamic viscosity, Âµ
=
1.408 * 105 Ns/m2
Fluid thermal Conductivity, kf
=
0.0311 W/mK
Specific heat, Cp
=
2335.2 J/kgK
e
e
R = Gs di
19.55 0.02216
=
=
1.4085 105
= 100000
Re > 2000 therefore flow inside tube is Transition and Turbulent.

Prandtle number,
r
r
P = Âµ Cp
kf
= 1.4085 10 5 2335 .2
0.0311
= 1.057

Nusselt number,
Nu = 0.023 * (Re)0.8 * (Pr)0.3
= 0.023 * (100000)0.8 * (1.057)0.3
= 233.86

Heat transfer coefficient,
o
o
h = Nu kf
di
= 233 .86 0.0311
0.02216
= 328.21 W/m2 – K

Overall heat transfer coefficient in Shell and tube heat exchanger,
The heat transfer is in radial direction, firstly the heat is transferred by hot fluid to inner wall of tube by convection . then through the wall of the tube by conduction and finally from the outer wall of tube to cold fluid by convection.
Surface Area of inner tube, 
Ai 
= 
2riL 
Surface Area of outer tube, 
Ao 
= 
2roL 
Hot fluid in
Cold fluid in ri
ro
Cold fluid out
Hot fluid out
i To
i To
T
Ri Rwall Ro
Fig. 3: Heat transfer in shell and tube heat exchanger
Total Thermal resistance,
R = Ri + Rwall + Ro
R =
1 +
+
2
1
1
1
= 2 +
+
2
1
2
Q =
= UAT = U A T = U A T
R i i o o
Where, U = Overall heat transfer coefficient in W/m2K
R = 1
= 1
= 1
Overall heat transfer coefficient based on outside surface area of tube can be expressed as:
Uo =
1 =
R Ao
1
+
+ 1
= 1
2
+ + 1
=
=
1
0.01408 + 0.01408 0.01408 + 1
0.01108 328 .21 225 0.01108 292 .35
= 136.89 W/m2K
Overall heat transfer coefficient based on inner surface area of tube can be expressed as:
Uo =
1 =
R Ai
1
+
+ 1
= 1
2
+
+ 1
=
=
1
0.01108 + 0.01108 0.01408 + 1
0.01408 292 .35 225 0.01108 328 .21
= 136.89 W/m2K
After comparing the overall heat transfer coefficient, I obtained from previous step with that I assumed in step 6. It is smaller to what I assumed, then I have a valid assumption, that tabulate my results such as total surface area of tubes, number of tubes, exchanger length and diameter and other design specification.
Following conclusion can be made after study:

Six different combination of No. of tubes and length of tubes were tried. Above design of heat exchanger was best, because our calculated dimensions are verified very accurately.

Due to agronomic consideration of design, the tube of very large diameter is not selected.

Smallest diameter is not selected because it is not practically feasible.

An appropriate heat exchanger is designed for multiple effect distillation unit to condense 45 Kg/hr steam. Dimension of heat exchanger is given below
Number of tubes, Nt
=
7
Length of tube, L
=
500 mm
Outer diameter of tube, do
=
28.16 mm
Thickness of the tube, t
=
6 mm
Inner diameter of tube, di
=
22.16 mm
Overall heat transfer coefficient,U
=
136.89 W/m2K
Tube pitch, pt
=
35.2 mm
Bundle diameter, Db
=
207.57 mm
Bundle diameter clearance, BDC
=
10 mm
Shell Diameter, Ds
=
217.57 mm
Baffle spacing, Bs
=
87.03 mm
Area for cross flow, As
=
3550.33 mm2
Material selected
=
Aluminum
Following Modification can be done for future work:

Other type of heat exchanger can be design.

Flow of steam and water can be reverse. That means the feed water may be taken inside tubes and condensing steam outside tubes.

Pressure drop inside shell and tube can be calculated.

Counter flow arrangement can be tried.

Design may done for different types of floating head.

We can use some other material than aluminum for better heat transfer.

Effect of corrosion can be considered.


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