Heat exchanger complete design

1. BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY
Dhaka-1000, Bangladesh February 2019
ME 310: Thermo-Fluid System Design
Submitted To
Department of Mechanical Engineering, Bangladesh University of Engineering and Technology
Submitted By
FAHIM MAHMUD DIPTHO (1510071)
Md. Habibur Rahman (1510072)
Shailee Mitra (1510079)

2.  Abstract
A double pipe concentric, counter-current, single-phase heat exchanger with working fluid as
water and oil is designed in this current project. This heat exchanger is going to be used in a
thermal power plant for heating up furnace oil. Different governing equations and empirical
relations are used for design purposes. The different dimension were chosen in the software
interface and several iterations were carried out in search of a suitable optimized design such as
given heat duty and allowable pressure drop. Finally 3x2 STD type M copper tubing pipe having a
length of 4.88 meter (16 feet) has been chosen. So a single hairpin with a length of 8 feet is enough
to carry out our purposes. Pressure drop due to U bend in the tube section has been neglected.
The main goal was to produce a thermally and hydraulically optimized double pipe heat exchanger
with a suitable number of material. Investigation on mechanical strength has also been carried
out. Economics analysis has been included. We prepared a simple scaled down prototype of our
designed double pipe heat exchanger made of Iron . Scaling has been chosen arbitrarily. Besides
software analysis has been done on this heat exchanger to ensure high service life and
performance.
 Introduction
Heat Exchangers are devices that facilitate transfer of heat between two or more fluids having a
different temperatures. They are widely used in space heating, refrigeration, air conditioning,
power stations, chemical plants, petrochemical plants, petroleum refineries, natural-gas
processing, and many other areas. There are many types of heat exchanger such as shell and
tube, double pipe heat exchanger, cooling tower, plate and frame heat exchanger. In this current
project we wish to design a double pipe heat exchanger for a thermal power plant that will be used
to heat furnace oil up to a certain temperature. The exchanger is counter flow and two passes.
The double pipe heat exchanger will bring together two fluid streams. As each fluid passes
through its temperature changes and heat transfer takes place.
*Problem statement
Furnace oil is to be heated from 32.2o
C to around 38o
C. Hot water is available at 65.6o
C. Allowable
pressure drops are:
∆𝐏𝐭(𝐚𝐥𝐥𝐨𝐰𝐚𝐛𝐥𝐞) = 𝟓 𝐊𝐩𝐚
∆𝐏𝐚(𝐚𝐥𝐥𝐨𝐰𝐚𝐛𝐥𝐞) = 𝟐 𝐊𝐩𝐚

3.  Objectives
 To meet certain required heat duty with maximum performance
 To meet required outlet temperature of working fluid
 To meet required allowable pressure drop for both shell and tube side
 To meet good mechanical strength so that it can carry maximum stress
 To construct a scale down prototype maintaining similarity with original design
 Method/ Design strategy
1. Selection of Tube Diameter
2. Calculation of the Film Coefficients
3. Calculation of the Overall Heat Transfer Coefficient, U
4. Calculation of the LMTD
5. Calculation of the Heat Transfer Area
6. Calculation of the Total Tube Length and Number of Tubes
7. Calculation of the Pressure Drop
8. Changes to the Original Design
 Design factors considered
 Are there any space limitations that need to be accounted for?
 Types of fluid we are working
 Do any fluid contains solid
 Are there any fluctuations in the flows or temperature
 Shell / annular side pressure drop should be lower
 Fluid placement should be based on either the hydraulic criterion (minimizing the pressure
drop) or the fouling criterion (easy mechanical cleaning of the heat exchanger).
 Inner tube in a DPHX should be of high thermal conductivity (copper is a good choice).The
material for the outer tube does not need to be made of an expensive material such as
copper

4.  3D Design drawings
Figure 1:3D drawing of designed DHTX
Figure 2:3D drawing of designed DHTX (Transparent view)

5.  Thermal hydraulic calculation
Nomenclature
 T refers to the temperature of the warmer fluid
 t refers to the temperature of cooler fluid
 w subscript refers to the warmer fluid
 h subscript refers to hydraulic diameter
 c subscript refers to cooler fluid
 a subscript refers to annular flow area or dimension
 p subscript refers to tube flow area or dimension
 1 subscript refers to an inlet condition
 2 subscript refers to an outlet condition
 e subscript refers to equivalent diameter
Fluid properties:
Water (annular side)
𝑚 𝑤 = 0.700 𝑘𝑔/𝑠 𝑃𝑟 = 2.5
𝜌 𝑤 = 982 𝑘𝑔/𝑚3
𝑇1 = 65.6 𝑜
𝐶
𝐾𝑓 = 0.656 𝑊/𝑚𝑘 𝐶𝑝 = 4187 𝐽/𝐾𝑔𝐾
𝜗 = 4.47𝑥10−7
𝑚2
/𝑠 𝛼 = 1.595𝑥10−7
𝑚2
/𝑠
Oil (tube side)
𝑚 𝑐 = 0.600 𝑘𝑔/𝑠 𝑃𝑟 = 4836
𝜌𝑐 = 878 𝑘𝑔/𝑚3
𝑡1 = 32.2 𝑜
𝐶
𝐾𝑓 = 0.144 𝑊/𝑚𝑘 𝐶𝑝 = 1943 𝐽/𝐾𝑔𝐾
𝜗 = 3.97𝑥10−4
𝑚2
/𝑠 𝛼 = 8.44𝑥10−8
𝑚2
/𝑠
Tube sizing:
𝐼𝐷 𝑎 = 0.0757 𝑚
𝑂𝐷 𝑝 = 0.0540 𝑚
𝐼𝐷 𝑝 = 0.0510 𝑚
Flow Areas:
𝐴 𝑝 =
𝜋𝐼𝐷 𝑝
2
4
= 0.0020 𝑚2
𝐴 𝑎 =
𝜋(𝐼𝐷 𝑎
2
− 𝑂𝐷 𝑝
2
)
4
= 0.0022 𝑚2

6. Fluid Velocities:
𝑉𝑎 =
𝑚
𝜌𝐴
= 0.377 𝑚/𝑠
𝑉𝑝 =
𝑚
𝜌𝐴
= 0.032 𝑚/𝑠
Annulus Equivalent diameter:
Friction: 𝐷ℎ = 𝐼𝐷 𝑎 − 𝑂𝐷 𝑝 = 0.0217 𝑚
Heat Transfer: 𝐷𝑒 =
𝐼𝐷 𝑎
2−𝑂𝐷 𝑝
2
𝑂𝐷 𝑝
= 0.0521 𝑚
Reynolds number:
Water: 𝑅𝑒 𝑎 =
𝑉𝑎 𝐷 𝑒
𝑣
= 44068
Oil: 𝑅𝑒 𝑝 =
𝑉𝑝 𝐼𝐷 𝑝
𝑣
= 4.15
Nusselt number:
𝑁𝑢 𝑎 = 0.023(𝑅𝑒)
4
5 𝑃𝑟0.3
= 162
𝑁𝑢 𝑝 = 1.86(
𝐼𝐷 𝑝 𝑅𝑒𝑃𝑟
𝐿
)
1
3 = 10.8
Convective heat transfer coefficient:
Water: ℎ 𝑎 =
𝑁𝑢 𝑎 𝐾 𝑓
𝐷 𝑒
= 2042 𝑤/𝑚2
𝑘
Oil: ℎ 𝑝∗ =
𝑁𝑢 𝑝 𝐾 𝑓
𝐼𝐷 𝑝
= 30.65 𝑤/𝑚2
𝑘 ℎ 𝑝 = ℎ 𝑝∗
𝐼𝐷 𝑝
𝑂𝐷 𝑝
= 28.95 𝑤/𝑚2
𝑘
Exchanger coefficient:
1
𝑈0
=
1
ℎ 𝑝
+
1
ℎ 𝑎
𝑈 𝑜 = 28.5 𝑤/𝑚2
𝑘
Outlet temperature Calculation:
𝐶 =
𝑚 𝑐 𝐶 𝑝𝑐
𝑚 𝑤 𝐶 𝑝𝑤
= 0.032
𝐴 𝑜 = 𝜋𝑂𝐷 𝑝 𝐿 = 0.82 𝑚2
𝑁 =
𝑈𝐴
(𝑚𝐶 𝑝) 𝑚𝑖𝑛
= 0.20

7. Counter flow effectiveness:
𝐸 =
1 − exp[−𝑁(1 − 𝐶)]
1 − 𝐶exp[−𝑁(1 − 𝐶)]
= 0.18
𝑡2 = 𝑡1 + 𝐸(𝑇1 − 𝑡1) = 38.2 𝑜
𝐶
𝑇2 = 𝑇1 − 𝐶(𝑡2 − 𝑡1) = 65.40 𝑜
𝐶
Log mean temperature difference:
Counter flow
𝐿𝑀𝑇𝐷 =
(𝑇1 − 𝑡2) − (𝑇2 − 𝑡1)
ln(
𝑇1 − 𝑡2
𝑇2 − 𝑡1
)
= 30.15 𝑜
𝐶
Heat Balance:
Water: 𝑞 𝑤 = 𝑚 𝑤 𝑐 𝑤𝑐(𝑇1 − 𝑇2) = 0.690 𝑘𝑤
Oil: 𝑞 𝑐 = 𝑚 𝑐 𝑐 𝑝𝑐(𝑡2 − 𝑡1) = 0.690 𝑘𝑤
LMTD 𝑞 = 𝑈𝑜𝐴𝑜𝐿𝑀𝑇𝐷 = 0.690 𝑘𝑤
Fouling factors and Design coefficient:
𝑅 𝑑𝑖 = 3.52𝑥10−4
1
𝑈
=
1
𝑈 𝑜
+ 𝑅 𝑑𝑖 + 𝑅 𝑑𝑜 = 27.90
Heat transfer area and tube length:
𝑞
𝑈 𝑜(𝐿𝑀𝑇𝐷)
= 𝐴 = 0.82 𝑚2
𝐿 =
𝐴
𝜋𝑂𝐷 𝑝
= 4.83 𝑚 ≈ 4.88 𝑚

8. Friction Factors:
Oil : 𝑓𝑝 =
64
𝑅𝑒 𝑝
= 15.40
Water: 𝑓𝑎 = (1.82𝑙𝑜𝑔10
𝑅𝑒 𝑎
− 1.64)−2
= 0.021
Pressure drop calculation:
∆𝑝 𝑝 =
𝑓𝑝 𝐿
𝐼𝐷 𝑝
𝑃𝑝 𝑉𝑝
2
2
= 0.68 𝑘𝑝𝑎
∆𝑝 𝑎 = (
𝑓𝑎 𝐿
𝐷ℎ
+ 1)
𝑃𝑎 𝑉𝑎
2
2
= 0.4 𝑘𝑝𝑎
Pumping power required:
Tube side: 𝑊𝑝 =
1
𝜂 𝑝
𝑚
𝜌
∆𝑝 = 0.111 𝑤𝑎𝑡𝑡 (𝑤𝑖𝑡ℎ 40 % 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦)
Shell side: 𝑊𝑝 =
1
𝜂 𝑝
𝑚
𝜌
∆𝑝 = 0.85 𝑤𝑎𝑡𝑡 (𝑤𝑖𝑡ℎ 40 % 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦)
Final result:
We selected a 3x2 STD type M tubing for our design with a length of
4.88 m long. So a single hairpin DPHTX with a length of 8 feet will be
enough.

9. Temperature variation over DPHTX:
Pressure variation over DPHTX:

10. Wall resolution over DPHTX:
Surface velocity magnitude over DPHTX:

11.  Economics Calculation
Chart 1: Comparison between Materials Suitable for DHTX:
Chart 2: Material Density & Cost
Material Density (kg/m3) Cost per Ton
(in US dollar,$)
Aluminum 2700 2500.00
Cupper 8960 6670.00
Steel 8050 600.00
PVC 1380 150.00
Iron 7874 305.00
Name Availability
(10)
Cost
(30)
Thermal
Resistance
(25)
Estimated
life span
(10)
Wear
Resistance
(10)
Temperature
Range
(15)
Total
(100)
Aluminum 08 11 11 06 06 05 47
Copper 05 05 13 04 03 10 31
Steel 08 08 23 09 08 13 69
PVC 07 25 03 08 10 03 56
IRON 10 19 17 07 05 13 71

12. Why Iron (For scaled down porotype)?
From chart 1 we can see that Iron & steel are the 2 most suitable metals for making our double
pipe heat exchanger For our HTX we prefer Iron .Because
 Steel prices are higher than Iron
 We get almost same thermal resistance in much lesser price
 Thermal conductivity for iron & steel are almost same (at 125o
C iron 68 W/mo
C ,Steel 51
W/mo
C)
 Density of iron is less than steel .So our HTX will be lighter.
 Construction price for Steel are much higher.
 Melting temperature for Iron (1482-1593)o
C, for steel (1425-1540)o
C. So our HTX operating
temperature within range of Iron.
 Iron has fair lifespan within our budget
Operating & Service Costing (For scaled down prototype)
Cost types Cost in taka
Installation cost
Labor Cost
Pumping cost
Others cost
Total Cost Calculation (for scaled down prototype)
 Length of the HTX =
 Tube material Iron
 Shell material Iron
 Volume of HTX =
 Mass of the HTX =
 Per kg Iron price in Bangladesh = taka
 Total cost for buying Iron =
 Total Cost after 1 year =

13.  Mechanical calculation:
For Tube (Copper)
Dimension and pressure
𝑟𝑜 = 0.0255 𝑚
𝑟𝑖 = 0.027 𝑚
𝑃𝑖 = 5 𝑘𝑝𝑎
𝑃𝑜 = 2𝑘𝑝𝑎
L = 4.88 m
Volume = area x length=
𝜋
4
(𝑑 𝑜
2
− 𝑑𝑖
2
)𝐿 = 0.0012𝑚3
Mass = 𝜌𝑉 = 10.80 𝑘𝑔
Weight = 105.8 N
Moment of inertia, 𝐼 =
𝜋
64∗2
(𝑑 𝑜
4
− 𝑑𝑖
4
) = 4.265𝑥10−8
𝑚4
𝐶 =
𝑟𝑜 − 𝑟𝑖
2
= 7.5𝑥10−4
𝑚
Principle stress calculation
First principle stress 𝜎1 = 𝜎1 = 𝜎𝑡|𝑟=𝑟 𝑖
=
𝑃 𝑖 𝑟𝑖
2
−𝑃𝑜 𝑟 𝑜
2−
𝑟 𝑖
2 𝑟 𝑜
2(𝑃 𝑜−𝑃 𝑖)
𝑟 𝑖
2
𝑟 𝑜
2−𝑟𝑖
2 = 50 𝑘𝑝𝑎
Second principle stress 𝜎2 = 𝜎2 = 𝜎𝑟|𝑟=𝑟 𝑖
=
𝑃 𝑖 𝑟𝑖
2
−𝑃𝑜 𝑟 𝑜
2+
𝑟 𝑖
2 𝑟 𝑜
2(𝑃 𝑜−𝑃 𝑖)
𝑟 𝑖
2
𝑟 𝑜
2−𝑟𝑖
2 = −5.5 𝑘𝑝𝑎
Third principle stress 𝜎3 = 𝜎𝑏𝑒𝑛𝑑𝑖𝑛𝑔(𝑡𝑢𝑏𝑒) = −
𝑀𝐶
𝐼
=
𝐹
𝑙
2
𝑐
2𝐼
= −465 𝑘𝑝𝑎
Von mises stress
𝜎′
= [
( 𝜎1 − 𝜎2)2
+ ( 𝜎2 − 𝜎3)2
+ ( 𝜎3 − 𝜎2)2
2
]
1
2 = 461 𝑘𝑝𝑎
Static factor of safety
𝑛 =
𝑆 𝑦
𝜎′
=
45000 𝑘𝑝𝑎(𝑆𝑜𝑓𝑡 𝑐𝑜𝑝𝑝𝑒𝑟 𝑦𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ)
461 𝑘𝑝𝑎
= 98 (𝑂𝑘!)

14. For Annuls (Copper)
Dimension and pressure
𝑟𝑜 = 0.03785 𝑚
𝑟𝑖 = 0.0385 𝑚
𝑃𝑖 = 2 𝑘𝑝𝑎
𝑃𝑜 = 0 𝑘𝑝𝑎
L = 4.88 m
Volume = area x length=
𝜋
4
(𝑑 𝑜
2
− 𝑑𝑖
2
)𝐿 = 7.60𝑥10−4
𝑚3
Mass = 𝜌𝑉 = 6.80 𝑘𝑔
Weight = 66.64 N
Moment of inertia, 𝐼 =
𝜋
64∗2
(𝑑 𝑜
4
− 𝑑𝑖
4
) = 5.86𝑥10−8
𝑚4
𝐶 =
𝑟𝑜 − 𝑟𝑖
2
= 3.25𝑥10−4
𝑚
Principle stress calculation
First principle stress 𝜎1 = 𝜎1 = 𝜎𝑡|𝑟=𝑟 𝑖
=
𝑃 𝑖 𝑟𝑖
2
−𝑃𝑜 𝑟 𝑜
2−
𝑟 𝑖
2 𝑟 𝑜
2(𝑃 𝑜−𝑃 𝑖)
𝑟 𝑖
2
𝑟 𝑜
2−𝑟𝑖
2 = 117 𝑘𝑝𝑎
Second principle stress 𝜎2 = 𝜎2 = 𝜎𝑟|𝑟=𝑟 𝑖
=
𝑃 𝑖 𝑟𝑖
2
−𝑃𝑜 𝑟 𝑜
2+
𝑟 𝑖
2 𝑟 𝑜
2(𝑃 𝑜−𝑃 𝑖)
𝑟 𝑖
2
𝑟 𝑜
2−𝑟𝑖
2 = −2 𝑘𝑝𝑎
Third principle stress 𝜎3 = 𝜎𝑏𝑒𝑛𝑑𝑖𝑛𝑔(𝑡𝑢𝑏𝑒) = −
𝑀𝐶
𝐼
=
𝐹
𝑙
2
𝑐
2𝐼
= −465 𝑘𝑝𝑎
Von mises stress
𝜎′
= [
( 𝜎1 − 𝜎2)2
+ ( 𝜎2 − 𝜎3)2
+ ( 𝜎3 − 𝜎2)2
2
]
1
2 = 470 𝑘𝑝𝑎
Static factor of safety
𝑛 =
𝑆 𝑦
𝜎′
=
45000 𝑘𝑝𝑎
470 𝑘𝑝𝑎
= 95(𝑂𝑘!)

15.  A similar approach has been done using software analysis to find out stress
distribution through the heat exchanger wall.
Governing equations used for stress study:
0=∇𝑠 + 𝐹𝑣
𝜀𝑙𝑖𝑛𝑒𝑙 = 𝜀 𝑜 + 𝜀 𝑒𝑥𝑡 + 𝜀𝑡ℎ + 𝜀ℎ𝑠 + 𝜀 𝑝𝑙 + 𝜀 𝑐𝑟 + 𝜖 𝑢𝑝
𝑆 𝑜 = 𝑆 𝑒𝑥𝑡 + 𝑆 𝑜 + 𝑆 𝑞
𝜖 =
1
2
[(∇𝑢) 𝑇
+ (∇𝑢)]
𝑐 = 𝑐(𝐸, 𝜗)
Figure 3: Meshed geometry

16.  Acknowledgement
We would like to express our gratitude to our course supervisors whose amicable and generous
guidance helped us to overcome various problems of our project and helped us by motivating
through our hard times. We are thankful to our honorable teachers-, Dr. Md. Ashiqur Rahman
(Assistant Professor, Department of Mechanical Engineering, BUET), Md. Rakib Hossain
(Lecturer, Department of Mechanical Engineering, BUET. Credit also goes to Dr. Md. Zahurul Haq
(Professor, Department of Mechanical Engineering, BUET).
Moreover, we are truly grateful to Md. Rakib Hossain for his continuous support and dedication to
solve our unwanted problems which helped us in the long run. He was always there when we were
in need of guidance and any kind of help.
 References
1. Heat and Mass Transfer – Fundamentals and Applications, Y. A. Cengel, A. J. Ghajar.
2. Principles of Heat Transfer, F. Kreith, R. M. Manglik, M. S. Bohn.
3. Heat Transfer in Process Engineering, E. Cao.
4. Design of Fluid Thermal Systems, W. S. Janna.
Figure 4: Surface von mises stress over DPHTX