This presentation shows the steps of shell and tube heat exchanger design

1. Design of shell and tube heat
exchanger
Submitted By:
Mahmoud Mohammed
Supervision by:
Dr. Mohammed El dalli
Dr. Asma Mustafa
Chemical & Petroleum Engineering
Department
Faculty of Engineering- Al-khoms
Elmerghib University

2. Gas oil at 200oC is to be cooled to 40oC. The oil flow-rate is 22,500 kg/h.
Cooling water is available at 30oC and the temperature rise is to be limited to
20oC. The pressure drop allowance for each stream is 100 kN/m2. Design a
suitable exchanger for this duty.
Description Symb
ol
Valu
e
Unit
Hot stream mass flow rate mh 22500 kg/h
Inlet
Temperature
Ti,h 200 oC
Outlet
Temperature
To,h 40 oC
Cold stream Inlet
Temperature
Ti,c 30 oC
Outlet
Temperature
To,c 50 oC
Pressure drop ΔP 100 kN/m
2
𝑄𝑐 = 𝑚𝑐𝐶𝑐(𝑇𝑜,𝑐 − 𝑇𝑖,𝑐)
𝑄ℎ = 𝑚ℎ𝐶ℎ (𝑇𝑖,ℎ − 𝑇𝑜,ℎ)
𝑄ℎ = 𝑄𝑐
𝑚ℎ𝐶ℎ 𝑇𝑖,ℎ − 𝑇𝑜,ℎ = 𝑚𝑐𝐶𝑐(𝑇𝑜,𝑐 − 𝑇𝑖,𝑐)
𝑚𝑐 =
𝑚ℎ𝐶ℎ(𝑇𝑖,ℎ − 𝑇𝑜,ℎ)
𝐶𝑐(𝑇𝑜,𝑐 − 𝑇𝑖,𝑐)
mc= 98181.82 kg/h
Step#1

3. Physical properties of water at 40 oC Gas oil at oC
120
Heat Capacity Cc= 4.18 kJ/kg.oC Ch= 2.28 kJ/kg.oC
Thermal conductivity kc= 6.31E-04 kW/m.oC kh= 0.125 W/m.oC
Viscosity μc= 6.71E-01 mN/m2 s μh= 0.17 mN/m2 s
Density ρc= 992.8 kg/m3 ρh= 850 kg/m3
Prandtl number Prc= 4.444976 Prh= 3.1008
℃
Number of material Material name Thermal conduct.
7 Stainless steel(18/8) 16 W/m.oC
Select Shell Type
Type of shell
One shell passes
Two shell passes
Divided-flow shell
Split- flow shell
Step#2
Step#3

4. To,h
Ti,c
ΔT1
Ti,h
To,c
ΔT2
∆𝑇𝑙𝑚 =
∆𝑇1 − ∆𝑇2
ln(
∆𝑇1
∆𝑇2
)
ΔTlm= 51.69771 oC
𝑅 =
𝑇𝑖,ℎ − 𝑇𝑜,𝑐
𝑇𝑜,𝑐 − 𝑇𝑖,𝑐
𝑆 =
𝑇𝑜,𝑐 − 𝑇𝑖,𝑐
𝑇𝑖,ℎ − 𝑇𝑖,𝑐
ΔTm= 48.59585 oC
F= 0.94
∆𝑇𝑚= 𝐹 ∗ ∆𝑇𝑙𝑚
Step#4

5. 𝐴𝑜 = 𝑞/(𝑈. ∆𝑇𝑙𝑚)
𝑞 = 𝑚ℎ𝐶ℎ(𝑇𝑖,ℎ − 𝑇𝑜,ℎ)
q= 2280 KW
Ao= 75.06814 m2
𝐴𝑠 = 𝜋 𝐷𝑜 𝐿
𝑁𝑡 =
𝐴
𝐴𝑠
𝐴𝑐 =
𝜋 𝐷𝑖
2
4
𝐴𝑡 = 𝑁𝑡 ∗ 𝐴𝑐
Surface area of one tube
Number of Tubes
Cross-sectional area " one tube "
Total tube area
Do= 16 mm
t= 1.6 mm
Di= 12.8 mm
L= 2.44 m
As =0.122648 m2
Nt=613
AC =0.000129 m2
At =0.078881 m2
Step#5 Step#6
Step#7

6. No. of passes= 4
Geometry of tube arrangement:
Triangular Pitch
K1= 0.175
n1= 2.285
Db= 0.569232 m
(Shell inside-bundle)diameter = 91.32309 mm
Shell inside diameter = 660.5554 mm
Pt= 20 mm
Step#8

7. 𝑢𝑡 =
𝑚
𝜌𝑤 ∗ 𝐴𝑡
Tube Water velocity, ut= 0.34825431 m/s
Tube velocity for Passes= 1.39301726 m/s
Heat transfer coefficient hi= 7069.86228 W/m2.oC
𝑑𝑒 =
𝐴
𝑑𝑜
(𝑃𝑡
2
− 𝐵𝑑𝑜
2
) de= 11.3608 mm
Baffle spacing lB= 0.132111m
Tube pitch Pt= 20mm
Total area As= 0.017453m2
Mass velocity Gs= 358.0977kg/m2.s
Linear velocity us= 0.421291m/s
A 1.1
B 0.917
Re= 23931.04
Pr= 3.1008
Step#9
Step#10

8. 1.00E-04
1.00E-03
1.00E-02
1.00E-01
1.00E+00
1.00E+01 1.00E+02 1.00E+03 1.00E+04 1.00E+05 1.00E+06
Heat
transfer
factor,
j
h
Reynolds number, Re
Buffle cut, percent
15
25
35
45
𝑗ℎ = 𝑐𝑅𝑒𝑚
jh= 0.0038935
Heat transfer coefficient in shell hs=
1494.9521 W/m2.oC
c= 0.4275
m= -0.466
do 16 mm
Di 12.8 mm
Kw 16 W/m .oC
Ho 1494.952 W/m2.oC
hi 7069.862 W/m2.oC
hod= 5000 W/m2.oC hid= 4255.319 W/m2.oC
Uo= 689.1578 W/m2.oC
Error%= 10.2653 %
Step#11

9. jf = 8.0722Re-1.004
R² = 1
jf = 0.0169Re-0.09
R² = 1
jf = 0.0725Re-0.296
R² = 1
jf = 0.0438Re-0.241
R² = 1
jf = 0.0303Re-0.206
R² = 1
0.001
0.01
0.1
1
10 100 1000 10000 100000 1000000
Heat
transfer
factor,
j
f
Reynold,Re
Reynold vs. Heat transfer factor
𝑗𝑓 = 𝑧𝑅𝑒𝑤
z= 0.0438
w= -0.241
ΔPt= 31.7643 kPa
Step#12

10. 𝑗𝑓 = 𝑥𝑅𝑒𝑦
0.01
0.1
1
10
1 10 100 1000 10000 100000 1000000
Heat
transfer
factor,
j
f
Reynold number, Re
15
25
35
45
x= 0.1939
y= -0.148
ΔPs= 28.2544 kPa
jf= 0.043601

11. 𝐸𝑥𝑐ℎ𝑎𝑛𝑔𝑒𝑟 𝐶𝑜𝑠𝑡, $1000 = 𝑒 ∗ 𝐴𝑟𝑒𝑎 𝑟
Total area = 75.06814 m2
Exchanger Cost= 78874.27 $
Pressure factor= 1.1
Type factor= 1
Purchased cost = (bare cost from figure) * Type factor * Pressure factor
Purchased cost= 86761.7 $
"Time base 2004"
𝐶𝑜𝑠𝑡 𝑎𝑡 2019 = 𝐶𝑜𝑠𝑡 𝑎𝑡 2004 ∗
𝐼𝑛𝑑𝑒𝑥 𝑣𝑎𝑙𝑢𝑒(𝑎𝑡 2019)
𝐼𝑛𝑑𝑒𝑥 𝑣𝑎𝑙𝑢𝑒(𝑎𝑡 2004)
Purchased cost= 131398.2 $
"Time base 2019"
Step#13
e= 2.2083
r= 0.828