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Design Of Shell & Tube Heat Exchanger & Its
CFD Analysis
Objective Of The Case
Study
 A shell & Tube heat exchanger is to be
designed which has the capacity to
raise the temperature of fresh water
from 40°C to 50°C at a flow rate of
15LPM. The other fluid available is
waste water at temperature 80°C and
flow rate of 50LPM.
 First of all, an analytical calculation
was performed to estimate the size of
heat exchanger required.
(Determination Of Heat Transfer Area)
 CFD simulation was performed later in
ANSYS Fluent to validate the
calculations.
Determination Of Heat Transfer Area Required
To determine the heat transfer area, following steps were followed as described below:
 Temperature of the hot fluid at the outlet after heat exchange process using the equation:
𝑇ℎ,𝑜𝑢𝑡 = 𝑇ℎ,𝑖𝑛 −
𝑄
𝑚ℎ×𝐶𝑝,ℎ
, where 𝑄 = 𝑚𝑐 × 𝐶𝑝,𝑐 × 𝑇𝑐,𝑜𝑢𝑡 − 𝑇𝑐,𝑖𝑛
Q= 10460 Watts & Th,out =77°C
The bulk mean temperature of tube and
shell side can be then calculated using
the equation:
𝑇𝑏 =
𝑇𝑖𝑛 + 𝑇𝑜𝑢𝑡
2
𝑇𝑏,tube = 45°C & 𝑇𝑏,shell = 78.5°C
Fluid properties at bulk temperature are
then used for calculation of Reynolds
and Prandtl number.
Determination Of Heat Transfer Area Required
Tube of inner diameter 0.05m and outer diameter 0.054m inside another concentric tube of inner diameter 0.1m was assumed first
to determine the heat transfer area required for this heat transfer purpose.
 Reynolds & Prandtl number calculation was performed for both sides. (With properties at bulk temperature)
 Then Nusselt number was predicted using Dittus-Boelter correlation.
𝑅𝑒𝑡𝑢𝑏𝑒 =
𝜌@𝑏𝑢𝑙𝑘 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 × 𝐷ℎ,𝑡𝑢𝑏𝑒 × 𝑣𝑖𝑛,𝑡𝑢𝑏𝑒
𝜇@𝑏𝑢𝑙𝑘 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒
𝑃𝑟𝑡𝑢𝑏𝑒 =
𝜇@𝑏𝑢𝑙𝑘 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 × 𝐶𝑝
𝑘
= 8847 = 5.0
𝑅𝑒𝑠ℎ𝑒𝑙𝑙 =
𝜌@𝑏𝑢𝑙𝑘 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 × 𝐷ℎ,𝑡𝑢𝑏𝑒 × 𝑣𝑖𝑛,𝑠ℎ𝑒𝑙𝑙
𝜇@𝑏𝑢𝑙𝑘 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒
= 15682 𝑃𝑟𝑠ℎ𝑒𝑙𝑙 =
𝜇@𝑏𝑢𝑙𝑘 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 × 𝐶𝑝
𝑘
= 3.1
𝑁𝑢 = 0.023 × 𝑅𝑒0.8 × 𝑃𝑟0.4
Dittus-Boelter correlation is given as:
𝑁𝑢𝑡𝑢𝑏𝑒 = 63
𝑁𝑢𝑠ℎ𝑒𝑙𝑙 = 82
ℎ =
𝑁𝑢 𝑋 𝑘
𝐷ℎ
Heat transfer coefficients are then calculated as:
ℎ𝑡𝑢𝑏𝑒 𝑜𝑟 ℎ𝑖 = 756 ℎ𝑠ℎ𝑒𝑙𝑙 𝑜𝑟 ℎ𝑜 = 1071.5
Once the heat transfer coefficients at the inner side
and outer side of the tube are calculated then the
overall heat transfer coefficient can be estimated
next.
 The overall heat transfer coefficient is estimated using the equation as shown below:
1
𝑈𝑜
=
1
ℎ𝑜
+
1
𝑘
𝑟𝑜
1
ln
𝑟𝑜
𝑟𝑖
+
1
𝑟𝑖
𝑟𝑜
ℎ𝑖
Determination Of Heat Transfer Area Required
 𝑈𝑜 was calculated as 434W/m2-k
 LMTD was calculated next.
𝐿𝑀𝑇𝐷 =
𝑇1 − 𝑇2
ln
𝑇1
𝑇2
Where, 𝑇1 = 𝑇ℎ,𝑜𝑢𝑡 − 𝑇𝑐,𝑖𝑛𝑙𝑒𝑡 and 𝑇2 = 𝑇ℎ,𝑖𝑛𝑙𝑒𝑡 − 𝑇𝑐,𝑜𝑢𝑡
𝑇1 = 77 − 40 = 37°𝐶
𝑇2 = 80 − 50 = 30°𝐶
Thus LMTD = 33.33°C
 Required heat transfer area is then calculated using the equation.
𝑄 = 𝑈𝑜 × 𝐴𝑜 × 𝐿𝑀𝑇𝐷
𝐴𝑜= 0.7m2 (Approximately)
L=4.32m of diameter assumed.
 If a single tube of diameter 0.054m (outer) and 0.05m (inner) is used, then the length of tube required will be 4.32m.
 Instead of using a single tube, we will use multiple tubes of smaller diameter.
 Let us assume the tube diameter be 0.03 (outer) & 0.025m (inner). (Length of tube = 1.0m)
 Then the number of tubes required will be 7 to get the same heat transfer area.
Designing A Shell & Tube Heat Exchanger
Instead of using single tube of 4.32m, a shell
tube heat exchanger design was selected as
shown.
Drawing Of Heat Exchanger
 Some important dimensions are shown here.
 Solidworks model of shell & tube heat exchanger was imported in ANSYS DM, and then the fluid volume was extracted.
 Due to symmetry in the CFD model, symmetry assumption was taken and the computational time was reduced to half.
 The model was then meshed in the ANSYS mesh software. Inflation layer was applied at the walls to capture the
boundary layer.
CFD model & Mesh
CFD MODEL AFTER
TAKING SYMMETRY
MESHED MODEL
The figure below shows the temperature contour in the XY plane or the plane of symmetry. It can be seen that the
temperature of tube fluid (i.e. cold water) has increased from 40°C to 50°C as required. This indicates the validity of
the calculations done above. On the other hand, the shell fluid shows temperature drop from 80°C to 77°C after
exchanging the heat with tube fluid.
CFD Results – Temperature Contour On XY Plane
 The results can also be seen in the
top view on ZX plane. The
temperature drop of 10°C can be
seen in the tube fluid.
 The heat exchange between the two
fluids along the heat exchanger
length can be seen clearly in the
figure below
CFD Results TEMPERATURE
CONTOUR ON ZX PLANE
TEMPERATURE CONTOURS
ON YZ PLANE
Conclusion
 In this case study, a shell and tube heat
exchanger was designed using analytical
calculations and then simulated using ANSYS
CFD.
 As theoretically calculated, the tube fluid
temperature shows temperature rise of 10°C
where as the shell fluid shows temperature
drop of 3°C.
 The overall heat transfer coefficient
calculated using CFD was 422 W/m2-K
against the theoretical calculation of 434
W/m2-K with a percentage error of 2.8%.

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Design & CFD Analysis of Heat Exchanger

  • 1. Design Of Shell & Tube Heat Exchanger & Its CFD Analysis
  • 2. Objective Of The Case Study  A shell & Tube heat exchanger is to be designed which has the capacity to raise the temperature of fresh water from 40°C to 50°C at a flow rate of 15LPM. The other fluid available is waste water at temperature 80°C and flow rate of 50LPM.  First of all, an analytical calculation was performed to estimate the size of heat exchanger required. (Determination Of Heat Transfer Area)  CFD simulation was performed later in ANSYS Fluent to validate the calculations.
  • 3. Determination Of Heat Transfer Area Required To determine the heat transfer area, following steps were followed as described below:  Temperature of the hot fluid at the outlet after heat exchange process using the equation: 𝑇ℎ,𝑜𝑢𝑡 = 𝑇ℎ,𝑖𝑛 − 𝑄 𝑚ℎ×𝐶𝑝,ℎ , where 𝑄 = 𝑚𝑐 × 𝐶𝑝,𝑐 × 𝑇𝑐,𝑜𝑢𝑡 − 𝑇𝑐,𝑖𝑛 Q= 10460 Watts & Th,out =77°C The bulk mean temperature of tube and shell side can be then calculated using the equation: 𝑇𝑏 = 𝑇𝑖𝑛 + 𝑇𝑜𝑢𝑡 2 𝑇𝑏,tube = 45°C & 𝑇𝑏,shell = 78.5°C Fluid properties at bulk temperature are then used for calculation of Reynolds and Prandtl number.
  • 4. Determination Of Heat Transfer Area Required Tube of inner diameter 0.05m and outer diameter 0.054m inside another concentric tube of inner diameter 0.1m was assumed first to determine the heat transfer area required for this heat transfer purpose.  Reynolds & Prandtl number calculation was performed for both sides. (With properties at bulk temperature)  Then Nusselt number was predicted using Dittus-Boelter correlation. 𝑅𝑒𝑡𝑢𝑏𝑒 = 𝜌@𝑏𝑢𝑙𝑘 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 × 𝐷ℎ,𝑡𝑢𝑏𝑒 × 𝑣𝑖𝑛,𝑡𝑢𝑏𝑒 𝜇@𝑏𝑢𝑙𝑘 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑃𝑟𝑡𝑢𝑏𝑒 = 𝜇@𝑏𝑢𝑙𝑘 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 × 𝐶𝑝 𝑘 = 8847 = 5.0 𝑅𝑒𝑠ℎ𝑒𝑙𝑙 = 𝜌@𝑏𝑢𝑙𝑘 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 × 𝐷ℎ,𝑡𝑢𝑏𝑒 × 𝑣𝑖𝑛,𝑠ℎ𝑒𝑙𝑙 𝜇@𝑏𝑢𝑙𝑘 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 = 15682 𝑃𝑟𝑠ℎ𝑒𝑙𝑙 = 𝜇@𝑏𝑢𝑙𝑘 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 × 𝐶𝑝 𝑘 = 3.1 𝑁𝑢 = 0.023 × 𝑅𝑒0.8 × 𝑃𝑟0.4 Dittus-Boelter correlation is given as: 𝑁𝑢𝑡𝑢𝑏𝑒 = 63 𝑁𝑢𝑠ℎ𝑒𝑙𝑙 = 82 ℎ = 𝑁𝑢 𝑋 𝑘 𝐷ℎ Heat transfer coefficients are then calculated as: ℎ𝑡𝑢𝑏𝑒 𝑜𝑟 ℎ𝑖 = 756 ℎ𝑠ℎ𝑒𝑙𝑙 𝑜𝑟 ℎ𝑜 = 1071.5 Once the heat transfer coefficients at the inner side and outer side of the tube are calculated then the overall heat transfer coefficient can be estimated next.
  • 5.  The overall heat transfer coefficient is estimated using the equation as shown below: 1 𝑈𝑜 = 1 ℎ𝑜 + 1 𝑘 𝑟𝑜 1 ln 𝑟𝑜 𝑟𝑖 + 1 𝑟𝑖 𝑟𝑜 ℎ𝑖 Determination Of Heat Transfer Area Required  𝑈𝑜 was calculated as 434W/m2-k  LMTD was calculated next. 𝐿𝑀𝑇𝐷 = 𝑇1 − 𝑇2 ln 𝑇1 𝑇2 Where, 𝑇1 = 𝑇ℎ,𝑜𝑢𝑡 − 𝑇𝑐,𝑖𝑛𝑙𝑒𝑡 and 𝑇2 = 𝑇ℎ,𝑖𝑛𝑙𝑒𝑡 − 𝑇𝑐,𝑜𝑢𝑡 𝑇1 = 77 − 40 = 37°𝐶 𝑇2 = 80 − 50 = 30°𝐶 Thus LMTD = 33.33°C  Required heat transfer area is then calculated using the equation. 𝑄 = 𝑈𝑜 × 𝐴𝑜 × 𝐿𝑀𝑇𝐷 𝐴𝑜= 0.7m2 (Approximately) L=4.32m of diameter assumed.
  • 6.  If a single tube of diameter 0.054m (outer) and 0.05m (inner) is used, then the length of tube required will be 4.32m.  Instead of using a single tube, we will use multiple tubes of smaller diameter.  Let us assume the tube diameter be 0.03 (outer) & 0.025m (inner). (Length of tube = 1.0m)  Then the number of tubes required will be 7 to get the same heat transfer area. Designing A Shell & Tube Heat Exchanger Instead of using single tube of 4.32m, a shell tube heat exchanger design was selected as shown.
  • 7. Drawing Of Heat Exchanger  Some important dimensions are shown here.
  • 8.  Solidworks model of shell & tube heat exchanger was imported in ANSYS DM, and then the fluid volume was extracted.  Due to symmetry in the CFD model, symmetry assumption was taken and the computational time was reduced to half.  The model was then meshed in the ANSYS mesh software. Inflation layer was applied at the walls to capture the boundary layer. CFD model & Mesh CFD MODEL AFTER TAKING SYMMETRY MESHED MODEL
  • 9. The figure below shows the temperature contour in the XY plane or the plane of symmetry. It can be seen that the temperature of tube fluid (i.e. cold water) has increased from 40°C to 50°C as required. This indicates the validity of the calculations done above. On the other hand, the shell fluid shows temperature drop from 80°C to 77°C after exchanging the heat with tube fluid. CFD Results – Temperature Contour On XY Plane
  • 10.  The results can also be seen in the top view on ZX plane. The temperature drop of 10°C can be seen in the tube fluid.  The heat exchange between the two fluids along the heat exchanger length can be seen clearly in the figure below CFD Results TEMPERATURE CONTOUR ON ZX PLANE TEMPERATURE CONTOURS ON YZ PLANE
  • 11. Conclusion  In this case study, a shell and tube heat exchanger was designed using analytical calculations and then simulated using ANSYS CFD.  As theoretically calculated, the tube fluid temperature shows temperature rise of 10°C where as the shell fluid shows temperature drop of 3°C.  The overall heat transfer coefficient calculated using CFD was 422 W/m2-K against the theoretical calculation of 434 W/m2-K with a percentage error of 2.8%.