1. Polytechnic University of Puerto Rico
Chemical Engineering Department
Winter 2016
Ashley M. Ramirez Quiles
Stephanie P. Rivera Ares
Edgared M. Troche Muñiz
George Sepulveda
December 20, 2016
Prof. Pablo Traverso

2.  Objectives
 Theory
 Equipment
 Procedure
 Data
 Calculations
 Security
 References

3.  Demonstrate
 Indirect heating or cooling by fluid to fluid heat transfer, streams separated by a solid
wall.
 Differences between cocurrent flow and countercurrent flow, its effects on heat
transferred, and temperature efficiencies.

4.  Investigate
 Effect of changes in hot and cold fluid flowrate on the Temperature Efficiencies and
Overall Heat Transfer Coefficient.
 Effect of the driving force with cocurrent and countercurrent flow.
 Heat loss from the Shell and Tube Heat Exchanger, and reduction in heat transfer
coefficient due to fouling of the heat transfer surfaces.

5.  Determine
 Overall Heat Transfer Coefficient for a Tubular Heat Exchanger by the Logarithmic
Mean Temperature Difference method.
 Perform
 An energy balance across a Shell and Tube Heat Exchanger and calculate the Overall
Efficiency at different fluid flowrates.

6. • Parallel tubes enclosed in a cylindrical shell allows two fluids
to be in thermal contact.
• Hot water will enter the tubes while cold water will enter the
shell in a co-current or counter-current manner.
• Baffles are placed inside the shell so that cold water flowing
outside the tubes will increase velocity and therefore the
rate of heat transfer.

7. Countercurrent has a more
effective heat transfer
rate.

8.  Energy Balance
Acc = In – Out + Gen
 In Theory, at SS:
Heat absorbed by cold fluid = Heat released by hot fluid
Qin = Qout
𝐦 𝐜 𝐂 𝐩 𝐜 𝐓𝐜 𝐨𝐮𝐭 − 𝐓𝐜 𝐢𝐧 = 𝐦 𝐡 𝐂 𝐩 𝐡 𝐓𝐡 𝐢𝐧 − 𝐓𝐡 𝐨𝐮𝐭
 But there is heat loss or gained to/from the atmosphere (Accumulation)
Qf = Qin – Qout
 Taking this into account, Overall Efficiency becomes
Ƞ=
𝑄𝑜𝑢𝑡
𝑄𝑖𝑛
*100

9. • One Conduction Resistance (Fourier’s
Law)
𝑅𝑡ℎ =
ln
𝑟0
𝑟 𝑖
2𝜋 𝑘 𝐿
• Two Convection Resistances (Newton’s
Law of Cooling)
𝑅𝑡ℎ =
1
ℎ𝑖 𝐴 𝑖
𝑅𝑡ℎ =
1
ℎ 𝑜 𝐴 𝑜

10. • A useful measure of the heat exchanger performance is the temperature
efficiency of each fluid stream
Temperature efficiency for hot fluid:
𝜂ℎ =
𝑇1 − 𝑇2
𝑇1 − 𝑇3
∗ 100
Temperature efficiency for cold fluid:
𝜂 𝑐 =
𝑇4 − 𝑇3
𝑇1 − 𝑇3
∗ 100
**Where :
𝑇1= hot fluid temperature (shell inlet)
𝑇2= hot fluid temperature (shell outlet)
𝑇3= cold fluid temperature (tube inlet)
𝑇4= cold fluid temperature (tube outlet)

11.  Fouling resistance can play in important
role increasing overall thermal
resistance and decreasing overall heat
transfer coefficient.
 Therefore, based on the outside area of
the tube:
𝑈 𝑜=
1
𝐴 𝑜 𝑅𝑡ℎ
=
1
𝑙𝑛
𝑟𝑜
𝑟𝑖
2𝜋𝑘𝐿
+
1
ℎ𝑖 𝐴𝑖
+
1
ℎ 𝑜 𝐴 𝑜
+ 𝑅𝑓𝑖 + 𝑅𝑓𝑜 𝐴 𝑜

12.  Overall Heat Transfer Coefficient (U) based on LMTD (Logarithmic mean temperature
difference) is used since the hot and cold fluid streams vary along the length of the heat
exchanger.
Δ𝑇𝐿𝑀=
𝑇2 − 𝑡1 − 𝑇1 − 𝑡2
ln
𝑇2 − 𝑡1
𝑇1 − 𝑡2
Where:
𝑇1= hot fluid temperature (shell inlet)
𝑇2= hot fluid temperature (shell outlet)
𝑡1= cold fluid temperature (tube inlet)
𝑡2= cold fluid temperature (tube outlet)
U =
𝑄
𝐴∗∆𝑇 𝐿𝑀
Where:
Q = heat power emitted from hot fluid
A = Heat transmission area

13.  Correction Factor (F) Needs to be calculated to correct for the true temperature difference
True Temperature Difference = F* Δ𝑇𝐿𝑀
Correction factor plot for one shell pass and two, four or any
multiple of tube passes

14. 1. Base
2. Thumb nuts
3. Hot water outlet
4. Cold water outlet
5. Cold water inlet
6. Thumb nuts
7. Flexible tube
8. Hot water inlet
9. Thermocouple temperature sensor

15. 1. Verify all inlets and outlets hoses are firmly installed and the direction of the inlet
and outlets for counter current arrangement.
2. Assure that the water level of the recirculating vessel is filled about 20mm with
demineralized water.
3. Make sure that the isolating valves are fully open allowing water through the heat
exchanger.
4. Set up the software for approximately periods of 30 minutes or when the
temperatures remain constants.
5. For the counter current arrangement used a hot flow of about 1 L/min and a cold flow
of 1 L/min and a set point temperatures of 40°C and 65°C.
6. The heat flow is going to be varied and the cold flow remains constant.
7. Repeat procedure for a hot flow of 3L/min and record measurements for 40°C and
65°C.
8. Record and plot all four runs for counter current arrangement.
9. Make sure all hoses are exchanged for co-current arrangement.
10. Repeat procedure steps 1-5 and 8 for the co-current arrangement.

16.  Counter Current Arrangement Co- Current Arrangement
qhot
T1 (IN)
T3 (IN)
qcold
T4 (OUT)
qhot
qcold
T4 (OUT)
T2 (IN)
T3 (IN)
T2
(OUT)
T1 (OUT)

17. Co-current Arrangement
Hot Temperature
(∘C)
Hot flow
(L/min)
Cold flow
(L/min)
T1
(∘C)
T2
(∘C)
T3
(∘C)
T4
(∘C)
Counter current Arrangement
Hot Temperature
(∘C)
Hot flow
(L/min)
Cold flow
(L/min)
T1
(∘C)
T2
(∘C)
T3
(∘C)
T4
(∘C)

18.  One Conduction Resistance (Fourier’s Law)
𝑅𝑡ℎ =
ln
𝑟0
𝑟 𝑖
2𝜋 𝑘 𝐿
 Two Convection Resistances (Newton’s Law of Cooling)
𝑅𝑡ℎ =
1
ℎ𝑖 𝐴 𝑖
𝑅𝑡ℎ =
1
ℎ 𝑜 𝐴 𝑜
 Overall heat transfer coefficient
𝑈 𝑜=
1
𝐴 𝑜 𝑅𝑡ℎ
 Temperature efficiency for hot fluid:
𝜂ℎ =
𝑇1 − 𝑇2
𝑇1 − 𝑇3
∗ 100
 Temperature efficiency for cold fluid:
𝜂 𝑐 =
𝑇4 − 𝑇3
𝑇1 − 𝑇3
∗ 100
 Logarithmic mean temperature difference:
Δ𝑇𝐿𝑀=
𝑇2−𝑡1 − 𝑇1−𝑡2
ln
𝑇2−𝑡1
𝑇1−𝑡2
U =
𝑄
𝐴∗∆𝑇 𝐿𝑀
 True Temperature Difference = F* ΔTLM

19.  Wear Proper Protective Equipment
IAW stablished SOP’s

20.  Correction Factor
 https://2ch204ns2013.files.wordpress.com/2014/04/lmtd-correction-factor.pdf
 Fouling
 http://www.hcheattransfer.com/fouling_factors2.html
 https://www.corrosionpedia.com/definition/540/fouling
 http://www.me.uprm.edu/o_meza/Inme4032/Heat%20Exchanger%20Analysis-3.pdf
 Armfield Instruction Manual, HT33 Shell and Tube Heat Exchanger (2005)
 INSTRUCTION MANUAL, HT33 SHELL AND TUBE HEAT EXCHANGER HT33-
ISSUE 1, NOVEMBER 2004