ECH155B– HUMG
TECHNICAL MEMORANDUM
Date: September12th, 2017
To: Dr. Nguyen Thi Linh
From: Nguyen Khac Duy, Group 3, K4 student
Subject: SHELL AND TUBE HEAT EXCHANGERS

Executive Summary:
This experiment consists 2 parts: the 1stpart is to determine the overall heat transfer coefficient for those two cases and the effectiveness of the heat exchanger in two cases; The 2ndpart is to determine the overall heat transfer coefficient by usingFourier’s Law and Newton’s Law of Cooling. In Part I, heat flow rate of hot flow is higher than that of cold flow. Heat flow rate of countercurrent is higher than that of co-current. Efficiency of countercurrent is higher than that of co-current. The fractional errors of Q,efficiencyand Utare small. The values of NTU, effectiveness and Ut for countercurrent case are higher than that of co-current case.Because the transfer of heat takes place due to temperature gradient. The gradient is a constant in the countercurrent, which leads to constant flow of heat at each point. Although the gradient is high initially, the gradient decreases by time in the co-current. So it leads to the difference in Q, ∆Tltmd and Ut.Ut of Part II is lower than that of Part I because we can choose incorrectly the formula of Nusselt number.
Parameter
Part I
Part II


Co-current
Countercurrent
Co-current
Countercurrent

Heat flow rateof hot flow (W)
1967.87 ± 48.31
2096.01 ± 34.94



Heat flow rateof cold flow (W)
1789.46 ± 37.46
2072.19 ± 32.61



Efficiency (%)
90.93 ± 0.03
98.86 ± 0.02



Number of transfer units,NTU(-)
0.71
0.74



Effectiveness, ε (-)
0.41
0.42



Heat transfer coe. Ut (W.m-2.K-1)
427.74 ± 10.76
442.46 ± 7.52
68.99 ±
69.37 ±

Introduction and Purpose:
The shell and tube heat exchanger is commonlyused in chemical processindustries. Heat is transferred between one fluid flowing through the tubes and anotherfluid flowing through the cylindrical shell around the tubes.In this experiment, we will investigate the difference between co-current and countercurrent;determine the overall heat transfer coe.for a shell and tube heat exchanger and investigate the flow rates on heatexchangerefficiencyandoverallheattransfer coefficient. Co-current and countercurrent shown in Figure 1.
/
Theory and Analysis:
Part 1: Overall energy balance
In this part, we base on the equations of energy (heat) transfer:

Q
duty hot
ṁ
c
c
pc∆T (1) Where ṁ is mass flow rate of hot stream (kg/sec
c
p
is the heat capacity of water (kJ/kg.K), ∆T
T
in
T
out(˚C or K).
The same for the cold stream equation is
Q
cold
m
h
c
ph∆T (2T
T
out
T
in

Next, we have to calculate the Logarithmic Mean Temperature Difference(∆Tltmd) by the equation∆T
lmtd∆t
2∆t
1
ln∆t
2∆t
1 (3)
For countercurrent flow: ∆t
1
T
h,i−T
c,o
∆t
2
T
h,o
T
c,i

For co-current (parallel) flow: ∆t
1=T
h,i
T
c,i
∆t
2=T
h,o
T
c,o

Finally, the overall heat transfer coefficient can be found using
U
t
Q
duty hot
A∗∆T
lmtd (4)
Where A=π
d
m
L ∀ L=n∗l
d
m=0.5
d
od
d
id

A (m2) is area of HEX; dm(m) is the arithmetic meandiameter of innertubes; L (m) is total heat transfer length; n is the number of tubes and l (m) is the length of one tube.
Part 2: Traditional Analysis of Conductive and Convective Heat Transfer
In this part, we based on the Fourier’s law and Newton’s Law of Cooling in order to get this equation to calculate Ut:
1
U
t
A
R
t
1
h
h
π
D
id
L
ln
D
od
D
id
2πLK
1
h
c
π
D