Thermal fluid Lab – MEC 554/LSRC/SCA Rev.01-2017
UNIVERSITI TEKNOLOGI MARA
FACULTY OF MECHANICAL ENGINEERING
Program
Mechanical Course
Course Code
Lecturer/Tutor Name
Group
: Bachelor of Engineering (Hons.)
: Thermalfluids Lab
: MEC 554
:MOHD SHAHAR MOHD SHAWAL
: EMD5M1A
Lab Report
CONCENTRIC TUBE HEAT EXCHANGER
Bil
Student Name
Student ID
Signature
MOHD FARHAN HAQIMI BIN ABDUL
HALIM
2020970471
MUHAMAD IQMAL BIN LOKMAN
2020825268
MUHAMMAD ALIF FARHAN BIN MOHD
RIDZUAN
2020898722
MUHAMMAD AFIQ BIN MOHD ZOHADI
2020816882
1.
2.
3.
4.
Lab Session
: 10 November 2021
Submission of Report : 17 November 2021
Approved by: ____________________
Received by:
____________________
1
FACULTY(OF(MECHANICAL(ENGINEERING(
UNIVERSITI)
TEKNOLOGI)
MARA)40450)SHAH)
ALAM) SELANGOR)
DARUL)EHSAN)
Tel:)03-55435190)Fax:)0355435160)
!
_____________________________________________________________________________________________
REPORT ASSESSMENT FORM
Experiment’s Title:
CONCENTRIC TUBE HEAT EXCHANGER
!
Prepared'by:'
!
Bil
Name
Matrix,No
1.
MOHD FARHAN HAQIMI BIN ABDUL HALIM
2020970471
2.
MUHAMAD IQMAL BIN LOKMAN
Signature,
2020825268
3.
MUHAMMAD ALIF FARHAN BIN MOHD RIDZUAN
2020898722
4.
MUHAMMAD AFIQ BIN MOHD ZOHADI
2020816882
!
!
2
Course Outcome
No
1
2
Technical Report
Introduction
Assessment
CO2
Experimental Procedures
C02
10
10
3
Data/result/sample of
calculation
CO2
20
4
Discussion and
Conclusion
CO2
20
TOTAL
60
Course Outcome
No
GIVEN MARK
FULL MARK
Writing Performance
/60
Ass essment
FULL MARK
GIVEN MARK
1
Structure
CO3
5
2
Graphics/Figures/Tables
C03
5
3
Mechanics
CO3
5
4
Formatting and
references
CO3
5
TOTAL
20
/20
TOTAL MARKS
80
/80
3
Table of Contents
1.
INTRODUCTION ........................................................................................................ 5
2.
OBJECTIVE ................................................................................................................. 6
3.
THEORY ...................................................................................................................... 9
4.
EXPERIMENTAL PROCEDURES .......................................................................... 11
5.
RESULTS ................................................................................................................... 14
•
Power Absorbed ...................................................................................................... 21
•
Power Lost .................................................................................................................. 21
•
Overall Efficiency, 𝜼 ................................................................................................ 21
6.
DISCUSSION............................................................................................................. 34
7.
CONCLUSION .......................................................................................................... 39
8.
REFERENCES ........................................................... Error! Bookmark not defined.
9.
APPENDICES ............................................................ Error! Bookmark not defined.
4
1. INTRODUCTION
One of the devices for transporting heat from one medium to another is the heat
exchanger. Heat can be exchanged between two fluids with different temperatures
flowing in and out. The heat exchanger's primary function is to either add heat to the
cold fluid or remove heat from the hot fluid. The fluid motion in a heat exchanger can
flow in a variety of directions, including parallel flow, counter flow, and cross flow.
Only the parallel and counter flow have been studied as a result of this experiment.
Parallel flow, also known as co-current flow, is characterized by the simultaneous
movement of hot and cold fluids in the same direction. Opposite for the counter flow,
both of hot and cold fluids are flowing in the different direction, these concepts can be
visualized in the figure below.
Figure 1 : Flow Direction of Parallel Flow and Counter Flow
5
We focused on the shell and tube heat exchanger in this experiment. Heat exchangers are used
in a variety of applications in everyday life, including air intercoolers and preheaters,
condensers, and boilers in steam plants. Recuperator, regenerator, and evaporative heat
exchangers are the three main types of heat exchangers. Consider the automobile radiator as an
illustration of how a heat exchanger works. Water is utilized to cool the engine before flowing
through the radiator. The cold air surrounding the engine will expose the radiator's fin, which
will help to remove some of the heat. The car radiator as one of the examples of heat exchanger
because it has transferred the heat from the water to the open air surrounding.
Figure 2: example of heat exchanger in air conditioning system in a car
2. OBJECTIVE
To demonstrate the effect flow rate variation on the performance characteristics
of a counter-flow and parallel-flow concentric tube heat exchanger.
6
3. APPARATUS
1. H900 Concentric Tube Heat Exchanger
Figure 1 : H900 Concentric Tube Heat Exchanger
2. Digital Clock
Figure 2: Digital Clock
3. Heat Exchanger Apparatus System Diagram
Figure 3: Heat Exchanger Apparatus System Diagram
7
4. Parallel Flow and Counter Flow of Valve Diagram
Figure 4: Parallel Flow and Counter Flow of Valve Diagram
8
4. THEORY
There are several important formulas or equations to calculate the performance
characteristics for both parallel-flow and counter-flow concentric tube heat exchangers.
The performance required are power emitted, power absorbed, power lost efficiency
(Ƞ), logarithmic mean temperature difference (∆𝑇 m), and overall heat transfer
coefficient (U).
The Efficiency for the Cold Medium is:
𝜂! =
The Efficiency for the Hot Medium is:
𝑇!,#$% − 𝑇!,&'
× 100
𝑇(,&' − 𝑇!,&'
𝜂( =
𝑇(,&' − 𝑇(,#$%
× 100
𝑇(,&' − 𝑇!,&'
The Mean Temperature Efficiency is:
𝜂)*+' =
𝜂 ! + 𝜂(
2
The Power Emitted is given below (where 𝑉-h is the Volumetric Flow Rate of the hot fluid):
𝑃𝑜𝑤𝑒𝑟3𝐸𝑚𝑖𝑡𝑡𝑒𝑑 = 𝑉-( 𝜌( 𝐶,( (𝑇(,&' − 𝑇(,#$% )
The Power Absorbed is given below (where 𝑉- c is the Volumetric Flow Rate of the cold fluid):
𝑃𝑜𝑤𝑒𝑟3𝐴𝑏𝑠𝑜𝑟𝑏𝑒𝑑 = 𝑉!- 𝜌! 𝐶,! (𝑇!,#$% − 𝑇!,&' )
The Power Lost is therefore:
𝑃𝑜𝑤𝑒𝑟3𝐿𝑜𝑠𝑡 = 𝑃𝑜𝑤𝑒𝑟3𝐸𝑚𝑖𝑡𝑡𝑒𝑑 − 𝑃𝑜𝑤𝑒𝑟3𝐴𝑏𝑠𝑜𝑟𝑏𝑒𝑑
The Overall Efficiency (𝜂) is:
𝜂=
𝑃𝑜𝑤𝑒𝑟3𝐴𝑏𝑠𝑜𝑟𝑏𝑒𝑑
× 100
𝑃𝑜𝑤𝑒𝑟3𝐸𝑚𝑖𝑡𝑡𝑒𝑑
9
The Logarithmic Mean Temperature Difference (∆𝑇m) is:
∆𝑇) =
∆𝑇- − ∆𝑇. E𝑇(,) − 𝑇!,#$% F − (𝑇(,#$% − 𝑇!,&' )
=
∆𝑇
(𝑇 − 𝑇!,#$% )
𝑙𝑛 C - D
𝑙𝑛 G (,&'
H
∆𝑇.
(𝑇(,#$% − 𝑇!,&' )
The Overall Heat Transfer Coefficient (U) is:
𝑈=
𝑃𝑜𝑤𝑒𝑟3𝐴𝑏𝑠𝑜𝑟𝑏𝑒𝑑
𝐴/ ∆𝑇)
10
5. EXPERIMENTAL PROCEDURES
a) Apparatus
Figure 3 : Heat exchanger
system
Figure 4 : Heat exchanger
apparatus system diagram
(schematic diagram)
11
Figure 3 : Cold fluid
volumetric flow rate
control
Figure 4 : Hot fluid
volumetric flow rate control
Figure 5 : Temperature
control
Figure 6 : Valve diagram for parallel flow and counter flow
12
b) Procedure
1. The experiment for the operation of a counter-flow heat exchanger had been set up.
With the decade switch, the required hot water inlet temperature was set to Th,in =
60c. The cold water volumetric flow rate (Vc) is also set to 2000 cm3/min.
2. The initial volumetric flow rate (Vh) for hot fluid was set to 1000 cm3/min. The
following table shows the six temperature readings that were recorded. The
volumetric flow rates of 2000, 3000, and 4000 cm3/min were measured again.
3. The values for density (
) for cold fluids at
and
) and constant pressure specific heat (
and hot fluids at
and
were discovered.
4. Using the data, the following heat exchanger performance factors were calculated
and documented in the tables: power emitted, power absorbed, power lost,
efficiency (ŋ), logarithmic mean temperature difference (Δ
), and total heat
transfer coefficient (U).
5. The effect of changing the volumetric flow rate of the hot fluid on each of these
heat exchanger performance variables was analyzed and discussed.
13
6. RESULTS
Parallel Flow
Hot water temperature = 64 °C
Fixed cold water flow rate = 2000cm3/min
1. Heat transmission length = 1.5 m
2. Heat transmission area = 0.067 m2
3. Tube outer diameter = 15 x 0.7 mm (thin wall)
4. Shell outer diameter = 22 x 0.9 mm (thin wall)
5. Insulation thickness = 20mm
Hot Temperature
Hot
water
in,°𝐶
Hot
Cold Temperature
Hot
water
mid,
°𝐶
water
〖Δ
out,
Coldwater
Coldwater
Coldwater
in, °𝐶
mid, °𝐶
out, °𝐶
〖Δ
°𝐶
64
47
39
25
27
29
30
3
64
48
41
23
27
30
31
4
64
48
41
23
27
30
32
5
64
49
42
22
27
31
33
6
Table 1: water temperature
14
Calculation
)³
1. To find QH ( ), TH,avg (°C)
/
Hot water flow
Rate, QH (
𝒄𝒎³
𝒎𝒊𝒏
Hot water flow
),
1000
2000
3000
4000
𝒎³
rate, QH ( ),
𝒔
Average hot
Average cold
temperature,
temperature,
𝑻𝑯,𝒂𝒗𝒈 (
𝑻𝑪, 𝒂𝒗𝒈(
℃)
℃)
1.667 × 10
−5
51.5
28.5
3.333 × 10
−5
52.5
29.0
5.000 × 10
−5
52.5
29.5
6.667 × 10
−5
53.0
30.0
Table 2: Average hot temperature and cold temperature at different flow rate
SAMPLE CALCULATION
At flow rate (hot), QH 1000(
•
!)³
)
)&'
𝒄𝒎³
!)³
QH = 1000(
)&'
)x(
QH = 1.667 x 10-5
•
𝒎³
Conversion of, QH (𝒎𝒊𝒏 ) to, QH ( 𝒔 )
-)
-66!)
)3 x (
-)&' 3
)
768/
)³
/
Average Hot temperature, TH,avg (°C)
TH,avg =
TH,avg =
89: (<=)?9:(@AB)8
.
7C?DE
.
TH,avg = 51.5 °C
15
•
Average cold temperature, Tc,avg (°C)
Tc,avg =
Tc,avg =
9F(<= )
?9F(@AB)
.
.G?D.
Tc,avg = 29°C
HI
2. To find density of the water at TH,avg and TC,avg ,( )
)³
Hot water flow rate, 𝝆 H ( 𝒌𝒈)
𝒎³
(
𝒄𝒎³
𝒎𝒊𝒏
𝝆C ( )
HI
)³
)
1000
987.22
995.74
2000
986.70
995.56
3000
986.70
995.38
4000
986.44
995.20
Table 3 : Density of water for different flow rate
Sample Calculation
•
Density of water at TH,avg , 𝝆 H (𝒎³)
𝒌𝒈
Using interpolation by referring to the Table A-3, at hot water flow rate, 𝑄𝐻
Is 1000(
!)³
)&'
)
16
Temperature,
𝒂𝒗𝒈
𝑻𝑯,
Density, 𝝆𝑯
50
988
51.5
𝜌𝐻
75
975
Table 4 : Interpolation data of density
GLML-.L
GLML6
=
EGLM8O:
EGLMEPP
𝜌𝐻 = 987.22
•
Density of water at TC,Avg , 𝝆 C
𝒌𝒈
𝒎³
𝒌𝒈
𝒎³
Using interpolation, at hot water flow rate, 𝑄𝐻 is1000(
Temperature,
𝒂𝒗𝒈
𝑻𝑪,
!)³
)
)&'
Density, 𝝆𝑪
25
997
28.5
𝜌𝐶
50
988
Table 5 : Interpolation data on density
L6M.P.L
L6M.L
=
EPPM8OF
EPPMEEG
𝜌c = 995.74
3. To find specific heat capacity at TH,avg (
Q
𝒌𝒈
𝒎³
) and TC, avg (
HI.R
Q
)
HI.R
17
Hot water flow rate, 𝝆 c ( Q )
HI.R
(
𝒄𝒎³
𝒎𝒊𝒏
C 𝝆 C (HI.R)
Q
)
1000
4180.6
4180
2000
4181
4180
3000
4181
4180
4000
4181.2
4180
Table 6 : Specific heat capacity for different flow rate
18
SAMPLE CALCULATION
•
𝑱
Specific heat capacity for TH,Avg , CPH (𝒌𝒈.𝑲)
!)³
Using interpolation, at hot water flow rate, QH is1000(
)&'
Temperature,
𝒂𝒗𝒈
𝑻𝑪,
)
Specific
heat
𝑱
capacity, CPH (𝒌𝒈.𝑲)
50
4180
51.5
𝐶𝑃𝐻
75
4190
Table 7 : interpolation data of specific heat capacity
GLML-.L
GLML6
=
C-E6M8UV:
C-E6MC-P6
CPH = 4180.6
•
𝑱
𝒌𝒈.𝑲
𝑱
Specific heat capacity for TC,avg , CPC (𝒌𝒈.𝑲)
!)³
Using interpolation, at hot water flow rate, QH is1000(
)&'
Temperature,
𝒂𝒗𝒈
𝑻𝑪,
)
Specific
heat
𝑱
capacity, CPH (𝒌𝒈.𝑲)
25
4180
28.5
𝐶𝑃𝐶
50
4180
L6M.P.L
L6M.L
=
C-E6M8UWF
C-P6MC-P6
𝑱
CPC = 4180 𝒌𝒈.𝑲
19
4. To calculate all the required data using all values from previous calculation
Hot
Power
Power
Power
Overall
Temperature
Overall
water
Emitted
Absorbed
Lost
Efficiency
Heat
flow rate
(W)
(W)
△ 𝑻𝒎(℃)
(W)
(W)
coefficient
𝒘
𝒎𝟐 . °𝑪
1000
1719.655
208.110
1511.545
12.102
21.124
147.040
2000
3162.801
554.859
2607.942
17.543
22.159
373.734
3000
4744.202
1040.172
3704.03
21.925
21.774
713.008
4000
6049.271
1663.974
4385.297
27.507
22.040
1126.810
Table 8: The required Data Using all values from previous calculation
20
SAMPLE CALCULATION
At how flow rate, QH is 1000
•
!)³
8)&'
Power emitted
Pe = QH pH CPH (TH,in – TH,out)
Pe = (1.667 x 10-5) (987.22) (4180.6) (64-39)
Pe = 1719.655 W
•
Power Absorbed
𝑃𝑎3 =3 𝑄𝐻𝜌𝐶𝐶𝑃𝐶 (𝑇𝐶,𝑜𝑢𝑡8 −3𝑇𝐶,𝑖𝑛 )3
𝑃𝑎 = (1.667 × 10−5) (995.74) (4180) (30 − 27)
𝑃𝑎 = 208.110 𝑊
•
Power Lost
𝑃𝑜𝑤𝑒𝑟 𝑙𝑜𝑠𝑡 = 𝑃𝑜𝑤𝑒𝑟 𝑒𝑚𝑖𝑡𝑡𝑒𝑑 − 𝑃𝑜𝑤𝑒𝑟 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑
𝑃𝑜𝑤𝑒𝑟 𝑙𝑜𝑠𝑡 = 1719.655 − 208.110
𝑃𝑜𝑤𝑒𝑟 𝑙𝑜𝑠𝑡 = 1511.545 𝑊
•
Overall Efficiency, 𝜼
𝜂3=3
𝑃𝑜𝑤𝑒𝑟3𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑3
𝑃𝑜𝑤𝑒𝑟3𝑒𝑚𝑖𝑡𝑡𝑒𝑑3
×31003
21
𝜂 = -G-E.7LL 3 × 100
.6P.--6
𝜂 = 12.102%
• Temperature ∆𝑻𝟏
∆𝑇1 = (𝑇𝐻, 𝑖𝑛 − 𝑇𝐶, 𝑜𝑢𝑡)
∆𝑇1 = (64 − 30)
∆𝑇1 = 34 ℃
• Temperature ∆𝑻𝟐
∆𝑇 2 = (𝑇𝐻, 𝑜𝑢𝑡 − 𝑇𝐶, 𝑖𝑛)
∆𝑇2 = (39 − 27)
∆𝑇2 = 12 ℃
22
•
∆ Tm =
∆𝑇𝑚8 =3
(∆𝑇18 −3∆𝑇2 )3
∆𝑇 28 3
∆ Tm =
(DCM-.)
!"
a=8(#$)
∆ Tm = 21.24 °C
•
Overall heat coefficient, U
𝑈=
𝑃𝑜𝑤𝑒𝑟3𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 3
∆𝑇) 3(𝑎𝑟𝑒𝑎)
𝑈 = 147.043(
𝑊
)
𝑚. . ℃
Overall Heat Coefficient (W/ (m².˚C) vs Hot water flow rate, (cm³/min)
1200
1000
800
600
400
200
0
0
500
1000
1500
2000
2500
3000
3500
4000
4500
Hot Water Flow Rate, (cm³/min)
23
Sample calculation for Counter Flow.
Counter Flow
Hot water temperature = 64°𝐶
Fixed cold water flow = 2000𝑐𝑚D /𝑚𝑖𝑛
Tube specification.
1) Heat Transmission Length = 1.5 m
2) Heat Transmission Area = 0.067 m²
3) Tube outer diameter = 15 x 0.7 mm (thin wall)
4) Shell outer diameter = 22 x 0.9 mm (thin wall)
5) Insulation thickness = 20 mm
Hot water
Cold
Cold
Cold
Hot water Hot water
Hot water
in, °𝐶
mid, °𝐶
out, °𝐶
water in, water
water out,
cm3/min
°𝐶
mid, °𝐶
°𝐶
1000
64
47
39
27
29
31
2000
64
50
42
27
31
34
3000
64
51
43
28
32
35
4000
64
50
43
28
32
35
flow rate,
Table 9 : Flow rate and Water Temperature
24
Sample Calculation
1. Find 𝑄e(m3/s), 𝑇e,+fI(℃) and 𝑇g,+fI(℃)
Hot water flow
Hot water flow
Average hot
Average
temperature,
rate, 𝑸𝑯 (cm3/min)
rate, 𝑸𝑯 (m3/s)
temperature,
1000
1.667 × 10−5
51.5
29.0
2000
3.333 × 10−5
53.0
30.5
3000
5.000 × 10−5
53.5
31.5
4000
6.667 × 10−5
53.5
31.5
𝑻𝑯,𝒂𝒗𝒈(℃)
cold
𝑻𝑪,𝒂𝒗𝒈(℃)
Table 10 : Average Hot Temperature and Average Cold Temperature at Different Flow Rate
SAMPLE CALCULATION
At hot water flow rate, 𝑄e = 10003 )&'
!) !
Convert (
!)!
)!
)&'
/
)to (
)
𝑄e = 10003(
𝑐𝑚D
13𝑚𝑖𝑛 D
13𝑚 D
)×d
e ×d
e
𝑚𝑖𝑛
60𝑠
1003𝑐𝑚
𝑄e = 1.667 × 10ML
•
𝑚D
𝑠
Average hot temperature, 𝑇e,+fI (℃)
𝑇e,+fI =
𝑇e,&' + 𝑇e,#$%
2
𝑇e,+fI =
64 + 39
2
𝑇e,+fI = 51.53℃
25
•
Average cold temperature. 𝑇g,+fI (℃)
𝑇g,+fI =
𝑇g,&' + 𝑇g,#$%
2
𝑇g,+fI =
27 + 31
2
𝑇g,+fI = 293℃
26
2. Find density of water at 𝑇e,+fI and 𝑇g,+fI, i)! j
HI
𝝆𝑯 (kg/m3)
𝝆𝑪 (kg/m3)
1000
987.22
995.56
2000
986.44
995.02
3000
986.18
994.66
4000
986.18
994.66
Hot
water
flow
rate,
(cm3/min)
Table 11 : Density of Water for Different Flow Rate
SAMPLE CALCULATION
•
Density of water at 𝑇e,+fI , 𝜌e (kg/m3)
Using interpolation from Table of Properties, at hot water flow rate 𝑄_𝐻 = 10003𝑐𝑚D /𝑚𝑖𝑛 .
Temperature, 𝑻𝑯, 𝒂𝒗𝒈
Density, 𝝆𝑯
50
988
51.5
𝜌𝐻
75
975
Table 12 :interpolation Data of Density, ρH
75 − 51.5
975 − 𝜌e
=
975 − 988
75 − 50
𝜌e = 987.223𝑘𝑔/𝑚D
27
Density of water at 𝑇g,+fI , 𝜌g (kg/m3)
•
Using interpolation from Table of Properties.
Temperature, 𝑻𝑪, 𝒂𝒗𝒈
Density, 𝝆𝑪
25
997
29
𝜌𝐶
50
988
Table 13 :interpolation Data of Density, ρC
50 − 29
988 − 𝜌g
=
50 − 25 988 − 997
𝜌_𝐶 = 995.563𝑘𝑔/𝑚D
3. Find specific heat capacity at 𝑇e,+fI (
Q
HI.R
Hot
water
flow
rate
) and 𝑇g,+fI ( HI.R)
Q
(cm /min)
𝑪𝑷𝑯 (J/kg. K)
𝑪𝑷𝑪 (J/kg. K)
1000
4180.6
4180
2000
4181.2
4180
3000
4181.4
4180
4000
4181.4
4180
3
Table 14 :Specific Heat Capacity for Different Flow Rate
28
SAMPLE CALCULATION
•
Specific heat capacity for 𝑇e,+fI , 𝐶he8 (
Q
HI.R
)
Using interpolation at flow rate is 1000 (cm3/min)
Temperature, 𝑻𝑯 ,𝒂𝒗𝒈
Specific heat capacity, 𝑪𝑷𝑯 (J/kg. K)
50
4180
51.5
𝐶𝑃𝐻
75
4190
Table 15 :Interpolation Data of Specific Heat Capacity, 𝑪𝑷𝑯
75 − 51.5 4190 − 𝐶𝑃𝐻3
=
4190 − 4180
75 − 50
𝐶he = 4180.6
•
Specific heat capacity for 𝑇g,+fI , 𝐶hg 3(
Q
HI.R
𝐽
𝑘𝑔. 𝐾
)
Using interpolation at flow rate is 1000 (cm3/min)
Temperature, 𝑻𝑪, 𝒂𝒗𝒈
Specific heat capacity, 𝑪𝑷𝑪 (J/kg. K)
25
4180
29
𝐶𝑃𝐶
50
4180
Table 16 : : Interpolation Data of Specific Heat Capacity, 𝑪𝑷𝑪
50 − 29
4180 − 𝐶hg
𝐽
=
= 𝐶hg = 4180
𝑘𝑔. 𝐾
50 − 25 4180 − 4180
29
4. Calculate the data
Hot
Power
Water
Power
Power
Emitted Absorbed Lost
Flow Rate (𝑾)
(𝑾 )
(𝑾)
Overall
Temperature Overall
Efficiency △ 𝑻𝒎(℃)
Heat
(%)
Coefficient
(W/m2.℃)
(cm3/min)
1000
1719.655 277.429
1442.226 16.133
20.759
199.465
2000
3024.635 970.476
2054.159 32.086
21.640
669.336
3000
4329.794
1455.188 2874.606 33.609
21.236
1022.736
4000
5773.058
1940.250 3832.808 33.609
21.236
1363.648
Table 17: Calculation summary table.
SAMPLE CALCULATION
At hot water flow rate, 𝑄e = 1000 )&'
!)!
•
Power Emitted
𝑃𝑒 = 𝑄e 𝜌e 𝐶he (𝑇e,&' − 𝑇e,#$% )
𝑃𝑒 = (1.667 × 10ML )(987.22)(4180.6)(64 − 39)
𝑃𝑒 = 17203𝑊
•
Power Absorbed
𝑃𝑎 = 𝑄e 𝜌g 𝐶hg (𝑇g,#$% − 𝑇g,&' )
𝑃𝑎 = (1.667 × 10ML )(995.96)(4180)(31 − 27)
𝑃𝑎 = 277.483𝑊
30
•
Power Lost
𝑃𝑜𝑤𝑒𝑟3𝑙𝑜𝑠𝑡 = 𝑃𝑜𝑤𝑒𝑟3𝑒𝑚𝑖𝑡𝑡𝑒𝑑 − 3𝑃𝑜𝑤𝑒𝑟3𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑
𝑃𝑜𝑤𝑒𝑟3𝐿𝑜𝑠𝑡 = 1720 − 277.48
𝑃𝑜𝑤𝑒𝑟3𝑙𝑜𝑠𝑡 = 1442.523𝑊
•
Overall Efficiency, 𝜂
𝜂=
𝑃𝑜𝑤𝑒𝑟3𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑
× 100
𝑃𝑜𝑤𝑒𝑟3𝑒𝑚𝑖𝑡𝑡𝑒𝑑
𝜂=
277.48
× 100 = 16.13%
1720
Temperature ∆𝑇∆𝑇- = (𝑇e,&' − 𝑇g,#$% )
∆𝑇- = (64 − 31)
∆𝑇- = 33℃
•
Temperature ∆T.
∆𝑇. = E𝑇e,#$% − 𝑇g,&' F
∆𝑇. = (39 − 27)
∆𝑇. = 12℃
•
Temperature ∆𝑇)
∆𝑇) =
(∆𝑇- − 3 ∆T. )
∆𝑇
ln i - j
∆T.
3∆𝑇) =
(33 − 12)
33
ln3(12)
3∆𝑇) = 20.76℃
31
•
Overall Efficiency, 𝑈
𝑈=
𝑃𝑜𝑤𝑒𝑟3𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 3
∆𝑇) 3(𝑎𝑟𝑒𝑎)
𝑈=
277.48
20.76(0.067)
𝑊
𝑈 = 199.493( . )
𝑚 .℃
32
Overall Heat Coefficient (W/m². ˚C) vs Hot water flow rate, (cm³/min)
1600
1400
1200
1000
800
600
400
200
0
0
500
1000
1500
2000
2500
3000
3500
4000
4500
Hot Water Flow Rate, (cm³/min)
Graph 2: Overall heat coefficient vs Hot water flow rate
33
6. DISCUSSION
MUHAMMAD AFIQ BIN MOHD ZOHADI
2020816882
Based on the experiment on the heat exchanger, we can identify the principles of the
heat exchanger. The experiment is separated into 2 types which is parallel and counter flow, as
for the medium used to conduct the experiment is water. The hot water temperature at the inlet
is kept constant 64°C and the water flow rate is also kept constant 2000𝑐𝑚D /𝑚𝑖𝑛 . The
temperature is then recorded for both hot and cold water.
For the first part which is parallel we gather the properties of the water through
interpolation on the table A-2, then we got the density of 987.22 )! for hot water and
HI
995.74
HI
)!
. after that we list out all the necessary properties inside the table to calculate the
specific heat capacity for both hot and cold water, for the hot water we got 4180.6
cold water 4180
𝑱
𝒌𝒈.𝑲
and
𝑱
𝒌𝒈.𝑲
And then we calculate the power emitted using the Pe = Q p CP (T,in – T,out) for both hot
and cold water by inputting the value gathered, we got 1719.655 W then we calculate the
power absorbed using 𝑃𝑎 = 𝑄𝐻𝜌𝐶𝐶𝑃𝐶 (𝑇𝐶,𝑜𝑢𝑡 − 𝑇𝐶,𝑖𝑛) and we got 208.110 W and power
lost using 𝑃𝑜𝑤𝑒𝑟3𝑙𝑜𝑠𝑡 = 𝑃𝑜𝑤𝑒𝑟3𝑒𝑚𝑖𝑡𝑡𝑒𝑑 − 𝑃𝑜𝑤𝑒𝑟3𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 and we got 1511.545 W and
lastly we can calculate the Overall Efficiency of 12.102%.
For the next part we calculate the Overall heat Efficiency using 𝑈 = h#i*j8+k/#jk*l8 ,
∆n 8(+j*+)
%
and 147.043( $ ), and then we plot the graph of Overall Heat Coefficient vs Hot water flow
) .℃
o
rate and we can deduce the relationship is directly increasing, as the hot water flow rate
increase, the overall heat coefficient also increases.
And then we repeat the process to the next part which is the counter flow. As for the
graph for counter flow (Overall heat coefficient vs hot water flow rate), the relationship also
similar to the parallel flow.
34
MUHAMMAD ALIF FARHAN BIN MOHD RIDZUAN
2020898722
The principles of the heat exchanger may be identified based on the heat exchanger
experiment. The experiment is divided into two types: parallel flow and counter flow, and the
medium utilised to conduct the experiment is water. The hot water temperature at the entrance
is kept constant at 64°C, as is the water flow rate of 2000cm3/min. The temperature of both
hot and cold water is then recorded.
The density of 987.22 kg/m3 for hot water and 995.74 kg/m3 for cold water was
obtained using interpolation on table A-2 for the first section, which is parallel. Following that,
we include all of the relevant parameters inside the table to compute the specific heat capacity
for both hot and cold water, yielding 4180.6 J/(kg.K) for hot water and 4180 J/(kg.K) for cold
water (kg.K).
The power emitted (Pe ) for both hot and cold water is 1719.655 W. The power absorbed
𝑃𝑎 = 𝑄𝐻𝜌𝐶𝐶𝑃𝐶 (𝑇𝐶,𝑜𝑢𝑡 − 𝑇𝐶,𝑖𝑛) is 208.110 W and power lost is 1511.545 W and lastly we
can calculate the Overall Efficiency of 12.102%.
Next, the Overall heat Efficiency by using 𝑈 = h#i*j8+k/#jk*l8 , is147.043( o ), and
)$ .℃
∆n 8 (+j*+)
%
the graph of Overall Heat Coefficient vs Hot water flow is directly proportional increasing, as
the hot water flow rate increase, the overall heat coefficient also increases.
35
MUHAMMAD IQMAL BIN LOKMAN
2020825268
We can identify the principles of the heat exchanger based on the heat exchanger
experiment. The experiment is divided into two types: parallel and counter flow, with water as
the medium utilised to conduct the experiment. The temperature of the hot water at the inlet is
maintained at 64°C, and the water flow rate is maintained at 2000cm3/min. The temperature
of both hot and cold water is then recorded.
We obtained the density of 987.22 )! for hot water and 995.74 )!for cold water by
HI
HI
approximating the attributes of the water on table A-2 for the first section, which is parallel.
Following that, we list all of the necessary properties inside the table to compute the specific
heat capacity for both hot and cold water, yielding 4180.6 𝒌𝒈.𝑲for hot water and 4180 𝒌𝒈.𝑲) for
𝑱
𝑱
cold water (kg.K).
Then we calculate the power emitted using Pe = Q p CP (T,in – T,out)for both hot and
cold water by inputting the value gathered, and we get 1719.655 W. Next, we calculate the
power absorbed using 𝑃𝑎 = 𝑄𝐻𝜌𝐶𝐶𝑃𝐶(𝑇𝐶,𝑜𝑢𝑡 − 𝑇𝐶,𝑖𝑛) and we get 208.110 W. Finally, we
3
=
3
−
3
calculate the power lost using 𝑃𝑜𝑤𝑒𝑟 𝑙𝑜𝑠𝑡 𝑃𝑜𝑤𝑒𝑟 𝑒𝑚𝑖𝑡𝑡𝑒𝑑 𝑃𝑜𝑤𝑒𝑟 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 , and we
get 1511.545 W.
In the next section, we calculate the Overall Heat Efficiency using 𝑈 =
h#i*j8+k/#jk*l8
∆n% 8(+j*+)
and 147.043( $ ), , and then plot the graph of Overall Heat Coefficient vs Hot Water Flow
) .℃
o
Rate. We can see that the relationship is directly increasing, with the overall heat coefficient
increasing as the hot water flow rate increases.
The process was repeated for the next part, which is the counter flow. The relationship
between the overall heat coefficient and the hot water flow rate is similar in the graph for
counter flow (Overall heat coefficient versus hot water flow rate).
36
MOHD FARHAN HAQIMI BIN ABDUL HALIM
2020970471
For this experiment, there are two principal flow configurations employed by heat exchangers
that we need identify which are parallel flow and counter flow. For both situations, we must
find the power lost for that system but in order to find that value, we must calculate the
power emitted and power absorbed by using this formula,
Qemitted =
H CpH (THin-THout
) (power emitted) and Qabsorbed =
absorbed)
CCpC
(TCout-TCin) (power
Then, we can use this formula to find power lost,
Power lost = Qemitted - Qabsorbed
After that we must find the efficiency of this system by using this formula,
h=
Q absorbed
x100%
Q emitted
And we also need to find the logarithmic mean temperature different using this formula,
∆Tm= (∆T1- ∆T2)/ ln (
∆9∆9.
)
Lastly, we must find the overall heat transfer coefficient with the formula of,
U= Power absorbed/ AS∆Tm.
From the data and the calculation for parallel flow that have been done, we can find
that the lower the hot water flow rate, the higher the efficiency of the system. The highest
37
efficiency is 24.27% at 1000 cm3/min while the lowest is 13.81% at 4000cm3/ min. For the
logarithmic mean temperature difference, the highest value is 22.04 oC at 4000 cm3/min while
the lowest is 21.12 oC at 1000cm3/min. Next, for overall heat transfer coefficient, the highest
value is 563.09 W/m2. oC at 4000 cm3/min and the lowest is 293.77 W/m2. oC at 1000 cm3/min.
From the data and the calculation for counter flow that have been done, we can find
that the lower the hot water flow rate, the higher the efficiency of the system. The highest
efficiency is 32.36% at 1000 cm3/min while the lowest is 16.89% at 4000cm3/ min. For the
logarithmic mean temperature difference, the highest value is 21.24 oC at 4000 cm3/min while
the lowest is 20.76 oC at 1000cm3/min. Next, for overall heat transfer coefficient, the highest
value is 681.76 W/m2. oC at 4000 cm3/min and the lowest is 398.51 W/m2. oC at 1000 cm3/min.
So, from the analysis for both systems, we can find out that the efficiency for counter
flow system is higher than the parallel flow system
38
7. CONCLUSION
MUHAMMAD AFIQ BIN MOHD ZOHADI
2020816882
In conclusion, the heat exchanger experiment was successfully conducted and the
objective of the experiment which is to demonstrate the effect flow rate variation on the
performance characteristics of a counter-flow and parallel-flow concentric tube heat
exchanger. We could identify the efficiency of the flow of the parallel and the counter flow,
we deduce that the counter flow has the higher efficiency than the parallel flow. Lastly the
experiment could also have a small error, where the water might have an impurity that could
affect the reading of the temperature of the reading, a way to avoid this error is, taking several
temperatures and calculate the average to reduce the error
39
MUHAMMAD ALIF FARHAN BIN MOHD RIDZUAN
2020898722
To conclude, the objective of the experiment was achieved where the flow rates affect
the performance parameters of a counter-flow and parallel-flow concentric tube heat
exchanger. And since the experiment were held by the supervision of lab staff from UiTM
itself make the result was gained properly. The counter flow overall heat transfer coefficient
had the highest value makes it the better heat exchanger compare parallel flow.
40
MUHAMMAD IQMAL BIN LOKMAN
2020825268
Finally, the heat exchanger experiment was completed successfully, with the goal of
showing the effect of flow rate variation on the performance parameters of a counter-flow and
parallel-flow concentric tube heat exchanger. We were able to determine the efficiency of the
parallel and counter flows, and we concluded that the counter flow is more effective than the
parallel flow. Finally, the experiment could have a slight error, such as when the water contains
a particle that affects the temperature measurement. One approach to eliminate this problem is
to take numerous temperatures and average them to reduce the error.
41
MOHD FARHAN HAQIMI BIN ABDUL HALIM
2020970471
For the conclusion, I think the experiment was carried out successfully and we also had
achieved all the objectives that stated before. First, we had study the operation of concentric
tube heat exchanger. Next, we also had achieved statistically significant trends for overall
heat transfer coefficient by advancing the set of experiments and to make inside heat transfer
coefficient as water supply. Lastly, we also had observed the difference between parallel flow
and counter flow operation of heat exchanger.
There are a few errors that can occur during this experiment such as parallax error and
systematic error. Some improvements need to be count during this experiment to make sure
that we can higher efficiency and can get more precise value than before.
42
8. REFERENCES
1. Yunus A. Cengal & Michael A. Boles, “Thermodynamics – An Engineering Approach”, 3rd
Edition, 4th Edition, McGraw Hill, 2002
2. Holman, J.P. “Heat Transfer,” McGraw-Hill Book Company, New York, 2001
3.
Perry,
J.H.(Ed.):”Chemical
Engineers
Handbook,”
4th
Edition,
McGraw-Hill
BookCompany, New York, 1963
4.
Heat
Exchanger.
Retrieved
from
www.real-world-physics-
problems.com/heatexchanger.htm
5.
Concentric
Tube
Heat
Exchanger,
https://acikders.ankara.edu.tr/pluginfile.php/59064/mod_resource/content/3/Experiment%2
010_Concentric%20Tube%20Heat%20Exchanger.pdf
6. How do heat exchangers work? (2020, March 26). Retrieved November 10, 2020, from
https://www.explainthatstuff.com/how-heat-exchangers-work.html
7. Physics problems, R. (2020). Heat Exchanger. Retrieved November 10, 2020, from
https://www.real-world-physics-problems.com/heat-exchanger.html
43
9. APPENDICES
44
CONCENTRIC TUBE HEAT
EXCHANGER DISCUSSION
AND CONCLUSION
by MUHAMMAD AFIQ MOHD ZOHADI
Submission date: 17-Nov-2021 12:14AM (UTC+0800)
Submission ID: 1704671597
File name: Discussion_and_Conclusion_Heat_Exchanger_AFIQ.pdf (131.83K)
Word count: 440
Character count: 2039
45
46
47
48
discussion and conclusion
(heat exchanger)
by alif Farhan
Submission date: 17-Nov-2021 02:18AM (UTC+0800)
Submission ID: 1704781316
File name: Discussion_n_conclusion_lab3.docx (13.9K)
Word count: 285
Character count: 1442
49
50
51
Heat Exchanger
by Iqmal Lokman
Submission date: 17-Nov-2021 02:10AM (UTC+0800)
Submission ID: 1704772863
File name: Discussion_and_Conclusion_Heat_Exchanger.pdf (108.48K)
Word count: 416
Character count: 1993
52
53
54
55
CONCENTRIC TUBE HEAT
EXCHANGER DISCUSSION
AND CONCLUSION
by Mohd Farhan Haqimi Bin Abdul Halim
Submission date: 17-Nov-2021 12:14AM (UTC+0800)
Submission ID: 1704671597
File name: Discussion_and_Conclusion_Heat_Exchanger_AFIQ.pdf (131.83K)
Word count: 440
Character count: 2039
56
DISCUSSION &
CONCLUSION_FARHAN
by Mohd Farhan Haqimi Bin Abdul Halim
Submission date: 17-Nov-2021 08:43PM (UTC+0800)
Submission ID: 1705527769
File name: DISCUSSION_CONCLUSION_FARHAN.docx (18.7K)
Word count: 480
Character count: 2363
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