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Designing a Double-Pipe Heat Exchanger according to the Customer specifications. Conducted all calculations required to meet the specification.Note that, this is the property of Daniel James Watkins.Faculty of Advance TechnologyBEng Mechanical EngineeringUniversity of GlamorganUnited Kingdom
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Module Code: NG3H237 [ ]
Introduction
Heat exchangers are devices that facilitate heat transfers between two or more fluids at different
temperatures (Ozisik, 1985). Heat exchangers may be classified (Bejan and Kraus, 2003) according to
transfer processes, number of fluids, construction, heat transfer mechanisms, surface compactness,
flow arrangement, number of fluid passes and types of surface. Various types of heat exchangers have
been developed for use at such varied levels of technological sophistication and sizes such as steam
power plant, chemical processing plants, building heating and air conditioning, household
refrigerators, car radiators, radiators for space vehicles, and so on (Bejan and Kraus, 2003).
Fraas (1989) suggested that in common types of heat exchangers such as double-pipe and shell-and-
tube, heat transfer is primarily by conduction and convection from a hot to a cold fluid, which are
separated by a metal wall. Double-pipe and shell-and-tube heat exchangers are available in a variety
of materials, and are suitable for a wide range of pressures and temperatures. Their construction
(Ozisik, 1985) allows a wide variation in design as factors such as tube size, number of tubes and tube
length can all be varied. Foumeny and Heggs (1991) justified that this allows the flexibility in
adapting designs to suit particular requirements; it also means that design optimisation is required to
achieve cost-effective designs.
Bejan and Kraus (2003) expressed that the design of heat exchangers is a complicated matter. Ozisik
(1985) evaluated that heat transfer and pressure drop analysis, sizing and performance estimation, and
the economic aspects play important roles in the final design of heat exchangers. For example,
although the cost considerations are very important for applications in large installations such as
power plants and chemical processing plants, the weight and size considerations become the dominant
factor in the choice of design for space and aeronautical application (Ozisik, 1985).
.
Daniel James Watkins 09032266 Page 1
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Background Theory
Double Pipe Heat Exchangers are a form of Tubular Heat Exchanger. A typical double pipe heat
exchanger (Kuppan, 2000) has two concentric pipes, usually in the form of a U-bend design as
illustrated in figure 1. The flow arrangement is pure counter-current (Ozisik, 1985). Kuppan (2000)
and Fraas (1989) suggested that a number of double pipe heat exchangers can be connected in series
or parallel as necessary and their usual application is for a small duties requiring, typically, less than
300 ft2 (27.87 m2). They are suitable for high pressures and temperatures, and thermally long duties.
This contain the advantages of flexibility since units can be added or removed as required, and the
design is easy to service and requires low inventory of spares due to its standardization (Kuppan,
2000). Kuppan (2000) added, either longitudinal fins or circumferential fins within the annulus on the
inner pipe wall are required to enhance the heat transfer from the inner pipe fluid to the annulus fluid.
Figure 1: Double pipe heat exchanger. (a) Single pass with counter-flow; and (b) multi-pass with
counter-flow (Kuppan, 2000, p. 4)
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Heat Exchanger Analysis with the use of the Log Mean Temperature Difference
Incropera and DeWitt (2002) suggested that in order to design or to predict the performance of a heat
exchanger, it is necessary to relate the quantities such as the inlet and outlet fluid temperatures, the
overall heat transfer coefficient, and the total surface area for heat transfer. Two such relations may be
obtained by applying overall energy balances to the hot and cold fluids as illustrated in figure 2.
Figure 2: Overall Energy balances for the hot and cold fluids of a two-fluid heat exchanger (Incropera
and DeWitt, 2002, p. 587)
The parallel-Flow Heat Exchanger
The hot and cold fluid temperature distributions associated with a parallel-flow heat exchangers are
illustrate in figure 3. It is important to note that, for such an exchanger, the outlet temperature of the
cold fluid never exceeds that of the hot fluid (Incropera and DeWitt, 2002). In figure 3, the sub scripts
1 and 2 designate opposite ends of the heat exchanger. The energy balances and the subsequent
analysis are subject to the following assumptions (Incropera and DeWitt, 2002).
1. The heat exchanger is insulated from its surroundings, in which case the only heat exchange
is between the hot and cold fluids.
2. Axial conduction along he tubes is negligible.
3. Potential and kinetic energy changes are negligible.
4. The fluid specific heats are constant.
5. The overall heat transfer coefficient is constant.
The specific heats may change as a result of temperature variations, and the overall heat transfer
coefficient may change because of variations in fluid properties and flow conditions (Incropera and
DeWitt, 2002). Incropera and DeWitt (2002) added, in many applications such variations are not
significant, and it is reasonable to work with average values of Cp,c , Cp,h , and U for the heat
exchanger.
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Figure 3: Temperature distributions for a parallel-flow heat exchanger (Incropera and DeWitt, 2002,
p. 588)
The Counter-flow Heat Exchanger
The hot and cold fluid temperature distributions associated with a counter-flow heat exchangers are
illustrated in figure 4. In contrast to the parallel-flow exchanger, this configuration provides for heat
transfer between the hotter portions of the two fluids at one end, as well as between the older portions
at the other (Incropera and DeWitt, 2002). Hence, the outer temperature of the cold fluid may exceed
the outlet temperature of the hot fluid (Incropera and DeWitt, 2002). Incropera and DeWitt (2002)
added, for the same inlet and outlet temperatures, the log mean temperature difference for cunter-flow
exceeds that for parallel flow. The surface area required to effect the heat transfer rate q is smaller for
the counter-flow than for the parallel-flow arrangement, assuming the same value of U.
Figure 4: Temperature distributions for a counter-flow heat exchanger (Incropera and DeWitt, 2002,
p. 590)
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Aims and Objectives
Aims
Fully develop the understanding of Heat Exchangers Design
Fully develop the understanding of different Heat Exchangers Calculation methods
Objectives
Design a double-pipe water-to-water heat exchangers to meets the specifications
provided
Evaluate the design by identifying what features and configurations could be explored
with the customer in order to develop more complete specifications
Design Assumptions
1. The fluid velocity is not greater than 5 m/s
2. The tube length should not exceed 12 m
3. The flow is a fully developed turbulent flow
4. The system kinetic and potential energy is dynamic
5. The heat exchanger is a Counter-flow system
6. The thermal resistance is negligible
7. The surrounding heat lost is negligible
8. Constant thermal-physical properties throughout the system
Design Approaches
1. Compute all calculations required to obtain the value of UA
2. Assume ranges of overall coefficient (U) in order to estimate the heat transfer area (A)
3. Once area (A) is obtained, estimate the tube diameters.
4. Determine the overall coefficient (U) of the selected tube diameters. Hence,
Correlation of cold and hot convection coefficient can be used.
5. Once the UA products of all the selected diameters are obtained, compare it to the
required (original) UA value.
6. If UA value from the first sets of selected diameters did not satisfy the required UA
value, Interpolation of the ranges of the tubes against the original UA value can be
compute to obtain the necessary tube diameter.
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Calculations and Results
Table 1: Given Data for the Design
FluidMass Flow Rate
m (kg/s)Inlet Temperature
Ti (°C) » (K)Outlet Temperature
TO (°C) » (K)
Hot 28 90°C = 363 K 65°C = 338 K
Cold 27 34°C = 307 K 60°C = 333K
Step 1:
Determination of the Average Mean Temperature (AMT):
Average Mean Temperature for Hot Fluid (AMTh) = 363+338
2 = 350.5 K
Average Mean Temperature for Cold Fluid (AMTC) = 307+333
2 = 320 K
From table A1 (Appendix 1), determine the CP, h value:
At, AMTh = 350.5 K, Interpolate to obtain Value of CP, h as illustrate in figure 5.
Figure 5: Graphical interpolation method to obtain CP, h
y− y0
x−x0
=y1− y0
x1−x0
350.5−350x−4.195
= 355−3504.199−4.195
CP,h = x = 4.1954 × 103 J/kg K
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and at AMTC = 320 K
CP, C = 4.180 × 103 J/kg K (Obtained directly from table A1 without interpolation)
Determinations of the Heat Capacity (C) for both fluids:
Heat Capacity of hot fluid (Ch) = mh ×C p ,h= (28 ) ×(4.1954 × 103)
Heat Capacity of hot fluid (Ch) = 117.4712 × 103 W/K
____________________________________________
Heat Capacity of cold fluid (CC) = mC× C p ,C = (27) × (4.180 × 103)
Heat Capacity of hot fluid (CC) = 112.86 × 103 W/K
____________________________________________
Determination of Heat Transfer Effectiveness(ε):
Heat Transfer Effectiveness (ε )=CC(T C ,O−T C ,i)Cmin (T h ,i−T C ,i)
Heat Transfer Effectiveness (ε )= (112.86×103 )×(333−307)(112.86×103 )×(363−307)
= 0.464
Determination of Heat Capacity Ratio (Cr):
Heat Capacity R atio (C r )=Cmin
Cmax
Heat Capacity Ratio ( Cr )=(112.86 ×103)(117.47 ×103)
= 0.96 → Cr ≤ 1
Since the Heat Capacity Ratio (Cr) is determined to be lesser than 1, therefore, this state that
the counter flow arrangement must be apply.
Determination for Number of Transfer Units (NTU) with counter flow arrangement:
NTU=1
Cr−1ln( ε−1
ε Cr−1 )NTU= 1
(0.96−1)ln( 0.464−1
(0.464 × 0.96)−1 )Daniel James Watkins 09032266 Page 7
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NTU = 0.851
Determination of the Maximum Heat Transfer Rate (qmax):
qmax=CC (T h i−T C i )
qmax=(112.86× 103)×(363−307)
qmax=6.32× 106 W
Determination of the Heat Transfer Rate (q):
heat transfer rate (q )=ε × qmax
heat transfer rate (q )=0.464 ×(6.32 ×106)
heat transfer rate (q )=2.932× 106W/K
Determination of the UA Value:
NTU=U × ACmin
UA=NTU ×Cmin=0.851 × (112.86× 103 )=96.04 ×103 W/K
Step 2: Estimate the area (A). The typical range of U for Water-to-Water exchangers is 850-1700
W/m2K as illustrated in table A2.
Therefore, with UA = 96.04 ×103 W/K, the ranges for Area (A) are as follow:
(850 ) A1=96.04 ×103
∴ A1=112.99≈ 113m2=Amax
(1700 ) A2=96.04 ×103
∴ A2=56.494 ≈ 56.5 m2=Amin
Now, consider: A=π Di ln → Ranges for the Number of Tubes can be obtained.
Where, L = the length of the tube and N = Number of Tubes
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Trial 1: Assuming that Di = 0.025 m and L = 10 m
From, A=π Di ln
Nmin=Amin
π × Di× L= 56.5
π× 0.025 ×10=71.938 ≈ 72 Tubes
Nmax=Amax
π × D i× L= 113
π× 0.025 ×10=143.876 ≈ 144 Tubes
Trial 2: Assuming that Di = 0.045 m and L = 10 m
Nmin=Amin
π × Di× L= 56.5
π× 0.045 ×10=39.96 ≈ 40 Tubes
Nmax=Amax
π × D i× L= 113
π× 0.045 ×10=79.93 ≈ 80 Tubes
Trial 3: Assuming that Di = 0.075 m and L = 10 m
Nmin=Amin
π × Di× L= 56.5
π× 0.075 ×10=23.979 ≈ 24 Tubes
Nmax=Amax
π × D i× L= 113
π× 0.075 ×10=47.95 ≈ 48 Tubes
Table 2: Dimensional properties and number of tubes results
Trial Di (m) L (m) N A (m2)
1 0.025 10 72 → 144 56.5 → 113
2 0.045 10 40 → 80 56.5 → 114
3 0.075 10 24 → 48 56.5 →115
Step 3: Estimate the overall coefficient, U. With the inner (hot) and outer (cold) fluids in the concentric tube arrangement, the overall coefficient is:
1U
≈1h1
+ 1h0
Where, hi = coefficient of hot side and ho = the coefficient of cold side
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The h are estimate using the Dittus-Boelter correlation assuming fully develop turbulent flow.
From table A1, determine the viscosity, μh value.
At, AMTh = 350.5 K, Interpolate to obtain Value of μh as illustrate in figure 6.
Figure 6: Graphical interpolation method to obtain μh
y− y0
x−x0
=y1− y0
x1−x0
350.5−350x−365
=355−350343−365
μh=x=362.8× 106 N ∙ S/m2
and at AMTC = 320 K,
μC=577 × 106 N ∙ S/m2 (Obtained directly from table A1 without interpolation)
From table A1, determine the Prandtl number (Pr)
At, AMTh = 350.5 K, Interpolate to obtain Value of Pr, h as illustrate in figure 7.
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Figure 7: Graphical interpolation method to obtain Pr, h
y− y0
x−x0
=y1− y0
x1−x0
350.5−350x−2.29
= 355−3502.14−2.29
Pr , h=x=2.275 ≈ 2.28
and at AMTC = 320 K
Pr , C=3.77(Obtained directly from table A1 without interpolation)
From table A1, determine the Thermal Conductivity, k (W/m.K)
At, AMTh = 350.5 K, Interpolate to obtain Value of kh as illustrate in figure 8.
Figure 8: Graphical interpolation method to obtain kh
y− y0
x−x0
=y1− y0
x1−x0
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350.5−350x−668
=355−350671−668
k h=x=668.3 ×103 W/m.K
and at AMTC = 320 K
k c=640 ×103W/m.K (Obtained directly from table A1 without interpolation)
Table 3: Results of Prandtl Number (Pr), Viscosity (µ) and Thermal Conductivity (k) for both
Hot/Cold fluids at Average Mean Temperature (AMT).
Fluids AMT (K) Pr µ (N•S/m2) k (W/.K)
Hot 350.5 2.28 362.8 × 106 668.3× 103
Cold 320 3.77 577 × 106 640 × 103
Study 1
Consider, Di = 0.025 m
1. mhi=mh
N=28
72=0.39 kg/ s
2. R eDi=4 × mhi
π× Di× μh
= 4 × 0.39π ×(0,025)×(362.8 × 10−6)
=54747.9
3. NUD=h i× D i
k h
=0.027 × R eDi
45 × Pr ,h
13…………… Cooling
NUD=h i× D i
k h
=0.027 ×(5474745 )×(2.28
13 )
∴219.5=hi ×0.025
668.3× 10−3
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∴hi=5867.67 W /m2 K
4. Aci=Aco →π × Di
2
4=
π ×(Do2−Di
2)4
π ×(0.0252)4
=π ×(Do
2−0.0252)4
∴D 0=0.035 m
5. mhi ,c=mc
N=27
72=0.375 kg /s
6. R eDo=4×mhi ,c
π × Do × μc
= 4× 0.375π ×(0.035)×(577 ×10−6)
=23642.72
7. NUD=ho× Do
kc
=0.024 × R eDo
45 × Pr ,c
25……………Heating
NUD=ho× Do
kc
=0.024 × ¿¿
∴128.74=ho× 0.035
640 ×10−3
∴ho=2354.1 W /m2 .K
8.1U
≈1hi
+ 1ho
1U
= 15867.67
+ 12354.1
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1U
=5.95 ×10−4
U=1680.06 W /m2. K
Study 2
Consider, Di = 0.045 m
1. mhi=mh
N=28
40=0.7 kg /s
2. R eDi=4 × mhi
π× Di× μh
= 4 ×0.7π ×(0,045)×(362.8 × 10−6)
=54591.92
3. NUD=h i× D i
k h
=0.027 × R eDi
45 × Pr ,h
13…………… Cooling
NU D=hi × Di
kh
=0.027 ×(54591.9245 )×(2.28
13)
∴218.97=h i× 0.045
668.3 ×10−3
∴hi=3251.95 W /m2 K
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4. Aci=Aco →π × Di
2
4=
π ×(Do2−Di
2)4
π ×(0.0252)4
=π ×(Do
2−0.0452)4
∴D 0=0.064 m
5. mhi ,c=mc
N=27
40=0.675 kg /s
6. R eDo=4×mhi ,c
π × Do × μc
= 4× 0.675π ×(0.064 )×(577 ×10−6)
=23273.31
7. NUD=ho× Do
kc
=0.024 × R eDo
45 × Pr ,c
25……………Heating
NUD=ho× Do
kc
=0.024 × ¿¿
∴127.1=ho ×0.064
640 × 10−3
∴ho=1271 W /m2 . K
8.1U
≈1hi
+ 1ho
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1U
= 13251.95
+ 11271
1U
=1.094 ×10−3
U=913.83 W /m2 . K
Study 3
Consider, Di = 0.075 m
1. mhi=mh
N=28
24=1.17 kg /s
2. R eDi=4 × mhi
π× Di× μh
= 4 ×1.17π ×(0.075)×(362.8 × 10−6)
=54747.9
3. NUD=h i× D i
k h
=0.027 × R eDi
45 × Pr ,h
13…………… Cooling
NUD=h i× D i
k h
=0.027 ×(54747.945 )×(2.28
13 )
∴219.5=hi ×0.075
668.3× 10−3
∴hi=1955.89 W /m2 K
4. Aci=Aco →π × Di
2
4=
π ×(Do2−Di
2)4
π ×(0.0752)4
=π ×(Do
2−0.0752)4
∴D 0=0.12 m
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5. mhi ,c=mc
N=27
24=1.125 kg /s
6. R eDo=4×mhi ,c
π × Do × μc
= 4× 1.125π ×(0.12)×(577 × 10−6)
=20687.38
7. NUD=ho× Do
kc
=0.024 × R eDo
45 × Pr ,c
25……………Heating
NUD=ho× Do
kc
=0.024 × ¿¿
∴115.69=ho× 0.12
640 ×10−3
∴ho=617.01 W /m2 .K
8.1U
≈1hi
+ 1ho
1U
= 11955.89
+ 1617.01
1U
=2.132× 10−3
U=469.04 W /m2. K
Table 4: Obtained physical properties from the studies
Study Di (m) A (m2) N hi (W/m2.K) ho (W/m2.K) U (W/m2.K) UA (W/K)
1 0.025 56.5 72 5867.67 2354.1 1680.06 94.92 × 103
2 0.045 56.5 40 3251.95 1271 913.83 51.63 × 103
3 0.075 56.5 24 1955.89 617.01 469.04 26.5 × 103
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The value of UA obtained from the 3 studies did not meet the original required UA value
(96.04 × 103 W/K). Since the results of UA values increases as the diameter decrease,
assumption for the suitable diameter can be made. From table, at the diameter of 0.025 m, its
UA value is the closest to the Original UA value. Therefore, let’s consider a new diameter to
be studied.
Study 4
Assuming, Di = 0.02 m and L = 10 m
From, A=π Di ln
Nmin=Amin
π × Di× L= 56.5
π× 0.02 ×10=89.9 ≈ 90 Tubes
Nmax=Amax
π × D i× L= 113
π× 0.02 ×10=179.84 ≈ 180 Tubes
--------------------------------------------------------------
1. mhi=mh
N=28
90=0.311kg /s
2. R eDi=4 × mhi
π× Di× μh
= 4× 0.311π ×(0.02)×(362.8 ×10−6)
=54572.42
3. NUD=h i× D i
k h
=0.027 × R eDi
45 × Pr ,h
13…………… Cooling
NUD=h i× D i
k h
=0.027 ×(54572.4245)×(2.28
13 )
∴218.9=hi ×0.02
668.3× 10−3
∴hi=7533.44 W /m2 K
4. Aci=Aco →π × Di
2
4=
π ×(Do2−Di
2)4
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π ×(0.022)4
=π ×( Do
2−0.022)4
∴D 0=0.028 m
5. mhi ,c=mc
N=27
90=0.3 kg /s
6. R eDo=4×mhi ,c
π × Do × μc
= 4 × 0.3π ×(0.028)×(577 ×10−6)
=23642.72
7. NUD=ho× Do
kc
=0.024 × R eDo
45 × Pr ,c
25……………Heating
NUD=ho× Do
kc
=0.024 × ¿¿
∴128.74=ho× 0.028
640 ×10−3
∴ho=2942.63 W /m2 . K
8.1U
≈1hi
+ 1ho
1U
= 17533.44
+ 12942.63
1U
=4.726 × 10−4
U=2116.07W /m2 . K
9. ∴UA=2116.07× 56.5=119.56 ×103 W / K
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With Di = 0.02 m, the value of UA obtain is119.56× 103W / K , from here, interpolation
between Di = 0.025 m and Di = 0.02 m in order to gain the required diameter to meet the
required UA value can be carried out as follow:
Figure 9: Graphical interpolation method to obtain required Di
y− y0
x−x0
=y1− y0
x1−x0
y−0.0296.04−119.56
= 0.025−0.0294.92−119.56
∴ y=0.0248 m=Di required to satisfy the original UA value.
Study 5 – Check obtained result
Where, Di = 0.0248 m and L = 10 m
From, A=π Di ln
Nmin=Amin
π × Di× L= 56.5
π× 0.0248 ×10=72.52 ≈ 73 Tubes
--------------------------------------------------------------
1. mhi=mh
N=28
73=0.386 kg/ s
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2. R eDi=4 × mhi
π× Di× μh
= 4 ×0.386π ×(0.0248)×(362.8 × 10−6)
=54623.367
3. NUD=h i× D i
k h
=0.027 × R eDi
45 × Pr ,h
13…………… Cooling
NUD=h i× D i
k h
=0.027 ×(54623.36745 )×(2.28
13 )
∴219.068=hi ×0.0248
668.3× 10−3
∴hi=5903.353 W /m2 K
4. Aci=Aco →π × Di
2
4=
π ×(Do2−Di
2)4
π ×(0.02482)4
=π ×(Do
2−0.02482)4
∴D 0=0.0351 m
5. mhi ,c=mc
N=27
73=0.372 kg/ s
6. R eDo=4×mhi ,c
π × Do × μc
= 4× 0.372π ×(0.0351)×(577 × 10−6)
=23386.764
7. NUD=ho× Do
kc
=0.024 × R eDo
45 × Pr ,c
25……………Heating
NUD=ho× Do
kc
=0.024 × ¿¿
∴127.62=ho× 0.0351
640 ×10−3
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∴ho=2326.974 W /m2 . K
8.1U
≈1hi
+ 1ho
1U
= 15903.353
+ 12326.974
1U
=5.991× 10−4
U=1669.065W /m2 . K
9. ∴UA=1669.065 ×56.5=94.036 ×103 W / K → satisfied the required UA value.
Discussion of Results
Firstly, all the calculations were conducted to obtain the value of UA (94.04 × 10 3 W/K). These
calculations include the determination of the Average Mean Temperature (AMT) for both fluids, CP, h
value, Heat Capacity (C) for both fluids, Heat Transfer Effectiveness(ε), Heat Capacity Ratio (Cr),
Number of Transfer Units (NTU → counter flow arrangement), the Maximum Heat Transfer Rate
(qmax) and the Heat Transfer Rate (q). The Heat Capacity Ratio (Cr) obtained was lesser than 1, this
state that counter flow arrangement must be apply. From here, the ranges for the area (A) can be
determined. With the ranges of Area (A), the length (L) and the number of tubes (N) can then be
determined according to the ranges of diameter assumed (0.025m, 0.045m and 0.075m). Hence, the
viscosity ( μh), Prandtl number (Pr) and Thermal Conductivity (k) can be determine from the table of
thermal-physical properties for saturated water. When all the properties are obtained, the studies for
all the selected diameters can then be compute. The studies for each case are divided into 8 steps to
obtain the UA value for each case as shown previously in the calculation section. From the 3 assumed
diameters, the obtained UA value did not satisfy the required UA value. As mentioned earlier, from
observation, the results of the UA values increases as the diameter decrease, therefore
assumption for the suitable diameter is then made. From here, a new study is then conducted
with the diameter value of 0.02m. The UA obtained is119.56× 103W / K , then interpolation
between Di = 0.025 m and Di = 0.02 m was conducted in order to gain the required diameter
to meet the required UA value. The required diameter obtained to satisfy the original UA
value is 0.0248m. Determination for the length (L) and number of tubes (N) required for this
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diameter is then conducted. With the use of 0.248m diameter tube, the minimum requirement
for the number of tubes is 73. Therefore, the specification for the heat exchanger design that
meets the requirement is Di =0.0248 m, Do = 0.0351 m, L = 10 m and N = 73 tubes.
Conclusion
Double Pipe Heat Exchangers are a form of Tubular Heat Exchanger. A typical double pipe heat
exchanger has two concentric pipes. A number of double pipe heat exchangers can be connected in
series or parallel as necessary. The required UA value was obtain through all the necessary
calculations. Several trials were conducted using different diameter sizes. The UA values for each
diameter were then obtained. The calculation shows that with the physical properties of Di =0.0248
m, Do = 0.0351 m, L = 10 m and N = 73 tubes, the design will be able to achieve the
necessary UA value that satisfy with the customer requirement.
Reference
Bejan, A. and Kraus, A. (2003) Heat Transfer Handbook. Toronto: John Wiley & Sons,
p.797-822.
Foumeny, E. and Heggs, P. (1991) Heat Exchange Engineering. Chichester: Ellis Horwood
Limited, p.58,67-77.
Fraas, A. (1989) Heat Exchanger design. 2nd ed. Toronto: John Wiley & Sons, Inc., p.2, 22,
54.
Incropera, F. and Dewitt, D. (2002) Introduction to Heat Transfer. 4th ed. New York: John
Wiley & Sons, p.606-643
Ozisik, M. (1985) Heat Transfer - A basic Approach. Singapore: McGraw-Hill, p.524-571.
Kuppan, T. (2000) Heat Exchanger Design Handbook. 2nd ed. New York: Marcel Dekker,
Inc, p.2-4.
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Appendix
Table A1: Thermophysical Properties of Saturated Water
Source: Incropera, F. and Dewitt, D. (2002) Introduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, p.844.
Table A2: Representative value of the overall Heat transfer coefficient
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Module Code: NG3H237 [ ]
Source: Incropera, F. and Dewitt, D. (2002) Introduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, p.586
Daniel James Watkins 09032266 Page 25