Double-Pipe Heat Exchanger Design

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


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)



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.



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)



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.



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



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


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



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


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


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



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.



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



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



∴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



1U

=5.95 ×10−4

U=1680.06 W /m2. K

Study 2


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


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




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


NUD=ho× Do

kc

=0.024 × ¿¿

∴127.1=ho ×0.064

640 × 10−3

∴ho=1271 W /m2 . K

8.1U

≈1hi

+ 1ho



1U

= 13251.95

+ 11271

1U

=1.094 ×10−3

U=913.83 W /m2 . K

Study 3


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


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


2

4=

π ×(Do2−Di

2)4

π ×(0.0752)4

=π ×(Do

2−0.0752)4

∴D 0=0.12 m



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


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



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


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


2

4=

π ×(Do2−Di

2)4



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


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



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



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


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


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


NUD=ho× Do

kc

=0.024 × ¿¿

∴127.62=ho× 0.0351

640 ×10−3



∴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



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.



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



Source: Incropera, F. and Dewitt, D. (2002) Introduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, p.586


Documents

Double-Pipe Heat Exchanger Design