Bachelor Research Project Report

1. INTRODUCTION In the scope of the Research Project, which is a 5th year course of Chemical Engineering, it was suggested to research and study the project of software of Heat Exchangers Design in the base of Visual Basic Application (VBA). Thus, this study although having an academic origin, its operation have designed to be mostly possible way of situations. The process of heat exchange between two fluids that are at different temperatures and separated by a solid wall occurs in many engineering applications. The device that is being used for this heat exchange is termed as heat exchanger. They are typically classified according to their flow arrangement and type of construction. The simplest heat exchanger is one for which the hot and cold fluids move in the same or opposite directions in a concentric tube construction.

Figure 1 - Operating principle of a parallel-flow concentric tube heat exchanger. As you see in the parallel flow (co-current) arrangement, the hot and cold fluids enter at the same end, flow in the same direction, and leave at the same end. In contrast, in the counter flow arrangement, the fluids enter at the opposite ends, flow in opposite directions, and leave at opposite ends. Another common configuration is the shell and tube heat exchanger. It can have some specific forms according to the number of shell and tube passes and the simplest form, which involves single tube and shell passes.

1

mshell

mtube

mpentane

mwater

Figure 2- Shell and tube heat exchanger with one shell pass and one tube pass (Crossparallel-flow mode of operation). In my research project there will be two different types of heat exchangers, which are concentric tube, and shell & tube heat exchangers. Both types were studied according to different flow arrangements in different problems. The problems will be given and the unknowns will be solved in VBA program in excel sheet. Until here, there was not any phase change but, it is also considered due to condensation by introducing in another problem. Below, different cases will be studied in different heat exchangers with the consequent results. At the end of the report, you can see the codes of the problems that were performed in Visual basic application under the title of “Modules.”

2

2. CONCENTRIC TUBE HEAT EXCHANGERS In this case, we have a concentric tube heat exchanger, which contains an inner and an annulus part that enables the fluid to flow in the surrounding of the exchanger. It can also be considered as 1-1 type shell and tube heat exchanger. Actually, the types with more tubes were considerable but they are mechanically and dynamically less effective comparing to the shell and tube type. Concentric type has also a shape of hollow cylinder from the cross sectional view.

2.1 Case 1: Concentric Tube Heat Exchanger with constant Heat Transfer Coefficients and without Phase Change water Di

oil

Do

Equations that are used in the concentric tube heat exchanger calculation are given as follows: -Heat rate equation;

Q  m  Cp  T

[1]

-LMTD calculation; Tln 

Tln 

(Ttube,inlet  Tannulus,outlet )  (Ttube ,outlet  Tannulus,inlet )  For counter-current flow type [2] (Ttube ,inlet  Tannulus,outlet ) ln (Ttube ,outlet  Tannulus,inlet ) (Ttube,inlet  Tannulus,inlet )  (Ttube ,outlet  Tannulus,outlet )  For co-current flow type (Ttube ,inlet  Tannulus,inlet ) ln (Ttube ,outlet  Tannulus,outlet )

[3]

-Overall heat transfer coefficient; U  (h 1tubein  h 1tubean ) 1

[4]

-Design equation; Q  U  Aexchanger  Tln

[5]

Aexchanger    Dout  Ltube

[6]

-NTU method, which is very useful in case of the two outlet temperatures are unknown;

3

NTU 

U  Aexchanger (m  Cpmin imum )

 For both flow type

[7]

(m  Cp) min imum indicates the minimum value of one of the fluid.

[8]

-Thermal efficiency (ε);





1  exp   NTU (1  C   1  C  exp   NTU (1  C  )  1  exp   NTU (1  C   1  C

C 

 Counter-current flow type

 Co-current flow type

Cmin Cmax

[10] [11]

Cmin  (m  Cp )min imum

[12]

Cmax  (m  Cp) max imum



[9]

Ttube ,inlet  Ttube, outlet Ttube,inlet  Tannulus ,inlet

[13]

Here you see an illustrative example, which is governed by the given equations above. Example 2.1 In a concentric tube heat exchanger, which is isolated thermally from the exterior, has a duty to cool hot oil by using cooling water. The hot oil enters the system into the fine metallic tube of 25mm diameter at a mass flow rate of 0.6 kg/s. The cooling water enters the exchanger counter-currently into the annulus section the external diameter of 42mm at the same mass flow rate. The inlet and outlet temperatures of hot oil are 420 K and 380 K, respectively. The convective heat transfer coefficients of the oil and water are 847.61 and 3367.22 W/m2 K, and the specific heats are 2508 and 4180J/kgK, respectively. a) Determine the length that the tube should have if the water enters at 290 K? b) Using the length of the tube (calculated in a) and admitting the water inlet temperature as 290 K, determine the outlet temperatures of both fluid if the system operates in cocurrent? Solution: Known: Fluid flow rates, inlet temperatures, and convective heat transfer coefficients for a counter flow, concentric tube heat exchangers of prescribed inner and outer diameter.

Find: a) Tube length to achieve a desired cold fluid outlet temperature.

4

b) Both outlet temperatures of the fluids using the length calculated in a. Ttube, inlet

T (x)

Ttube, outlet

Tannulus, outlet

Tannulus, inlet

x

Counter-current temperature profile

Assumptions: 1) Heat loss to the surroundings is negligible. 2) Thermodynamic properties are constant. 3) Tube wall thermal resistance and fouling factors are negligible. Part a): Counter-current flow First of all heat released from oil is calculated from the heat rate equation: Q  m  Cpoil  Toil Secondly, outlet water temperature was calculated via heat rate equation of cooling water by equating to oil heat rate ( Qoil  Qwater ): Q Twater ,out   Twater ,in m  Cpwater Then, Tln is calculated: (Ttube,inlet  Tannulus,outlet )  (Ttube ,outlet  Tannulus,inlet ) Tln  (T  Tannulus,outlet ) ln tube ,inlet (Ttube ,outlet  Tannulus,inlet ) U  (h 1tubein  h 1tubean ) 1 Q  U  Aexchanger  Tln

Aexchanger 

Q U  Tln

Aexchanger    Dout  Ltube Ltube 

Aexchanger

  Dout

5

Co-current flow Ttube, inlet

T (x)

Ttube, outlet Tannulus, outlet Tannulus, inlet x

Co-current temperature profile

Q  m  Cpoil  Toil (Ttube,inlet  Tannulus,inlet )  (Ttube ,outlet  Tannulus,outlet ) Tln  (T  Tannulus,inlet ) ln tube ,inlet (Ttube ,outlet  Tannulus,outlet ) U  (h 1tubein  h 1tubean ) 1

Aexchanger 

Q U  Tln

Aexchanger    Dout  Ltube Ltube 

Aexchanger

  Dout Results: Concentric Tubes Heat Exchanger

Concentric Tubes Heat Exchanger

Type

Type

countercurrent

cocurrent

Variables

inner (oil)

annular(water)

Variables

inner (oil)

annular(water)

m [kg/s]

0,6

0,6

m [kg/s]

0,6

0,6

Cp [J/(kg K)]

2508

4180

Cp [J/(kg K)]

2508

4180

Tin [K]

420

290

Tin [K]

420

290

Tout [K]

380

314

Tout [K]

380

314

Diameter [m]

0,025

0,042

Diameter [m]

0,025

0,042

h [J/(K m s)]

847,61

3367,22

U [J/(K m2 s)]

677,15

2

h [J/(K m s)]

847,61

U [J/(K m2 s)]

677,15

2

2

3367,22

2

A [m ]

0,91

A [m ]

0.94

Ltube [m]

6.9

Ltube [m]

7.14

Duty [J/s]

60192

Duty [J/s]

60192

6

Part b): Counter-current flow Since both outlet temperatures are unknown in this part NTU method will be applied as follows: U  Aexchanger NTU  (m  Cpmin imum ) Aexchanger    Dout  L (m  Cpmin imum )oil (m  Cpmin imum )water

C 





 Yields the minimum value

Cmin Cmax 1  exp   NTU (1  C  

1  C  exp   NTU (1  C  ) 

 Counter-current flow expression

Ttube ,inlet  Ttube, outlet Ttube,inlet  Tannulus ,inlet

Q  m  Cpoil  Toil Co-current flow: NTU 

C 





U  Aexchanger (m  Cpmin imum )

Cmin Cmax

1  exp   NTU (1  C   1  C

 Co-current flow expression

Ttube ,inlet  Ttube, outlet Ttube,inlet  Tannulus ,inlet

Q  m  Cpoil  Toil

7

Results: Concentric Tubes Heat Exchanger

Concentric Tubes Heat Exchanger

Type

Type

countercurrent

cocurrent

Variables

inner (oil)

annular (water)

Variables

inner (oil)

annular (water)

m [kg/s]

0,6

0,6

m [kg/s]

0,6

0,6

Cp [J/(kg K)]

2508

4180

Cp [J/(kg K)]

2508

4180

Tin [K]

420

290

Tin [K]

420

290

Tout [K]

379,95

314

Tout [K]

380

314

Diameter [m]

0,025

0,042

Diameter [m]

2

h [J/(K m s)]

847,61

2

U [J/(K m s)]

677,15

2

3367,22

0,025

0,042

2

847,61

3367,22

2

677,15

h [J/(K m s)] U [J/(K m s)] 2

A [m ]

0,91

A [m ]

0,94

Ltube [m]

6,90

Ltube [m]

7,14

Duty [J/s]

60260,3

Duty [J/s]

60218,5

Ntu

0,41

Ntu

0,42

Eps

0,31

Eps

0,31

C

x

0,6

C

x

0,6

2.2 Case 2: Concentric Tube Heat Exchanger with Phase Change In this case, we have steam condensation flows through the annulus part of the exchanger. Schematically representation of the exchanger is given below. Steam

Di

Water

Do

The equations that are going to be used in the calculation are as follows: For tube section Laminar Regime -Reynolds Number:

Re water 

4  m water  2000   Di   water

[14]

-Nusselt Number: Sieder and Tate Correlation:

8

Nu  1.86  Re water

1/ 3

 Prwater

1/ 3

 D   i  Ltube

  

1/ 3

    s

  

0.14

[15]

Transition Regime -Reynolds Number:

2000  Re water 

4  m water  10000   Di   water

[16]

-Nusselt Number:

Nu  0.116  (Re water

2/3

 125)  Prwater

1/ 3

  D  1   i   Ltube

  

2/3

         s

  

0.14

[17]

Turbulent Regime -Reynolds Number:

Re water 

4  m water  10000   Di   water

[18]

-Nusselt Number: The Dittus-Boelter Correlation: 0.8

Nu  0.023  Re water  Prwater

n

[19]

Where n=0.4 for heating n=0.3 for cooling - Heat transfer coefficient hi 

Nu  k fluid

[20]

Di

For annular section Laminar Regime -Reynolds Number

Re steam 

2  msteam  1800  steam  Ltube

[21]

9

-Heat Transfer Coefficient

 g   L  (  L   v )  k L3    ho  0 .729      L  (Tsat  Tsurface )  D o 

1/ 4

[22]

-Surface Temperature

Tsurface

T T  Tsat   tube,inlet tube ,outlet  2    2

[23]

Turbulent Regime -Reynolds Number

Re steam 

2  msteam  1800  steam  Ltube

[24]

-Heat Transfer Coefficient 1/ 3

 g   L2  k L3  ho  0.023     L2  

 2  msteam      L  Ltube 

1/ 4

1/ 2

 C  L    p ,L   kL 

[25]

These equations are illustrated with an example problem below.

Example 2.2 A thin walled concentric tube heat exchanger of 15m length is to be used to heat deionised water, which enters to the system at 40oC at a flow rate of 1.5 kg/s, flows through the inner tube of 30mm diameter. In contrast saturated steam is supplied at 1atm (100oC) to the annulus section formed with the outer tube of 60mm diameter. Determine the outlet temperature of deionised water (Twater,out) and mass flow rate of condensation (mcondensation)? Thermo physical properties of water @ 50 oC : Density, [kg/m3]

982.3

Heat capacity, [J/kgK]

4181

Viscosity, [kg/ms]

548x10-6

Prandtl number, [-]

3.56

Thermal conductivity, [W/mK]

0.643

10

Thermo physical properties of saturated steam @ 100 oC and 1 atm: Liquid density, [kg/m3]

957.8

Vapour density, [kg/m3]

0.6

Heat capacity, [J/kgK]

4217

Heat of condensation, [J/kg]

2257000

Viscosity, [kg/ms]

280x10-6

Prandtl number, [-]

1.76

Thermal conductivity, [W/mK]

0.68

Solution: Known: Water flow rate, inlet temperatures, thermo physical properties of both fluid, and dimensional properties of the system. Find: Outlet temperature of deionised water, mass flow rate of condensed steam, heat transfer coefficients for inner and annulus section, and duties by heat rate and by design equation.

In the system, two heat duties (Q1 and Q2) are defined to be determined from different ways. By doing that, we will be able to check the results in the iterative procedure. Q1  mwater  Cp  Twater Twater,outlet = ? Assumption: Twater,outlet=Ttube,outlet=360 K Then Q1 is obtained. Since; Q1 = Q2 msteam 

Q2 is determined. 

Then, to satisfy the assumed temperature is right, the difference between Q1 and Q2 is expected to be zero, Q  Q1  Q2

Then other parameters are obtained as follows: U  (hi 1  ho 1 )1

11

For hi:

Re water 

4  m water  2000  Laminar Regime   Di   water

Sieder and Tate Correlation:

Nu  1.86  Re water

2000  Re water 

1/ 3

 D   i  Ltube

  

1/ 3

    s

2/3

 125)  Prwater

1/ 3

0.8

n

Where n=0.4 for heating n=0.3 for cooling Then;

Nu kwater Di

For ho:

2  msteam  1800  Laminar Regime  steam  Ltube

 g   L  (  L   v )  k L3    ho  0 .729      L  (Tsat  Tsurface )  D o 

Tsurface

  

2/3

4  m water  10000  Turbulent Regime   Di   water

Nu  0.023  Re water  Prwater

Re steam 

0.14

  D  1   i   Ltube

The Dittus-Boelter Correlation:

hi 

  

4  m water  10000  Transition Regime   Di   water

Nu  0.116  (Re water

Re water 

 Prwater

1/ 3

1/ 4

T T  Tsat   tube,inlet tube ,outlet  2    2

12

         s

  

0.14

Re steam 

2  msteam  1800  Turbulent Regime  steam  Ltube 1/ 3

 g   L2  k L3  ho  0.023     L2  

 2  msteam      L  Ltube 

1/ 4

1/ 2

 C  L    p ,L   kL 

And finally; U  (hi 1  ho 1 )1 is obtained.

Tln 

(Tannulus,inlet  Ttube ,outlet )  (Tannulus, outlet  Ttube ,inlet ) (T  Ttube ,outlet ) ln annulus,inlet (Tannulus,outlet  Ttube ,inlet )

Aexchanger    Do  Ltube

U

Q Aexchanger  Tln

13

Results: Concentric Tubes Heat Exchanger Type

countercurrent

Variables

inner (water)

annular (steam)

m [kg/s]

1,5

0,13

 [J/kg]

2257000

Cp [J/(kg K)] 3

4180

4217

ρ [kg/m ]

982,3

957,8

μ[kg/m.s]

0,00055

0,00028

prandtl [-]

3,56

1,76

reynolds [-]

77447,66

62,13

nusselt [-]

311,54

kfluid [W/m.k]

0,643

0,68

d [m]

0,045

0,09

h [W/m .K]

4451,57

7914,04

Tin [K]

313,00

373,00

Tout [K]

359,80

373,00

dhydraulic [m]

0,045

ktube [W/m.k]

80,2

ltube [m]

15

2

2

areaexchanger[m ]

4,24

ΔTln [K]

30,91 2

uexchanger [W/m .K]

2849,02

uexchanger [W/m2.K]

2237,98

dutyexchanger [W]

293410,00

dutyexchanger [W]

293410,00

Tsat [K]

373

Tsurface [K]

354,70 2

gravity [m/s ]

9,81

Difference

0,00

14

0,6

3. SHELL AND TUBE HEAT EXCHANGERS Shell and tube heat exchangers are commonly used as oil coolers, power condensers, preheaters and steam generators in both fossil fuel and nuclear-based energy production applications. They are also widely used in process applications and in the air conditioning and refrigeration industry. Although they are not specially compact, their robustness, and shape make them well suited for high pressure operations. In view of the diverse engineering applications and basic configuration of shell and tube heat exchangers, the thermal analysis and design of such exchangers form an integral part of the undergraduate mechanical and chemical engineering curricula of most universities.

3.1 Case: Shell and Tube Heat Exchanger without Phase Change mshell

mtube

mshell

mtube

In this case, shell and tube heat exchanger rate will be studied in case of a deposit formation inside of the tubes due to the flowing fluid without any phase change. Besides, fouling factor (Rfi) is also going to be studied. Here are the equations used in this application; -Heat rate:

Q  m   and Q  m  Cp  T

[25]

-Design Equation: [26]

Q  U  Aexchanger  F  Tln

-Overall heat exchange coefficient:

15

U  ( Rtotal  Aexchanger ) 1

[27]

-Total resistance inside the tubes:

Rtotal 

1 x 1   hi  Ai ktube  Aln ho  Ao

[28]

-Fouling factor:

  1 x 1 R fi   Rtotal      Aint erior hwater  Aint erior ktube  Aln hpen tan e  Aexterior  

[29]

-Logarithmic area: Aln 

Aexterior  Aint erior A ln exterior Aint erior

[30]

-Internal and external area: Aint erior    Din  Ltube  N tube

[31]

Aexterior    Dout  Ltube  N tube

[32]

Below, you can see an illustrative example related to the equations above. Example 3.1 A shell and tube heat exchanger, which possesses 140 steel tubes of 2m length, 19mm of external and 16mm of internal diameter, isolated thermally from the exterior. The thermal conductivity of the steel tube is 53,4W/mK. It is intended to use to condense 2.8 kg/s of saturated pentane at 2atm in the shell side. The saturation temperature of the pentane is 60oC and the latent heat of condensation is 313500 J/kg. In contrast, cooling water, enters at 15oC , circulates inside the tubes. The convective heat transfer coefficients are given by the following expressions: For water hwater =290 x (mwater )0.8 For pentane hpentane=1861.7 x (mpentane) 0.25 mwater and mpentane are the mass flow rates of water and pentane in kg/s; hwater and hpentane are in W/m2 K. a) Determine the flow rate of water necessary to condense 2.8 kg/s of pentane? b) Due to the formation of deposits inside the tubes, the flow rate of pentane was decreased to 1.9 kg/s. Assuming that the flow rate of water is 20.83 kg/s, determine the thermal resistance of the formed deposits?

16

Solution: Known: Flow rate of shell side fluid, inlet temperatures, and convective heat transfer coefficients as functions of mass flow rates, dimensions of shell and tube heat exchanger. Find: Required water mass flow rate to condense pentane. Thermal resistance of the formed deposits. Schematic: mpentane

mwater

mpentane

mwater

Part a): Heat released by pentane: Q  m pen tan e  

Trial-error procedure is required due to not to know the outlet temperatures of water and mass flow rate. Assuming Twater,out=300 K

mwater 

Q Cpwater  T

hwater  290  (17.5)0.8  2864.27 watt / m2 K

The procedure will continue until the difference between heat of condensation and design equation becomes zero. Otherwise new assumption is required. Q  U  Aexchanger  Tln U  ( Rtotal  Aexchanger ) 1

Rtotal 

1 x 1   hwater  Aint erior ktube  Aln hpen tan e  Aexterior

17

Aexchange r  Aexterior

Aint erior    Din  Ltube  N tube Aexterior    Dout  Ltube  N tube Aln 

Aexterior  Aint erior A ln exterior Aint erior

Difference  Q  Qdesign  Qcondensation Results: Shell & Tube Heat Exchanger Type Variables

1-shell pass

m [kg/s]

tube side (water) 17,50

shell side (pentane) 2,80

Cp [J/(kg K)], [J/kg]

4180

313500

Tin [K]

288

333

Tout [K]

300,00

333

h [J/(K m2 s)]

2864,27

2408

U [J/(K m2 s)]

1122,39

Din [m]

0,016

Dout [m]

0,019

Atubein [m2]

17,02

Atubeout [m2]

20,21

Aln [m2]

18,57

Ntube [-]

140

Ltube [m]

2,42

x [m]

0,003

ktube [J/s m K]

53,41

Tln [K]

38,69

Rtotal [K s/J]

0,0000441

F [-]

1

Duty [J/s]

877800,00

Design Duty [J/s]

877800,00

Difference

0

Part b): Q  m pen tan e  

Ttube,outlet 

Q  Ttube ,inlet mwater  Cp

18

Tln 

(Ttube,inlet  Tshell ,outlet )  (Ttube ,outlet  Tshell ,inlet ) (T  Tshell ,outlet ) ln tube ,inlet (Ttube ,outlet  Tshell ,inlet )

Aexchanger    Dout  L  N tube

U

Q Aexchanger  Tln

Rtotal  (U  Aexchanger ) 1

  1 x 1 R fi   Rtotal      Aint erior hwater  Aint erior ktube  Aln hpen tan e  Aexterior  

Results: Shell & Tube Heat Exchanger Type

1-shell pass

Variables

tube side(water)

shell side(pentane)

m [kg/s]

20,83

1,90

Cp [J/(kg K)],  [J/kg]

4180

313500

Tin [K]

288

333

Tout [K]

294,84

333

h [J/(K m s)]

3292,57

2186

U [J/(K m2 s)]

859,08

Din [m]

0,016

Dout [m]

0,019

2

Atubein [m2]

14,07

2

16,71

Atubeout [m ] 2

Aln [m ]

15,36

Ntube [-]

140

Ltube [m]

2

x [m]

0,003

ktube [J/s m K]

53,41

Tln [K]

41,49

Rtotal [K/W]

0,00007

2

Rfi [m K/W]

0,0004

F [-]

1

Duty [J/s]

595650,00

Duty [J/s]

595650,00

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4. NOMENCLATURE Aexchanger: Exchanger Area

[m2]

Ainterior: Internal area

[m2]

Aexterior: External area

[m2]

Aln: Logarithmic area

[m2]

Q: Heat transfer rate

[W]

U: Overall heat transfer coefficient

[W/m2.K]

hi: Inside heat transfer coefficient

[W/m2.K]

ho: Outside/exterior heat transfer coefficient

[W/m2.K]

Re: Reynolds number

[-]

Resteam: Reynolds number for the condensed steam

[-]

Rewater: Reynolds number for the water

[-]

Nu: Nusselt number

[-]

Pr: Prandtl number

[-]

PrL: Prandtl number for the liquid

[-]

Prwater: Prandtl number for the water

[-]

Di: Internal diameter

[m]

Do: External diameter

[m]

Dh: Hydraulic diameter

[m]

V: Velocity of the fluid

[m/s]

g: Gravitational acceleration

[m/s2]

ktube: Thermal conductivity of the tube

[W/m.K]

msteam: Mass flow rate of condensate

[kg/s]

 L : Viscosity of condensate

[kg/m.s]

 s : Viscosity at the surface temperature

[kg/m.s]

L: Length of the tube

[m]

mwater: Mass flow rate of water

[kg/s]

moil: Mass flow rate of oil

[kg/s]

Cp,L: heat capacity of condensate evaluated at film temperature

[J/kg.K]

kL: Thermal conductivity of the condensate

[W/m.K]

 : Latent heat of condensation

[J/kg]

 L : Density of the condensate

[kg/m3]

 v : Density of the vapour

[kg/m3] 20

Vv: Velocity of the vapour in the tube

[m/s]

 v : Viscosity of the vapour in the tube

[kg/m.s]

Tsat: Saturation temperature of the vapour

[K]

Tsurface: Surface temperature of the tube

[K]

Tm: Mean temperature

[K]

Ts,o: Outlet surface temperature of the tube

[K]

N: Number of the tubes

[-]

Rtotal: Total resistance

[K/W]

Rfi: Fouling factor

[m2K/W]

NTU: Number of transfer units

[-]

ε : Thermal Efficiency

[-]

21

5. REFERENCES [1] P., Incropera, Frank, P., De Witt, David, Introduction to heat transfer, 4th edition, John Wiley & Sons. [2] The heat transfer notes from Prof. Luis Melo. [3] Heat Exchangers, Selection Rating & Design, Sadik Kakac & Hongtan Liu, CRC Press, 2nd Edition, 2002 [4] Shell & Tube Heat Exchanger Design Software for Educational Applications. Int. J. Engng. Ed. Vol. 14, No. 3, p 217-224, 1998 K.C. Leong, K.C. Toh, Y.C. Leong [5] Wolverine Tube Heat Transfer Data Book, www.wolverine.com.

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6. MODULES Here are the modules of the illustrated problems, which were programmed in VBA. Example 2.1: Have two parts; Part_A and Part_B and meanwhile both parts have two different stream types; co-current and counter-current flow. Part_A: Sub main_Part_A_cocurrent() Declaration of variables Dim name_sheet As String Dim m_tubein As Double Dim m_tubean As Double Dim cp_tubein As Double Dim cp_tubean As Double Dim h_tubein As Double Dim h_tubean As Double Dim d_tubein As Double Dim d_tubean As Double Dim tin_tubein As Double Dim tout_tubein As Double Dim tin_tubean As Double Dim tout_tubean As Double Dim u_exchanger As Double Dim l_exchanger As Double Dim duty_exchanger As Double Dim delt_ln As Double Dim area_exchanger As Double Dim l_tube As Double Inputs name_sheet = "Part_A" m_tubein = Sheets(name_sheet).Cells(4, 3).Value m_tubean = Sheets(name_sheet).Cells(4, 4).Value cp_tubein = Sheets(name_sheet).Cells(5, 3).Value cp_tubean = Sheets(name_sheet).Cells(5, 4).Value tin_tubein = Sheets(name_sheet).Cells(6, 3).Value tout_tubein = Sheets(name_sheet).Cells(7, 3).Value tin_tubean = Sheets(name_sheet).Cells(6, 4).Value d_tubein = Sheets(name_sheet).Cells(8, 3).Value d_tubean = Sheets(name_sheet).Cells(8, 4).Value h_tubein = Sheets(name_sheet).Cells(9, 3).Value h_tubean = Sheets(name_sheet).Cells(9, 4).Value If IsEmpty(Sheets(name_sheet).Cells(6, 3).Value) = True Then tin_tubean = Sheets(name_sheet).Cells(6, 4).Value tout_tubein = Sheets(name_sheet).Cells(7, 3).Value 23

tout_tubean = Sheets(name_sheet).Cells(7, 4).Value duty_exchanger = m_tubean * cp_tubean * (tout_tubean - tin_tubean) tin_tubein = tout_tubein + duty_exchanger / (m_tubein * cp_tubein) Sheets(name_sheet).Cells(6, 3).Value = tin_tubein ElseIf IsEmpty(Sheets(name_sheet).Cells(6, 4)) = True Then tin_tubein = Sheets(name_sheet).Cells(6, 3).Value tout_tubein = Sheets(name_sheet).Cells(7, 3).Value tout_tubean = Sheets(name_sheet).Cells(7, 4).Value duty_exchanger = m_tubein * cp_tubein * (tin_tubein - tout_tubein) tin_tubean = tout_tubean - duty_exchanger / (m_tubean * cp_tubean) Sheets(name_sheet).Cells(6, 4) = tin_tubean ElseIf IsEmpty(Sheets(name_sheet).Cells(7, 4)) = True Then tin_tubein = Sheets(name_sheet).Cells(6, 3).Value tout_tubein = Sheets(name_sheet).Cells(7, 3).Value tin_tubean = Sheets(name_sheet).Cells(6, 4).Value duty_exchanger = m_tubein * cp_tubein * (tin_tubein - tout_tubein) tout_tubean = tin_tubean + duty_exchanger / (m_tubean * cp_tubean) Sheets(name_sheet).Cells(7, 4) = tout_tubean Else tin_tubein = Sheets(name_sheet).Cells(6, 3).Value tin_tubean = Sheets(name_sheet).Cells(6, 4).Value tout_tubean = Sheets(name_sheet).Cells(7, 4).Value duty_exchanger = m_tubean * cp_tubean * (tout_tubean - tin_tubean) tout_tubein = tin_tubein - duty_exchanger / (m_tubein * cp_tubein) Sheets(name_sheet).Cells(7, 3).Value = tout_tubein End If

Calculations u_exchanger = 1 / ((1 / h_tubein) + (1 / h_tubean)) delt_ln = ((tin_tubein - tin_tubean) - (tout_tubein - tout_tubean) / Log((tin_tubein tin_tubean) / (tout_tubein - tout_tubean))) area_exchanger = duty_exchanger / (u_exchanger * delt_ln) l_tube = area_exchanger / (d_tubein * Application.WorksheetFunction.Pi())

Print Results Sheets(name_sheet).Cells(7, 4).Value = tout_tubean Sheets(name_sheet).Cells(7, 4).Font.ColorIndex = 3

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Sheets(name_sheet).Cells(7, 4).Font.Bold = True

Sheets(name_sheet).Cells(10, 3).Value = Round(u_exchanger, 2) Sheets(name_sheet).Cells(10, 3).Font.ColorIndex = 3 Sheets(name_sheet).Cells(10, 3).Font.Bold = True

Sheets(name_sheet).Cells(11, 3).Value = Round(area_exchanger, 2) Sheets(name_sheet).Cells(11, 3).Font.ColorIndex = 3 Sheets(name_sheet).Cells(11, 3).Font.Bold = True

Sheets(name_sheet).Cells(12, 3).Value = Round(l_tube, 2) Sheets(name_sheet).Cells(12, 3).Font.ColorIndex = 3 Sheets(name_sheet).Cells(12, 3).Font.Bold = True

Sheets(name_sheet).Cells(13, 3).Value = duty_exchanger Sheets(name_sheet).Cells(13, 3).Font.ColorIndex = 3 Sheets(name_sheet).Cells(13, 3).Font.Bold = True

End Sub

Sub main_Part_A_countercurrent()

Declaration of variables Dim name_sheet As String Dim m_tubein As Double Dim m_tubean As Double Dim cp_tubein As Double Dim cp_tubean As Double Dim h_tubein As Double Dim h_tubean As Double Dim d_tubein As Double Dim d_tubean As Double Dim tin_tubein As Double Dim tout_tubein As Double Dim tin_tubean As Double Dim tout_tubean As Double

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Dim u_exchanger As Double Dim l_exchanger As Double Dim duty_exchanger As Double Dim delt_ln As Double Dim area_exchanger As Double Dim l_tube As Double Inputs name_sheet = "Part_A" m_tubein = Sheets(name_sheet).Cells(4, 3).Value m_tubean = Sheets(name_sheet).Cells(4, 4).Value cp_tubein = Sheets(name_sheet).Cells(5, 3).Value cp_tubean = Sheets(name_sheet).Cells(5, 4).Value tin_tubein = Sheets(name_sheet).Cells(6, 3).Value tout_tubein = Sheets(name_sheet).Cells(7, 3).Value tin_tubean = Sheets(name_sheet).Cells(6, 4).Value d_tubein = Sheets(name_sheet).Cells(8, 3).Value d_tubean = Sheets(name_sheet).Cells(8, 4).Value h_tubein = Sheets(name_sheet).Cells(9, 3).Value h_tubean = Sheets(name_sheet).Cells(9, 4).Value If IsEmpty(Sheets(name_sheet).Cells(6, 3).Value) = True Then tin_tubean = Sheets(name_sheet).Cells(6, 4).Value tout_tubein = Sheets(name_sheet).Cells(7, 3).Value tout_tubean = Sheets(name_sheet).Cells(7, 4).Value duty_exchanger = m_tubean * cp_tubean * (tout_tubean - tin_tubean) tin_tubein = tout_tubein + duty_exchanger / (m_tubein * cp_tubein) Sheets(name_sheet).Cells(6, 3).Value = tin_tubein

ElseIf IsEmpty(Sheets(name_sheet).Cells(7, 3).Value) = True Then tout_tubean = Sheets(name_sheet).Cells(7, 4).Value tin_tubein = Sheets(name_sheet).Cells(6, 3).Value tin_tubean = Sheets(name_sheet).Cells(6, 4).Value duty_exchanger = m_tubean * cp_tubean * (tout_tubean - tin_tubean) tout_tubein = tin_tubein - duty_exchanger / (m_tubein * cp_tubein) Sheets(name_sheet).Cells(7, 3).Value = tout_tubein ElseIf IsEmpty(Sheets(name_sheet).Cells(6, 4)) = True Then tin_tubein = Sheets(name_sheet).Cells(6, 3).Value tout_tubean = Sheets(name_sheet).Cells(7, 4).Value tout_tubein = Sheets(name_sheet).Cells(7, 3).Value duty_exchanger = m_tubein * cp_tubein * (tin_tubein - tout_tubein) tin_tubean = tout_tubean - duty_exchanger / (m_tubean * cp_tubean) Sheets(name_sheet).Cells(6, 4).Value = tin_tubean Else

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tin_tubein = Sheets(name_sheet).Cells(6, 3).Value tout_tubein = Sheets(name_sheet).Cells(7, 3).Value tin_tubean = Sheets(name_sheet).Cells(6, 4).Value duty_exchanger = m_tubein * cp_tubein * (tin_tubein - tout_tubein) tout_tubean = tin_tubean + duty_exchanger / (m_tubean * cp_tubean) Sheets(name_sheet).Cells(7, 4).Value = tout_tubean End If Calculations

u_exchanger = 1 / (1 / h_tubein + 1 / h_tubean)

delt_ln = ((tin_tubein - tout_tubean) - (tout_tubein - tin_tubean)) / _ Log((tin_tubein - tout_tubean) / (tout_tubein - tin_tubean))

area_exchanger = duty_exchanger / (u_exchanger * delt_ln) l_tube = area_exchanger / (d_tubein * Application.WorksheetFunction.Pi())

Print results Sheets(name_sheet).Cells(7, 4).Value = tout_tubean Sheets(name_sheet).Cells(7, 4).Font.ColorIndex = 5 Sheets(name_sheet).Cells(7, 4).Font.Bold = True

Sheets(name_sheet).Cells(10, 3).Value = Round(u_exchanger, 2) Sheets(name_sheet).Cells(10, 3).Font.ColorIndex = 5 Sheets(name_sheet).Cells(10, 3).Font.Bold = True

Sheets(name_sheet).Cells(11, 3).Value = Round(area_exchanger, 2) Sheets(name_sheet).Cells(11, 3).Font.ColorIndex = 5 Sheets(name_sheet).Cells(11, 3).Font.Bold = True

Sheets(name_sheet).Cells(12, 3).Value = Round(l_tube, 1) Sheets(name_sheet).Cells(12, 3).Font.ColorIndex = 5 Sheets(name_sheet).Cells(12, 3).Font.Bold = True

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Sheets(name_sheet).Cells(13, 3).Value = duty_exchanger Sheets(name_sheet).Cells(13, 3).Font.ColorIndex = 5 Sheets(name_sheet).Cells(13, 3).Font.Bold = True

End Sub

Example 2.2: In the problem, we have a concentric tube heat exchanger. The objective is to heat the deionised water by condensing the saturated steam. The heat transfer coefficients are going to be determined but before mass flow rate of steam and water outlet temperature are to be found. The “Goal Seek” method is applied for the calculation procedure.

Range("c29").GoalSeek Goal:=0, ChangingCell:=Range("c16")

Example 3.1: Part_A: In the first part of this problem, there is a trial-error procedure. In normal case, problem could be solved by using several numerical methods such as “Newton-Raphson, Bisection” methods etc. But for simplicity, it was solved by using “Goal Seek” method. This method was applied to the problem as follows: Firstly, water outlet temperature is assumed with an arbitrary value between water inlet and saturated pentane inlet temperature. Heat duty of the exchanger has already calculated by the mass flow rate of sat’d pentane and heat of condensation. Then, mass flow rate of water is calculated by the heat rate equation. Finally, the procedure is let to proceed until the difference between the heat duties becomes zero.

Part_A: Sub main_Part_A() Declaration of variables ' Dim name_sheet As String ' Dim m_tube As Double ' Dim m_shell As Double ' Dim cp_tube As Double ' Dim cp_shell As Double ' Dim h_tube As Double

28

' Dim h_shell As Double ' Dim d_tubein As Double ' Dim d_tubeout As Double ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' '

Dim tin_tube As Double Dim tout_tube As Double Dim tin_shell As Double Dim tout_shell As Double Dim u_exchanger As Double Dim R_total As Double Dim area_tubein As Double Dim area_tubeout As Double Dim area_ln As Double Dim del_x As Double Dim l_exchanger As Double Dim duty_exchanger As Double Dim delt_ln As Double Dim area_exchanger As Double Dim l_tube As Double Dim N_tube As Double Dim k_tube As Double Dim f_correction As Double

Inputs ' name_sheet = "Part_A"' 'm_tube = Sheets(name_sheet).Cells(4, 3).Value 'm_shell = Sheets(name_sheet).Cells(4, 4).Value 'cp_tube = Sheets(name_sheet).Cells(5, 3).Value 'cp_shell = Sheets(name_sheet).Cells(5, 4).Value 'tin_tube = Sheets(name_sheet).Cells(6, 3).Value 'tin_shell = Sheets(name_sheet).Cells(6, 4).Value 'tout_shell = Sheets(name_sheet).Cells(7, 4).Value 'h_tube = Sheets(name_sheet).Cells(8, 3).Value 'h_shell = Sheets(name_sheet).Cells(8, 4).Value 'u_exchanger = Sheets(name_sheet).Cells(9, 3).Value 'd_tubein = Sheets(name_sheet).Cells(10, 3).Value 'd_tubeout = Sheets(name_sheet).Cells(11, 3).Value 'area_tubein = Sheets(name_sheet).Cells(12, 3).Value 'area_tubeout = Sheets(name_sheet).Cells(13, 3).Value 'area_ln = Sheets(name_sheet).Cells(14, 3).Value 'N_tube = Sheets(name_sheet).Cells(15, 3).Value 'l_tube = Sheets(name_sheet).Cells(16, 3).Value 'del_x = Sheets(name_sheet).Cells(17, 3).Value 'k_tube = Sheets(name_sheet).Cells(18, 3).Value 'f_correction = Sheets(name_sheet).Cells(21, 3).Value

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Calculations 'duty_exchanger = m_shell * cp_shell 'area_tubeout = Application.WorksheetFunction.Pi() * d_tubeout * l_tube * N_tube 'area_tubein = Application.WorksheetFunction.Pi() * d_tubein * l_tube * N_tube 'area_ln = (area_tubeout - area_tubein) / Log(area_tubeout / area_tubein)

'h_shell = 207 * m_shell ^ 0.25 'm_tube = duty_exchanger / (cp_tube * (tout_tube - tin_tube)) 'h_tube = 0.357 * m_tube ^ 0.8 'u_exchanger = 1 / (R_total * area_tubeout) 'R_total = (1 / (h_tube * area_tubein)) + (del_x / (k_tube * area_ln)) + _ '(1 / (h_shell * area_tubeout)) 'delt_ln = ((tin_shell - tout_tube) - (tout_shell - tin_tube)) / _ ' Log((tin_shell - tout_tube) / (tout_shell - tin_tube)

Print results ' Sheets(name_sheet).Cells(4, 3).Value = m_tube ' Sheets(name_sheet).Cells(4, 3).Font.ColorIndex = 5 ' Sheets(name_sheet).Cells(4, 3).Font.Bold = True 'Sheets(name_sheet).Cells(7, 3).Value = tout_tube 'Sheets(name_sheet).Cells(7, 3).Font.ColorIndex = 5 'Sheets(name_sheet).Cells(7, 3).Font.Bold = True ' Sheets(name_sheet).Cells(8, 3).Value = h_tube ' Sheets(name_sheet).Cells(8, 3).Font.ColorIndex = 5 ' Sheets(name_sheet).Cells(8, 3).Font.Bold = True ' Sheets(name_sheet).Cells(9, 3).Value = u_exchanger ' Sheets(name_sheet).Cells(9, 3).Font.ColorIndex = 5 ' Sheets(name_sheet).Cells(9, 3).Font.Bold = True 'Sheets(name_sheet).Cells(19, 3).Value = delt_ln 'Sheets(name_sheet).Cells(19, 3).Font.ColorIndex = 5 'Sheets(name_sheet).Cells(19, 3).Font.Bold = True ' Sheets(name_sheet).Cells(20, 3).Value = R_total ' Sheets(name_sheet).Cells(20, 3).Font.ColorIndex = 5 'Sheets(name_sheet).Cells(20, 3).Font.Bold = True

30

Range("c24").GoalSeek Goal:=0, ChangingCell:=Range("c16")

End Sub

Part_B: In the second part, it is aimed to obtain the thermal resistance (Rf) inside the tubes of the formed deposits. These deposits occur due to the some contaminants of the fluid.

Sub main_Part_B() Declaration of variables Dim name_sheet As String Dim m_tube As Double Dim m_shell As Double Dim cp_tube As Double Dim cp_shell As Double Dim h_tube As Double Dim h_shell As Double Dim d_tubein As Double Dim d_tubeout As Double Dim tin_tube As Double Dim tout_tube As Double Dim tin_shell As Double Dim tout_shell As Double Dim u_exchanger As Double Dim R_total As Double Dim R_fi As Double Dim area_tubein As Double Dim area_tubeout As Double Dim area_ln As Double Dim del_x As Double Dim l_exchanger As Double Dim duty_exchanger As Double Dim delt_ln As Double Dim area_exchanger As Double

Dim l_tube As Double Dim N_tube As Double Dim k_tube As Double Dim f_correction As Double 31

Inputs name_sheet = "Part_B" m_tube = Sheets(name_sheet).Cells(4, 3).Value m_shell = Sheets(name_sheet).Cells(4, 4).Value cp_tube = Sheets(name_sheet).Cells(5, 3).Value cp_shell = Sheets(name_sheet).Cells(5, 4).Value tin_tube = Sheets(name_sheet).Cells(6, 3).Value tin_shell = Sheets(name_sheet).Cells(6, 4).Value tout_shell = Sheets(name_sheet).Cells(7, 4).Value h_shell = Sheets(name_sheet).Cells(8, 4).Value d_tubein = Sheets(name_sheet).Cells(10, 3).Value d_tubeout = Sheets(name_sheet).Cells(11, 3).Value area_tubein = Sheets(name_sheet).Cells(12, 3).Value area_tubeout = Sheets(name_sheet).Cells(13, 3).Value area_ln = Sheets(name_sheet).Cells(14, 3).Value N_tube = Sheets(name_sheet).Cells(15, 3).Value l_tube = Sheets(name_sheet).Cells(16, 3).Value del_x = Sheets(name_sheet).Cells(17, 3).Value k_tube = Sheets(name_sheet).Cells(18, 3).Value f_correction = Sheets(name_sheet).Cells(22, 3).Value Calculations duty_exchanger = m_shell * cp_shell tout_tube = duty_exchanger / (cp_tube * m_tube) + tin_tube area_tubeout = Application.WorksheetFunction.Pi() * d_tubeout * l_tube * N_tube area_tubein = Application.WorksheetFunction.Pi() * d_tubein * l_tube * N_tube area_ln = (area_tubeout - area_tubein) / Log(area_tubeout / area_tubein) h_shell = 207 * (m_shell * 3600) ^ 0.25 * 4180 / 3600 h_tube = 0.357 * (m_tube * 3600) ^ 0.8 * 4180 / 3600 delt_ln = ((tin_shell - tout_tube) - (tout_shell - tin_tube)) / _ Log((tin_shell - tout_tube) / (tout_shell - tin_tube)) u_exchanger = duty_exchanger / (delt_ln * area_tubeout) R_total = 1 / (u_exchanger * area_tubeout) R_fi = R_total - (1 / (h_tube * area_tubein)) + (del_x / (k_tube * area_ln)) + _ (1 / (h_shell * area_tubeout) * area_tubein)

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Print results Sheets(name_sheet).Cells(7, 3).Value = tout_tube Sheets(name_sheet).Cells(7, 3).Font.ColorIndex = 5 Sheets(name_sheet).Cells(7, 3).Font.Bold = True Sheets(name_sheet).Cells(9, 3).Value = u_exchanger Sheets(name_sheet).Cells(9, 3).Font.ColorIndex = 5 Sheets(name_sheet).Cells(9, 3).Font.Bold = True Sheets(name_sheet).Cells(19, 3).Value = delt_ln Sheets(name_sheet).Cells(19, 3).Font.ColorIndex = 5 Sheets(name_sheet).Cells(19, 3).Font.Bold = True Sheets(name_sheet).Cells(20, 3).Value = Round(R_total, 5) Sheets(name_sheet).Cells(20, 3).Font.ColorIndex = 5 Sheets(name_sheet).Cells(20, 3).Font.Bold = True Sheets(name_sheet).Cells(21, 3).Value = Round(R_fi, 5) Sheets(name_sheet).Cells(21, 3).Font.ColorIndex = 5 Sheets(name_sheet).Cells(21, 3).Font.Bold = True

End Sub

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7. CONCLUSION In this research report, I tried to learn how to programme the basic design applications of heat exchangers based on VBA. During one year period, I was supervised by Ass. Prof. Fernando Martins. With his useful information about programming, I have gained different point of view for an engineer in the programming language area. And, Prof. Luis Melo; with his valuable distributions, we have found our right way in some challenging points. I would like to thank especially to these two people, for their patience and for their interests in any situation. I felt myself very lucky to be studied with them.

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