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UNIVERSITI TEKNOLOGI MARAFAKULTI KEJURUTERAAN KIMIA

PROCESS ENGINEERING LABORATORY II (CPE554)NAME : NURWANI BINTI HUSSINGROUP : 5AEXPERIMENT : LAB 6 : SHELL AND TUBE HEAT EXCHANGERDATE : 24 OCTOBER 2014PROG/CODE : EH242SUBMIT TO : MOHAMAD SUFIAN BIN SOAIB

No Title Allocated Marks (%) Marks1 Abstract 52 Introduction 5

3 Objectives 54 Theory 5

5 Procedures/Methodology 10

6 Apparatus 57 Results 108 Calculation 10

9 Discussion 2010 Conclusion 10

11 Recommendations 5

12 References 513 Appendices 5

TOTAL 100

Remarks:

Checked by: Rechecked by:

Date: Date


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I. ABSTRACT

Heat exchanger is an equipment used to reduce or heat up between liquid and liquid.

Therefore, one experiment has been conducted to find the most efficient process. from the

result, the counter current perform the most suitable process in transferring heat. Either tocool or heat process. the LMTD value for counter current are higher means has higher

temperature drop and Reynold number decide that counter current having laminar flow

rather that turbulence.

II. INTRODUCTION

Heat exchanger has been widely used in industries in order to control the temperature of

the chemical substance in a processing. Therefore, the design of the heat exchangers is now

has been highly developed. There are many equipment that involve heating and cooling in

chemical processing but heat exchanger only perform liquid-liquid heat transfer. Other liquid

gas heat transfer may be used cooling tower.

Basically, the two hot and cold fluid enter the heat exchanger and attract to each other

indirectly by transferring heat. There is two type of flow in the heat exchanger, which is co-

current and counter-current. Co-current mainly has the parallel flow where the hot and coldfluid enter at one side and counter-current enter and different side. The illusion shown the

appendix part. Most of the time, counter-current flow perform better heat transfer. The

function in industries it is either to cool the hot fluid entered or to heat up the cold water

entered.

The cold water flow in the shell and hot flow in the tube, so there will be in indirect contact

of heat. Both convection and conduction occur in this scenario. Convection is occur between

water with two different temperature and conduction is between water and the tube wall.

One shell consist of many tubes. Therefore, the cold water flow in the shell and hot water

flow in the tube.


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III. OBJECTIVES

The objective of the shell and tube heat exchanger is to evaluate and study the

performance of the shell and tube heat exchanger at various operating conditions.

Therefore, the heat load, head balance, LMTD, overall heat transfer coefficient U,

Reynolds number, type of flow, heat transfer coefficient in tube and shell side must

be determined.

IV. THEORY

Heat exchanger perform a good heat transfer by conduction and convection between liquidand liquid. The treatment presented in this report is based on the work done by Incropera

and Dewitt (1996). Heat is conducted from the hot water in the tubes through the inner wall

tube and convection occur between the hot water in tube and cold water outside tubes.

Convection is a mode of heat transfer that involves motion of some fluid that either absorbs

heat from a source or gives heat to some surrounding. Conduction is a mode of heat transfer

in which the heat is moving through a stationary object or fluid.

There are two type of flow in the heat exchangers, which is parallel or co-current flow and

counter or cross-current flow. Basically, these two types of flow bring some significant result

to the cooling or heating process. The difference between these two flow can be

differentiate by calculating the log mean temperature difference (LMTD).


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Graph of Parallel flow Graph of Counter Flow

LMTD = (T1 - T2) / ln(T1/T2) where T1= T, HotIn T,ColdIn and T2= T, HotOut

T.ColdOut , for co-current and T1= T_Hot_In - T_Cold_Out and T2= T_Hot_Out - T_Cold_In

for counter current. But, the LMTD

LMTDco-current ;

LMTDcounter-current ;

The equation above shown for calculating LMTD value which means the true average

temperature drop for the system. From the figure, it is obviously shown the difference

between the parallel and counter flow and its LMTD difference.

The overall heat transfer coefficient will be using the value of LMTD and heat load. The

overall heat transfer coefficient, or U-value, refers to how well heat is conducted over a

series of mediums. Its units are the W/(m2C) [Btu/(hr-ft2F)]. Therefore, the equation is

given as;

U =

Where, h is the heat transfer coefficient.

Later, Reynolds number was determined to find the type of flow that divided into three

type. These are laminar, transitional and turbulence. The volumetric flow rate of water

determine the type of flow. As the velocity is high, the possibility of getting turbulence flow

also high and resulting to a efficient heat transfer. (Richardson)


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V. PROCEDURES/METHODOLOGY

A) GENERAL SET UP

1. Quick inspection was performed ensuring all equipment in proper working condition2. All valves was left closed except V1 and V23. Hot water tank was filled up by water supply hose connected to V27. Valve was latter

closed once tank was full.4. V28 was opened to fill up cold-water tank and the valve was leaved opened for

continuous water supply.5. Drain hose was connected to the cold water drain point.6. Main power, heater for hot water tank was switched on and temperature controller

was set point to 50 7. Water temperature in hot water tank was allowed to reach the set-point.8. Experiment was latter conducted.

B) EXPERIMENT: DIFFERENTIATING FLOW IN HEAT EXCHANGER AND HOW IT EFFECTSEFFECIENCY OF EXCHANGER.

1. Experiment was first conducted on co-current flow for shell and tube heat exchanger.2. For co-current flow V16 and V17 was opened.3. Consequently, FT1 for hot water was left constant at 10 L/min while FT2 for cold water

was change from 2 till 10 L/min.4. Reading for TT 1 to TT4 together with DPT 1 & DPT2 was recorded every 3 minutes FT2

changing.5. Experiment continues by adjusting the FT1 and kept FT2 constant (increasing flow rate).6. Next, for counter current flow, V15 and V18 was opened and step 3, 4 and 5 was

repeated.7. All the data was recorded and tabulated at results below.


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VI. APPARATUS

1) Valve

2) Rotameter

3) Monitoring panel4) Shell and tube heat exchanger

5) Water tank


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VII. RESULTS

A. Co-current

i. Constant FT1

FT1,cold (L/min) FT2,hot (L/min) TT1coldout ( )

TT2,coldin( )

TT3,hotout ( )

TT4hotin ( ) DPT1( ) DPT2( )

10 2 41.2 28.1 47.5 49.4 54.0 118.010 4 37.1 28.4 46.3 49.2 53.0 127.010 6 34.5 28.7 45.4 49.5 117.0 121.010 8 33.6 29.0 45.0 48.9 178.0 120.010 10 33.5 29.3 44.7 48.7 265.0 121.0

Run Cold water (tube) Hot water (shell) LMTD

( C)

U

(W/m 2

.K

Qc

(J/s)

Re Nu h i

(W/m 2.K)

Qh

(J/s )

Re Nu h o

(W/m 2.K)

1 1.83 42136.02 191.47 18751.9 1.32 1675.81 14.56 206.34 12.10 204.09

2 2.43 40582.63 185.81 18197.6 2.16 3406.89 25.60 363.03 14.30 355.93

3 2.43 41077.50 187.62 18374.8 2.86 5010.13 34.86 494.35 15.32 481.40

4 2.56 41329.55 188.54 18465.0 2.72 6632.88 43.63 618.72 15.26 598.66

5 2.93 41584.67 189.47 18556.0 2.79 8517.22 53.29 755.71 14.93 726.14


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ii. Constant FT2

FT1,cold (L/min)

FT2,hot (L/min)

TT1,coldout ()

TT2,cold in( )

TT3,hotout ()

TT4,hotin ()

DPT1()

DPT2()

2 10 30.7 29.5 39.2 49.5 207.0 -5.04 10 31.9 29.6 43.5 49.4 208.0 -5.06 10 32.6 29.7 44.1 48.8 206.0 -5.08 10 33.4 29.8 44.5 48.8 209.0 61.0

10 10 33.8 29.9 45.0 49.2 207.0 125.0


( C)

U

(W/m2

.K)

Q c

(J/ a)

Re Nu h i

(W/m 2.K)

Q h

(J/ a)

Re Nu h o

(W/m 2.K

)

1 0.84 8420.89 52.81 5172.02 1.44 8534.26 53.37 756.84 13.44 660.23

2 1.60 16491.35 90.40 8853.46 1.64 8549.80 53.45 757.74 15.34 698.00

3 2.02 24950.80 125.91 12331.19 1.96 8411.90 52.76 748.19 14.98 705.39

4 2.51 33683.58 160.08 15677.68 2.40 7955.66 50.46 715.57 14.70 684.34

5 2.72 42262.97 191.94 18797.93 3.62 8457.37 52.99 751.45 14.88 722.56


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B. Counter-current

i. Constant FT1

FT1,cold (L/min) FT2,hot (L/min) TT1( ) TT2( ) TT3( ) TT4( ) DPT1( ) DPT2( )10 2 30.2 37.8 47.3 49.1 -5.0 124.0

10 4 30.4 36.0 46.8 49.5 -5.0 125.010 6 30.0 36.4 46.2 49.2 70.0 121.010 8 29.8 33.7 45.7 49.0 -5.0 122.010 10 29.9 33.3 45.3 48.9 -5.0 121.0


( C)

U

(W/

m 2.K

Q c(W)

Re Nu h i

(W/m 2.K

)

Q h(W)

Re Nu h o (W/m 2.K)

1 5.29 42.21 0.73 727.68 0.25 1.69 0.06 1.18 14.000 1.18

2 3.90 41.43 0.72 717.71 0.75 3.41 0.10 1.96 14.762 1.95

3 4.46 41.33 0.71 707.75 1.25 5.11 0.13 2.55 14.247 2.54

4 2.72 42.11 0.73 727.68 1.84 6.74 0.17 3.33 15.549 3.31

5 2.37 42.16 0.76 757.59 2.51 8.44 0.20 3.93 15.500 3.91

ii. Constant FT2


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FT1,cold (L/min) FT2,hot (L/min) TT1( ) TT2( ) TT3( ) TT4( ) DPT1( ) DPT2( )2 10 29.8 31.0 38.3 48.9 -5.0 -5.04 10 29.8 31.6 43.9 49.5 -5.0 -5.06 10 29.7 32.2 44.7 49.6 -5.0 -5.08 10 29.8 32.8 45.0 49.5 -5.0 64.0

10 10 29.8 33.3 45.8 49.3 -5.0 122.0

Cold water (tube) Hot water (shell) LMTD

( C)

U

(W/m 2

.K

Q c

(W)

Re Nu h i (W/m2.K) Q h

(W)

Re Nu h o

(W/m 2.K)

0.17 8.47 0.20 199.36 7.38 8.44 0.20 3.93 12.622 3.85

0.51 16.95 0.35 348.89 3.90 8.47 0.20 3.93 15.925 3.89

1.05 25.45 0.48 478.47 3.41 8.50 0.20 3.93 16.170 3.90

1.67 33.89 0.61 608.06 3.14 8.49 0.20 3.93 22.014 3.90

2.44 42.10 0.73 727.68 2.44 8.50 0.20 3.93 15.698 3.91

VIII. CALCULATION

Co-current flow

1) Constant FT1


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FT1,cold (L/min) FT2,hot (L/min) TT1coldout ( )

TT2,coldin( )

TT3,hotout ( )

TT4hotin ( ) DPT1( ) DPT2( )

10 2 41.2 28.1 47.5 49.4 54.0 118.010 4 37.1 28.4 46.3 49.2 53.0 127.010 6 34.5 28.7 45.4 49.5 117.0 121.0

10 8 33.6 29.0 45.0 48.9 178.0 120.010 10 33.5 29.3 44.7 48.7 265.0 121.0

Run 1 Run 2 Run 3 Run 4 Run 5

CW HW CW HW CW HW CW HW CW HW

Nominalflow

(L/min)

10 2 10 4 10 6 10 8 10 10

Flow rate

(L/min)

10 2 10 4 10 6 10 8 10 10

TempInlet(C)

T2: 28.1 T4: 49.4 28.4 49.4 28.7 49.5 29.0 48.9 29.3 48.7

TempOutlet

(C)

T1: 41.2 T3: 47.5 37.1 46.3 34.5 45.4 33.6 45.0 33.5 44.7

PressureDrop(kPa)

DP :54.0

DP :118.0

53.0 127.0 117.0 121.0 178.0 120.0 265.0 121.0

TempChange,

(C)

T1 T2

= 13.1

T4 T3

= 1.9 = 8.7 = 3.1 = 5.8 = 4.1 = 4.6 = 3.9 = 4.2 = 4.0

AverageTemp

(C)

= 34.65

=48.45

=32.75 =47.85 =31.6 =47.45 =31.3 =46.95 =31.4 =46.7

Q, HeadLoad,J/min

109.52 79.42 145.46 129.58 145.46 171.38 153.82 163.02 175.56 167.20

LMTD 12.21 14.30 15.32 15.26 14.93


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2) Constant FT2FT1,cold (L/min

)

FT2,hot (L/min

)

TT1,coldout (

)

TT2,col

d in( )

TT3,hotout (

)

TT4,hotin (

)

DPT1(

)

DPT2(

)2 10 30.7 29.5 39.2 49.5 207.0 -5.04 10 31.9 29.6 43.5 49.4 208.0 -5.06 10 32.6 29.7 44.1 48.8 206.0 -5.08 10 33.4 29.8 44.5 48.8 209.0 61.0

10 10 33.8 29.9 45.0 49.2 207.0 125.0




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NominalFlow, (L/min)

FC: 2 FH: 10 4 10 6 10 8 10 10 10

Actual Flow,(L/min) FC: 2 FH: 10 4 10 6 10 8 10 10 10

Temp, C,Inlet

T2: 29.5 T4: 49.5 29.6 49.4 29.7 48.8 29.8 48.8 29.9 49.2

Temp, COutlet

T1: 30.7 T3: 39.2 31.9 43.5 32.6 44.1 33.4 44.5 33.8 44.0

PressureDrop, mmH2O

DP :207.0

(Shell)

DP : -5.0

(Tube)208 -5.0 206.0 -5.0 209.0 61.0 207.0 125.0

TempChange,

(C/K)

T1 T2

= 1.2

T4 T3

=10.3 = 2.3 = 5.9 =2.9 =4.7 = 3.6 =4.3 =3.9 = 5.2

Average

Temp, (C/K)

=30.1

=44.35=30.75 =46.45 =31.15 =46.45 =31.6 =46.65 =31.85 =46.60

Q, HeadLoad, J/min

QC:

50.16

QH:

86.11

96.14 98.65 121.22 117.88 150.48 143.79 163.02 217.36

LMTD 13.44 15.34 14.98 14.70 14.88

Counter-current


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1) Constant FT1

FT1cold, (L/min) FT2,hot (L/min) TT1( ) TT2( ) T T3( ) TT4( ) DPT1( ) DPT2( )

10 2 30.2 37.8 47.3 49.1 -5.0 124.0

10 4 30.4 36.0 46.8 49.5 -5.0 125.010 6 30.0 36.4 46.2 49.2 70.0 121.010 8 29.8 33.7 45.7 49.0 -5.0 122.010 10 29.9 33.3 45.3 48.9 -5.0 121.0



NominalFlow, L.min

2 10 4 10 6 10 8 10 10 10

Actual Flow,L/min

FC: 2 FH: 10 4 10 6 10 8 10 10 10

Temp, C,Inlet

T1= 30.2 T4=49.1 36.0 49.5 36.4 49.2 33.7 49.0 33.3 48.9

Temp, C

Outlet

T2= 37.8 T3=47.3 46.8 30.4 46.2 30.0 45.7 29.8 45.3 29.9

PressureDrop, mm

H2O

DP :

-5.0

DP :124.0

(Tube)

-5.0 125.0 70.0 121.0 -5.0 122.0 -5.0 121.0

TempChange, T1 and T2Cfor LMTDcalculation

T1 =T4 T1

= 17.1

T2= T3 T2

=11.3=16.1 =13.5 =15.8 =12.8 =15.8 =15.3 =15.4 =15.6

AverageTemp, C

=34.0

=48.2

=33.2 =48.2 =433.2 =47.7 =31.8 =47.4 =31.6 =47.1

TempChange, T for cold andhot water.

T2-T1

=7.6

T3-T4=1.8

=5.6 =2.7 =6.4 =3.0 =3.9 =3.3 =3.4 =3.6

2) Constant FT2


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FT1(L/min) FT2(L/min) TT1( ) TT2( ) TT3( ) TT4( ) DPT1( ) DPT2( )2 10 29.8 31.0 38.3 48.9 -5.0 -5.04 10 29.8 31.6 43.9 49.5 -5.0 -5.06 10 29.7 32.2 44.7 49.6 -5.0 -5.08 10 29.8 32.8 45.0 49.5 -5.0 64.0

10 10 29.8 33.3 45.8 49.3 -5.0 122.0



NominalFlow,

FC: 10 FH: 2 10 4 10 6 10 8 10 10

Actual Flow,

FC: 10 FH: 2 10 4 10 6 10 8 10 10

Temp, C,Inlet

T1: 29.4 T4:48.9 29.8 49.5 29.7 49.6 29.8 49.5 29.9 49.3

Temp, COutlet

T2: 31.0 T3: 38.3 31.6 43.9 32.2 44.7 32.8 45.0 33.3 45.8

PressureDrop, mmH2O

DP :-5

(Shell)

DP : -5.0

(Tube)-5.0 -5.0 -5.0 -5.0 -5.0 64.0 -5.0 122.0

TempChange, T1 and T2Cfor LMTDcalculation

T1 =T4 T1

= 8.5

T2= T3 T2

=17.9=14.1 =17.9 =15.0 =17.4 =15.2 =16.7 =15.4 =16.0

AverageTemp, C/F

=30.4

=43.6

=30.7 =46.7 =31.0 =47.2 =31.3 =47.3 =31.2 =47.8

TempChange, T for cold andhot water.

T2-T1

=1.2

T3-T4=10.6

=1.8 =5.6 =2.5 =4.9 =3.0 =4.5 =3.5 =3.5


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A. Temperature change

i. Temperature change of cold water (CW);

T = T- T = 13.1 C

ii. Temperature change of hot water (HW):

T = T T = 1.9 C

B. Average Temperature

i. Average temperature of cold water;

: 22 = 34.65C

ii. Average temperature of hot water;

4: 32 = 48.45 C

C. Amount of heat, Q

1) Find the area of the tube with diameter 100mm

Area, A =

4 =

4 =7.8549 X 10 -5 m 2

2) Mass flow rate of cold,mcold and hot water,mhot since the density is assumed to be constant

that is 1kg/m

i. Mcold = V

= ( )( ) ( ) = 0.01 kg/min

ii. Mhot = V

= ( )( ) ( )

= 0.002 kg/min

3) Amount of heat load, Qi. Qc = m Cold CpCold (TCold_out TCold_in )


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= (0.002kg/min)(4.18kJ/Kg.K)(13.1K)

= 109.52J/min

ii. Qh = mHot C pHot (THot_in THot_out )

= (0.01kg/min)(4.18kJ/Kg.K)(1.9K)

= 79.42 J/min

D. Log mean temperature difference (LMTD)where T1= T, HotIn T,ColdIn and T2= T, HotOut T.ColdOut ;

LMTD = (T1- T2) / ln(T1/T2)

< ; ; ;

< 49 4;28 4; 47 5 ; 4 2

= 12.21

E. Reynolds number for tube (cold) side and shell (hot) side

a. Co-current flow (Run 1)

According to the references, and , and = viscosity for fluid by referring at their temperature for

every run at table A.6 (Principles of Heat and Mass Transfer textbook, 7 th edition).

1) For shell side(hot)

Information;

i. = 2 x 6 = 0.033 kg/sii. T = 322.4 K,iii. = 553.48 x 10 -6 ,iv. D0 = 0.039 m,v. Di = 0.0063m,vi. Dh = 0.0453m

Re = 4


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= 4 33453553 48 ;6 = 1675.81

2) For tube side (cold)i. = 10 x 6 = 0.167 kg/s

ii. Di = 0.0063m,iii. = 801x 10 -6

Re = 4

= 4 67638 ;6 = 42136.02

b. Counter-current flow (Run 1)

According to the references, and , and = viscosity for fluid by referring at their temperature for

every run at table A.6 (Principles of Heat and Mass Transfer textbook, 7 th edition).

1) Shell side

Information;

i. ii. iii.

Re = 4

= 4 333 39 635 574 = 1679.1488

2) Tube side

According to the references, , and = viscosity for fluid byreferring their temperature at every run at table A.6 (Principles of Heat and Mass Transfertextbook, 7 th edition). (P.Incropera)

Re =4

= 4 33363696 = 9672.9


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F. Nusselts number

To obtain the Nusselts number by taking the 6339 = 0.16 and referring at the table 8.2, linearinterpolation (Principles of Heat and Mass Transfer textbook, 7 th edition) the Nusselts number (Nu i) atevery run for counter-current and co-current flow is 9.884 and the values of Pr for fluid were referred at

the table A.6 (Principles of Heat and Mass Transfer textbook, 7th

edition).

a. Tube side in co-current flow ,

Nu= 0.023Re 0.8 Pr0.3

Nu=0.023 (12 614) 0.8 (5.4394) 0.3

Nu=72.96

b. Tube side in counter-current flow ,

8 3 As Re obtained at the previous was 9672.9

8 4 = 65.46

G. Heat transfer coefficient, ha. Co-current flow (Run 1)1) shell side,

Nu i =

h0 = 9 884 642 4 453 = 140.16 W/m2.K

2) tube side,

h i = Nu (k/D)

h i = 72.96 x 617 x 10-3/0.0063

hi = 7145 W/m 2.K

b. Counter-current flow (Run 1)1) shell/annulus side,

Nu i =

From the previous Nu i was 9.884, Dh = Do Di = 0.039m 0.0063m = 0.0327m


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9.884 = 327

to obtain k by referring table A.6 (Principles of Heat and Mass Transfer textbook, 7 th edition) atrecorded temperature for every run; k at run 1 is 6.42 x 10 -4

9.884 = 3276 42

ho = 503.4374 W/m 2.K

2) tube side,

hi =

hi = 65 46628 63

= 6577 W/m 2.K

H. Overall heat transfer coefficient,Ua. Co-current flow (Run 1)

U = : = : = 137.46 W/m2.K

b. Counter-current flow (Run 1)

U = = 467.6417 W/m2.K

I. Pressure drop , P a. Co-current flow

1) For shell side,

P = 64

= 648 2 5

= 0.0787 Pa


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2) For tube side,

P = 4 2

f = [ 1.58 In (12614) 3.28] 2

= 7.28 x10 -3

A = (0.0063) 2

4

= 3.117 x10 -5 m 2

V = 2LPM 60s

= 33.33 m 3/s

P = 4 (7.38 x10 -3) x (1.07 x10 6)

0.0063 ( 2(9.81))

= 1.393 x10 10 Pa

b. Counter-current flow1) For shell side,

P = 64

= 649672 9

= 6.6164 x 10 -3 Pa

2) For tube side,

P = 4 2

f = [ 1.58 In (12614) 3.28]2

= 7.794 x10-3

A = (0.0063) 2

4

= 3.117 x10 -5 m2

V = 2LPM 60s


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= 33.33 m 3/s

P = 4 (7.794 x10 -3) x (1.07 x10 6)

0.0063 ( 2(9.81))

= 3.09 x106 Pa

IX. DISCUSSION

Based on the data tabulated in result section, both co-current and counter current has change in

term of temperature difference. Hot water flow in reduce its temperature transfer its heat to the cold

water in. therefore, the cold water in increase its temperature during flow out of the tube. Some

equation has been used to calculate and differentiate these two type flow. The calculated value has

been analize and it has been found that the heat load for counter-current is more preferable for cooling

or heating process. Heat value shown by counter current mostly convenience for the process.

Other than that, the Reynolds value has been calculated and shows that co-current using the

turbulence flow whereas the counter-current just using the laminar flow. The range of Reynolds number

can decide the type of flow of the fluid such as laminar, transitional or turbulence. It has been prove that

the turbulence flow has the most efficient cooling or heating rate. But, the counter current can perform

better with laminar flow. Based on the theory, the amount of the loss of the heat from hot water should

equally to the heat gain by cold water. However, this can be achieved in real life. This is because of the

heat loss to the surrounding.

LMTD is calculated to find the average temperature drop of the system and counter-current

found to has the better value compare to co-current. Based on the theory, the amount of the loss of the

heat from hot water should equally to the heat gain by cold water. However, this can be achieved in real

life. This is because of the heat loss to the surrounding. But, there is some mistake in the calculation, the

overall heat loss doesnt match with the statement. The experiment went well and no mistake was doneunless in the calculation part. There is error in getting the value of Pr and k value.


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X. CONCLUSION

Overall, counter-current has the most efficient heat transfer. It perform better even with laminar flow,

unlike co-current. Heat load in the counter current higher that parallel flow. The convection and

conduction is more preferable in this flow. The LMTD value easily proving the difference. Basically,

higher flow rate perform better heat transfer.

XI. RECOMMENDATIONS

i. Before starting the experiment, one must do the start up to stabilize the equipment.

ii. It is also recommended that during the process of data collection that the user adjusts the flow rate

of only one stream per setup. If this is not done the graphs of the data becomes very difficult to

read and understand.

iii. In an effort to reduce the heat loss to the surroundings, it is recommended that the heat exchanger

be well insulated. (Vaughn & Ware, September 19, 2000)

iv. The temperature reading was taken in certain time intervals to make the data more accurate and

standardize.

XII. REFERENCES

1. P.Incropera, F. (n.d.). Principle of heat and mass transfer 7th edition.

2. Richardson, J. C. (n.d.). Chemical Engineering Volume 1. Elsivier.

3. Vaughn, D., & Ware, T. (September 19, 2000). Heat Exchanger Analysis. Cunningham.


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XIII. APPENDICES

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