CPE533 Shell and Tube Heat Exchanger Full Lab Report

UNIVERSITI TEKNOLOGI MARA FAKULTY OF CHEMICAL ENGINEERING PROCESS ENGINEERING LABORATORY 2 (CPE553) NAME : MOHAMAD FAZRUL BIN BASRI (2013331897) GROUP : EH2414 (GROUP1) EXPERIM : LAB : SHELL AND TUBE HEAT EXCHANGER ENT DATE : 30 OCTOBER 2015 PROG/CO : EH241 DE SUBMIT : MDM LIM YING PEI TO N Title Allocated Marks o Marks (%) 1 Abstract 5 2 Introduction 5 3 Objectives 5 4 Theory 5 5 Procedures/Methodology 10 6 Apparatus 5 7 Results 10 8 Calculation 10 9 Discussion 20 10 Conclusion 10 11 Recommendations 5 12 References 5 13 Appendices 5 TOTAL 100 Remarks:

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ABSTRACT This experiment was conducted to evaluate and study the performance of the shell and tube heat exchanger heat load and heat balance, LMTD, overall heat transfer coefficient, Reynolds shell side and tube side, heat transfer coefficient and pressure drop at shell side and tube side. Every run will be using different flow rate. From the result, the pressure drop depends on the flow rate not the temperature. Heat exchanger is a device that are design to transfer or exchange heat from one matter to another in efficiently. There are several types of heat exchanger and one of it that are mostly use in industrial applications is a shell and tube heat exchanger. Fluids that flow in this device are in counter-current flow where two fluids flow against each other, maintaining a maximum temperature difference between the hot and cold streams which allows for maximum heat transfer. We assumed that internal, potential and kinetic energy was negligible in this process, so Q H must be equal to Q C. In experiment 1, FT1 was constant at 10LPM. The highest efficiency of heat transfer was 99.58% at FT1=10 and FT2=6. It means that heat transfer occur efficiently at equal volumetric flowrates of hot and cold water. the results was different than the basic theory where the amount of heat release by hot water was not equal to the amount of heat absorb by cold water, Q H ≠ QC due to some errors and the recommendation and precautions were made to improve this experiment.

INTRODUCTION Shell-and-tube heat exchangers are commonly used in oil refineries and other large-scale chemical processes. Heat exchanger is a device that are design to transfer or exchange heat from one matter to another in efficiently. It means that matter that release heat will decrease in temperature while the other matters that gain heat will increase in temperature. A heat exchanger is a device that is used to transfer thermal energy between two or more fluids, between a solid surface and a fluid a fluid, or between solid particulates and a fluid at different temperatures and in thermal contact In this model, two separated fluids at different temperatures flow through the heat exchanger: one through the tubes (tube side) and the other through the shell around the tubes (shell side). Heat exchangers can be in cross-flow, parallel-flow or countercurrent. Cross flow is the flow where the cold and the hot fluid flow axis is at an angle to each other. Hence, the fluids will cross each other in this arrangement. Mostly, this type of flow has the angle between axes as 90 degree. Parallel flow or co-current flow is the flow where the hot and the cold fluid is flow in the same direction. The most effective flow in the heat exchanger is a countercurrent flow where the fluid paths flow in opposite directions, with exiting and the other enters. This results in faster heat exchange. Heat exchangers are classified according to transfer process, number of fluids, degree of surface contact, design features, flow arrangements and heat transfer mechanism. Several design parameters and operating conditions influence the optimal performance of a shell-andtube heat exchanger. The main purpose of this model is to show the basic principles for setting up a heat exchanger model. It can also serve as a starting point for more sophisticated applications, such as parameter studies or adding additional effects like corrosion, thermal stress, and vibration. . There are several types of heat exchanger and one of it that are mostly use in industrial applications is a shell and tube heat exchanger.

Figure 1: Shell and tube heat exchanger

It contain a large number of tubes (sometimes several hundred) packed in a shell with their axes parallel to that of the shell. Heat transfer takes place as one fluid flows inside the tubes while the other fluid flows outside the tubes through the shell. Shell-and-tube heat exchangers are further classified according to the number of shell and tube passes involved.

OBJECTIVES 1. To evaluate and study the heat load and head balance, LMTD and overall heat transfer coefficient. 2. To calculate the Reynolds numbers at the shell and tubes sides. 3. To measure and determine the shell and tube sides pressure drop.

THEORY The main function of heat exchanger is to either remove heat from a hot fluid or to add heat to the cold fluid. The direction of fluid motion inside the heat exchanger can normally categorized as parallel flow, counter flow and cross flow. In this experiment, we study only counter-current flow. For counter-current flow, both the hot and cold fluids flow in the opposite direction. Both the fluids enter and exit the heat exchanger on the opposite ends. In this experiment, we focused on the shell and tube heat exchanger. Heat load and heat balance

QH This part of the calculation is to use the data in Table 1 to check the heat load

QC

QC and to select the set of values where

HW Hot water flow rate (

)

FH  Cp H  (t1  t 2 )

QH =

QH is closest to

.

and

CW Hot water flow rate (

)

FC  CpC  (T2  T1 )

QC =

Where:

QH = Heat load for hot water flow rate

QC = Heat load for cold water flow rate

FH  Hot water mass flow rate

FC  Cold water mass flow rate

t1  Hot water inlet temperature

t2  Hot water outlet temperature

T1  Cold water inlet temperature

T2  Cold water outlet temperature LMTD Calculations of log mean temperature difference (LMTD).

LMTD 

(t1 T 2 )  (t 2  T1 ) (t  T2 ) ln 1 (t 2  T1 )

Where, all variables are same with the above section:

R

 (t1  T2 )  (t2  t1 )

S

 (t2  t1 )  (T1  t1 )

Both equations would determine the value of correction factor

obtained from the graph with respect to

R

FT

. Practically,

FT

value

S

and

value. In this case, the correction factor would

apply to enhance the LMTD value. So, equation below show the corrected LMTD can be determined.

LMTD  FT  LMTD

U

Overall heat transfer coefficient,

UD Overall heat transfer coefficient at which equivalent to

below. In this case, the value of total heat transfer area

can be calculated by using equation

A

has been given and equal to 31.0 ft2

U

Q A  LMTD  FT

Where:

Q Heat rate with respect to the average head load

FT 

Correction factor

Reynolds Number Calculation

CW

Re(s) 

Shell-side

for

Re( s ) 

De.Gs 

Where:

De 

de 12

2

do 4(1 / 2 PT  0.86 PT  1 / 2 . ) 4 de  1 / 2 .do

At which:

PT 

do 

Pitch = 0.81inch

Tube outside diameter, inch

 Viscosity, taken at average fluid temperature in the shell, lbmft-1hr-1

Gs 

Ws 

As 

Ws As

(lbmft-2hr-1)

Flow rate in (lbmhr-1)

0.029 ft2

HW

Re(t ) 

Tube-side

for

Re( t ) 

D.Gt 

Where:

D

Tube ID = 0.04125 ft

 Viscosity, taken at average fluid temperature in the tube, lbmft-1hr-1

Gt 

Wt 

At 

Wt At

(lbmft-2hr-1)

Flow rate in lbmhr-1

0.02139 ft2

Pressure drop This part would determine the following:

H w : The measured tube-inside pressure drop DP (tube) which will be corrected and is expected

to be more than calculated tube-side pressure drop.

CW :

The measured shell-inside pressure drop DP (shell) which will be corrected and is

expected to be more than calculated tube-side pressure drop. Notice that, both calculated pressure and also measured pressure are considered in unit mmH2O. In this case, since calculated pressure drop in both of shell and tube side have been obtained during the experiment, so it’s only required conversion factor to change the value into unit of mmH2O. Conversion factor:

x.bar 

1  105 Pa 1mmH2O  1bar (9.81) Pa .

Where ‘x’ is the calculated pressure value in the unit bar.

APPARATUS

Figure 2: SOLTEQ Heat Exchanger Training Apparatus (Model HE 158C)

PROCEDURE General start-up procedures 1. A quick inspection is performed to make sure that the equipment is in a proper working condition. 2. All valve are initially closed, except V1 and V12. 3. Hot water tank is filled up via a water supply hose connected to valve V27. The valve is closed after the tank is full. 4. The cold water tank is filled up by opening valve V28 and leaves the valve opened for continuous water supply. 5. A drain hose is connected to the cold water drain point. 6. Main power is switched on and heater for the hot water also switched on and set the temperature controller to 50°C. 7. The water temperature in the hot water tank is allowed to reach the set point. 8. The equilibrium is already set up. General Shut-down 1. The heater is switched off. The hot water temperature drops is wait until below 40°. 2. The pump P1 and P2 is switched off. 3. Main power is switched. 4. All the water in process lines is drain off. All valves is closed.

Experiment 1: Counter-current Concentric Heat Exchanger 1. The general start-up procedure is performed. 2. The valve is switched to counter-current Concentric Heat Exchanger arrangement. 3. The pumps P1 and P2 are switched on. 4. The valve V3 and V14 is opened and adjusted to obtain the desired flowrates for hot water and cold water stream. 5. The system is allowed to reach steady state for 10 minutes. 6. FT1, FT2, TT1, TT2, TT3 and TT4 is recorded. 7. The pressure drop measurement for shell-side and tube side also recorded for pressure drop studies. 8. The steps 4 to 7 is repeated for different combination of flowrates FT1 and FT2 as in the result sheet. 9. The pumps P1 and P2 is switched off after the experiment is completed. The next experiment is proceed.

RESULTS Experiment 1 Table 1: Counter-current Shell and Tube Heat Exchanger at constant FT1 FT 1

FT 2

TT 1

TT 2

TT 3

TT 4

DPT 1

DPT 2

(LPM)

(LPM)

(0C)

(0C)

(0C)

(0C)

(mmH2O)

(mmH2O)

10

2

44.7

29.5

47.5

48.8

95

5

10

4

37.5

27.5

46.9

50.0

98

17

10

6

34.9

28.6

45.3

49.1

86

84

10

8

33.8

28.5

44.8

49.5

91

125

10

10

33.2

29.2

44.2

49.5

93

215

Experiment 2 Table 2: Counter-current and Tube Heat Exchanger at constant FT2 FT 1

FT 2

TT 1

TT 2

TT 3

TT 4

DPT 1

DPT 1

(LPM)

(LPM)

(0C)

(0C)

(0C)

(0C)

(mmH2O)

(mmH2O)

2

10

30.7

28.7

39.5

48.3

5

209

4

10

31.3

29.3

42.8

48.5

5

211

6

10

32.2

28.9

43.5

48.8

28

212

8

10

32.7

28.6

43.7

49.8

57

213

10

10

33.2

29.2

44.2

48.4

93

215

CALCULATIONS Experiment A: Counter-Current Flow

Hot Water Density:

988.18 kg/m3

Heat Capacity:

4175.00 J/kg.K

Thermal condition:

0.6436 W/m.K

Viscosity:

0.0005494 Pa.s

Cold Water Density:

995.67 kg/m3

Heat Capacity:

4183.00 J/kg.K

Thermal condition:

0.6155 W/m.K

Viscosity:

0.0008007 Pa.s

1. Calculation of heat transfer and heat lost Hot Water Flowrate = 10.0 LPM

Cold water flowrate = 2, 4, 6, 8 & 10 LPM

1)

Q hot (W )=m h C p ∆ T =10.0

Qcold ( W )=mh C p ∆ T =2.0

L 1m 3 1min kg J × × ×988.18 3 ×4175 × ( 48.8−47.5 ) ℃=893.89W min 1000 L 60 s kg ∙ ℃ m

L 1 m3 1 min kg J × × × 995.67 3 × 4183 × ( 44.7−29.5 ) ℃=2110.21 W min 1000 L 60 s kg ∙℃ m

Heat Lost Rate=Qhot −Qcold =( 893.89−2110.21 ) W =−1216.32W

ε=

Q 893.89 = ×100 =42.36 Qmax 2110.21

2)

Qhot (W )=mh C p ∆=10.0

L 1 m3 1 min kg J × × ×988.18 3 × 4175 ×(50.0−46.9)℃=2131.59 W min 1000 L 60 s kg ∙ ℃ m

Q cold ( W )=m h C p ∆ T =4.0

L 1 m3 1 min kg J × × × 995.67 3 × 4183 × (37.5−27.5 ) ℃=2776.59 W min 1000 L 60 s kg ∙ ℃ m

Heat Lost Rate=Qhot −Qcold =( 2131.59−2776.59 ) W =−645 W

ε=

Q 2131.59 = ×100 =76.77 Qmax 2776.59

3)

Q hot (W )=mh C p ∆=10.0

L 1 m3 1 min kg J × × ×988.18 3 × 4175 × ( 49.1−45.3 ) ℃=2612.91 W min 1000 L 60 s kg ∙ ℃ m

Qcold ( W )=mh C p ∆=6.0

L 1m3 1min kg J × × ×995.67 3 × 4183 × ( 34.9−28.6 ) ℃=2623.88 W min 1000 L 60 s kg ∙ ℃ m

Heat Lost Rate=Qhot −Qcold =( 2162.91−2623.88 ) W =−10.97 W

ε=

Q 2612.91 = ×100 =99.58 Qmax 2623.88

4) 3

Qhot (W )=mh C p ∆=10.0

L 1m 1 min kg J × × ×988.18 3 × 4175 × ( 49.5−44.8 ) ℃=3231.76 W min 1000 L 60 s kg ∙ ℃ m

Q cold ( W )=m h C p ∆ T =8.0

L 1 m3 1 min kg J × × × 995.67 3 × 4183 × ( 33.8−28.5 ) ℃=2943.19 W min 1000 L 60 s kg ∙℃ m

Heat Lost Rate=Qhot −Qcold =( 3231.76−2943.19 ) W =288.57 W

ε=

Q 2943.19 = ×100 =91.07 Qmax 3231.76

5)

Q hot (W )=mh C p ∆=10.0

L 1 m3 1 min kg J × × ×988.18 3 × 4175 × ( 49.5−44.2 ) ℃=3644.33 W min 1000 L 60 s kg ∙ ℃ m

Qcold ( W )=mh C p ∆ T =10.0

L 1m3 1 min kg J × × ×995.67 3 × 4183 × ( 33.2−29.2 ) ℃=2776.59W min 1000 L 60 s kg ∙ ℃ m

Heat Lost Rate=Qhot −Qcold =( 3644.33−2766.59 ) W =877.74 W

ε=

Q 2766.59 = ×100 =75.91 Qmax 3644.33

2. Calculation of Log Mean Temperature Difference (LMTD)

∆ T lm=

[ ( Th ,¿−Tc ,out ) −( Th ,out −Tc ,¿ ) ] ln ⁡[

( Th, ¿−Tc ,out ) ] ( Th, out −Tc ,¿ )

1)

∆ T lm=

[ ( 48.8−44.7 ) −( 47.5−29.5 ) ] =9.40 ℃ ( 48.8−44.7 ) ln ⁡[ ] ( 47.5−29.5 )

2)

∆ T lm=

[ (50.0−37.5 )− ( 46.9−27.5 ) ] =15.70℃ ( 50.0−37.5 ) ln ⁡[ ] ( 46.9−27.5 )

3)

∆ T lm=

[ ( 49.1−34.9 ) −( 45.3−28.6 ) ] =15.42 ℃ ( 49.1−34.9 ) ln ⁡[ ] ( 45.3−28.6 )

4)

∆ T lm=

[ ( 49.5−33.8 )−( 44.8−28.5 ) ] =16.00 ℃ ( 49.5−33.8 ) ln ⁡[ ] ( 44.8−28.5 )

5)

∆ T lm=

[ ( 49.5−33.2 )−( 44.2−29.2 ) ] =15.64 ℃ ( 49.5−33.2 ) ln ⁡[ ] ( 44.2−29.2 )

3. Calculate of the tube and shell heat transfer coefficient 0.8

At tube side (hot water-cooling process): Nu=0.023 × ℜ × Pr

0.33

3 L 1 m3 ´ 1 min −4 m V´ =10 × × =1.67 ×10 min 1000 L 60 s s

2

2

π d π ×(0.02664) 2 A= = =0.000557 m 4 4

V´ 1.67 ×10−4 m v= = =0.299 A 0.000557 s

ℜ=

ρvd = μ

Pr=

988.18

μCp = k

kg m × 0.299 × 0.02664 m 3 s m =14327 ( turbulent flow ) 0.0005494 Pa ∙ s

( 0.0005494 Pa ∙ s ) ×( 4175 W 0.6436 m∙K

0.8

Nu=0.023 × ℜ × Pr

h=

Nuk = d

J ) kg ∙ K

0.33

=3.564

0.8

=0.023 ×14327 ×3.564

0.33

=73.55

W m∙ K W =1776.91 2 0.02664 m m ∙K

73.55 × 0.6436

At shell side (cold water-heating process):

At the shell side: 2 LPM

0.8

Nu=0.023 × ℜ × Pr

0.4

3 L 1m 3 ´1 min −5 m V´ =2 × × =3.33× 10 min 1000 L 60 s s

0.085 ¿ (¿ ¿ 2−( 0.0334 )2) ¿ π ׿ π (d 2s−d 2o) A= =¿ 4

−5 V´ 3.33 ×10 m v= = =0.0069 A 0.0048 s

ℜ=

ρv ( d s−d o ) μ

955.67 =

kg m × 0.0069 × ( 0.085−0.0334 m) 3 s m 0.0008007 Pa ∙ s

¿ 425 ( laminar flow )

μCp Pr= = k

( 0.0008007 Pa ∙ s ) ×(4183 W 0.6155 m∙ K

J ) kg ∙ K

=5.49

Nu=0.023 × ℜ0.8 × Pr0.4 =0.023 × 4250.8 ×5.490.4 =5.76

W 5.76 ×0.6155 Nuk m∙ K W h= = =68.68 2 d (0.085 m−0.0334 m) m ∙K

At the shell side: 4 LPM

3 L 1m 3 ´1 min −5 m V´ =4 × × =6.67 ×10 min 1000 L 60 s s

0.085 ¿ (¿ ¿ 2−( 0.0334 )2) ¿ π ׿ π (d 2s−d 2o) A= =¿ 4

−5 V´ 6.67 ×10 m v= = =0.0139 A 0.0048 s

ℜ=

ρv ( d s−d o ) μ

955.67 =

kg m × 0.0139 × ( 0.085−0.0334 m) 3 s m 0.0008007 Pa ∙ s

¿ 856 ( laminar flow )

μCp Pr= = k

( 0.0008007 Pa ∙ s ) ×(4183 W 0.6155 m∙ K

J ) kg ∙ K

=5.49

Nu=0.023 × ℜ0.8 × Pr0.4 =0.023 × 8560.8 ×5.490.4 =10.80

W 10.80 ×0.6155 Nuk m∙ K W h= = =120.26 2 d (0.085 m−0.0334 m) m ∙K

At the shell side: 6 LPM

3 L 1 m3 ´ 1 min −4 m V´ =6 × × =1 ×10 min 1000 L 60 s s

0.085 ¿ (¿ ¿ 2−( 0.0334 )2) ¿ π ׿ π (d 2s−d 2o) A= =¿ 4

−4 V´ 1 ×10 m v= = =0.0208 A 0.0048 s

ℜ=

ρv ( d s−d o ) μ

955.67 =

kg m × 0.0208 × ( 0.085−0.0334 ) 3 s m 0.0008007 Pa ∙ s

¿ 1281 ( laminar flow )

μCp Pr= = k

( 0.0008007 Pa ∙ s ) ×(4183 W 0.6155 m∙ K

J ) kg ∙ K

=5.49

Nu=0.023 × ℜ0.8 × Pr0.4 =0.023 ×12810.8 ×5.49 0.4=13.91

W 12.35 ×0.6155 Nuk m∙ K W h= = =166.03 2 d (0.085 m−0.0334 m) m ∙K

At the shell side: 8 LPM

3 L 1 m3 1´ min −4 m V´ =8 × × =1.333 ×10 min 1000 L 60 s s

0.085 ¿ (¿ ¿ 2−( 0.0334 )2) ¿ π ׿ π (d 2s−d 2o) A= =¿ 4

−4 V´ 1.333 ×10 m v= = =0.0278 A 0.0048 s

ℜ=

ρv ( d s−d o ) μ

955.67 =

kg m × 0.0278 × ( 0.085−0.0334 ) 3 s m 0.0008007 Pa ∙ s

¿ 1712 ( laminar flow )

μCp Pr= = k

( 0.0008007 Pa ∙ s ) ×(4183 W 0.6155 m∙ K

J ) kg ∙ K

=5.49

Nu=0.023 × ℜ0.8 × Pr0.4 =0.023 ×17120.8 ×5.49 0.4=17.55

W 17.55 ×0.6155 Nuk m∙ K W h= = =209.38 2 d (0.085 m−0.0334 m) m ∙K

At the shell side: 10 LPM

3 ´1 min L 1 m3 −4 m V´ =10 × × =1.667 ×10 min 1000 L 60 s s

0.085 ¿ (¿ ¿ 2−( 0.0334 )2) ¿ π ׿ π (d 2s−d 2o) A= =¿ 4

V´ 1.667 ×10−4 m v= = =0.0347 A 0.0048 s

ℜ=

ρv ( d s−d o ) μ

955.67 =

kg m × 0.0347 × ( 0.085−0.0334 ) 3 s m 0.0008007 Pa∙ s

¿ 2137 ( laminar flow )

μCp Pr= = k

( 0.0008007 Pa ∙ s ) ×(4183 W 0.6155 m∙ K

J ) kg ∙ K

=5.49

Nu=0.023 × ℜ0.8 × Pr0.4 =0.023 ×2137 0.8 ×5.490.4 =20.96

W 20.96 ×0.6155 Nuk m∙ K W h= = =250.02 2 d (0.085 m−0.0334 m) m ∙K

Overall heat transfer coefficient:

Total exchange area , A=π × tube od ×length=π × 0.02664 m× 0.5 m=0.05 m2

1.

U=

Qhot 893.89 W W = =1901.89 2 2 A ∆ T lm 0.05 m × 9.40 ℃ m ∙K

2.

U=

Qhot 2131.59W W = =2715.40 2 2 A ∆ T lm 0.05 m × 15.70℃ m ∙K

3.

U=

Qhot 2612.91W W = =3388.99 2 A ∆ T lm 0.05 m 2 × 15.42℃ m ∙K

4.

U=

Qhot 3231.76 W W = =4039.7 2 A ∆ T lm 0.05 m2 × 16.00℃ m ∙K

5.

11 U=

Qhot 3644.44 W W = =4660.41 2 2 A ∆ T lm 0.05 m ×15.64 ℃ m ∙K

DISCUSSION This experiment is conducted by using SOLTEQ Heat Exchanger Training apparatus which is used as cooling devices. Some of the purposes of this experiment are to cool the hot streams until both cold and hot streams have the same temperature, to study the working principle of counter flow heat exchanger and to evaluate and study the overall heat transfer coefficient, LMTD and heat transfer and heat loss for energy balance. In this shell and heat pump exchanger, cold water flows through the outer pipe (the shell) while hot water will flows through the inner pipe (in the tube). Heat will be transfer from high temperature (hot water stream) to low temperature (cold water stream). This causes hot water to decrease in temperature while cold water to increase in temperature until both hot and cold water streams have the same temperature. We need also calculate Reynolds’s number at the shell and tube heat exchanger and measure and determine the shell and tube side pressure drop. During the experiment, we carried out Run III and Run IV experiment. Every run consist of three set of data which need to be considered. It is found that the calculated values of QH and QC are not really satisfied the theory since supposedly, the ratio of QC/QH is unity means the ideal condition is the value of QC should be closed to the value of QH. But in the calculated results, it is found that there are some deviations in the value but it is normal because it is impossible to have an ideal system in real life. For LMTD, the calculations consist of the use of graph which called as correction factor graph. This graph is used to obtain a more accurate LMTD as the calculated LMTD values may deviated from the actual one. The correction factor, FT is obtained from the graph by finding the values of R and S. In the experiment, volumetric flowrates of hot water is constant which is 10 LPM while volumetric flowrates of cold water is change every 10 minutes from 2 LPM to 10 LPM. Heat transfer of hot water, QH is higher than heat transfer of cold water, QC. However, QH keep decreasing while QC keeps increasing as volumetric flowrate of cold water increases. The highest efficiency in experiment is 99.58% at FT1=10 LPM and FT2=6 LPM where its QH=2612.91 W, QC=2623.88 W, heat loss rate is 10.97 W, LMTD=15.42° and heat transfer coefficient, U=3388.99 kg/s2. It means that heat transfer occur efficiently at almost equal volumetric flowrates of hot and cold water. At the end of the experiments, all objectives are met although maybe there are some errors. Presence of air bubbles in the tube also is one of the factors that cause inaccurate results.

CONCLUSION Based on the experiment, the main objective is to evaluate and study the overall heat transfer coefficient, LMTD, heat transfer and heat loss for energy balance as well as to evaluate and study the performance of shell and tube heat exchanger at various operating condition. . In this shell and tube heat exchanger, the fluids flow in counter-current flow which results in faster heat exchange. The experiment shows that the flow rate of one of the stream is directly proportional to the rate of heat transfer since the rate of heat transfer is increases as the flow rate of fluid increases. The basic theory in this experiment is Q H=QC, which the amount of heat release by hot water is equal to the amount of heat absorb by cold water. However, the results is different than the basic theory where the amount of heat release by hot water is not equal to the amount of heat absorb by cold water, QH ≠ QC. This is due to some errors during conducting this experiment which are the presence of bubbles in tube where the hot water flows. The presence of these bubbles can cause corrosion and disturb the process of heat transfer. Although the results are not following the basic theory, this experiment can be said as successful because objectives of this experiment were already achieved. RECOMMENDATIONS There are a few recommendations and precautions that need to be taken during conducting this experiment in order to get an accurate value and success results. Make sure that the equipment is in good condition so that the flow of the experiment does not disturb by the inconstant data. To ensure the data obtained is accurate, make sure there is no air bubbles in the tube during experiment. The readings of FT1, FT2, DPT1, and DPT2 must be taken when the system is stabilized and reach its steady state to get good results in calculations. The eye must be perpendicular to the reading scale of volumetric flowrates of hot and cold water to avoid parallax error during changing this flowrates. Besides that, the heat exchanger must be well insulated in order to reduce the heat loss to the surroundings. The last set of temperature readings should be taken when all the temperatures are fairly steady. While recording the data, make sure that the pressure and temperature is at constant value because this can affect the calculation made. Lastly, to improve the system of shell and tube heat exchanger, it is recommended that the shell and tube heat exchanger have alert sign or alarm that can give a sign to the engineer who handles the equipment to take the readings at the correct time in order to get accurate readings.

REFERENCES 

Concentric

 

https://www.scribd.com/doc/27156908/CONCENTRIC-TUBE-HEAT-EXCHANGER Coulson and Richardson. Chemical Engineering; Volume 1, 6th edition. Kessler, D.P., Greenkorn, R.A. (1999). Momentum, Heat, and Mass Transfer



Fundamentals, New York: Marcel Dekker Inc., pp (768-828). Shell and Tube Heat Exchanger(2015)



http://www.ejbowman.co.uk/products/ShellandTubeHeatExchangers.html Yunus A.Cengel, 2006, Heat and Mass Transfer: A Practical Approach. Mc Graw Hill,,



3rd Edition Yunus A. Cengel, Afshin J. Ghajar.(2011). Heat and Mass Transfer: Fundamentals &

Tube

Heat

Exchanger,

Applications Fourth Edition McGraw-Hill.

APPENDICES

by

amirhazwan,

Retrieved

retrieved

from

from