Please copy and paste this embed script to where you want to embed

UNIVERSITI TEKNOLOGI MARA FAKULTI KEJURUTERAAN KIMIA PROCESS ENGINEERING LABORATORY II (CPE554) NAME STUDENT NO GROUP EXPERIMENT DATE PERFORMED SEMESTER PROGRAMME / CODE SUBMIT TO NO 1 2 3 4 5 6 7 8 9 10 11 12 13

: MOHAMAD FAHMI BIN ABD RASED : 2013879302 :3 : CONCENTRIC HEAT EXCHANGER : 24 MARCH 2015 :4 : EH221 : MISS HABSAH ALWI

Title Abstract/Summary Introduction Aims Theory Apparatus Methodology/Procedure Results Calculations Discussion Conclusion Recommendations Reference Appendix TOTAL MARKS Remarks :

Allocated Marks (%) 5 5 5 5 5 10 10 10 20 10 5 5 5 100

Marks

Checked by :

Rechecked by:

------------------------------

------------------------------

Date:

Date:

ABSTRACT/SUMMARY Heat exchanger is a device that allows heat from a fluid (a liquid or a gas) to pass to a second fluid (another liquid or gas) without the two fluids having to mix together or come into direct contact. Thus, a device named Heat Exchanger Training Apparatus (Model; HE 158C) was used to conduct Shell and Tube Heat Exchanger experiment. The objectives of this experiment are to evaluate and study the heat load and head balance, LMTD and overall heat transfer coefficient, to calculate the Reynolds numbers at the shell and tubes sides and to measure and determine the shell and tube sides pressure drop. To do that, we vary the hot water and cold water flow rates and record the inlet and outlet temperatures of both the hot water and cold water streams at steady state. The flow of hot and cold water is counter-current flow. This experiment consists of five runs. For each of the run, three sets of data are obtained. A set of data from each of the run is selected based on the best convergence of QC and QH (the ratio of QC/QH is nearest to 1.0). INTRODUCTION Heat exchanger is a device that transfers heat from one fluid to another or from or to a fluid and the environment. These are a few types of shell and tube heat exchanger.

Figure 1: Heat exchanger with fixed tube plates (four tubes, one shell-pass)

Figure 2: Heat exchanger with floating head (two tube-pass, one shell pass)

Figure 3: Heat exchanger with hairpin tubes

Shells in the device are used to transport cold water while tubes are used to transport hot water across the device. Baffles are used in the heat exchanger to support the tubes and allow water to flow across the tubes other than providing a higher transfer rate due to increase of turbulence. Shell and tube heat exchangers have the ability to transfer large amounts of heat in relatively low cost, serviceable designs. They can provide large amounts of effective tube surface while minimizing the requirements of floor space, liquid volume and weight.

There are a few considerations of mechanical arrangement in the heat exchanger need to be made. This is important because different arrangement gives different efficiency and practicality. The four basic considerations are: 1. Methods of controlling fluid flow through the shell. 2. Consideration for differential thermal expansion of tube and shell. 3. Consideration for ease of maintenance and servicing. 4. Means of directing fluid through the tubes.

Applications of Heat Exchanger Shell and tube heat exchangers represent the most widely used vehicle for the transfer of heat in industrial process applications. They are frequently selected for such duties as:

1. 2. 3. 4. 5. 6. 7. 8.

Process liquid or gas cooling Process or refrigerant vapor or steam condensing Process liquid, steam or refrigerant evaporation Process heat removal and preheating of feed water Thermal energy conservation efforts, heat recovery Compressor, turbine and engine cooling, oil and jacket water Hydraulic and lube oil cooling Many other industrial applications

Advantages of Heat Exchanger The

main

advantages

of

shell-and-tube

heat

exchangers

are:

1. Condensation or boiling heat transfer can be accommodated in either the tubes or the shell, and the orientation can be horizontal or vertical. 2. The pressures and pressure drops can be varied over a wide range. 3. Thermal stresses can be accommodated inexpensively. 4. There is substantial flexibility regarding materials of construction to accommodate corrosion and other concerns. The shell and the tubes can be made of different materials. 5. Extended heat transfer surfaces (fins) can be used to enhance heat transfer. 6. Cleaning and repair are relatively straightforward, because the equipment can be dismantled for this purpose.

AIMS 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

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

is closest to

QH

.

QH

and

HW Hot water flow rate (

) QH

=

FH Cp H (t1 t 2 )

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

T1

Hot water outlet temperature

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

and

S

FT

. Practically,

FT

value

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

Overall heat transfer coefficient,

U

Overall heat transfer coefficient at which equivalent to

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

U

UD

A

has been given and equal to 31.0 ft2

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

can be calculated by using equation

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 mmH 2O. 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:

1 105 Pa 1mmH2O x.bar 1bar (9.81) Pa . Where x is the calculated pressure value in 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 leave 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 is 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. 10. The next experiment is proceed.

RESULTS Experiment A: Counter-current Flow i.

FT1 (HOT) Constant=5 LPM

FT1

FT2

TT1

TT2

TT3

TT4

DPT1

DPT2

(LPM) 10 10 10 10 10

(LPM) 2 4 6 8 10

(oC) 40.0 39.6 36.8 35.6 34.8

(oC) 31.6 31.5 31.1 30.5 30.1

(oC) 44.9 46.4 45.9 45.8 45.6

(oC) 49.1 48.7 49.5 49.4 49.4

(mmH20) 32 38 73 123 138

(mmH20 79 80 81 86 95

ii.

FT2 (COLD) Constant = 5 LPM FT1

FT2

TT1

TT2

TT3

TT4

DPT1

DPT2

(LPM) 2 4 6 8 10

(LPM) 10 10 10 10 10

(oC) 31.6 32.6 33.2 34.5 35.3

(oC) 30.2 30.4 30.5 30.6 30.7

(oC) 41.6 43.4 43.8 45.2 45.0

(oC) 48.7 49.3 48.9 49.4 48.9

(mmH20) 155 153 150 149 147

(mmH20 8 9 5 24 28

CALCULATIONS Experiment A: Counter-Current Flow Hot Water Density:

988.18 kg/m3

Heat Capacity:

4175.00 J/kg.K

Thermal cond:

0.6436 W/m.K

Viscosity: Cold Water

0.0005494 Pa.s

Density:

995.67 kg/m3

Heat Capacity:

4183.00 J/kg.K

Thermal cond:

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

L 1m 3 1min kg J × × ×988.18 3 ×4175 × ( 49.1−44.9 ) ℃=2887.96 W min 1000 L 60 s kg ∙℃ m

3

Qcold ( W )=mh C p ∆ T =2.0

L 1m 1 min kg J × × × 995.67 3 × 4183 × ( 40.0−31.6 ) ℃=1166.17 W min 1000 L 60 s kg ∙℃ m

Heat Lost Rate=Qhot −Qcold =( 2887.96−1166.17 ) W =1721.79 W

ε=

Q 1166.17 = ×100 =40.38 Qmax 2887.96

2)

Q hot (W )=mh C p ∆=10.0

L 1 m3 1 min kg J × × ×988.18 3 × 4175 ×(48.7−46.4)℃=1581.50 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 × (39.6−31.5 ) ℃=2249.04 W min 1000 L 60 s kg ∙℃ m

Heat Lost Rate=Qhot −Qcold =( 1581.50−2249.04 ) W =667.54 W

ε=

Q 1581.50 = ×100 =70.32 Qmax 2249.04

3)

Qhot (W )=mh C p ∆=10.0

L 1 m3 1 min kg J × × ×988.18 3 × 4175 × ( 49 .5−45.9 ) ℃=2475.39 W min 1000 L 60 s kg ∙ ℃ m

L 1m 3 1min kg J ( ) Q cold W =mh C p ∆=6.0 × × ×995.67 3 × 4183 × ( 36.8−31.1 ) ℃=2373.99W min 1000 L 60 s kg ∙ ℃ m

Heat Lost Rate=Qhot −Qcold =( 2475.39−2373.99 ) W =101.4 W

ε=

4)

Q 2373.9 9 = ×100 =95.88 Qmax 2475.39

Q hot (W )=mh C p ∆=10.0

L 1 m3 1 min kg J × × ×988.18 3 × 4175 × ( 49.4−45.8 ) ℃=2475.39W min 1000 L 60 s kg ∙℃ m

3 L 1m 1 min kg J ( ) Qcold W =mh C p ∆ T =8.0 × × × 995.67 3 × 4183 × ( 35.6−30.5 ) ℃=2832.12 W min 1000 L 60 s kg ∙℃ m

Heat Lost Rate=Qhot −Qcold =( 2475.39−2832.12 ) W =356.73 W

ε=

Q 2475.39 = ×100 =8 7.40 Qmax 2832.12

5)

Q hot (W )=mh C p ∆=10.0

L 1 m3 1 min kg J × × ×988.18 3 × 4175 × ( 49.4−45.6 ) ℃=2612.91W 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 × ( 34.8−30.1 ) ℃=3262.50W min 1000 L 60 s kg ∙ ℃ m

Heat Lost Rate=Qhot −Qcold =( 2612.91−3262.50 ) W =649.59W

ε=

Q 2612.91 = ×100 =80.09 Qmax 3262.50

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=

[ ( 49.1−40.0 )−( 44.9−31.6 )] =11.07 ℃ ( 49.1−40.0 ) ln [ ] ( 44.9−31.6 )

2)

∆ T lm=

[ ( 48.7−39.6 )−( 46.4−31.5 ) ] =11.76 ℃ ( 48.7−39.6 ) ln [ ] ( 46.4−31.5 )

3)

∆ T lm=

[ ( 49.5−36.8 )−( 45.9−31.1 ) ] =1 3.72℃ ( 49.5−36.8 ) ln [ ] ( 45.9−31.1 )

4)

∆ T lm=

[ ( 49.4−35.6 )− ( 45.8−30.5 ) ] =1 4.54 ℃ ( 49.4−3 5.6 ) ln [ ] ( 4 5.8−30.5 )

5)

∆ T lm=

[ ( 49.4−34.8 )−( 45.6−30.1 )] =15.04 ℃ ( 49.4−34.8 ) ln [ ] ( 45.6−30.1 )

3. Calculate of the tube and shell heat transfer coefficient

0.8 0.33 At tube side (hot water-cooling process): Nu=0.023 × ℜ × Pr

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):

Nu=0.023 × ℜ0.8 × Pr0.4

For (2 LPM)

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

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

ρv ( d s−d o ) ℜ= = μ

955.67

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

¿ 425 ( laminar flow )

Pr=

μCp = k

( 0.0008007 Pa ∙ s ) ×(4183 0.6155

0.8

0.4

J ) kg ∙ K

W m∙ K

0.8

=5.49

0.4

Nu=0.023 × ℜ × Pr =0.023 × 425 ×5.49 =5.76

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

At shell side : ( 4 LPM ) L 1m 3 ´1 min m3 V´ =4 × × =6.67 ×10−5 min 1000 L 60 s s

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

V´ 6.67 ×10−5 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

0.8

0.4

0.8

0.4

Nu=0.023 × ℜ × Pr =0.023 × 856 ×5.49 =10.80

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

At 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

V´ 1 ×10−4 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

J ) kg ∙ K

W 0.6155 m∙ K

0.8

0.4

=5.49

0.8

0.4

Nu=0.023 × ℜ × Pr =0.023 ×1281 ×5.49 =13.91

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

At shell side : ( 8 LPM) L 1 m3 1´ min m3 V´ =8 × × =1.333 ×10−4 min 1000 L 60 s s

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

V´ 1.333 ×10−4 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 Nuk m∙ K W h= = =209.38 2 d (0.085 m−0.0334 m) m ∙K 17.55 ×0.6155

At 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

ρv ( d s−d o ) ℜ= = μ

955.67

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

¿ 2137 ( laminar flow )

Pr=

μCp = k

( 0.0008007 Pa ∙ s ) ×(4183 0.6155

0.8

0.4

J ) kg ∙ K

W m∙ K

0.8

=5.49

0.4

Nu=0.023 × ℜ × Pr =0.023 ×2137 ×5.49 =20.96

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

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 2887.96W W = =5 217.63 2 2 A ∆ T lm 0.05 m × 11.07 ℃ m ∙K

2.

U=

Qhot 1581.50W W = =2689.63 2 2 A ∆ T lm 0.05 m × 11.76 ℃ m ∙K

3.

U=

Qhot 2475.39W W = =3608.44 2 2 A ∆ T lm 0.05 m × 13.72℃ m ∙K

4.

U=

Qhot 2475.39 W W = =3404.94 2 A ∆ T lm 0.05 m 2 × 1 4.54 ℃ m ∙K

5.

11 U=

Qhot 2612.91W W = =3474.61 2 A ∆ T lm 0.05 m 2 ×1 5.04 ℃ m ∙K

DISCUSSION In this experiment, the objectives are to evaluate and study the heat load and head balance, LMTD and overall heat transfer coefficient, to calculate the Reynolds numbers at the shell and tubes sides and to measure and determine the shell and tube sides pressure drop. At the end of the experiments, all objectives are met although maybe there are some errors.

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. The most irrelevant data for QC/QH is in run 1, set 3 where the ratio is 2.11. The margin is big when compare to the ideal condition where QC/QH = 1.0. The irrelevant value of this ratio is maybe caused by the unstable conditions of shell and tube heat exchanger where this phenomenon occurs at the beginning of the experiment. 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. The overall heat transfer coefficients are also calculated in this experiment to determine the total thermal resistance to heat transfer between two fluids. The resistance can be reduced by increasing the surface area, which will lead to a more efficient heat exchanger The calculated Reynolds Number is to determine whether the flow of water in shell and tube heat exchanger is turbulent flow or laminar flow. After the Reynolds Number are obtained, we can determine whether the flow is turbulent or laminar as for Re4200, the flow is turbulent flow. For this experiment, based on the calculated results, the water flow is turbulent at the tube sides of heat exchanger as Reynolds Number that we obtained all exceeded 4200.

CONCLUSION In conclusion, shell and tube heat exchanger follows the basic law of Thermodynamics and fulfilled the study of Heat Transfer. Every objectives of this experiment had been achieved. Although there might be errors, the objectives of this experiment still can be achieved. In countercurrent flow configuration, the exit temperature of the hot fluid is also higher than the exit temperature of the cold fluid. 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. Furthermore, the amount of heat loss form the hot water is not equal to the heat gain by the cold water due to the heat loss to the surrounding.

RECOMMENDATIONS 1. Make sure that the equipment is in good condition so that the flow of the experiment does not disturb by the inconstant data. 2. Time taken to collect the data is punctually followed. 3. All the temperature and flow rate readings are taken simultaneously as CW

inlet

temperature

is

increasing

gradually

and

CW

outlet

temperature varies together with the HW inlet/outlet temperature. 4. The last set of temperature readings should be taken when all the temperatures are fairly steady. 5. While recording the data, make sure that the pressure and temperature is at constant value because this can affect the calculation made.

REFERENCES

1. Yunus A.Cengel, 2006, Heat and Mass Transfer: A Practical Approach. Mc Graw Hill,, 3rd Edition 2. Coulson and Richardson; Chemical Engineering; Volume 1, 6th edition. 3. Rase, Howard F; Chemical Reactor Design and for Process and plants; Volume 1; 1st edition. 4. CONCENTRIC TUBE HEAT EXCHANGER, by amirhazwan, Retrieved from https://www.scribd.com/doc/27156908/CONCENTRIC-TUBE-HEATEXCHANGER APPENDIX

View more...
: MOHAMAD FAHMI BIN ABD RASED : 2013879302 :3 : CONCENTRIC HEAT EXCHANGER : 24 MARCH 2015 :4 : EH221 : MISS HABSAH ALWI

Title Abstract/Summary Introduction Aims Theory Apparatus Methodology/Procedure Results Calculations Discussion Conclusion Recommendations Reference Appendix TOTAL MARKS Remarks :

Allocated Marks (%) 5 5 5 5 5 10 10 10 20 10 5 5 5 100

Marks

Checked by :

Rechecked by:

------------------------------

------------------------------

Date:

Date:

ABSTRACT/SUMMARY Heat exchanger is a device that allows heat from a fluid (a liquid or a gas) to pass to a second fluid (another liquid or gas) without the two fluids having to mix together or come into direct contact. Thus, a device named Heat Exchanger Training Apparatus (Model; HE 158C) was used to conduct Shell and Tube Heat Exchanger experiment. The objectives of this experiment are to evaluate and study the heat load and head balance, LMTD and overall heat transfer coefficient, to calculate the Reynolds numbers at the shell and tubes sides and to measure and determine the shell and tube sides pressure drop. To do that, we vary the hot water and cold water flow rates and record the inlet and outlet temperatures of both the hot water and cold water streams at steady state. The flow of hot and cold water is counter-current flow. This experiment consists of five runs. For each of the run, three sets of data are obtained. A set of data from each of the run is selected based on the best convergence of QC and QH (the ratio of QC/QH is nearest to 1.0). INTRODUCTION Heat exchanger is a device that transfers heat from one fluid to another or from or to a fluid and the environment. These are a few types of shell and tube heat exchanger.

Figure 1: Heat exchanger with fixed tube plates (four tubes, one shell-pass)

Figure 2: Heat exchanger with floating head (two tube-pass, one shell pass)

Figure 3: Heat exchanger with hairpin tubes

Shells in the device are used to transport cold water while tubes are used to transport hot water across the device. Baffles are used in the heat exchanger to support the tubes and allow water to flow across the tubes other than providing a higher transfer rate due to increase of turbulence. Shell and tube heat exchangers have the ability to transfer large amounts of heat in relatively low cost, serviceable designs. They can provide large amounts of effective tube surface while minimizing the requirements of floor space, liquid volume and weight.

There are a few considerations of mechanical arrangement in the heat exchanger need to be made. This is important because different arrangement gives different efficiency and practicality. The four basic considerations are: 1. Methods of controlling fluid flow through the shell. 2. Consideration for differential thermal expansion of tube and shell. 3. Consideration for ease of maintenance and servicing. 4. Means of directing fluid through the tubes.

Applications of Heat Exchanger Shell and tube heat exchangers represent the most widely used vehicle for the transfer of heat in industrial process applications. They are frequently selected for such duties as:

1. 2. 3. 4. 5. 6. 7. 8.

Process liquid or gas cooling Process or refrigerant vapor or steam condensing Process liquid, steam or refrigerant evaporation Process heat removal and preheating of feed water Thermal energy conservation efforts, heat recovery Compressor, turbine and engine cooling, oil and jacket water Hydraulic and lube oil cooling Many other industrial applications

Advantages of Heat Exchanger The

main

advantages

of

shell-and-tube

heat

exchangers

are:

1. Condensation or boiling heat transfer can be accommodated in either the tubes or the shell, and the orientation can be horizontal or vertical. 2. The pressures and pressure drops can be varied over a wide range. 3. Thermal stresses can be accommodated inexpensively. 4. There is substantial flexibility regarding materials of construction to accommodate corrosion and other concerns. The shell and the tubes can be made of different materials. 5. Extended heat transfer surfaces (fins) can be used to enhance heat transfer. 6. Cleaning and repair are relatively straightforward, because the equipment can be dismantled for this purpose.

AIMS 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

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

is closest to

QH

.

QH

and

HW Hot water flow rate (

) QH

=

FH Cp H (t1 t 2 )

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

T1

Hot water outlet temperature

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

and

S

FT

. Practically,

FT

value

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

Overall heat transfer coefficient,

U

Overall heat transfer coefficient at which equivalent to

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

U

UD

A

has been given and equal to 31.0 ft2

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

can be calculated by using equation

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 mmH 2O. 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:

1 105 Pa 1mmH2O x.bar 1bar (9.81) Pa . Where x is the calculated pressure value in 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 leave 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 is 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. 10. The next experiment is proceed.

RESULTS Experiment A: Counter-current Flow i.

FT1 (HOT) Constant=5 LPM

FT1

FT2

TT1

TT2

TT3

TT4

DPT1

DPT2

(LPM) 10 10 10 10 10

(LPM) 2 4 6 8 10

(oC) 40.0 39.6 36.8 35.6 34.8

(oC) 31.6 31.5 31.1 30.5 30.1

(oC) 44.9 46.4 45.9 45.8 45.6

(oC) 49.1 48.7 49.5 49.4 49.4

(mmH20) 32 38 73 123 138

(mmH20 79 80 81 86 95

ii.

FT2 (COLD) Constant = 5 LPM FT1

FT2

TT1

TT2

TT3

TT4

DPT1

DPT2

(LPM) 2 4 6 8 10

(LPM) 10 10 10 10 10

(oC) 31.6 32.6 33.2 34.5 35.3

(oC) 30.2 30.4 30.5 30.6 30.7

(oC) 41.6 43.4 43.8 45.2 45.0

(oC) 48.7 49.3 48.9 49.4 48.9

(mmH20) 155 153 150 149 147

(mmH20 8 9 5 24 28

CALCULATIONS Experiment A: Counter-Current Flow Hot Water Density:

988.18 kg/m3

Heat Capacity:

4175.00 J/kg.K

Thermal cond:

0.6436 W/m.K

Viscosity: Cold Water

0.0005494 Pa.s

Density:

995.67 kg/m3

Heat Capacity:

4183.00 J/kg.K

Thermal cond:

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

L 1m 3 1min kg J × × ×988.18 3 ×4175 × ( 49.1−44.9 ) ℃=2887.96 W min 1000 L 60 s kg ∙℃ m

3

Qcold ( W )=mh C p ∆ T =2.0

L 1m 1 min kg J × × × 995.67 3 × 4183 × ( 40.0−31.6 ) ℃=1166.17 W min 1000 L 60 s kg ∙℃ m

Heat Lost Rate=Qhot −Qcold =( 2887.96−1166.17 ) W =1721.79 W

ε=

Q 1166.17 = ×100 =40.38 Qmax 2887.96

2)

Q hot (W )=mh C p ∆=10.0

L 1 m3 1 min kg J × × ×988.18 3 × 4175 ×(48.7−46.4)℃=1581.50 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 × (39.6−31.5 ) ℃=2249.04 W min 1000 L 60 s kg ∙℃ m

Heat Lost Rate=Qhot −Qcold =( 1581.50−2249.04 ) W =667.54 W

ε=

Q 1581.50 = ×100 =70.32 Qmax 2249.04

3)

Qhot (W )=mh C p ∆=10.0

L 1 m3 1 min kg J × × ×988.18 3 × 4175 × ( 49 .5−45.9 ) ℃=2475.39 W min 1000 L 60 s kg ∙ ℃ m

L 1m 3 1min kg J ( ) Q cold W =mh C p ∆=6.0 × × ×995.67 3 × 4183 × ( 36.8−31.1 ) ℃=2373.99W min 1000 L 60 s kg ∙ ℃ m

Heat Lost Rate=Qhot −Qcold =( 2475.39−2373.99 ) W =101.4 W

ε=

4)

Q 2373.9 9 = ×100 =95.88 Qmax 2475.39

Q hot (W )=mh C p ∆=10.0

L 1 m3 1 min kg J × × ×988.18 3 × 4175 × ( 49.4−45.8 ) ℃=2475.39W min 1000 L 60 s kg ∙℃ m

3 L 1m 1 min kg J ( ) Qcold W =mh C p ∆ T =8.0 × × × 995.67 3 × 4183 × ( 35.6−30.5 ) ℃=2832.12 W min 1000 L 60 s kg ∙℃ m

Heat Lost Rate=Qhot −Qcold =( 2475.39−2832.12 ) W =356.73 W

ε=

Q 2475.39 = ×100 =8 7.40 Qmax 2832.12

5)

Q hot (W )=mh C p ∆=10.0

L 1 m3 1 min kg J × × ×988.18 3 × 4175 × ( 49.4−45.6 ) ℃=2612.91W 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 × ( 34.8−30.1 ) ℃=3262.50W min 1000 L 60 s kg ∙ ℃ m

Heat Lost Rate=Qhot −Qcold =( 2612.91−3262.50 ) W =649.59W

ε=

Q 2612.91 = ×100 =80.09 Qmax 3262.50

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=

[ ( 49.1−40.0 )−( 44.9−31.6 )] =11.07 ℃ ( 49.1−40.0 ) ln [ ] ( 44.9−31.6 )

2)

∆ T lm=

[ ( 48.7−39.6 )−( 46.4−31.5 ) ] =11.76 ℃ ( 48.7−39.6 ) ln [ ] ( 46.4−31.5 )

3)

∆ T lm=

[ ( 49.5−36.8 )−( 45.9−31.1 ) ] =1 3.72℃ ( 49.5−36.8 ) ln [ ] ( 45.9−31.1 )

4)

∆ T lm=

[ ( 49.4−35.6 )− ( 45.8−30.5 ) ] =1 4.54 ℃ ( 49.4−3 5.6 ) ln [ ] ( 4 5.8−30.5 )

5)

∆ T lm=

[ ( 49.4−34.8 )−( 45.6−30.1 )] =15.04 ℃ ( 49.4−34.8 ) ln [ ] ( 45.6−30.1 )

3. Calculate of the tube and shell heat transfer coefficient

0.8 0.33 At tube side (hot water-cooling process): Nu=0.023 × ℜ × Pr

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):

Nu=0.023 × ℜ0.8 × Pr0.4

For (2 LPM)

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

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

ρv ( d s−d o ) ℜ= = μ

955.67

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

¿ 425 ( laminar flow )

Pr=

μCp = k

( 0.0008007 Pa ∙ s ) ×(4183 0.6155

0.8

0.4

J ) kg ∙ K

W m∙ K

0.8

=5.49

0.4

Nu=0.023 × ℜ × Pr =0.023 × 425 ×5.49 =5.76

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

At shell side : ( 4 LPM ) L 1m 3 ´1 min m3 V´ =4 × × =6.67 ×10−5 min 1000 L 60 s s

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

V´ 6.67 ×10−5 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

0.8

0.4

0.8

0.4

Nu=0.023 × ℜ × Pr =0.023 × 856 ×5.49 =10.80

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

At 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

V´ 1 ×10−4 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

J ) kg ∙ K

W 0.6155 m∙ K

0.8

0.4

=5.49

0.8

0.4

Nu=0.023 × ℜ × Pr =0.023 ×1281 ×5.49 =13.91

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

At shell side : ( 8 LPM) L 1 m3 1´ min m3 V´ =8 × × =1.333 ×10−4 min 1000 L 60 s s

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

V´ 1.333 ×10−4 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 Nuk m∙ K W h= = =209.38 2 d (0.085 m−0.0334 m) m ∙K 17.55 ×0.6155

At 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

ρv ( d s−d o ) ℜ= = μ

955.67

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

¿ 2137 ( laminar flow )

Pr=

μCp = k

( 0.0008007 Pa ∙ s ) ×(4183 0.6155

0.8

0.4

J ) kg ∙ K

W m∙ K

0.8

=5.49

0.4

Nu=0.023 × ℜ × Pr =0.023 ×2137 ×5.49 =20.96

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

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 2887.96W W = =5 217.63 2 2 A ∆ T lm 0.05 m × 11.07 ℃ m ∙K

2.

U=

Qhot 1581.50W W = =2689.63 2 2 A ∆ T lm 0.05 m × 11.76 ℃ m ∙K

3.

U=

Qhot 2475.39W W = =3608.44 2 2 A ∆ T lm 0.05 m × 13.72℃ m ∙K

4.

U=

Qhot 2475.39 W W = =3404.94 2 A ∆ T lm 0.05 m 2 × 1 4.54 ℃ m ∙K

5.

11 U=

Qhot 2612.91W W = =3474.61 2 A ∆ T lm 0.05 m 2 ×1 5.04 ℃ m ∙K

DISCUSSION In this experiment, the objectives are to evaluate and study the heat load and head balance, LMTD and overall heat transfer coefficient, to calculate the Reynolds numbers at the shell and tubes sides and to measure and determine the shell and tube sides pressure drop. At the end of the experiments, all objectives are met although maybe there are some errors.

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. The most irrelevant data for QC/QH is in run 1, set 3 where the ratio is 2.11. The margin is big when compare to the ideal condition where QC/QH = 1.0. The irrelevant value of this ratio is maybe caused by the unstable conditions of shell and tube heat exchanger where this phenomenon occurs at the beginning of the experiment. 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. The overall heat transfer coefficients are also calculated in this experiment to determine the total thermal resistance to heat transfer between two fluids. The resistance can be reduced by increasing the surface area, which will lead to a more efficient heat exchanger The calculated Reynolds Number is to determine whether the flow of water in shell and tube heat exchanger is turbulent flow or laminar flow. After the Reynolds Number are obtained, we can determine whether the flow is turbulent or laminar as for Re4200, the flow is turbulent flow. For this experiment, based on the calculated results, the water flow is turbulent at the tube sides of heat exchanger as Reynolds Number that we obtained all exceeded 4200.

CONCLUSION In conclusion, shell and tube heat exchanger follows the basic law of Thermodynamics and fulfilled the study of Heat Transfer. Every objectives of this experiment had been achieved. Although there might be errors, the objectives of this experiment still can be achieved. In countercurrent flow configuration, the exit temperature of the hot fluid is also higher than the exit temperature of the cold fluid. 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. Furthermore, the amount of heat loss form the hot water is not equal to the heat gain by the cold water due to the heat loss to the surrounding.

RECOMMENDATIONS 1. Make sure that the equipment is in good condition so that the flow of the experiment does not disturb by the inconstant data. 2. Time taken to collect the data is punctually followed. 3. All the temperature and flow rate readings are taken simultaneously as CW

inlet

temperature

is

increasing

gradually

and

CW

outlet

temperature varies together with the HW inlet/outlet temperature. 4. The last set of temperature readings should be taken when all the temperatures are fairly steady. 5. While recording the data, make sure that the pressure and temperature is at constant value because this can affect the calculation made.

REFERENCES

1. Yunus A.Cengel, 2006, Heat and Mass Transfer: A Practical Approach. Mc Graw Hill,, 3rd Edition 2. Coulson and Richardson; Chemical Engineering; Volume 1, 6th edition. 3. Rase, Howard F; Chemical Reactor Design and for Process and plants; Volume 1; 1st edition. 4. CONCENTRIC TUBE HEAT EXCHANGER, by amirhazwan, Retrieved from https://www.scribd.com/doc/27156908/CONCENTRIC-TUBE-HEATEXCHANGER APPENDIX

Thank you for interesting in our services. We are a non-profit group that run this website to share documents. We need your help to maintenance this website.