heat exchanger

UNIVERSITI TEKNOLOGI MARA FAKULTI KEJURUTERAAN KIMIA PROCESS ENGINEERING LABORATORY II

(CPE554)

NO Title Allocated Marks (%) Marks

1 Abstract/Summary 5 2 Introduction 5 3 Aims 5 4 Theory 5 5 Apparatus 5 6 Methodology/Procedure 10 7 Results 10 8 Calculations 10 9 Discussion 20 10 Conclusion 10 11 Recommendations 5 12 Reference 5 13 Appendix 5 TOTAL MARKS 100 Remarks :

Checked by : Rechecked by:

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---Date: Date:

NAME : MOHAMAD FAHMI BIN ABD RASED

STUDENT NO : 2013879302

GROUP : 3

EXPERIMENT : CONCENTRIC HEAT EXCHANGER DATE PERFORMED : 24 MARCH 2015

SEMESTER : 4 PROGRAMME / CODE : EH221

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 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.

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. Process liquid or gas cooling

2. Process or refrigerant vapor or steam condensing 3. Process liquid, steam or refrigerant evaporation 4. Process heat removal and preheating of feed water 5. Thermal energy conservation efforts, heat recovery

6. Compressor, turbine and engine cooling, oil and jacket water 7. Hydraulic and lube oil cooling

8. 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 H Q

and

C Q

and to select the set of values where

C Q

is closest to H Q

Hot water flow rate ( W H ) H Q = ) (t₁ t₂ Cp F_H  _H  

Hot water flow rate ( W C ) C Q = ) (T₂ T₁ Cp F_C _C   Where: H Q

= Heat load for hot water flow rate

C Q

= Heat load for cold water flow rate 

H

F

Hot water mass flow rate 

C F

Cold water mass flow rate 

1 t

Hot water inlet temperature 

2 t

Hot water outlet temperature 

1 T

 2 T

Cold water outlet temperature

LMTD

Calculations of log mean temperature difference (LMTD).

) ( ) ( ln ) ( ) ( 1 2 2 1 1 2 2 1 T t T t T t T t LMTD      

Where, all variables are same with the above section:

) ( ) ( 1 2 2 1 t t T t R      ) ( ) ( 1 1 1 2 t T t t S     

Both equations would determine the value of correction factorFT. Practically, FTvalue

obtained from the graph with respect to Rand S 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 D U

can be calculated by using equation

below. In this case, the value of total heat transfer area A has been given and equal to 31.0 ft2

FT LMTD A Q U    Where:  Q

Heat rate with respect to the average head load 

FT

Correction factor

Reynolds Number Calculation

 Shell-side ) Re(s for W C

 Gs De s) . Re(  Where: 12 de De do do PT PT de . 2 / 1 ) 4 . 2 / 1 86 . 0 2 / 1 ( 4 2      At which:  PT Pitch = 0.81inch  do

Tube outside diameter, inch 



Viscosity, taken at average fluid temperature in the shell, lbmft-1_hr-1

As Ws Gs (lbmft-2_hr-1₎  Ws Flow rate in (lbmhr-1₎  As 0.029 ft2

 Tube-side ) Re(t for W H  Gt D t) . Re(  Where:  D Tube ID = 0.04125 ft  

Viscosity, taken at average fluid temperature in the tube, lbmft-1_hr-1

At Wt Gt (lbmft-2_hr-1₎  Wt Flow rate in lbmhr-1  At 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.

W C

: 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: Pa O mmH bar Pa bar x ) 81 . 9 ( 1 1 10 1 . 2 5    . Where x is the calculated pressure value in unit bar.

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 (LPM) FT2 (LPM) TT1 (o_C) TT2 (o_C) TT3 (o_C) TT4 (o_C) DPT1 (mmH20) DPT2 (mmH20 10 2 40.0 31.6 44.9 49.1 32 79 10 4 39.6 31.5 46.4 48.7 38 80 10 6 36.8 31.1 45.9 49.5 73 81 10 8 35.6 30.5 45.8 49.4 123 86 10 10 34.8 30.1 45.6 49.4 138 95

ii. FT2 (COLD) Constant = 5 LPM FT1 (LPM) FT2 (LPM) TT1 (o_C) TT2 (o_C) TT3 (o_C) TT4 (o_C) DPT1 (mmH20) DPT2 (mmH20 2 10 31.6 30.2 41.6 48.7 155 8 4 10 32.6 30.4 43.4 49.3 153 9 6 10 33.2 30.5 43.8 48.9 150 5 8 10 34.5 30.6 45.2 49.4 149 24 10 10 35.3 30.7 45.0 48.9 147 28 CALCULATIONS

Experiment A: Counter-Current Flow

Hot Water Density: Heat Capacity: Thermal cond: Viscosity: 988.18 kg/m3 4175.00 J/kg.K 0.6436 W/m.K 0.0005494 Pa.s Cold Water Density: Heat Capacity: Thermal cond: Viscosity: 995.67 kg/m3 4183.00 J/kg.K 0.6155 W/m.K 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_hC_p∆ T =10.0 L min× 1m3 1000 L× 1min 60 s ×988.18 kg m3×4175 J kg ∙℃× (49.1−44.9)℃=2887.96 W Q_cold(W )=m_hC_p∆ T =2.0 L min× 1 m3 1000 L× 1 min 60 s × 995.67 kg m3× 4183 J kg ∙℃×(40.0−31.6)℃=1166.17 W

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

ε= Q Q_max= 1166.17 2887.96×100 =40.38 2) Q_hot(W )=m_hC_p∆=10.0 L min× 1 m3 1000 L× 1 min 60 s ×988.18 kg m3× 4175 J kg ∙℃×(48.7−46.4)℃=1581.50 W

Q_cold(W )=m_hC_p∆ T =4.0 L min× 1 m3 1000 L× 1 min 60 s × 995.67 kg m3× 4183 J kg ∙℃× (39.6−31.5)℃=2249.04 W

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

ε= Q Q_max= 1581.50 2249.04×100 =70.32 3) Q_hot(W )=m_hC_p∆=10.0 L min× 1 m3 1000 L× 1 min 60 s ×988.18 kg m3× 4175 J kg ∙℃× (49 .5−45.9)℃=2475.39 W Qcold(W )=mhCp∆=6.0 L min× 1m3 1000 L× 1min 60 s ×995.67 kg m3× 4183 J kg ∙℃× (36.8−31.1)℃=2373.99W

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

ε= Q

Q_max=

2373.9 9

2475.39 ×100 =95.88

Q_hot(W )=m_hC_p∆=10.0 L min× 1 m3 1000 L× 1 min 60 s ×988.18 kg m3× 4175 J kg ∙℃× (49.4−45.8 )℃=2475.39W Q_cold(W )=m_hC_p∆ T =8.0 L min× 1 m3 1000 L× 1 min 60 s × 995.67 kg m3× 4183 J kg ∙℃×(35.6−30.5)℃=2832.12 W

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

ε= Q Q_max= 2475.39 2832.12×100 =8 7.40 5) Q_hot(W )=m_hC_p∆=10.0 L min× 1 m3 1000 L× 1 min 60 s ×988.18 kg m3× 4175 J kg ∙℃× (49.4−45.6 )℃=2612.91W Q_cold(W )=m_hC_p∆ T =10.0 L min× 1m3 1000 L× 1 min 60 s ×995.67 kg m3× 4183 J kg ∙℃×(34.8−30.1 )℃=3262.50W

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

ε= Q

Q_max=

2612.91

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 )] ln ⁡[(49.1−40.0 ) (44.9−31.6)] =11.07℃ 2) ∆ T_lm=[(48.7−39.6)−( 46.4−31.5)] ln ⁡[(48.7−39.6 ) (46.4−31.5)] =11.76℃ 3) ∆ T_lm=[(49.5−36.8)−(45.9−31.1)] ln ⁡[(49.5−36.8) (45.9−31.1)] =1 3.72℃

4) ∆ Tlm= [(49.4−35.6)−(45.8−30.5)] ln ⁡[(49.4−3 5.6 ) (4 5.8−30.5 )] =1 4.54℃ 5) ∆ T_lm=[(49.4−34.8)−( 45.6−30.1)] ln ⁡[(49.4−34.8 ) (45.6−30.1)] =15.04℃

3. Calculate of the tube and shell heat transfer coefficient

´ ´ V =10 L min× 1 m3 1000 L× 1 min 60 s =1.67 ×10 −4m3 s A=π d 2 4 = π ×(0.02664)2 4 =0.000557 m 2 v =V´ A= 1.67 ×10−4 0.000557 =0.299 m s ℜ=ρvd μ = 988.18kg m3× 0.299 m s × 0.02664 m 0.0005494 Pa ∙ s =14327 (turbulent flow) Pr=μ Cp k = (0.0005494 Pa ∙ s )×(4175 J kg ∙ K) 0.6436 W m ∙ K =3.564 Nu=0.023 × ℜ0.8_{× Pr}0.33_{=0.023 ×14327}0.8_×3.5640.33_=73.55 h=Nuk d = 73.55 × 0.6436 W m∙ K 0.02664 m =1776.91 W m2∙ K

For (2 LPM) ´ ´ V =2 L min× 1m3 1000 L× 1 min 60 s =3.33× 10 −5m3 s 0.085 ¿ (¿¿2−(0.0334 )2) ¿ π ׿ A=π (ds 2 −do2) 4 =¿ v =V´ A= 3.33 ×10−5 0.0048 =0.0069 m s ℜ=ρv

(

ds−do

)

μ = 955.67kg m3× 0.0069 m s × (0.085−0.0334 m) 0.0008007 Pa ∙ s ¿425 (laminar flow ) Pr=μ Cp k = (0.0008007 Pa ∙ s )×(4183 J kg ∙ K) 0.6155 W m∙ K =5.49 Nu=0.023 × ℜ0.8× Pr0.4=0.023 × 4250.8×5.490.4=5.76

h=Nuk d = 5.76 ×0.6155 W m∙ K (0.085 m−0.0334 m)=68.68 W m2_{∙ K} At shell side : ( 4 LPM ) ´ ´ V =4 L min× 1m3 1000 L× 1 min 60 s =6.67 ×10 −5m3 s 0.085 ¿ (¿¿2−(0.0334 )2) ¿ π ׿ A=π (ds 2 −do 2 ) 4 =¿ v =V´ A= 6.67 ×10−5 0.0048 =0.0139 m s ℜ=ρv

(

ds−do

)

μ = 955.67kg m3× 0.0139 m s × (0.085−0.0334 m) 0.0008007 Pa ∙ s ¿856 (laminar flow ) Pr=μ Cp k = (0.0008007 Pa ∙ s )×(4183 J kg ∙ K) 0.6155 W m∙ K =5.49

Nu=0.023 × ℜ0.8× Pr0.4=0.023 × 8560.8×5.490.4=10.80 h=Nuk d = 10.80 ×0.6155 W m∙ K (0.085 m−0.0334 m)=120.26 W m2∙ K At shell side : ( 6 LPM) ´ ´ V =6 L min× 1 m3 1000 L× 1 min 60 s =1 ×10 −4m3 s 0.085 ¿ (¿¿2−(0.0334 )2) ¿ π ׿ A=π (ds 2 −do2) 4 =¿ v =V´ A= 1 ×10−4 0.0048 =0.0208 m s

ℜ=ρv

(

ds−do

)

μ = 955.67kg m3× 0.0208 m s × (0.085−0.0334 ) 0.0008007 Pa ∙ s ¿1281(laminar flow) Pr=μ Cp k = (0.0008007 Pa ∙ s )×(4183 J kg ∙ K) 0.6155 W m∙ K =5.49 Nu=0.023 × ℜ0.8× Pr0.4=0.023 ×12810.8×5.490.4=13.91 h=Nuk d = 12.35 ×0.6155 W m∙ K (0.085 m−0.0334 m)=166.03 W m2∙ K At shell side : ( 8 LPM) ´ ´ V =8 L min× 1 m3 1000 L× 1 min 60 s =1.333 ×10 −4m3 s

0.085 ¿ (¿¿2−(0.0334 )2) ¿ π ׿ A=π (ds 2 −do 2 ) 4 =¿ v =V´ A= 1.333 ×10−4 0.0048 =0.0278 m s ℜ=ρv

(

ds−do

)

μ = 955.67kg m3× 0.0278 m s × (0.085−0.0334 ) 0.0008007 Pa ∙ s ¿1712(laminar flow) Pr=μ Cp k = (0.0008007 Pa ∙ s )×(4183 J kg ∙ K) 0.6155 W m∙ K =5.49 Nu=0.023 × ℜ0.8× Pr0.4=0.023 ×17120.8×5.490.4=17.55 h=Nuk d = 17.55 ×0.6155 W m∙ K (0.085 m−0.0334 m)=209.38 W m2∙ K

At shell side : ( 10 LPM) ´ ´ V =10 L min× 1 m3 1000 L× 1 min 60 s =1.667 ×10 −4m3 s 0.085 ¿ (¿¿2−(0.0334 )2) ¿ π ׿ A=π (ds 2 −do2) 4 =¿ v =V´ A= 1.667 ×10−4 0.0048 =0.0347 m s ℜ=ρv

(

ds−do

)

μ = 955.67kg m3× 0.0347 m s × (0.085−0.0334 ) 0.0008007 Pa∙ s ¿2137 (laminar flow) Pr=μ Cp k = (0.0008007 Pa ∙ s )×(4183 J kg ∙ K) 0.6155 W m∙ K =5.49 Nu=0.023 × ℜ0.8× Pr0.4=0.023 ×21370.8×5.490.4=20.96

h=Nuk d = 20.96 ×0.6155 W m∙ K (0.085 m−0.0334 m)=250.02 W m2_{∙ K}

Overall heat transfer coefficient:

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

1. U= Qhot A ∆ Tlm = 2887.96W 0.05 m2× 11.07℃=5 217.63 W m2∙ K 2. U= Qhot A ∆ Tlm = 1581.50W 0.05 m2× 11.76℃=2689.63 W m2∙ K

3. U= Qhot A ∆ Tlm = 2475.39W 0.05 m2× 13.72℃=3608.44 W m2∙ K 4. U= Qhot A ∆ Tlm = 2475.39 W 0.05 m2_{× 1 4.54℃}=3404.94 W m2_{∙ K} 5. 11U= Qhot A ∆ Tlm = 2612.91W 0.05 m2_{×1 5.04℃}=3474.61 W m2_{∙ 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 Re<2100, the flow is laminar flow and for Re>4200, 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.

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-HEAT-EXCHANGER