Concentric Heat Exchanger

Abstract/Summary

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.

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). From the selected set, the heat load and head balance, LMTD and

overall heat transfer coefficient, Reynolds numbers and pressure drop are determined. From

the results obtained, an ideal set to choose based on the ratio is majorly from set 3.

Introduction

A heat exchanger can be defined as any device that transfers heat from one fluid to another or

from or to a fluid and the environment. There are several 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

Basic Considerations in Choosing a Mechanical Arrangement of Heat Exchanger

There are four basic considerations in choosing a mechanical arrangement that provides for

efficient heat transfer between the two fluids or vapors, while taking care of such practical

matters as preventing leakage from one into the other. They are:

Consideration for differential thermal expansion of tube and shell

Means of directing fluid through the tubes

Methods of controlling fluid flow through the shell

Consideration for ease of maintenance and servicing

Advantages of Heat Exchanger

The main advantages of shell-and-tube heat exchangers are:

Condensation or boiling heat transfer can be accommodated in either the tubes or the

shell, and the orientation can be horizontal or vertical.

The pressures and pressure drops can be varied over a wide range.

Thermal stresses can be accommodated inexpensively.

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.

Extended heat transfer surfaces (fins) can be used to enhance heat transfer.

Cleaning and repair are relatively straightforward, because the equipment can be

dismantled for this purpose.

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:

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

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.

Aims/Objectives

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.

To measure and determine the shell and tube sides pressure drop.

Theory

Heat load and heat balance

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

to select the set of values where QC is closest toQH .

Hot water flow rate (HW )

QH = FH×CpH×(t1−t2 )

Hot water flow rate (CW )

QC = FC×CpC×(T 2−T1 )

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

T 1=Cold water inlet temperature

T 2= Cold water outlet temperature

LMTD

Calculations of log mean temperature difference (LMTD).

LMTD=( t1−T 2 )−( t2−T1 )

ln( t1−T 2)( t2−T 1)

Where, all variables are same with the above section:

R=−( t1−T 2)−( t2−t1 )

S=−( t2−t1 )−(T1−t1)

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

obtained from the graph with respect to R and 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 U D 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

U= QA×LMTD×FT

Where:

Q= Heat rate with respect to the average head load

FT=Correction factor

Reynolds Number Calculation

Shell-side Re( s )for CW

Re( s )=De .Gsμ

Where:

De=de12

de=4 (1/2PT×0 .86 PT−1 /2 π .

do4

2

)

1/2π .do

At which:

PT=Pitch = 0.81inch

do=Tube outside diameter, inch

μ= Viscosity, taken at average fluid temperature in the shell, lbmft-1hr-1

Gs=WsAs (lbmft-2hr-1)

Ws= Flow rate in (lbmhr-1)

As= 0.029 ft2

Tube-side Re( t ) for HW

Re( t )=D .Gtμ

Where:

D= Tube ID = 0.04125 ft

μ= Viscosity, taken at average fluid temperature in the tube, lbmft-1hr-1

Gt=WtAt (lbmft-2hr-1)

Wt= Flow rate in lbmhr-1

At= 0.02139 ft2

Pressure drop

This part would determine the following:

HW : 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×1×105Pa

1bar×

1mmH 2O

(9 . 81)Pa .

Where x is the calculated pressure value in unit bar.

6) 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.

Experiment 2: Co-Current Concentric Heat Exchanger

1. The valves is switched to Co-Current Concentric Heat Exchanger arrangement.

2. The pumps P1 and P2 is switched on.

3. The valves V3 and V14 is opened and adjusted to obtain the desired flowrates for hot water and cold water streams.

4. The system is allowed to reach steady state for 10 minutes.

5. FT1, FT2,TT1, TT2, TT3 and TT4 is recorded.

6. The pressure drop measurement for shell-side and tube side is recorded for pressure drop studies.

7. Steps 4 to 7 is repeated for different combinations of flowrate FT1 and FT2 as in result sheet.

8. Pumps P1 and P2 is switched off after the experiment is completed.

9. The equipment is shut-down.

7) RESULT

FT 1 – Hot Water Flow rate

FT 2 – Cold Water Flow rate

TT 1 – Cold water Outlet temperature

TT 2 – Cold water Inlet temperature

TT 3 – Hot water Outlet temperature

TT4 - Hot water Inlet temperature

Experiment 1: Counter current

FT 1

(LPM)

FT 2

(LPM)

TT 1

(°C)

TT 2

(°C)

TT 3

(°C)

TT 4

(°C)

DPT 1

(mmH2O)

DPT 1

(mmH2O)

10.0 2.0 33.5 29.4 48.4 49.1 33 81

10.0 4.0 31.9 29.9 48.3 49.0 40 82

10.0 6.0 31.9 30.7 48.8 49.7 77 81

10.0 8.0 32.2 31.2 48.7 49.4 130 87

10.0 10.0 32.6 31.7 48.5 49.2 140 96

FT 1

(LPM)

FT 2

(LPM)

TT 1

(°C)

TT 2

(°C)

TT 3

(°C)

TT 4

(°C)

DPT 1

(mmH2O)

DPT 1

(mmH2O)

2.0 10.0 32.4 31.8 47.5 50.1 140 10

4.0 10.0 32.4 31.8 47.6 49.3 140 18

6.0 10.0 32.6 31.9 48.2 49.3 140 36

8.0 10.0 32.6 31.9 48.4 49.3 140 54

10.0 10.0 32.8 31.7 48.5 49.3 140 60

Experiment 2 : Co-Current Concentric Heat Exchanger

FT 1

(LPM)

FT 2

(LPM)

TT 1

(°C)

TT 2

(°C)

TT 3

(°C)

TT 4

(°C)

DPT 1

(mmH2O)

DPT 1

(mmH2O)

10.0 2.0 32.4 34.9 48.7 49.2 -5 81

10.0 4.0 32.4 34.1 48.9 49.8 -5 81

10.0 6.0 32.1 33.5 48.0 48.8 -5 83

10.0 8.0 31.8 33.0 48.7 49.5 90 82

10.0 10.0 31.7 32.7 48.4 49.2 113 83

FT 1

(LPM)

FT 2

(LPM)

TT 1

(°C)

TT 2

(°C)

TT 3

(°C)

TT 4

(°C)

DPT 1

(mmH2O)

DPT 1

(mmH2O)

2.0 10.0 31.7 32.3 46.4 49.2 137 10

4.0 10.0 31.8 32.4 48.1 50.1 133 18

6.0 10.0 31.8 32.6 47.7 49.2 137 35

8.0 10.0 31.8 32.7 47.9 49.0 136 54

10.0 10.0 31.8 32.6 47.7 48.8 136 83

Notes

FT 1 – Hot Water Flow rate TT 1 – Cold water Outlet temperature

FT 2 – Cold Water Flow rate TT 2 – Cold water Inlet temperature

TT 3 – Hot water Outlet temperature

TT4 - Hot water Inlet temperature

8) Calculation

Exp 1: 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)

Qhot (W )=mhC p∆T=10.0Lmin

×1m3

1000 L×

1min60 s

×988.18kgm3 ×4175

Jkg ∙℃

× (49.1−48.4 )℃=481.33W

Qcold (W )=mhC p∆T=2.0Lmin

×1m3

1000L×

1min60 s

×995.67kgm3 ×4183

Jkg ∙℃

× (33.5−29.4 )℃=563.84W

Heat Lost Rate=Qhot−Qcold=( 481.33−563.84 )W=82.51W

ε= QQmax

=481.33563.84

×100 %=85.37 %

2)

Qhot (W )=mhC p∆=10.0Lmin

×1m3

1000 L×

1min60 s

×988.18kgm3 ×4175

Jkg ∙℃

× (49.0−48.3 )℃=481.33W


×1m3

1000 L×

1min60 s

×995.67kgm3 ×4183

Jkg ∙℃

× (31.9−29.9 )℃=555.32W

Heat Los t Rate=Qhot−Qcold=( 481.33−555.32 )W=73.99W

ε= QQmax

=481.33555.32

×100 %=86.68 %

3)


×1m3

1000 L×

1min60 s

×988.18kgm3 ×4175

Jkg ∙℃

× (49.7−48.8 )℃=618.85W

Qcold (W )=mhC p∆=6.0Lmin

×1m3

1000 L×

1min60 s

×995.67kgm3 ×4183

Jkg ∙℃

× (31.9−30.7 )℃=499.79W

Heat Lost Rate=Qhot−Qcold=(618.85−499.79 )W=119.06W

ε= QQmax

=499.79618.85

×100 %=80.76 %

4)


×1m3

1000 L×

1min60 s

×988.18kgm3 ×4175

Jkg ∙℃

× (49.4−48.7 )℃=481.33W


×1m3

1000 L×

1min60 s

×995.67kgm3 ×4183

Jkg ∙℃

× (32.2−31.2 )℃=555.32W


ε= QQmax

=481.33555.32

×100 %=86.68 %

5)


×1m3

1000 L×

1min60 s

×988.18kgm3 ×4175

Jkg ∙℃

× (49.2−48.4 )℃=481.33W


×1m3

1000L×

1min60 s

×995.67kgm3 ×4183

Jkg ∙℃

× (32.6−31.7 )℃=624.73W


ε= QQmax

=481.33624.73

×100 %=77.05 %

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=[ (49.1−33.5 )−( 48.4−29.4 )]

ln [( 49.1−33.5 )(48.4−29.4 )

]=17.24℃

2)

∆T lm=[ (49.0−31.9 )− (48.3−29.9 )]

ln [(49.0−31.9 )(48.3−29.9 )

]=17.74℃

3)

∆T lm=[ (49.7−31.9 )− (48.8−30.7 )]

ln [(49.7−31.9 )(48.8−30.7 )

]=18.1℃

4)

∆T lm=[ (49.4−32.2 )−(48.7−31.2 )]

ln [(49.4−32.2 )(48.7−31.2 )

]=17.35℃

5)

∆T lm=[ (49.2−32.6 )−( 48.5−31.7 )]

ln [(49.2−32.6 )(48.5−31.7 )

]=16.70℃

3. Calculate of the tube and shell heat transfer coefficient

At tube side (hot water-cooling process):Nu=0.023×ℜ0.8×Pr0.33

˙V̇=10

Lmin

×1m3

1000 L×

1min60 s

=1.67×10−4 m3

s

A=π d2

4=π ×(0.02664)2

4=0.000557m2

v= V̇A

=1.67×10−4

0.000557=0.299

ms

ℜ= ρvdμ

=988.18

kg

m3×0.299

ms×0.02664m

0.0005494 Pa ∙ s=14327 (turbulent flow )

Pr=μC p

k=

(0.0005494 Pa ∙ s )×(4175J

kg ∙ K)

0.6436Wm∙ K

=3.564

Nu=0.023×ℜ0.8×Pr0.33=0.023×143270.8×3.5640.33=73.55

h=Nukd

=73.55×0.6436

Wm∙ K

0.02664m=1776.91

W

m2 ∙K

At shell side (cold water-heating process): Nu=0.023×ℜ0.8×Pr0.4

For (2 LPM)

˙V̇=2

Lmin

×1m3

1000 L×

1min60 s

=3.33×10−5 m3

s

A=π (ds

2−do2)

4=π × ¿¿

v= V̇A

=3.33×10−5

0.0048=0.0069

ms

ℜ=ρv (d s−do )

μ=

955.67kg

m3×0.0069

ms× (0.085−0.0334m )

0.0008007 Pa∙ s

¿425 ( laminar flow )

Pr=μC p

k=

(0.0008007 Pa∙ s )×(4183J

kg ∙K)

0.6155Wm∙ K

=5.49

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

h=Nukd

=5.76×0.6155

Wm∙ K

(0.085m−0.0334m)=68.68

W

m2 ∙ K

At shell side : ( 4 LPM)

˙V̇=4

Lmin

×1m3

1000 L×

1min60 s

=6.67×10−5 m3

s

A=π (ds

2−do2)

4=π × ¿¿

v= V̇A

=6.67×10−5

0.0048=0.0139

ms

ℜ=ρv (d s−do )

μ=

955.67kg

m3×0.0139

ms× (0.085−0.0334m )

0.0008007 Pa∙ s

¿856 (laminar flow )

Pr=μC p

k=

(0.0008007 Pa∙ s )×(4183J

kg ∙K)

0.6155Wm∙ K

=5.49

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

h=Nukd

=10.80×0.6155

Wm∙ K

(0.085m−0.0334m)=120.26

W

m2 ∙ K


˙V̇=6

L

m∈¿× 1m3

1000L×

1min60 s

=1×10−4 m3

s

¿

A=π (ds

2−do2)

4=π × ¿¿

v= V̇A

=1×10−4

0.0048=0.0208

ms

ℜ=ρv (d s−do )

μ=

955.67kg

m3×0.0208

ms× (0.085−0.0334 )

0.0008007 Pa ∙ s


Pr=μC p

k=

(0.0008007 Pa∙ s )×(4183J

kg ∙K)

0.6155Wm∙ K

=5.49

Nu=0.023×ℜ0.8×Pr0.4=0.023×12810.8×5.490.4=13.91

h=Nukd

=12.35×0.6155

Wm∙ K

(0.085m−0.0334m)=166.03

W

m2 ∙ K


˙V̇=8

Lmin

×1m3

1000L×

1min60 s

=1.333×10−4 m3

s

A=π (ds

2−do2)

4=π × ¿¿

v= V̇A

=1.333×10−4

0.0048=0.0278

ms

ℜ=ρv (d s−do )

μ=

955.67kg

m3×0.0278

ms× (0.085−0.0334 )

0.0008007 Pa ∙ s


Pr=μC p

k=

(0.0008007 Pa∙ s )×(4183J

kg ∙K)

0.6155Wm∙ K

=5.49

Nu=0.023×ℜ0.8×Pr0.4=0.023×17120.8×5.490.4=17.55

h=Nukd

=17.55×0.6155

Wm∙ K

(0.085m−0.0334m)=209.38

W

m2 ∙ K


˙V̇=10

Lmin

×1m3

1000 L×

1min60 s

=1.667×10−4 m3

s

A=π (ds

2−do2)

4=π × ¿¿

v= V̇A

=1.667×10−4

0.0048=0.0347

ms

ℜ=ρv (d s−do )

μ=

955.67kg

m3×0.0347

ms× (0.085−0.0334 )

0.0008007 Pa ∙ s

¿2137 ( laminar flow )

Pr=μC p

k=

(0.0008007 Pa∙ s )×(4183J

kg ∙K)

0.6155Wm∙ K

=5.49

Nu=0.023×ℜ0.8×Pr0.4=0.023×21370.8×5.490.4=20.96

h=Nukd

=20.96×0.6155

Wm ∙K

(0.085m−0.0334m)=250.02

W

m2 ∙K

Overall heat transfer coefficient:

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

1.

U=Qhot

A ∆T lm

= 481.33W0.05m2×17.24℃

=558.39W

m2∙ K

2.

U=Qhot

A ∆T lm

= 481.33W0.05m2×17.74℃

=542.65W

m2∙ K

3.

U=Qhot

A ∆T lm

= 681.85W0.05m2×18.1℃

=753.43W

m2 ∙K

4.

U=Qhot

A ∆T lm

= 481.33W0.05m2×17.35℃

=554.84W

m2∙ K

5.

U=Qhot

A ∆T lm

= 481.33W0.05m2×16.70℃

=576.44W

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, every objectives of this experiment had been achieved. Although there

might be errors, students still can achieve the objectives of this experiment. At the end of the

experiment, students are able to evaluate and study the heat load and head balance, LMTD and

overall heat transfer coefficient, as well as to calculate the Reynolds numbers at the shell and

tubes sides and also to measure and determine the shell and tube sides pressure drop.

Students also are able to learn the fundamentals of shell and tube heat exchanger, as well as

the applications and advantages of it. All the calculated data for this experiment can be referred

to the table in calculation section.

Recommendations

Follow safety regulations such as wearing a goggle, appropriate clothes, and gloves to

avoid any over-exposure to the substances which can be harmful.

All the temperature and flowrate readings are taken simultaneously as CW inlet

temperature is increasing gradually and CW outlet temperature varies together with the

HW inlet/outlet temperature.

The last set of temperature readings should be taken when all the temperatures are

fairly steady.

Whenever the annunciator TAH3 is activated during the course of the experiment, press

the red acknowledge button to silence the buzzer.

The first set of data must be taken right away after the process is started.

References

1. Coulson and Richardson; Chemical Engineering; Volume 1, 6th edition. 2. Max S. Peter & Klaus D. Timmerhaus; Plant Design and Economic for Chemical

Engineering; 4th edition; Page 576.3. Rase, Howard F; Chemical Reactor Design and for Process and plants; Volume 1; 1 st

edition.4. G.C DRYDEN; The Efficient Use of Energy; 1st edition.5. Frank P. Incropera, David P. DeWitt, 2002, Fundamental of Heat and Mass Transfer,

United State of America, 5th Edition, John Wiley & Sons, Inc.

Documents

Concentric Heat Exchanger