Double Pipe Heat Exchanger

MIDDLE EAST TECHNICAL UNIVERSITY

DEPARTMENT OF CHEMICAL

ENGINEERING

ChE 410, CHEMICAL ENGINEERING

LABORATORY II

“HEAT TRANSFER CHARACTERISTICS OF A

DOUBLE PIPE HEAT EXCHANGER”

Date of Experiment : 14.11.2004

Submission of Report: 23.12.2004

Submitted to : Dr.Görkem KIRBAŞ

Submitted by : TUE 12

Bener Yeginoğlu

Egemen Uçar

Onur Aksoy

Sinan Önceler

ANKARA, 2004

ABSTRACT

In double pipe heat exchanger experiment, it is desired to

investigate the effect of Reynold’s Number on the individual heat transfer

coefficients and to the performances of two different types of flow (co-

current and counter current) during the heat exchange in the double pipe

system.

During the experiment, firstly co-current flow is examined and

different temperatures at different positions in the pipes, are collected as

raw data. After that, the type of flow is changed by reversing the valves

and the temperatures are recorded again. At the same time, the flowrate

of the condensate which comes from the steam condenser, is recorded.

These raw data can be seen in appendix.

The main equipments are a steam condenser which supplies the hot

water for the double pipe exchanger which consists of a cold stream

flowing through the inner tube, and a hot stream flowing through the outer

one, and the control panel. Temperatures at the specified positions and

the valves can be obtained from the control panel.

Main results obtained from this experiment can be summarized as,

with the increasing pressure drop, mass flow rate, Reynold’s Number, log

mean temperature, individual heat transfer coeffcient and the turbulance

of the flow increase for the cold stream whereas all these parameters are

kept constant for hot stream as the pressure drop in the hot stream is kept

constant during the experiment. It is known that heat transfer is directly

proportional with the overall heat transfer coefficient and log mean

temperature. So, it can be concluded that with the increasing turbulance

of the flow, the amount of heat transfer increases. The quantitative data

can be obtained in “results and discussion” section.

2

TABLE OF CONTENTS

Page #

1.Nomenclature...............................................1

2.Introduction..................................................2

3.Experimental Methods....................................3

4.Results&Discussion.........................................5

5.Conclusions...................................................11

6.References....................................................12

7.Appendices....................................................13

3

NOMENCLATURE

ΔPC Pressure Difference in the cold stream

ΔPH Pressure difference in the hot stream

ΔTLM Log Mean Temperature Difference

U Overall heat transfer coefficient

Reh Reynolds Number of the hot stream

Rec Reynolds Number of the cold stream

Prh Prondtl Number of the hot stream

Prc Prondtl Number of the cold stream

Nuh Nusselt Number of the hot stream

Nuc Nusselt Number of the cold stream

THin Inlet Temperature of the hot stream

THout Outlet Temperature of the hot stream

TCin Inlet temperature of the cold stream

TCout Outlet temperature of the cold stream

Vcond Volume of the condansate

ρcond Density of the condansate

Mcond Mass flow rate of the condansate

Mc Mass flow rate of the cold stream

Mh Mass flow rate of the hot stream

THmean Mean temperature of the hot stream

TCmean Mean temperature of the cold stream

µC Viscosity of the cold stream

µH Viscosity of the hot stream

kh Thermal conductivity of hot stream

kc Thermal conductivity of cold stream

hh Heat transfer coefficient of hot stream

hc Heat transfer coefficient of cold stream

Q Heat

4

INTRODUCTION

Heat exchangers are classified in accordance to their flow

configurations, mixing and their construction. A double pipe type is used

both co- and counter- flow manner. This equipment is a simple closed type

heat exchanger, in which one pipe is placed in the other. The one used in

experiment has three passes.

The aims of the experiment are to observe effect of Reynolds

number on the individual heat transfer coefficient and compare the effect

of co-current and counter-current.

In the experimental procedure, both in co-current and counter-

current, the cold stream’s flow rate is changed where hot stream’s kept

constant in the turbulent region. Orifice pressure readings are used to

relate determine cold and hot stream flow rates with the calibration

equation formerly obtained. Also, condensate flow rate is measured to

obtain total energy used in the experiment as both energy transferred to

cold stream and energy lost. With these measurements, individual heat

transfer coefficients and performances of co-current and counter-current

flow arrangements.

As additional background, the double pipe heat exchanger is closed

type exchanger. In this type of exchanger, the hot and cold fluids do not

come into direct contact with each other. Theory of the equipment is that,

the energy exchanged flows from one fluid to the outer surface of the pipe

by forced convection, through the pipe wall by conduction, and then from

the inside surface of the pipe to the second fluid by forced convection. The

flow can either be co-current or counter current. The double pipe heat

exchanger is a simple but useful apparatus, because it can be constructed

from standard parts.

5

The understanding of this experiment is as follows, experience the

application of the heat transfer theory, make the necessary calculations,

see the effect of flow rate to the transferred heat and compare co-current

and counter-current flow in the aspect of performance.

EXPERIMENTAL METHODS

In the double pipe heat exchanger experiment, as told before, the

main aim is to investigate the effect of Reynolds number on the individual

heat transfer coefficients and to the performances of co-current and

counter current flow heat transfer in a double pipe heat exchanger.

While doing these, the double pipe heat exchanger that is

constructed in the Tarık Somer’s Lab (Unit Operations Lab) is used. There

are 3 double pipes in the heat exchanger. Each of the pipes is 3.15 meters

in length. The inner pipe dimesion is 1” Sch. 40 steel pipe and the outer

pipe dimension is 2” Sch. 40 steel pipe. The detailed diagram of the

equipment can be seen from the photograph that is taken in the

laboratory in the appendix.

Cold water passes in the inner pipe and the hot steam passes in the

outer pipe of the exchanger. The cold water at 15˚C approximately is

supplied from the universities own supplies and the hot steam at 62˚C

approximately is supplied from the heating system of the university are

entered to the equipment. Some small changes can be seen in the

temperatures of these utilities due to the changes in ambient conditions.

In the first part of the experiment, the equipment is set to co-current

flow. The flowrate of the cold water that enters to the exchanger is

changed by the valve that is located just prior of the inlet. Four separate

runs are made for different ΔP values (10,15,20 and 25). The manometer

readings for both orificemeters during each run at suitable time intervals

are taken. The mass flowrates for the streams are calculated by the

6

calibration equation which can be found in the sample calculations part.

The attainment for steady state is waited during each runtime after

changing the flow rate of the cold stream. The temperatures are changed

while reading them from the control panel. One decimal point can be seen

in the control panel so the steady-state is checked by waiting for the

fluctuations change only one decimal. Also the steam flowrate is

determined by measuring the flowrate of the condensate from the steam

heater.

In the second part of the experiment, the equipment is set to

counter current flow. The same procedure that is done for the co-current

flow is followed. The flowrates are set with the same ΔP values that are set

for the first part.

7

RESULTS and DISCUSSION

1. OBTAINING RAW DATA

At the beginning of the experiment, the type of the flow is

determined as co-current. After that, by changing the pressure drop

through the pipe that the cold stream flows, raw data is collected and

stated below in table 1. Later on the same procedure is followed to obtain

the raw of counter current flow and table 2 is formed.

Table 1. Effect of Pressure Drop on Temperature(co-current flow)

For Co-Current FlowTemp(°C) @

ΔP=10Temp(°C) @

ΔP=15Temp(°C) @

ΔP=20Temp(°C) @

ΔP=25T1 61,7 62,9 63 62,3T2 58 58,5 58,6 58,1T3 55 55,2 55,3 54,7T4 51,7 51,4 51,4 50,8T5 14,8 14,9 15 15,1T6 20,5 19,7 19,4 19,1T7 24,9 23,5 22,8 22,1T8 27,7 26,2 25,3 24,6

Table 2. Effect of Pressure Drop on Temperature (counter current flow)

For Counter Current Flow

Temp(°C) @ ΔP=10

Temp(°C) @ ΔP=15

Temp(°C) @ ΔP=20

Temp(°C) @ ΔP=25

T1 63,2 62,8 62,8 62,3

T2 59,8 59 58,8 58,2

T3 56,6 55,8 55,4 54,8

T4 52,1 51,1 50,6 50,2

T5 29,6 27,2 25,8 25

T6 25 23,3 22,3 21,7

T7 20,6 19,6 19 18,7

T8 15,5 15,1 15,3 15,6

8

These tables shows that as pressure drop increases through the

pipe, the temperature difference in the hot stream increases and the

temperature difference in the cold stream decreases.

9

Figure 2. Temperature profile vs. tube length graph for co-current

flow

Temperature Profile for P = Δ10

0

10

20

30

40

50

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70

0 2 4 6 8 10 12

Position (m)

Tem

peratu

re (

°C

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Position (m)

Te

mp

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(°C

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Position (m)

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10

Figure 3. Temperature profile vs. tube length graph for counter-current flow


0

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70

0 2 4 6 8 10 12

Position (m)

Tem

peratu

re (

°C

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0

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Position (m)

Tem

peratu

re (

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Position (m)

Tem

peratu

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0 2 4 6 8 10 12

Position (m)

Tem

peratu

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°C

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11

Hot streams are indicated as red lines in these figures while the cold

streams are indicated as dark blue. Temperature profiles in both of the

diagrams are just as they are expected. It can be summarized from these

diagrams that, the temperature of both the hot and cold streams in

counter current flow increase much more than they do in co-current flow.

3. CALCULATION OF INDIVIDUAL AND OVERALL HEAT

COEFFICIENTS

After the raw data for both type of flow is obtained, by using the

correlation M = 8.114 (P)0.51 , mass flow rates of cold and hot streams

are found. The mean temperatures are obtained due to Tm = (Tin + Tout)/2

for the streams. Physical properties of water(kc, c, Pr) are obtained

according to this mean temperature and Reynold’s Number (Re = 4M / ( *

Di * )) are found for cold and streams are determined. Then with the

appropriate ranges of these physical properties, Nusselt Number(Nu =

0.023 Re0.8 Pr0.4) is found to go further and find individual heat transfer

coefficients(h = Nu * k / Di).

Finally, overall heat transfer coefficient is found by using the

individual heat transfer coefficients and diameters of the pipes(U =

1/( 1/hh + Do,inner/( Di,inner *hc )) where subscripts h and o indicate hot and

cold streams respectively. The results of these parameters can be found at

the end of the “results and discussion” section in table 3.

4. HEAT FLUX CALCULATIONS

For each run, the inlet and outlet temperatures of the streams are

found from table1 and Tlm= ((Th,in - Tc,in )– (Th,out - Tc,out)) / ln((Th,in - Tc,in ) /

(Th,out - Tc,out)) correlation and the overall heat transfer coefficent (U) is

used, to obtain heat flux at the specific positions of the pipes. (q' = U

Tlm). This heat flux can also be calculated by using the mass flow rates (q’

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= M* Cpav*T) where subscript av indicates that average of inlet and outlet

temperatures is used for the heat capacity of the water. These results can

also be found in table3.

5. CALCULATION OF HEAT LOSS TO THE SURROUNDINGS

As it is mentioned earlier, heat fluxes of both cold and hot streams

can be found either by using (q' = U Tlm) or (q’ = M* Cpav*T). If qh’

indicates the heat flux of hot stream where qc’ indicates the heat flux of

cold stream, then heat loss to the surroundings can be found by q loss' = qh'

- qc' . After that percantage of the heat loss are found easily by

proportioning the heat loss to the heat flux of hot stream (qloss' / qh' *100).

Next step is the determination of the amount of heat loss. It can be

said that the heat required for heating the feed water to the heater from

room temperature to an higher temperature is equal to the heat released

from condensate minus the heat loss to the surroundings(Mh Cpav,h Th =

Mcond Hlatent - Qloss). The mass flow rate of the hot stream is found in earlier

calculations and Cp value is obtained at the average of the inlet and outlet

temperatures of the hot stream. Hlatent is obtained from the literature and

Mcond = Vcond / , where is the specific volume the water at 1000C and

found from literature again. Finally Qloss = Mcond Hlatent+ Mh Cpav,h Th

equation is used and amount of heat loss is determined. All of these

quantitative results can be seen in table 3 below.

Table 3. Whole Results of the experiment

Cocurrent Countercurrent

1.run 2.run 3.run 4.run 1.run 2.run 3.run 4.run

PC (cmHg) 10,000 15,000 20,000 25,000 10,000 15,000 20,000 25,000Ph (cmHg)

15,000 15,000 15,000 15,000 15,000 15,000 15,000 15,000T1(C) 61,700 62,900 63,000 62,300 63,200 62,800 62,800 62,300T2(C) 58,000 58,500 58,600 58,100 59,800 59,000 58,800 58,200T3(C) 55,000 55,200 55,300 54,700 56,600 55,800 55,400 54,800

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T4(C) 51,700 51,400 51,400 50,800 52,100 51,100 50,600 50,200T5(C) 14,800 14,900 15,000 15,100 29,600 27,200 25,800 25,000T6(C) 20,500 19,700 19,400 19,100 25,000 23,300 22,300 21,700T7(C) 24,900 23,500 22,800 22,100 20,600 19,600 19,000 18,700T8(C) 27,700 26,200 25,300 24,600 15,500 15,100 15,300 15,600VCond(ml) 250,000 250,000 250,000 250,000 250,000 250,000 250,000 250,000time(s) 3,950 3,780 3,820 4,030 4,050 4,200 4,150 4,040Vcond(ml/s) 63,291 66,138 65,445 62,035 61,728 59,524 60,241 61,881ρcond(kg/m3)

986,200 986,200 986,200 986,200 986,200 986,200 986,200 986,200Mcond(kg/s) 0,062 0,065 0,065 0,061 0,061 0,059 0,059 0,061MC(kg/s) 0,438 0,538 0,623 0,698 0,438 0,538 0,623 0,698Mh(kg/s) 0,538 0,538 0,538 0,538 0,538 0,538 0,538 0,538Tcmean(K) 294,400 293,700 293,300 293,000 295,700 294,300 293,700 293,450kC(W/mK) 0,606 0,604 0,603 0,601 0,607 0,606 0,604 0,604C(Ns/m2)

0,001 0,001 0,001 0,001 0,001 0,001 0,001 0,001PrC 6,600 6,800 6,900 7,100 6,400 6,600 6,800 6,800ReC 21797,598 25477,983 28931,300 31800,942 23779,198 26528,622 29504,197 33390,989NuC 144,635 165,830 184,655 201,457 153,164 169,246 186,483 205,890Th,mean(K) 329,850 330,300 330,350 329,700 330,800 330,100 329,850 329,400kh(W/mK) 0,650 0,650 0,650 0,650 0,650 0,650 0,650 0,650h(Ns/m2)

0,000 0,000 0,000 0,000 0,000 0,000 0,000 0,000Prh 21,700 21,700 21,700 21,700 21,700 21,700 21,700 21,700Reh 16319,942 16319,942 16319,942 16319,942 16319,942 16319,942 16319,942 16319,942Nuh 184,708 184,708 184,708 184,708 184,708 184,708 184,708 184,708hC 3290,112 3759,812 4179,698 4544,890 3489,893 3849,954 4228,060 4668,072hh 6285,888 6285,888 6285,888 6285,888 6285,888 6285,888 6285,888 6285,888U 2159,699 2352,624 2510,429 2637,730 2244,023 2387,604 2527,796 2678,755∆Tlm 34,181 35,384 35,945 35,676 35,079 35,800 36,143 35,933QC'(J/kgm2)

73820,557 83245,517 90237,152 94103,108 78717,224 85475,344 91362,968 96255,964QC'(J/kgm2)

23809,037 25647,031 27071,644 27978,545 26023,831 27462,750 27597,307 27684,034Cpc(J/kgK) 4180,000 4180,000 4180,000 4180,000 4180,000 4180,000 4180,000 4180,000Cph(J/kgK) 4600,000 4600,000 4600,000 4600,000 4600,000 4600,000 4600,000 4600,000Qh'(J/kgm2)

24754,156 28467,280 28714,821 28467,280 27477,113 28962,363 30200,071 29952,529Qloss'(J/kgm2)

945,120 2820,249 1643,177 488,735 1453,283 1499,613 2602,763 2268,495%heat loss 3,818 9,907 5,722 1,717 5,289 5,178 8,618 7,574

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CONCLUSIONS

For the first part of the experiment co-current flow and later on

counter current flow are examined. The sequence of the pressure drops in

the cold stream are 10,15,20, and 25 mmHg respectively while the

pressure drop of the hot stream is kept constant during the experiment at

15 mmHg. These adjustments of pressure drops surely and continuously

changed the results in inlet and outlet temperatures of the streams, the

mass flow rates of the cold stream, so the Reynold’s Numbers, the

individual and overall heat transfer coefficients, heat fluxes of the streams,

and finally the heat loss to the surroundings.

According to the changes in the pressure drop through the pipe,the

varying Reynold’s Number for the cold stream are found as 21.798,

25.478, 28.931, 31.801 for the co-current flow and 23.779, 26.529,

29.504, 33.391 for the counter current flow which points out that the type

of flow remained constant at turbulent region. The individual heat transfer

coefficients of the cold stream also varied and the results are 3290, 3760,

4180, 4545 for the co-current flow and 3490, 3850, 4228, 4668 for the

counter current flow, in the units of W/m2K. As the mass flow rate of the

hot stream is kept constant, the Reynold’s Number and the individual heat

transfer coefficient for the hot stream did not change and are kept

constant at 16.320 and 6286 respectively.

It can be concluded that, an increase in pressure drop increase the

mass flow rate, Reynold’s Number so the turbulance of the flow, ∆T lm,

individual and overall heat transfer coefficients. When the equation

Q=U*∆Tlm is thought, the increase in the U and ∆T lm increase the heat

15

transfer. Finally it can be said that, the more the flow gets turbulent, the

more heat transfer can be obtained.

REFERENCES

1. Mc Cabe, W.L., J.C. Smith, and P. Harriott, “Unit Operations of Chemical

Engineering”, 4th ed., Mc Graw Hill, N.Y. (1986) Chapter 11, Principles of

Heat Flow in Fluids.

2. F. P. Incropera, D. P. De Witt, “Fundamentals of Heat and Mass

Transfer”, 5th ed., Wiley, N.J. (2002) Chapter 11, Heat Exchangers

3. Perry, R.H. and D. Green, “Perry’s Chemical Engineers’ Handbook”, 7 th

ed., Mc Graw Hill, N.Y. (1997) Section 11, Heat Transfer Equipment.

16

APPENDICES

1. Raw Data

Raw data obtained in the experiment is;

For Co-Current FlowTemp(°C) @

ΔP=10Temp(°C) @

ΔP=15Temp(°C) @

ΔP=20Temp(°C) @

ΔP=25T1 61,7 62,9 63 62,3T2 58 58,5 58,6 58,1T3 55 55,2 55,3 54,7T4 51,7 51,4 51,4 50,8T5 14,8 14,9 15 15,1T6 20,5 19,7 19,4 19,1T7 24,9 23,5 22,8 22,1T8 27,7 26,2 25,3 24,6

For Counter-Current Flow

Temp(°C) @ ΔP=10

Temp(°C) @ ΔP=15

Temp(°C) @ ΔP=20

Temp(°C) @ ΔP=25

T1 63,2 62,8 62,8 62,3T2 59,8 59 58,8 58,2T3 56,6 55,8 55,4 54,8T4 52,1 51,1 50,6 50,2T5 29,6 27,2 25,8 25T6 25 23,3 22,3 21,7T7 20,6 19,6 19 18,7T8 15,5 15,1 15,3 15,6

Mcond(kg/s) (co 0,062 0,065 0,065 0,061 Mcond (kg/s)(counter)

2. Sample Calculation

For the first run of co-current flow

**Calculation of individual and overall heat transfer coefficients

0,061 0,059 0,059 0,061

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The mass flow rate of the cold stream:

Mc = 8.114 (Pc)0.51

The value of Pc was measured as 10 cmHg

Mc = 8.114 (10)0.51 = 0.438 kg/s

Tin and Tout for the cold streams were measured as 14.8C and 27.7C,

respectively.

Tm = ( Tin + Tout )/2 = 294.4 K

The inner diameter of inner pipe which is 1'' Sch. 40 steel pipe is 26.64

mm (1

For water at 294.4K, (2)

kc = 0,606 W/mK

c = 0,001Ns/m2

Pr = 6,600

Re = 4Mc / ( Di c ) = 4 * 0.438 / ( * 26.64E-3 * 0.001 ) = 21797,598

Nuc = 0.023 Rec0.8 Pr0.4

Nuc = 0.023 * (21797,598)0.81 (6.6)0.4 = 144,635

Finally we can calculate hc,

hc = Nuc kc / Di = 144,635* 0.606 / 0.02664 = 3290,112 W/m2K

in a similar way hh can be calculated but;

Reh = 4 Mh / [ (Di + Do) h ] for the hot fluid.

Reh =16319,942

hh = 6285,888 W/m2K

Then overall heat transfre can be calculate as;

U = 1/( 1/hh + 1/ hc ) = 1/( 1/3290,112 + 1/6285,888) = 2159,699 W/m2K

**Calculation of Heat flux

The heat flux can be calculated as;

q' = U Tlm

Tlm = (Tin - Tout ) / ln(Tin/Tout)

where Tin = Th,in - Tc,in and Tout = Th,out - Tc,out

For the first run of cocurrent flow, the temperatures of the streams were

measured as,

Th,in = 61.7C Th,out = 51.7C

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Tc,in = 14.8C Tc,out = 27.7C

Tlm = 34,181 K

qc' = U Tlm = 2159,699 * 34,181 = 73820,557 W/m2

qc' can be also calculated by the heat transfer equation,

q = M Cpav T

Cpav at (14.8+27.7)/2 = 294.4 K is taken as 4180J/kgK (2)

Tc = Tc,out - Tc,in = 27.7 - 14.8 = 12.9 K

qc' = qc / A = Mc Cpav,c T / ( * 0.0334 * L )

where L = 3 * 3.15 m

qc' = 0.438 * 4180 * 12.9 / ( * 0.0334 * 3 * 3.15 ) = 23809,037 W/m2

**Calculation of heat fluxes to the surroundings

qh' = Mh Cpav,h Th

qh' = 24754,156W/m2

qloss' = qh' - qc' = 24754,156 - 23809,037 = 945,120 W/m2

% heat loss = qloss' / qh' *100 = % 3,818

**Energy balance around the steam heater

The heat required for heating the feed water to the heater from room

temperature to an higher temperature is equal to the heat released from

condensate minus the heat loss to the surroundings.That means;

Mh Cpav,h Th = Mcond Hlatent - Qloss

Hlatent = 2257 kJ/kg

The volume of the condensate for the first run was measured as 63,291ml/s

Mcond = Vcond *ρ where is the specific volume

Mcond = 63,291ml/s*0.000996kg/ml = 0,062kg/s

Cpav,h = 4200J/kgK (2)

Mh was calculated as 0,538 kg/s

Th = 61.7 – 14.8 = 46.9 K

Mh Cpav,h Th = Mcond Hlatent - Qloss

0.538 * 4200 * 46.9 = 0,062* 2257000 - Qloss

Qloss = 33958 W

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3. Flow Diagram and Control Panel of DPHE

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Documents

Double Pipe Heat Exchanger