Bangladesh University of Engineering & Technology

Chemical Engineering Laboratory- III

ChE-304

Experiment No: 01
Experiment Name: Study Of Heat Transfer Coefficient In A Double Pipe
Heat Exchanger

Prepared By:
Abdullah Jubaer
ID:2002035
Level/Term:3/2
Lab Group:1
Partner’s Id:
2002031
2002032
2002033
2002034 .

Submitted To
Hasan Tasnim Salehin
Lecturer
Department of Chemical Engineering, BUET

Date of Performance: 10-9-24

Date of Submission: 16-9-24

Summary
The main objectives of this experiment are to obtain individual and overall heat transfer
coefficients, to observe the variation of heat transfer coefficient with Reynolds number (Re)
and fluid velocity (v).Comparison of experimental and estimated heat transfer coefficients from
the obtained individual and overall heat transfer coefficients was also an important observation
in this experiment.Double pipe heat exchanger was used to conduct the experiment. Tube side
and shell side passing fluids were cold water and saturated steam respectively. Steam pressure
was changed three times from 5 to 15 psig and for each pressure water flowrate was varied
three times as well as condensates were collected for the volumetric flowrate calculation. Due
to convection in each fluid and conduction through the wall of inner pipe heat was transferred
from steam to water. Using this data experimental overall heat transfer coefficient was
calculated. The steam side heat transfer coefficient was calculated by using Nusselt equation
and the water side heat transfer coefficient was calculated by using Dittus-Boelter equation.
The theoretical over all heat transfer coefficient was determined by neglecting the resistance
of metal tube.Logarithmic plot of ‘Nusselt number vs Reynolds number’ as well as ‘Water side
heat transfer coefficient vs Velocity’ and Wilson graphs were plotted for the three-consecutive
pressure. In the experiment, the overall theoretical heat transfer coefficient was found to be
varied from 968 Wm-2K-1 to 2338 Wm-2K-1 while the experimental values varied from 2342.29
Wm-2K-1 to 3428,42 Wm-2K-1. Brief discussion on errors and possible reasons behind them
were discussed in the results and discussion section.
Experimental Set up

Apparatus

1. Double Pipe

2. Flowmeter

3. Pressure Gauge

4. Steam Trap

5. Thermometer

6. Bucket

7. Weight Machine

Figure 1: Schematic diagram of experimental set up of double pipe heat exchanger

Experimmental Procedure
The thermometers were set in proper places and the necessary data for this experimental system
was recorded. Water was passed through the inner pipe. Steam was allowed to flow through
the annulus. Valve opening determined the steam pressure at the inlet. The pressure was fixed
at 5 psig and the inlet and outlet water temperature were recorded. Amount of water and
condensate collected for a known interval of time was noted. Keeping the steam pressure fixed,
readings were obtained for different water flow rates. Step 4,5,6 were repeated for different
steam pressures. (10 and 15 psig)
 Observed Data
Inner tube specifications:
Tube length = 7' 4''
Nominal diameter =1''
Schedule no. 40
Table 01: Observed data for study of double pipe heat exchanger
Water temperature
Steam Water flowrate Condensate flowrate
Observation (oC)
pressure, P
no. Volume
(psig) Inlet, T1 Outlet, T2 Time (s) Mass (kg) Time (s)
(L)

1
29 57 1 7.93 0.26 10.35

2 5
29 52 1 3.17 0.334 10.56

3
29 43 1 2.05 0.41 10

1
29 55 1 5.65 0.282 10.28

2 10
29 46 1 1.99 0.38 10.35

3
29 43 1 1.73 0.464 10.41

1
29 56 1 7.27 0.33 10.19

2 15
29 51 1 2.97 0.39 10.31

3
29 45 1 1.86 0.446 10.25
Calculated Data

Table 02: Water properties at average water temperature

Temperature
difference of Average
inlet and Specific Viscosity,
water Density, ρ Thermal
Observation outlet water heat, Cpm μ× 104 conductivity,
no temperature
(kg/m3) Km (W/m.K)
∆T (OC)
(J/kg.K) (Pa.s)

(OC)

1
28 43 4174 990.6 6.16 0.637

2
23 40.5 4174 991.9 6.49 0.633

3
14 36 4174 993.6 7.11 0.621

4
26 42 4174 990.3 6.19 0.635

5
17 37.5 4174 993 6.82 0.63

6
14 36 4174 993.6 7.11 0.621

7
27 42.5 4174 990.3 6.19 0.635

8
22 40 4174 991.7 6.47 0.632

9
16 37 4174 993 6.82 0.63
Table 03: Calculated data for saturation temperature, latent heat of condensation, mass flow
rate of water and condensate, heat given by steam and heat taken by water

Mass Heat Heat

Saturation flow Mass flow given up taken up
Steam Latent heat of
temperature rate of rate of by by
Obs. no pressure condensation,
of steam water, condensate, steam, water,
(psig) λ (kJ/kg) Qc Qw
(OC) MW MC (kg/s)
(kg/s) (W) (W)

1
0.12491 0.02512077 56067.0 14599.4

2 5 109.3 2231.9
0.31290 0.03162878 70592.2 30039.2

3
0.48468 0.041 91507.9 28322.9

1
0.17527 0.02743190 60780.8 19021.4

2 10 115.2 2215.7
0.49899 0.03671497 81349.3 35407.6

3
0.57433 0.04457252 98759.3 33561.8

1
0.13621 0.03238469 71298.1 15351.4

2 15 120.2 2201.6
0.33390 0.03782735 83280.6 30661.8

3
0.53387 0.04351219 95796.4 35654.0
Table 04: Calculated data for Mean rate of heat, experimental overall heat transfer
coefficient, Wall temperature, velocity and Reynolds number.

Experimental
Mean rate overall heat Wall
Steam Velocity,
Obs. of heat, LMTD transfer temperature, Reynolds
pressure v
no QM coefficient, TW no. Re
(OC) UOE
(psig) (m/s)
(W) (OC)
(W/m2.K)

1
35333.236 65.302581 2342.292323 76.15 0.226927 9707.0141

2 5
50315.764 68.154408 3195.936508 74.9 0.567675 23078.372

3
59915.415 73.076626 3549.344094 72.65 0.877820 32630.923

1
39901.173 72.423837 2385.020749 78.6 0.318501 13554.044

2 10
58378.528 77.389051 3265.589869 76.35 0.904287 35022.975

3
66160.601 78.993339 3625.742745 75.6 1.040191 38666.700

1
43324.778 76.956372 2437.13609 81.35 0.247528 10533.748

2 15
56971.296 79.740234 3092.905246 80.1 0.605903 24703.632

3
65725.243 82.989880 3428.428423 78.6 0.967490 37470.817
Table 05: Calculated data for Prandtl no., Water side heat transfer coefficient, Nusselt no.,
Film

Temperature and water density, viscosity, thermal conductivity at film temperature.

Water side Water Water Thermal

Film
Steam Prandtl heat transfer Nusselt viscosity
Densit conductivity
Obs. pressure no. coefficient, no. Temperature, y at Tf, at Tf, μf ×
no hi Tf ρf 104 at Tf, kf
(psig) Pr Nu
O
(W/m2.K) ( C) (kg/m3) (Pa.s) (W/m.K)

1
4.0363 1356.59456 56.649 84.4375 968.9 3.38 0.674

2 5
4.2795 2748.32481 115.49 83.5 969.8 3.41 0.673

3
4.7789 3690.27156 158.06 81.8125 971.5 3.49 0.665

1
4.0688 1771.03996 74.188 87.75 966.6 3.27 0.675

2 10
4.5185 3888.50746 164.18 86.0625 967.1 3.35 0.674

3
4.7789 4226.92491 181.05 85.5 967.8 3.36 0.675

1
4.0688 1447.57005 60.638 91.0625 964.35 3.14 0.677

2 15
4.2730 2896.06455 121.89 90.125 964.9 3.15 0.676

3
4.5185 4104.45068 173.29 89 965.2 3.17 0.675
Table 06: calculated data for Steam side heat transfer coefficient, theoretical overall heat
transfer coefficient, experimental 1/U, theoretical 1/U, and (1/v)0.8

Steam side heat Theoretical overall

Steam Experimental Theoretical
Obs. transfer (1/v)0.8
pressure heat transfer 1/U× 104 1/U× 104
no coefficient, hO coefficient, UOT
(s/m)0.8
(psig) 2 (m2.K/W) (m2.K/W)
(W/m .K) (W/m2.K)

1
8287.228413 968.6231146 4.269321939 10.3239328 3.27560

2 5
8187.459216 1747.723394 3.128973299 5.72172921 1.57296

3
7947.632041 2169.260747 2.817421962 4.60986537 1.10987

1
8136.320335 1217.617162 4.192835641 8.21276203 2.49753

2 10
7960.781058 2253.815074 3.0622339 4.43692125 1.08381

3
7928.591564 2388.138483 2.758055577 4.18736186 0.96896

1
8093.186794 1022.508329 4.103176692 9.77987143 3.05562

2 15 8016.38066 1812.180479 3.233206065 5.51821417 1.49306

3 7922.992158 2338.87145 2.916788326 4.27556632 1.02679

Graphical Representation

Nusselt Number Vs Reynolds number

1000

y = 0.0248x0.8418
100
Nu

10

1
1 10 100 1000 10000 100000
Re

Figure 1: Log-log plot of Nusselt number vs. Reynolds number at 5 psig steam pressure

Nusselt Number Vs Reynolds Number

1000

y = 0.0238x0.8453
100
Nu

10

1
1 10 100 1000 10000 100000
Re

Figure 2: Log-log plot of Nusselt number vs. Reynolds number at 10 psig steam pressure.
 Nusselt Number Vs Reynold Number
1000

100
Nu

y = 0.0287x0.8263

10

1
1 10 100 1000 10000 100000
Re

Figure 3: Log-log plot of Nusselt number vs. Reynolds number at 15 psig steam pressure.

Water-side Heat Transfer Coefficient (hi) vs velocity

10000

y = 4115.4x0.7444
R² = 0.9991
hi

1000
0.1 1
velocity

Figure 4: Log-log plot of water side heat transfer coefficient vs. velocity at 5 psig steam pressure.
 Water-side Heat Transfer Coefficient (hi) vs velocity for 10 psig
10000

y = 4144.6x0.742
hi

1000
0.1 1
velocity

Figure 6: Log-log plot of water side heat transfer coefficient vs. velocity at 10 psig steam pressure.

Water-side Heat Transfer Coefficient (hi) vs velocity

10000

y = 4225.9x0.7659
hi

1000
0 0.2 0.4 0.6 0.8 1 1.2
velocity

Figure 7: Log-log plot of water side heat transfer coefficient vs. velocity at 15 psig steam pressure.
 1/U vs (1/v)^0.8
0.0012

0.001 y = 0.0003x + 0.0002

0.0008

experimental
1/U

0.0006
theoretical

y = 7E-05x + 0.0002 Linear (experimental)

0.0004
Linear (theoretical)

0.0002

0
0 0.5 1 1.5 2 2.5 3 3.5
1/v^0.8

Figure 8: Wilson plot comparing the variation of reciprocal of the experimental and theoretical overall
heat transfer coefficients concerning with 1/v0.8 at 5 psig.

1/U Vs 1/v^0.8
0.0009

0.0008 y = 0.0003x + 0.0002

0.0007

0.0006
experimental
0.0005
1/U

theoretical
0.0004 y = 9E-05x + 0.0002
Linear
0.0003 (experimental)
Linear
0.0002 (theoretical)

0.0001

0
0 0.5 1 1.5 2 2.5 3
1/v^0.8

Figure 9: Wilson plot comparing the variation of reciprocal of the experimental and theoretical overall
heat transfer coefficients concerning with 1/v0.8 at 10 psig.
 1/U Vs 1/v^0.8
0.0012

0.001 y = 0.0003x + 0.0001

0.0008

experimental
1/U

0.0006
Theoretical
y = 6E-05x + 0.0002
Linear (experimental)
0.0004
Linear (Theoretical)

0.0002

0
0 0.5 1 1.5 2 2.5 3 3.5
1/v^0.8

Figure 10: Wilson plot comparing the variation of reciprocal of the experimental and theoretical
overall heat transfer coefficients concerning with 1/v0.8 at 15 psig.
Result and Discussion

Table 07 Data for experimental overall heat transfer coefficient, theoretical overall heat
transfer coefficient and dirt factor

Theoretical overall
Steam Experimental overall heat
Obs. heat transfer coefficient, Dirt factor
pressure transfer coefficient, UOE
no UOT Rd
2
(psig) (W/m K)
(W/m2K)

1 2342.292323 968.6231146

2 5 3195.936508 1747.723394 0

3 3549.344094 2169.260747

1 2385.020749 1217.617162

2 10 3265.589869 2253.815074 0

3 3625.742745 2388.138483

1 2437.13609 1022.508329

2 15 3092.905246 1812.180479 0.0001

3 3428.428423 2338.87145

Figure 2 to figure 4 are logarithmic plot of Nusselt number Vs Reynolds number for three
consecutive pressure. The graphs demonstrates the variation of heat transfer coefficient with
Reynolds number. Trends of the graphs seems to be linear with increasing Nusselt number
relative to the increase of Reynold number. Figure 5 to figure 7 are logarithmic plot of water
side heat transfer coefficient (hi) velocity for the above pressure. Trends of these graphs also
similar to linear and water side heat transfer coefficient increases with the increase of velocity.
Wilson plot is shown on figure 8 to figure 10 for three consecutive pressure.These graphs have
showed the relation of theoretical to experimental heat transfer coefficient.For each pressure
Wilson plot for theoretical heat transfer coefficient in our experiment is higher than
experimental heat transfer coefficient which we is opposite according to our
assumption.Reason of this discrepancy have been discussed in the errors.

Applicability of Dittus-Boelter equation

Figure 02, 03 and 04 showed the relationship between Nusselt number and Reynolds number
for the experiment. Theoretically it is known from the Dittus-Boelter equation that

Nu = 0.023(Re)0.8(Pr)1/3

For 5 psig, the trend line of the plot of Nu vs. Re is y = 0.0248x0.8418 (Figure 2)

For 10 psig, the trend line of the plot of Nu vs. Re is y = 0.0238x0.8453 (Figure 3)

For 15 psig, the trend line of the plot of Nu vs. Re is y = 0.0287x0.8263 (Figure 4)

Theoretically the power of ‘x’ should be 0.8. For all three pressures it is close to 0.8. So, it can

be said that the Dittus –Boelter equation is applicable.

Plot of water side heat transfer coefficient (hi) vs. velocity

hi Di
Nu = ; putting this value in Dittus-Boelter equation and solving for hi
k

k 0.8 1
hi = 0.023 Re Pr 3
Di

vρDi
Re = μ

k vρDi 0.8 1
⸫hi = 0.023 D ( ) Pr 3
i μ

So, the plot of hi against v in a logarithmic scale will yield a straight line with slope of 0.8.

And the trend line should be in the form of y=ax0.8

From Figure 5, 6 and 7 the exponent of x is 0.744, 0.746 and 0.765 for 5 psig, 10psig and 15
psig respectively. So, it can be said that experimental data is reasonably identical with
theoretical data.

Wilson Plot

Wilson plot is drawn by plotting 1/U against (1/v) 0.8 for both theoretical and experimental
values. U is overall heat transfer coefficient. So, the reciprocal of U is the resistance to heat
transfer. For all three pressures the experimental values of 1/U should be greater than
theoretical values. That is because while calculating U by using equations theoretically fouling
factor was not considered. So, the extra resistance is due to fouling and is called dirt factor. But
in our experiment we got the experimental values of 1/U is greater than theoretical values
because of discrepencies in the experiment for various reasons. Here, from the difference of
experimental and theoretical values a combined dirt factor was calculated for all three
pressures.For pressure 5 psig and 10 psig the Y-intercept of the linear equation for both
theoretical and experimental line was same.So the dirt factor calculated for these pressures was
zero.We got 0.0001 dirt factor for 15 psig.

Assumptions on possible errors

Though during the experiment greater effort was imposed to control the erroneous result to a
least amount. But there were some reasons which couldn’t be traced or controlled that led to
the results with some errors. Some possible reasons are

● During the experiment pressure gauge meter reading fluctuated continuously but, in the
calculation, it is considered constant for the particular time of its operation.

● The amount of heat lost by steam should be equal to the amount of heat gain by water
for the experiment. But in the experiment, it was seen that, in all observation Qc > Qw.

● The process was not completely adiabatic so some heat was lost due to convection and
radiation to the surrounding atmosphere.

● Frictional loss on pipe wasn’t considered in the calculation.

● Weight machine used in the experiment was old and provided different readings for the
same quantity.

● Manual error during flowrate controlling might be one of the reasons for erroneous
results.
Sample Calculation

Length of the pipe, L= 7 ft 4 inch = 7.33 ft. = 2.235 m

For nominal size 1 inch and schedule no. 40 of inner pipe,

Outer diameter, DO = 1.315 inch = 0.0334 m

Inner diameter, Di = 1.049 inch = 0.0266 m

(Reference: J. P. Holman. ‘Heat Transfer’. McGraw – Hill. 10th Edition. Table A-11, page
665)

Outside surface area of inner pipe, AO = πDOL

= 3.1416 × 0.0334× 2.235

= 0.2345 m2

𝜋𝐷𝑖2
Water flow area, Ai = = 5.57 × 10-4 m2
4

For observation no. 1 at 5 psig pressure

Water inlet temperature, T1 = 29 oC

Water outlet temperature, T2 = 57 oC

Temperature difference, T2-T1 = = 28 oC = 28 K

29+57
Mean temperature, Tm =( ) °C
2

= 43oC

At, 43oC mean temperature,

Specific Heat of capacity, Cpm = 4174 J.Kg-1K-1

Density of water, ρm = 990.6 Kgm-3

Thermal conductivity, km = .637 Wm-1.K-1

Dynamic viscosity, μm = 0.000616 Pa.s

(Reference: J. P. Holman. ‘Heat Transfer’. McGraw – Hill. 10th Edition. Table A-9, page 662)

Collected volume of water = 1 L

Time of collection = 6.93 s

.001 × 994.9
Mass flow rate of water, Mw = = 0.1249 Kg/s
6.93

Collected mass of condensate = 0.26 Kg

Time of collection = 10.35 s

0.26
Mass flow rate of condensate, Mc = 10.36 = 0.0251 kg/s

Rate of heat taken up by water, Qw = MwCpm(T2-T1)

= 0.1249 × 4174 × 28

= 14599.40 W

At 5 psig steam pressure,

Saturation temperature of steam, TS = 109.3 0C

Latent heat of vaporization, λS = 2231.9 kJkg-1

Rate of heat given up by steam, Qs = Mcλs

= 0.0251 × 2231.9 × 1000

= 56067.043 W

𝑄𝑤 +𝑄𝑠
Mean rate of heat flow, Qm = 2

14599.4+56067.043
= 2

= 35333.236 W

Calculation of experimental overall heat transfer coefficient, UOE

Temperature difference at inlet, ∆T1 = TS – T1 == 80.3 0C

Temperature difference at outlet, ∆T2 = TS – T2 = 52.3 0C

Logarithmic mean temperature difference,

80.3−52.3
= 80.3
𝑙𝑛𝑙𝑛
52.3

= 65.3 oC

𝑄𝑚
Experimental overall heat transfer coefficient, UOE = 𝐴
𝑜 ∆𝑇𝑙𝑚

35333.236
= 0.231 ×65.3 W.m-2.K-1

= 2342.292 W.m-2.K-1

Calculation of velocity (v), Reynolds number (Re) and Prandtl Number (Pr)

𝑇𝑠 +𝑇𝑚
Tube wall temperature, Tw = 2

109.3+43
= 2

= 76.15 oC

𝑀𝑤
Velocity, v = 𝜌
𝑚 𝐴𝑖

0.226
= 990.6×5.57 ×10−4

= 0.227 m/s

𝐷𝑖 𝑣𝜌𝑚
Reynolds Number, Re = 𝜇𝑚

0.0266 ×0.227 ×990.6

= 0.000616

= 9707.146

𝜇𝑚 𝐶𝑝𝑚
Prandtl Number, Pr = 𝑘𝑚

0.000616×4174
= = 4.036
0.637
Calculation of water side heat transfer coefficient (hi) and Nusselt Number (Nu)

For turbulent flow, according to Dittus-Boelter equation,

1
0.023×𝑘𝑚(𝑅𝑒 0.8 (𝑃𝑟)3
hi = 𝐷𝑖

1
0.023 × 0.637 × 97070.8 × 4.0363
= 0.0266

= 1356.595 Wm-2K-1

ℎ𝑖 𝐷 𝑖
Nusselt Number, Nu = 𝑘𝑚

1356.595 ×0.0266
= = 56.54
0.637

Calculation of steam side heat transfer coefficient (hO) and theoretical overall heat
transfer coefficient (UOT):

Film temperature, Tf = TS – 0.75(TS-TW)

= 109.3– 0.75 × (109.3 – 76.15) = 84.43 oC

At 84.43 oC film temperature,

Density of water, ρf = 968.9 Kg/m3

Thermal conductivity, kf = 0.675 W/m.K

Dynamic viscosity, μf = 0.000338 Pa.s

Nusselt equation for film type condensation,

1/ 4
 k f 3  2f g s 
 
 D (T  Tw ) f 
ho = 0.725  0 s 

1/4

= 0.725 × 0.67263 ×968.62 ×9.81 ×2215.5 ×1000

( 0.0334 ×(115.25 − 73.625)×0.000329 )

= 8287.228 Wm-2K-1
 1 𝐷𝑜 𝑥 𝐷 −1
Theoretical overall heat transfer coefficient, UOT = (ℎ + 𝐷 + 𝑘 𝑊𝐷 𝑜 )
𝑜 1 ℎ1 𝑀 𝑙𝑚

The term for conduction can be neglected, then it can be presented as

1
UOT = 1 𝐷
+ 𝑜
ℎ𝑜 𝐷𝑖 ℎ𝑖

1
= 1 0.0334
+
8287.228 0.0266 ×1356.595

= 968.623 Wm-2K-1

Calculation of dirt factor, Rd

From the graph of Wilson plot i.e. 1/U vs (1/v) 0.8

For 5 psig steam pressure in figure 08

The intercept for dirty tube = 0.0002 m2K/W

The intercept for clean tube = 0.0002 m2K/W

Dirt factor, Rd = 0.0002 – 0.0002 = m2K/W

For 10 psig steam pressure in figure 09

The intercept for dirty tube (experimental overall heat transfer coefficient) = 0.0002 m2K/W

The intercept for clean tube (theoretical overall heat transfer coefficient) = 0.0002 m2K/W

Dirt factor, Rd = 0.0002 – 0.0002 = 0 m2K/W

For 15 psig steam pressure in figure 10

The intercept for dirty tube (experimental overall heat transfer coefficient) = 0.0002 m2K/W

The intercept for clean tube (theoretical overall heat transfer coefficient) = 0.0001 m2K/W

Dirt factor, Rd = 0.0002 – 0.0001 = 0.0001 m2K/W

References

[1] Cengel, Yunus, and Ghajar, Afshin J. Heat and mass transfer: fundamentals and
applications (5th edition). McGraw-Hill Higher Education Page-918, Table A-9.

[2] Kern, Donald Q. Process heat transfer. Tata McGraw-Hill Education, 1997, Page-844,
Table 11.

[3] J. P. Holman. ‘Heat Transfer’. McGraw – Hill. 10th Edition. Table A-9, page 662
 Marking Scheme

Experiment no. 01
STUDY OF HEAT TRANSFER COEFFICIENT IN A DOUBLE PIPE HEAT EXCHANGER

Name: Abdullah Jubaer

Student number: 2002035

Section and % marks allocated Marks

Summary (10%)

Experimental set up (20%)

Observed data (5%)

Calculated data (10%)

Sample calculation (20%)

Result and discussion (20%)

Quality of tables and figures (10%)

Overall presentation (5%)

Total (100%)