ht 207_report_b7(a)

8/6/2019 HT 207_Report_B7(a)

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HT 207

Heat transfer in Laminar Flow

Date of Experiment : 7th February

Date of Presentation : 9th

February

Group B7(a)

Members

Chaitanya Talnikar (09002013) (Presentation)

Vivek Kumar Pandey (09d02032) (Report)

Rupak Kumar (09d02033) (Ppt)

8/6/2019 HT 207_Report_B7(a)


INTRODUCTION:

The objective of the experiment is to determine the overall heat transfer coefficient using the concept

of logarithmic mean temperature difference and using it to determine individual film transfer

coefficients and finally verify Sieder Tate equation for laminar flow.

MOTIVATION:

The motivation behind the experiment was to get an overall idea of heat transfer that occur in an

industrial process where the flow of the constituent liquid is laminar .

The laminar flow is characterized by small Reynold’s no. Therefore, we also get a fair idea of

dependence of heat transfer on Reynold’s no from the experiment.

THEORY:

Heat exchanger is used in various chemical processes to take away the heat produced in any process.

In heat exchanger, heat transfer occurs from hot to cold liquid . The hot and cold liquid are separated

by a metal wall between them. The heat transfer between hot liquid and metal wall is via convection

and through wall is via conduction and again by convection between cold fluid and wall. In most

cases, the conduction resistance is negligible as compared to convection resistance

Laminar flow is characterized by Re < 2100. In laminar flow fluid flows in order manner along

generally parallel “filament like” stream which don’t mix. It is observed that in this type of flow the

heat transfer to and through the fluid takes place via conduction.

The equivalent resistance for heat flow is given by the formula:

1 = 1 + x + 1 .

UiAi hiAi KAlm hoAo

Or,

1 = 1 + ∆xAi + Ai .

Ui hi KAlm hoAo

Once the heat exchanger and its geometry is fixed the conduction resistance X/KAlm is fixed.

Similarly if the flow rate of the cold fluid is fixed and its mean temperature does not change much

for different flow rates of the hot fluid the resistance of the cold fluid is also fixed. Thus the overall

heat transfer coefficient depends on inside film coefficient alone.

The Seider Tate equation to predict inside heat transfer coefficient in a laminar flow is given by

Nu = 1.86 (Re)1/3

(Pr)1/3

If the bulk mean temperature remains more or less same, then the physical properties remains nearly

same and hence above equation becomes

Nu = constant*(velocity)1/3

8/6/2019 HT 207_Report_B7(a)


PROCEDURE:

The experimental set up consists of double wall heat exchanger made of stainless steel with facility

to measure the inlet and outlet temperature of hot fluid up to a accuracy of 0.1oC.Hot fluid is

circulated via a pump and is heated with the help of a heater kept in steel chamber. The flow rate of the hot liquid can be varied. In our case the hot fluid is mono ethylene glycol and cold fluid is tap

water.

To start the experiment, the mains are switched on and temperature of heater is set at 65oCThe flow

rate of water is fixed at 380l/h. The flow rate of the hot fluid is varied using a knob. The time taken

by hot fluid to rise from zero level to mark of 10 is observed. For given flow rate of hot fluid, the

temperature is measured after steady state has reached. The process is repeated for 8 different flows

and the flow rate of the hot fluid is calculated and steady state temperature is measured.

SCHEMATIC DIAGRAM

8/6/2019 HT 207_Report_B7(a)


PROCESS FLOW DIAGRAM:

Switch on the mains and set the temperature of the

heater between 65-70 0C.

Set the flow rate of the tap water at 380 l/hr.

Record the temperature of inlet and outlet and findzero error .

Set a flow rate of hot fluid using the knob and wait tillthe cylinder is filled upto mark of 10 .

Measure the tiem taken and allow the temperature tostabilise and take readings of temperature.

Repeat the procedure for 8 readings and everytimemeasure the time taken and take readings of the steady

state temperature.

8/6/2019 HT 207_Report_B7(a)


OBSERVATIONS, CALCULATIONS & UNCERTAINITIES:

Inside diameter of inner tube (d1) : 1.0 cm

Outside diameter of inner tube (d2) : 1.27 cm

Inside diameter of outer tube (D1) : 2.20 cm

Length of heat exchanger (L) : 85 cm

Inner heat transfer area of Heat exchanger (A) : 0.0267 cm2

Zero error of hot fluid digital thermometers : 0.30C

Zero error of cold fluid digital thermometers : -0.20C

Diameter of measuring cylinder : 8.75 cm

Height of measuring cylinder : 10 cm

TABLE 1:

Obs.No.

Hot fluidInlet (T1)

Hot fluidOutlet (T2)

Cold fluidInlet (t1)

Cold fluidOutlet (t2)

Time req. for hot fluidlevel to rise between

two marks (t)

1 49.2 43.3 26.7 28.0 66

2 51.4 45.5 26.5 27.7 35

3 51.4 45.3 26.6 28.0 39

4 51.5 44.6 26.8 28.1 46

5 51.5 45.6 26.9 28.1 34

6 51.2 46.6 26.6 28.1 28

7 51.5 45.6 26.7 28.2 38

8 50.0 44.2 26.8 28.1 53

8/6/2019 HT 207_Report_B7(a)


TABLE 2:

Obs.

No.

Volumertic

flow rate of

hot fluid

(10^(-6)m

3 /sec)

Amount of heat

transferred

Q(Kw)

Velocity of hot

fluid

u (m/sec)

LMTD

∆Tlm0C

Overall heat

transfer coefficient

U (Kw m2 0C)

1 9.11 146.03 0.12 18.67 292.92

2 17.18 275.38 0.22 21.15 487.54

3 15.42 255.96 0.20 20.83 460.26

4 13.07 246.94 0.17 20.33 454.98

5 17.69 283.48 0.23 20.85 509.97

6 21.48 264.31 0.27 21.40 462.40

7 15.82 253.64 0.20 20.88 454.85

8 11.35 178.60 0.14 19.43 344.15

CALCULATIONS:

Volumetric Flow Rate: V=3.14*8.75^2*10^(-4)/(660*4)=9.11*10^(-6)m3 /s

Amount of heat transferred (Q) = ρ * V * Cp * (T1 – T2)

= 1085*2638*(49.2-43.3)*9.11^10(-6)

=146.03 Kw

Velocity of hot fluid (u)= V/A=9.11*10^(-6)*4/3.14*8.752= 0.12 m/s

LMTD= T1 – T2 - t1 + t2) / ln((T1 - t2)/ (T2 - t1))

= ((49.2-26.7)-(43.3-28))/ln((49.2-26.7)/ (43.3-28))

=18.67oC

Overall Heat Transfer Coefficient= Q/LMTD*A

=146.03/(18.67*3.14*.085*.85)

= 292.92 Kw m2 0C

8/6/2019 HT 207_Report_B7(a)


TABLE 3:

Obs.

No.

1/(V1/3

) 1/U Inside film heat transfer

coefficient hi (KW m2 0C)

Nusselt No.

Nu

Reynold’s No.

Re

1 2.05 3.41 174.18 6.71 242.05

2 1.66 2.05 215.19 8.29 456.43

3 1.72 2.17 207.57 8.00 409.62

4 1.82 2.20 196.454 7.57 347.28

5 1.64 1.96 217.28 8.34 469.85

6 1.54 2.16 231.80 8.93 570.54

7 1.71 2.20 209.37 8.07 420.40

8 1.91 2.91 187.40 7.22 301.42

Inside film heat transfer coefficient hi = (V1/3

)/slope of the 1/U vs 1/(V1/3

)

= 1/0.0028*2.05

= 174.18 KW m2 0C

Nusselt’s no Nu = hi * d1 / K

= 174.18/(100*.2596)

= 6.71

Reynold’s no = ρ * u * d1 / µ

= 1085 * .12 * 10-2

/ (5.2 * 10-3

)

= 242.05

8/6/2019 HT 207_Report_B7(a)


GRAPH 1:

GRAPH 2:

y = 0.3333x + 0.2323

1.85

1.9

1.95

2

2.05

2.1

2.15

2.2

2.25

4.9 5 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 6

l n

N u

ln Gr

Nu vs Gr log-log scale

y = 0.0028x - 0.0025

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0.0035

0.004

0 0.5 1 1.5 2 2.5

1 / U

1/(u^1/3)

1/U vs 1/(u^1/3)

8/6/2019 HT 207_Report_B7(a)


ERROR ANALYSIS:

Least count of Temperature sensor = 0.1 K

Least count of measuring cylinder = 1 mm

From Seider Tate equation

Intercept = 2.1

Error in intercept = 88.9 %

RESULTS:

The slope of the log Nu vs log Gr curve is 0.3333 which is equal to slope obtained using Seider Tate

equation.

The value of the intercept as obtained from the curve is 0.2323 while SeiderTate equation gives the

value to be 2.1. Hence the error in the intercept is 88.9%

CRITIQUE:

There are possibilities of heat loss through the heat exchanger which can result in errors in the

experiment. The heat loss should be minimized .

ht 207_report_b7(a)

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