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Applications Of Forced Convection Engineering Essay

The experiment was carried out to verify the relationship between Nusselt figure, Reynolds figure and Prandtl Number utilizing the different constructs of convection. Relative treatments and decisions

were drawn including the assorted factors impacting the truth of the deliberate consequences.

The chief aim of this experiment was to verify the undermentioned heat transportation relationship:

Therefore, the experiment is conducted by an setup where hot ait from warmer is generated and flow through Cu tubing. Different values of temperatures and force per unit area were taken and recorded in order to cipher. Besides, graphs plotted and analysed to hold a better apprehension of convection heat transportation.

Therefore a Laboratory experiment was conducted where hot air from a warmer was introduced through a Cu tubing with the aid of a blower. Thermocouples were fixed in placed at assorted locations along the length of the Cu tubing. The different values of temperature and force per unit area were measured along with the assorted subdivisions of the tubing and other required values were recorded and calculated. Graphs were besides plotted with the informations obtained and so analysed.

Introduction

Heat transportation scientific discipline trades with the clip rate of energy transportation and the temperature distribution through the thermic system. It may be take topographic point in three manners which is conductivity, convection and radiation. Theory of convection is presented since this experiment is concerned about convective heat transportation. Convective is the manner of energy transportation between a solid surface and the next liquid or gas that is in gesture due to a temperature difference. It involves the combined effects of conductivity and unstable gesture.

There are two major type of convective

Forced convection is known as unstable gesture generated by blowing air over the solid by utilizing external devices such as fans and pumps.

The other type is natural convection which meant by a phenomenon that occurs in fluid sections and facilitated by the perkiness consequence. It is less efficient than forced convection, due to the absence of unstable gesture. Hence, it depends wholly on the strength of the perkiness consequence and the fluid viscousness. Besides, there is no control on the rate of heat transportation.

Forced Convection

Force convection is a mechanism of heat transportation in which unstable gesture is generated by an external beginning like a pump, fan, suction device, etc. Forced convection is frequently encountered by applied scientists planing or analysing pipe flow, flow over a home base, heat money changer and so on.

Convection heat transportation depends on fluids belongingss such as:

Dynamic viscousness ( Aµ )

Thermal conduction ( K )

Density ( I? )

Specific heat ( Cp )

Velocity ( V )

Type of fluid flow ( Laminar/Turbulent )

Newton ‘s jurisprudence of chilling

Where

H = Convection heat transportation ( W/ ( m2.A°C )

A = Heat transportation country

= Temperature of solid surface ( A°C )

= Temperature of the fluid ( A°C )

The convective heat transportation coefficient ( H ) is dependent upon the physical belongingss of the fluid and the physical state of affairs.

Applications of Forced Convection

In a heat transportation analysis, applied scientists get the speed consequence by executing a fluid flow analysis. The heat transportation consequences specify temperature distribution for both the fluid and solid constituents in systems such as fan or heat money changer. Other applications for forced convection include systems that operate at highly high temperatures for maps for illustration transporting liquefied metal or liquefied plastic. Therefore, applied scientists can find what fluid flow speed is necessary to bring forth the coveted temperature distribution and prevent parts of the system from neglecting. Engineers executing heat transportation analysis can merely snap an option to include unstable convection effects and stipulate the location of the unstable speed consequences during apparatus to give forced convection heat transportation consequences.

Typical APPLICATIONS

Computer instance chilling

Cooling/heating system design

Electric fan simulation

Fan- or water-cooled cardinal processing unit ( CPU ) design

Heat money changer simulation

Heat remotion

Heat sensitiveness surveies

Heat sink simulation

Printed Circuit Board ( PCB ) simulation

Thermal optimisation

Forced Convection through Pipe/Tubes

In a flow in tupe, the growing of the boundary bed is limited by the boundary of the tubing. The speed profile in the tubing is characterized by a maximal value at the center line and nothing at the boundary.

For a status where the tubing surface temperature is changeless, the heat transportation rate can be calculated from Newton ‘s chilling jurisprudence.

Reynolds Number

Reynolds figure can be used to find type of flow in fluid such as laminar or turbulent flow. Laminar flow occurs at low Reynolds Numberss, where syrupy forces are dominant. The status of flow is smooth and changeless fluid gesture. Meanwhile, turbulent flow occurs at high Reynolds figure and is dominated by inertial forces and it produce random Eddies, whirls and other flow fluctuations.

Reynolds figure is a dimensionless figure. It is the ratio of the inactiveness forces to the syrupy forces in the fluids. Equation for Reynolds Number in pipe or tubing is as below:

Where

I? = Fluid denseness ( kg/m3 )

V = Fluid speed ( m/s )

D = Diameter of pipe

I? = The dynamic viscousness of the fluid ( PaA·s or NA·s/mA? )

I? = Kinematic viscousness ( I? = I? / I? ) ( mA?/s )

Q = Volumetric flow rate ( mA?/s )

A = Pipe cross-sectional country ( M2 )

EXPERIMENT OVERVIEW

Apparatus

Figure 1: Apparatus being used

The experimental setup comprises of a Cu pipe, which is supplied with air by a centrifugal blower and warmer as figure 1. The trial subdivision of the pipe is wound with a warming tape, which is covered with dawdling. Six Cu Eureka thermocouples are brazed into the wall of the trial subdivision. Another six thermocouples extend into the pipe to mensurate the fluxing air temperature. In add-on five inactive force per unit area tapping are positioned in the tubing wall. A BS 1042 criterion opening and differential manometer step the air mass flow rate though the pipe.

Experimental Procedure

Fully near the valve which commanding the air flow rate.

Measure the everage intermal diameter ( D ) of the trial subdivision pipe by utilizing a vernier caliper.

Adjust the disposition angle of the manometer package I± to 30A° .

Get down the blower and turn the valve to the to the full unfastened place bit by bit,

Adjust the power input to the warming tape to its maximal valve and let the setup to achieve thermic equilibrium.

Take down the information and record

Pressure bead through the metering opening

Pressure and temperature downstream of the opening

Ammeter and voltmeter readings

Tube wall temperature along the testing subdivision

Air temperature along the trial subdivision

Air force per unit area along the trial subdivision

Ambient temperature and force per unit area.

Repeat the foregoing process for another four different flow rate and adjust the warmer input to give about the same wall temperature at each flow rate.

DATA AND MEASUREMENT TABLE

Property

Symbol

Unit of measurements

Value

Barometric Pressure

Lead

millimeter Hg

741.60

Diameter of the trial subdivision pipe

Displaced person

m

0.038

Density of H2O ( Manometer ‘s fluid )

I?

Kg/m3

1000

Angle of the manometers bundle

grade

30

Property

Symbol

Unit of measurements

Trial

1

2

3

4

5

Pressure bead across opening

I”H

millimeter H2O

685

565

460

360

260

Pressure bead d/s opening to atmosphere

I”P

millimeter H2O

178

152

120

93

68

Air temperature downstream opening

T

A°C

35

38

38

38

39

EMF ( Voltage ) across tape

Volt

Volts

230

200

165

142

129

Current through tape warmer

I

Amperes

7.3

6.3

5.5

5.0

4.0

Flowing air temperature

t1

A°C

35.0

36.9

38.2

40.0

41.4

Flowing air temperature

t2

A°C

36.1

37.7

38.9

40.6

41.9

Flowing air temperature

t3

A°C

43.1

43.6

43.4

44.4

45.6

Flowing air temperature

t4

A°C

42.2

42.4

42.4

43.5

44.6

Flowing air temperature

t5

A°C

49.6

48.6

47.0

47.3

48.1

Flowing air temperature

t6

A°C

63.2

59.6

55.7

54.3

54.6

Tube wall temperature

t7

A°C

38.9

40.0

40.6

41.9

43.0

Tube wall temperature

t8

A°C

81.20

73.6

65.9

62.2

61.2

Tube wall temperature

t9

A°C

99.8

89.1

77.5

71.5

69.5

Tube wall temperature

t10

A°C

105.9

93.9

81.3

74.6

72.4

Tube wall temperature

t11

A°C

106.5

94.5

81.8

75.1

73.1

Tube wall temperature

t12

A°C

108.1

95.5

82.3

75.0

72.5

Air inactive gage force per unit area ( I”l.sin I± )

P1

millimeter H2O

385

324

255

195

145

Air inactive gage force per unit area ( I”l.sin I± )

P2

millimeter H2O

264

223

175

132

99

Air inactive gage force per unit area ( I”l.sin I± )

P3

millimeter H2O

210

181

141

108

79

Air inactive gage force per unit area ( I”l.sin I± )

P4

millimeter H2O

108

97

81

57

42

Air inactive gage force per unit area ( I”l.sin I± )

P5

millimeter H2O

23

31

20

16

14

Air inactive gage force per unit area ( I”l.sin I± )

P6

millimeter H2O

a‰?0

a‰?0

a‰?0

a‰?0

a‰?0

Sample Calculations

Based on 1st set informations,

Power Input to the tape warmer:

Power = = ( 230 x 7.3 ) /1000 = 1.679

Absolute Pressure downstream of the opening:

741.60 + ( 178/13.6 ) =754.69 mmHg

Absolute Temperature downstream of the opening:

T = t + 273 = 365+ 273 = 308 K

The Air Mass Flow Rate:

air =5.66x = = 231.88

231.88 Kg/hr = 0.06441 Kg/sec,

Since 1 Kg/hr = Kg/sec

Average Wall Temperature:

= ( 38.9+81.2+99.8+105.9+106.5+108.1 ) /6 =90.07

Average Air Temperature:

= ( 35+36.1+43.1+42.2+49.6+63.2 ) /6 = 44.87

The Bulk Mean Air ( arithmetic norm of average air ) Temperature:

= ( 35+63.2 ) /6 =49.1

The Absolute Bulk Mean Air ( arithmetic norm of average air ) Temperature:

49.1+273 =322.10 K

The Properties of Air at Tb:

Using the tabular arraies provided in “ Fundamentalss of Thermal-Fluid Sciences by Yunus A.Cengel ”

From the table A-18 ( Page958 ) , Properties of Air at 1atm force per unit area at K

Density, I? = 1.1029 kg/m3

Specific Heat Capacity, Cp = 1.006 kJ/ ( kg.K )

Thermal Conductivity, k = 0.0277 kW/ ( m.K )

Dynamic Viscosity, Aµ = 1.95 ten 10-5 kg/ ( m.s )

Prandtl Number, Pr = 0.7096

The Increase in Air Temperature:

63.2-35 = 28.2

The Heat Transfer to Air:

( 231.88/3600 ) x 1.006 ten 28.2 =1.827

Where: = Heat Transfer to air

= Mass flow rate

= Specific heat capacity

= Increase in air temperature

The Heat Losingss:

1.679-1.827 = -0.148

Where: = Heat losingss

= Heat Transfer to air

The Wall/Air Temperature Difference:

90.07-44.87 = 45.2

Where: = Wall/Air temperature difference

= Average air temperature

The Heat Transfer Coefficient:

= ( ( 231.88/3600 ) x 1.006 x 28.2 ) / ( 3.14 x.0382 x 1.69 x 45.2 ) = 0.199 kW/ ( m^2.k )

Where:

= Mass flow rate

= Specific heat capacity

= Increase in air temperature

= Average Diameter of the Copper pipe.

= Length of the tubing

= Wall/Air temperature difference

The Mean Air Velocity:

= ( 4 x ( 231.88/3600 ) ) / ( 1.1029 x 3.14 tens ( 0.0382 ^2 ) = 50.9575 m/s

Where:

= Mean air speed

= Mass flow rate

= Density

= Average Diameter of the Copper pipe.

The Reynolds Number:

The Nusselt Number:

= Nusselt Number

= Average Diameter of the Copper pipe.

= Thermal conduction

The Stanton Number:

Where:

St = Stanton Number

= Nusselt Number

= Prandtl figure

Re = Reynolds figure

The Pressure Drop across the proving subdivision:

at Tb = 320.1 K

= Pressure bead across the testing subdivision

= Absolute force per unit area downstream of opening.

= Barometric Pressure

The Friction Factor:

Consequence

Power

Power

kilowatt

1.679

1.260

0.908

0.710

0.516

Absolute Pressure downstream of the opening

Phosphorus

millimeter Hg

754.69

752.78

750.42

748.44

746.60

Absolute temperature downstream of the opening

Thymine

K

308

311

311

311

312

Pressure bead across the opening

a?†H

millimeter H20

685

565

460

360

260

Air mass flow Rate

air

231.88

209.31

188.57

166.60

141.18

Average wall Temperature

tw

90.07

81.1

71.57

66.72

65.28

Average air temperature

tair Ab

44.87

44.80

44.27

45.02

46.03

Bulk Mean air temperature

terbium

49.1

48.25

46.95

47.15

48.0

Absolute majority mean air temperature

Terbium

K

322.1

321.25

319.95

320.15

321.0

Density at Tb

I?

1.1029

1.1058

1.1102

1.1095

1.1066

Specific Heat Capacity at Tb

Cp

1.0060

1.0060

1.0060

1.0060

1.0060

Thermal Conductivity at Tb

K

2.77

2.76

2.75

2.75

2.76

Dynamic Viscosity at Tb

I?

1.95

1.95

1.94

1.94

1.95

Prandtl Number at Tb

Praseodymium

0.7096

0.7096

0.7100

0.7100

0.7098

Addition in air temperature from t1 to t6

a?†t a

28.2

22.7

17.5

14.3

13.2

Heat transportation to air

air

Tungsten

1.827

1.328

0.922

0.666

0.521

Heat losingss

losingss

Tungsten

-0.148

-0.068

-0.015

-0.044

-0.005

Wall/Air temperature difference

a?†t m

45.2

36.3

27.3

21.7

19.25

Heat transportation Coefficient

H

0.199

0.180

0.167

0.151

0.133

Mean air speed

Centimeter

50.9575

45.877

41.167

36.394

30.922

Reynolds ‘s Number

Rhenium

110096.353

99380.

144

89994.

330

79509.

225

67204.

418

Nusselt Number

Nu

274.4

249

232

209.8

184.1

Stanton Number

St

0.00351

0.00353

0.00363

0.0037

0.0039

Pressure Drop across the testing subdivision

a?†P

1746.42

1491.59

1176.73

912.57

667.08

Friction Factor

degree Fahrenheit

0.01378

0.0145

0.0141

0.0141

0.0143

Consequences

Plot A

Experiment

1

2

3

4

5

Y=ln ( Nu x Pr-0.4 )

5.75

5.65

5.58

5.48

5.35

X=ln ( Re0.8 )

9.29

9.21

9.13

9.03

8.89

Y-X

-3.54

-3.56

-3.55

-3.55

-3.54

Plot B

Experiment

1

2

3

4

5

Y=Nu

274.4

249

232

209.8

184.1

X=Re x Pr

78124.37

70520.15

63895.97

56451.55

47701.69

Stanton figure:

Reynolds Analogy:

Experiment

1

2

3

4

5

Friction factor

0.01378

0.0145

0.0141

0.014

0.0143

Reynolds Analogy

0.00689

0.00725

0.00705

0.007

0.00715

Stanton figure

0.00351

0.00353

0.00363

0.0372

0.0386

Discussion

In order to acquire more accurate consequences, there are some suggestions like cleaning the manometer, look intoing the insularity on the pipe and doing certain the valve is closed tightly.

An extra manner to turn out the heat transportation equation is by re-arranging it.

Nu = 0.023 x ( Re0.8 x Pr 0.4 )

Substituting in the experimental values into the above equation from subdivision 5.0 returns the undermentioned consequences below:

Experiment

1

2

3

4

5

Y=Nu

274.4

249

232

209.8

184.1

X=Re0.8 x Pr0.4

9415.08

8674.51

8014.48

7258.34

6344.14

Y/X

0.029

0.0287

0.0289

0.0289

0.029

Comparing this to the heat transportation invariable, it shows that there is a small difference merely which can be negligible.

It can besides be done by taking the gradient of the line from the secret plan Nu against ( Re0.8 x Pr0.4 )

as shown below:

Decision

A better apprehension of the heat transportation was achieved through carry oning the experiment. Theoretical amounts and experimental values were found to be about similar and the different beginnings of mistake have been identified.

The chief aim of this experiment was to verify the undermentioned heat transportation relationship:

Nu = 0.023 x ( Re0.8 x Pr 0.4 )

Therefore, relation of forced convective heat transportation in pipe is cleared and the aims were completed.

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