compressible flow(1)

7/30/2019 Compressible Flow(1)

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1.0 TITLE

Flow Characteristic through the Convergent-Divergent Duct

2.0 OBJECTIVE

1. To study the pressure-mass flow rate characteristic for convergent-divergent duct

2. To demonstrate the phenomena of choking


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3.0 INTRODUCTION

Converging-diverging nozzles designed for the accurate measurement and control of

all gaseous flow rates. This situation can be found in many engineering application including

steam and gas turbine, aircraft and spacecraft propulsion system, and even industrial

blasting nozzle and torch nozzle. A flow is considered to be a compressible flow if the

change in density of the flow with respect to pressure is non-zero along a streamline.

In general, this is the case where the Mach number in part or all of the flow exceeds

0.3. The Mach .3 value is rather arbitrary, but it is used because gas flows with a Mach

number below that value demonstrate changes in density with respect to the change in

pressure of less than 5%. Furthermore, that maximum 5% density change occurs at the

stagnation point of an object immersed in the gas flow and the density changes around the

rest of the object will be significantly lower.

The factor that distinguishes a flow from being compressible or incompressible is the

fact that in compressible flow the changes in the velocity of the flow can lead to changes thatthe temperature which are not negligible. On the other hand in case of incompressible flow,

the changes in the internal energy such as temperature are negligible even if the entire

kinetic energy of the flow is converted to internal energy like the flow is brought to rest.

The Mach number of the flow is high enough so that the effects of compressibility can

no longer be neglected. For subsonic compressible flows, it is sometimes possible to model

the flow by applying a correction factor to the answers derived from incompressiblecalculations or modelling. For many other flows, their nature is qualitatively different to

subsonic flows. A flow where the local Mach number reaches or exceeds 1 will usually

contain shock waves. A shock is an abrupt change in the velocity, pressure and temperature

in a flow; the thickness of a shock scales with the molecular mean free path in the fluid which

form because information about conditions downstream of a point of sonic or supersonic flow

cannot propagate back upstream past the sonic point.


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The behaviour of a fluid changes radically as it starts to move above the speed of

sound in that fluid which is when the Mach number is greater than 1. For example, in

subsonic flow, a stream tube in an accelerating flow contracts. But in a supersonic flow, a

stream tube in an accelerating flow expands. Consider that steady flow in a tube that has asudden expansion where the tube's cross section suddenly widens, so the cross-sectional

area increases. In subsonic flow, the fluid speed drops after the expansion. In supersonic

flow, the fluid speed increases. The mass flux is conserved but because supersonic flow

allows the density to change, the volume flux is not constant.


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4.0 THEORETICAL BACKGROUND

Figure 1: Convergent-Divergent Duct

From the figure above, air is suck into the converging diverging section by a vacuum pump.The atmospheric air can be considered stagnant. The energy equation for the flow between

the stagnant atmospheric condition (0) and the throat (2) can be written as;

)1(22

2222

2

200

0

0

022

f wwU gzV P

qU gzV P

The differences between (0) and (2) are negligible and the flow can be approximated to be

frictionless and adiabatic, thus isentropic. As such;

000 0V gzwq

Since the air can be treated as ideal gas, following ideal gas relationship can also be used,

manipulated and substituted into the energy equation.

V

PV P C

C RC C

RP

T

;;


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Therefore equation

RT P But

CvT V P

CvT P

becomes Equation

,

)2(2

0

)1(

22

2

20

0

0

So, )3( RP

T

)4(1

1

RCv

CV R

CvCp

RCvCp

Substitute (3) and (4) into (2),

With some algebraic manipulation the throat velocity, can be derived

)5(1

2

211

21

111

11

121

2

2

0

02

2

2

2

0

0

2

2

2

0

0

2

22

2

2

2

0

0

0

0

2

2

PPV

V PP

V PP

RP RV P

RP RP


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For isentropic flow (Using isentropic flow equation relation for ideal gas)

)6(

1

2

20

2

2

0

0

PP

PP

Substitute (6) into (5)

)7(11

2

1

2

1

2

1

0

02

0200

20

0

02

0200

20

0

02

1

r P

V

PPP

PPPV

PPP

PPPV i

Where0

2

PP

r

The mass flow rate in the section can be computed by m = 111 AV and can be shown to be.

2

22022

1

0

20222 r V AV A

PP

V Am

= )8(1

2 12

0

020

r r

P A

.

m


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5.0 APPARATUS

Figure 2: Compressible Flow Bench

Figure 3: Inclined Manometer


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Figure 3: DC Motor Speed Controller

Figure 4: Dynamometer Figure 5: U-Tube Manometer


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6.0 EXPERIMENTAL PROCEDURE

1. Make sure the plug is switching off before conduct the experiment.

2. Connect all the tube according to instruction given by lab instructor.

3. Connect respective tube to the compressor inlet.

4. While machine is running, make sure there is no blockage/object around the duct that

may interfere the air flow into the duct.

5. The choking valve of compressor exhaust has been closed to stop the flow of the

manometer fluid from entering to compressor

6. The inclined manometer tube is connected to read the value of (P o P 1 ) and U-tube

manometer to read (P o -P 2 ) and (P o -P 3 ).

7. Set the speed control to 0 and the speed of motor is increase slowly by rotating the

control knob clockwise until the level needed.

8. Continue the experiment by adjusting the motor speed to 30 sets all of them.

9. The reading of all 30 sets of speed is taken by calculating the different in h from the

barometer.


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7.0 RESULT AND DATA ANALYSIS

No.1P

(kPa)2h

(mm)3h

(mm)2P

(kPa)3P

(kPa)0 1P P

(kPa)0 2P P

(kPa)0 3P P

(kPa)

2

0

Pr

P

m (kg/s)

1. 0.05 0.25 0.25 0.033 0.033 101.275 101.292 101.292 0.0003 0.0002

2. 0.10 0.75 0.25 0.100 0.033 101.225 101.225 101.292 0.0010 0.0004

3. 0.15 1.00 0.25 0.133 0.033 101.175 101.192 101.292 0.0013 0.0005

4. 0.20 1.50 0.25 0.200 0.033 101.125 101.125 101.292 0.0020 0.0007

5. 0.25 1.75 0.25 0.233 0.033 101.075 101.092 101.292 0.0023 0.0008

6. 0.30 2.00 0.50 0.267 0.067 101.025 101.058 101.258 0.0026 0.0008

7. 0.35 2.25 0.50 0.300 0.067 100.975 101.025 101.258 0.0030 0.0009

8. 0.40 2.75 0.75 0.367 0.100 100.925 100.958 101.225 0.0036 0.0010

9. 0.45 3.25 0.75 0.434 0.100 100.875 100.891 101.225 0.0043 0.0012

10. 0.50 3.50 0.75 0.467 0.100 100.825 100.858 101.225 0.0046 0.0012

11. 0.55 4.00 0.75 0.534 0.100 100.775 100.791 101.225 0.0053 0.0014

12. 0.60 4.25 1.00 0.567 0.133 100.725 100.758 101.192 0.0056 0.0014

13. 0.65 5.00 1.00 0.667 0.133 100.675 100.658 101.192 0.0066 0.0016

14. 0.70 5.50 1.25 0.734 0.167 100.625 100.591 101.158 0.0072 0.0017

15 0.75 5.75 1.25 0.767 0.167 100.575 100.558 101.158 0.0076 0.0017

16. 0.80 6.00 1.25 0.800 0.167 100.525 101.158 101.158 0.0079 0.0018

17. 0.85 6.50 1.25 0.867 0.167 100.475 100.458 101.158 0.0086 0.0019

18. 0.90 7.00 1.25 0.934 0.167 100.425 100.391 101.158 0.0092 0.0020

19. 0.95 8.00 1.50 1.067 0.200 100.375 100.258 101.125 0.105 0.0022

20. 1.00 8.25 1.50 1.101 0.200 100.325 100.224 101.125 0.0109 0.002221. 1.05 9.75 2.00 1.301 0.267 100.275 100.024 101.058 0.0128 0.0025

22. 1.10 10.0 2.00 1.334 0.267 100.225 99.991 101.058 0.0132 0.0025

23. 1.15 10.50 2.25 1.401 0.300 100.175 99.924 101.025 0.0138 0.0026

24. 1.20 11.25 2.25 1.501 0.300 100.125 99.824 101.025 0.0148 0.0027

25. 1.25 12.00 2.25 1.601 0.300 100.075 99.724 191.025 0.0158 0.0028

Table 1: Data calculated from the experiment


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Sample Calculation

The calculation can refer this information:

3

0

2

22 295

13600

@ 101.325

@ 1.4

287 .

9.81

o

mercury

atm

T C K kg m

P P kPa

k

R J kg K

g m s

For data no. 1

h 2 = 0.25 mm = 0.0025 m

h 3 = 0.25 mm = 0.0025 m

P 1 = 0.05 kPa

2P gh

= 13600 (9.81) (0.0025)

= 0.033 kPa

3P gh

= 13600 (9.81) (0.0025)

= 0.033 kPa


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P 0 P 1 = 101.325 0.050 = 101.275 kPa

P 0 P 2 = 101.325 0.033 = 101.292 kPa

P 0 P 3 = 101.325 0.033 = 101.292 kPa

2

0

Pr

P

=

= 0.0003

0

P RT

3

3

101.325 10287 295

1.197 kg m

2

2 4d

A

2

5 2

(0.0095 )

4

7.088 10 m


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For mass flow rate,

2 1

00 2

0

2

1

Pm A r r

=

( ) ( )

= 0.0002 kg/s


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Graph

Figure 6: Mass flow rate, m (kg/s) against P 0 P2 (kPa)

Figure 7: Mass flow rate, m (kg/s) against P 2 (kPa)

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

99.5 100 100.5 101 101.5

M a s s

f l o w r a t e

, m (

k g / s )

P0 P2 (kPa)

(Mass flow rate, m (kg/s) against P0 P2(kPa))

(kg/s)

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0 0.5 1 1.5 2

M a s s

f l o w r a t e

, m (

k g / s )

P 2 (kPa)

Mass flow rate, m (kg/s) against P 2 (kPa)


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Figure 8: Mass flow rate, m (kg/s) against P 0 P3 (kPa)

Figure 9: Mass flow rate, m (kg/s) against P 3 (kPa)

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0 50 100 150 200 250 M a s s

f l o w r a t e

, m (

k g / s )

P 0 P 3 (kPa)

Mass flow rate, m (kg/s) against P 0 P 3 (kPa)

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 M a s s

f l o w r a t e

, m (

k g / s )

P 3 (kPa)

(Mass flow rate, m (kg/s) against P 3 (kPa)


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Figure 10: P0 P2 (kPa) against P 0 P3 (kPa)

99.6

99.8

100

100.2

100.4

100.6

100.8

101

101.2

101.4

0 50 100 150 200 250

P 0

P

2 ( k P a

)

P 0 P 3 (kPa)

P 0 P 2 (kPa) against P 0 P 3 (kPa)


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8.0 DISCUSSION AND CONCLUSION


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9.0 REFERENCES

1. Fluid Mechanics, Fundamental and Applications by Yunus A. Cengel and John. M.

Cimbala First Edition in SI Units

2. Fluid Mechanics by Frank M.White

3. Fluids Power with applications, Sixth Edition, Anthony Esposito

4. Fluid Mechanics, J. F. Douglas, J. M. Gasiorek, J. A. Swaffield, Third Edition,

Longman Scientific & Technical

5. Introduction to Fluid Mechanics, Robert W. Fox, Alan McDonald, Second Edition,

John Wiley & Sons.

compressible flow(1)

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