TITLE : COMPRESSIBLE FLOW

1.0 INTRODUCTION

The aim of this experiment is to investigate compressible flow in a convergent-divergent

nozzle. But before starting the experiment, a brief knowledge on this convergent-divergent

nozzle should be introduce first so that students know what this experiment all about is.

Converging-Diverging or "de Laval" Nozzles have been widely used over the last few

decades in many engineering contexts, from civil and mechanical up to aerospace uses. It is

a tube that is pinched in the middle, making a carefully balanced, asymmetric hourglass

shape. They are designed to accelerate fluids to supersonic speeds at the nozzle exit.

Because of this, the nozzle is widely used in some types of steam turbines and rocket engine

nozzles. It also sees use in supersonic jet engines.

Their operation relies on the different properties of gases flowing at subsonic and

supersonic speeds. The speed of a subsonic flow of gas will increase if the pipe carrying it

narrows because the mass flow rate is constant. The gas flow through a de Laval nozzle is

isentropic (gas entropy is nearly constant). In a subsonic flow the gas is compressible, and

sound will propagate through it. At the "throat", where the cross-sectional area is at its

minimum, the gas velocity locally becomes sonic (Mach number = 1.0), a condition called

choked flow. As the nozzle cross-sectional area increases, the gas begins to expand, and the

gas flow increases to supersonic velocities, where a sound wave will not propagate backwards

through the gas as viewed in the frame of reference of the nozzle (Mach number > 1.0).

The purpose of this report is to gain an understanding of the nature of this flow by

investigating the pressure ratios effects on the mass flow rate of air through the system and

the differing pressure distributions that occur at varying lengths into the nozzle.

1.1 OBJECTIVES

i. To study about the characteristics of pressure-flow rate of convergence-divergence

tube

ii. To demonstrate the phenomenon of choking.

1.2 THEORY

Figure 1. Covergent-Divergent duct

Referring to the figure above, the steady energy equation between 0 and 2 is given by:

𝑃0𝜌0+𝑣0

2

2+ 𝑔𝑧0 + 𝑢0 + 𝑞 =

𝑃2𝜌2+𝑣2

2

2+ 𝑔𝑧2 + 𝑢2 +𝑤𝑠 + 𝑤𝑓 …………… .……… .. (1)

For gas with small elevation differences, ∆𝑔𝑧 ≈ 0

For isentropic flow where there is no work is transferred, 𝑞 = 𝑤 = 0, ‘0’ is showing the stagnation

conditions, so 𝑣0 = 0.

Therefore, equation 1 now is,

𝑃0𝜌0+ 0 + 𝐶𝑣𝑇𝑜 =

𝑃2𝜌2+𝑣22+ 𝐶𝑣𝑇2 ………………… . (2)

But 𝑃 = 𝜌𝑅𝑇, so 𝑇 =𝑃

𝜌𝑅 …………………………(3)

𝐶𝑝 = 𝐶𝑣 + 𝑅

So, 𝛾 =𝐶𝑝

𝐶𝑣= 1 +

𝑅

𝐶𝑣

𝐶𝑣 =𝑅

𝛾 − 1 ………………………… . . (4)

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

𝑃0𝜌0+

𝑅

(𝛾 − 1)

𝑃0𝜌0𝑅

=𝑃2𝜌2+𝑣2

2

2+

𝑅

(𝛾 − 1)

𝑃2𝜌2𝑅

𝑃0𝜌0(1 +

1

𝛾 − 1) =

𝑃2𝜌2(1 +

1

𝛾 − 1) +

𝑣22

2

𝛾

𝛾 − 1

𝑃0𝜌0=

𝛾

𝛾 − 1

𝑃2𝜌2+𝑣2

2

2

𝑣2 = √2𝛾

(𝛾 − 1)(𝑃0𝜌0−𝑃2𝜌2) …………………… (5)

For isentropic flow,

𝑃0

𝜌0𝛾 =

𝑃2

𝜌2𝛾

𝜌2 = 𝜌0 (𝑃2𝑃0)

1𝛾 ……………………… (6)

Substitute (6) in (5),

𝑣2 =

√

2𝛾

(𝛾 − 1)

(

𝑃0𝜌0−

𝑃0𝑃2

𝑃0𝜌0 (𝑃2𝑃0)

1𝛾 )

= √2𝛾

(𝛾 − 1)

𝑃0𝜌0(1 −

𝑃21−

1𝛾

𝑃01−

1𝛾

)

= √2𝛾

(𝛾 − 1)

𝑃0𝜌0(1 − 𝑟

𝛾−1𝛾 ) , 𝑤ℎ𝑒𝑟𝑒 𝑟 =

𝑃2𝑃0 ……………(7)

�̇� = 𝜌2𝐴2𝑣2 = 𝜌0 (𝑃2𝑃0)

1𝛾𝐴2𝑣2 = 𝜌0𝐴2𝑣2

√𝑟2𝛾

�̇� = 𝜌0𝐴2√2𝛾

(𝛾 − 1)

𝑃0𝜌0(𝑟

2𝛾 − 𝑟

𝛾+1𝛾 ) ……… . (8)

1.3 APPARATUS

Figure 1 Inclined Manometer

Figure 4 Frequency Controller / DC Motor Controller

Figure 3 Convergent Divergent Nozzle

Figure 2 Pressure Reading

2.0 PROCEDURE

i. Check all the electrical supply to make sure it is switch off.

ii. All the respective tube connected to the compressor inlet.

iii. The apparatus setup was checked to make sure there is no blockage/object around the

duct that may interfered the air flow into the duct.

iv. The throat valve at the compressor exhaust was closed to avoid unnecessary manometers

fluid drawn into the compressor.

v. The inclined manometer connected to read 1oP P and U-tube manometer to read

2oP P and 3oP P .

vi. The speed control was set to 0. Then, start the motor by push the “run” button (the indicator

will be turn on). The speed control knob rotated clockwise slowly to increase the motor

speed until it reached the desired level.

vii. The motor speed or the exhaust valve adjusted to give approximately 20 sets or readings.

viii. All the reading of barometric pressure and the atmospheric temperature recorded and

tabulated in the table.

3.0 DATA AND RESULT

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 0.1 0.750 0.250 1.000 0.333 101.225 100.324 100.991 0.010 0.207

2 0.2 1.250 0.375 1.668 0.500 101.125 99.657 100.825 0.016 0.289

3 0.3 2.250 0.500 3.002 0.667 101.025 98.3231 100.658 0.030 0.422

4 0.4 2.750 0.625 3.668 0.833 100.925 97.656 100.491 0.036 0.478

5 0.5 3.750 0.750 5.003 1.000 100.825 96.321 100.324 0.049 0.579

6 0.6 4.750 1.000 6.337 1.334 100.725 94.987 99.990 0.063 0.668

7 0.7 5.250 1.083 7.004 1.444 100.625 94.320 99.880 0.069 0.708

8 0.8 6.000 1.250 8.004 1.667 100.525 93.320 99.657 0.079 0.766

9 0.9 7.250 1.500 9.673 2.001 100.425 91.652 99.324 0.095 0.853

10 1.0 8.250 1.625 11.007 2.168 100.325 90.318 99.157 0.109 0.918

11 1.1 10.250 2.000 13.675 2.668 100.225 87.649 98.657 0.135 1.032

12 1.2 11.750 2.375 15.676 3.168 100.125 85.648 98.156 0.155 1.108

13 1.3 13.000 2.625 17.344 3.502 100.025 83.980 97.823 0.171 1.166

14 1.4 14.000 2.875 18.678 3.835 99.925 82.646 97.489 0.184 1.209

15 1.5 15.250 3.125 20.346 4.169 99.825 80.979 97.156 0.201 1.259

16 1.6 16.750 3.375 22.347 4.502 99.725 78.977 96.822 0.221 1.315

17 1.7 18.250 3.750 24.348 5.003 99.625 76.976 96.322 0.241 1.366

18 1.8 20.500 4.000 27.350 5.336 99.525 73.974 95.989 0.270 1.433

19 1.9 21.750 4.500 29.018 6.003 99.425 72.307 95.321 0.286 1.467

20 2.0 23.750 4.880 31.683 6.510 99.325 69.638 94.814 0.313 1.515

21 2.1 25.750 5.255 34.355 7.011 99.225 66.970 94.314 0.340 1.556

22 2.2 28.750 6.000 38.357 8.005 99.125 62.967 93.320 0.379 1.608

GRAPHS

i. m

Versus 2( )oP P

Figure 11: Graph for m

Versus 2( )oP P

ii. m

Versus 2P

Figure 12: Graph for m

Versus 2P

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 20 40 60 80 100 120

Mas

s fl

ow

rat

e (k

g/s)

Po - P2 (KPa)

Mass Flow Rate against ( Po - P2 )

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 10 20 30 40 50

Mas

s Fl

ow

Rat

e (K

g/s)

P2 (kPa)

Mass Flow Rate against P2

iii. m

Versus 3( )oP P

Figure 13: Graph for m

Versus 3( )oP P

iv. m

Versus 3P

Figure 14: Graph for m

Versus 3P

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

92 94 96 98 100 102

Mas

s Fl

ow

Rat

e (k

g/s)

Po - P3 (kPa)

Mass Flow Rate against (Po - P3)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 2 4 6 8 10

Mas

s Fl

ow

Rat

e (k

g/s)

P3 (kPa)

Mass Flow Rate against P3

v. 2( )oP P Versus 3( )oP P

Figure 15: Graph for 2( )oP P Versus 3( )oP P

0

20

40

60

80

100

120

92 93 94 95 96 97 98 99 100 101 102

Po

-P

2 (

kPa)

Po - P3 (kPa)

(Po - P2) against (Po - P3)

Sample calculation

ℎ2 = 7.25 cm

𝑃2 = ρgh

= 13600 x 9.81 x (7.25 x 10^-2 m)

= 9.67266 kPa

ℎ3 = 0.375 cm

𝑃3 = ρgh

= 13600 x 9.81 x (0.375 x 10^-2 m)

= 0.50031 kPa

𝑷𝒐 - 𝑷𝟏 (kPa)

= 101.325 kPa – 0.1 kPa

= 101.225 kPa

𝑷𝒐 - 𝑷𝟐 (kPa)

= 101.325 kPa – 1.00062 kPa

= 100.32438 kPa

𝑷𝒐 - 𝑷𝟑 (kPa)

= 101.325 kPa – 0.33354 kPa

= 100.99146 kPa

r = 𝑃2

𝑃𝑜

r = 1.00062

101.325

r = 0.009875

𝐴2 =𝜋 × 0.0952

4= 7.088 × 10−3𝑚2

𝜌0=

𝑃𝑅𝑇

=101.325 × 103

287 × 295= 1.197 𝑘𝑔/𝑚3

�̇� = ρ𝑜𝐴2 [√2𝑘

𝑘−1 𝑃𝑜

ρ𝑜 ( 𝑟

2

𝑘 − 𝑟𝑘+1

𝑘 )] …. (kg/s)

�̇� = 1.197 × 7.088 × 10−3√2 × 1.4

1.4 − 1[101.325 × 103

1.197] [0.00987

21.4 − 0.00987

1.4+11.4 ]

0.20685 = 𝑘𝑔/𝑠

Percentage error calculation

𝑇ℎ𝑒𝑜𝑟𝑖𝑡𝑖𝑐𝑎𝑙 − 𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙

𝑇ℎ𝑒𝑜𝑟𝑖𝑡𝑖𝑐𝑎𝑙 x 100%

Theoretical value Mass flow rate

�̇�𝑚𝑎𝑥= 𝐴∗𝑃𝑜√𝑘

𝑅𝑇𝑜(

2

𝑘+1)

𝑘+1

2(𝑘−1)

= (7.0882 x 10^-3 𝑚2) (101.325 x 10^3 Pa) √1.4

287(299.15𝐾)(

2

1.4+1)

1.4+1

2(1.4−1)

= 1.678 Kg/s

Percentage error (maximum Mass Flow Rate)

1.678−1.608

1.678 x 100%

= 4.17%

Percentage error (minimum Mass Flow Rate of Po - P2)

1.678−0.207

1.678 x 100%

= 87.66%

4.0 DISSCUSION AND CONCLUSION

5.0 REFERENCES

Yunus A.Cengel. John M. Cimbala. Fluid Mechanics Fundamental and

Applications : 3rd Edition. Singapore. McGraw Hill Education.

Allan D. Kraus. James R. Welty. Abdul Aziz. 2011. Introduction to Thermal and

Fluid Engineering. CRC Press.

Bruce R. Munson. Alric P. Rothmayer. Theodore H. Okiishi. Wade W. Huebsch.

2012. Fundamentals of Fluid Mechanics. Wiley.