question 1 until 5

7/28/2019 Question 1 until 5

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TABLE OF CONTENTS

Solution Q1: ................................................................................................................................................................................ 1

Solution Q3 ................................................................................................................................................................................. 4Solution Q3: ................................................................................................................................................................................ 7

Solution Q4: .............................................................................................................................................................................. 11

Solution Q5 ............................................................................................................................................................................... 16

Solution Q6 ............................................................................................................................................................................... 19


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Q U E ST I O N 1

SOLUTION Q1:

The speed of sound,a is same at 1 and 2.

a1=a2= = =340.2 m/s

Strength of shock wave and expansion wave, which are pressure and velocity should be equal in the

region between 1 and 2. Then, P3 = P4.

First try and error approach, let

P3 = P4= 150kPa

First Step: Consider expansion wave

P3/P1 = [ ]

0.5 = [ ]

Step 2: Consider flow across the shock wave.

P4 = 150kPa

P4/P2 = 150/3 =5

Step 3 : Use Table Appendix C (Normal Shock Tube) or use software,

From Table where, P4/P2 =5,

1 3 4 2

P4=300kPa

T4=288K

P1=30 kPa

T4=288K


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2

By interpolating,

M2 = 2.11 , M4 = 0.56062 , T4/T2 = 1.7789

From M2 = 2.11,

M2 = V2 or Vs/a

M2 = Vs/a2

Vs = M2 x a2

= 2.11 x 340.2

= 717.82 m/s

From T4/T2 = 1.7789(table)

T4 = 1.7789 X T2

= 1.7789 x 288

=512 K

From equation M4 = (Vs V4)/a4

V4= Vs M4a4

V4= 463.7

The comparing V4=463.7m/s and V3=340.2 m/s , obvious difference between those values whichshould be the same. Then, these steps above repeated and iterated using Excel. The data of

iterations are shown in Table 1 below.

Summarized solution : P3 =P4 = 85.5kPa , V3 = V4=277 m/s


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Table 1 : Data of iteration to solve trial and error approach.

Assumption From table Appendix C

p3=p4

(kPa)p4/p2 M2 M4 T4/T2 T4 (K) a4(m/s) vs v4 v3 v4-v3

150 5 2.1044 0.56062 1.774192 5 10.967296 453.1077792 715.91688 461.8955968 160.3640471 301.5315498

80 2.666666667 1.558475 0.681367 1.359538 391.546944 396.6403435 530.193195 259.9355541 292.6848644 -32.74931038

90 3 1.647480519 0.6546829 1.4210425 409.26024 405.5129646 560.472873 294.9904689 268.7878622 26.20260676

85 2.833333333 1.60346 0.6674139 1.3903665 400.425552 401.1121873 545.497092 277.7892427 280.4349559 -2.645713195

88 2.933333333 1.6299566 0.6596525 1.40876972 405.7256794 403.7580686 554.511235 288.1712362 273.3784746 14.79276152

85.5 2.85 1.607986157 0.6660721 1.421398262 409.3626995 405.563722 547.036891 276.902209 279.2442023 -2.341993385

From the iteration of trial and error approach, it is found that the value of P3 and P4 is estimated 85.5kPa and the velocity at region of 3

and 4 is around 277 m/s (276.90 < V3 or V4 < 279.24).


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Q U E ST I O N 2

SOLUTION Q2

The speed of sound in the undisturbed air which is in section 1 and 2 is given by

The strength of the shock wave and expansion wave must be such that the pressure and velocity inthe region between the shock wave and the expansion wave is everywhere the same. In other

words, and .By using the simplest approach which is try and error approach, let

First consider the expansion wave. Across this wave

( ) () For the specified conditions

Hence for , this equation gives

(

) (

)

Next consider the flow across the shock wave. For For this pressure ratio, from Appendix C (Normal Shock) gives (by interpolating)

1 3 4 2

P4=500kPa

T4=293K

P1=100kPa

T4=293K


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However, is the shock Mach number

Hence for the guessed pressure

Also

()

Hence

But

from which it follows that

Hence, when it is guessed that , it is found that and .Calculations of this type have been carried out for a number of other values of the guessed

pressures, the result of some of these calculations being given in the following table.


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GUESS VALUE FROM INTERPOLATION

p3=p4 p4/p2 m2 m4 t4/t2 a4/a2 a4 Us V4 V3 V4-v3

150 1.50 1.19500 0.84500 1.12500 1.06066 363.913 410.005 102.498 271.079 -168.58

300 3.00 1.64748 0.65468 1.42104 1.19207 409.001 565.251 297.485 120.73 176.755

225 2.25 1.43924 0.72375 1.28017 1.13144 388.199 493.804 212.844 184.942 27.9015

210 2.10 1.39383 0.74236 1.25072 1.11836 383.708 478.224 193.377 199.954 -6.5769

213 2.13 1.40304 0.73845 1.25666 1.12101 384.618 481.383 197.361 196.880 0.48111

By interpolation between these results it can be deduced that when and that at this pressure

which the pressure and velocity of air between the mobbing shock wave and the moving expansion wave are 213kPa and 197

m/s respectively


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7

Q U E ST I O N 3

A shock tube containing air has initial pressures on two sides of the diaphragm of 400 kPa and

10 kPa, the temperature of the air being 25 C in both sections. If the diaphragm separating then

two sections is suddenly ruptured, find the velocity, pressure, and temperature of the air between

the moving shock wave and the moving expansion wave that are generated.

SOLUTION Q3:

In section 1 & 2 :

CTT 2521

K298

21 aa

RT

)298)(287(4.1

sm03.346 .

We know that 43 PP & 43 VV based on the graph theory of shock tube.

We assume kPaPP 10043

..

1

3

P

P=

400

100=0.25

2

1

1

3

P

P=

a

V3

2

11

P1 = 400kPa

T1 = 25 C

P2 = 10kPa

T2 = 25 CVs1

3 4

2


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4.12

14.1

25.0 =03.3462

14.11 3

V

0..820 = 03.346

2.01 3V

03.346

2.0 3V

= 820.01

= 18.0

32.0 V = )03.346(18.0 = 62.285

3V =2.0

285.62

= 311.43s

m

.. 43.31143 VV sm , kPaPP 10043

In shock wave region, 2 & 4 :-

4VVs sV

2

4

P

P=

10

100=10

refer shock wave table :-

102

4 P

PM2 = 2.95, M4 = 0.478, 623.2

2

4 T

T

a

VM

s2 aMVs 2


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= 2.95(346.03)

= 1020.79s

m .

623.22

4 T

T

24 623.2 TT

= 2.623(298) = 781.65K

44 RTa

= )65.781)(287(4.1 = 560.4s

m

4

44

a

VVM

s

0.478 =4.560

79.1020 4V

.. )4.560478.0(79.10204 V = 752.9 sm

Hence, when it is guessed that kPaPP 10043 , it is found that 3V = 311.43 sm and

4V = 752.9 sm .

Calculations of this type have been carried out for a number of other values of

guessed pressures, the result of some of these calculations being given in table below.


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)(43 kPaPP

2

4

P

P

2M 4M 2

4

a

a

)(4s

ma

)(s

mVs

)(4 smV )(3 s

mV )(34 smVV

100 10.0 2.95 0.478 1.620 560.4 1020.8 752.9 311.4 441.5

80 8.0 2.65 0.501 1.512 523.2 915.2 653.1 355.4 298

60 6.0 2.30 0.534 1.395 482.7 795.5 537.7 410.7 127

47 4.7 2.04 0.570 1.312 454.1 705.90 447.06 456.36 -9.3

45 4.5 2.00 0.577 1.299 449.5 692.0 432.6 463.8 -31

40 4.0 1.89 0.598 1.265 437.7 654.0 392.6 484.9 -92

By interpolation between these result, the pressure and velocity of air between the

moving shock wave and the moving expansion wave are :-

kPaPP 4743

45243 VV sm

The temperature of air between the moving shock wave and the moving expansion

wave is :-

1

3

T

T=

1

1

3

P

P=

4.114.1

400

47

1

3

T

T= 0.542

3T = 0.542 1T = 0.542 (298) = 161.5K


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Q U E ST I O N 4

The air pressures in the high and low pressure sections of a constant diameter shock tube are 600

kPa and 20 kPa respectively. The temperature in both sections are 30 C. After the diaphragm

that separates the two sections is ruptured, a shock wave propagates into the low pressure section

and an expansion wave propagates into the high pressure section. Find the air velocity and

temperature between the two shock waves and the velocity of the shock tube.

SOLUTION Q4:

In section 1 & 2 :

CTT 3021

K303

21 aa

RT

)303)(287(4.1

sm92.348 .

We know that 43 PP & 43 VV based on the graph theory of shock tube.

We assume kPaPP 30043

..

1

3

P

P=

600

300=0.5

P1 = 600kPa

T1 = 30 C

P2 = 20kPa

T2 = 30 CVs1

3 4

2


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2

1

1

3

P

P=

a

V3

2

11

4.12

14.1

5.0 = 92.3482

14.1

1

3V

0..906 = 92.348

2.01 3V

92.348

2.0 3V

= 906.01

= 094.0

32.0 V = )92.348(094.0 = 32.798

3V =2.0

798.32

= 163.99s

m

.. 99.16343 VV sm , kPaPP 30043

In shock wave region, 2 & 4 :-

4VVs sV

2

4

P

P =20300 =15

refer shock wave table :-

152

4 P

PM2 = 3.605, M4 = 0.447, 461.3

2

4 T

T


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13

a

VM

s2 aMVs 2

= 3.605(348.92)

= 1257.86s

m .

461.32

4 T

T

24 461.3 TT

= 3.461(303) = 1048.68 K

44 RTa

= )68.1048)(287(4.1 = 649.12s

m

4

4

4 a

VVM

s

0.447 =12.649

1257.86 4V

.. )12.649447.0(1257.864 V = 967.7 sm

Hence, when it is guessed that kPaPP 30043 , it is found that 3V = 163.99 sm and

4V = 967.7 sm .


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Calculations of this type have been carried out for a number of other values of

guessed pressures, the result of some of these calculations being given in table below.

)(43 kPaPP

2

4

P

P

2M 4M 2

4

a

a

)(4s

ma

)(s

mVs

)(4 smV )(3 s

mV )(34 smVV

300 15.0 3.605 0.447 1.860 649.12 1257.86 967.7 163.99 803.71

200 10.0 2.952 0.478 1.620 565.10 1030.01 759.89 254.71 505.18

100 5.0 2.104 0.561 1.332 464.73 734.13 473.42 393.95 79.47

88 4.40 1.978 0.581 1.292 450.90 690.16 428.19 418.36 9.83

87 4.35 1.968 0.583 1.289 449.75 686.50 424.30 420.97 3.33

86 4.30 1.957 0.585 1.286 448.60 682.84 420.41 423.58 -3.17

82 4.1 1.912 0.593 1.272 443.83 667.14 403.95 432.11 -28.16

80 4.0 1.890 0.598 1.265 441.35 659.46 395.53 436.74 -41.21

50 2.5 1.51 0.697 1.152 402.09 526.87 246.61 521.75 -275.14

By interpolation between these result, the velocity of air between two waves is :-

42243 VV sm

The velocity of the shock wave,s

V = 682.84s

m

The temperature of air between two waves is :-

1

3

T

T=

1

1

3

P

P=

4.114.1

600

86


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1

3

T

T= 0.573

.. 3T = 0.573 1T = 0.573 (303) = 173.62K

44 RTa

4)287(4.160.448 T

.. 4T = 500.85 K


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Q U E ST I O N 5

SOLUTION Q5

The speed of sound,a is same at 1 and 2.

a1=a2= = =347.189 m/s

Strength of shock wave and expansion wave, which are pressure and velocity should be equal in the

region between 1 and 2. Then, P3 = P4.

By using try and error (Check Table 5), the P4 and P3 is found around 38kPa and the velocity V3 and

V4 is 615m/s.

M2=2.563635

M4=0.5070847

T2/T4= 2.201253108

P3=P4= 38kPa

V4= 615 m/s

To find T4,

T4 = 2.2013 x 300K

= 660.39 K

To find speed of sound at region 4,

a4 = = =515.12 m/s

To get V4, V2 has to be determined first

1 3 4 2

P4=8 atm or

810kPa

T4=300K

P1=0.05 atm or

5.066 kPa

T4=300K


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V4 = V2 M4a4

V2 = M2 x a2

=2.56 x 347.189

=888.88 m/s

Then,

V4 = 888.88 0.5071(515.12)

= 627.586 m/s

V4 = V3 = 627.586 m/s

To get T3, it can be deduced from the equation

()

( )

M3 = 2.7408

From M3 = V3/a3

a3 = V3/M3

T3 = 130.49 K

Summarized solution, P3= P4= 38kPa , T3= 130.49K, T4 = 660.39K, V3=V4=627.586 m/s


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Table 5: Iteration for trial and error for Question 5

Assumption From table Appendix C

p3=p4 (kPa) p4/p2 M2 M4 T4/T2 T4 (K) a4(m/s) vs v4 v3 v4-v3

30 5.92183182 2.284436051 0.5362932 1.932582333 556.5837119 472.9009785 777.165145 523.5515476 651.8797492 -128.3282016

35 6.90880379 2.462650319 0.5166476 2.100784663 605.0259829 493.0511535 837.793639 583.0599345 627.7421808 -44.68224627

37 7.303592578 2.530412904 0.5101216 2.167793317 624.3244753 500.8528468 860.84647 605.3506144 618.9096447 -13.55903039

40 7.89577576 2.628797484 0.501421 2.2680818 653.2075584 512.3073267 894.316904 637.4352464 606.3992795 31.03596686

37.1 7.323332017 2.5337545 0.5098108 2.171138336 625.2878408 501.239119 861.983281 606.4461526 618.4788557 -12.03270307

38 7.500986972 2.563635 0.5070847 2.201253108 633.9608951 504.703366 872.148627 616.2212862 614.645903 1.575383168


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Q U E ST I O N 6

SOLUTION Q6

The speed of sound in the undisturbed air which is in section 1 and 2 is given by

The strength of the shock wave and expansion wave must be such that the pressure and velocity

in the region between the shock wave and the expansion wave is everywhere the same. In other

words, and .By using the simplest approach which is try and error approach, let

First consider the expansion wave. Across this wave

( ) () For the specified conditions

Hence for , this equation gives

( ) ()

Next consider the flow across the shock wave. For

1 3 4 2

P4=600kPa

T4=300K

P1=20kPa

T4=300K


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For this pressure ratio, from Appendix C (Normal Shock) gives (by interpolating)

However,

is the shock Mach number

Hence for the guessed pressure

Also

(

)

Hence

But

from which it follows that

Hence, when it is guessed that , it is found that and .Calculations of this type have been carried out for a number of other values of the guessed

pressures, the result of some of these calculations being given in the following table.


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GUESS

VALUEFROM INTERPOLATION

p3=p4 p4/p2 m2 m4 t4/t2 a4/a2 a4 Us V4 V3 V4-v3

150 7.50 2.56347 0.50710 2.20109 1.48361 515.108 890.037 628.826 274.318 354.508

60 3.00 1.64748 0.65468 1.42104 1.19207 413.888 572.005 301.04 453.657 -152.62

80 4.00 1.88980 0.59764 1.59999 1.26491 439.176 656.137 393.67 399.858 -6.188

82 4.10 1.91234 0.59324 1.61757 1.27184 441.582 663.965 401.999 395.137 6.86224

81 4.05 1.90112 0.59540 1.60879 1.26838 440.382 660.07 397.865 397.485 0.37971

By interpolation between these results it can be deduced that when and that at this pressure

which the pressure and velocity of air between the mobbing shock wave and the moving expansion wave are 81kPa and 397

m/s respectively.


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By using try and error (check table 6), the is found around 81kPa and the velocity is 397 m/s.

In order to find

To find speed of sound at region 4

To get has to be determined first since

Thus

Then,

is


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Summarized solution,

question 1 until 5

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