E Rathakrishnan Gas Dynamics Solutions

Solutions Mannual for the fourth edition of

Gas Dynamics Ethirajan Rathakrishnan

Preface This manual gives the detailed solution for all the problems given at the end of di fferent chapters of the 3rd edition of Gas Dynamics. My sincere thanks to my doctoral and masters students who helped me in checking and keying in the solutions of this manual. My sincere thanks to the Continuing Education Centre of Indian Institute of Technology Kanpur for the financial support to prepare this manual. E Rathakrishnan

i

ii

Contents 1 SomePreliminaryThoughts

1

2 Basic Equations of Compressible Flow

3

3 WavePropagation

23

4 One-DimensionalFlow

25

5 NormalShockWaves

79

6 Oblique Shock and Expansion Waves

119

7 Potential Equation for Compressible Flow

157

8 SimilarityRules

161

9 Two Dimensional Compressible Flows

165

10Prandtl-MeyerFlow

169

11 Flow with Friction and Heat Transfer

173

MOC 12

205

13 Measurements in Compressible Flow

iii

207

Chapter 1

Some Preliminary Thoughts

1

2

Some Preliminary Thoughts

Chapter 2

Basic Equations of Compressible Flow 2.1 In the reser voir, the air is at stagn ation state . So, the entropy relation would be T02 p02 s2 s1 = c p ln R ln T01 p01

 −  

−

But, T 01 = T02 for adiabatic process. Therefore, ∆s

= R ln = R ln =

    p01 p02

p01 1 2 p01

= R ln 2

198.933 J/(kg K)

Note: It should be noted that, for entropy only subscripts 2 and 1 are used;

since entropy is not defined like static or stagnation entropy. 2.2 Let the initial state be denoted by subscript 1 and expanded state by subscript 2. (a) Since the cylinder is insulated, preventing any heat transfer what-so-ever, the process is adiabatic. The governing equation for this process is given by γ p1 Vγ1 = p 2=Vconstant 2

(1)

Also, from ideal gas state equation p1 V1 p2 V2 = =R T1 T2 3

(2)

4

Basic Equations of Compressible Flow

From Eqs. (1) and (2), we get p1 = p2

γ

    V2

T1 T2

=

V1

γ /(γ

−1)

Therefore, T2

= T1

∆T

=



10(γ −1) = 557.35 K

− 842.65 K

(b) Work = Also,



pdv =

 − − dh

pv γ = constant

du

vdp

(3)

from equation (1)

Differentiating equation (1), we have, pγ v γ −1 dv + v γ dp = 0 Dividing throughout by v γ −1 and integrating, we get



pγ dv +





vdp =

vdp = 0

−γ w

(4)

Substituting equation (4) in equation (3) and simplifying, we get (1

− γ) w w

= R ∆T =

R∆ T 287 ( 842.65) = 1 γ ( 0.4)

=

6.04

×− −

−

× 105 J/kg

Note: Since the process undergone is expansion from a high pressure, the work

removed is positive, i.e, work is done by the gas. (c) Also, from equation (1) p1 = p2 Therefore,

  V2 V1

γ

= 10 1.4 = 25.1189

The pressure ratio =

25.1189

5 2.3 p 1 v1γ = p2 v2γ , where v is specific volume, i.e. volume per unit mass = Therefore, p1 Also,

V1

=

V2

=

V

γ

    V1

V2

= p2

m1

γ

m2

= volume of the tank.

γ

p2

=

p1

=

5

=

m2 m1

 

× 105 ×

 1 2

1.4

× 105 Pa

1.8946

From equation of state for a calorically perfect gas, p1 p2

=

ρ1 T1 ρ2 T2

T2

=

   p2 p1

1.8946 5

=



=

T1

× 2 × 500



378.92 K

2.4

 

p1 = p2

T1 T2

(a) Therefore, T2

m1 m2

=



=

61/3.5

p2 p1

γ /(γ

(γ −1)/γ

−1)

T1

× 290 = 483.868K

The change in the temperature is ∆T

= =

T2

− T1 = 483.868 − 290

193.868K

(b) By first law of thermodynamics, we have du + d(pe) + d(ke) = dq + dw

V/m.

6


Here, velocity changes are neglected. Therefore, d(ke) = 0 Also, assuming d(pe) = 0 The first law of thermodynamics reduces to du = dq + dw But the process is isentropic, thus dq = 0. Therefore, du

=

dw = c v ∆T = 717.5

=

1.39

× 193.868

× 105 J/kg

(c) The work don e is negative, i.e. work is done on the gas . It has bee n computed in (b) above. 2.5 Work done by the weight on the piston goes towards increasing the internal energy of the gas. From the first law of thermodynamics E2

− E1 = Q + W

where, E , Q, and W are respectively the internal energy, heat transfered, and work done. Since no heat is transfered, Q = 0. Therefore, E2

− E1 = W =



F .ds

where, F is force and ds is distance. At the new equilibrium position, the force acting on the piston face is F = p2 Ap , Ap is the area of the piston fac e. The distance traveled by the piston is ds = (V1 V2 )/Ap , V1 and V2 are the initial and final volumes. Thus we have,

−

E2

− E1

=

p2 . Ap (V1

− V2 )/Ap

=

−p2 (V2 − V1 )

For unit mass, e2

− e1 = −p2 (V2 − V1 )

For calorically perfect gas, e = c v T . Therefore

RT2 cv (T2 cv R

− T1 ) T2 −1 T1

   

T2 cv 1+ T1 R

= = =

− p2 p2 − − TT21 + pp21



RT 1 p1



cv + λ (where λ = p2 /p1 ) R

7 But

cv 1 = . Thus, R γ 1

−

T2

γ γ

= λ+

− 1 T1 T2 T1

1 γ

−1

1 + (γ

1)λ

−

=

γ

Entropy change for a perfect gas can be written as ∆s

∆s

= ln

R s2

− s1

Let λ = 1 + , where  ∆s

=

1

p2 p1

R ln

1 + (γ

λ

γ

1)λ

γ −1

γ

1

1 + (γ

λ

1)λ

γ

γ −1

γ

 1. Therefore, γ

−1 γ ln γ−1 γ

−1

γ

1 + (γ

ln

γ

=

T2 T1

= R ln

=

R

 −    −    −  

= cp ln

 

− 1)(1 + ) γ

1+γ

ln 1 +

γ

ln(1 + )

 −

− 1 + (γ − 1) − ln(1 + ) γ



− 1  − ln(1 + ) γ

Expanding the RHS, and retaining only upto second order terms, we get, ∆s

R

=

γ γ

−1



γ

− 1  − ( γ − 1)2 2 −  −  2 γ 2γ 2 2

  

2

=

=



2

− γ 2−γ 1 2 −  + 2 = − 2

+

2

2γ

+

2

2

2 2γ

Note: Work has been done by the weight which is equal to

The weight has moved by a distance of ds. Therefore, 2.6 Since it is an open system, Work done =

−cp (T2 − T1 )

∆E

p2 Ap on the gas. = W . ds = p2 Ap . ds.

8


=

−cp

 −

=

−cp

p2 p1

=

T2 T1

1 T1 γ −1 γ

−1

  ×

−1004.5

=

21/3.5

T1



− 1 × 303

− 66.66 kJ/kg

2.7 Work done is given by W

= =

p (V2

− V1 ) = 101325 × 6 (2 − 0.3)

1.0335 MJ

since 1 atm = 101325 Pa. 2.8 The compression process is given as isentropic. Let subscripts 1 and 2 refer to initial and final states, respectively. By isentropic process relatio n, we have p1 p2 = γ γ ρ1

ρ2

ρ2



= =

p2 p1

1/γ

ρ1 =

  690 150

1/1.3

× 1.5

4.85kg /m3

2.9 As we know, the relation between temperature and pressure for isentropic change of state may be written as T2 = T1

 p2 p1

(γ −1)/γ

where subscripts 1 and 2 refer to the initial and final states, respectively.

T2

= =

T1

 p2 p1

519.9 K

0.4/1.4

= 298

 7 1

0.286

9 2.10 By isentropic relation, T2 = T1

 v1 v2

(γ −1)

where subscripts 1 and 2 refer to the initial and final states and

v is specific

volume. For air γ = 1.4. Therefore, T2

=

T1

 v1 v2

0.4

= (30 + 273 .15)(30) 0.4 = 1181.7 K 908.55◦ C

=

2.11 (a) We have c p =

γ

− 1 R, therefore, the gas constant R = 0.4 × 1000 = 285.7 J/(kg K) R= 1.4 γ

γ

− 1 cp γ

Also,

R 8314 R = Mu = M where M is the molecular weight and R u is universal gas constant. Thus, M=

8314 = 29.1 285.7

(b) By ideal gas state equation, we have p1 V1 = mRT1 p2 V2 = mRT2 where subscripts 1 and 2 refer to initial and final state s, respectively. But p1 = p 2 and therefore, V2 V1

=

T2 50 + 273.15 323.15 = = T1 200 + 273.15 473.15

=

0.683

2.12 For an ideal gas, the speed of sound a may be expressed as a=



γ RT

10


where γ is the ratio of specific heats and R is the gas constant. For the given gas, R =

Ru /M = 8314/29 = 286 .7 J/(kgK)

Therefore, 400 = =

γ

γ

×

286.7

× 373.15

4002 = 1.5 286.7 373.15

×

The specific heat c p and c v can be written as cp

=

cv

=

γ

γ

−1R R

γ

−1

Therefore, cp

= (1.5/0.5)

=

× 286.7

860.1 J/(kg K)

cv = 286 .7/0.5 =

573.4 J/(kg K)

Note: The ratio of specific heats γ = cp /cv . For the pres ent case γ = 1.5 =

860.1/573.4 is corre ct. This way the answer obtained for cp and cv may be checked. 2.13 At the nozzle exit, V = 390m/s and corresponding speed of sound is a =



γ RT =

T = 28 + 273 .15 = 301 .15K. The

√1.4 × 287 × 301.15

= 347 .85 m/s Thus, M

=

V 390 = a 347.85

=

1.12

11 By isentropic relation, we have T0 T

=

γ

1+

− 1 M2 2

= 1 + 0 .2 T0

= 1.25

× 1.122 = 1.25

× 301.15 = 376 .44 K

103 .29◦ C

=

For the flow, the stagnation temperature is T 0 = 376.44K. The static temperature

T = 92.5◦ C = 92.5 + 273.15 = 365 .65 K

T0 376.44 = = 1.03 = 1 + 0 .2 M 2 T 365.65 Thus, M2

=

0.03 = 0.15 0.2

M

=

0.387

This is the Mach number at the station where temperature is 92.5 ◦ C. 2.14 For hydrogen, the gas constant R = 8314/2.016 = 412 4 J/(kg K). By isentropic relation, we have T2 T1

=

=

  p2 p1

(γ −1)/γ

0.286

1 7

= 0.573

Therefore, T2

= 0.573

× T1

= 0.573 By energy equation, we have h2 +

300 = 171.9 K

× V 22 = h 1 , since V1 = 0 2 V22 = 2 (h1

− h2 )

12 But


h = cp T ,

therefore, V2 =



cp

2cp (T1

− T2 )

γ

=

1

γ

R=

1.4 0.4

× 4124

−

= 14434J/(kg K) Thus, V2

=

×

=

1923m/s

2

14434(300

− 171.9)

The speed of sound is given by a2

=



γ RT2 =

√1.4 × 4124 × 171.9

= 996 .23 m/s Thus, M2 =

V2 1923 = = 1.93 a2 996.23

The mass flow rate is given by ˙ = ρ 2 A2 V2 m ρ2

=

p2 , by state equation RT2

ρ2

=

101325 , since 1 atm =101325 Pa 4124 171.9

=

0.143kg/m 3

×

Thus, m ˙ = =

0.143

× 10 × 10−4 × 1923

0.275kg/s

2.15 The given process is a polytropic process with index n = 1.32. Since air is given as an ideal gas with constant specific heats, we have from isentropic relations the change of entropy as s2

− s1 = cp ln

 −   T2 T1

R ln

p2 p1

13 For a polytropic process, T2 = T1

 p2 p1

n−1 n

Combining these two equations, we obtain

− s1 =

s2 since c p =

γ

γ

  − −    cp

n

n

1

R ln

p2 p1

− 1 R for an ideal gas, the above equation may be written as (n − γ ) R p2 s2 − s1 = ln n (γ − 1) p1



With the given data, γ

R=

− 1 cp = 0.4 × 1004 = 287 J/(kg K) 1.4

γ

Therefore, s2

− s1

=

(1.32 1.4)287 1100 ln 1.32 0.4 101

=

−103.8J/(kgK)

−

×

Note: Since the entropy of the gas decreases for this internally reversible process,

heat must be removed from the gas. This is why the cylinder used for suc h a compression process is usually water jacketed. Also, we know that the entropy of the system and surrounding cannot decrease. But in this problem, the entrop y decreases. It shoul d be noted that, what decreases is entropy of the syst em alone and not the combined entropy of the system and surrounding. 2.16 For oxygen, molecular weight M = 32. The gas constant R = 8314/32 = 259.8 J/(kg K). Therefore, cp

=

cv

=

γ

γ

− 1 R = 909.3J/(kgK)

cp γ

= 649.5J/(kgK)

The increase in internal energy is

u

= cv (T2 =

− T1 ) = 649.5 (125 − 25)

64950 J/(kg K)

14


The increase in enthalpy is

h

= cp (T2 =

− T1 ) = 909.3 (125 − 25)

90930 J/(kgK)

2.17 Let the subscripts 1 and 2 refer to the inlet and exit states. At state 1, p1 = 100 kPa, ρ1 = 1.175 kg/m 3 . Therefore, T1 = At state 2,

p1 100 103 = = 296.5 K R ρ1 287 1.175

× ×

p2 = 500 kPa, ρ 2 = 5.875 kg/m 3 . Therefore, T2 =

p2 500 103 = = 296.5 K R ρ2 287 5.875

× ×

Assuming air to be a perfect gas, the enthalpy di h2

− h1

= cp (T2 =

fference

− T1 ) = cp (296.5 − 296.5)

0

2.18 The entropy change is given by ds =

δq

T

where δ q is the reversible heat addition per unit mass. For an ideal gas δ q = dh vdp

−

where dh is the enthalpy change and dh = c p dT δ q = c p dT

− vdp

Using the above relation, we get ds = c p dT T By state equation

is given by

− Tv dp

pv = RT , v R = T p

15 Therefore, ds = c p Taking log of state equation and di

dT T

− Rp dp

fferentiating,

dT

dp

we get

dv

T = p + v Substituting for

dT is ds expression, we obtain T ds = c p



ds = c p But

cp

− cv = R,

cp

dp dv + p v

−

dv + (cp v

− R) dpp

R

dp p

− R = cv . Thus, ds = c p

dv dp + cv v p

For an isentropic change of state, we have = cp dv + cv dp v p

0 dv v cp dv cv v cp

But

=

−cv dpp

=

− dpp

cp = γ . Therefore, cv γ

dv = v

− dpp

Integrating both sides, we get γ ln v

=

− ln p + constant

γ

ln p + ln v

= constant

γ

ln (pv ) = constant or pv γ = constant

16


2.19 For an isothermal process, the change in entropy

s = s2 − s1 = R ln

 p1 p2

s is given by

where the subscripts 1 and 2 stand for the initial and final states, respectively. By state equation p1 V1

=

mRT1

p2 V2

=

mRT2

But T 1 = T 2 , therefore, p 1 V1 = p 2 V2 = mRT2 . Hence, p2

p1 V1

=

V2

=

=

0.7

× 106 × 0.014 0.084

0.117MPa

Thus, the change of entropy is

s

= R ln =

 p1 p2

= 287 ln

  0.7 0.117

513 .4J/(kgK)

2.20 Let subscripts 1 and 2 refer to initial and final states, respectively. For process 1

p2 = 700 kPa.

By state equation, we have p1 V = mRT1 p2 V = mRT2 Therefore, T2

=

p2 p1

× T1 = 700 × 308.15 = 616 .3 K 350

The change in entropy is given by

s

= cp ln

T2 T1

R ln

p2 p1

 −    −  

For air c p = 1004.5 J/(kg K) and R = 287 J/(kg K). Thus,

s

= 1004 .5 ln

616.3 308.15

= 497 .33J/(k g K)

287ln

700 350

17 For 0.3 kg of air,

s = 0.3 × 497.33 =

149.2J/K

For process 2 V

p

=

1

mRT1

pV2 = mRT2 Therefore, T2

V2

=

V1

× T1

Initial volume is given by V1

=

mRT1 0.3 287 308.15 = = 0.0758 m3 p 350 103

T2

=

0.2289 0.0758

×

× ×

× 308.15 = 930.55 K

The entropy change is given by

s

=

cp ln

  T2 T1

= 1004.5 ln

930.55 308.15

 

= 1110.163J/(kg K) For 0.3 kg of air,

S = 0.3 × 1110.163 =

333 J/K

2.21 Given flow is adiabatic and frictionless and therefore, isentropic. By energy equation, we have h1 +

V 12 V2 = h2 + 2 2 2

For perfect gas h = c p T , thus, cp T1 + V 12 = c p T2 + V 22 2 2 For air c p = 1004.5 J/(kg K). Therefore, 1004.5

× 273.15 + 9002

2

= 1004 .5 T 2 +

300 2 2

18


Solving this, we get T2 = 631.54 K. Therefore, the temperature increase becomes

T

=

T2

− T1 = 631.54 − 273.15

=

358 .39 K

By isentropic relation, we have p2 = p1

  T2 T1

γ γ −1

Thus, p2

= p1 =

p

T2 T1

3.5

= 140



631.54 273.15



3.5

= 2631 kPa

2.631 MPa

= p2 =

2.22 For nitrogen,

  − p1

2.491 MPa

R = R u /M = 8314/28 = 297 J/(kg K).

For reversible process, the change in entropy may be expressed as

s = cp ln

 −   T2 T1

R ln

p2 p1

But T2 = T 1 , therefore,

s

= =

− R ln

 p2 p1

=

− 297 ln

  300 100

− 0.3263 kJ/(kg K)

Note: We know that the entropy of the system and surrounding cannot decrease.

But in this problem, the entropy decreases. It shoul d be noted that, what decreases is entropy of the system alone and not the combined entropy of the system and surrounding. 2.23 By isentropic relation, p2 = p1

  T2 T1

γ γ −1

19 where subscripts 1 and 2 refer to initial and final states. That is, T2 T1

=



T2

=

T1

p2 p1

γ −1 γ

p2 p1

0.4 1.4

  

= 300

550 110

0.286

= 475.4 K

The change in enthalpy is ( h2

− h1 ), and h = cpT . Therefore, h2 − h1 = c p (T2 − T1 )

Also, cp =

γ

γ

−1R

and R = 287 J/(kg K), for air. cp =

1.4 0.4

× 287 = 1004 .5 J/(kgK)

Thus, h2

− h1

= 1004 .5 (475.5 =

− 300)

176 .19 kJ/kg

2.24 Let the initial and final states of air are designated by subscripts 1 and 2, respectively. By perfect gas state equation, we have p1 V =

mRT1

p2 V =

mRT2

Dividing one by other, we get

p1 T1 = p2 T2

The change in entropy for a perfect gas is given by T2 s = cp ln R ln T1



=

(cp

 −     − R) ln

T2 T1

p2 p1

20


Given, T1 = 50 + 273 .15 = 323.15 K T2 = 125 + 273 .15 = 398.15 K cp

γ

=

γ

− 1 R = 1004.5J/(kgK)

Thus,

s

= (1004 .5 =

− 287) ln



398.15 323.15



149 .75J/(k g K)

2.25 By energy equation, we have h0 = h +

V2 2

where subscript 0 refers to stagnati on condition. Assuming air to be a perfect gas, we can express h = c p T . Therefore, c p T0 T0

= cp T + =

−T

V2 2

V2 2 cp

From standard atmosphere table, at 10000 m, T = 223.15 K. Therefore, the speed of sound becomes a =



γRT =

√1.4 × 287 × 223.15

= 299 .436m/s The flight speed is V =M a= 2

× 299.436 = 598.872m/s

Thus, the temperature becomes 2

∆T

− T = 2598.872 × 1004.5

=

T0

=

178 .52

21 2.26 Let subscripts 1 and 2 refer to the initial and final states. Given, v1 = 0.06 m3 ,

T1 = 15◦ C = 288 .15 K,

v2 = 0.12 m3 ,

By perfect gas state equation, pv = RT Therefore,

T1 T2 = v1 v2

since p 1 = p 2 . Thus, T2

=

T1 288.15 v2 = v1 0.06

=

303 .15◦ C

× 0.12 = 576.3 K

p1 = p 2

22


Chapter 3

Wave Propagation

23

24

Wave Propagation

Chapter 4

One-Dimensional Flow 4.1 Total temperature at “1” is T01 = 300 K. This much temperature is required as static temperature at the test-section. Therefore, TT = 300K T0T TT

= 1+

T0T 300

= 1+

γ

− 1M2 2

0.4 2

× 2.52

T0T = 675K Hence, the temperature rise required, ∆T

=

T0T

− TT = 675 − 300

=

375K

4.2 Let the test-section conditions be denoted by subscript 2, and the sonic conditions by superscript *. T 02 T2

=

1

373 T2

= 1 + 0 .2

γ

1+

−2

× 22

T2 = 207 .2 K 25

M2

26

One-Dimensional Flow

Further, since we know that T∗ T0

= 0.8333

T ∗ = 373

× 0.8333 = 310 .82 K

√1.4 × 287 × 310.82 = 353 .4 m/s = V ∗

a∗

=

p0

= 3.12

ρ0

=

p0 RT0

=

3.16 287

× 101325 = 3 .16 × 105 × 105 × 373

= 2.95 kg/m3 ρ∗ ρ0

= 0.634

ρ∗

= 1.8703 kg/m

3

m ˙ = ρ∗ A∗ V ∗ = 1.8703 =

× 80 × 10−4 × 353.4

5.288 kg/s

4.3 Let subscripts 0, 1, and e refer to stagnation state and states at the nozzle entrance and exit, respectively. We know that, Te = 0.8333 T0 This gives,

T0 =

293 = 351 .6 K 0.8333

Also, T0 T1

=

1+

γ

− 1M2 2

27

T1

=

351.6 1 + 0.2 9

=

125 .57 K

×

pe p0

= 0.5283

p0

=

p0 p1

= =

p1

V

0.8 = 1.5143 atm 0.5283

 

γ

1+

− 1M2 2

1 + 0.2

γ γ −1

× 32 3.5



=

1.5143 36.73

=

0.04123 atm

= a =



√1.4

287

293 = 343 .11 m/s

e

e

Ae

= 40

ρe

=

pe RTe

=

0.8 101325 = 0.964kg/m 3 287 293

×

×

× 10−4 m2 ×

×

m ˙ = ρe Ae Ve = 0.964 =

× 40 × 10−4 × 343.11

1.323 kg/s

Note: It should be noted that the calculation made with equations for pressure,

density, and temperature in the problem can also be done using gas tables. In fact that procedure will result in considerable time saving. 4.4 Given, A1

= 0.6

× 0.4 = 0.24 m2

28


× 0.4 = 0.12 m2

A∗

=

0.3

A2

=

h2

M2

=

2.5

T2

=

− 10◦ C

p2

=

0.15atm

× 0.4 = 0.4 h2 m2

For M 2 = 2.5, isentropic table gives A2 = 2.6367 A∗

and

T2 = 0.4444 T02

This gives, A2 = 2.6367 Thus, h2

T02

× 0.12 = 0 .3164 m2

=

0.3164 0.4

=

0.791m

=

263 0.4444

=

592K

A1 0.24 = =2 A∗ 0.12 A1 From subsonic part of isentropic table, for A = 2, we get ∗

M2 = 0.3

T1 = 0.9823 T01

and

For isentropic flow, T 01 = T 02 . Therefore, T1

= 0.9823 =

V1

× 592

581 .5 K



= M1 a1 = 0.3 γ R T =

145m /s

29 4.5 For M 1 = 0.5, from isentropic table, we have p1 p0

=

0.84

Therefore, p2 p0

=

× pp10 = 1.0 × 0.84 6.5

p2 p1

= 0.129 For p/p 0 = 0.129, from isentropic table, we have M2

=

2.0

A∗ A2

=

0.6

Mass flow rate is given by m ˙ =

ρ1 V1 A1 = p 1 M1 A1



γ

RT1 1.4

= =

6.5

×

× 101325 × 0.5 × 0.016

17.54 kg/s

287.4

× 440

4.6 If p 0 is the stagnation pressure and A ∗ is the critical throat area, p∗ p0

=

A A∗

=



1 + 0.2M 2

1 M





−3.5

5 + M2 6



3

With these relation we obtain, p1 p2 A1 A2

=

=

1 + 0.2M12

−3.5

2

1 + 0.2M2 M2 5 + M12 M1 5 + M22

  

3

where subscripts 1 and 2 refer to entrance and throat of venturi. Substituting proper values into the above two relations, we get

30


1.5 1.2



=

4

1 + 0.2M12 1 + 0.2M22

5 + M12

M2

=



−3.5 3

2

3

M1 M2 M1

=



5 + M2 1 + 0.2M12 1 + 0.2M22



3.0

These two simultaneous equations can be solved to get M 1 and M 2

4 3 Therefore,

1.5 1.2

3.0/3.5

1.5 1.2

3.0/3.5

   

1 + 0.2M12 1 + 0.2M22

=



=

M1 M2



6/7

M2 M1

=

4 3



5 4

=



1 + 0.2M12 −3.5 1 + 0.2M22

From above two equations, we get,

5 4

−3.0

= 1.61



M1

=

0.46

M2

=

0.74

4.7 p 0 is the total pressure and p ∞ is the static pressure. 1 2 ρV q∞ = 2 p0

− p∞

= 490 mm of Hg = 0 .653

× 105 Pa

× 105 = 1.3632 × 105 Pa

p∞

=

(0.35 + 1.0132)

p0

=

(0.653 + 1.3632)

T0

=

25◦ C = 298K

× 105 = 2.0162 × 105 Pa

31 Therefore, T∞

=

  p∞ p0

T0

0.286

1.3632

= 298

×

But,

0.286

2.0162

T0

=

2 T∞ 1 + 0.2M∞

2 M∞

=

T0 T∞ 298 = 0.2T∞ 0.2

M∞

=

0.776

a∞

=





−

= 266 K





− 266 × 266 = 0.602

√1.4 × 287 × 266 = 326 m/s

γ RT∞ =

V∞ = 253m /s 4.8 pi

=

1.0atm

pf

=

6.0atm

T1 = 29 0 K

For isentropic compression, we have p2 = p1

γ

    ρ2 ρ1

=

T2 T1

γ γ −1

(a) Tf = Ti

  pf pi

γ γ −1

0.4

= (6) 1.4 = 1.67

Tf = 290

× 1.67 = 484 K ∆T = 484 − 290 = 194 K (b) Change in the internal energy is given by, de = = 1.39

× 105 (N m)/kg

cv dT =

287 0.4

× 194

32


(c) Since the process is isentropic, from first law of thermodynamics, we have de = δ q + δ w, and δ q = 0. Hence, work imparted to the air becomes, δ w = δ e = 1.39

× 105 N m/kg

4.9 a = M

=



γ RT = 347.21m/s

180 = 0.518 347.19

The maximum pressure that can be achieved is the isentropic stagnation pressure. Therefore, p0 p

=



p0

=

1.013

=

1+

γ

− 1M2 2



γ γ −1

= 1.201

× 105 × 1.201Pa 105 pa

1.217

× 4.10 (a) The critical pressure ratio for the sonic condition at the nozzle exit is p∗ = 0.528 p0 If the nozzle is choked, then p2

= p∗ = 0.528 =

3.64

× 6.895 × 105

× 105 Pa

since, p2 > patm , the flow is choked. Hence, the pres sure in the exit plane is p2 = p∗ (b) The minimum stagnation pressure for choked flow occurs when p2 = p ∗ = p atm Thus, p0min =

p∗ patm = = 1.92 0.528 0.528

× 105 Pa

33 (c) For p 0 = 1.724 105 Pa, p∗ = 0.528 p0 = 0.91 105 Pa. Since p ∗ < patm flow will not be choked. Further, subsonic flow is always correctly expanded, hence, p2 = p atm = 1.014 105 Pa

×

×

×

×

ff

4.11 Let the reaction force acting on the di user be R. Then, R + p1 A1 p2 A2 = mV ˙ 2 mV ˙ 1

−

−

R

=

m(V ˙ 2

− V1 ) − (p1 A1 − p2 A2 )

where subscripts 1 and 2 refer to di ffuser inlet and exit, respectiv ely. Given, p1 = 0.35 105 Pa, V1 = 200m/s , T1 = 23 0 K, m ˙ = 25 kg/s

×

From the given data, we get ρ1

=

A1

= M1

p1 0.35 = RT1 287 m ˙ ρ1 V 1

=

=

× 105 3 × 230 = 0.53kg/m

25 = 0.236m 2 0.53 200

×

200 1.4 287

√ ×

× 230 = 0.66

From isentropic table for M 1 = 0.66, we have p1 T1 = 0.7465, = 0.9199 p01 T01 Thus, p01

=

0.35 105 = 0.468 0.7465

×

× 105 Pa

230 = 25 0 K 0.9199 For M 2 = 0.2, from isentropic table, we have p2 T2 = 0.9725, = 0.9921 p02 T02 T01

p2

=

=

0.9725 p01 (since for isentropic flow p 01 = p02 )

=

0.45

T2

=

0.9921 T01 = 24 8 K

V2

=

M2 a2 = 0.2 1.4

× 105 Pa √ × 287 × 248 = 63m/s

34


Therefore, R

= 25 =

× (63 − 200) − (0.35 × 105 × 0.236 − 0.45 × 105 × 0.5)

−10.815 kN

4.12 Let subscripts 1 and 2 refer to entrance and exit of the tank. By energy equation we have, 1 cp T1 + u21 2 T1

1 = cp T2 + u22 2

− T2

T2

=

u22 u21 4 = 2cp

≈

15◦

=

T1

−

× 104 − 1 × 104 2 × 1004

− 15◦

4.13 Work

u1

u2

Figure S4.13 Schematic of the work delivering machine.

Work delivered per unit mass is given by u21

− u22 + cp (T1 − T2) = work delivered 2

Given, T1 = 373K

u1 = 200 m/s

T2 = 288K

cp = 1004 (N m)/(kg K)

Using this we get, 2

20000

− u22 + 1004 × 85

= 100000

35

2

20000

− u22 +85340

= 100000

u2 = 103 .3m/s When the machine is idling, u22 = 20000 + 85340 = 105350 2 u2idling = 459m /s

4.14 Let jets “1” and “2” be denoted by the subscripts 1 and 2, and let T0 denote the temperature in the reservoir. For q = 0, adiabatic energy equation gives, u22 u21 m + c p T2 + m + cp T1 = 2mcp T0 2 2



 



This gives, T0

T1 + T2 u 21 + u22 + 4cp 2

=

= 300 + =

(1 + 9) 104 4 1004

×

×

324 .9 K

4.15 Let T 0 denote the temperature in the tyre. Since the process is adiabatic, we have u2 2

=

cp T0

− T)

=

u2 2

u2

=

2cp (T0

cp T + cp (T0

u = =

− T)

√2 × 1004 × 37 272 .57 m/s

36


4.16 At 15000 m, we have T

Speed of sound

−56.5◦ C = 216.5 K

=

p =

1.206

a =



× 104 Pa

γ RT

√1.4 × 287 × 216.5

=

= 294 .94m/s Speed of the airplane is, u = 800 km/hr = 800

× 1000 3600

= 222 .22 m/s This gives the Mach number as M

=

u 222.22 = a 294.86

=

0.753

(a) Maximum possible temperature of the airplane skin will be the stagnation temperature, at the nose of the airplane. Thus, it is the total temper ature of the air, T0 T

= 1.1134

T0

= 1.1134 =

× 216.5

241 .05 K

(b) Maximum possible pressure that can be felt by the airplane cannot exceed the stagnation pressure. This will be felt at the place where air comes to complete rest, i.e., at the nose of the airplane and other similar places. Thus, p0 p = 1/0.6866 p0

=

1.206 104 0.6866

=

1.756

×

× 104 Pa

37 (c) Critical velocity of air relative to the airplane is a∗

=

a∗

=



2 γ+1

√γ RT

0

a0 =

√γ RT

√1.2 =

0

√1.2

284 .1m/s

(d) Vmax

= =

Vmax

 

2

γ

− 1 ao

γ+1 ∗ a γ 1

−

=

√

=

695 .9m/s

6a∗

4.17

M = 0 .6 Area = A

Figure S4.17 Schematic of convergent channel.

The mass flow rate is given by m ˙ = ρAV = =

p RT A M

γ RT

   γ p p0 R p0

T0 1 MA T T0

From the problem and theory we know that,

√

38


T0 = 55 0 K

p p0

=

0.7840

× 105 Pa

T T0

=

0.9328

=

1.0354

=



p0 = 2

 

T0 T

m ˙ = 29.188 kg/s M = 0.6

γ

R

= =

γ 1 cp γ − γ

γ

 √

cp (γ

− 1)

1.4 = 0.0694 1017 0.4

×

This gives m ˙

A =

γ p

R p 0 p0

=



T0 T

√1 M T0

0.10125 m2



4.18

1 m2 2m2

Ae = 4 m 2


At the mouth of the duct p =

patm = 1.013

T0 = 288K

× 105 Pa

(at sea level)

39 (a) Maximum mass flow rate is given by m ˙ max

=

1 p0 ∗ A 24.741 T0

√

1

=

1.013

1

√288 ×

24.741 =

× 105

241 .26 kg/s

(b) Ae A∗

A∗ = 0.25 Ae

= 4,

For this area ratio, from isentropic table, we get pe = 0.0298 p0

Me

=

2.94,

pe

=

0.0298

=

3018 .7 Pa

× 1.013 × 105 Pa

4.19

p, T, ρ

A

∗

M=2

1 m2


Here we have to find m, ˙ p ∗ , T ∗ , ρ ∗ , A, p, T , and ρ . ρ0

=

p0 7 105 = kg/m3 RT0 287 313

=

7.8 kg/m3

× ×

m ˙ =

0.6847 p0 A∗ RT0

=

1599 .133kg/s

√×

40


Fluid properties at the throat T∗ = 0.8333 T0 p∗

=

= 0.5283

=

ρ∗ = 0.6339 ρ0

=

p0

⇒

T ∗ = 261K 105 Pa

p∗ = 3.7

⇒ ⇒

×

ρ∗ = 4.944 kg/m3

For the test-section Mach number 2, from isentropic table, we have T = 0.5556 T0

=

⇒

T = 173 .9 K

p = 0.1278 p0

=

⇒

p = 8.946

ρ = 0.2300 ρ0

=

⇒

ρ = 1.794kg/m 3

1 A = 0.5926 A∗

=

⇒

× 104 Pa

A = 1.6875 m2

4.20

60 m/s

245 m/s 1

2

Figure S4.20 Schematic of divergent channel.

V1 = 245m/ s p1 = 1 105 Pa m ˙ = 13.6 kg/s

V2 = 60m/ s T1 = 300K

×

(a) ρ1

=

p1 1 105 = RT1 287 300

× ×

= 1.1614 kg/m3 a1

=



γ RT1 = 347.2m/s

41

M1

=

245 = 0.7056 347.2

≈ 0.706

p1 p0

=

0.7171

=

p0 = 1.39

T1 T0

=

0.909

=

A1

=

A1

m ˙

⇒

=

ρ1 V1

× 105 Pa

T0 = 33 0 K

⇒

13.6 m 2 = 0.0478 m2 1.614 245

×

=

1 π D12 4

=

0.2467 m

=

D1

⇒

(b) At the outlet, the energy equation is c p T2 +

V 22 = c p T0 2

Therefore, V22 T2

= T0

a2

=

M2

=



602

− 2cp = 330 − 2 × 1004 = 328.2 K γ RT2 = 363.1m/s

V2 = 0.165 a2

For M 1 = 0.706, A 1 /A∗ = 1.09 and for M 2 = 0.165, A2 /A∗ = 3.57. Thus gives A2 = 0.1566 m2 and D 2 = 0.4465 m (c) The rise in static temperature is T 2

− T1 =

28.2 K

4.21 p02

= 2.0

× 105 Pa 5

p1 = 0.15 By Rayleigh supersonic pitot formula, p1 p02

=



2γ γ +1



M 12

γ +1

2

−

γ 1 γ +1

M12

−





× 10

1/(γ −1)

γ /(γ +1)

Pa

42


p02 p1

=



γ+1

2

γ+1

=

2

Mi2



γ /(γ

−1)

γ +1

2 M1

2



2γ γ +1

1



2γ M12 γ +1

−

γ 1 γ +1

−

1



1/(γ −1)

γ −1

M12

M 12

−

γ 1 γ +1

−



Select M1 and calculate p02 for given p1 . Check that

p1 p01

is less than

p1 p02 .

Trial and error method gives M1 = 3.16 . Also, check p 02 /p01 calculated with normal shock relation and isentropic relation agree exactly . For this problem p02 /p01 = 0.286. 4.22 m ˙ max = 241 kg/s A = 3 A∗ For this area ratio from isentropic table, M = 2.64. This is the Mach nu mber just upstream of the shock. Let the conditions just upstream and downstream of the shock be represented by subscript s 1 and 2. From Normal shock table, for M 1 = 2.64, we have M2 = 0.5

p02 = 0.4452 p01

p2 = 7.9645 p1

Therefore, p02

= 0.4452

× p01 = 0.4452 × 1.0133 × 105 = 45110 Pa

For M 2 = 0.5, from isentropic table, we have A2 = 1.3398 A∗2 Therefore, A∗2

=

A2 3 = = 2.239m 2 1.3398 1.3398

Ae A∗2

=

4 = 1.786 2.239

Ae From isentropic table, for A∗2 = 1.786, we have Me

=

0.35

pe p0e

=

0.9187

43 But p 0e = p02 . Therefore, pe

= 0.9187 =

× 45110

41442.5 Pa

4.23 p1 = 0.7 T1 = 30 0 K A1 = 0.15 m2

× 105 Pa V1 = 240m/s V2 = 120 m/s

(a) m ˙ =

ρ1 A1 V1 =

p1 A1 V1 RT1

× 105 × 0.15 × 240 × 300

=

0.7 287

=

29.26 kg/s

(b) a1

=

M1

=



γ RT1 = 347.5m/s

V1 = 0.69 A1

Thus, p1 p01

= 0.72735

T1 T01

= 0.9131

p01

= 0.9624

× 105 Pa

T01 = 328 .6 K Thus, stagnation pressure at the exit p 02 = p01 = 0.9624

× 105 Pa.

(c) From (b) above, the stagnation temperature at the exit is 328.6 K

T02 = T01 =

44


(d) 1 cp T2 + V22 2

=

cp T0

T2

=

T0



V22

1202 )

= 328.6

− 2cp

A2

=

M2

=

V2 = 0.334 A2

p2 p02

=

0.9257

= 321.4 K

− 2 × 1004

γ RT2 = 359 m/s

Therefore, the static pressure at the exit is

p2 = 8.91

× 105 Pa

(e) Entropy change across the diffuser is zero since the flow is isentropic. (f) For M1 = 0.69, the exit area,

A1 A∗

= 1.1018, and for M2 = 0.334,

A2 A∗

= 1.850. This gives

A2 A∗ A2 = ∗ A1 = 0.252m 2 A A1

4.24 The pressure, density, and temperature in the settling chamber can be taken as stagnation quantities. Therefore, p0 = 1.014

× 105 Pa,

ρ0 = 1.44kg /m3 ,

T0 = 35 + 273 = 308K

Since the e ffects of viscosity is neglected, the flow in the working section can be treated as isentropic. For M = 0.8, from isentropic table, we have p = 0.656, p0

ρ = 0.740, ρ0

T = 0.886 T0

Hence, p = = ρ

= =

0.656 p0 = 0.656 0.665

× 1.014 × 105

× 105 Pa

0.740 ρ0 = 0.740 0.847 kg/m2

× 1.144

45

T

= 0.886 T0 = 0.886 =

× 308

273K

4.25 Shock 1 2

M1 = 3

Ath

Ae

Figure S4.25 Schematic of convergent-divergent channel.

Given,

Ae = 11.91, M1 = 3.0 At M1 = 3.0 gives, M2 = 0.4752 downstream of the normal shock, and

×

p02

105

p02 p01

=

T02

0.3281. This gives, = 2.2981 Pa. Since the flow is adiabatic, = T01 = 500K. And sinc e the flow is isentropic between the down stream of the normal shock and the nozzle exit, T 0e = T 02 , and p 0e = p 02 . A1 A2 1 For M1 = 3.0, A = A At = 4.2346, and for M2 = 0.4752, A2 = 1.390(from 1 isentropic table). Also, since the shock is very thin, the area before and after the shock can be taken as the same, i.e. A1 = A2 . Therefore, ∗

∗

  A A∗

=

Ae A∗2

=

Ae At A2 At A1 A∗2

e

= 3.9094 For this area ratio, from isentropic table, we get Me

= 0.15

Te T0e

= 0.9955

pe p0e

= 0.9844

46


Thus, pe

=

× 105 Pa

2.2623

Te = 497 .8 K 4.26 M Ats

= =

p0 = 7 105 Pa T0 = 27 + 273 = 300 K

2.5 1 m2

×

(a) For M = 2.5 we have, A A∗

=

A∗

=

2.637 0.38 m2

(b) T∗ T0

= 0.833

T ∗ = 0.833 × 300 = 249.9 K =

− 23.1 ◦ C

(c) M

=

a = V

V a



γ RT = 20.04

×

√

T = 231.5m/s

= M a = 578 .75m/s

(d) m ˙ A

=

p0 24.743 T0

=

7 105 24.743 300

√ ×

× √ ×

1 A A∗

1 × 2.637

47 = 620 m ˙ =

620kg /s

ffi

4.27 By definition, we have the pressure coe p p∞ CP = q∞

cient as

−

CP∗

=

=

p∞



p∗ p∞ q∞





−1

p∗ 1 p∞ 2 γ p∞ M ∞

2p∞

−



 −    −   −     × −  −   − p∗ p0 p0 p∞

=

2 2 γ M∞

=

2 2 γ M∞

γ+1

2 2 γ M∞

2 + (γ

=

=

1

−

γ

2 + (γ

γ −1

2

γ +1

2

2 1)M∞

γ

γ −1

2

2 + (γ

2

2 1)M∞

γ γ −1

γ

γ −1



−1

γ

γ −1

1

γ

γ −1

2 1)M∞ /(γ + 1) 2 /2 γ M∞

γ

γ −1

1

4.28 p0 = 500 kPa, T0 = 30◦ C = 30 + 273 .15 = 303 .15 K 101325 The nozzle has to choke since pe /p0 = = 0.203, which is well below 500 103 the critical pressure ratio of 0 .528. Therefore,

×

m ˙ =m ˙ max =

0.6847 p0 ∗ A RT0

√

where R = 28 7 m2 /(s2 K) for air. Thus, m ˙ = =

0.6847 500 103 287 303.15

√ ×× ×

0.058kg/s

× 0.5 × 10−4

48


4.29 The pressure ratio 0.1429, we get

pe 1 pe = = 0.1429. From isentropic table, for = p0 7 p0 Me = 1.93,

Te = 0.57307 T0

T0 = 180 + 273.15 = 453.15 K Therefore, Te = 259.7 K. The speed of sound ae =

√γ RT

e

= 323 m/s.

The exit velocity is Ve

= Me ae = 1.93 =

× 323

623 .4m/s

4.30 The pressure ratio p/p0 across the nozzle is 1 /5. This is well belo w the critical pressure ratio and therefore the flow is chocked at the exit. The flow is adiabatic and frictionless, therefore, the maximum mass flow rate is given by m ˙ = =

× 5 × 105 6.5 10−4 √287 × 288.15 × ×

0.6847

0.774 kg/s

Thus, from isentropic table, for M = 1. we get T T0 T

= 0.83333 = 240 .12 K

4.31 The mass flow rate may be expressed as m ˙ =

0.6847 p0 Ath RT0

√

where p 0 and T 0 are the stagnation pressure and temperature, respectively. pe 91.4 pe The pressure ratio = = 0.90495. From isentropic table, for = 0.905, p0 101 p0 we get Ae Me = 0.38 and = 1.6587 Ath

49 Therefore, A th =

0.033 = 0.0199 m2 . Thus, 1.6587 0.6847

m ˙ = = At the section with

101 × 103 × 0.0199 √×287 kg/s × 293.15

4.74 kg/s

A = 0.022m 2 , the area ratio is

A 0.022 = = 1.1055 Ath 0.0199 The corresponding Mach number is M = 0.69 and p = 0.72735 p0 Thus, the pressure at A = 0.022 m 2 is

p = 7 3 .5kPa

4.32 Nozzle is adapted, therefore the exit pressure altitude. From standard atmospheric table, we get

pe = p16000 at 16000 m

p16000 = 10.299kPa = pe Therefore, pe p0

10299

=

15

× 101325 ,

since 1 atm = 101325 Pa

= 0.0067762 By isentropic relation, p0 = pe



1+

γ

− 1 M2 e

2

1 + 0.2 M e2

=



Me

=

3.98



γ γ −1

1 0.0067762



0.286

For M e = 3.98, from isentropic table, we get A T e e A∗ = 10.53 , T0 = 0.23992 Te

=

0.23992

× (2600 + 273.15)

=

689.33 K

50


The exit velocity becomes Ve

=

Me ae = M e

=

3.98 1.4

=

2094 .6m/s

Thrust = 9000 =



γ RTe

√ × 287 × 689.33

m ˙ e Ve = (ρe Ae Ve ) Ve ρe Ae Ve2

By state equation, we have pe 10299 = = 0.052kg/m 3 RTe 287 689.33

ρe =

×

Therefore, Ae =

9000

0.052

Thus,

× (2094.6)2 = 0.0394

0.0394m 2

2

A∗ = Ath = 10.53 = 0.00374 m 4.33 Given, V = 200 m/s and T = 15+273 .15 = 288 .15 K. The speed of sound is given by



a =

γ RT =

√1.4 × 287 × 288.15

= 340 .3m/s The Mach number becomes M=

V = 0.59 a

From isentropic table, for M = 0.59, we get p p0

ρ T = 0.79013, T0 = 0.93491, ρ0 = 0.84514

Thus, p0

=

101 0.79013

51

T0

=

127 .8kPa

=

288.15 0.93491

=

308 .2 K

=

ρ0

=

ρ

0.84514 287

=

p RT

1 × 0.84514

101 × 103 × 288.15 × 0.84514

1.445 kg/m3

= ρ0 is also given by ρ0

=

p0 127.8 103 = R T0 287 308.2

=

1.445kg/m 3

×

×

Note: This problem may also be solved using the isentropic relations directly,

instead of table. 4.34 The sound speed at station 1 is a1

=

M1

=



γ RT1 =

√1.4 × 287 × 303.15 = 349m/s

V1 90.5 = = 0.26 a1 349

For this flow with M1 = 0.26 to choke, the area ratio tropic table) is

A1 required (from isenA∗

A1 = 2.317 A∗ The present area ratio 2.317 = 1.60.

A2 6.9 = = 0.69. Therefore, A1 10

A2 A2 A1 = = 0.69 A∗ A1 A∗

For this area ratio, from isentropic table, we get M2 = 0.4 ,

p2 = 0.89561, T2 = 0.96899 p02 T02

For M 1 = 0.26, from isentropic table, we get p1 T1 = 0.95408, = 0.98666 p01 T01

×

52


p01

=

T01

=

For isentropic flow, T 01 = T 02

100 = 104.8kPa 0.95408

303.15 = 307.25 K 0.98666 and p 01 = p 02 . Therefore,

p2

= 0.89561

× 104.8 =

T2

= 0.96888

× 307.25 =

93 .86 kPa 297.72 K

4.35 For CO2 the molecular weight is 44 and γ = 1.3. The gas co nstant for CO2 is 8314 R= = 189 J/(kg K) 44 The stagnation pressure p0 = 6 atm and the back pressu re pa = 1atm. Therefore, the pressure ratio pa 1 = = 0.167 p0 6 which is well below the critical pressure ratio of 0 .54573, required for the flow to choke. Hence, the flow is choked at the orifice. That is, M = 1 at the orifice. From the isentropic table, for M = 1, we get T = 0.86957 T0 Thus, T

= =

0.86957

× (30 + 273.15) = 263.6 K

−9.55◦ C

This is the temperature with which CO flow rate m ˙ = ρ AV . By state equation, ρ

=

2

comes out of the orifi ce. The mass

p 0.54573 6 101325 = RT 189 263.6

× × ×

= 6.66kg /m3 3 2

A = V

π 1

10− 4

= Ma = M

= 254 .5m/s

=π 4

× 10−6 m2 √ γ R T = 1 × 1.3 × 189 × 263.6

×  

53 Thus, m ˙ = 6.66 =

× π4 × 10−6 × 254.5

0.00133 kg/s

Aliter:

The mass flow rate (for the gas with γ = 1.3) can also be expressed as m ˙ =

0.6672

√×RpT0 × A

th

0

0.6672 = =

−6

× 6 × 101325 × π × 410 √189 × 303

0.00133kg/s

4.36 Let subscripts 1 and 2 refer to inlet and exit of the nozzle. V1

=

m ˙

0.7

=

ρ1 A1

8

× 12 × 10−4

= 72.92m/s M1

=

V1 = a1

1 √γVRT

1

72.92 1.4 287

√ × × 400 = 0.18 p1 = ρ 1 RT1 = 8 × 287 × 400 = 918 .4kPa =

From isentropic table, for M 1 = 0.18, we get p1 p01

= 0.97765

T1 T01

= 0.99356

p01

=

918.4 = 939.4kPa 0.97765

T01

=

400 = 402.59 K 0.99356

54


p2

=

ρ2 RT2 = 4

=

344 .4kPa

× 287 × 300

p2 p2 = , since the flow is isentropic. Therefore, p02 p01 p2 344.4 = = 0.3666 p01 939.4 p2 = 0.3666, we get p01

From isentropic table, for

M2 = 1.29 Therefore, V2

= M2 a2 = M2



γ R T2

√ × 287 × 300

= 1.29 1.4 =

447 .87m/s

Mass flow rate is m ˙ = A2

= =

ρ2 A2 V2

m ˙ ρ2 V2

=

0.7 4

× 447.87

3.9 cm2

4.37 From energy equation, we have T0

=

T1 +

V12 2 cp

where the subscripts 0 and 1 refer to stagnation and inlet conditions. T0 = 400 + = 405K

1002 2 1004.5

×

55 By isentropic relation, we have p0 p1

  T0 T1

=

γ γ −1

×

pe p0

3.5

405

103

p0 = 200

= 209 kPa

400

 

150 = 0.718 209

=

By isentropic relation, we have p0 pe

=

1 + 0.2Me2



=

(1.393) 3.5

1 + 0.2 Me2 1



3.5

Solving we get,

The speed of sound

Me

=

0.7

T0 Te

=

1+

Te

=

405 = 368 .85 K 1.098

ae

√ = γ RT

e

γ

− 1 M 2 = 1.098 2

e

= 385 m/s.

Thus, the exit velocity is Ve

= Me ae = 0.7 =

× 385

269 .5m/s

By continuity, ρ1 A1 V1 = ρ e Ae Ve . or p1 pe A1 V1 = Ae Ve RT1 RTe p1 Te V1 Ae

=

pe T1 Ve A1

=

200 150

=

100 × 368.85 × 269.5 ×π 400

0.00202m

2



75

× 10−3 2 4



56


Thus, de

× 4

= =

0.00202 π

= 0.0507m

50.7 mm

The mass flow rate is m ˙ = ρ1 A1 V1

=

p1 π 75 10−3 RT1 4

=

200 287

=

0.77 kg/s



×



2

× 100

× 103 × π × 752 × 10−6 × 100 × 400 4

With the exit conditions, m ˙ = = =

ρe Ae Ve

× 103 0.00202 269.5 × 287 × 368.85 × 150

0.77 kg/s

4.38 The flow process is described by the relation pVγ = constant. Per unit mass of air, pv γ = constant. Therefore, p1 V1γ

 v2 v1

By state equation, we have v1

=

=

p2 V2γ

=

p1 p2

γ

pv = RT . Therefore, R T1 287 = p1

× (400 + 273.15) 3 × 106

= 0.0644 m3 /kg Thus, v2

= v1

 p1 p2

1 γ

= 0.0644

1 3 1.4 0.5

 

57

T2

=

0.2316 m3 /kg

=

0.5 p2 v2 = R

× 106 × 0.2316 287

= 403 .4 K (a) By energy equation h2 +

V 22 2

= h1 +

V 12 2

where V1 and V2 are the velocity at the entrance and exit of the nozzle. Also, it is reasonable to assume that V 1 is very small and hence can be taken as zero. Thus, V2 h2 + 2 = h1 2 Further, for air being a perfect gas, h = c p T . Therefore, V22

=

2 (h1

=

2

− h2 ) = 2cp (T1 − T2 )

× 1004.5 (673.15 − 403.4)

V2 = 736m/ s a2

=



γ RT2 =

√1.4 × 287 × 403.4

= 402 .6m/s Therefore, M2

=

V2 736 = a2 402.6

=

1.83

(b) SinceThat the flow the nozzleM exit=is1. supersonic, at the throat the flow choked. is, atat the throat From isentropic table, for M =is1, we th get p∗ p0

=

0.52828

58


T∗ T0

= 0.83333

p∗

= 0.52828

T∗

= 0.83333

a∗

=



× 3 × 106 = 1.58484 MPa 673.15 = 560.96 K

× γ RT ∗ = 474.76 m/s

m ˙ = ρ∗ Ath a∗ ρ∗

=

p∗ 1.584 106 = RT ∗ 287 560.96

×

×

= 9.8 kg/m3 Thus, Ath

=

9.8

×

3415m m 2

= (c) The mass flow rate is

15.89 474.76

˙m = ρ2 A2 V2 . ρ2 =

1 = 4.318 kg/m3 v2

Therefore, m ˙ = 4.318 =

× 5000 × 10−6 × 736

15.89 kg/s

4.39 First of all, let us check whether the flow is isentropic. The change in entropy (assuming air to be calorically perfect gas) is s2

− s1

= cp ln

T2 T1

p2 p1

 −    −  

= 1004.5 ln

R ln

290 300

= 162 J/(kg K)

287 ln

101 200

59 Hence, the flow is nonisentropic. Consider the inner surface of the nozzle as the contro l volume. Let F be the force acting on the nozzle, due to the flow. By momentum balance, we have p1 A1

− p2 A2 − F

=

V1

=

ρ1

=

p1 200 = RT1 287

=

2.32kg /m3

V1

m ˙ (V2 m ˙

ρ1 A1

2.32

By energy equation, we have

2

− h2

=

× 0.2 = 10.78 m/s V22 2

− T2 )

= V22

× 1004.5 (300 − 290) + 10.782

= V22

2 cp (T1

× 103 × 300

5

=

h1

− V1 )

2

− V21

− V12 = V22 − 10.782

V2 = 142 .15m/s The exit area is given by A2 =

m ˙ ρ2 V 2

where ρ2

=

p2 101 = RT2 287

× 103 × 290

= 1.214 kg/m3 A2

=

5

1.214

× 142.15 = 0.029m

2

Using these values in the momentum equation, we get 200

× 103 × 0.2 − 101 × 103 × 0.029 − F = 5(142 .15 − 10.78)

Thus, F = 40000

− 2929 − 656.85

60


=

36414 .15 N

4.40 Let us solve the problem by using relations and also by using gas tables. The speed of sound at the flight altitude is given by a =



γ RT =



1.4

× 287 × (−15 + 273.15)

= 322 .1m/s The flight speed V = M a = 0.8

× 322.1 = 257.68 m/s.

The stagnation enthalpy h 0 is given by h0 = h + For an ideal gas

h0 = c p T +

V2 2

V2 and therefore, 2 c p T0 = c p T +

V2 2

or

2

T0 = T +

V 2cp

Thus, T0 = 258 .15 + =

(257.68)2 2 1004.5

×

291 .2 K

By isentropic relation, we have γ

=

p0

= 44 =

Using gas tables

 −  

p0 p

T0 T

γ

1

291.2 258.15

67.08 kPa

3.5

61 For M = 0.8, from isentropic table, we get p T = 0.65602 = 0.88652 p0 T0 Thus, p0

=

44 = 67.07 kPa 0.65602

T0

=

258.15 = 291 .2 K 0.88652

Note: From the above solutio n it is seen that, use of the gas tables has tremen-

dous advantage over using equations directly. 4.41 Static temperature T = 216 K. Total temperature T0 = 40 + 273.15 = 313.15 K. From isentropic relation, we have T0 γ 1 2 = 1+ M T 2 2 1.45 = 1 + 0 .2 M

−

M2

=

M

=

0.45

= 2.25

0.2 1.5

The pressure ratio p 0 /p given by isentropic relation is p0 p

= =

p0

 

1+

γ

 

− 1 M2

1 + 0.2

2

γ γ −1

× 1.52 3.5 = 3.671

= (3.671)(0.55) = 2 .019 atm

Note: This problem can be solved by using gas tables instead of solving the

isentropic relations, as follows. For

T = 0.689655, from isentropic tables, we get T0 M = 1.5 and p/p0 = 0.2724

Therefore, p0

=

0.55 = 2.019 atm 0.2724

62


4.42 By energy equation, we have T01

=

T02 = T1 +

V12 2cp

= (70 + 273 .15) +

1002 2 1004.5

×

= 348 .13 K Similarly, 2

T2

= T02

V2 − 2c p 2

− 2 ×400 = 268.49 K 1004.5 √ γ RT = 1.4 × 287 × 268.49

= 348 .13 a2



=

2

= 328 .45m/s Therefore, V2 M2

Vmax

400

=

a2

= 328.45

=

1.218

=



=

836 .3m/s

2 cp T01 =

√2 × 1004.5 × 348.13

By isentropic relation,

Thus,

p01

= p1 1 + 0.2 M12

M1

=



1 √γVRT

= 0.269 1

p01

= 2.5 1 + 0.2

p02



= p2

3.5



0.2692

3.5

× 1 + 0.2 × (1.218)2



= 2.63 atm

3.5

p02 1.24 = = 0.471 p01 2.63

= 1.24atm

63 4.43 Given, p0 = 350 kPa, T0 = 42 0 K Ae = 0.22 m2 , Ve = 525m/s By energy equation we have 2

T0 = Te + 2c Ve p 2

Te

2

− 2Vcep = 420 − 2 ×525 1004.5

=

T0

=

282 .8 K

√γ RT . That is, e √ = 1.4 × 287 × 282.8 = 337.01 m/s

Therefore, the speed of sound at nozzle exit a e = ae Thus, Me

pe

ρe

=

Ve 525 = ae 337.01

=

1.56

=

p0 (1 + 0.2Me2 )3.5

=

350 4.007

=

87.35 kPa

=

pe 87.35 103 = RTe 287 282.8

=

1.076 kg/m3

×

×

m ˙ = ρe Ae Ve = 1.076 =

× 0.22 × 525

124 .3 kg/s

From isentropic table, for M e = 1.56, we get Ae = 1.219 A∗ Thus, A∗ =

0.22 = 0.18 m2 1.219

64


4.44 The speed of sound at 1 is a1



=

γ RT1 =

√1.4 × 287 × 315

= 355 .76 m/s 1 The Mach number at 1 is M 1 = V 150 = 0.42. a1 = 355.76

Now, for M1 = 0.42, the corresponding area ratio A1 /A∗ may be calculated using the area–Mach number relation for isentropic flow, or the value of A 1 /A∗ may be read directly from isentr opic table. From isentropic table for M 1 = 0.42, we get A1 /A∗ = 1.52891 Thus, A1 = 1.52891 The mass flow rate m ˙ = ρ 1 A1 V1 ρ1

× 25 =

38 .22cm 2

× 103 × 315

=

p1 152 = RT1 287

=

1.681 kg/m3

Therefore, m ˙ = 1.681

× 38.22 × 10−4 × 150 =

0 .9637 kg/s

4.45 T0 = 40◦ C = 40 + 273 .15 = 313 .15 K, γ = 1.4 The exit velocity

√

Ve = 200m/s = Me γ RTe .

By isentropic relation we have p0 = 1 + 0.2 M e2 pe



By energy relation, we have T0

= Te +

ae

=



3.5

Ve2 2cp 2002

Te = 313 .15 The speed of sound is



= 293.24 K

− 2 × 1004.5 √ γ RT = 1.4 × 287 × 293.24 e

= 343 .25 m/s

65 The Mach number at the exit is Me

p0 pe

=

Ve 200 = ae 343.25

=

0.584



=

1 + 0.2

× 0.5832

3.5



= 1.259

For the convergent nozzle,pexit has to be equal to the atmospheric pressure, since the subsonic flow exiting a convergent nozzle will always be correctly expanded. pe

=

× 103 × 9.81 × 22.7 × 10−3

ρgh = 13.6

= 3028.54Pa

(gauge)

= 101325 + 3028 .54 = 104353.54Pa (absolute) Thus,

pa = 104353 .54Pa

The storage pressure is given by p0

= 1.259

× 104353.54

=

131381.11Pa

=

224 .7mm of Hg

(gauge)

Note: The standard atmospheric pressure is patm = 101325 Pa = 760 mm of

Hg. 4.46 Let the first and second sections are represented by subscripts 1 and 2, respectively. At section 1, M1

= =


1 √γVRT

= 1

365 √1.4 × 287 × 305.15

1.04 M1 = 1.041, we get p1 p0

=

0.50389

T1 T0

=

0.82215

66


Therefore, p 0 = 158.76kPa ,

T0 = 371.16K.

At station 2, p2 p0

120 = 0.7559 158.76

=


p2 /p0 = 0.7559, we get T2 M2 = 0.64 = 0.92428 T0 T2 = 343 .06 K = 69 .91◦ C a2

=

Therefore,



γ RT2 = 371.27m/s

V2 = M 2 a2 = 237 .6m/s 4.47 For M = 3.0, from isentropic table, we get p = 0.027224, p0

T = 0.35714, T0

A = 4.23456 A∗

Thus, A∗

=

0.05 = 0.0118m 2 4.23456

p0

=

0.2 101325 = 744 .4kPa 0.027224

T0

=

300 = 840K 0.35714

×

The mass flow rate is given by m ˙ = = =

0.6847 p0 A∗ RT0

√

0.6847

× √7.444 × 105 × 0.0118 287 × 840

12.25 kg/s

4.48 The velocity at nozzle entrance is very low. Hence, the pressure and temperature at the entrance can be taken as the stagnation pressure and stagnation temperature. That is, p0 = 1MPa ,

T0 = 30 0 K

67 From isentropic table, for M = 2, we have p = 0.1278, p0

T = 0.55556, T0

ρ = 0.23005 ρ0

Therefore, p =

0.1278

× 1 × 106 =

127.8kPa

T

=

0.55556

× 300 =

ρ

=

0.23005

× ρ0 = (0.23005)( p0/RT0 )

=

0.23005



The speed of sound fore,

a=

V

166.67 K

106 287 300

×



= 2.672 kg/m3

√γ RT = √1.4 × 287 × 166.67 = 258.78 m/s. There-

= M a = 517 .56 m/s

m ˙ = ρAV = 2.672

0.15

× =

517.56

×

207 .43 kg/s

4.49 Given, p0 = 3.5MPa , T0 = 500◦ C = 773.15 K pe = 0.7MPa ,

m ˙ = 1.3 kg/s

At the nozzle exit, the pressure ratio is pe 0.7 = = 0.2 p0 3.5 From isentropic table, for this pressure ratio, we get Me = 1.71 ,

Te Ae = 0.63099, = 1.3471 T0 Ath

Therefore, Te = 0.63099

× 773.15 = 487.85 K

The corresponding speed of sound a e is ae =



γ RTe =

√1.4 × 287 × 487.85 = 442.74m/s

68


The exit velocity Ve = M e ae = 757m/s ρe =

pe , by state equation RTe

Therefore, 7

ρe = 287 0.7 487.85 10 = 4.9 kg/m3

The mass flow rate

××

m ˙ = ρ e Ae Ve . Thus, Ae

=

Ath

m ˙ ρe Ve

=

4.9

1.3 757

×

=

3.5 cm2

=

Ae 3.5 = 1.3471 1.3471

=

2.6 cm2

4.50 Given, T0 = 25◦ C = 298.15 K, Ae = 15 cm2 . Applying momentum analysis to the control volume considered, we get F = 100 N = ρ e Ae Ve2 Assuming air to be an ideal gas, Ae ρe Ve2 =

Ae pe 2 Ae γ pe 2 V = V = Ae γ pe Me2 RTe e γ RTe e

Also, γ Ae pe Me2 = 10 0 N

The nozzle exit flow is subsonic and hence has to be correctly expanded with pe = p atm = 101325 Pa. Thus, Me2

=

Me

=

100

1.4

× 15 × 10−4 × 101325 = 0.47

0.69

From isentropic table, for M e = 0.69, we get Te = 0.91306, T0

pe = 0.72735 p0

69 Thus, Te = 298 .15

p0

Ve

× 0.91306 = 272.23 K

=

−0.92◦ C

=

101325 = 139 .3kPa 0.72735

=

1.37 atm

=

Me ae = M e

=

0.69

=

228 .2m/s



γ RTe

× √1.4 × 287 × 272.23

4.51Given, ˙m = 1kg/s, pe = 101.325 kPa. The mass flow rate may be expressed as m ˙ =

0.6847 p0 ∗ RT0 A

√

Thus, A∗

At the exit,

=

√RT

0

0.6847 p0

=

√287 × 325 0.6847 × 700 × 103

=

6.37cm 2

pe 101325 = = 0.14475 p0 700 103

×

From isentropic table, the corresponding Mach number ratio are the following. Me

=

Te T0

=

0.57561

Te

=

0.57561

1.92

× 325 = 187 K

Me and temperature

70


The speed of sound Thus,

√γ RT

ae =

Ve

e

= 274.1m/s .

=

Me ae = 1.92

=

526 .27 m/s

× 274.1

4.52 Assuming air to be a perfect gas, we have γ = 1.4 and R = 287 J/(kg K). The speed of sound at nozzle inlet is a1

=



γ RT1 =

√1.4 × 287 × 300

= 347 .2m/s The inlet Mach number is M1 =

V1 100 = = 0.29 a1 347.2

From isentropic table, for M1 = 0.29, we get p1 p01

= 0.94329

T1 T01

= 0.98346

p01

=

101325 = 107416.6 Pa 0.94329

T01

=

300 = 30 5 K 0.98346

The flow is supersonic at the exit, therefore at the throat from isentropic table we have p∗ p0

=

0.52828

T∗ T0

=

0.8333

M ∗ = 1. For M = 1,

Thus, p∗

=

0.52828

× 107416.6,

=

56746Pa

since p ∗0 = p 01

71

T∗

= 0.56 atm = 0.8333 305

×

=

254K

Since the flow is isentropic, T02

= T01 = 305 , K

p02

= p01 = 107416.6 Pa =

1.06 atm

The mass flow rate is m ˙ =

ρ1 A1 V1 = ρ t At Vt

=

p1 101325 A1 V1 = RT1 287 300

=

0.0588 kg/s

At

=

0.0588 ρt V t

ρt

=

pt 56746 = RTt 287 254

×

× 5 × 10−4 × 100

Thus,

×

= 0.78kg /m3 Vt

= at =



γ RTt =

√1.4 × 287 × 254

= 319 .5m/s Therefore, At =

0.0588 = 2.36cm 2 0.78 319.5

×

4.53 Let subscripts 01 and 02 refer to properties at reservoirs 1 and 2, respectively. Pressure p02 is the back pressure for the nozzle. Therefore, the nozzle pressure ratio (NPR) becomes p02 3 = = 0.5 p01 6

72


Also,

 

p∗ = p01

2 γ+1

(γ )γ − 1

= 0.528

where p∗ is the pressure corresponding to sonic speed. The NPR is less than the critical pressure ratio of 0.528. Therefore, the nozzle must experience sonic condition at the exit . Thus, the mass flow thr ough the nozzle m ˙ is m ˙ = ρ ∗ V ∗ A∗ where ρ ∗ and V ∗ are the density and velocity at M = 1. From isentropic table, for M = 1, we get T∗ T01

= 0.8333

Thus, T∗

=

0.8333(30 + 273.15) = 252 .6 K

V∗

=

a∗ =



γ RT ∗ =

√1.4 × 287 × 252.6

= 318 .58 m/s From state equation, we have ρ∗

=

p∗ 6 = RT ∗

=

4.428 kg/m3

× 101325 × 0.528 287 × 252.6

Thus, A∗

=

m ˙ ρ∗ V ∗

=

4.428

0.5 318.58

×

= 0.0003544 m2 = 3.544 cm2 Aliter:

This problem can also be solved without going into the details at the nozzle exit, as follows. We know that in terms of reservoir pressure and temperature, the mass flow rate m ˙ is given for a gas with γ = 1.4 as m ˙ =

√0.6847 p01 A∗ RT 01

73 Thus, A∗

=

m ˙ 0.6847 p01

=

0.0003543 m2 = 3.543cm 2

4.54 The overall pressure ratio



pe /p0 =

RT01

100 103 = 0.1. 106

×

This is well below the critical press ure ratio of 0.528. Therefore, the flow is a choked flow. The area ratio

Ae /Ath = 13.5/8 = 1.6875.

From isentropic table, for Ae /Ath = 1.6875, the corresponding exit Mach number Me = 2.0 4.55 Given, p0 = pa = 101325 Pa and T0 = 288.15 K. For the flow to choke, pb /p0 0.528. Therefore, the pressure in the tank p b has to be

≤

pb

0.528

≤

101325

×

53.5kPa

≤

This implies that p bmax = 53.5 kPa. The mass flow rate is given by m ˙ =

0.6847 p0 Ath RT0

√

=

0.6847 287

=

0.303kg/s

√

× 101325 π 2 × 288.15 × 4 (0.04)

4.56 Given, Ae /Ath = 4 and Me > 1. Therefore, the super sonic solution of the area Mac h number relation is the answer for the problem. That is, the supersonic value of M given by the following equation is the exit Mach number. Ae 1 = Ath M



5 + M2 6



3

This equation may be solved directly to get Me or the value of Me corresponding to A e /Ath = 4 can be read directly from isentropic table. From isentropic table, we get

74


Me = 2.94 and

pe = 0.029795 p0

This is the pressure ratio required. 4.57 (a) For M e = 1.63, from isentropic table, we have Ae pe = 1.275 = 0.22501 A∗ p0 For correct expansion, p e = pb . Therefore, pb = 0.22510

× p0 = 0.22501 × 10 × 101325 Pa

since 1 atm = 101325 Pa. Thus,

pb = 228 kPa

(b) The flow will remain supersonic at the exit for all back pressures below 228 kPa (c) For choking, M = 1 at the throat. After choking the flow can expan d as a subsonic flow at the divergent portion of the nozzle. From isentropic table, for Ae /A∗ = 1.275, we have Me = 0.54 p/p 0 = 0.82 (approx) Therefore, the nozzle will remain choked for all back pressures below pb = 0.82

× 10 × 101325 =

830.87 kPa

4.58 From normal shock theory, we have ρ2 ρ1

M2

= 1+

=



2 M12 1 M12 (γ 1) + 2



 −  −  −  −

γ M12 + 1

M12 1 2 (γ M12 + 1) (γ + 1)

1/2

−

where subscripts 1 and 2 refer to states upstream and downstream of the shock, respectively. Given, M 1 is very large. Also, γ = 1.4 for air. The density ratio can be written as 2 1 M12 ρ2 1 = 1+ ρ1 (γ 1) + M22

−  −

when M 1

→ ∞, we get

1

75

ρ2 ρ1

= 1+ = 1+ =

2

γ

1

− 1 = 1 + 1.4 − 1

2 = 1+5 0.4

6

Similarly,

  −  −    −  −  1 M12

γ+

M2 = In the limit M 1

2 γ+

1 M12

1

1/2

γ +1

1 M12

M12

→ ∞, it reduces to M2

1

γ

=

2γ

=

=

0.4 2.8

0.378

4.59 From isentropic table, for M = 2.0, we get p = 0.1278 p0 Therefore, p =

0.1278

× 3 × 101325 = 38848 Pa

since, 1 atm = 101325 Pa. Given, 101325 Pa = 760 mm Hg, therefore, p =

38848 101325

× 760

= 291 .384mmof Hg This is absolute pressure. The gauge pressur e, shown by the manometer, will be pg

= =

pabs

− patm = 291.384 − 760

− 468.61mm

The negative sign indicates that the measured pressure is below the atmospheric pressure or subatmospheric.

76


4.60 The di fference between the measured pressures is 500 mbar. That is, ∆p

= 0.5bar

We know that, 1 bar = 10 5 Pa Therefore, ∆p

= 0.5

Also, ∆p

× 105 Pa

= ρ air g h

where h is the vertical height climbed. Therefore, h =

0.5 105 9.81 1.1

=

4633 .49 m

× ×

4.61 Let subscript ‘0  refer to stagnation state. Given that the total pressure of air is p 0 . For maximum velocity, the limiting pressure is zero. Assuming the flow as incompressible, by Bernoulli equation, we have 1 p + ρV 2 = p0 2 For V max , p = 0, thus,

1 2 ρV = p 0 2 Also, for incompressible flow, ρ = ρ 0 , thus, Vmax, incomp. =



2

p0 ρ0

For compressible flow, the Bernoulli equation, we have γ

γ

Therefore,

p

−1ρ+

γ p0 V2 = γ 1 ρ0 2

−

Vmax, comp. = That is,



2

γ γ

p0

− 1 ρ0

γ

Vmax, comp.

= = =

 

γ

− 1 Vmax, incomp.

1.4 V max, incomp. 1.4 1

−

1.87 Vmax, incomp.

77 That is, the error in treating this compressible flow as incompressible is 87 percent. 4.62 Let the inlet and the exit of the nozzle be denoted by subscripts 1 and 2, respectively. (a) By energy equation, we have h1 + 3025

V 12 2

× 103 + 602

=

h2 +

2

V 22 2 2

= 2790

× 103 + V22

V22 = 473600 V2

=

688.2m/s

(b) The mass flow rate is m ˙ = ρAV =

A1 V1 v1

=

0.1 60 0.19

=

31.58 kg/s

×

(c) Mass flow rate is also given by m ˙ = ρ 2 A2 V2 =

A2 V2 v2

Thus, A2

=

m ˙ v2 V2

=

31.58 0.5 688.2

=

0.0229m 2

×

78


Chapter 5

Normal Shock Waves 5.1

Vg

500 m/s

500 − Vg

(a) Stationary observer

(b) Observer moving with the shock

Figure S5.1

Mach number of the air stream, M 1 is given by M1 =

√1.4 ×500 = 1.51 287 × 273

From Normal shock table, for M 1 = 1.51, we have p2 p1

=

2.493,

T2 = 1.327 T1

Therefore, p2

=

1.745 atm

T2 = 362 .27 K Also, 2 V V1

500

− Vg

500 m/s

=

500500 Vg = 1.879 1

−

= 266 .1m/s

Vg = 233 .9m/s 79

80

Normal Shock Waves

Since the velocity of the observer does not a

ffect the

pb

= p2 = 1.745atm

Tb

= T2 = 362.3 K

static properti es,

The Mach number of the flow behind the shock wave is Mb

=

g √γVRT

= b

233.9 381.5

= 0.613 From isentropic table, for M = 0.613, we have p pt

= 0.7760,

T = 0.9301 Tt

Therefore, after passage of shock the stagnation pressure is ptb

Ttb

=

1.745 0.7760

=

2.25 atm

=

362.3 0.9301

=

389 .5 K

Note:

• For a stationary observer the stagnation temperature after passage of the wave is greater than that before passage of the wave. • For an observer sitting on the wave , however, there is no change of stagnation temperature across the wave.

81 5.2

up

u 2 = C s − up

Cs 2

u1 = C s

1

Figure S5.2 Shock wave motion in a tube.

a1

=

u1 u2

=

M1

=



γ RT1 = 347m/s

Cs

Cs

=

− up

(γ + 1)M12 (γ 1)M12 + 2

−

Cs a1

Therefore, Cs Cs

− up

(γ + 1) = (γ

− 1)

2

(γ

− 1) Ca2s +2

= ( γ + 1)

1

(γ 2Cs2 Cs2

− 1)Cs2 + 2a21

− (γ + 1)upCs − 2a21

−

 

M12

γ+1

Cs a1

Cs (Cs a21

2

+2

− up )

− (γ + 1)upCs

= 0 = 0

− γ +2 1 ua1p M1 − 1

= 0

M1

1 2

up Cs

Cs a1 2



= (γ + 1)Cs2

− a21

2



=

 

γ + 1 up

2

a1

±



= 1.19

γ + 1 up

2

a1

   2

+4

Positive sign is taken here, since M 1 cannot be less tha n 1. Hence, C s = M 1 a1 = 413m/s . From Normal shock table, for M1 = 1.19, pp12 = 1.485. Thus, the pressure on the face of the piston is p2 = 1.485 1.0133 105 = 1.505

×

×

× 105 Pa

82

Normal Shock Waves

5.3 up

2

1

Figure S5.3a The flow field.

(a) p2 p1

=

a1

=

−  1

− 1 |up|

γ

2

a1



2γ γ −1

√

γ RT1 = 20.04 300

= 347m/ s Therefore, p2 p1 p2

= =

1



= 0.606



0.606p1

=

7

− 0.2 120 347

0.606atm

(b)

up

u2 = C s

Cs 2

−

1

Figure S5.3b Schematic of the flow field.

u1 u2

= =

Cs Cs

− up

(γ + 1)M12 (γ 1)M12 + 2

−

up

u1 = C s

83 (γ M12

−

− 1)M12 +2

γ + 1 up

2

a1

= (

−1

M1

γ + 1 up

2 Therefore, M 12

γ + 1)M12

− (γ + 1) uap M1 1

= 0 120

= 1.2

= 0.415

× 347

a1

− 0.415 M1 − 1 = 0. Solving for positive value of M1

M 1 , we get

= 1.228

p2 p1

2γ M12 γ+1



= 1+

−1



= 1.595 Therefore, p2 = 1.595 atm 5.4 Cs

up

Cs

u1 = C s + up

Gas at rest

u2 = 0

u2 = C s

Figure S5.4 Schematic of the flow field.

The velocity of the wave relative to the pipe = C s . Velocity of air entering the normal shock wave relative to the shock wave is u1

=

Cs + up

M1

=

C s + up a1

u1 u2

=

(γ + 1)M12 2 + ( γ + 1)M12

84

Normal Shock Waves

Cs + u Cs

= Cs From this we get

= M1 a1

(γ + 1)M12 2 + ( γ 1)M12

=

M1 a1 M1 a1 up

=

M1 u M1 ap1

−1

=

0

a1

=



−

M12

− γ +2 1

  up a1

M1

− up −

−

γ RT1 = 347m/s

Solving for M 1 , we get M 1 = 1.2924, taking only the positive sign, since M 1 is supersonic. Hence, Cs

= M 1 a1 =

− up = 1.2924 × 347 − 150

298 .5m/s

From shock tables, for M 1 = 1.294, we have p2 p1

T2 = 1.185 T1

= 1.775,

Therefore, p2

T2

= 1.775

× 1.5 × 105

=

× 105 Pa

2.66

= 1.185 =

× 300

355 .5 K

Also, since the gas is at rest,

5.5

p02

=

p2

T02

=

T2

85

Vp 1

u2 = C s

u1 = C s + Vp

Cs 2


u1 u2

=

(γ + 1)M12 2 + ( γ 1)M12

u1

=

Cs

M1

=

Cs a1

u2

=

Cs

−

− Vp

Therefore,

2 + (γ

Cs a1

1)



Cs Cs (γ + 1) a1 a1 Solving for

Cs a1 ,

Cs a1

−

Vp a1

=

2

Cs Cs

up

−

=



Cs a1 Cs a1

− Va

p 1

= 2+( γ

− 1)

  Cs a1

2

we get,

Cs = a1 In the limit as

2

 −  

(γ + 1)

Vp a1

2 1/2

       γ+1

4

Vp 1 + 1+ a1 4

2

γ+1

2

→ ∞, Ca → ∞. In the limit as s 1

Vp a1

5.6 a4

Cs 3

up = 300 m/s

4

Vp a1

→ 0, Ca → 1. s 1

86

Normal Shock Waves

Figure S5.6a Schematic of the flow field.

(a) up = 300m/ s p3

=

− −  − ×   × − − 

p4 =

1

=

Therefore, a 3 =

5 6

p3

=

a3 a4

=

1 |up |

γ

1

2

a4

0.2

300 360

5 6

7

5 6

7

2γ γ −1

2.8 0.4

1 = 0.2791 atm

γ

1

2

1 |up | 5 = a4 6

× 360 = 300 m/s. Slope of the terminating characteristic is dx dt

=

C3 = a 4

= 360

− γ +2 1 |up|

− 2.4 × 300 = 0 2

For the pressure on the face of the piston, a3

 −  p2 p3

up

=

up

= a3 = 300m/s

γ

1

2γ γ +1 p2 γ 1 p3 + γ +1

1 2

−



p2 p3

1 + γγ − +1

Therefore,

 − p2 p3

p2 p3

Therefore,

2

 −

3.68

1

2

= γ2

p2 + 0.72 = 0 p3

p2 p3

= 3.473

p3

= 0.2791 atm



2γ γ +1



87 Therefore, pressure at the piston face p 2 = 3.473

× 0.2791 =

0.969 atm

(b) The velocity of the shock with respect to a stationary observer is γ

Cs = a 3

1 2

− 1 + γ + 1 p2

2γ

= 529.88m/s

2γ p3





Therefore, time for the shock to hit the terminating characteristic after the piston has stopped is 30 = 0.0566s t1 = 529.88 (c) t Cs

0

Piston path

300 m/s

300 m/s 0 = d /

t

t

x d

k oc Sh

dx /d t

Pi sto =

|u

np ath

p

dx

g c n ti ti si a r in te c m r a e ra Th c

|

t= /d

a4

x 30 m

Figure S5.6c

5.7 Given, MS = 5.0, T1 = 27c ircC = 300K , p1 = 0.01 atm, T4 = 300K. From normal shock table, for M S = 5, we have p2 p1

= 29.0

p4 p1

=

2γ4

−

2γ1 MS2

= 2.26



− (γ1 − 1) 1 − γ4 − 1 γ1 + 1 γ4 + 1

× 106



MS

− M1S



γ4 −1

Cs 0 p2

p3

88

Normal Shock Waves

p4

=

2.26

× p1 =

2.26

× 104 atm

Note: Instead of above equation for p 4 /p1 , equation (5.57) also can be used for

solving this problem. 5.8 (a) p 2 /p1 = 29. Therefore, static pressure behind the shock is, p2 = 29p1 = 0.29 atm. From normal shock table for M 1 = 5, we get T2 = 5.8 T1 Therefore, static temperature behind the shock T 2 is, T2 = 5.8

× 300 =

1740K

(b) Particle speed u 2 is given by, u2



2 a1 MS γ1 + 1

=

− M1S



= 1388.71 m/s a2

=

20.04

×



T2

= 836m/ s M2

=

1388 836

=

1.66

From isentropic tables, for M 2 = 1.66, we have T2 p02

= 0.6447

p2 p02

= 0.215

Thus, T02 = 2699 K and p02 = 1.349atm (c) Testing time available is ∆t

=

t2

− t1 = u82 − C8s

89

u2 = 1388 .71 m/s

Cs

=

Ms a1 = 5

347 = 1735 .9m/s

×

t t2

dx

t1

t= /d

u2

t dx/d

=

C

s

x

8m

Figure S5.8c

Therefore, ∆t

=

8 1388

8 − 1735

=

1.15

× 10−3 s

(d) 1 1 µ = sin −1 = sin −1 M2 1.66 =

37d eg

5.9 up

2

up

2

2

Cs

1

uR

up + u R

5

uR

5

90

Normal Shock Waves


(a) uR Cs

=

uR up up Cs

up + uR uR

=

ρ5 , ρ2

=

ρ5 ρ1

=

ζ η

from continuity

× ρρ1 2

Therefore, up uR

ζ

=

−η η

Also, (refer to Example 5.5) up Cs

=

(1

− η1 )

From this, we get uR Cs

= =

η

× ζ −η η ζ

−  1

1

η

−1 −η

(b)

−  − 

p2 p1

= 1 + γ1 MS2 1 = 1 + γ1 MS2 1

ρ1 ρ2

1

Similarly, p5 p2

= 1 + γ2

(up + uR )2 a22

= 1 + γ2

a21 a22



  − 

× 1 − ρρ25

up + uR a1

2

1

η ζ

η

91

= 1 + γ2

a21 a22

= 1 + γ2

γ 1 p1 ρ 2 ρ 1 γ 2 p2

= 1 + γ1



up u R Cs + a1 C s a1

η

1)2

ζ

ζ

2

− η MS

p5 p2 p2 p1

=

p2 (η 1)2 ζ + γ1 MS2 η ζ η p1

−

1

+ γ1 MS2

η

1

η

−η ζ

−

−  − 

= 1 + γ1 MS2 1

ζ

 

=

= 1 + γ1 MS2

2

η

 − −

p1 (η η p2 η2

η ζ

1

− 1 MS + η − 1 MS

η ζ

Therefore, p5 p1

2

− 

η

1+

(η (ζ

(η

− 1)2 η

− 1)ζ − η)

ζ ζ

−η



− 1 η(ζ − 1) η ζ −η 2 (η − 1)(ζ − 1) 1 + γ1 MS (ζ − η )

= 1 + γ1 M 2

η

S

= (c)

Therefore,

h2 h1

=

1+

h5 h2

=

1+

=

1+

h2

=

h1

=

γ1

S

2 γ2

−1

2 γ2

−   − 

− 1M2

1

1

η2

up + uR a2

2

− 1 a21 M 2 (η − 1)2 S

a22

2

η2

1

ζ2 (ζ η )2

a22 γ2

−1

a21 γ1

−1 a21 a22

=

γ1 γ2

ρ 22 ρ25

− 1 h2 − 1 h1

−

η2 ζ2

−  1

92

Normal Shock Waves

Therefore,

− 1 h1 M 2 (η − 1)2 ζ + η h2 S η 2 ζ−η

h5 h2

= 1+

h5 h1

=

h5 h2 h2 h1

=

h2 γ1 1 2 (η 1)2 ζ + η + MS h1 2 η2 ζ η

= 1+

γ1

2

−

γ1

−

− 1M2 η − 1 S

2

= 1 + ( γ1

η2

−



(η + 1) + ( η

− 1) ζζ +− ηη



ζ − 1) − 1)MS2 (η −η(1)( ζ − η)

5.10 Given that, the stagnation pressure and temperature are p0 = 1 atm = 101325 Pa,

T0 = 15◦ C = 288.15 K

From isentropic table, for M = 3.0, we have p p0 = 0.02722 Therefore, the static pressure at the test-section is p

= 0.02722 p0 = 0.02722 =

× 101325

2758P a

The pitot pressure measured by a pitot tube placed in the test-section is the pressure behind a normal shock. From normal shock table, for M = 3.0, we have p02 = 0.3283 p01 Thus, the pressure that a pitot tube at the test-section will measure is p02

= 0.3283 p01 = 0.3283 =

× 101325

33.265 kPa

93 5.11 From normal shock table, for M 1 = 2.5, we have p2 = 7.125, p1

T2 = 2.1375, T1

ρ2 = 3.3333, ρ1

p02 = 8.5261 p1

Therefore, M2

=

0.51299

p2

= (7.125)(1) = 7.125atm

ρ2

= (3.3333)(1.225) = 4.083 kg/m3

p02 T2

= (8.5261)(1) = 8.5261 atm = (2.1375)(T1 )

From state equation, we have T1

p1

=

ρ1 R

=

101325 1.225 287

×

= 288 .2 K Therefore, T2

= (2.1375)(288.2) = 616.03 K

a2

=

V2

= M2 a2 = (0.51299)(497.51)



=

γ R T2 = 497.51m/s

255 .22

From isentropic table, for M 2 = 0.51299, we have T2 = 0.950 T02 Therefore, T02

=

T2 616.03 = 0.950 0.950

=

648 .45 K

94

Normal Shock Waves

5.12 For nitrogen, molecular weight is 28.02 and γ = 1.4. Thus, the gas constant is γ R = 8314/28.02 = 297 J/(kg K) and cp = γ − R = 1.4 297 = 1039 .5 J/(kg 1 0.4 K).

×

The speed of sound upstream of the shock is



a =

√1.4 × 297 × 303

γRT =

= 354 .95m/s Therefore, the upstream Mach number M 1 is M1

=

V1 923 = a1 354.95

=

2.6

Also, since the flow process across a normal shock is adiabatic, T 01 = T02 . Now, from normal shock table, for M 1 = 2.6, we have p2 = 7.72 p1

T2 = 2.2383 T1

M2 = 0.504

Therefore, p2 = 7.72 a2

=



× 300 = 2 .316 MPa √ γ R T = 1.4 × 297 × 2.2383 × 303

= 531 .03 m/s V2

=

M2 a2 = 0.504

=

267 .64 m/s

× 531.03

When the flow slows down isentropically from V 1 to V 2 , by energy relation, we have T2

=

T1 +

= 303 + = 678.32 K

V 12 V22 2cp

−

923 2 267.642 2 1039.5

− ×

95 By isentropic relation, we have p2 p1

=

  T2 T1

γ γ −1

3.5

678.32 =



303

= 16.786



Thus, p2 = 5.036MPa 5.13 For blunt nosed model at Mach 3, there will a detached bow-shock standing in front of the nose. This shock can be approximated to a normal shock at the nose of the model aroun d the stagnation point . Therefore, pressure at the stagnation point is the total pressure behind the normal shock. From normal shock table, for M 1 = 3, we have p02 = 0.32834 p01 Thus, p02

= 0.32834 p01 = 0.32834

× 10

= 3.2834 atm =

332 .69 kPa

The flow process across the shock is adiabatic. Hence, T 02 = T 01 . Therefore, T02 = 315K After the normal shock, the flow decelerates isentropically to stagnation condition at the nose. Hence, the stagnation density ρ 02 can be expressed as ρ02

=

p02 332690 = R T02 287 315

=

3.68kg /m3

×

5.14 Let us make the shock stationa ry and look at the field. The flow field with stationary shock will look like that shown in Fig. s5.14.

96

Normal Shock Waves T1 = 300 K

p2 = 5000 kPa

p1 = 101 kPa V1

V2 Shock

Figure S5.14

p2 p1

= 49.5 = 1+

2γ M12 γ+1



= 1 + 1 .167 M12



−1

− 1.167

Thus, M12

=

49.5 + 0.167 = 42.55 1.167

M1

=

6.52

The velocity becomes V1

√ × 287 × 330

= M1 a1 = 6.52 1.4 =

2374 .16 m/s

By normal shock relation, we have M2

T2 T1

= 1+

2( γ

− 1)



=



=

0.4

γ M12 + 1

(γ + 1) 2 M12

1 2 1 + γ− 2 M1 γ −1 2 γ M1 2

−



M12

−1





1/2

= 1+

48.41 244.86

×

T2 = 9.205 330 = 3037 .65 K The speed of sound downstream of the shock is a2

=



γ R T2 =

= 1104.9m/s

√1.4 × 287 × 3038.34

× 41.5 = 9.205

97 Thus, the velocity V 2 is V2

= M2 a2 = 0.4

× 1104.8

= 441 .92m/s = Cs

− Vg

Vg = 2374 .16 =

− 441.92

1932 .24 m/s

Aliter

The above problem can also be solved using gas tables, as follows. From normal shock tables, for pp21 = 49.5, we have M1

≈

6.5

M2

≈

0.4

T2 T1

≈

9.2

T2

≈

3036 K

a2 = 1104 .47m/s V2

= M2 a2 = 0.4

× 1104.47

= 441 .8m/s a1

=

√1.4 × 287 × 330

= 364 .13m/s V1

= M1 a1 = 6.5 = 2366.84m/s

× 364.13

98

Normal Shock Waves

Vg = 2366 .84

− 441.8

= 1925.04m/s

5.15 Upstream of the shock, the speed of sound a 1 is γ RT1 = 1.4 287 300 a1 =

√ ×



×

= 347 .2m/s The Mach number

M1 =

V1 412 = = 1.19. a1 347.2

From isentropic table, for M 1 = 1.19, we get p1 T1 = 0.41778, = 0.77929 p01 T01 Therefore, p01

=

p1 92 = 0.41778 0.41778

= 220 .21 kPa T01

=

T1 300 = 0.77929 0.77929

= 384 .96 K From normal shock table, for M 1 = 1.19, we get p2 T2 p02 = 1.4854, = 1.1217, = 0.99372, M2 = 0.84846. p1 T1 p01 Therefore, p2

= 1.4854 =

T2

136 .66 kPa

= 1.1217 =

× 92

× 300

336 .51 K

99

T02

=

T01 = 384 .96 K , since flow proce ss across a shock is adiabatic

p02

=

0.99372

=

218 .83 kPa

× 220.21

The speed of sound is a 2 = fore, V2

√γ R T = √1.4 × 287 × 336.51 = 367.7m/s. There=

M2 a2 = 0.84846

=

311 .98m/s

× 367.7

The entropy rise across a normal shock is given by ∆s

=

R ln

    p01 p02

= 287ln =

220.21 218.83

1.8042 J/(kg K)

5.16 Let the subscripts 1 and 2 refer to conditions upstream and downstream of the normal shock, respectively. From normal shock table, for M1 = 2.5, we get p2 p02 = 7.1250, = 8.5261. p1 p1 The static pressure in the flow just downstream of the shock is, p2 = 7.125

× 1.0 = 7.125atm

If a normal shock has to be positioned at the nozzle exit, the back pressure to which the nozzle discharges has to be equal to the total pressure downstream of the shock. The total pressure downstream of the shock is p02

= 8.5261 =

× 1.0 = 8.5261 atm

863 .91 kPa

i.e. the back pressure has to be 863.91 kPa to position a normal shock at the nozzle exit.

100

Normal Shock Waves

5.17 From isentropic table, for M e = 2.5, we have pe = 0.058528, p0e where subscripts 0, e, and The throat area is

Ae = 2.637 A∗

∗ refer to stagnation, exit and throat, respectively. A∗

Ae 4 = 2.637 2.637

=

1.517 cm2 1 p0e = = 17.09 atm 0.058528 For the normal shock, the upstream Mach number is 1.5. From isentropic table, for M 1 = 1.5, =

A1 = 1.176, A∗

p1 = 0.2724, p01

T1 = 0.68966 T01

Thus,

× 1.517 = 1.784 cm2

A1

=

1.176

p1

=

0.2724

T1

=

0.68966

17.09 = 4 .655 atm

× × 500 = 344 .83 K

From normal shock table, for M 1 = 1.5, p2 = 2.40583, p1

T2 = 1.3202 T1

p02 = 3.4133, p1

M2 = 0.70109

The back pressure required is p 02 , thus, p02 = 3.4133

× 4.655 =

15.89atm

Downstream of the shock, the flow is isentropic upto the nozzle exit. However, for this flow the e ffective throat area is not the same as A ∗ , since p 02 < p01 . Let the e ffective throat area downstream of shock to be table, for M 2 = 0.70, A2 = 1.09437 A∗2

A∗2 . From isentropic

where A 2 is the area at the shock location, which is same as 1.784 A∗2 = = 1.63cm 2 1.09437 Ae A∗2

=

4 = 2.454 1.63

A 1 . Thus,

101 From isentropic table for

Ae A2 ∗

= 2.454, we have

T2 = 0.98, Te

Me = 0.245

Thus, Te = 0.98 The speed of sound is ae =

× 500 =

490 K

√1.4 × 287 × 490 = 443.71m/s

The flow velocity is Ve

=

Me

× ae = 0.245 × 443.71

=

108 .71 m/s

5.18 Since the Mach number upstream of the shock is 2 .32, the area ratio corresponding to this Mach number will give the area at the shock location. (a) For M 1 = 2.32, from isentropic table, A1 = 2.23 A∗ Thus, area at shock location is A 1 = 2.23

×5=

11 .15cm 2

(b) The Mach number downstream of the shock M2 given by normal shock table for M 1 = 2.32 is M2 = 0.53 For M 2 = 0.53, from isentropic table, we have A2 = 1.29 A∗ Since A 1 = A 2 = area at the shock location, we have A∗2 =

A2 11.15 = = 8.64cm 2 1.29 1.29

Therefore, Ae 12.5 = = 1.447 A∗2 8.64 From isentropic table, for

Ae = 1.447, the exit Mach number M e = 0.45 A∗2

102

Normal Shock Waves

(c) For the given nozzle, the area ratio

Ae is Ath

Ae Ae 12.5 = ∗ = = 2.5 Ath A 5 Ae From isentropic table, for A∗ = 2.5, we have p2 Me = 2.44 and = 0.064261 p02 For complete isentropic flow, p02 = p 0 = 700 kPa. Thus, 44.98 kPa.

p2 = 0.064261 700 =

×

The back pressure range for the flow to be completely isentropic is pb

≤

44.98kPa .

5.19 Given, T1 = 22 K and T01 = 400 K, where subscripts 1 and 2 refer to conditions ahead of and behind the shock, respectively. For

T1 22 = = 0.055, from isentropic table, T01 400 M1 = 9.2

From normal shock table, for M 1 = 9.2, we have M2

=

T2 T1

=

0.3893 17.4

T2 = 382 .8 K

5.20 Upstream of the normal shock, the Mach number M 1 is M1

=

V1 = a1

√1.4 ×500 287 × 300

= 1.44 From normal shock table for M 1 = 1.44, we get M2 = 0.72345,

p2 = 2.2525, p1

T2 = 1.2807, T1

Thus, p2

= 2.2525

× 100 =

225.25 kPa

p02 = 0.9476 p01

103

T2

=

1.2807

× 300 =

384.21 K

V2

=

M2 a2 = 0.72345

=

284 .25 m/s

× √1.4 × 287 × 384.21

The entropy increase is ∆s

= R ln =

  p01 p02

= 28 7 ln(1 .0552)

15.432J/(kg K)

5.21 The flow velocity at nozzle entrance is low. Therefore, the pressure and temperature of the flow at the entry can be treated as the stagnation quantities. Thus, p01 = 1MPa , T01 = 30 0 K From isentropic table, for M 1 = 2, we have p1 p01 = 0.1278

and

T1 T01 = 0.55556

Therefore, p1

= (0.1278)(1) = 0.1278 MPa

T1

= (0.55556)(300) = 166.67 K

From normal shock table, for M 1 = 2, we get p2 T2 p02 = 4.5, = 1.6875, = 0.72087 p1 T1 p01

and

M2 = 0.57735

Thus, p2

=

4.5

× 0.1278 =

0.5751 MPa

T2

=

1.6875

p02

=

0.72087

a2

=

√1.4 × 287 × 281.3 = 336.19m/s

× 166.67 = ×1=

281.3 K

0.72087 MPa

104

Normal Shock Waves

V2

= M2 a2 = 0.57735 =

× 336.19

194 .1m/s

5.22 Given, p0 = 101 kPa and T0 = 30 + 273.15 K. A normal shock at the nozzle exit implies that the entire nozzle flow is isentropic and also the flow is choked at the throat. The area ratio is Ae 0.0724 = = 2.896 A∗ 0.025 Ae A∗


= 2.896, we have p1 = 0.050115, p0

M1 = 2.6,

T1 = 0.42517 T0

This is the Mach number upstream of the shock. Thus, M1

=

p1

=

0.050115

T1

=

0.42517

p01 =

2.6

× 101 =

5 .06 kPa

× 303.15 =

128 .9 K

101kP a

Let the conditions downstream of the shock be referred to by subscript 2. From normal shock table, for M 1 = 2.6, we get M2

=

p2 p1

=

0.504 T2 = 2.2383, T1

7.72,

p2

=

39.06kPa

T2 = 288 .5 K p02

=

46.47kPa

p02 = 0.46012 p01

105 5.23 Given, p0 = 200 kPa and T0 = 350 K. Let subscripts 1 2 and 3 refer to locations upstream and downstream of the shock wave and the nozzle exit, respectively. Ath = A∗1 = 0.2 m2 At the shock location, A 1 = 0.6 m2 , thus, A1 = 0.6 = 3.0 A∗1 0.2 From isentropic table, for

A1 = 3.0, we get A∗1 M1 = 2.64,

p1 = 0.04711 p01

Up to the shock the flow in the nozzle is isentropic and therefore, Thus,

p01 = p0 .

p01 = 200k Pa p1 = 200

× 103 × 0.04711

= 9.422kPa where, p1 and p 01 are the static and total pressures, respectively, ahead of the shock. Let subscript 2 refer to condition behind the shock. Now, from normal shock table, for M 1 = 2.64, we have p2 = 7.9645, p1

M2 = 0.50,

p02 = 0.44522 p01

Thus, p2

= =

p02

= =

7.9645

× 9.422

75.04 kPa 0.44522

× 200

89.04 kPa

Also, from normal shock theory we have (Section 4.6) p01 A2 ∗ = = 2.25 p02 A1 ∗

106

Normal Shock Waves

Thus, A∗2 = 2.25

× 0.20 = 0 .45 m2

A∗2 may also be obtained from M 2 . From isentropic table, for M 2 = 0.5, A2 = 1.34 A∗2 Thus,

0.6 = 0.448m 2 1.34 This A∗2 is the equivalent throat area for the flow downstream of the shock. Therefore, at the nozzle exit, A2 ∗ =

A3 0.8 = = 1.786 A∗3 0.448 Now, from isentropic table, for M3 = 0.35,

A3 = 1.786, A∗3

p3 = 0.91877, p03

T3 = 0.97609 T03

Here, p 03 = p02 and T 03 = T 02 = T 0 . Thus, p3 p03 T3

= 0.91877 =

× 89.04 =

81 .81 kPa

89.04 kPa

= 0.97609

× 350 =

341.63 K

For area ratio A3 /A∗3 the subsonic solution from the isentropic table was used since after a normal shock the flow becomes subsonic and this flow is further decelerated in the divergent portion of the duct. 5.24 Given, p 01 = 5 atm and p 02 = 3.6 atm. Therefore, p02 3.6 = = 0.72 p01 5 Assuming γ = 1.4, from normal shock table, for M1 = 2.0,

p02 = 0.72, we have p01

p2 = 4.5 p1

Now, from isentropic table, for M 1 = 2.0, we get p1 = 0.1278 p01

107 Hence, the pressure just behind the normal shock at the nozzle exit is p2

=

4.5 p1 = 4.5

=

2.8755 atm

× 0.1278 × 5

5.25 The pitot tube will read the actual total pressure in a subsonic stream. But in a supersonic flow, the pressure measured by a pitot probe is the total pressure downstream of a detached shock which stands at the nose of the pitot tube. Therefore, it is essential to find out whether the flow is subson ic or supersonic. It can be easily seen from the isentropic relations that for M = 1, the pressure p 0.95 ratio pp0 = 0.528. Hence, p 0 = = = 1.8 atm. 0.528 0.528 Thus, when p 0 < 1.8 atm, the flow is subsonic, and when p 0 > 1.8 atm, the flow is supersonic. (i) p0 = 1.1 atm. The flow is subsonic and hence the pitot tube is measuring the actual total pressure of the flow. p 0.95 = = 0.8636 p0 1.1 From isentropic table, for

p = 0.8636, we get p0 M = 0.465

(ii) p0 = 2.5 atm. The flow is supersonic and the pitot tube measu res p02 behind a normal (detached) shock. p02 2.5 = = 2.63 p1 0.95 p02 From normal shock table, for = 2.63, we get p1 M = 1.275 (iii) p 0 = 10 atm. The flow is supersonic. p02 = 10 = 10.526 0.95 p1 p02 From normal shock table, for = 10.526, we have p1 M = 2.79

108

Normal Shock Waves

5.26 The shock will be only at the divergent portion of the nozzle, since only after the throat the flow becomes supersonic. The Mach number M 1 just upstream of the shock will be given by the area ratio Ashock . Ath Ashock 2000 = =2 Ath 1000 From isentropic table, for area ratio 2, we have M1 = 2.2 ,

p1 = 0.093522 p01

Up to the shock, the stagnation pressure does not change and therefore, p01 = 200 kPa p1 = 200

× 0.093522 = 18.7 kPa

Now, from normal shock table, for M 1 = 2.2, we get p2 = 5.48, p1

M2 = 0.55,

p02 = 0.62814 p01

where subscript 2 refer to condition just downstr eam of the shock. Therefore,

× 18.7 = 102 .48 kPa

p2

=

5.48

p02

=

0.62814

× 200 = 125 .63 kPa

For M 2 = 0.55, from isentropic table, we get A2 = 1.255 Ath2 where A th2 is the throat area required for the flow downstream of the shock to choke. A2 Ath2 = 1.255 But A 2 = A1 = Ashock = 2000mm2 , therefore, Ath2 = Thus,

2000 = 1593.63mm 2 1.255

Ae 3000 = = 1.8825 Ath2 1593.63

For this area ratio, from isentropic table (subsonic part), we get Me = 0.325

109 Therefore, p02 pe

=

pe

=

p02 125.63 = 1.076 1.076

=

116 .76



1 + 0.2

× 0.3252

3.5



= 1.076

The pressure loss occurs only across the shock and the loss of pressure ∆ p0

= =

p01

∆ p0

is

− p02 = (200 − 125.63)

74.37 kPa

5.27 Let the subscripts i, 1, 2 and e refer to the inlet, just upstream and just downstream, and the nozzle exit, respectively. It can be shown for this flow that, p01 A∗1 = p02 A∗2 (1) For M i = 2.0, from isentropic table, we get A∗i = 1.6875 A i Therefore, A1 A∗1

=

A1 Ai Ai A∗1

=

A1 Ai , since A∗1 = A ∗i Ai A∗i

=

2

× 1.6875 = 3 .375

For this area ratio, from isentropic table, we get M1 = 2.76 For M 1 = 2.76, from normal shock table, we have p02 = 0.40283 p01 Using this in Eq. (1), we get A∗1 = 0.40283 A∗2

110

Normal Shock Waves

Ae A∗2

=

Ae Ai A∗1 Ai A∗1 A∗2

=

4

× 1.6875 × 0.40283 = 2.7191

For this area ratio, from isentrop ic table (subsonic solution), we get Me = 0.22 The exit pressure p e may be expressed as pe

=

pe p02 p01 pi p02 p01 pi

=

pe p02 p0i pi , since p01 = p0i p02 p01 pi

= 0.96684 =

1 × 0.40283 × 0.1278 × 80

243 .8kPa

Thus, the back pressure required is 243.8 kPa. 5.28 When a pitot tube is placed in a supersonic stream, there will be a detached shock at its At the nose where pressure tap tube is located, the shock maystanding be treated as anose. normal shock and hencethe what the pitot measures is the pitot pressure p 02 downstream of the shock. The wall pressure measured by a pressure tap may be treated as the actual static pressure of the stream. Thus we may take the static pressure upstr eam of the shock as p 1 = 112 kPa. Thus, p02 2895 = = 25.848 p1 112 p02 = 25.848, we get p1 T2 M1 = 4.44 , = 4.7706 T1 Now from isentropic table, for M 1 = 4.44, we get From normal shock table, for

T1 T01

=

0.2023

T1

=

0.2023

a1

=



× 500 = 101.15 K

γ R T1 = 201.6m/s

111 Thus, V1

= M1 a1 = 895 .1 m/s

5.29 At 10,000 m altitude, from standard atmospheric table, we have p = 26.452kPa , T = 223.15 K During steady–state operation, mass flow through the test–section is given by m ˙ =

ρ AV =

p AM RT

=

26.452 103 287 223.15

=

14.57 kg/s

×

×



γ RT 2

× π (0.25) × 2.4 √1.4 × 287 × 223.15 4

The stagnation temperature during steady–state operation is T0 =

T0 T T

For, M = 2.4,

T0 1 = = 2.152 T 0.46468

Thus, T0 = 2.152

× 223.15 = 480.22 K

For the present geometry of fixed angle di ffuser, the optimum conditio n for steady state operation is a normal shock at the di ffuser throat. The di ffuser throat area is A∗ A∗ = A A From normal shock table, for M 1 = 2.4, M 2 = 0.52. From isentropic table, for M 2 = 0.52, we have A2 = 1.3 A

∗2 Here A∗2 is the area of the second throat and

A is the area of test-section. A2 = A∗2

Therefore, looking in to the corresponding supersonic Mach number for 1.3, we get M = 1.66

112

Normal Shock Waves

This is the Mach number just upstream of the second throat with a shock. From normal shock table, for M 1 = 1.66, we have p02 = 0.872 p01 This pressure loss must be compensated by the compressor. (a) The power required for the compressor is given by Power = h 0 hi = C p (T0 Ti )

−

−

where the subscripts 0 and i refer to outlet and inlet conditions. For an isentropic compression, T0 Ti T0

− Ti

=

γ −1

    −    − p0 pi

= Ti

γ

γ −1

p0 pi

γ

1

1 0.872

= 480 .22

0.286

1 = 19.18 K

Thus, the power input required for mbox mass of air becomes Power = 1004.5

19.18 = 19.266 kJ/kg

×

The total power required for the compressor is m ˙ 19266 Power = 746 =

376 .3 hp

This is the running power required for the compressor. (b) During start-up, M1 = 2.4 in the test -section. The corr esponding total pressure ratio across the normal shock is p02 = 0.5401 p01 The isentropic work required for the compressor during start-up to drive the shock out of the test-section per mbox mass of air is Power =

Cp T0i

1 0.5401

= 1004.5

0.286

   × 480.22

= 92929.87 J/kg

1

− 1 0.5401

 

0.286



−1

113 The power required to start the tunnel is m ˙

Power =

× 92929.87 = 14.57 × 92929.87 746

=

746

1815h p

5.30 Let subscripts 1 and 2 refer to conditions just upstream and downstream of the normal shoc k. The Mach number just upstream of the shoc k is 1.85. From isentropic table, for M 1 = 1.85, we get p1 A1 = 0.1612, ∗ = 1.495 p01 A1 p1 = 0.1612 A1 = 1.495

× 400 = 64.48 kPa × 10 = 14.95cm 2

Now from normal shock table, for M 1 = 1.85, we have

p02 = 0.79023 p01

M2 = 0.60,

Again from isentropic table, for M 1 = 0.6, we get A2 A∗2 = 1.1882 But A 2 = A 1 = 14.95cm 2 . Therefore, A∗2 =

14.95 = 12.58cm 2 1.1882

We want the Mach number and pressure at a state 3 where 2 14.95 = 29.9 cm2 .

A3 = 2 A1 =

×

Downstream of the shock, A ∗2 = A∗3 . Thus, A3 A∗3

=

29.9 = 2.377 12.58

≈ 2.4

From isentropic table, for this area ratio (from subsonic solution), M3 = 0.25

and

p3 = 0.95745 p03

But p 03 = p 02 = 316.1 kPa. Therefore, p3 = 0.95745 316.1 = 302 .65

×

Note: This problem has been solved using isentropic table and hence the results

obtained are only approximate. For exact results, we have to use the actual relations.

114

Normal Shock Waves

5.31 (a) Given, m ˙ = 0.9 kg/s, p0 = 4 101325 = 405300 Pa, T0 = 30+273 = 303 ∆p K and = 2. p1

×

The pressure ratio across the shock is p2 p1

−

=

2

p1 p2 p1

= 2+1 =

For

3

p2 = 3, from normal shock table, p1 M1 = 1.65, M2 = 0.654,

T2 = 1.4228 T1

From isentropic table, for M 1 = 1.65, T1 A = 0.6475, ∗ = 1.292 T0 A Therefore, T1

= 0.6475T0 = 0.6475

× 303

= 196 .19 K T2

= 1.4228T1 = 1.4228

× 196.19

= 279 .14 K The mass flow rate is 0.9 kg/s, therefore, m ˙ = 0.9 =

0.6847 p0 Ath RT0

√

√

Ath

= =

0.9 RT0 0.6847 p0

√

0.9 287 303 0.6847 405300

×

= 9.56cm 2

×

115 Therefore, the exit area becomes Ae

= 1.292 Ath = 1.292

× 9.56

12.35cm 2

= (b) The exit velocity is V2

=

M2 a2 = M 2

=

0.654

=

219m /s



γ RT2

× √1.4 × 287 × 279.14

5.32 a) Given, (p2 p1 )/p1 = 3.5, where p1 and p2 are the pressures ahead of and behind the shock. Therefore, the pressure ratio across the shock is p 2 /p1 = 4.5.

−

For p 2 /p1 = 4.5, from normal shock table, M2 = 2,

ρ2 = 2.67 ρ1

Therefore, the shock speed becomes Cs

=

M1 a1 = 2

=

2

×



γ RT

× √1.4 × 287 × 300

= 694 .4m/s The piston speed is given by Vp

= V1

 

− V2 = V1 1 − VV21

−  × − 

= Cs 1

ρ1 ρ2

= 694 .4

1

= 694 .4 =

1 2.67

× (1 − 0.375)

434m /s

116 (b) Given ( p2

Normal Shock Waves

− p1 )/p1 = 7. Therefore, p 2 /p1 = 8.

For p 2 /p1 = 8, from normal shock table, M2 = 2.65,

ρ2 = 3.5047 ρ1

Therefore, the shock speed becomes Cs

= 2.65

× √1.4 × 287 × 300

= 920 .05m/s Thus, Vp = 920 .05 = 920 .05 =



1 × 1 − 3.5047



× 0.7147

657 .56m/s

5.33 (a) Let subscripts 1 and 2 refer to states ahead of and behind the shock, respectively and subscript e refer to the nozzle exit. Given, p01 p02 100 = 12 .4 p01

−

×

1

− pp02

= 0.124

p02 p01

= 0.876

01

From normal shock table, for p 02 /p01 = 0.876, M1 = 1.65 , M2 = 0.654,

T2 = 1.4228 T1

(b) For M 1 = 1.65 from isentropic table, we have T1 T01

= 0.6475

T1

= 0.6475

× T01

= 0.6475

× 330

= 213 .7 K

117 Therefore, T2 = 1.4228

× 213.7 = 304K

By energy equation, we have he +

V e2 2

=

h0e

c p Te +

V e2 2

=

cp T0e

But the flow process across the shock is adiabatic, therefore, Hence the flow speed at the nozzle exit becomes

Ve

= = =

× × 2

cp (T0e

− Te )

2

1004.5

× (330 − 300)

245 .5m/s

The flow speed behind the shock is V2

= =

× γ R T2 √ 0.654 × 1.4 × 287 × 304 M2 a2 = M 2

=



228 .57m/s

(c) The mass flow rate is m ˙ = = =

0.6847 p01 Ath RT01

√

0.6847

× (5 ×√ 101325) × (5 × 10−4 ) 287 × 330

0.5636 kg/s

T 01 = T 02 = T 0e .

118

Normal Shock Waves

Chapter 6

Oblique Shock and Expansion Waves 6.1 Given, M1 = 2 and β = 40◦ , therefore, M1n = 2.0sin40 = 1 .29. From normal shock tables, for M 1n = 1.29, we have p2 p1

= 1.775

T2 p1

= 1.185 and

M2n

= 0.7911

Therefore, p2

= =

T2

0.5

× 1.775 × 105

0.8875

× 105 Pa

=

1.185

× 273

=

323 .5 K

For the adiabatic process, Tt1 = Tt2 . From isentropic tables, for Mac h 2, we have T 1 /Tt = 0.556. Thus, T1t = T2t =

273 = 491K 0.556

From isentropic table, for T 2 /T2t = 0.659, we have M2 = 1.61 119

120

Oblique Shock and Expansion Waves

Also, sin( β

− θ) = wu22

=

0.791 0.791 = = 0.4913 M2 1.61

−θ

= 29.43

⇒θ

= 10.57 deg

β

=

the wedge angle = 2 θ = 21.14◦

Thus,

6.2 For M1 = 2.0, from isentropic table, we have ν1 = 26.38◦ . Prandtl-Meyer function after expansion is ν2 = ν 1 + θ = 26.38 + 5 = 31 .38 deg

Therefore, for ν2 = 31.38◦ the corresponding Mach number from isentropic table is M 2 2.18 . From isentropic tables for M 2 = 2.18, we have

≈

p2 p02 T2 Tρ02 2 ρ02

=

0.0965;

=

0.5127;

=

p2

6.3 (a)

Tρ01 1

0.1882;

ρ01

0.0965 × 0.0965 = 98 × 0.1278 0.1278

=

p1

=

74k Pa

T2 = 300

ρ2

p1 p01 T1

× 0.5127 0.5556

=

276 .8 K

=

74 287

=

0.9315 kg/m3

× 103 × 276.8

=

0.1278

=

0.5556

=

0.2300

121 pe = 1 atm Me = 3

p0 = 70 × 105 kPa

Figure S6.3a

M1 = 3.0

=

⇒

p1 = 0.02722 p01

With p 1 = pe = p atm , p01 =

1.01325 105 = 37.2 0.02722

×

× 105 Pa.

Therefore, the supply pressure for which oblique shock wave will first appear in the exhaust jet is

≤ 37.2 × 105 Pa (b)

0.45 D p0 = 70 × 105 kPa 0.1 D 0.45 D

1.0

97

D

D 0.45 D

D

Figure S6.3b

sin β

=

0.45 1.097 = 0.41

M1n

=

M1 sin β = 3

p2 p1

=

1.5984; p2 = p e

× 0.41 = 1 .23

122


Therefore, p1

M1

=

pe atm = 1.5984 1.5984

=

0.634

=

3.0 =

× 105 Pa

⇒ pp011 = 0.02722

Therefore, p01

=

0.634 105 0.02722

=

23.3

×

× 105 Pa

Therefore, minimum supply pressure for obtaining the desired test region is 23.3 105 Pa.

×

(c) pe = 1 atm M1 = 3

p0 = 70 × 105 kPa 1

2

Figure S6.3c

M1 p2 p1 p1 p01

=

3.0

=

10.33

=

0.02722

p1 p2 p01

patm = = 0.098105 Pa 10.333 = pe = patm 0.098 105 = = 3.6 105 Pa 0.02722

×

×

Therefore, theexit minimum supply pressure for which a normal shock will appear at the nozzle is 3.6 105 Pa

×

Note: The static pressure after the shock has to be equal to the back pressure,

namely the atmos pheric press ure. This is because, subsonic jets are always correctly expanded. Thus, the total pressure of this subsonic flow is higher than

123 its static pressure. Hence, the flow will move some distance downstream of the nozzle exit before coming to rest. 6.4 The relation between θ and β is 2cot β M12 sin 2 β tan θ =

1

M12 ( γ + cos 2 β ) + 2

−





For θ = 15◦ , M1 = 2.0, solution by iteration yields β = 79.8◦ . This is the strong shock solution. (a) β = 79.8deg

(b) p2 p1

=

2γ M 2 sin 2 β γ+1 1

=

4.354

− γγ +− 11

(c) T2 T1

p2 ρ 2 =

p1 / ρ 1

=

1.662

(d) β

−θ

=

64.8◦

ρ2 ρ1

=

tan β tan( β θ)

=

2.615

−

(e) M12 sin 2 (β

− θ)

=

1 2 2 1 + γ− 2 M1 sin β

γ M12 sin 2 β

γ

−1 2

× 4 ×− (sin 79.8◦ × sin79 .8◦) × 4 × (sin 79.8◦ )2 − 0.2

=

1 + 0.2 1.4

=

1.775 = 0.34 5.224

124


M22

=

0.34 0.34 = = 0.415 (sin 64.8)2 0.819

M2

=

0.644

Weak shock

(a) Solving θ

− β − M relation we can obtain

β weak = 45.3◦

(b) p2 p1

=

2γ M 2 sin2 β γ+1

=

2.191

− γγ +− 11

(c) T2 T1

=

1.267

(d) β

−θ

=

45.3

− 15 = 30.3deg

ρ2 ρ1

=

tan45 .3◦ tan30 .3◦

=

1.729

(e) M2 = 1.446

Aliter:

Using Normal shock tables: M1n = M 1 sin β

Strong shock solution lution

 1.97 Weak shock so-

125

γ = 1.4

M1n = 1.97

γ = 1.4

M1n = 2sin 45.3 deg = 1 .42

p2 p1

= 4.361

M2n

T2 T1

= 1.663

p2 p1

= 2.186

M2n

= 0.583

ρ2 ρ1

= 1.724

ρ2 ρ1

= 2.622

T2 T1

= 1.268

M2

=

M2n sin( β θ )

=

0.7314 = 1.45 sin30 .3

M2n M2

= M2 sin( β

− θ)

=

M2 n sin( β θ )

=

0.583 = 0.644 sin(79.8 15)

−

= 0.7314

−

−

Note: The solution obtained with oblique shock relations may also be obtained

using oblique shock tables, which will result in considerable time saving. 6.5 2

1

Figure S6.5

From isentropic tables, for M 1 = 2.0, we have p1 = 0.1278. Since, pt1 0.75 0.1278 = 0.0599. 1.60

pt1 = pt2 for this isentro pic expansion,

×

For this pressure ratio, from isentropic table, we get

M2 = 2.48 .

p2 = pt2

126


From isentropic table, the Prandtl-Meyer function for Mach 2.0 and 2.48, respectively are ν1

=

26.38◦

ν2

=

38.655◦

Therefore, the flow turning angle becomes ν12 = ν 2

− ν1 =

12.275◦

6.6 (a) Angles for which the oblique shock remains attached to the wedge (from oblique shock table) are At

M1

= =

2.0 22◦

θmax

= 3.0 =

34◦

θd

M1min

15◦◦ 25 40◦

1.65 2.11 4.45

θmax

At

M1

(b)

6.7 θ

M1 = 3.5

M2

45

◦

θ

Figure S6.7

M1

= 3.5

β = 45de g

127 =

28.158◦

⇒θ

=

M1n

=

M1 sin β = 3.5 sin45 = 2 .47

M2n

=

0.51592

M2

=

M2n sin( β θ)

=

1.78

−

6.8 M2

=

4.0

ν2

=

65.785◦

ν2

=

ν1 + |∆θ |

ν1

= ν2

Therefore,

− | ∆θ |

(a) |∆θ | = 60de g

−15 deg = 45 deg

Therefore, ν1

=

⇒ M1

= 65.785deg =

−45 deg = 20.785deg

1.8022

(b) |∆θ| = 60de g

−30 deg = 30 deg

Therefore, ν1

=

⇒ M1

= 65.785deg =

2.360

−30 deg = 35.785deg

128


(c) |∆θ | = 0d eg Therefore, nu1

= ν2 = 65.785deg

For this value of Prandtl Meyer function, from isentropic table, we get 4.0 .

M1 =

(d) |∆θ| = 60 deg +15 deg = 75de g ν1

=

65.785 deg 75 deg =

−

− 9.215 deg

A negative ν is not possible. The flow downstream can exist only upto 65.785 deg for which ν1 = 0 = M1 = 1.0.

| ∆θ | =

⇒

6.9 p1 , M1 p2 ,

M

2

θ

Figure S6.9

For M 1 = 2.0, from isentropic table, we have p 1 /p0 = 0.1278. Therefore, p2 p2 = p0 p1

× pp10 = 12 × 0.1278 = 0 .0639

For p 2 /p0 = 0.0639, from isentropic table, we have M 2 = 2.444 . M1

=

2.0 =

M2

=

2.444 =

ν2

=

ν1 + |∆θ | = ν 1 + θ2

θ2

=

ν2

=

⇒ ν1 = 26.38deg ⇒ ν2 = 37.81 deg

− ν1 = 37.81 − 26.38

11.43deg

129 6.10

M1 , p1

p0

θ

Me

Figure S6.10

(a) M1

=

3.0

⇒ pp10

=

0.02722

pe p0

=

1.01325 105 70 105

=

0.0145 <

=

and

×

×

p1

p0 Hence, there will be expansion at the nozzle exit to reduce the pressure from to p e . From isentropic table, for ppe0 = 0.145 = Me = 3.431.

⇒

⇒ ν1 = 49.75◦ Me = 3.431 =⇒ νe = 57.42◦ M1 = 3.0 =

For expansion,

νe = ν 1 + |∆θ |

Therefore, |∆θ| = 57.42

− 49.75 = 7 .67◦

(b) For θ e = 0,

pe = p 1 = 1.0133 105 Pa p1 pe = = 0.02722 p0 p0 Therefore, stagnation pressure required for θe = 0 is νe = ν 1 ,

p0

×

= =

1.0133 105 0.02722

×

37.226

× 105 Pa

p1

130


6.11 From isentropic table, for M 1 = 1.5, we have p1 p01

= 0.2724

T1 T01

= 0.6897

Therefore, p2 p01

=

p2 p1

=

0.13 0.30

⇒ M2

=

2.05

T2 T02

=

0.5433

=

× pp011 × 0.2724 = 0.118

(a) From Prandtl-Meyer function table, we have M1 =

⇒ νM

1

M2 =

⇒ νM

2

The deflection angle required is

= 1.5 = 12◦ = 2.05 = 27.75◦

θ = ν M1

− νM

2

=

−15.75◦

(b) M2 = 2.05 (c) T2 T2

6.12

T 01

×

=

T02

T1

=

275 .7K

0.5433

× T1 = 0.6897 × 350

131 1 1

2

2

3

A1

3

A2 5

4

A3

◦

10

◦

A4

◦

15

Figure S6.12

Flow deflection in regions (2), (3) and (4) as measured from flow deflection in region (1) are 5deg, 15 deg and 30 deg. For expansion, ν2 = ν 1 + |∆θ|

−

Region (1) (2) (3) (4)

−

−

ν | ∆θ | M µ (deg) (deg) (deg) 0 2 6.38 2.000 29.928 5 3 1.38 2.187 27.200 10 4 1.38 2.60 22.600 15 56.38 3.370 17.260

A Fan angle (deg) A (µ2 ∆θ) 7.73 1.688 14.60 1.980 20.34 2.896 6.012

∗

ξ1 = µ 1

−

−

Therefore, A1 : A 2 : A 3 : A4

= 1.688 : 1.980 : 2.896 : 6 .012 = 1 : 1 .173 : 1.716 : 3.562

6.13 From oblique shock table, for M1 = 2.4 and θ = 7deg, we get 30.25 deg.

M1n

= =

M1 sin β 2.4sin30 .25 = 1 .21

From normal shock table for M1n = 1.21, we have p2 p1 T2 T1 a2 a1 p02 p01

= 1.539 = 1.134 = 1.065 = 0.9918

β =

132


M2n

=

M2

=

From normal shock table, for

0.830 M2n = 2.12 sin( β θ )

−

M2 = 2.12, we have

pp3 2 T3 T2 a3 a2 p03 p02 M3

=

5.077

=

1.7875

=

1.337

=

0.6649

=

0.5583

p2 T2 p3 T3

= = = =

1.57 p1 = 0.77 105 Pa 1.134 280 = 317 .52 K 4.881 p 2 = 3.91 105 Pa 1.7875 317.52 = 567.57 K

× ×

×

×

×

(a) ρ3 A3 V3

m ˙ = Therefore, Ai

= A3 =

ρ3

=

V3

m ˙ ρ3 V 3

p3 RT3 = 2.4 kg/m3 = M3 a3 = 0.5583 = 266 .64m/s

× 477.6

Hence, Ai

= =

20 2.385

× 266.64

0.03125 m2

(b) At the exit, M e = V e /ae . By energy equation, we have Ve2 a2e a2 + = 0e 2 γ 1 γ 1

−

−

133 ae



=

a0e

a20e

− 0.2Ve2



= a03 = a 01 = 20.04 T01

From isentropic table, for M 1 = 2.4, 0.061 T1 T01

=

T01

=

Thus, a 0e

=

Hence a e

=

Me

=

0.4647 280 = 602.5K 0.4647 20.04 602.5 = 492 m/s

√



4922 0.2 30 = 0.061 492

−

× 302 = 492m/s

For M e = 0.061, pe p T03 e T03 For M3 p3 p03 T3 p03

=

0.9975

=

0.9993

=

0.5583

=

0.809

=

0.941

Therefore, pe

= = =

T

= e

ρe

Hence, A e

pe p03 p3 p03 p3 0.9975 3.91 0.809

× × 105 4.82 × 105 Pa

0.9993

567.57 = 602.7 K 0.940 pe 4.75 105 = = RTe 287 595 = 2.79 kg/m3 20 = = 0.240m 2 2.78 30

×

×

× ×

for

Me =

134


6.14 (a) For normal shock diffuser, at M 1 = 3.0,

p02 p01

= 0.3283

(b) For the wedge shaped di ffuser, M1 = 3.0, θ = 8◦ Therefore, β

M2 θ

= = =

25.5◦ 2.6 8◦

= =

29.0◦ , 2.26

For M 2 = 2.6 and θ = 8◦ , we have β

M3 Therefore, M1n

M2n

= = = =

M1 sin β 3.0sin25 .5 1.3 2.6 sin 29 = 1 .26

From Normal shock table, for M 1n = 1.3, p02 p01 M2n p03 p01 M3 p04 p03

=

0.9794,

=

1.26

=

0.986,

=

2.26

=

0.60105

Therefore, the overall stagnation pressure ratio is p04 = 0.9794 p01

× 0.986 × 0.60105 =

0.5804

Note: The first di ffuser with a single normal shock, the pressure loss is 67.2%

but for the wedge shaped di ffuser it is only 41.96%.

135 6.15 The wave pattern over the plate will be as shown in the Fig. S6.15a. (a)

M1

1

2 θ=α β = 42 ◦

3

Figure S6.15a

M1

= 2.0

β = 42de g

=

⇒θ

= α = 12.3◦

(b)

1 2 4

M1 β δ

Slipstream

3 5

Figure S6.15b

M1n = M 1 sin β = 1.338

For M1 = 1.338, from normal shock table, for M 1 = 2.0, p3 p1 M3n Therefore, p 3

=

1.928

= =

0.7664 1.928atm

From isentropic table,

M3n = 1.55 sin( β θ ) ◦ ν1 = 26.38 |∆θ | = 12.3◦ M3

=

−

For expansion, ν2

= ν1 + |∆θ| = 26.38 + 12.3 = 38 .68◦

The corresponding Mach number from isentropic table is

136


M2 For M 1 p1 p0

= =

2.48 2.0

=

0.1278

For M 2 p2 p0 Therefore, p2

= 2.48 = 0.0604 = =

p2 p0

p1 p0 p1 0.473atm

(c) Trial

1

2

Trial

δ 12de g 12 .5deg β

M2n = M2 sin β M4 p4 p1 p4

ν5 = ν 3 + |∆θ |

34 34 .5 1.387 1 .405 0.744 0 .740 2.09

2.12

0.989

1 .003

M5 p5 p0 p3 p5 p5

p5(1)

p5(2)

0.041

∆p

p4(2)

p4(1)

12

12.5

δ

◦

◦

Fig. S6.15c

∆p

δ

− 0.041 − 12

For

∆p

=

−0.002 − 0.041

=

0

=

12.5◦

0.5

δ

For which p 4 = p 5 = 1at m 6.16

1

2

25.38 25 .88 1.96 1.98 0.136 0 .132 0.534 0 .521 1.03

1.005

137

1

2

M1 15◦

Figure S6.16

For θ = 15◦ , M 1 = 3.0 =

⇒ β = 32.2deg. M1n

= M1 sin β = 1.60

M2n

= 0.6684

p2 p1

= 2.82

ρ2 ρ1

= 2.03

T2 T1

= 1.388

p02 p01

= 0.8952

From isentropic table, for M 1 = 3.0, we have p1 p01

=

0.0272

T1 T01

=

0.3571

= 1.904

× 105 Pa

Therefore, p1

T1 = 107 .13 K ρ1

=

a1

=

p1 = 6.19 kg/m3 RT1



γ RT1 = 207.5m/s

Using ratios of flow properties, we obtain,

138


(a)

× 105 Pa

p2

=

5.37

ρ2

=

12.57kg/m 3

T2 = 148 .7 K (b) p02

=

× 105 Pa

62.66

(c) M2

M2n = 2.26 sin ( β θ )

=

−



a2

=

γ RT2 = 244.4m/s

V2

= M2 a2 = 552 .3m/s

6.17 λu

= = =

λl

= =

CL

= =

  dz dx

u

1 3

− α 0 ≤ x ≤ 0.3c − 17 − α 0.3c ≤ x ≤ c

  −  −   −  −  − −  −   −  − dz dx

l

α

2 βc 2 βc

c

(λu + λl ) dx

0

0.3c

1 3

0

α dx

c

+ +

α dx

17

0.3c c

αdx

0

=

2

βc

1 3

α

0.3c

139

 

1 + α 0.7c 7

− =

−

−

4α

= For

β

α = 0 deg, C L = 0 and for

CD

= =

α = 2deg, C L =

c

2 βc

λ2u + λ2l dx

0

0.3c

2

+ 2 = = CD λu λl λc

α2 dx ]

0

2

π

0.3c

90

π 1 + 7 90

c

dx +

2

0.7c +

π

2

90

c]

β [0.02672 + 0.02212 + 0.0012185]

2 β

× 0.05006

= λt + λ c α λt + λc α = ( λu + λ l ) = +α 2

−

−

=

Cmac

2

α

= 0.0354

=

λc

1 3

[

dx

α

1 7

0.3c

βc

2

1 3

0

c

=

− √48 × 90π = 0.049365

   −  − −   − −     

2 [ βc +

− αc ]

2 [ 2αc] βc

=

1 3

−

−− α

α

2

+α

0

≤ x ≤ 0.3c

1 6

1 7

α

α

− − −  2

1 14

=

−

=

4 β c2



c

λc xdx 0

+α

0.7c

x

≤ ≤

c

140


= = = = xcp c

= =

xcp =

c 0.3c 4 1 1 [ xdx xdx ] β c2 6 14 0 0.3c 4 1 1 2 (0.3c)2 c (0.3c)2 β c2 12 28 4 0.09 1 (1 0.09) 8 12 28 0.0354



√ −



−



−

−





−

−





− CCmL + 12 − (−0.0354) + 1 ac

0.049365

2

1.217 c

6.18 1 2

M1 = 3

3 

1

2

3

Figure S6.18

Upper Surface

For M 1 = 3.0, from isentropic table, we have ν1

=

49.757deg

Therefore, ν2

= 49.757 + 12 = 61 .757deg

For ν 2 = 61.757◦ , from isentropic table, we get M2 =

3.71

Flow Mach number normal to the oblique shock is M2n

= M2 sin β

where β is shock angle. From oblique shock chart for M 2 = 3.71 at θ = 12 deg, we have β2 = 26de g

141 Therefore, M2n

=

3.71sin26 = 1 .626

From normal shock table, for M 2n = 1.63, we get M3n M3

= =

0.6596 M3n Therefore, 0.6596 = sin( β θ ) sin14

=

2.726

−

Lower Surface

For M 1 = 3.0, and θ = 12◦ , from oblique shock chart, we get M1 n

= M1 sin β



β1

= 29.25 deg



Therefore, M1 n = 3s in2 9 .25 = 1 .47 

From normal shock table, for M 1 n = 1.47, we have 

M2 n 

= 0.712

Therefore, M2



0.712 = 2.4 sin (29.25 12)

=

−

From isentropic table, for M 2 = 2.4, we have 

=

ν2



36.75 deg

Therefore, ν3



=

36.75 + 12 = 48 .75deg

The corresponding Mach number is M3



6.19

= 2.95

142

Oblique Shock and Expansion Waves 1 2

12 M1 = 3 ◦

3 10◦

4 2 



3

4

Figure S6.19

θ12 = 2d eg

= 51.75 deg M2 = 3.105 p2 = 0.02326 p01 θ23 = 20de g ν3 = 71.75 deg ν2

3 M p3 = 4.493 = 0.003484 p01 θ34 = 22de g β34 = 33.5deg M3n4 = 2.48 p4 = 7.009 p3 p4 = 0.0245 p01 M4n = 0.5149 0.5149 M4 = = 2.58 sin11.5 θ12 = 22de g 12 = 40de g 



M1n p2 p1 p2 p01 p02 p01 M2 n1 







M2



ν2



θ2 3



ν3





M3 p3 p02 p3 p01



= 1.93 = 4.179 = 0.115 = 0.7535 = 0.5899 0.5899 = sin18 = 1.91 = 23.85 = 20 = 43.87deg = 2.71 = 0.0423





θ3 4



ν4





M4 p4 p02 p4 

= 0.0319 = 2d eg = 45.87deg = 2.8058 = 0.0366





p01 Comparing 6.20

p4 p01

and

p4 p01 ,

= 0.0276

it can be seen that, the slipstream is very weak.

143 p2

p3

M1

10◦

p1

Figure S6.20

M1 p1 θ β

M2 p2 p1 ν2

= = = = =

3.0 1.0133 10◦ 27.38◦ 2.505

=

2.054

=

39.24◦

θ3

× 105 Pa

ν3

M3 M2 p2 p02 M3 p3 p03

= = = = = =

20◦ ν2 + θ3 39.24◦ + 20 ◦ 59.24◦ 3.545 2.505

=

0.05808

=

3.545

=

0.0123

Therefore, p3 p1

p2 p3 p02 p1 p03 p2 0.01230 = 2.055 = 0.435 0.05808 =

×

L = CL

−p2 1 × 2c − p3 1 × 2c

=

L q1 c

=

γ

= = = =

   

+ p1 [1

L 2 2 M1 p1 c

2

L

γ M12 cp1

2

p2 + p 3 1 + 1 p1 p1 2 2 1 1 (2.055 + 0.435) 1.4 9 2 γ M12

×

−    − 

− 0.0389

× c]

144


Drag, D

c c p2 tan10 ◦ p3 tan 10◦ 2 2 p1 c p2 p 3 tan10 ◦ 2 p1 p1

=

−



=

CD

=

−



1 γ D2 2 M1 p1 c





1 p2 p 3 tan10 ◦ γ M12 p1 p1 1 (1.62 0.1763) 1.4 9

= =

−

×

×

=

0.02267

6.21 Let subscripts 1 and 2 refer to states upstream and downstream of the expansion fan. We know that the flow across the Prandtl-Meyer expansion fan is isentropic. Therefore, p01 = p02 . From isentropic table, for M1 = 2.0, we have p1 = 0.1278, µ1 = 30◦ , and ν1 = 26.37◦ p01 Therefore, p2 = 0.5 = p1

2 pp02 p1 p01

Thus,

p2 = 0.5 0.1278 = 0 .0639 p02 p2 Again from isentropic table, for = 0.0639, we have p02

×

µ2 = 24.19◦ ,

M2 = 2.44,

and

ν2 = 37.71◦

Thus, the flow turning angle is θ = ν2

− ν1 = (37.71 − 26.37) = 11 .34◦

For the first Mach line the angle relative to the freestream is the last Mach line the angle relative to the freestream is µ2

− θ = 24.19 − 11.33 =

µ1 = 30◦ . For

12 .86◦

6.22 From oblique shock table, for M 1 = 2.2 and θ = 5◦ C, we have β = 31.09719

and

p2 = 1.3397 p1

145 M1n = M 1 sin β = 1.136. From normal shock table, for M 1n = 1.136, we have p02 = 0.99726 p01 6.23 Assuming air to be a perfect gas, γ = 1.4. From oblique shock table, for M1 = 3.0 and θ = 8◦ , from gas tables, we have the shock angle as β = 25.61◦

p2 = 1.7953, where p 2 is the pressure behind the shock, which is also the p1 pressure at the cone surface. Thus, Also,

p2

= 0.05

× 1.7953 = 0 .0898 atm

= 0.0898 =

× 101.325

9.1kPa

6.24 Basically the given stream passes through an expansion fan and a oblique shock at the convex and concave corners, respectively, as shown in Fig. S6.24. 1

2 3

15◦ 15◦

Figure S6.24

From isentropic table, for M 1 = 3.0, we have ν1 = 49.757◦ ,

p1 = 0.02722 p01

Given, θ 1 = 15◦ , thus, ν2 = ν 1 + 15 ◦ = 64.757◦

For ν 2 = 64.757◦ , from isentropic tables, we have M2 = 3.92,

p2 = 0.0073316 p02

From oblique shock chart, for M2 = 3.92 and θ2 = 15◦ , we have β = 28◦ . Therefore, M2n = M2 sin β = 3.92sin28 = 1 .84

146


From normal shock table, for M 2n = 1.84, we have p3 M3n = 0.6078, = 3.7832 p2 Thus, M3

=

M3n 0.6078 sin ( β θ) = sin (28 15)

=

2.7

−

p3 =

−

p3 p2 p1 p2 p1

Now, p2 p1

=

p2 p02 p01 p02 p01 p1

=

p2 p01 p02 p1

since p 01 = p02 across an expansion fan. Therefore, p2 p1

=

0.0073316 0.02722

=

0.269346

Thus, p3

= (3.7832)(0.269346)(1) =

1.019atm

6.25 By Ackerets theory, CL = Also, CL = where A = 1

 1 2

4α 2 M∞

−1

L

γ M2 A p

× 1 = 1m 2 is the wing area. Thus, CL α

M2 4∞

1

=

− √ 0.228 1.62 − 1

=

0.0712 radians = 4 ◦

=



4

147 Again by Ackerets theory, CD = Thus,



4 α2 = α CL 2 M∞ 1

−

CD = (0.228)(0.0712) = 0 .0162 The aerodynamic e fficiency of the plate is CL 0.228 = = 14 CD 0.0162 6.26 For M 1 = 2.4 and β = 33◦ , from oblique shock table, we get θ = 10◦

M1n = M 1 sin β = 1.31 From normal shock table, for M 1n = 1.31, we have p2 = 1.8354 p1 T2 T1 M2n M2

=

1.1972

= 0.78093 =

M2n 0.78093 = sin( β θ ) sin23 ◦

=

2.00

−

6.27 The pressure ratio across an oblique shock is given by p2 2γ =1 + M12 sin 2 β p1 γ+1



That is, p2 p1

−1

=

p1

=

p2

− = p2

− p1

ρ1

=

2γ M12 sin 2 β γ+1 2γ



−1



p 1 M12 sin2 β

γ+1 2γ u2 p 1 21 γ+1 a1

2 γ p1 u21 γ + 1 ρ1 a21

1 2

− M1 1 sin2 β − 2 M

 

1

2

sin β



−1

 

, since M12 =

− M1 2 1

u21 a21

148 But

Oblique Shock and Expansion Waves γp ρ

= a 2 , from isentropic relations. Therefore, p2 1 2

− p1 =

ρ 1 u21

4 γ+1



sin2 β

− M1 2 1



6.28 Given, p2 /p1 = 5. From normal shock table, for p2 /p1 = 5, we get the Mach number normal to the oblique shock at the compression corner as M1n = 2.1. But, M1n = M 1 sin β where β is the shock angle. Thus, β = sin −1

= sin −1

    M1n M1 2.1 3.0

= 44.427◦

From oblique shock chart, for M 1 = 3.0 and β = 44.427◦ , we get θ = 25.5◦

6.29 From oblique shock table, for M 1 = 2 and θ = 10◦ , we get β = 39.3◦

Therefore, the Mach number normal to the shock becomes M1n = M 1 sin β1 = 1.27 From normal shock table, for M 1n = 1.27, we get M2n = 0.80167 Therefore, M2 =

M2n = 1.64 sin ( β1 θ)

−

For M 2 = 1.64 and θ = 10◦ , from oblique shock chart, we get β3 = 49.5◦ Thus, the angle of the reflected shock relative to the flat wall is φ = β3

− θ = 49.5 − 10 =

39 .5◦

M2n = M 2 sin 49.5 = 1.25

149 From normal shock table, for M 2n = 1.25, we have M3n = 0.81264 Thus, M3n M3

=

sin ( β3

=

1.28

0.81264

− θ) = sin39 .5◦

6.30 For M 1 = 2 and θ = 7◦ , from oblique shock table, we get M2 = 1.75 For M 2 = 1.75, from oblique shock table, we have θmax = 18◦

This is the maximum of θ up to which the second shock will remain attached. 6.31 The given flow is through an obliqu e shock. Therefore, there are two solutions possible for the flow. One is called the weak solut ion, for which the flow downstream of the shock will continue to be supersonic with a reduces Mach number. The second is strong solution for which the Mach number downstream of the shock will be subsonic. I. Weak solution

Let subscripts 1 and 2 refer to conditions upstream downstream of the shock. For M 1 = 2 and flow turning angle θ = 15◦ , from oblique shock table, we have β = 45.34 deg and M2 = 1.45

The Mach number normal to the shock M 1n = M 1 sin β . M1n = 2 sin 45.34 = 1 .42 Now, the shock may be treated as a normal shock with upstream Mach number 1.42. From normal shock relations, the change in entropy across the shock is given by ∆s

= R ln

p01 p02

 

From normal shock table, for M 1n = 1.42, we have p01 = 1.049 p02

150


Thus,

s

= 287ln( 1 .049) =

13.73J/(k g K)

II. Strong solution

For M 1 = 2 and flow turning angle θ = 15◦ , from oblique shock table, we have β

=

79.83◦

M2

=

0.64

M1n = 2 sin (79.83◦ ) = 1.97 From normal shock table, for M 1n = 1.97, we get p01 = 1.36 p02 Thus,

s

= 287ln( 1 .36) =

88.25J/(k g K)

The shock will be attached up to θ max = 22.97◦ 6.32 Since the flow is turned at a compression corner, the problem is e ffectively getting the flow field downstream of a oblique shock. From oblique shock table(or chart), for M 1 = 3.0 and θ = 10◦ , we have β = 27.38◦

The Mach normal to the shock is M1n = M 1 sin β = 1.38 From normal shock table, for M 1n = 1.38, we get M2n = 0.74829,

p2

T2

p02

p1 = 2.0551,

T1 = 1.2418,

p01 = 0.96304

Thus, M2

=

M2n sin ( β θ )

−

151

p2

=

0.74829 sin (17.38◦ )

=

2.5

= (2.0551)(1) =

T2

2.0551 atm

= (1.2418)(200) =

248 .36 K

For M 1 = 3, from isentropic table, p1 = 0.027224, p01

p02

=

T1 = 0.35714 T01

p02 p01 p1 p01 p1

= (0.96304) (1/0.027224) (1) = T02

35.37atm

= T01 =

T01 T1 T1

=

200 0.35714

=

560K

6.33 Note that for a given Mach number and turning angle, the oblique shock can have a weak and a strong solutions. Further, the Mach number downstream of the shock is supersonic for weak solution and subsonic for the strong solution. Let us solve this problem with oblique shock table as well as with oblique shock charts. From oblique shock table, for γ = 1.4 and M 1 = 3.0 and θ = 10◦ , we have For weak solution

152


β = 27.38256◦

M2 = 2.505 ,

p2 = 2.0545 p1

For strong solution β = 86.40836◦

M2 = 0.48924 ,

pp21 = 10.292

From oblique shock chart I, for M 1 = 3.0 and θ = 10◦ , we have β = 27.4◦

as weak solution

β = 86.3◦

as strong solution

Using oblique shock chart II, the pressure ratio across and Mach number downstream of the shock are obtained as For weak solution

M2 = 2.5,

p2 = 2.05 p1

For strong solution

M2 = 0.49,

p2

= 10.3 p1 From the above solutions the elegance of oblique shock tables is obvious. Further, the solutions obtained with oblique shock charts are only approximate whereas, the oblique shock table gives accurate results. 6.34 (a) From oblique shock table, for M 1 = 2.2 and θ = 6◦ , we have β = 31.98◦

Therefore, the Mach number normal to the shock is M1n = M 1 sin β = 1.16 From normal shock table, for M 1n = 1.16, we have p02 = 0.99605, p01 Therefore,

M2 =

M2n = 0.86816

M2n = 1.98 sin ( β θ )

−

From normal shock table, for M 2n = 1.98, we have p03 = 0.73021 p02

153 Thus, p03 p02

=

p03 p02 p02 p01

=

(0.73021)(0.99605) = 0 .727

Thus, Pressure loss = (1 =

− 0.727) × 100%

27.3 per cent

(b) From oblique shock table, for M 2 = 1.98 and θ = 6◦ , β = 35.8◦

Therefore, the Mach number normal to the shock is M2n = M 2 sin β = 1.1582

≈ 1.16

From normal shock table, for M 2n = 1.16, we have p03 = 0.99605, p02 Therefore, M3 =

M3n = 0.86816

M3n = 1.7468 sin ( β θ )

−

From normal shock table, for M 3 = 1.75, we have p04 = 0.83457 p03 Thus, p04 p01

=

p04 p03 p02 p03 p02 p01

= (0.83457)(0.99605)(0.99605) = 0 .8279 Thus, Pressure loss = (1 =

− 0.8279) × 100%

17.2 per cent

6.35 For M 1 = 2.4 and β 12 = 30◦ , from oblique shock chart, we have θ12 = 6.2◦

154


The Mach number normal to the incident shock is M1n = M 1 sin β12 = 2.4

× sin30 ◦ = 1.2

From normal shock table, for M 1n = 1.2, we have M2n = 0.84217 Therefore, M2

=

M2n sin ( β θ )

=

0.84217 sin (30 6.2)

=

2.09

−

−

For M 2 = 2.09 and θ 23 = 6.2◦ , from oblique shock chart, we have β23 = 34◦

The Mach number normal to the reflected shock is M2rn = M 2 sin β23 = 2.09sin34 ◦ = 1.17 From normal shock table, for M 2rn = 1.17, we get M3n M3

= 0.86145 =

M3n sin ( β θ )

=

0.86145 sin (34 6.2)

=

1.84

−

−

Hence the Mach numbers upstream and downstream of the reflected shock are 2.09 and 1.84, respectively. 6.36 Let the state ahead of and behind the first shock be represented by subscripts 1 and 2, respectively, and those behind the second shock by 2 and 3, respectively. Given M1 = 2.3 and θ1 = 8◦ . For this Mach number and turning angle, from oblique shock table, we get β1 = 32.42◦ , M2 = 1.99

155 Therefore, M1n

=

M1

× sin β

=

2.3

× sin (32.42◦ )

=

1.23

From normal shock table for M n1 = 1.23, p02 = 0.9896 p01 The pressure loss caused by the second shock is 1 2

−  1

p 02 p01

1 (1 2

=

− 0.9896)

= 0.0052 Thus,

p03 =1 p02

− 0.0052 = 0 .9948

For p 03 /p02 = 0.9948, from normal shock table, M2n = 1.18 Thus, β2 = sin −1

=

  M2n M2

= sin −1

  1.18 1.99

36.37◦

For M 2 = 1.99 and β2 = 36.37◦ , from oblique shock chart I, θ2

≈

7◦

156


Chapter 7

Potential Equation for Compressible Flow 7.1 (a) We know from Eq. (7.6) that, T

ds dn

dh 0 dn

=

uζ +

=

uζ + cp dn

dT0

T ds =

dh

i.e., T 0 ds =

T0

ds dn

ds 0 dn

=

− ρ dp

dh0

− ρ1 dp0 0

=

cp dT0

=

cp

T0

− ρ10 dp0

dT0 dn

− ρ10 dpdn0

uζ +

T

T0 T

c p

dT0

(3)

dn

From Eqs. (2) and (3), we get cp

dT0 dn

(2)

T0 , we get T

Multiplying Eq. (1) by T

(1)

1

− ρ10 dpdn0

=

T0 T 0 dT0 uζ + cp T T dn 157

158

Potential Equation for Compressible Flow

− ρ10 dpdn0

i.e.,

 −  − 

T0 uζ + T

=

=

1+

T0 T

1

γ

2

1 cp

dT0 dn γ

M 2 uζ +

− 1 M 2 cp dT0 2

dn

7.2 For supersonic flows,by Eq. (7.37), we have



or

2 M∞

−1



∂ 2φ ∂ x2

2

− 1r ∂φ − ∂∂ rφ2 = 0 ∂r

∂ 2 φ 1 ∂φ + ∂ r2 r ∂r



− β2 = 0

2 where, β = M∞ 1. The solution for the above equation, valid over a slender body of revolution with a closed nose and arbitrary (smooth) meridional section (Ref. Liepmann and Roshko) is,

−

φ(x, r) =

− 2Uπ S (x) ln

 −  2

U 2π

βr

x

S  (ξ )ln( x 0

− ξ )dξ

where, S (x) = π R2 is the cross sectional area of the body at x and ξ = x Therefore, u = U∞

∂φ = U∞ ∂x

−

S  (x) ln 2π

 − 2 βr

1 d 2π dx



x

S  (ξ )ln( x 0

− ξ )dξ

β r.

−



∂φ R dR and, v = U∞ = U∞ i On the body, r = R, the pressure coe fficient is ∂r r dx given by

Cp =

S  (x) π

ln

 

For the given body, R =

2 λR

+

 x3/2

1 d π dx 0



x

S  (x)ln( x 0

− ξ)dξ −

≤x≤1

Therefore, S = π R2 = π2 x3 S  (x) =

dS (x) 2 2 dx = 3 π x

S  (x) = 6π 2 x dR dx

=

3 1/2 x 2

  dR dx

2

159 Substituting the above relation into the C P expression, we obtain = 62 x ln

Cp Cp

d dx



ln x3/2 + 6

d

+ 6 2

β

x

ξ ln (x 0

− ξ)dξ

x 0

d dx

ξ ln (x

= i6

x

d dx

  − 

Now consider the term 6



2 β x3/2 2

= 6x ln

2





dx



− ξ) dξ − 94 2 x

x

ξ ln (x



9

ξ ) dξ

−

0

x

−4

− ξ ) dξ. Express this as x

− 

d dx

ξ ln (x 0

(x

− ξ )ln( x − ξ ) dξ +

0



x

x ln (x 0

− ξ ) dξ

Now use the rule,



a

f (a 0

a



− t)dt =

f (t)dt

0

Therefore, we have 6

d dx



x

ξ ln (x 0

− ξ ) dξ

−6 dxd

= =

6

since lim ξ 6

d dx



→ 0 both

ξ2

2

0

0

ξ2



x 2

ln ξ



x

ln ξ dξ 0



x

ξ2

2

Cp 2

− ξ)dξ



=

−6

x ln x +

=

6x ln x

  −  −  −

=

6x ln

=

6x ln

=

ξ )0

−4 − x2 ln x − − x2 ln x + x2 4 0

x 2

 

− x2 − x − 2x ln x + 2x

− 6x

Hence, Cp

x

x (ξ ln ξ

−

2

2

−x

ln ξ and ξ ln ξ tend to zero. Therefore,

x

ξ ln (x

ξ ln ξ dξ

d

− dx −6 dxd

=

x



6x ln

2

β

2

β

2

β



ln x3/2 + 6x ln x

3 6x ln x + 6x ln x 2 3x ln x

− 334 x

− 6x − 94 x

− 334 x





160

Potential Equation for Compressible Flow

Drag Coe fficient The drag coe fficient can be expressed as, CD S (L) = S

=



L

Cp (x)S  (x)dx 0

3π 2 x2

Therefore, CD S (L) 3π 4

L

  

=

6x3 ln

0

6 4 2 x ln β 4

=

2

− 3x3 ln x − 334 x3 β −3

3 3 2 L L ln 2 β

=



x4 ln x 4

4



dx 4 L



− x16 − 334 x4

− L2 ln L + L8 − 118



π 2 x3

S (x) =

Therefore, S (L) = π 2 L3 Thus, CD π 2 L3 3π 4

But, β =

=



3 3 2 L L ln β 2

With L =

1, we get,

CD 32

=

3 2 ln 2 β

CD

=

32





− L2 ln L + L8 − 118

− 108

3 2 ln β 2



− 158

√M 2 − 1 = 1 and  = 0.1. Therefore, CD

= =

0.03[1 .5ln 20 0.0786



− 1.875]





0

Chapter 8

Similarity Rules 8.1 By Prandtl-Glauert rule, we have dCL dα

  dCL dα

= =

inc

(dCL /dα)inc 2 1 M∞ dCL = 0.108deg dα M =0.2

  −

∞

Therefore, dCL dα

M∞ 0.2 0.3 0.4 0.5 0.6 0.7 0.75

dCL dα measured

 

0.108 0.113 0.115 0.124 0.130 0.127 0.100

0.108 per degree 2 1 M∞

=

−  

dCL dα P −G rule

0.1080 0.1132 0.1178 0.1247 0.1350 0.1512 0.1633

8.2 For hypersonic flow, M ∞ ta n (θ + α) When ( θ + α) is small, M (θ + α)

∞

≥ 0.5.

0.5.

≥

where, θ is the half angle at the vertex of the model and α is the angle of attack. For hypersonic similarity, M∞1 (θ1 + α1 ) = M∞2 (θ2 + α2 ) M∞1 = 10 161

162

Similarity Rules

α1 = 3d eg θ1 = 3d eg

M∞1 (θ1 + α1 ) = 10

π × (3 + 3) × 180 >1

Hence the flow is hypersonic . When the test Mach num ber is small, the angle involved will be large. (a) M∞2

=

3.0

θ2 = 12de g

M∞1 ta n (θ1 + α1 ) M∞2 ta n (θ2 + α2 ) tan (12 + α2 ) 12 + α2

= 10ta n6 deg = 1 .05 = 3t an( 12 + α2 ) = 1.05 = 0.35 = 19.3

Therefore, α2 = 7.3deg

(b) M∞3

= 3.0

θ3 = 3d eg

M∞3 ta n (θ3 + α3 ) = 3.0 tan (3 + α3 ) = 1.05 tan(3 + α3 ) = 0.35 3 + α3 = 19.3 Therefore, α3 = 16.3deg

8.3 Hypersonic similarity factor K = M θ . In ord er to appl y the results of model to the missile, K should be same for the model and the missile. That is, K1 = K 2 that is, M1 θ1 = M 2 θ2 Given, M 1 = 12, θ1 = 4 deg and M 2 = 2.5. Therefore, θ2

=

M1 θ1 M2

163

12 4 2.5

×

=

= 19.2deg That is, the semi -vertex angle of model = 19.2deg 8.4 For supersonic flow past thin wedge with semi-wedge angle δ , the linearized theory yields, CPu

2

=

β

2

=

β

  dz dx

tan δ

u

≈ β2 δ

Therefore, (γ + 1) 1/3 CP δ 2/3

=

2δ 1/3 (γ + 1) 1/3 2 M∞ 1

=

2

 

−

[(γ + 1) δ ]2/3 M2 1

∞−

or 2 CP (γ + 1) M∞



where, χ =

δ 2/3



1/3

=

2 χ

2 M∞ −1 2 ]3/2 [δ (γ +1)M∞

∗ = 0.3. Therefore, by Equation (8.57), we have the minimum 8.5 (a) Given M ∞ Cp over the profile as Cp

=

   − ×    × − 2 + 0.2.32 2.4

2

1.4

0.32

=

15.873

=

−7.22

2.018 2.4

3.5

1

3.5

1

∗ = 0.4. Therefore, (b) Given M ∞ Cp

=

2 1.4

×

0.42



2 + 0.2.42 2.4

 − 3.5

1

164

Similarity Rules

= 8.929 =

×

−3.94

   − 2.032 2.4

3.5

1

Chapter 9

Two Dimensional Compressible Flows 9.1 The linearized perturbation velocity potential equation for two-dimensional supersonic flow Eq.(9.1) is β 2 φxx 0= φyy (1)

−

2 where β = M∞ 1 and φ is the perturbation velocity poten tial. Equation (1) may also be written as,

−



∂2φ ∂ x2

2

− M 2 1− 1 0∂∂=yφ2

(2)

∞

2 Keeping in mind that ( M∞ 1) > 0 for supersonic flow, it can be visualized that equation (2) is also of the form of the classical wave equation. Hence, a solution to equation (2) can be expressed as,

−

φ(x, y) = f (x

−



2 M∞

− 1y) + g(x +



2 M∞

− 1y)

(3)

For the problem under consideration, only left running waves are present and therefore, 2 g(x + M∞ 1y) = 0



Thus,

−

φ(x, y) = f (x

−

and ∂φ = f (x ∂y



By boundary condition, at the wall dyw = k(1 dx



∂φ ∂y



2 M∞

− 1y)



− β y) (−β ) w = U ∞ dy dx

− xl ) − k xl = k − 2k xl 165

(4) (5)

166

Two Dimensional Compressible Flows

Therefore, ∂φ = U∞ k ∂y

Substituting this into Eq.(5), we get, U

k

∞

2k

x



β f (x

−

l



2k

=

− 

f (x) =

x l

− 

− Uβ∞

β y)

(6)

−

x l

−  k

2k

(7)

Integrating Eq.(7) with respect to its argument, [Note that the argument is (x β y), but with y = 0], we have

−



− Uβ∞

f (x) =

2

kx

− k xl



+constant

(8)

At x = 0, f = 0, which gives constant = 0. Therefore, f (x) =

− Uβ∞



2

− k xl

kx



(9)

Since f (x) is defined throughout the flow, not just at the wall, and because it has the form of Eq.(8a), where x represents the argument of f , Eq.(4) can be written as φ(x, y) = f (x

− β y) = − U

∞

β

k



(x

− β y) 1 − x−l y

0

≤ x ≤ 0.7c

β



9.2 λu

= =

λL

CL

= = = =

  dz dx

= u

1 7

−α

− 13 − α

0.7c

≤x≤c

  −  −   −   − −   −  −  −   −  − dz dx

=

c

2

βc

2 βc 2 βc

α

l

(λu + λL ) dx

0

0.7c

0

1 7

1 7

c

α dx +

α 0.7c

0.7c

1 3

1 + α 0.3c 3

c

α dx +

αc

0

−αdx



167 = β

=

4α β



M2

α = 2d eg =

− 1 = 2.29 2π

= 0.0349

180 Thus, CL

=

4

2.29

= CD

=

× 0.0349 0.06096

2 βc

c

      − λ2u + λ2L dx

0

2

2 βc

=

2 [0.0081583 + 0.0406787 + 0.001218] 2.29

=

2

=

1 7

0.7c +

α

1 +α 3

2

=

0.3c + α2 c



× 0.050055 2.29

0.04372

9.3 0

≤ x ≤ 0.3c λu

0.3c

 

= λl

 

0.1 α 0.3 0.333 α 0.03 α 0.3 0.1 α

=

= =

− − −

−

α = 2de g = 2 β

=

√

≤x≤c λu

= =

λl

= =

π

180

8 = 2.83

≈ 0.035 radian

− − − −

 

0.1 α 0.7 0.143 α 0.03 α 0.7 0.043 α

− − − −

168

Two Dimensional Compressible Flows

φ(x, y)

=

f (x

− βy)

∂φ | ∂ y y=0

=

−βf  (x) = U∞ λ

Therefore,

− Uβ∞ λ f  (x − β y)

f  (x) =

For 0

φx (x, y)

=

φx (x, 0)

=

f  (x) =

Cp

=

−2 Uφ∞x = 2 βλ

− Uβ∞ λ

≤ x ≤ 0.3c Cpu

Cpl

For 0.3

= 2

λu 0.298 2 = β 2.83

=

0.211

= 2

λl 0.065 2 = β 2.83

=

0.046

×

×

≤x≤c Cpu

=

2

= Cpl

= =

λu = β

×2 − 0.178 2.83

− 0.1258 2

λl = β

×2 − 0.078 2.83

−0.0551

Chapter 10

Prandtl-Meyer Flow 10.1 Let subscripts 1 and 2 refer to conditions upstream and downstream of the Prandtl-Meyer fan. p0

=

33.5

M1

=

2.0

× 105 Pa

From isentropic table, for M 1 = 2.0, ν1

= 26.38 deg

p1 p0

= 0.1278

p2 p0

=

1.0133 = 0.03025 33.5

where subscript 2 refer to conditio ns downstream of the expansion fan. From p2 isentropic table, for = 0.03025, p0 M2

=

2.93

ν2

=

48.388deg

But ν2

=

ν1 + ∆θ

Therefore, ∆θ

= =

48.388

− 26.38

22d eg

169

170

Prandtl-Meyer Flow

That is, after initial expansion, the flow direction is 22 deg with respect to nozzle axis. 10.2 For M 1 = 2.3 and θ = 12 deg, from oblique shoc k chart, β = 36.5deg. M1n

= M1 sin β = 2.3

sin36 .5deg

×

= 1.368 For M1n = 1.368, normal shock table gives: Thus, M2

T2

M2n = 0.7527, and

=

M2n sin( β θ )

=

1.815

= 1.235.

−

= 1.235 =

T2 T1

× 500

617 .5K

For M2 = 1.815, from isentropic table, ν2 = 21 deg. Also, |θ | = 24 deg. Therefore, = ν2 + |θ | = 21 + 24

ν3

= 45de g For ν 3 = 45 deg, isentropic table gives, M 3 = 2.764 , and fore,

T3 = 0.396. ThereT0

T3 T3 T02 = T2 T02 T2

But T 02 = T03 , thus, we have T3 T2

=

0.396 = 0.657 0.603

T3

=

0.657

=

× 617.5

405 .7 K

For M3 = 2.764 and θ = 12 deg, oblique shock chart give s, β = 32deg, and M3n = 2.764 sin 32 = 1 .465. Now, for M3n = 1.465, the normal shock table

×

171 gives M 4n = 0.7157, and

T4 = 1.2938. Thus, T3 M4

=

T4 =

0.7157 = 2.09 sin20deg 524 .9 K

Note: In this problem, oblique shock table may be used in places where oblique

shock chart is used. 10.3 Let the subscripts 1 and 2 refer to conditions upstream and downstream of expansion. M1

=

2.0

p1 p01

=

0.1278

For isentropic compression p 02 = p 01 . Therefore, p2 p02 =

760 = 0.1278

⇒ M2 =

×

2.44

=

= 26.38 deg

=

= 37.71 deg

⇒ ν1 ⇒ ν2



or, the flow has been turned through ν 2 Note: Compare this problem with 6.8.

240

−

760 + 275

− ν1 =



= 0.0642

11.33deg .

172

Prandtl-Meyer Flow

Chapter 11

Flow with Friction and Heat Transfer 11.1

M1 , p1

M =1 ∗

p

∗

p01

p02

Figure S11.1

This problem has to be solved by trial and error. The pressure ratio written as

Given,

p1 can be p∗

p1 p1 p01 p02 = p∗ p01 p02 p∗

p01 p02 = 10. We want ∗ = 1.893 (critical pressure ratio). Thus, p02 p p1 p∗

=

18.93

p1 p01

or p1 /p∗ p1 /p01

=

18.93

p1 p1 Calculate ∗ and for di fferent M 1 and check for the ratio to be 18.93. For p p01 that, use the relations p1 p∗

=

1 M1



1.2 1 + 0.2M12

173



1 2

174

Flow with Friction and Heat Transfer

p1 p01

=

Trial 1 2 3 4 5



1+

γ

− 1M2 1

2



−

γ γ −1

  p1 p 1 p∗ / p01

M1 0.1 0.06 0.055 0.059 0.058

10.944/0.99 18.25/0.977 19.90/0.998 18.56/0.998 18.88/0.998

= = = = =

11.05 18.29 19.95 18.60 18.92

It is seen that M 1 = 0.058 is the required Mach number. Now use the relation 4f¯L∗ 1 M12 γ + 1 = + ln D γ M12 2γ



4f¯L∗ = 2 11 .6 + 0.857 ln D



−

γ +1

2

M12

1 2 1 + γ− 2 M1



to calculate L ∗ , 0.0040368 1.0006728



= 211 .6 L∗

=

− 4.72 = 206.88 206.88 × 0.1 4 × 0.005

=

1034 .4 m

p1 T1 p2 m ˙ L f¯

= 3.5 105 mboxPa = 300K = 1.4 105 mboxPa = 0.09 kg/s = 600m = 0.004

11.2

× ×

m ˙ = ρ1 V 1 A p1 RT1 3.5 287



π 2 γ RT1 M1 4 D

× 10 × 20.04√300 × π × M1 D2 × 300 4 5

M1 D2

= 0.090 = 0.090 = 8.12

× 10−5

175 For an isothermal flow with friction, it can be shown that 4f L D 4

0.004

×

D

     −

1 1 γ M12

=

600

1

× 9.6 D

2

p2 p1

+ ln

2

1.4

=

γ M12 1

=

0.6

=

9.6 D

= 9.10

1 M12

3.5

1.4 + ln

− 1.832

0.6 D4 (8.12)2 10−10

×

− 1.832

× 107 D4 − 1.832

Solving equation (3), We obtain D = 0.0402m . 11.3 Energy equation gives, h+

V2 = h 0 = constant 2

Also, for a perfect gas, a2

V2

−1+ 2 γ −1 2 1+ M γ

a2

 

2

T1 1 +

γ

− 1M2 2

 

=

a20 γ

−1

=

a20

=

T0

From continuity equation for a perfect gas, G =

m ˙ = ρ 1 V1 = ρ 1 a1 M1 A

=

p1 RT1

=





γ

R

γ RT1 M1

√1T

p1 M1 1

From Eq. (2), we have T1

2

 −    

From Eqs. (1) and (2), we obtain 9.6 D

p2 p1

=

γ p21 M12

RG2

3.5

2

176


Using this in Eq. (1), we get γ p21 M12



R G2

− 1M2

γ

1+

1

2

M1 1 +

γ

R

T0

γ

− 1M2 1

2

T0

=

1

2

 

=

− 1M2

γ

M12 1 +



   1 2

=

G2 2

p1 R γ

T0

G p1

Now, G

=

m ˙ m ˙∗ = A A

=

√γΓRT

p0 0

A∗ A

where Γ

= γ

  2 γ+1

γ +1 2(γ −1)

Using Eq. (4) in Eq. (3), we get,



M1 1 +

γ

− 1M2 1

2



1 2



=

R γ

T0

Γ

γ RT0

p0

A∗ 1 A1 p1

p0 A∗ γ p1 A1

Γ

= i.e. γ Γ

p1 p0

=

A A∗ Therefore,



M1 1 +

2.38 67.3

− 1M2 2

1



1 2

=

p0 A∗ p1 A1

× 105 = 0.0354 × 105

D =

γ

2

2

0.0127

D∗

=

0.0061

p1 A p0 A∗

=

0.0354

2

   

= 2.08 = 4.33

× 4.33 = 0 .1533

177 =

⇒ M1

  ⇒ 4f L∗ D

=

= 2.51 = 0.4341

1

p2

105

4.85

p0

=

67.3

× 105 = 0.072

Therefore, p2 p0

× AA∗

=

0.072

⇒ M2

=

1.53

=

0.147

= =

⇒

  4fL∗ D

× 4.33 = 0 .312

2

Therefore,

  4f L∗ D

= 0.4341

− 0.147

∆

DL

= 29.60

− 1.75 = 27.85

Hence, the average friction coefficient is f¯ = =

0.2871 4(27.85) 0.002577

11.4 (a)

Ae /A = 2 ∗

1

p0 T0

pb

M=1 A

Shock

∗

Figure S11.4a

L

178


Normal shocks can exist only for M 1 > 1. Therefore, the interpretation that a normal shock stands in the throat implies M = 1 at the throat and downstream is subsonic. Therefore, A t = A∗ . A∗ Ae

M1 0.306

0.50

ρ1 ρ0

T1 T0

0.9547

0.9816

p1 p0

a1 a0

0.9907

0.9371

p0 7 105 = = 8.13kg/m 3 RT0 287 300

× × √ = 1.4 × 287 × 300 = 347m/s

ρ0

=

a0

=



ρ1

=

(0.9547)ρ0 = (0.9547)

× (8.13) = 7.7617 kg/m3

a1

=

(0.9907)a0 = (0.9907)

× (347) = 343 .8m/s

p1

=

(0.9371)p0 = (0.9371)

γ RT0

 × ¯ 4fL D

M1 0.306

(7

× 105 ) = 6.5597 × 105 Pa

∗

p1 p∗

M1

5.0776

3.547

p1 Therefore, p ∗ < pB =

⇒

5

p∗ = 3.547 = 1.849 10 Pa duct length L < L ∗ and p2 = pB at L.

×

p2 p∗

pB 2.8 105 = = 1.514 ∗ p 1.849 105

× ×

=

  4f¯L∗ D

p2 p∗

1.514

0.220

M2

Therefore, 4f¯L D

=

4f¯L D

= 5.0776 = G =

4f¯L D

  −  M1

M2

− 0.220

4.8576 ρ1 V1 = ρ 1 a1 M1

=

(7.7617)(343.8)(0.306)

=

816 .5 kg/m2 s

179 (b)

1 Shock M =1 A∗ L

Figure S11.4b

Upstream of normal shock wave

A∗ Ae

p p0

ρ ρ0

T T0

0.5

0.09396

0.1847

0.5088

p = = = T = a = ρ

a a0

M

0.7133

(0.09396)(7 105 ) = 0.6577 (0.1847)(8.13) 1.5 kg/m3 (0.5088)(300) = 152.6 K (0.7132)(347) = 247.5m/s

×

2.197

× 105 Pa

Downstream of normal shock wave

This zone is referred by subscript 1.

M 2.197

p1 p 5.465

ρ1 ρ

2.947

T1 T 1.854

a1 a 1.362

M1 0.5475

Therefore, p1 ρ1

T1 a1

= = = =

(5.465)(0.6577 105 ) = 3.5943 (2.947)(1.5) = 4 .4205 kg/m3 (1.854)(152.6) = 282 .9 K (1.362)(247.5) = 337 .1m/s

×

M1 0.5475

  4f¯L∗ D

M1

0.7451

p1 p∗

1.9434

× 105 Pa

180


Thus,

p1 3.5943 105 = = 1.85 105 Pa 1.9434 1.9434 duct length L < L∗ and p 2 = pB at L.

×

p∗ = Since p ∗ < pB =⇒

×

p2 pB 2.8 105 = = = 1.514 p p 1.85 105

∗

∗

p2 p∗

1.514

× × 4f¯L

  ∗

D

0.220

M2

Therefore, 4f¯L D

4f¯L 4f¯L D M1 D = 0.7451 0.220 = 0.5251

  − 

=

M2

−

G = =

ρ1 V1 = ρ 1 a1 M1

(4.4205)(337.1)(0.5475) 815.9 kg/m 2 s

= (c)

1

Shock M =1 A ∗

Figure S11.4c

Properties at station 1 are the same as properties in Part-b. Upstream of normal shock wave

Therefore, p1 ρ1

T1 a1 M1

×

= 0.6577 105 Pa = 1.5kg/m 3 = 152 .6 K = 247 .5m/s = 2.197

181 M1 2.197

  ¯ ∗ 4fL D

p1 p∗

M1

0 .36011

0 .35566

p1 = 1.849 105 Pa. 0.35566 Therefore, p ∗ < pB = duct length L < L∗ and p2 = p a at L downstream of normal shock wave. Thus, p =

∗

×

⇒

From the diagram in Fig. s11.4d, it is clear that the Mach num ber Mx at point x jumps.

h =

0 28

a kP

pb

*

1

s

Figure S11.4d

The normal shock pressure jump and the Fanno line equation are given by p3 px

=

2γ M2 γ+1 x

px p∗

=

1 Mx

2.8 105 1.849 105

=

1 1.1667Mx2 Mx

− 0.1667

1.2 1 + 0.2Mx2

Let f (Mx ) =

1 1.1667Mx2 Mx

− 0.1667

1.2 1 + 0.2Mx2

× ×



− γγ +− 11 γ +1

2 1 2 1 + γ− 2 Mx

 



1/2

Use Secant method to find M x as follows. xj+1 = x j Trial 1

x1

=

1.6

− f (xj )



xj f (xj )

 

− xj − 1 − f (xj−1 )



 

1/2

1/2

− 1.5143

182


x0

= 1.5

x2

= 1.6

− 0.0559

x2 = 1.53 f (x2 ) = 0.000752 x3

= 1.53

x3

= 1.53,



1.6 1.5 0.0559 ( 0.0236)

− −−

− 0.000752



−−

1.53 1.6 0.000752 0.0553

Therefore, M x = 1.53. ¯ ∗ 4fL D

 

M2 1.53

M2

0.14699

Therefore, 4f¯L

=

4f¯L

4f¯L M1

  − 

D

D

= 0.36011 =

G

D

M2

− 0.14699

0.21312

= ρ1 V1 = ρ 1 a1 M1 = (2.197)(247.5)(1.5) =

815 .6 kg/(m2 s)

11.5 The speed of sound a 1 is given by, a1 M1





√

γ RT1 = 20.04 333.3 = = 366m/ s V1 73.15 = = a1 366 = 0.2



183 From tables Rayleigh flow tables, for M 1 = 0.2, we have T01 T0∗ T1 T∗ pp01 ∗ 0 p1 p∗ V1 V∗

=

0.17355

=

0.20611

=

1.2346

=

2.2727

=

0.09091

From isentropic tables, for M 1 = 0.2, we have p1 = 0.97250 p01 Therefore, p01

=

0.5516 105 0.9725 5.672 104 Pa

=

0.99206

=

T1 T01

×

×

Therefore, T01

=

333.3 = 33 6 K 0.99206

(a) T02 = stagnation temperature after combustion ∆h = cP (T02 T01 )

−

Therefore, T02

= T01 +

∆h

cP 1395.5 103 = 336 + 1004.5 = 1725 .2 K

×

(b) T02 T01

=

1725.2 = 5.1345 336.0

184


T02 T0∗

=

⇒ M2 T2 T∗ p02 p∗0 p2 p∗ V2 V∗

=

T01 T02 T0∗ T01

=

(0.17355)(5.1345) = 0 .89109

=

0.68

=

0.98144

=

1.0489

=

1.4569

=

0.67366

(c) T2

=

T2 T ∗ T1 T ∗ T1

=

0.98144 0.20661

=

1583 .2 K

× 333.3

(d) p02

∆ p0

=

p02 p∗ p01 p∗ p01

=

1.0489 1.2346

= p02 =

× 5.672 × 104 = 4.819 × 104 Pa

− p01 = (4.819 − 5.672) × 104

−8530 Pa

(e) s2

− s1

=

cp ln

T02 T01

R ln

= 1004.5 ln (5 .1345) =

p02 p01

 −     −

1690 .1 J/kg.K

287 ln

4.819 5.672

185 (f ) V2

= = =

V2 V ∗ V1 V ∗ V1 0.67366

73.15

×

0.09091 542 .1m/s

(g) The initial conditions can be maintained till the flow is choked at the duct T01 exit after combustion. That is, M2 = 1 and TT02 = 1 and since = 0.17355, 0 T0∗ and T 01 = 336K, we have ∗

T02

=

T02 T0∗ T01 T0∗ T01

=

336.0 0.17355

= 1936 K Maximum heat of reaction q is given by q

=

cP ∆T0 = 1004.5(1936

=

1607 .2kJ/kg

− 336)

11.6 The flow process is adiabatic and therefore, it can be treated as a Fanno flow. The velocity at station 1 is V1

=

M1 a1 = M 1

=

0.2 1.4



γ R T1

√ × 287 × 300 = 69 .44 m/s

From Fanno flow table, for M 1 = 0.2, we have 1 pp∗ = 5.4554,

1 TT ∗ = 1.1905,

1 VV ∗ = 0.21822,

pp01 ∗ = 2.9635 0

Again, from Fann o flow table, for M 2 = 0.5, we have p2 = 2.1381, p∗

T2 = 1.1429, T∗

V2 = 0.53452, V∗

p02 = 1.3398 p∗0

186


Thus, p2

T2

V2

p02

=

p2 p∗ 2.1381 p1 = p∗ p1 5.4554

=

198 .558 kPa

=

T2 T ∗ 1.1429 T1 = T ∗ T1 1.1905

=

288K

=

V2 V ∗ 0.53452 V1 = V ∗ V1 0.21832

=

170 .09 m/s

=

p02 p∗0 1.3398 p01 = p∗0 p01 2.9635

=

235 .52 kPa

× 5 × 101325

× 300

× 69.44

× 520.951

11.7 For, M 1 = 0.2, from isentropic table, we have, T1 = 0.99206 T01 Therefore,

T1 72 + 273.15 = = 347 .9 K 0.99206 0.99206 Since the tube is perfectly insulated, T 01 = T 02 , thus, T01 =

T02 = 347.9 K The initial density is ρ1

=

p1 2 101325 = R T1 287 345.15

=

2.046 kg/m3

× ×

since, 1 atm = 101325 Pa. Thus, the mass flow rate through the tube is m ˙ = =

ρ1 A1 V1 = ρ 1 A1 M1 a1

2.046

× (0.1 × 0.1) × 0.2 ×



γ R T1

187

√

= 0.004092 1.4 =

× 287 × 345.15

1.524 kg/s

Now, assuming a control force balance equation asvolume between the sections 1 and 2, we can write the ( p1

− p2 ) A + F = ρ1 A1 V1 (V2 − V1 )

where F is the frictional drag and A is the cross-sectional area of the tube. For M 2 = 0.76, from isentropic table, we have T2 = 0.89644 T02 Therefore, T2

=

0.89644 T02 = 0.89644

× 347.9

= 311 .87 K a

=

√1.4

287

×

2

311.87

×

= 354m/ s V2

=

M2 a2 = 0.76

× 354

= 269 .04 m/s For obtaining p 2 , let us use the Fanno flow table. For M 1 = 0.2 and M 2 = 0.76, respectively, from Fanno flow table, we have p1 = 5.4554, p∗

p2 = 1.3647 p∗

Therefore, p2

=

p2 p∗ p1 p p

=

1.3647 5.4554

∗

1

Thus, F

= ρ1 A1 V1 (V2

− V1 ) − (p1 − p2 ) A

× 2 = 0.5atm

188


=

1.524(269.04

=

− 1223.43 N

− 74.52) − (2 − 0.5) × 101325 × (0.1 × 0.1) = 296.45 − 1519.88

Hence, Drag = 1223.43 N 11.8 Given, L/D = 50, V1 = 195 m/s, T1 = 310 K and Me = 1. The speed of sound at the entrance is a1 = γ RT1 . For carbon dioxide, γ = 1.3 and molecular weight is 44, thus,

√

R

Ru 8314 = M 44 = 188 .95 J/(kg K) =

Therefore, a1 M1

√ ×

= 1.3 188.95 = 275 .95 m/s V1 195 = = a1 275.95 = 0.71

× 310

From Fanno flow table for γ = 1.3, for M 1 = 0.71, we have 4 f Lmax D

=

0.20993

Thus, f

= =

0.20993 50 4 0.00105

×

11.9 From Fanno flow table, for M 2 = 0.8, we have 4 fL∗2 D p2 p∗ T2 T∗ 4 fL∗1 D

=

0.07229

=

1.2893

=

1.0638

= =

4 f L∗2 4 f L + D D 4 0.005 51 0.07229 + = 40.87229 0.025

×

×

189 From Fanno flow table, for M1

4 fL 1∗ D

=

= 40.87229, we have

0.13

p1 p∗

= 8.457

T1 T∗

= 1.1959

p1

T1

=

p1 p∗ 8.457 p2 = p∗ p2 1.2893

=

6.56 atm

=

T1 T ∗ 1.1959 T2 = T ∗ T2 1.0638

=

303 .52 K

×1

× 270

11.10 By energy equation, we have h0 = h 1 +

V 12

2 Treating air as a perfect gas, we can express h = c p T and hence c p T0 = c P T1 +

V 12 2

For air c p = 1004.5 J/(kg K). Therefore, 2

T1

= T0 =

a1

=

 

V1 − 2c p 2



− 2 ×135 K = 349 .9 K 1004.5 √ γ RT = 1.4 × 287 × 349.9

359

1

= 374 .95m/s Therefore, 1 M1 = V = 135 = 0.36 a1 374.95

(a) For M 1 = 0.36, from Fanno flow table, we have 4fLmax = 3.1801 D

190


Thus, Lmax

× 5 × 10−2 × 0.02

=

3.1801 4

=

1.99 m

This is the minimum length of the tube for the flow to choke. (b) For L 2 = 0.6 m,

    −  4f L D

=

2

4f L D

1

4f L D

12

where 1 and 2 stands for the inlet and exit of the tube

  4f L D

=

3.1801

× 0.6 − 4 ×5 0.02 × 10−2

=

3.1801

− 0.96 = 2 .22

2

The corresponding Mach number, from Fanno flow table, is p02 and p∗0 = 1.5587 p02 For M 1 = 0.36, from Fanno flow table, ∗ = 1.7358. Therefore, p0 M2

≈ 0.41 p02

=

p02 p∗0 p01 p∗0 p01

=

1.5587 1.7358

=

121 .23 kPa

× 135

For M 1 = 0.36, from isentropic table, p1 p01

= 0.91433

p1

= 0.91433

ρ1

=

× 135 = 123 .43 kPa p1 123.43 × 103 = RT1 287 × 349.9

= 1.229 kg/m3

191 Thus, the mass flow rate th rough the tube m ˙ is m ˙ = ρ1 A1 V1 = 1.229 =

−4 2 × 135 × π × 5 4× 10

0.326 kg/s

11.11 For hydrogen, M = 2.016 and γ = 1.4 R=

8314 = 4124 J/(kg K) 2.016

At the tube inlet, the Mach number is M1

= =

V1 = a1

1 √γVRT

= 1

√1.4 × 200 4124 × 303

0.15

From Fanno flow table, for M 1 = 0.15, we get 4f¯Lmax D p1 p∗

=

27.932

=

7.2866

Thus, the tube length required for the flow to choke is Lmax

pexit

=

27.932D 4f¯

=

27.932 4

=

5.82 m

= p∗ =

× 25 × 10−3 × 0.03

p1 7.286

=

250 7.2866

=

34.31kPa

11.12 At the pipe entrance, Mach number M 1 = V1 = 200m/s

V1 a1 ,

where

192


and a1



=

γ RT1 =

√1.4 × 287 × 303.15

= 349m/ s Thus,

200 = 0.57 349 From Fanno flow table, for M 1 = 0.57, we have M1 =

4f Lmax = 0.62288 D Therefore, Lmax =

0.62288 4

× 20 × 10−3 = 0.156m × 0.02

Thus, the length of the pipe at whic h the flow would be sonic is 15 .6 cm 11.13 For methane, γ = 1.3, R

=

Ru 8314 = M 16.04

= 518 J/(kg K) M1

=

V1 = a1

1 √γVRT

=

√1.3 ×

=

0.0538

25 518

1

25

× 320 = 464.2

From Fanno flow equations, we have 4f¯L∗1 1 M12 γ + 1 (γ + 1) M12 = + ln 1 2 D γ M12 2γ 2 1 + γ− 2 M1

−

=

 

0.997 + 0.8846 ln 0.00376

0.00666 2



= 265.16

= 260 .114 The pipe length at which the flow chokes is L∗

=

260.114 25 10−3 4 0.004

× × ×

= 406 .4 m



− 5.046

193 Again by Fanno flow relations, we have p1 p∗

1 M1

=



γ+1 1 2 1 + γ− M 12 2

=

1 0.0538

=

19.93

   



1 2

1 2

2.3 2

Thus, p∗

T1 T∗

=

p1 1 106 = 19.93 19.93

=

50.18 kPa

=

γ+1 2.3 = 1 2 2 2 1 + γ− 2 M1

×





= 1.15 Thus, T∗

=

T1 320 1.15 = 1.15

= 278 .16 K V∗

= a∗ = =



γ RT ∗ =

√1.3 × 518 × 278.26

432 .87 m/s

11.14 For argon, γ = 1.67. The given flow is a Fanno flow. From Fanno flow table, for M 1 = 0.6, we have



4f Lmax D



=

0.4908

p1 p∗

=

1.76336

T1 T∗

=

1.1194

1

For the given duct, 4f L 4 = D

× 0.02 × 1.1194 = 0.2984 0.3

194


Also, 4f L D

= =



4f Lmax D

 − 1

4f Lmax D



2

0.2984

where subscript 2 refers to duct exit. From the above equation, we have

 For



4f Lmax D





4f Lmax D

=

0.4908

=

0.1924

2

− 0.2984

= 0.1924, from Fanno flow table, we get 2

M2

=

0.73

p2 p∗

=

1.426

T2 T∗

=

1.084

Thus, p2

T2

=

p2 p∗ 1.426 p1 = p∗ p1 1.76336

=

72.81 kPa

=

T2 T ∗ 1.084 T1 = T ∗ T1 1.1194

=

290 .62 K

× 90

× 300

11.15

L

=

D 4f

     − 4f l D

=

D 4f (14.533

=

50 × 10−3

=

30.277m

4

M1

− 0)

× 0.006 × 14.533

4f l D

M2

195 11.16 π

A = ρ

× 0.12 = 78.539 × 10−4m2 4

p = 1.94 kg/m3 RT

=

Therefore, the inlet velocity becomes V

=

M

=

m ˙ ρA

= 177.2m/s

0.49

From Fanno flow table, for M = 0.49, we have 4f l D

= 1.1539

p p∗

= 2.1838

T T∗

= 1.145

Thus, L =

p∗

T∗

1.1539 40.1 4 0.006

×

×

=

4.8 m

=

1.8 105 2.1838

=

82.4kPa

=

323.15 = 282.23 K 1.145

=

9.08◦ C

×

At half way before the chocking location, L = 2.4 m. Thus, 4f L 4 = D From Fanno flow table, for

4fL D

× 0.006 × 2.4 = 0.576 0.1

= 0.576, we get

p p∗

= 1.8282

196


T T∗

=

p = T

1.1244 150 .6kPa 44.19◦ C

=

11.17 (a) From Fanno flow table, for M = 0.2, we have 4f Lmax = 14.533 D where L max the distance from the Mach 0.2 location at which the Mach number becomes unity. Therefore, Lmax

=

14.533 0.05 4 0.00375

=

48.44 m

×

×

(b) This problem has to be solved by finding the chocking location for initial Mach numbers 0.2 and 0.6. LM =0.6 = (Lmax )M =0.2

− (Lmax)M =0.6

From Fanno flow table, for M = 0.6, we have 4f Lmax D

=

0.49082

Lmax

=

0.49082 0.05 = 1.64 m 4 0.00375

×

×

Thus, LM =0.6 = 48.44

− 1.64 =

46 .8 m

11.18 (a) Given, T 0 = 380 K. By energy equation, we have V2 h0 = h 1 + 1 2 where h 0 is the stagnation enthalpy and h 1 and V 1 are the static enthalpy and velocity, respectively, at the duct entrance. Treating air to be a perfect gas, we

197 have, h 0 = c p T0 and h 1 = c p T1 . Therefore, the energy equation becomes, T0

=

T1 +

T1

=

T0

V12 2 cp V12

302

= 380

− 2 cp

− 2 × 1004.5

= 379 .6 K

since c p = 1004.5 J/(kg K) for air. The speed of sound is a1 =



M1 =

γ R T1 = 390.54 im/s

V1 30 = a1 390.54

From Fanno flow table, for M 1

≈ 0.08

0.08, we have

≈  4f L D

= 106.72 1

Thus, 4 D

0.02

55

×106.72×

= =

m

4.12cm

(b) The inlet velocity V 1 = 90 m/s. Therefore, 2

T1 = 380 a1

=

M1

=



− 2 ×901004.5 = 37 6 K

γ R T1 = 388.7m/s

90 = 0.23 388.7

From Fanno flow table, for M 1 = 0.23, we have 4f L D

=

D

= =

10.416 4

× 0.02 × 55 1046

0.422m

198


(c) The inlet velocity V 1 = 425 m/s. Therefore, 2

T1 = 380 a1

=

− 2 ×425 = 290.1 K 1004.5

γ R T1 = 341.4m/s



425 = 1.24 341.4 From Fanno flow table, for M 1 = 1.24, we have M1

=

4fL D D

= 0.04547 4

=

× 0.02 × 55 0.04547

=

96.77 m

Note: For the supersonic flow at the duct entrance, the diameter comes out to

be more than the length, for the present data. 11.19 The hydraulic diameter of the duct is Dh

4

=

× cross-sectional area = 4 × 0.03 × 0.03 perimeter 4 × 0.03

= 0.03 m

At the duct entrance the flow velocity is V1 = 1000 m/s. The local speed of sound is a1

=



γ RT1 =

√1.4 × 287 × 350

= 375m/ s Thus, the Mach number at duct entrance is M1

=

V1 1000 == = 2.67 a1 375

For M 1 = 2.67, from Fanno flow table, we have 4fLmax D



= 0.46619



1

Therefore, the duct length required to decelerate the flow to Mach 1.0 is Lmax

= =

0.46619 0.03 4 0.0025

×

1.40 m

×

199 11.20 For mass flow rate m ˙ to be maximum, the exit Mach number M be unity, i.e. the flow is choked. For the given duct, 4f L 4 0.023 0.25 = D 0.025

×

=

2

should

×

0.92

For this value of 4fL D , if the flow at the exit has to choke, from Fanno flow table, we have M1 = 0.52. From isentropic table, for M 1 = 0.52, we have p1 = 0.83165 p01 T1 T01

=

0.94869

p1

=

0.83165

× 50 × 101325 = 4 .21 MPa

T1

=

0.94869

× 320 = 303 .58 K

Therefore, m ˙ max

= ρ1 A1 V1 = ρ 1 A1 M1 a1

× π4 25 × 10−3 2 × 0.52 × √ 0.012334 1.4 × 285 × 303.58



= 48.32 = =





γ RT1

4.3 kg/s

Further, p∗ p1

=

1 = 0.487 2.0519

Thus, p∗

=

0.487

× 4.21 = 2 .05 MPa

Therefore, the mass flow rate will remain maximum for the back pressure range 0 < pb < 2.05 MPa 11.21 (a) Let subscripts 1 and 2 refer to duct entry and exit conditions. The duct length ∆L required to accelerate the from Mach 0.2 to 0.5 can be determined from f ∆L f L∗ f L∗ = D D M =0.2 D M =0.5

 

−

 

200


where f is friction factor, D is duct diameter and M is Mach number. From Fanno flow table, we have

   

Thus,

f L∗ D

M =0.2

fD L∗

M =0.5

0.025∆L = 30 10−3 ∆L =

=

14.533

=

1.0691

14.533

×

− 1.0691

16.157m

(b) From Fanno flow table, we have

  f L∗ D

≈

M =1

0

Thus, f ∆L D

=

∆L

=

14.533 14.533

× 30 × 10−3 0.025

=

17.44 m

11.22 For the given pipe, we have 4f Le De

= =

4

× 0.02 × 18 5 × 10−2

28.8

e For 4fL De = 28.8, from Fanno flow table, M e = 0.15. The speed of sound at pipe exit is ae = 1.4 287 468 = 433 .74 m/s

√ ×

×

The exit velocity and density are Ve

= 0.15

ρe

=

× 433.74 = 65.062m/s

pe 101325 = = 0.754kg/m 3 RTe 287 468

×

201 Therefore, 2

m ˙ =

ρe Ae Ve = 0.754

=

× π × 40.05 × 65.062

0.096kg/s

11.23 Let subscripts 1 and 2 refer to conditions at inlet and exit of the tube, respectively. Given, p e = 0 Pa, T01 = 300 K and p01 = 6 101325 Pa, since 1 atm = 101325 Pa.

×

(a) L = 0 m and pp012 is much lower than the choking limit of 0.48667(for 1.67), therefore, the flow is choked. Thus, the mass flow rate is 0.7266

m ˙ =

6 × 101325 π √×RT 01



30

× 10−2 4



γ =

2

For argon, molecular weight is 39.944, gas const ant R is 208 J/(kg K) and γ = 1.67. Therefore, the mass flow rate becomes m ˙ = 125 kg/s (b) L = 2.22 m. Therefore, 4f Lmax D

= =

4

× 0.005 × 2.22 30 × 10−2

0.148

From Fanno flow table, for γ = 1.67 and

4f Lmax D

= 0.148, we have M 1 = 0.71.

Now, from isentropic table, for M 1 = 0.71( γ = 1.67), we have p1 = 0.67778 p01 T1 T01

=

0.85552

p1

=

6

× 101325 × 0.67778 = 412056.35Pa

T1 = 300 The density at the inlet is ρ1

0.85552 = 256.66 K

× =

p1 412056.35 = RT1 208 256.66

×

3

= 7.72 kg/m

202


Thus, the mass flow rate is m ˙ =

ρ1 A1 V1 = ρ 1 A1 M1 a1 2 × π0.3 × 0.71√1.67 × 208 × 256.66 4

=

7.72

=

115 .68 kg/s

11.24 Given, p 1 /p2 = 15 and p 1 = 150 atm. Therefore, p2 =

150 = 10 atm 15

where p 1 is the storage tank static pressure and p 2 is the settling chamber static pressure. Let p 2 be the pressure at which the flow chokes, i.e. p2 = p∗ . Thus, p1 = 15 p∗ From Fanno flow table, for p 1 /p∗ = 15, we have 4fL∗ D = 140 .66 Thus, L∗

=

140.66 0.1 4 0.005

=

703 .3 m

×

×

That is, the pipe will choke at a length of 703.3 m. 11.25 For the given piping system, p01 600 = = 13.33 p4 45 This pressure ratio is much more than the pressure ratio required for the flow to choke. Therefore, at the exit, the Mach number M4 can be taken as unity. Also,

  4f L D

= B

=

4

0.013 1.1 3.10 10−2

×

1.845

×

×

203 For this value, from Fanno Table, we get M 3 M3 = 0.43, we have A3 A∗

=

≈ 0.43. From isentropic table, for

1.5

Also, 2

A2 A3

=

 

A2 A2 ∗

=

A2 A3 , A3 A3 ∗

6 3.1

= 3.75 since A 2 ∗ = A 3 ∗

Thus, A2 A2 ∗

=

3.75

× 1.5 = 5.625

From isentropic table, for AA22 = 5.625, we have M 2 flow table, for M 2 = 0.1, we have ∗

≈ 0.1 and from from Fanno

4fL = 66.922 D For tube A,

  4fL D

4

= A

× 0.015 × 0.9 6 × 10−2

= 0.9 Also,

  4f L D

= 1−2

  −  4f L D

4f L D

1

2

Therefore,

  4f L D

= 1

=

    4f L D

+

1−2

4f L D

= 0.9 + 66.922 2

67.822

4f L From Fanno flow table, for this D isentropic table, for M 1 = 0.1, we get

 

= 67.822, we have M1

0.1. From

≈

ρ1 ρ01

=

0.99502

T1 T01

=

0.998

204


The stagnation density is ρ0

=

p0 600 = RT0 287

=

5.36 kg/m3

× 103 × 390

Thus,

× 5.36 = 5 .33 kg/m3

ρ1

= 0.99502

T1

= 0.998

a1

=

V1

= M1 a1 = 39.55m/s

A1

=

 π

× 390 = 389.22 K

γ RT = 395.5m/s

× 0.062 = 28.27 × 10−4 m2 4

Thus, the mass flow rate is 4

m ˙ = =

ρ1 A1 V1 = 5.33

0.596kg/s

× 28.27 × 10− × 39.55

Chapter 12

MOC

205

206

MOC

Chapter 13

Measurements in Compressible Flow 13.1 (a) p = (760

− 500) = 260 mm of Hg

p0 = (760

− 350) = 410 mm of Hg

p p0

=

260 = 0.634 410

From isentropic table the corresponding Mach number is M = 0.835 (b) p0 = 760 + 275 = 1035 mm of Hg p p0

=

260 = 0.251 1035

From isentropic table the corresponding Mach number is M = 1.56 13.2 pgauge patm

= 3.6

× 104 Pa

= 0.756

× 9.81 × 13.6 × 103

= 1.0086

× 105 Pa 207

208

Measurements in Compressible Flow

p0

−p

= 50cm of Hg = 0 .5 =

6.67

× 9.81 × 13.6 × 103

× 104 Pa

T0 = 300K

× 105 Pa

p0,gauge

=

1.027

p0abs

=

(1.027 + 1.0086)

× 105 Pa

Therefore, ρ0 =

p0 RT0

2.0356 105 287 300

× ×

=

= 2.36 kg/m3 (a) By compressible Bernoulli equation, we have 2

γ

u2 + γ

γ

− 1 pρ ρ ρ0

=

γ

  p p0

=

u2 2

1 γ

1.3686 2.0356

= ρ

− 1 ρp00

= 0.753 = 3.5

u =



× 105 × 105



1 1.4

= 0.753

× 2.36 = 1 .78kg/m 3 2.0356 2.36



− 1.3686 × 105 1.78

256 .06 m/s

(b) For incompressible flow, ρ = ρ 0 . Therefore, u

=



=

237 .55 m/s

2 ( p0

− p) =

ρ

× 2

6.67 104 2.364

×

209 Note: The error committed in assuming the flow to be incompressible in this

problem is 7.23%. 13.3 From isentropic table, for M 1 = 0.9, we have p1 = 0.5913 p 0

For M 2 = 0.2, again from isentropic table, we have p2 = 0.9725 p0 Therefore, p2 p1

=

0.9725 = 1.6448 0.5913

p2

= 1.6447

× 4.15 × 105 = 6.826 × 105 Pa

Therefore, p2

− p1 = (6.826 − 4.15) × 105 =

2.676

× 105 Pa

13.4 (a) T = 500 ◦ C = 773 K. Therefore,

√ a =

1.4

× 287 × 773 = 557.3m/s

V 400 M = = = 0.718 a 557.3 From isentropic table, for M = 0.718, we have pp0 = 1.4098. Therefore, p0

= 1.4098

× 1.01325 × 105

=

× 105 Pa

1.428

(b) T

=

a =

−50◦ C = 273 − 50 = 223 K √1.4 × 287 × 223 = 299 .33 m/s 400 299.33 = 1.336

M

=

p0

= p 1 + 0.2 (1.336)2

p0

=



2.949

× 105 Pa



3.5

210


Note: Note the time saving in using tables instead of actual relations.

13.5 (a) From Standard atmospheric table, at 10, 000 m altitude, T∞ = 223.15K. The speed of sound is given by, a∞ =

√1.4

287

×

The aeroplane velocity is

223.15 = 299.44m/s

×

V∞ =

900 = 250 m/s 3.6

The flight Mach number is M∞ =

V∞ = 0.835 a∞

From isentropic table, for M ∞ = 0.835, we get T0 = 1.139 T∞ Thus, the temperature at the stagnation region is T0 = 1.139

× 223.15 =

254.17 K

(b) The temperature caused by impact is given by, ∆T

= T0

− T = 254.17 − 223.15 =

31 .02 K

13.6 The test section Mach number is M = 4. At 1650 m, fro m standard atmospheric table, ρ0 = 1.0425 kg/m3 . From isentropic table, for M = 4, we have ρ = 0.0277 ρ0

Therefore, the test-section density is ρts = 0.0277

× 1.0425 =

0.0289 kg/m3

13.7 The probe measures the stagnation temperature T 0 as 100 ◦ C. That is, T0 = 100 + 273.15 = 373.15 K By energy equation, we have h0

=

h+

V2 2

211

c p T0

= cp T +

T0

= T+

V2 2

V2 2 cp 2

T

= T0

− 2Vcp

where T is the actual temperat ure(static temperature) of the air. For air c p = 1004.5 J/(kg K). Theref ore, 2

T = 373 .15

− 2 ×250 = 373.15 − 31.11 1004.5

= 342 .04 K = 68 .9◦ C

13.8 Let the subscripts ‘ TS’, ‘i’ and ‘0’ refer to the test-section, nozzle inlet and stagnation state, respectively. Given M TS = 2.5 and p TS = 100 kPa. From isentropic table, for M TS = 2.5, pTS ATS = 0.0585, = 2.637 p0 Ath where A th is the nozzle throat area. Therefore, Ai Ath

p0

=

2

ATS Ath

=

2

× 2.637

=

5.274

=

100 0.0585

=

1.71 MPa

For A i /Ath = 5.274, from isentropic table (subsonic solution), Mi

≈

0.11 ,

pi = 0.9916 p0

212


Thus, the pressure at the nozzle inlet is pi

= =

0.9916

× 1.71

1.696MPa

E Rathakrishnan Gas Dynamics Solutions

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