16...16 compressible flow 16.1. introduction. 16.2. basic equations of compressible fluid flow....

16Compressible Flow

16.1. Introduction. 16.2. Basic equations of compressible fluid flow. 16.3. Propagation of disturbancesin fluid and velocity of sound. 16.4. Mach number. 16.5. Propagation of disturbances in compressiblefluid. 16.6. Stagnation properties. 16.7. Area-velocity relationship and effect of variation of area forsubsonic, sonic and supersonic flows. 16.8. Flow of compressible fluid through a convergent nozzle.16.9. Variables of flow in terms of Mach number. 16.10. Flow through Laval nozzle (convergent-divergent nozzle). 16.11. Shock waves. Highlights—Objective Type Questions—TheoreticalQuestions—Unsolved Examples.

16.1. INTRODUCTION

A compressible flow is that flow in which the density of the fluid changes during flow. Allreal fluids are compressible to some extent and therefore their density will change with change inpressure or temperature. If the relative change in density ∆ρ/ρ is small, the fluid can be treated asincompressible. A compressible fluid, such as air, can be considered as incompressible with con-stant ρ if changes in elevation are small, acceleration is small, and/or temperature changes arenegligible. In other words, if Mach’s number U/C, where C is the sonic velocity, is small, com-pressible fluid can be treated as incompressible.

• The gases are treated as compressible fluids and study of this type of flow is often referredto as ‘Gas dynamics’.

• Some important problems where compressibility effect has to be considered are :(i) Flow of gases through nozzles, orifices ;

(ii) Compressors ;(iii) Flight of aeroplanes and projectiles moving at higher altitudes ;(iv) Water hammer and accoustics.• ‘Compressibility’ affects the drag coefficients of bodies by formation of shock waves,

discharge coefficients of measuring devices such as orificemeters, venturimeters and pitot tubes,stagnation pressure and flows in converging-diverging sections.

16.2. BASIC EQUATIONS OF COMPRESSIBLE FLUID FLOW

The basic equations of compressible fluid flow are : (i) Continuity equation, (ii) Momentumequation, (iii) Energy equation, and (iv) Equation of state.

16.2.1. Continuity EquationIn case of one-dimensional flow, mass per second = ρAV

(where ρ = mass density, A = area of cross-section, V = velocity)Since the mass or mass per second is constant according to law of conservation of mass,

therefore,ρAV = constant ...(16.1)

857

858 ENGINEERING THERMODYNAMICS

dharm\M-therm\Th16-1.pm5

Differentiating the above equation, we getd(ρAV) = 0 or ρd(AV) + AVdρ = 0

or ρ(AdV + VdA) + AVdρ = 0 or ρAdV + ρVdA + AVdρ = 0Dividing both sides by ρAV and rearranging, we get

d dAA

dVV

ρρ

+ + = 0 ...(16.2)

Eqn. (16.2) is also known as equation of continuity in differential form.

16.2.2. Momentum EquationThe momentum equation for compressible fluids is similar to the one for incompressible

fluids. This is because in momentum equation the change in momentum flux is equated to forcerequired to cause this change.

Momentum flux = mass flux × velocity = ρAV × VBut the mass flux i.e., ρAV = constant ...By continuity equationThus the momentum equation is completely independent of the compressibility effects and

for compressible fluids the momentum equation, say in X-direction, may be expressed as :ΣFx = (ρAVVx)2 – (ρAVVx)1 ...(16.3)

16.2.3. Bernoulli’s or Energy EquationAs the flow of compressible fluid is steady, the Euler equation is given as :

dpρ + VdV + gdz = 0 ...(16.4)

Integrating both sides, we get

� dpρ + � VdV + � gdz = constant

or � dpρ + V2

2 + gz = constant ...(16.5)

In compressible flow since ρ is not constant it cannot be taken outside the integration sign.In compressible fluids the pressure (p) changes with change of density (ρ), depending on the type ofprocess. Let us find out the Bernoulli’s equation for isothermal and adiabatic processes.

(a) Bernoulli’s or energy equation for isothermal process :In case of an isothermal process,

pv = constant or pρ = constant = c1 (say)

(where v = specific volume = 1/ρ)

∴ ρ = pc

Hence �dpρ = � � �= =

dpp c

c dpp

cdpp/ 1

11 = c1 loge p =

pρ loge p

� cp

1 =��

��ρ

Substituting the value of �dpp in eqn. (16.5), we get

pρ loge p +

V2

2 + gz = constant

COMPRESSIBLE FLOW 859


Dividing both sides by g, we get

pgρ loge p +

Vg

2

2 + z = constant ...(16.6)

Eqn. (16.6) is the Bernoulli’s equation for compressible flow undergoing isothermal process.(b) Bernoulli’s equation for adiabatic process :In case of an adiabatic process,

pvγ = constant or p

ργ = constant = c2 (say)

∴ ρ γ = pc2

or ρ = pc2

1/�

��

��

γ

Hence �dpρ =

dp

p cc

pdp c p dp

( / )( ) ( )

21/ 2

1/1/ 2

1/ 1/1γ

γγ

γ γ= = =� � � −

= ( )( ) ( )

( ) ( )cp c p

c p21/

11

21/

1

21/

1

11

1 1γ

γ γγ

γγ

γγ

γγ

γ

γγ

− +−�

�� −�

��

− +��

��

�

�

�

=−�

��

=−

= γ

γ ργ

γ γγ

−��

��

��

��

−��

��

1

1/ 1p

p( ) � cp

2 =�

��

��ργ

= γ

γρ

γ

γγ

γγ

−��

��

�

�

��

�

�

��×

−��

��

1

1/

1

1p

p( )

= γ

γ ργ

γ ρ

γγ

γ

−��

��

=−

��

��

+−�

��

1 1

1 1

( )p p

Substituting the value of �dpp in eqn. (16.6), we get

γγ ρ−��

��

+1 2

2p V + gz = constant

Dividing both sides by g, we get

γγ ρ−��

��

+1 2

2pg

Vg + z = constant ...(16.7)

Eqn. (16.7) is the Bernoulli’s equation for compressible flow undergoing adiabatic process.

Example 16.1. A gas with a velocity of 300 m/s is flowing through a horizontal pipe at asection where pressure is 78 kN/m2 absolute and temperature 40°C. The pipe changes in diam-eter and at this section, the pressure is 117 kN/m2 absolute. Find the velocity of the gas at thissection if the flow of the gas is adiabatic. Take R = 287 J/kg K and γ = 1.4.

(Punjab University)



Sol. Section 1 : Velocity of the gas, V = 300 m/sPressure, p1 = 78 kN/m

2

Temperature, T1 = 40 + 273 = 313 KSection 2 : Pressure, p2 = 117 kN/m

2

R = 287 J/kg K, γ = 1.4

Velocity of gas at section 2, V2 :Applying Bernoulli’s equations at sections 1 and 2 for adiabatic process, we have

γγ ρ−��

��

+1 2

1

1

12p

gV

g = z1 = γ

γ ρ−��

��

+1 2

2

2

22p

gV

g + z2 [Eqn. (16.7)]

But z1 = z2, since the pipe is horizontal.

∴γ

γ ρ−��

��

+1 2

1

1

12p

gV

g = γ

γ ρ−��

��

+1 2

2

2

22p

gV

gCancelling ‘g’ on both sides, we get

γ

γ ρ ρ−��

��

−�

��

��= −

1 2 21

1

2

2

22

12p p V V

or, γ

γ ρ ρρ

−��

��

− ×�

��

��= −

11

2 21

1

2

2

1

1

22

12p p

pV V

∴ γ

γ ρρρ−

��

��

− ×�

��

��= −

11

2 21

1

2

1

1

2

22

12p p

pV V

...(i)

For an adiabatic flow : p1

1ργ =

p2

2ργ or

pp

1

2

1

2=�

��

��ρρ

γ

or ρρ

γ1

2

1

2

1

=�

��

��pp

Substituting the value of ρρ

1

2 in eqn (i), we get

γγ ρ

γ

−��

��

− ×�

��

��

�

��

��

�

��

��1

111

2

1

1

2

1

p pp

pp

= V V2

212

2 2−

γ

γ ργ

−��

��

−�

��

��

�

��

��

�

��

��

−

111

1

2

1

11

p pp

= V V2

212

2 2−

or,γ

γ ρ

γγ

−��

��

−�

��

��

�

��

��

�

��

��

= −−

11

21

1

2

1

1

22

12p p

pV V

...(ii)

At section 1 :p1

1ρ = RT1 = 287 × 313 = 89831

pp

2

1 =

11778

= 1.5, and V1 = 300 m/s



Substituting the values in eqn. (ii), we get

1.41.4 −��

��1

× 89831 11

−��

��

��

��

−

( )1.51.4

1.4 = V2

2

2 –

3002

2

314408.5 (1 – 1.1228) = V2

2

2 – 45000 or – 38609.4 =

V22

2 – 45000

or, V22 = 12781.2 or V2 = 113.05 m/s. (Ans.)

Example 16.2. In the case of air flow in a conduit transition, the pressure, velocity andtemperature at the upstream section are 35 kN/m2, 30 m/s and 150°C respectively. If at thedownstream section the velocity is 150 m/s, determine the pressure and the temperature if theprocess followed is isentropic. Take γ = 1.4, R = 290 J/kg K.

Sol. Section 1 (upstream) : Pressure, p1 = 35 kN/m2,

Velocity, V1 = 30 m/s,Temperature, T = 150 + 273 = 423 KVelocity, V2 = 150 m/s

R = 290 J/kg K, γ = 1.4Section 2 (downstream) :Pressure, p2 :Applying Bernoulli’s equation at sections 1 and 2 for isentropic (reversible adiabatic) process,

we have

γ

γ ρ−��

��

+1 2

1

1

12p

gV

g + z1 = γ

γ ρ−��

��

+1 2

2

2

22p

gV

g + z2

Assuming z1 = z2, we have

γγ ρ−��

��

+1 2

1

1

12p

gV

g = γ

γ ρ−��

��

+1 2

2

2

22p

gV

gCancelling ‘g’ on both the sides, and rearranging, we get

γ

γ ρρρ−

��

��

− ×�

��

��111

1

2

1

1

2

p pp =

V V22

12

2 2− ...(i)

For an isentropic flow : p1

1ργ =

p22ρ

γ orpp

1

2 =

ρρ

γ1

2

�

��

�� or

ρρ

1

2 =

pp

1

2

1�

��

��γ


1

2 in eqn. (i), we have

γγ ρ ρ

γ

−��

��

− ×�

��

��

�

��

��

�

��

��1

111

2

1

1

2

1

p pp

p =

V V22

12

2 2−

γ

γ ργ

−��

��

−�

��

��

�

��

��

�

��

��

−

111

1

2

1

11

p pp

= V V2

212

2 2−



γ

γ ρ

γγ

−��

��

−�

��

��

�

��

��

�

��

��

−

111

1

2

1

1

p pp =

V V22

12

2 2−

Substituting the values, we get

1.41.4 − 1 × 122670

1150

230

22

1

12 2

−�

��

��

�

��

��

�

��

��

= −

−pp

1.41.4

= 10800

�

pRT1

11 290 423 122670ρ

= = × =�

��

��

429345 1 21

0 2857

−�

��

��

��

��

��

pp

.

= 10800

or pp

2

1

0 2857�

��

��

.

= 1 – 10800429345

= 0.9748

orpp

2

1 = (0.9748)1/0.2857 = (0.9748)3.5 = 0.9145

or p2 = 35 × 0.9145 = 32 kN/m2 (Ans.)

Temperature, T2 :For an isentropic process, we have

TT

2

1 =

pp

2

1

11 1

1 0 28570 9145 0 9145 0 9748�

��

��= = =

−−

γγ

( . ) ( . ) ..4

.4 .

∴ T2 = 423 × 0.9748 = 412.3 Kor t2 = 412.3 – 273 = 139.3°C (Ans.)

16.3. PROPAGATION OF DISTURBANCES IN FLUID AND VELOCITY OF SOUND

The solids as well as fluids consist of molecules. Whereas the molecules in solids are closetogether, these are relatively apart in fluids. Consequently whenever there is a minor disturbance,it travels instantaneously in case of solids ; but in case of fluid the molecules change its positionbefore the transmission or propagation of the disturbance. Thus the velocity of disturbance in caseof fluids will be less than the velocity of the disturbance in solids. In case of fluid, the propagationof disturbance depends upon its elastic properties. The velocity of disturbance depends upon thechanges in pressure and density of the fluid.

The propagation of disturbance is similar to the propagation of sound through a media. Thespeed of propagation of sound in a media is known as acoustic or sonic velocity and depends uponthe difference of pressure. In compressible flow, velocity of sound (sonic velocity) is of paramountimportance.

16.3.1. Derivation of Sonic Velocity (velocity of sound)Consider a one-dimensional flow through long straight rigid pipe of uniform cross-sectional

area filled with a frictionless piston at one end as shown in Fig. 16.1. The tube is filled with acompressible fluid initially at rest. If the piston is moved suddenly to the right with a velocity, apressure wave would be propagated through the fluid with the velocity of sound wave.



PistonWave frontRigid

pipe

dx = Vdt (dL – dx)

(dL = Cdt)

CV

Fig. 16.1. One dimensional pressure wave propagation.

Let A = cross-sectional area of the pipe,V = piston velocity,p = fluid pressure in the pipe before the piston movement,ρ = fluid density before the piston movement,

dt = a small interval of time during which piston moves, andC = velocity of pressure wave or sound wave (travelling in the fluid).

Before the movement of the piston the length dL has an initial density ρ, and its total mass= ρ × dL × A.

When the piston moves through a distance dx, the fluid density within the compressedregion of length (dL – dx) will be increased and becomes (ρ + dρ) and subsequently the total massof fluid in the compressed region = (ρ + dρ) (dL – dx) × A

∴ ρ × dL × A = (ρ + dρ) (dL – dx) × A ...by principle of continuity.But dL = C dt and dx = Vdt ; therefore, the above equation becomes

ρCdt = (ρ + dρ) (C – V) dtor, ρC = (ρ + dρ) (C – V) or ρC = ρC – ρV + dρ . C – dρ . Vor, 0 = – ρV + dρ . C – dρ . V

Neglecting the term dρ.V (V being much smaller than C), we get

dρ . C = ρV or C = ρ

ρV

d ...(16.8)

Further in the region of compressed fluid, the fluid particles have attained a velocity whichis apparently equal to V (velocity of the piston), accompanied by an increase in pressure dp due tosudden motion of the piston. Applying inpulse-momentum equation for the fluid in the compressedregion during dt, we get

dp × A × dt = ρ × dL × A (V – 0) (force on the fluid) (rate of change of momentum)

or, dp = ρ dLdt

V = ρ × Cdtdt

× V = ρCV (� dL = Cdt)

or, C = dpVρ ...(16.9)

Multiplying eqns. (16.8) and (16.9), we get

C2 = ρ

ρV

d ×

dpVρ =

dpdρ



∴ C = dpdρ ...(16.10)

16.3.2. Sonic Velocity in Terms of Bulk ModulusThe bulk modulus of elasticity of fluid (K) is defined as

K = dpdvv��

...(i)

where, dv = decrease in volume, and v = original volume(– ve sign indicates that volume decreases with increase in pressure)

Also, volume v ∝ 1ρ , or vρ = constant

Differentiating both sides, we get

vdρ + ρdv = 0 or – dvv

d= ρρ

Substituting the value of – dvv

dpK

=��

��

from eqn. (i), we get

dpK

= dρρ

or dpdρ

= Kρ

Substituting this value of dpdρ in eqn. (16.10), we get

C = Kρ ...(16.11)

Eqn. (16.11) is applicable for liquids and gases.

16.3.3. Sonic Velocity for Isothermal Process

For isothermal process : pρ = constant

Differentiating both sides, we get

ρ ρ

ρ. .dp p d−

2 = 0 ordp p dρ

ρρ

− . 2 = 0

or, dpρ =

p d. ρρ2

ordpd

pρ ρ

= = RT ...(16.12)

pRT

ρ=

��

��

...equation of state

Substituting the value of dpdρ

in eqn. (16.10), we get

C = pρ = RT ...(16.13)



16.3.4. Sonic Velocity for Adiabatic Process

For isentropic (reversible adiabatic) process : p

ργ = constant

or p . ρ–γ = constantDifferentiating both sides, we have p(– γ) . ρ–γ–1 dρ + ρ–γ dp = 0Dividing both sides by ρ–γ, we get

– p γ ρ–1 dρ + dp = 0 or dp = p γ ρ–1 dρ

or, dpdρ

= pρ

γ = γRT �p

RTρ

=��

��

Substituting the value of dpdρ

in eqn. (16.10), we get

C = γRT ...(16.14)The following points are worth noting :(i) The process is assumed to be adiabatic when minor disturbances are to be propagated

through air ; due to very high velocity of disturbances/pressure waves appreciable heat transferdoes not take place.

(ii) For calculation of velocity of the sound/pressure waves, isothermal process is consideredonly when it is mentioned in the numerical problem (that the process is isothermal). When noprocess is mentioned in the problem, calculation are made assuming the process to be adiabatic.

16.4. MACH NUMBER

The mach number is an important parameter in dealing with the flow of compressiblefluids, when elastic forces become important and predominant.

Mach number is defined as the square root of the ratio of the inertia force of a fluid to theelastic force.

∴ Mach number, M = Inertia forceElastic force

= ρAVKA

2

= VK

V

K

VC

2

/ /ρ ρ= = [� K C/ρ = ...Eqn. (16.11)]

i.e. M = VC

...(16.15)

Thus, M = Velocity at a point in a fluid

Velocity of sound at that point at a given instant of timeDepending on the value of Mach number, the flow can be classified as follows :1. Subsonic flow : Mach number is less than 1.0 (or M < 1) ; in this case V < C.2. Sonic flow : Mach number is equal to 1.0 (or M = 1) ; in this case V = C.3. Supersonic flow : Mach number is greater than 1.0 (or M > 1) ; in this case V > C.When the Mach number in flow region is slightly less to slightly greater than 1.0, the flow

is termed as transonic flow.The following points are worth noting :(i) Mach number is important in those problems in which the flow velocity is comparable

with the sonic velocity (velocity of sound). It may happen in case of airplanes travelling at veryhigh speed, projectiles, bullets etc.



(ii) If for any flow system the Mach number is less than about 0.4, the effects of compressibilitymay be neglected (for that flow system).

Example 16.3. Find the sonic velocity for the following fluids :(i) Crude oil of specific gravity 0.8 and bulk modulus 1.5 GN/m2 ;

(ii) Mercury having a bulk modulus of 27 GN/m2.Sol. Crude oil : Specific gravity = 0.8 (Delhi University)∴ Density of oil, ρ = 0.8 × 1000 = 800 kg/m3

Bulk modulus, K = 1.5 GN/m2

Mercury : Bulk modulus, K = 27 GN/m2

Density of mercury, ρ = 13.6 × 1000 = 13600 kg/m3

Sonic velocity, Coil, CHg :Sonic velocity is given by the relation :

C = Kρ [Eqn. (16.11)]

∴ Coil = 1.5 10

800

9× = 1369.3 m/s (Ans.)

CHg = 27 1013600

9× = 1409 m/s (Ans.)

Example 16.4. An aeroplane is flying at a height of 14 km where temperature is – 45°C.The speed of the plane is corresponding to M = 2. Find the speed of the plane if R = 287 J/kg Kand γ = 1.4.

Sol. Temperature (at a height of 14 km), t = – 45°C. T = – 45 + 273 = 228 K

Mach number, M = 2Gas constant, R = 287 J/kg K

γ = 1.4Speed of the plane, V :Sonic velocity, (C) is given by,

C = γRT (assuming the process to be adiabatic) ...[Eqn. (16.14)]

= 1.4 × ×287 228 = 302.67 m/s

Also M = VC

...[Eqn. (16.15)]

or, 2 = V

302 67.

or, V = 2 × 302.67 = 605.34 m/s = 605 34 3600

1000. ×

= 2179.2 km/h (Ans.)

16.5. PROPAGATION OF DISTURBANCE IN COMPRESSIBLE FLUID

When some disturbance is created in a compressible fluid (elastic or pressure waves are alsogenerated), it is propagated in all directions with sonic velocity (= C) and its nature of propagationdepends upon the Mach number (M). Such disturbance may be created when an object moves in arelatively stationary compressible fluid or when a compressible fluid flows past a stationary object.



Consider a tiny projectile moving in a straight line with velocity V through a compressiblefluid which is stationary. Let the projectile is at A when time t = 0, then in time t it will movethrough a distance AB = Vt. During this time the disturbance which originated from the projectilewhen it was at A will grow into the surface of sphere of radius Ct as shown in Fig. 16.2, which alsoshows the growth of the other disturbances which will originate from the projectile at every t/4interval of time as the projectile moves from A to B.

Let us find nature of propagation of the disturbance for different Mach numbers.Case I. When M < 1 (i.e., V < C). In this case since V < C the projectile lags behind the

disturbance/pressure wave and hence as shown in Fig. 16.2 (a) the projectile at point B lies insidethe sphere of radius Ct and also inside other spheres formed by the disturbances/waves started atintermediate points.

Case II. When M = 1 (i.e., V = C). In this case, the disturbance always travels with theprojectile as shown in Fig. 16.2 (b). The circle drawn with centre A will pass through B.

Case III. When M > 1 (i.e., V > C). In this case the projectile travels faster than thedisturbance. Thus the distance AB (which the projectile has travelled) is more than Ct, and hence

Ct Ct34 Ct

34 Ct1

2 Ct1

2 Ct1

4 Ct

14 Ct

14 Ct

A B AB

Ct 34 Ct 12 Ct

a

A

aV

a = Mach angle

B

Machcone

ZONE

OF

ACTION

ZONE

OF

SILENCE

ZONE

OF

SILENCE

ZONE

OF

ACTION

( ) M < 1 (V < C)a

( ) M > 1 (V > C)c

( ) M = 1 (V = C)b

C

Subsonic motion Sonic motion

Supersonic motion

Fig. 16.2. Nature of propagation of disturbances in compressible flow.



the projectile at point ‘B’ is outside the spheres formed due to formation and growth of disturbanceat t = 0 and at the intermediate points (Fig. 16.2 (c)). If the tangents are drawn (from the point B)to the circles, the spherical pressure waves form a cone with its vertex at B. It is known as Machcone. The semi-vertex angle α of the cone is known as Mach angle which is given by,

sin α = CtVt

CV M

= = 1 ...(16.16)

In such a case (M > 1), the effect of the disturbance is felt only in region inside the Machcone, this region is called zone of action. The region outside the Mach cone is called zone of silence.

It has been observed that when an aeroplane is moving with supersonic speed, its noise isheard only after the plane has already passed over us.

Example 16.5. Find the velocity of a bullet fired in standard air if its Mach angle is 40°.Sol. Mach angle, α = 40°

γ = 1.4For standard air : R = 287 J/kg K, t = 15°C or T = 15 + 273 = 288 KVelocity of the bullet, V :

Sonic velocity, C = γRT = 14 287 228. × × = 340.2 m/s

Now, sin α = CV

or, sin 40° = 340 2.

V or V =

340 240.

sin ° = 529.26 m/s (Ans.)

Example 16.6. A projectile is travelling in air having pressure and temperature as88.3 kN/m2 and – 2°C. If the Mach angle is 40°, find the velocity of the projectile. Take γ = 1.4 andR = 287 J/kg K. [M.U.]

Sol. Pressure, p = 88.3 kN/m2

Temperature, T = – 2 + 273 = 271 KMach angle, M = 40°

γ = 1.4, R = 287 J/kg KVelocity of the projectile, V :

Sonic velocity, C = γRT = 14 287 271. × × ~− 330 m/s

Now, sin α = CV

or sin 40° = 340V

or, V = 330

40sin ° = 513.4 m/s (Ans.)

Example 16.7. A supersonic aircraft flies at an altitude of 1.8 km where temperature is4°C. Determine the speed of the aircraft if its sound is heard 4 seconds after its passage over thehead of an observer. Take R = 287 J/kg K and γ = 1.4.

Sol. Altitude of the aircraft = 1.8 km = 1800 mTemperature, T = 4 + 273 = 277 KTime, t = 4 s



Speed of the aircraft, V :Refer Fig. 16.3. Let O represent the observer and

A the position of the aircraft just vertically over theobserver. After 4 seconds, the aircraft reaches the posi-tion represented by the point B. Line AB represents thewave front and α the Mach angle.

From Fig. 16.3, we have

tan α = 18004

450V V

= ...(i)

But, Mach number, M = CV

= 1

sin α

or, V = C

sin α ...(ii)

Substituting the value of V in eqn. (i), we get

tan α = 450 450

( /sin )sin

C Cαα=

or,sincos

αα =

450 sin αC

or cos α = C

450...(iii)

But C = γRT , where C is the sonic velocity. R = 287 J/kg K and γ = 1.4 ...(Given)

∴ C = 14 287 277. × × = 333.6 m/sSubstituting the value of C in eqn. (ii), we get

cos α = 333 6450

. = 0.7413

∴ sin α = 1 1 0 74132 2− = −cos .α = 0.6712Substituting the value of sin α in eqn. (ii), we get

V = C

sin.

.α= = ×333 6

0 6712497 m/s =

497 36001000

= 1789.2 km/h (Ans.)

16.6. STAGNATION PROPERTIES

The point on the immersed body where the velocity is zero is called stagnation point. Atthis point velocity head is converted into pressure head. The values of pressure (ps), temperature(Ts) and density (ρs) at stagnation point are called stagnation properties.

16.6.1. Expression for Stagnation Pressure (ps) in Compressible FlowConsider the flow of compressible fluid past an immersed body where the velocity becomes

zero. Consider frictionless adiabatic (isentropic) condition. Let us consider two points, O in the freestream and the stagnation point S as shown in Fig. 16.4.

Let, p0 = pressure of compressible fluid at point O,V0 = velocity of fluid at O,ρ0 = density of fluid at O,T0 = temperature of fluid at O,

a

4 V

1800

m

O

AB = Vt = 4V

BA

Fig. 16.3



and ps, Vs, ρs and Ts are corresponding values ofpressure, velocity density, and temperature at pointS .

Applying Bernoulli’s equation for adiabatic(frictionless) flow at points O and S, (given byeqn. 16.7), we get

γγ ρ−��

��

+1 2

0

0

02p

gV

g + z0 =

γγ ρ−��

��

+1 2

2pg

Vg

s

s

s + zs

But z0 = zs ; the above equation reduces to

γ

γ ρ−��

��

+1 2

0

0

02p

gV

g =

γγ ρ−��

��

+1 2

2pg

Vg

s

s

s

Cancelling ‘g’ on both the sides, we have

γ

γ ρ−��

��

+1 2

0

0

02p V

= γ

γ ρ−��

��

+1 2

2p Vss

s

At point S the velocity is zero, i.e., Vs = 0 ; the above equation becomes

γ

γ ρ ρ−��

��

−�

��

��= −

1 20

0

02p p Vs

s

or,γ

γ ρ ρρ

−��

��

− ×�

��

��110

0

0

0

p pp

s

s = –

V02

2

or,γ

γ ρρρ−

��

��

− ×�

��

��110

0 0

0p pp

s

s = –

V02

2...(i)

For adiabatic process : p00ρ

γ = pssργ or

pps

0 = ρρ

γ

γ0

s or

ρρ

0

s =

pps

0

1�

��

��γ

...(ii)


0

s in eqn. (i), we get

γγ ρ

γ

−��

��

− ×�

��

��

�

�

�

= −1

12

0

0 0

0

1

02p p

ppp

Vss

or, γγ ρ

γ

−��

��

−�

��

��

�

��

��

�

��

��

−

110

0 0

11

p pp

s = – V0

2

2

or, 10

1

−�

��

��

�

�

�

−pp

s

γγ

= – V0

2

2

γγ

ρ−��

��

1 00p

or, 1 + V0

2

2

γγ

ρ−��

��

1 00p

= pp

s

0

1�

��

��

−γγ

...(iii)

O

Streamlines

S (stagnation point)

Body

Fig. 16.4. Stagnation properties.



For adiabatic process, the sonic velocity is given by,

C = γ γ ρRT

p= � p RTρ

=��

��

For point O, C0 = γ ρp0

0 or C0

2 = γ p0

0ρ

Substituting the value of γρp00

= C02 in eqn. (iii), we get

1 + V0

2

2 (γ – 1) ×

1

02

0

1

Cpp

s=�

��

��

−γγ

or, 1 + VC0

2

022 (γ – 1) =

pp

s

0

1�

��

��

−γγ

1 + M0

2

2 (γ – 1) = pp

s

0

1�

��

��

−γγ

�

V

CM0

2

02 0

2=�

��

��

or, pp

s

0

1�

��

��

−γγ

= 11

2 02+ −�

��

�

�

�

γM

or, pp

s

0 = 1

12 0

2 1+ −��

��

�

�

�

−γγ

γM ...(iv)

or, ps = p0 11

2 02 1+ −�

��

�

�

�

−γγ

γM ...(16.17)

Eqn. (16.17) gives the value of stagnation pressure.

Compressibility correction factor :If the right hand side of eqn. (16.17) is expanded by the binomial theorem, we get

ps = p0 12 8

2480

20

40

6+ + + −�

��

γ γ γ γM M M

( )

= p0 1 21

4224

02 2

04+ + + − +

�

��

��

�

�

γ γM MM ...

or, ps = p0 + p M M

M0 02

02

04

21

4224

γ γ+ + − +�

��

��... ...(16.18)

But, M02 =

V

C

V

p

Vp

02

02

02

0

0

02

0

0=�

��

��

=γρ

ργ

� Cp

02 0

0=

�

��

��γρ



Substituting the value of M02 in eqn. (16.18), we get

ps = p0 + p V

pM

M0 02

0

0

02

04

21

4224

γ ργ

γ× + + − +�

��

��...

or, ps = p0 + ρ γ0 0

202

04

21

4224

V MM+ + − +

�

��

��... ...(16.19)

Also, ps = p0 + ρ0 0

2

2V

(when compressibility effects are neglected) ...(16.20)

The comparison of eqns. (16.19) and (16.20) shows that the effects of compressibility areisolated in the bracketed quantity and that these effects depend only upon the Mach number. The

bracketed quantity i eM

M. ., ...14

224

02

04+ +

−+

�

��

��

�

�

γ may thus be considered as a compressibility

correction factor. It is worth noting that :� For M < 0.2, the compressibility affects the pressure difference (ps – p0) by less than 1 per cent

and the simple formula for flow at constant density is then sufficiently accurate.� For larger value of M, as the terms of binomial expansion become significant, the

compressibility effect must be taken into account.� When the Mach number exceeds a value of about 0.3 the Pitot-static tube used for

measuring aircraft speed needs calibration to take into account the compressibilityeffects.

16.6.2. Expression for Stagnation Density (ρs)From eqn. (ii), we have

ρρ

γ0 0

1

s s

pp

=�

��

�� or

ρρ

γs spp0 0

1

=�

��

�� or ρs = ρ0

pp

s

0

1�

��

��γ

Substituting the value of pp

s

0

�

��

�� from eqn. (iv), we get

ρs = ρ0 11

2 02 1

1

+−�

��

��

��

�

�

�

−γγ

γγ

M

or, ρs = ρ0 11

2 02

11

+ −��

��

�

�

�

−γ γM ...(16.21)

16.6.3. Expression for Stagnation Temperature (Ts)

The equation of state is given by : pρ = RT

For stagnation point, the equation of state may be written as :

pssρ

= RTs or Ts = 1R

pssρ



Substituting the values of ps and ρs from eqns. (16.17) and (16.18), we get

Ts = 1

11

2

11

2

0 02 1

0 02

11

R

p M

M

+ −��

��

�

�

�

+ −��

��

�

�

�

−

−

γ

ρ γ

γγ

γ

= 1R

p

M00

02 1

11

11

2ργ

γγ γ

+ −��

��

�

�

�

−−

−��

��

= 1R

p

M00

02

11

11

2ργ

γγ

+ −��

��

�

�

�

−−

��

��

or, Ts = T0 11

2 02+ −�

��

�

�

�

γM ...(16.22) �

pRT0

00ρ

=�

��

��

Example 16.8. An aeroplane is flying at 1000 km/h through still air having a pressure of78.5 kN/m2 (abs.) and temperature – 8°C. Calculate on the stagnation point on the nose of theplane :

(i) Stagnation pressure, (ii) Stagnation temperature, and(iii) Stagnation density.Take for air : R = 287 J/kg K and γ = 1.4.

Sol. Speed of aeroplane, V = 1000 km/h = 1000 1000

60 60××

= 277.77 m/s

Pressure of air, p0 = 78.5 kN/m2

Temperature of air, T0 = – 8 + 273 = 265 KFor air : R = 287 J/kg K, γ = 1.4The sonic velocity for adiabatic flow is given by,

C0 = γRT0 = 14 287 265. × × = 326.31 m/s

∴ Mach number, M0 = VC

0

0

277 77326 31

= ..

= 0.851

(i) Stagnation pressure, ps :The stagnation pressure (ps) is given by the relation,

ps = p0 11

2 02 1+ −�

��

�

�

�

−γγ

γM ...[Eqn. (16.17)]

or, ps = 78.5 114 1

20 8512

11 1

+ −��

��

×�

�

�

−..

.4.4

= 78.5 (1.145)3.5 = 126.1 kN/m2 (Ans.)



(ii) Stagnation temperature, Ts :The stagnation temperature is given by,

Ts = T0 11

2 02+ −�

��

�

�

�

γM ...[Eqn. (16.22)]

= 265 114 1

20 8512+ − ×�

��

.. = 303.4 K or 30.4°C (Ans.)

(iii) Stagnation density, ρs :The stagnation density (ρs) is given by,

pssρ

= RTs or ρs = p

RTs

s

or, ρs = 126 1 10287 303 4

3..

×× = 1.448 kg/m

3 (Ans.)

Example 16.9. Air has a velocity of 1000 km/h at a pressure of 9.81 kN/m2 in vacuumand a temperature of 47°C. Compute its stagnation properties and the local Mach number. Takeatmospheric pressure = 98.1 kN/m2, R = 287 J/kg K and γ = 1.4.

What would be the compressibility correction factor for a pitot-static tube to measure thevelocity at a Mach number of 0.8.

Sol. Velocity of air, V0 = 1000 km/h = 1000 1000

60 60×× = 277.78 m/s

Temperature of air, T0 = 47 + 273 = 320 KAtmospheric pressure, patm = 98.1 kN/m

2

Pressure of air (static), p0 = 98.1 – 9.81 = 88.29 kN/m2

R = 287 J/kg K, γ = 1.4

Sonic velocity, C0 = γRT0 = 14 287 320. × × = 358.6 m/s

∴ Mach number, M0 = VC

0

0

277 78358 6

= .. = 0.7746

Stagnation pressure, ps :The stagnation pressure is given by,

ps = p0 11

2 02 1+ −�

��

�

�

�

−γγ

γM ...[Eqn. (16.17)]

or, ps = 88.29 11

20 77462

1+ − ×�

��

−1.41.4

1.4.

= 88.29 (1.12)3.5 = 131.27 kN/m2 (Ans.)Stagnation temperature, Ts :

Ts = T0 11

2 02+ −�

��

�

�

�

γM ...[Eqn. (16.22)]

or, Ts = 320 114 1

20 77462+ − ×�

��

.. = 358.4 K or 85.4°C (Ans.)



Stagnation density, ρs :

ρs = p

RTs

s= ×

×131.27 10287 358.4

3

= 1.276 kg/m3 (Ans.)

Compressibility factor at M = 0.8 :

Compressibility factor = 1 + M0

2

4224

+ − γ M04 + ...

= 1 + 0 8

42 14

24

2. .+ − × 0.84 = 1.1702 (Ans.)

Example 16.10. Air at a pressure of 220 kN/m2 and temperature 27°C is moving at avelocity of 200 m/s. Calculate the stagnation pressure if

(i) Compressibility is neglected ; (ii) Compressibility is accounted for.For air take R = 287 J/kg K, γ = 1.4.Sol. Pressure of air, p0 = 200 kN/m

2

Temperature of air, T0 = 27 + 233 = 300 KVelocity of air, V0 = 200 m/sStagnation pressure, ps :(i) Compressibility is neglected :

ps = p0 + ρ0 0

2

2V

where ρ0 = p

RT0

0

3220 10287 300

= ×× = 2.555 kg/m

3

∴ ps = 220 + 2 555 200

2

2. × × 10–3 (kN/m2) = 271.1 kN/m2. Ans.

(ii) Compressibility is accounted for :The stagnation pressure, when compressibility is accounted for, is given by,

ps = p0 + ρ γ0 0

202

04

21

4224

V MM+ + − +

�

��

��... ...[Eqn. (16.19)]

Mach number, M0 = VC RT

0

0 0

200 20014 287 300

= =× ×γ .

= 0.576

Whence, ps = 220 + 2 555 200

210 1

0 5764

2 1424

0 5762

32

4. . . .× × + + − ×

�

��

��−

or, ps = 220 + 51.1 (1 + 0.0829 + 0.00275) = 275.47 kN/m2 (Ans.)

Example 16.11. In aircraft flying at an altitude where the pressure was 35 kPa andtemperature – 38°C, stagnation pressure measured was 65.4 kPa. Calculate the speed of theaircraft. Take molecular weight of air as 28. (UPSC, 1998)

Sol. Pressure of air, p0 = 35 kPa = 35 × 103 N/m2

Temperature of air, T0 = – 38 + 273 = 235 KStagnation pressure, ps = 65.4 kPa = 65.4 × 10

3 N/m2



Speed of the aircraft, Va :

p0V = mRT0 = m × RM

0�� T0 or ρ0 =

mV

p MR T

= 00 0

where R = characteristic gas constant, R0 = universal gas constant = 8314 Nm/mole K. M = molecular weight for air = 28, and ρ0 = density of air.Substituting the values, we get

ρ0 = ( )35 10 28

8314 235

3× ××

= 0.5 kg/m3

Now, using the relation : ps = p0 + ρ0

2

2Va ...[Eqn. (16.20)]

or, Va = 2 2 65 4 10 35 10

0 50

3 3( ) ( . ).

p ps − = × − ×ρ = 348.7 m/s (Ans.)

16.7. AREA-VELOCITY RELATIONSHIP AND EFFECT OF VARIATION OF AREA FORSUBSONIC, SONIC AND SUPERSONIC FLOWS

For an incompressible flow the continuity equation may be expressed as : AV = constant, which when differentiated gives

AdV + VdA = 0 ordAA

dVV

= − ...(16.23)

But in case of compressible flow, the continuity equation is given by, ρAV = constant, which can be differentiated to give

ρd(AV) + AVdρ = 0 or ρ(AdV + VdA) + AVdρ = 0or, ρAdV + ρVdA + AVdρ = 0

Dividing both sides by ρAV, we getdVV

dAA

d+ + ρρ = 0 ...(16.24)

or,dAA

dVV

= − – dρρ ...[16.24 (a)]

The Euler’s equation for compressible fluid is given by,

dpρ + VdV + gdz = 0

Neglecting the z terms the above equation reduces to, dpρ + VdV = 0

This equation can also be expressed as :

dpρ ×

dd

ρρ + VdV = 0 or

dpd

dρ

ρρ

× + VdV = 0

Butdpdρ = C

2 ...[Eqn. (16.10)]

∴ C2 × dρρ + VdV = 0 or C

2 dρρ = – VdV or

dρρ = –

VdV

C2



Substituting the value of dρρ in eqn. (16.24), we get

dVV

+ dAA

– VdV

C2 = 0

or,dAA

= VdV

C2 –

dVV

= dVV

V

C

2

2 1−�

��

��

∴ dAA

= dVV

(M2 – 1) � MVC

=��

�� ...(16.25)

This important equation is due to Hugoniot.

Eqns. (16.23) and (16.25) give variation of dAA

��

�� for the flow of incompressible and com-

pressible fluids respectively. The ratios dAA

��

�� and

dVV��

�� are respectively fractional variations in

the values of area and flow velocity in the flow passage.Further, in order to study the variation of pressure with the change in flow area, an expres-

sion similar to eqn. (16.25), as given below, can be obtained.

dp = ρV2 1

1 2−�

��

��MdAA ...(16.26)

From eqns. (16.25) and (16.26), it is possible to formulate the following conclusions of prac-tical significance.

(i) For subsonic flow (M < 1) :

dVV

> 0 ; dAA

< 0 ; dp < 0 (convergent nozzle)

dVV

< 0 ; dAA

> 0 ; dp > 0 (divergent diffuser)

V < Vp > p

>T > T

2 1

2 1

2 1

2 1

r r

V > Vp < p

<T < T

2 1

2 1

2 1

2 1

r rV2V1 V1 V2

( ) Convergent nozzle.a ( ) Divergent diffuser.b

Fig. 16.5. Subsonic flow (M < 1).

(ii) For supersonic flow (M > 1) :dVV

> 0 ; dAA

> 0 ; dp < 0 (divergent nozzle)

dVV

< 0 ; dAA

< 0 ; dp > 0 (convergent diffuser)



ThroatA = A2 1

Fig. 16.7. Sonic flow (M = 1).

Large tank

Convergentnozzle

2V2

p , , T2 2 2r

p1

r1

T1

V = 01

1

V < Vp > p

>T > T

2 1

2 1

2 1

2 1

r r

V > Vp < p

<T < T

2 1

2 1

2 1

2 1

r rV2V1 V1 V2

( ) Convergent nozzle.b( ) Divergent nozzle.a

Fig. 16.6. Supersonic flow (M > 1).

(iii) For sonic flow (M = 1) :

dAA

= 0 (straight flow passage

since dA must be zero)and dp = (zero/zero) i.e., indeterminate, but

when evaluated, the change of pressurep = 0, since dA = 0 and the flow is fric-tionless.

16.8. FLOW OF COMPRESSIBLE FLUID THROUGH A CONVERGENT NOZZLE

Fig. 16.8 shows a large tank/vessel fitted with ashort convergent nozzle and containing a compressiblefluid. Consider two points 1 and 2 inside the tank andexit of the nozzle respectively.

Let p1 = pressure of fluid at the point 1,V1 = velocity of fluid in the tank (= 0),T1 = temperature of fluid at point 1,ρ1 = density of fluid at point 1, and p2, V2, T2 and

ρ2 are corresponding values of pressure, velocity, tem-perature and density at point 2.

Assuming the flow to take place adiabatically,then by using Bernoulli’s equation (for adiabatic flow),we have

γγ ρ−��

��

+1 2

1

1

12p

gV

g + z1 = γ

γ ρ−��

��

+1 2

2

2

22p

gV

g + z2 [Eqn. (16.7)]

But z1 = z2 and V1 = 0

∴ γ

γ ρ− 11

1

pg =

γγ ρ−��

��

+1 2

2

2

22p

gV

g

or,γ

γ −��

��1

pg

pg

1

1

2

2ρ ρ−

�

�

�

=

Vg2

2

2or

γγ − 1

p p1

1

2

2ρ ρ−

�

�

�

V22

2

Fig. 16.8. Flow of fluid through aconvergent nozzle.



or V2 = 2

11

1

2

2

γγ ρ ρ( )−

−�

��

��p p

or V2 = 2

111

1

2

2

1

1

γγ ρ ρ

ρ( )−

− −�

��

��p p

p...(1)

For adiabatic flow : p11ργ =

p2

2ργ or

pp

1

2

1

2=�

��

��ρρ

γ

or ρρ

γ1

2

1

2

1

=�

��

��pp

...(i)


1

2 in eqn. (1), we get

V2 = 2

111

1

2

1

1

2

1

γγ ρ

γ

( )−− ×

�

��

��

�

�

�

p pp

pp

= 2

111

1

2

1

11

γγ ρ

γ

( )−−�

��

��

�

�

�

−p p

p

or V2 = 2

111

1

2

1

1

γγ ρ

γγ

( )−−�

��

��

�

�

�

−p p

p ...(16.27)

The mass rate of flow of the compressible fluid,m = ρ2A2V2, A2 being the area of the nozzle at the exit

= ρ2A2 2

111

1

2

1

1

γγ ρ

γγ

( )−−�

��

��

�

�

�

−p p

p , [substituting V2 from eqn. (16.27)]

or m = A2 2

111

12

2 2

1

1

γγ ρ

ρρ

γγ

( )−× −

�

��

��

�

�

�

−p p

From eqn. (i), we have ρ2 = ρ ργ

γ1

1 21/ 1

2

1

1

( / )p ppp

=�

��

��

∴ ρ 22 = ρ1

2 pp

2

1

2�

��

��γ

Substituting this value in the above equation, we get

m = A2 2

111

112 2

1

2

2

1

1γ

γ ρρ

γγ

γ

−×

�

��

��−�

��

��

�

�

�

−p p

ppp

= A2 2

1 1 12

1

22

1

1 2

γγ

ργ γ

γ γ

−�

��

��−�

��

��

�

�

�

− +

ppp

pp

/



m = A2 2

1 1 12

1

22

1

1

γγ

ργ γ

γ

−�

��

��−�

��

��

�

�

�

+

ppp

pp

/

...(16.28)

The mass rate of flow (m) depends on the value of pp

2

1 (for the given values of p1 and ρ1 at

point 1).

Value of pp

2

1 for maximum value of mass rate of flow :

For maximum value of m, we have d

dpp

m2

1

��

( ) = 0

As other quantities except the ratio pp

2

1 are constant

∴d

dpp

m2

1

��

( ) = pp

pp

2

1

22

1

1�

��

��−�

��

��

�

�

�

+/γ γ

γ

= 0

or,2 2

1

21

γγp

p�

��

��

−

– γ

γ

γγ+�

��

��

��

+−

1 21

11

pp

= 0

or, pp

2

1

21

�

��

��

−γ

= γ γ+��

��

��

��1

22

1

1

pp

orpp

2

1

2�

��

��

− γγ

= γ γ+ �

��

��1

22

1

1

pp

or, pp

2

1

2�

��

��

− γ

= γ γ+��

��

��

��1

22

1

pp

or, pp

2

1

2 1�

��

��

− −γ

= γ γ+��

��

12

or pp

2

1

11

2�

��

��= +��

��

− γ γγ

or, pp

2

1

1�

��

��

−γ

= 2

1γ

γ

+��

��

or, pp

2

1

�

��

�� =

21

1

γ

γγ

+��

��

−...(16.29)

Eqn. (16.29) is the condition for maximum mass flow rate through the nozzle.It may be pointed out that a convergent nozzle is employed when the exit pressure is equal

to or more than the critical pressure, and a convergent-divergent nozzle is used when the dis-charge pressure is less than the critical pressure.

For air with γ = 1.4, the critical pressure ratio,

pp

2

1 =

24 1

44 1

1.

1.1.

+��

��

− = 0.528 ...(16.30)



Relevant relations for critical density and temperature are :

ρρ

2

1 =

21

11

γγ

+��

��

− ...[16.30 (a)]

TT

2

1 =

21γ + ...[16.30 (b)]

Value of V2 for maximum rate of flow of fluid :


2

1 from eqn. (16.29) in eqn. (16.27), we get

V2 = 2

11

21

21

12

11

1

11

1

1

γγ ρ γ

γγ ρ γ

γγ

γγ

−−

+��

��

�

�

�

=−

−+

��

��

−× −

p p

= 2

11 2

12

111

1

1

1

1

γγ ρ

γγ

γγ ρ

γγ−

+ −+

��

��

=−

−+

��

��

p p

or V2 = 2

11

1

γγ ρ+

p (= C2) ...(16.31)

Maximum rate of flow of fluid through nozzle, mmax :


2

1 from eqn. (16.30) in eqn. (16.28), we get

mmax = A2 2

12

12

11 11

21

1

γγ

ργ γ

γγ γ

γγ

γγ

−��

�� +

��

��

−+

��

��

�

�

�

+×

−× +

p

= A2 2

12

12

11 1

21

11γ

γρ

γ γγ

γγ

−��

�� +

��

��

−+

��

��

�

�

�

−+−

p

For air, γ = 1.4,

∴ mmax = A2 2

12

12

11 1

21

11×

− +��

��

−+

��

��

�

�

�

−+−1.4

1.4 1.4 1.41.4

1.41.4

( )p ρ

= A2 7 0 4018 0 33481 1p ρ ( . . )−

or mmax = 0.685 A2 p1 1ρ ...(16.32)

Variation of mass flow rate of compressible fluid with pressure ratio pp

2

1

�

��

�� :

A passage in which the sonic velocity has been reached and thus in which the mass flowrate is maximum, is often said to be choked or in choking conditions. It is evident from eqn.(16.28) that for a fixed value of inlet pressure the mass flow depends on nozzle exit pressure.



Fig. 16.9. depicts the variation of actual and

theoretical mass flow rate versus pp

2

1 . Following

points are worthnoting :(i) The flow rate increases with a decrease in

the pressure ratio pp

2

1 and attains the maximum

value of the critical pressure ratio pp

2

1 = 0.528 for

air.(ii) With further decrease in exit pressure

below the critical value, the theoretical mass flowrate decreases. This is contrary to the actual re-sults where the mass flow rate remains constantafter attaining the maximum value. This may beexplained as follows :

At critical pressure ratio, the velocity V2 atthe throat is equal to the sonic speed (derived below).For an accelerating flow of a compressible fluid in a convergent nozzle the velocity of flow withinthe nozzle is subsonic with a maximum velocity equal to the sonic velocity at the throat : Thusonce the velocity V2 at the throat has attained the sonic speed at the critical pressure ratio, it

remains at the same value for all the values of pp

2

1

�

��

�� less than critical pressure ratio, since the flow

in the nozzle is being continuously accelerated with the reduction in the throat pressure below thecritical values and hence the velocity cannot reduce. Thus, the mass flow rate for all values of

pp

2

1

�

��

�� less than critical pressure ratio remains constant at the maximum value (indicated by the

solid horizontal line in Fig. 16.9). This fact has been verified experimentally too.Velocity at outlet of nozzle for maximum flow rate :The velocity at outlet of nozzle for maximum flow rate is given by,

V2 = 2

11

1

γγ ρ+��

��

p ...[Eqn. (16.31)]

Now pressure ratio,pp

2

1 =

21

1

γ

γγ

+��

��

−

∴ p1 = p

p2

1

21

21

21

γ

γγγ

γγ

+��

��

=+

�

��

��−

−

For adiabatic flow :p1

1ργ =

p2

2ργ or

pp

1

2 =

ρρ

γ1

2

�

��

�� or

ρρ

1

2 =

pp

1

2

1�

��

��γ

= pp

2

1

1�

��

��

−γ

Actual

Chokedflow

Subsonicflow

mAmax

2

Theoretical

0.528

Pressure ratiopp

2

1m

/A2

Fig. 16.9. Mass flow rate througha convergent nozzle.



∴ ρ 1 = ρ2 pp

2

1

1�

��

��

−γ

or ρ2 2

11

1

γ

γγ γ

+��

��

−×

or ρ2 2

1

11

γγ

+��

��

−

Substituting the values of p1 and ρ1 in the above eqn. (16.31), we get

V2 = 2

12

11 2

121

2

11γ

γ γ ρ γ

γγ γ

+��

��

×+

��

��

× ×+

��

��

�

��

��

�

��

��

− −p

= 2

12

12

2

11

1γγ ρ γ

γγ γ

+��

��

× ×+

��

��

−+

−p =

21

21

2

2

1γ

γ ρ γ+��

��

×+

��

��

−p

or V2 = 2

11

22

2

γγ ρ

γ+

��

��

× +��

��

p =

γρp22

= C2

i.e., V2 = C2 ...(16.33)Hence the velocity at the outlet of nozzle for maximum flow rate equals sonic velocity.

16.9. VARIABLES OF FLOW IN TERMS OF MACH NUMBER

In order to obtain relationship involving change in velocity, pressure, temperature anddensity in terms of the Mach number use is made of the continuity, perfect gas, isentropic flow andenergy equations.

For continuity equation, we haveρAV = constant

Differentiating the above equation, we get ρ[AdV + VdA] + AVdρ = 0

Dividing throughout by ρAV, we havedVV

dAA

d+ + ρρ

= 0

From isentropic flow, we have p

ργ = constant or

dpp

= γ dpp

For perfect gas, we have p = ρRT ordpp

= dρρ

+ dTT

From energy equation, we have cpT + V2

2 + constant

Differentiating throughout, we get

cpdT + VdV = 0 orγ

γR−

��

��1

dT + VdV = 0 � cR

p = −��

��

γγ 1

or,γ

γR− 1

dT

V 2 +

dVV

= 0 ...(i)

Also, sonic velocity, C = γRT ∴ γR = CT

2



Substituting the value of γR = CT

2 in eqn. (i), we get

C

TdT

V

dVV

2

21( )γ −× + = 0

or,11 2( )γ −

× +M

dTT

dVV

= 0 � MVC

=��

�� ...(16.34)

From the Mach number relationship

M = V

RTγ (where γRT = C)

dMM

= dVV

– 12

dTT

...(16.35)

Substituting the value of dTT

from eqns. (16.34) in eqn. (16.35), we get

dMM

= dVV

– 12

− × −�

��

dVV

M( )γ 1 2

= dVV

+ 12

dVV

× (γ – 1) M2

or, dMM

= dVV

11

22+ −�

��

γM or

dVV

= 1

11

22+ −�

��

�

�

�

γM

dMM

...(16.36)

Since the quantity within the bracket is always positive, the trend of variation of velocityand Mach number is similar. For temperature variation, one can write

dTT

= − −

+ −��

��

�

�

�

( )γγ

1

11

2

2

2

M

M dMM

...(16.37)

Since the right hand side is negative the temperature changes follow an opposite trend tothat of Mach number. Similarly for pressure and density, we have

dpp

= −

+ −

�

�

�

γγ

M

M

2

211

2

dMM

...(16.38)

and,dρρ

= −

+ −��

��

�

�

�

M

M

2

211

2γ

dMM ...(16.39)

For changes in area, we have

dAA

= − −

+ −

�

�

�

( )1

11

2

2

2

M

Mγ

dMM ...(16.40)



The quantity within the brackets may be positive or negative depending upon the magni-tude of Mach number. By integrating eqn. (16.40), we can obtain a relationship between thecritical throat area Ac, where Mach number is unity and the area A at any section where M 1

AAc

= 1 2 1

1

21

2 1

MM+ −

+�

�

�

+−( ) ( )γ

γ

γγ

...(16.41)

Example 16.12. The pressure leads from Pitot-static tube mounted on an aircraft wereconnected to a pressure gauge in the cockpit. The dial of the pressure gauge is calibrated to readthe aircraft speed in m/s. The calibration is done on the ground by applying a known pressureacross the gauge and calculating the equivalent velocity using incompressible Bernoulli’s equa-tion and assuming that the density is 1.224 kg/m3.

The gauge having been calibrated in this way the aircraft is flown at 9200 m, where thedensity is 0.454 kg/m3 and ambient pressure is 30 kN/m2. The gauge indicates a velocity of152 m/s. What is the true speed of the aircraft ? (UPSC)

Sol. Bernoulli’s equation for an incompressible flow is given by,

p + ρV2

2 = constant

The stagnation pressure (ps) created at Pitot-static tube,

ps = p0 + ρ0 0

2

2V

(neglecting compressibility effects) ...(i)

Here p0 = 30 kN/m2, V0 = 152 m/s, ρ0 = 1.224 kg/m

3 ...(Given)

∴ ps = 30 + 1224 152

2

2. × ×10–3 = 44.139 kN/m2

Neglecting compressibility effect, the speed of the aircraft whenρ0 = 0.454 kg/m

3 is given by [using eqn. (i)],

44.139 × 103 = 30 × 103 + 0 454

20

2. × V

or V02 =

( . ).

44 139 30 10 20 454

3− × × = 62286.34

∴ V0 = 249.57 m/s

Sonic velocity, C0 = γRT0 = γ ρp0

0 = 1.4

0.454× ×30 10

3 = 304.16 m/s

Mach number, M = VC

0

0

249 57304 16

= .. = 0.82

Compressibility correction factor = 140

2+

�

��

��M

, neglecting the terms containing higher powers

of M0 (from eqn 16.19).

= 10 82

4+�

��

. = 1.168

∴ True speed of aircraft = 249 57

1168

.

. = 230.9 m/s

Hence true speed of aircraft = 230.9 m/s (Ans.)



Example 16.13. (a) In case of isentropic flow of a compressible fluid through a variableduct, show that

AAc

= 1M

1 12

( 1) M

12

( 1)

2

12( 1)+ −

+

�

�

�

+−γ

γ

γγ

where γ is the ratio of specific heats, M is the Mach number at a section whose area is A and Ac isthe critical area of flow.

(b) A supersonic nozzle is to be designed for air flow with Mach number 3 at the exitsection which is 200 mm in diameter. The pressure and temperature of air at the nozzle exit areto be 7.85 kN/m2 and 200 K respectively. Determine the reservoir pressure, temperature and thethroat area. Take : γ = 1.4. (U.P.S.C. Exam.)

Sol. (a) Please Ref. to Art. 16.9.(b) Mach number, M = 3Area at the exit section, A = π/4 × 0.22 = 0.0314 m2

Pressure of air at the nozzle, (p)nozzle = 7.85 kN/m2

Temperature of air at the nozzle, (T)nozzle = 200 KReservoir pressure, (p)res. :

From eqn. (16.17), (p)res. = (p)nozzle 11

22 1+ −�

��

�

�

�

−��

��γ

γγ

M

or, (p)res. = 7.85 114 1

232

11 1

+ −��

��

×�

�

�

−��

��.

.4.4

= 288.35 kN/m2 (Ans.)

Reservoir temperature, (T)res. :

From eqn. (16.22), (T)res. = (T)nozzle 11

22+ −�

��

�

�

�

γM

or, (T)res. = 200 114 1

232+ −�

��

×�

�

�

. = 560 K (Ans.)

Throat area (critical), Ac :

From eqn. (16.41), AAc

= 1 2 1

1

21

2 1

MM+ −

+�

�

�

+−( ) ( )γ

γ

γγ

or, 0 0314.

Ac =

13

2 14 1 3

14 1

21 1

2 1 1+ −+

�

�

�

+−( . )

.

.4( .4 )

or0 0314 1

3.Ac

= (2.333)3 = 4.23

or, Ac = 0 03144 23.

. = 0.00742 m2 (Ans.)

16.10. FLOW THROUGH LAVAL NOZZLE (CONVERGENT-DIVERGENT NOZZLE)

Laval nozzle is a convergent-divergent nozzle (named after de Laval, the swedish scientistwho invented it) in which subsonic flow prevails in the converging section, critical or transonicconditions in the throat and supersonic flow in the diverging section.



Let p2 (= pc) = pressure in the throat when the flow is sonic for given pressure p1.— When the pressure in the receiver, p3 = p1, there will be no flow through the nozzle, this

is shown by line a in Fig. 16.10 (b).

d e

cba

f

Shock wavefronts

Criticalp = 0.528 pc 1

pc

p1

Inlet(p )1 1

Flow 2

Exit

3Receiver

Throat

( )a

( )b

p3

j

Inlet Throat Exit

Fig. 16.10. (a) Laval nozzle (convergent-divergent nozzle) ; (b) Pressure distribution througha convergent-divergent nozzle with flow of compressible fluid.

— When the receiver pressure is reduced, flow will occur through the nozzle. As long as thevalue of p3 is such that throat pressure p2 is greater than the critical pressure 0.528 p1,the flow in the converging and diverging sections will be subsonic. This condition isshown by line ‘b’.

— With further reduction in p3, a stage is reached when p2 is equal to critical pressurepc = 0.528 p1, at this line M = 1 in the throat. This condition is shown by line ‘c’. Flow issubsonic on the upstream as well the downstream of the throat. The flow is also isentropic.

— If p3 is further reduced, it does not effect the flow in convergent section. The flow inthroat is sonic, downstream it is supersonic. Somewhere in the diverging section ashock wave occurs and flow changes to subsonic (curve d). The flow across the shock isnon-isentropic. Downstream of the shock wave the flow is subsonic and decelerates.

— If the value of p3 is further reduced, the shock wave forms somewhat downstream (curve e).— For p3 equal to pj, the shock wave will occur just at the exit of divergent section.— If the value of p3 lies before pf and pj oblique waves are formed at the exit.

Example 16.14. A large tank contains air at 284 kN/m2 gauge pressure and 24°Ctemperature. The air flows from the tank to the atmosphere through a convergent nozzle. If thediameter at the outlet of the nozzle is 20 mm, find the maximum flow rate of air.

Take : R = 287 J/kg K, γ = 1.4 and atmospheric pressure = 100 kN/m2.(Punjab University)



Sol. Pressure in the tank, p1 = 284 kN/m2 (gauge)

= 284 + 100 = 384 kN/m2 (absolute)Temperature in the tank, T1 = 24 + 273 = 297 KDiameter at the outlet of the nozzle, D = 20 mm = 0.02 m

∴ Area, A2 = π4

× 0.022 = 0.0003141 m2

R = 287 J/kg K, γ = 1.4(Two points are considered. Point 1 lies inside the tank and point 2 lies at the exit of the

nozzle).Maximum flow rate, mmax :

Equation of state is given by p = ρRT or ρ = p

RT

∴ ρ 1 =p

RT1

1 =

384 10287 297

3××

= 4.5 kg/m3

The fluid parameters in the tank correspond to the stagnation values, and maximum flowrate of air is given by,

mmax = 0.685 A2 p1 1ρ ...[Eqn. (16.32)]

= 0.685 × 0.0003141 384 10 4 53× × . = 0.283 kg/sHence maximum flow rate of air = 0.283 kg/s (Ans.)

Example 16.15. A large vessel, fitted with a nozzle, contains air at a pressure of 2500 kN/m2 (abs.) and at a temperature of 20°C. If the pressure at the outlet of the nozzle is 1750 kN/m2,find the velocity of air flowing at the outlet of the nozzle.

Take : R = 287 J/kg K and γ = 1.4.Sol. Pressure inside the vessel, p1 = 2500 kN/m

2 (abs.)Temperature inside the vessel, T1 = 20 + 273 = 293 KPressure at the outlet of the nozzle, p2 = 1750 kN/m

2 (abs.) R = 287 J/kg K, γ = 1.4

Velocity of air, V2 :

V2 = 2

111

1

2

1

1γ

γ ρ

γγ

−��

��

−�

��

��

�

�

�

−p p

p ...[Eqn. (16.27)]

where, ρ1 = p

RTp

RT11

From equation of state :ρ

=��

��

= 2500 10287 293

3×× = 29.73 kg/m

3

Substituting the values in the above equation, we get

V2 = 2 1414 1

2500 1029 73

117502500

31 1

1×−

��

��

××

− ��

��

�

�

�

−.

. .

.4.4



= 7 84090 1 0 903× −( . ) = 238.9 m/s

i.e., V2 = 238.9 m/s (Ans.)

Example 16.16. A tank fitted with a convergent nozzle contains air at a temperature of20°C. The diameter at the outlet of the nozzle is 25 mm. Assuming adiabatic flow, find the massrate of flow of air through the nozzle to the atmosphere when the pressure in the tank is :

(i) 140 kN/m2 (abs.), (ii) 300 kN/m2

Take for air : R = 287 J/kg K and γ = 1.4, Barometric pressure = 100 kN/m2.Sol. Temperature of air in the tank, T1 = 20 + 273 = 293 KDiameter at the outlet of the nozzle, D2 = 25 mm = 0.025 mArea, A2 = π/4 × 0.025

2 = 0.0004908 m2

R = 287 J/kg K, γ = 1.4(i) Mass rate of flow of air when pressure in the tank is 140 kN/m2 (abs.) :

ρ1 = p

RT1

1 =

140 10287 293

3××

= 1.665 kg/m3

p1 = 140 kN/m2 (abs.)

Pressure at the nozzle, p2 = atmospheric pressure = 100 kN/m2

∴ Pressure ratio, pp

2

1 =

100140

= 0.7143

Since the pressure ratio is more than the critical value, flow in the nozzle will be subsonic,hence mass rate of flow of air is given by eqn. 16.28, as

m = A2 2

1 1 12

1

2

2

1

1

γγ

ργ

γγ

−�

��

��−�

��

��

�

�

�

+

ppp

pp

= 0.0004908 2 1414 1

140 10 1665 0 7143 0 714332

11 1

1×−

��

��

× × × −�

�

�

+.

.. ( . ) ( . ).4

.4.4

= 0.0004908 1631700 0 7143 0 71431 17142( . ) ( . ).4285 .−

or m = 0.0004908 1631700 0 6184 0 5617( . . )− = 0.1493 kg/s (Ans.)

(ii) Mass rate of flow of air when pressure in the tank is 300 kN/m2 (abs.) :p1 = 300 kN/m

2 (abs.)p2 = pressure at the nozzle = atmospheric pressure = 100 kN/m

2

∴ Pressure ratio,pp

2

1 =

100300

= 0.33.

The pressure ratio being less than the critical ratio 0.528, the flow in the nozzle will besonic, the flow rate is maximum and is given by eqn. (16.32), as

mmax = 0.685 A2 p1 1ρ

where, ρ1 = p

RT1

1 =

300 10287 293

3×× = 3.567 kg/m

3

∴ mmax = 0.685 × 0.0004908 300 10 3 5673× × . = 0.3477 kg/s (Ans.)



Example 16.17. At some section in the convergent-divergent nozzle, in which air is flow-ing, pressure, velocity, temperature and cross-sectional area are 200 kN/m2, 170 m/s, 200°C and1000 mm2 respectively. If the flow conditions are isentropic, determine :

(i) Stagnation temperature and stagnation pressure,(ii) Sonic velocity and Mach number at this section,

(iii) Velocity, Mach number and flow area at outlet section where pressure is 110 kN/m2,(iv) Pressure, temperature, velocity and flow area at throat of the nozzle.Take for air : R = 287 J/kg K, cp = 1.0 kJ/kg K and γ = 1.4.Sol. Let subscripts 1, 2 and t refers to the conditions at given section, outlet section and

throat section of the nozzle respectively.Pressure in the nozzle, p1 = 200 kN/m

2

Velocity of air, V1 = 170 m/sTemperature, T1 = 200 + 273 = 473 KCross-sectional area, A1 = 1000 mm

2 = 1000 × 10–6 = 0.001 m2

For air : R = 287 J/kg K, cp = 1.0 kJ/kg K, γ = 1.4(i) Stagnation temperature (Ts) and stagnation pressure (ps) :

Stagnation temperature, Ts = T1 + V

cp12

2 ×

= 473 + 170

2 10 1000

2

× ×( . ) = 487.45 K (or 214.45°C) (Ans.)

Also, pp

s

1 =

TT

s

1

11

1 1487 45473

�

��

��= ��

��

− −

γγ .

.4.4

= 1.111

∴ Stagnation pressure, ps = 200 × 1.111 = 222.2 kN/m2 (Ans.)

(ii) Sonic velocity and Mach number at this section :

Sonic velocity, C1 = γRT1 = 14 287 473. × × = 435.9 m/s (Ans.)

Mach number, M1 = VC

1

1

170435 9

=. = 0.39 (Ans.)

(iii) Velocity, Mach number and flow area at outlet section where pressure is110 kN/m2 :

Pressure at outlet section, p2 = 110 kN/m2 ...(Given)

From eqn (16.17), pp

s

1 = 1

12 2

2 1+ −��

��

�

�

�

−γγ

γM

222 2110

. = 1

14 12 0

2

11 1

+ −��

��

�

�

�

−..4

.4M = (1 + 0.2 M0

2)3.5

or, (1 + 0.2 M22) =

222 2110

13 5. .�

�� = 1.222

or, M2 = 1222 1

0 2

1/2..

−��

�� = 1.05 (Ans.)



Also,TTs

2 =

pps

2

1�

��

��

−γγ

= 110

222 2

1 11

.

.4.4�

��

−

= 0.818

or, T2 = 0.818 × 487.45 = 398.7 KSonic velocity at outlet section,

C2 = γRT2 = 14 287 398 7. .× × = 400.25 m/s

∴ Velocity at outlet section, V2 = M2 × C2 = 1.05 × 400.25 = 420.26 m/s. Ans.Now, mass flow at the given section = mass flow at outlet section (exit)

......continuity equation

i.e., ρ1A1V1 = ρ2A2V2 orp

RT1

1 A1V1 =

pRT

2

2 A2V2

∴ Flow area at the outlet section,

A2 = p A V TT p V1 1 1 2

1 2 2

200 0 001 170 398 7473 110 420 26

= × × ×× ×. .

. = 6.199 × 10–4 m2

Hence, A2 = 6.199 × 10–4 m2 or 619.9 mm2. Ans.

(iv) Pressure (pt), temperature (Tt), velocity (Vt), and flow area (At) at throat ofthe nozzle :

At throat, critical conditions prevail, i.e. the flow velocity becomes equal to the sonic veloc-ity and Mach number attains a unit value.

From eqn. (16.22),TT

s

t = 1

12

2+ −��

��

�

�

�

γMt

or, 487 45.

Tt = 1

14 12

12+ −��

��

×�

�

�

. = 1.2 or Tt = 406.2 K

Hence Tt = 406.2 K (or 133.2°C). Ans.

Also, pT

t

s =

TT

t

s

�

��

��−γ

γ 1 or

pt222 2.

= 406 2487 45

11 1.

.

.4.4�

��

− = 0.528

or, pt = 222.2 × 0.528 = 117.32 kN/m2. Ans.

Sonic velocity (corresponding to throat conditions),

Ct = γRTt = 14 287 406 2. .× × = 404 m/s

∴ Flow velocity, Vt = Mt × Ct = 1 × 404 = 404 m/sBy continuity equation, we have : ρ1A1V1 = ρt AtVt

or,p

RT1

1 A1V1 =

pRT

t

t AtVt

∴ Flow area at throat, At = p A V TT p V

t

t t

1 1 1

1

200 0 001 170 406 2473 117 32 404

= × × ×× ×. .

. = 6.16 × 10–4 m2

Hence, At = 6.16 × 10–4 m2 or 616 mm2 (Ans.)



16.11. SHOCK WAVES

Whenever a supersonic flow (compressible) abruptly changes to subsonic flow, a shockwave (analogous to hydraulic jump in an open channel) is produced, resulting in a sudden rise inpressure, density, temperature and entropy. This occurs due to pressure differentials and whenthe Mach number of the approaching flow M1 > 1. A shock wave is a pressure wave of finitethickness, of the order of 10–2 to 10–4 mm in the atmospheric pressure. A shock wave takes place inthe diverging section of a nozzle, in a diffuser, throat of a supersonic wind tunnel, in front ofsharp nosed bodies.

Shock waves are of two types :1. Normal shocks which are almost perpendicular to the flow.2. Oblique shocks which are inclined to the flow direction.

16.11.1. Normal Shock WaveConsider a duct having a compressible sonic flow (see Fig. 16.11).Let p1, ρ1, T1, and V1 be the pressure, density, temperature and velocity of the flow (M1 > 1)

and p2, ρ2, T2 and V2 the corresponding values of pressure, density, temperature and velocity aftera shock wave takes place (M2 < 1).

V > C1 1 V < C2 2

p , , T1 1 1r p , , T2 2 2r

Normal shock wave

M > 1 M < 1

Fig. 16.11. Normal shock wave.

For analysing a normal shock wave, use will be made of the continuity, momentum andenergy equations.

Assume unit area cross-section, A1 = A2 = 1.Continuity equation : m = ρ1V1 = ρ2V2 ...(i)Momentum equation : ΣFx = p1A1 – p2A2 = m (V2 – V1) = ρ2A2V2

2 – ρ1A1V12

for A1 = A2 = 1, the pressure drop across the shock wave, p1 – p2 = ρ2V2

2 – ρ1V12 ...(ii)

p1 + ρ1V12 = p2 + ρ2V2

2

Consider the flow across the shock wave as adiabatic.

Energy equation : γ

γ ρ−��

��

+1 2

1

1

12p V

= γ

γ ρ−��

��

+1 2

2

2

22p V

...[Eqn. (16.7)]

(z1 = z2, duct being in horizontal position)

or,γ

γ − 1 p p V V2

2

1

1

12

22

2ρ ρ−

�

��

��= − ...(iii)

Combining continuity and momentum equations [refer eqns. (i) and (ii)], we get

p1 + ( )ρ

ρ1 1

2

1

V = p2 +

( )ρρ

2 22

2

V...(16.42)



This equation is ‘known as Rankine Line Equation.Now combining continuity and energy equations [refer eqns. (i) and (iii)], we get

γγ − 1

p1

1ρ�

��

�� +

( )ρρ

1 12

122

V =

γγ − 1

p2

2ρ�

��

�� +

( )ρρ

2 22

222

V...(16.43)

This equation is called Fanno Line Equation.

Further combining eqns. (i), (ii) and (iii) and solving for pp

2

1 , we get

pp

2

1 =

γγ

ρρ

γγ

ρρ

+−

��

��

−

+−

��

��

−

11

1

11

2

1

2

1...(16.44)

Solving for density ratio ρρ

2

1 , the same equations yield

ρρ

2

1 =

VV

1

2 =

111

11

2

1

2

1

+ +−

��

��

+−

��

��

+

γγ

γγ

pp

pp

...(16.45)

The eqns. (16.44) and (16.45) are called Ranking-Hugoniot equations.

One can also express pp

2

1 ,

VV

2

1 ,

ρρ

2

1 and

TT

2

1 in terms of Mach number as follows :

pp

2

1 =

2 11

12γ γγ

M − −+

( )( ) ...(16.46)

VV

1

2 =

ρρ

2

1 =

( )( )

γγ

+− +

11 2

12

12

MM ...(16.47)

TT

2

1 =

[( ) ] [ ( )]( )

γ γ γγ

− + − −+

1 2 2 11

12

12

212

M MM ...(16.48)

By algebraic manipulation the following equation between M1 and M2 can be obtained.

M22 =

( )( )

γγ γ

− +− −

1 22 1

12

12

MM

...(16.49)

Example 16.18. For a normal shock wave in air Mach number is 2. If the atmosphericpressure and air density are 26.5 kN/m2 and 0.413 kg/m3 respectively, determine the flow condi-tions before and after the shock wave. Take γ = 1.4.

Sol. Let subscripts 1 and 2 represent the flow conditions before and after the shock wave.Mach number, M1 = 2Atmospheric pressure, p1 = 26.5 kN/m

2

Air density, ρ1 = 0.413 kg/m3



Mach number, M2 :

M22 =

( )( )

γγ γ

− +− −

1 22 1

12

12

MM

...[Eqn. (16.49)]

= ( )

( )1.41.4 1.4

3.611.2 0.4

− × +× × − −

=−

1 2 22 2 1

2

2 = 0.333

∴ M2 = 0.577 (Ans.)Pressure, p2 :

pp

2

1 =

2 11

12γ γγ

M − −+

( )( )

...[Eqn. (16.46)]

= 2 14 2 14 1

14 1112 0 4

2 4

2× × − −+

= −. ( . )( . )

. ..

= 4.5

∴ p2 = 26.5 × 4.5 = 119.25 kN/m2 (Ans.)

Density, ρ2 :

ρρ

2

1 =

( )( )

γγ

+− +

11 2

12

12

MM ...[Eqn. (16.47)]

= ( . )( . )

..

14 1 214 1 2 2

9 616 2

2

2+

− +=

+ = 2.667

∴ ρ2 = 0.413 × 2.667 = 1.101 kg/m3 (Ans.)

Temperature, T1 :

Since p1 = ρ1 RT1, ∴ T1 = pR1

1ρ =

26 5 100 413 287

3..

×× = 223.6 K or – 49.4°C (Ans.)

Temperature, T2 :

TT

2

1 =

[( ) ] [ ( )]( )

[( ) ] [ ( )]( )

γ γ γγ

− + − −+

= − + × × − −+

1 2 2 11

1 2 2 2 2 11 2

12

12

212

2 2

2 2M M

M1.4 1.4 1.4

1.4

= ( . )( . .

.16 2 112 0 4)

23 04+ −

= 1.6875

∴ T2 = 223.6 × 1.6875 = 377.3 K or 104.3°C (Ans.)Velocity, V1 :

C1 = γRT1 = 14 287 223 6. .× × = 299.7 m/s

Since VC

1

1 = M1 = 2 ∴ V1 = 299.7 × 2 = 599.4 m/s (Ans.)

Velocity, V2 :

C2 = γRT2 = 14 287 377 3. .× × = 389.35 m/s

Since VC

2

2 = M2 = 0.577 ∴ V2 = 389.35 × 0.577 = 224.6 m/s (Ans.)



16.11.2. Oblique Shock WaveAs shown in Fig. 16.12, when a supersonic flow

undergoes a sudden turn through a small angle α (posi-tive), an oblique wave is established at the corner. Incomparison with normal shock waves, the oblique shockwaves, being weaker, are preferred.

The shock waves should be avoided or made asweak as possible, since during a shock wave conversionof mechanical energy into heat energy takes place.

16.11.3. Shock StrengthThe strength of shock is defined as the ratio of pressure rise across the shock to the upstream

pressure.

i.e. Strength of shock = p p

ppp

2 1

1

2

1

− = – 1

= 2 1

11

2 1 11

12

12γ γ

γγ γ γ

γM M− −

+− = − − − +

+( ) ( ) ( )

= 2 1 1

12 2

12

112

12γ γ γ

γγ γ

γγ

γM M− + − −

+= −

+=

+ (M12 – 1)

Hence, strength of shock = 2

1γ

γ + (M12 – 1) ...(16.50)

Example 16.19. In a duct in which air is flowing, a normal shock wave occurs at a Machnumber of 1.5. The static pressure and temperature upstream of the shock wave are 170 kN/m2

and 23°C respectively. Determine :(i) Pressure, temperature and Mach number downstream of the shock, and

(ii) Strength of shock.Take γ = 1.4.Sol. Let subscripts 1 and 2 represent flow conditions upstream and downstream of the

shock wave respectively.Mach number, M1 = 1.5Upstream pressure, p1 = 170 kN/m

2

Upstream temperature, T1 = 23 + 273 = 296 K γ = 1.4

(i) Pressure, temperature and Mach number downstream of the shock :

pp

2

1 =

2 11

12γ γγ

M − −+

( ) ...[Eqn. (16.46)]

= 2 14 15 14 1

14 16 3 0 4

2 4

2× × − −+

16...16 compressible flow 16.1. introduction. 16.2. basic equations of compressible fluid flow....

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