16
Compressible Flow
16.1. Introduction. 16.2. Basic equations of compressible fluid flow. 16.3. Propagation of disturbances
in fluid and velocity of sound. 16.4. Mach number. 16.5. Propagation of disturbances in compressible
fluid. 16.6. Stagnation properties. 16.7. Area-velocity relationship and effect of variation of area for
subsonic, 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 (convergentdivergent nozzle). 16.11. Shock waves. Highlights—Objective Type Questions—Theoretical
Questions—Unsolved Examples.
16.1. INTRODUCTION
A compressible flow is that flow in which the density of the fluid changes during flow. All
real fluids are compressible to some extent and therefore their density will change with change in
pressure or temperature. If the relative change in density ∆ρ/ρ is small, the fluid can be treated as
incompressible. A compressible fluid, such as air, can be considered as incompressible with constant ρ if changes in elevation are small, acceleration is small, and/or temperature changes are
negligible. In other words, if Mach’s number U/C, where C is the sonic velocity, is small, compressible fluid can be treated as incompressible.
• The gases are treated as compressible fluids and study of this type of flow is often referred
to 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) Momentum
equation, (iii) Energy equation, and (iv) Equation of state.
16.2.1. Continuity Equation
In 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
Differentiating the above equation, we get
d(ρAV) = 0 or ρd(AV) + AVdρ = 0
or
ρ(AdV + VdA) + AVdρ = 0 or ρAdV + ρVdA + AVdρ = 0
Dividing both sides by ρAV and rearranging, we get
dρ dA dV
+
+
=0
ρ
A
V
Eqn. (16.2) is also known as equation of continuity in differential form.
...(16.2)
16.2.2. Momentum Equation
The 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 force
required to cause this change.
Momentum flux = mass flux × velocity = ρAV × V
But the mass flux i.e., ρAV = constant
...By continuity equation
Thus 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 :
...(16.3)
ΣFx = (ρAVVx)2 – (ρAVVx)1
16.2.3. Bernoulli’s or Energy Equation
As the flow of compressible fluid is steady, the Euler equation is given as :
dp
+ VdV + gdz = 0
ρ
Integrating both sides, we get
z
z
dp
+
ρ
z
VdV +
z
...(16.4)
gdz = constant
dp
V2
+
+ gz = constant
...(16.5)
ρ
2
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 of
process. 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,
or
pv = constant or
p
= constant = c1 (say)
ρ
(where v = specific volume = 1/ρ)
∴
Hence
ρ=
p
c
z z z
z
dp
=
ρ
Substituting the value of
dp
=
p/c1
c1dp
= c1
p
z
dp
in eqn. (16.5), we get
p
p
V2
loge p +
+ gz = constant
ρ
2
dharm
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FG
H
dp
p
p
= c1 loge p =
loge p 3 c1 =
p
ρ
ρ
IJ
K
859
COMPRESSIBLE FLOW
Dividing both sides by g, we get
p
V2
loge p +
+ z = constant
...(16.6)
ρg
2g
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,
p
= constant = c2 (say)
ργ
pvγ = constant or
∴
Hence
z z
dp
=
ρ
ργ =
dp
p
or ρ =
c2
= = (c2 )1/ γ
( p/c2 )1/ γ
z
FG p IJ
Hc K
1/ γ
2
1
p1/ γ
dp = (c2 )1/ γ
FG
LM
OP
H
p
(
c
)
(
p
)
P
=
= (c ) M
MM F − 1 + 1I PP F γ − 1I
MN GH γ JK PQ GH γ JK
FG IJ
F
γ I F pI
H K
= G
H γ − 1JK GH ρ JK ( p)
F I FG IJ
F
JJ ( p)H K
γ IG p
= G
J
G
H γ − 1K G J
Hρ K
−
2
1/ γ
1
+1
γ
1/ γ
2
1/ γ
γ −1
γ
γ −1
γ
γ
1/ γ
1
γ
γ ×
=
FG 1 + γ − 1IJ
F γ I ( p)H γ γ K
GH γ − 1JK
Substituting the value of
z
ρ
IJ
K
z
p−1/ γ dp
F γ − 1IJ
γ K
GH
γ
=
(c2 )1/ γ ( p)
γ −1
F3 c = p I
GH
J
ρ K
2
γ
γ −1
γ
=
FG γ IJ p
H γ − 1K ρ
dp
in eqn. (16.6), we get
p
FG γ IJ p + V + gz = constant
H γ − 1K ρ 2
2
Dividing both sides by g, we get
FG γ IJ p + V + z = constant
H γ − 1 K ρg 2 g
2
...(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 a
section where pressure is 78 kN/m2 absolute and temperature 40°C. The pipe changes in diameter and at this section, the pressure is 117 kN/m2 absolute. Find the velocity of the gas at this
section if the flow of the gas is adiabatic. Take R = 287 J/kg K and γ = 1.4.
(Punjab University)
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860
ENGINEERING THERMODYNAMICS
Sol. Section 1 : Velocity of the gas, V = 300 m/s
Pressure,
p1 = 78 kN/m2
Temperature,
T1 = 40 + 273 = 313 K
Section 2 : Pressure, p2 = 117 kN/m2
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
FG γ IJ p + V = z = FG γ IJ p + V + z
H γ − 1K ρ g 2 g
H γ − 1K ρ g 2 g
2
1
1
2
2
2
1
1
2
2
[Eqn. (16.7)]
But z1 = z2, since the pipe is horizontal.
FG γ IJ p + V = FG γ IJ p + V
H γ − 1K ρ g 2 g H γ − 1K ρ g 2 g
2
1
1
∴
2
2
2
1
2
Cancelling ‘g’ on both sides, we get
or,
∴
FG γ IJ FG p − p IJ = V − V
H γ − 1K H ρ ρ K 2 2
FG γ IJ p FG 1 − p × ρ IJ = V − V
H γ − 1K ρ H ρ p K 2 2
FG γ IJ p FG 1 − p × ρ IJ = V − V
H γ − 1K ρ H p ρ K 2 2
1
2
1
2
1
2
1
1
2
1
1
2
1
1
1
2
For an adiabatic flow :
p1
ρ1γ
Substituting the value of
=
2
2
2
1
2
2
1
2
2
1
2
2
p2
ρ2 γ
FG IJ or ρ = FG p IJ
H K ρ Hp K
p
ρ
or 1 = 1
p2
ρ2
At section 1 :
dharm
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γ
1
1
2
2
1
γ
ρ1
in eqn (i), we get
ρ2
FG γ IJ p R|S1 − p × FG p IJ U|V = V − V
H γ − 1K ρ | p H p K | 2 2
T
W
FG γ IJ p R|S1 − FG p IJ U|V = V − V
H γ − 1K ρ | H p K | 2 2
T
W
FG γ IJ p R|S1 − FG p IJ U|V = V − V
H γ − 1K ρ | H p K | 2
T
W
1
2
1
1
1
2
1−
or,
...(i)
1
2
1
1
1
2
1
1
1
γ
2
1
γ
γ −1
γ
2
2
2
2
2
1
2
2
1
2
1
p1
= RT1 = 287 × 313 = 89831
ρ1
p2
117
=
= 1.5, and V1 = 300 m/s
p1
78
...(ii)
861
COMPRESSIBLE FLOW
Substituting the values in eqn. (ii), we get
FG 1.4 IJ × 89831 R|S1 − (1.5)
H 1.4 − 1K
T|
1.4 − 1
1.4
U| V 300
V| = 2 – 2
W
2
2
2
V22
V2
– 45000 or – 38609.4 = 2 – 45000
2
2
V22 = 12781.2 or V2 = 113.05 m/s. (Ans.)
314408.5 (1 – 1.1228) =
or,
Example 16.2. In the case of air flow in a conduit transition, the pressure, velocity and
temperature at the upstream section are 35 kN/m2, 30 m/s and 150°C respectively. If at the
downstream section the velocity is 150 m/s, determine the pressure and the temperature if the
process 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 K
Velocity,
V2 = 150 m/s
R = 290 J/kg K, γ = 1.4
Section 2 (downstream) :
Pressure, p2 :
Applying Bernoulli’s equation at sections 1 and 2 for isentropic (reversible adiabatic) process,
we have
FG γ IJ p + V + z = FG γ IJ p + V + z
H γ − 1K ρ g 2 g
H γ − 1K ρ g 2 g
2
1
1
2
2
2
1
1
2
2
Assuming z1 = z2, we have
FG γ IJ p + V = FG γ IJ p + V
H γ − 1K ρ g 2 g H γ − 1K ρ g 2 g
2
1
1
2
2
2
1
2
Cancelling ‘g’ on both the sides, and rearranging, we get
FG γ IJ p FG 1 − p × ρ IJ = V − V
H γ − 1K ρ H p ρ K 2 2
Fρ I
ρ
p
p
Fp I
p
=
or
= G J or
= G J
For an isentropic flow :
ρ
ρ
p
Hρ K
ρ
Hp K
1
2
1
1
1
2
2
2
1
1
1
1
1
2
2
2
2
2
γ
1
γ
1
2
γ
2
ρ
Substituting the value of 1 in eqn. (i), we have
ρ2
FG γ IJ p R|S1 − p × FG p IJ U|V = V − V
H γ − 1K ρ | p H ρ K | 2 2
T
W
FG γ IJ p R|S1 − FG p IJ U|V = V − V
H γ − 1K ρ | H p K | 2 2
T
W
1
2
1
1
1
2
1−
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\M-therm\Th16-1.pm5
1
2
1
1
1
γ
2
1
γ
2
2
2
1
2
2
1
...(i)
1
γ
862
ENGINEERING THERMODYNAMICS
FG γ IJ p R|S1 − FG p IJ
H γ − 1K ρ | H p K
T
1
2
1
1
γ −1
γ
Substituting the values, we get
R| F I
S| GH JK
T
1.4
p2
× 122670 1 −
1.4 − 1
p1
U|
V| = V − V
W 2 2
1.4 − 1
1.4
2
2
2
1
U| 150 30
V| = 2 − 2 = 10800
W
FG3 p = RT = 290 × 423 = 122670IJ
H ρ
K
2
2
1
R| F p I
S|1 − GH p JK
T
FG p IJ
Hp K
0.2857
429345
2
1
or
0.2857
=1–
1
or
Temperature, T2 :
For an isentropic process, we have
∴
or
U|
V| = 10800
W
2
or
F I
GH JK
1
1
10800
= 0.9748
429345
p2
= (0.9748)1/0.2857 = (0.9748)3.5 = 0.9145
p1
p2 = 35 × 0.9145 = 32 kN/m2 (Ans.)
γ −1
1.4 − 1
T2
p2 γ
= (0.9145) 1.4 = (0.9145)0.2857 = 0.9748
=
T1
p1
T2 = 423 × 0.9748 = 412.3 K
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 close
together, 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 position
before the transmission or propagation of the disturbance. Thus the velocity of disturbance in case
of fluids will be less than the velocity of the disturbance in solids. In case of fluid, the propagation
of disturbance depends upon its elastic properties. The velocity of disturbance depends upon the
changes in pressure and density of the fluid.
The propagation of disturbance is similar to the propagation of sound through a media. The
speed of propagation of sound in a media is known as acoustic or sonic velocity and depends upon
the difference of pressure. In compressible flow, velocity of sound (sonic velocity) is of paramount
importance.
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 a
compressible fluid initially at rest. If the piston is moved suddenly to the right with a velocity, a
pressure wave would be propagated through the fluid with the velocity of sound wave.
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863
COMPRESSIBLE FLOW
Piston
Rigid
pipe
Wave front
V
dx = Vdt
C
(dL – dx)
(dL = Cdt)
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, and
C = 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 compressed
region of length (dL – dx) will be increased and becomes (ρ + dρ) and subsequently the total mass
of 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) dt
or,
ρC = (ρ + dρ) (C – V) or ρC = ρC – ρV + dρ . C – dρ . V
or,
0 = – ρV + dρ . C – dρ . V
Neglecting the term dρ.V (V being much smaller than C), we get
ρV
...(16.8)
dρ
Further in the region of compressed fluid, the fluid particles have attained a velocity which
is apparently equal to V (velocity of the piston), accompanied by an increase in pressure dp due to
sudden motion of the piston. Applying inpulse-momentum equation for the fluid in the compressed
region during dt, we get
dp × A × dt
= ρ × dL × A (V – 0)
(force on the fluid) (rate of change of momentum)
dρ . C = ρV or C =
or,
or,
dL
Cdt
V=ρ×
× V = ρCV
dt
dt
dp
C=
ρV
Multiplying eqns. (16.8) and (16.9), we get
dp = ρ
C2 =
dharm
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dp
dp
ρV
×
=
dρ
ρV
dρ
(3 dL = Cdt)
...(16.9)
864
ENGINEERING THERMODYNAMICS
∴
dp
dρ
C=
...(16.10)
16.3.2. Sonic Velocity in Terms of Bulk Modulus
The bulk modulus of elasticity of fluid (K) is defined as
K=
dp
FG dvIJ
H vK
...(i)
where, dv = decrease in volume, and v = original volume
(– ve sign indicates that volume decreases with increase in pressure)
1
, or vρ = constant
ρ
Differentiating both sides, we get
Also, volume v ∝
vdρ + ρdv = 0 or –
FG
H
IJ
K
dv dρ
=
v
ρ
dv
dp
=
from eqn. (i), we get
v
K
Substituting the value of –
K
dp
dρ
dp
=
or
=
ρ
K
ρ
dρ
Substituting this value of
dp
in eqn. (16.10), we get
dρ
C=
K
ρ
...(16.11)
Eqn. (16.11) is applicable for liquids and gases.
16.3.3. Sonic Velocity for Isothermal Process
p
= constant
ρ
Differentiating both sides, we get
For isothermal process :
ρ . dp − p . dρ
ρ2
or,
Substituting the value of
or
dp p . dρ
−
=0
ρ
ρ2
dp
p . dρ
=
ρ
ρ2
or
dp p
=
= RT
dρ ρ
...(16.12)
FG p = RT ...equation of stateIJ
Hρ
K
dp
in eqn. (16.10), we get
dρ
C=
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\M-therm\Th16-1.pm5
=0
p
=
ρ
RT
...(16.13)
865
COMPRESSIBLE FLOW
16.3.4. Sonic Velocity for Adiabatic Process
For isentropic (reversible adiabatic) process :
p
ργ
= constant
or
p . ρ–γ = constant
Differentiating both sides, we have p(– γ) . ρ–γ–1 dρ + ρ–γ dp = 0
Dividing both sides by ρ–γ, we get
– p γ ρ–1 dρ + dp = 0 or dp = p γ ρ–1 dρ
dp
p
=
γ = γRT
dρ
ρ
dp
Substituting the value of
in eqn. (16.10), we get
dρ
or,
C=
FG3 p = RT IJ
H ρ K
...(16.14)
γRT
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 transfer
does not take place.
(ii) For calculation of velocity of the sound/pressure waves, isothermal process is considered
only when it is mentioned in the numerical problem (that the process is isothermal). When no
process 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 compressible
fluids, 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 the
elastic force.
∴ Mach number,
V2
=
K /ρ
=
i.e.
M=
ρAV 2
KA
Inertia force
=
Elastic force
M=
V
C
V
K /ρ
=
V
C
[3
K /ρ = C ...Eqn. (16.11)]
...(16.15)
Velocity at a point in a fluid
Velocity of sound at that point at a given instant of time
Depending 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 very
high speed, projectiles, bullets etc.
Thus,
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M=
866
ENGINEERING THERMODYNAMICS
(ii) If for any flow system the Mach number is less than about 0.4, the effects of compressibility
may 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=
K
ρ
Coil =
1.5 × 109
= 1369.3 m/s (Ans.)
800
[Eqn. (16.11)]
27 × 109
= 1409 m/s (Ans.)
13600
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 K
and γ = 1.4.
Sol. Temperature (at a height of 14 km), t = – 45°C.
T = – 45 + 273 = 228 K
Mach number,
M=2
Gas constant,
R = 287 J/kg K
γ = 1.4
Speed of the plane, V :
Sonic velocity, (C) is given by,
CHg =
γRT (assuming the process to be adiabatic)
C=
1.4 × 287 × 228 = 302.67 m/s
=
M=
V
C
or,
2=
V
302.67
or,
V = 2 × 302.67 = 605.34 m/s =
Also
...[Eqn. (16.14)]
...[Eqn. (16.15)]
605.34 × 3600
= 2179.2 km/h (Ans.)
1000
16.5. PROPAGATION OF DISTURBANCE IN COMPRESSIBLE FLUID
When some disturbance is created in a compressible fluid (elastic or pressure waves are also
generated), it is propagated in all directions with sonic velocity (= C) and its nature of propagation
depends upon the Mach number (M). Such disturbance may be created when an object moves in a
relatively stationary compressible fluid or when a compressible fluid flows past a stationary object.
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867
COMPRESSIBLE FLOW
Consider a tiny projectile moving in a straight line with velocity V through a compressible
fluid which is stationary. Let the projectile is at A when time t = 0, then in time t it will move
through a distance AB = Vt. During this time the disturbance which originated from the projectile
when it was at A will grow into the surface of sphere of radius Ct as shown in Fig. 16.2, which also
shows the growth of the other disturbances which will originate from the projectile at every t/4
interval 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 inside
the sphere of radius Ct and also inside other spheres formed by the disturbances/waves started at
intermediate points.
Case II. When M = 1 (i.e., V = C). In this case, the disturbance always travels with the
projectile 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 the
disturbance. Thus the distance AB (which the projectile has travelled) is more than Ct, and hence
Ct
3
4 C
Ct
ZONE
t
3
4 C
ZONE
t
1
4 C
t
t
t
t
OF
B
A
1
2 C
1 C
4
1
2 C
A
B
ACTION
(a) M < 1 (V < C)
Z
O
N
E
OF
SILENCE
Mach
cone
Subsonic motion
(b) M = 1 (V = C)
Sonic motion
Z
O
N
E
1
4 Ct
Ct
3
4 C
1
2 C
t
t
B
A
A
C
T
I
O
N
a
a = Mach angle
V
C
a
OF
S
I
L
E
N
C
E
(c) M > 1 (V > C)
Supersonic motion
Fig. 16.2. Nature of propagation of disturbances in compressible flow.
dharm
\M-therm\Th16-1.pm5
OF
868
ENGINEERING THERMODYNAMICS
the projectile at point ‘B’ is outside the spheres formed due to formation and growth of disturbance
at 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 Mach
cone. The semi-vertex angle α of the cone is known as Mach angle which is given by,
Ct C
1
=
=
Vt V M
sin α =
...(16.16)
In such a case (M > 1), the effect of the disturbance is felt only in region inside the Mach
cone, 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 is
heard 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 = 1.4
For standard air : R = 287 J/kg K, t = 15°C or T = 15 + 273 = 288 K
Velocity of the bullet, V :
Sonic velocity, C =
sin α =
Now,
or,
sin 40° =
γRT =
1.4 × 287 × 228 = 340.2 m/s
C
V
340.2
340.2
or V =
= 529.26 m/s (Ans.)
sin 40°
V
Example 16.6. A projectile is travelling in air having pressure and temperature as
88.3 kN/m2 and – 2°C. If the Mach angle is 40°, find the velocity of the projectile. Take γ = 1.4 and
R = 287 J/kg K.
[M.U.]
Sol. Pressure,
p = 88.3 kN/m2
Temperature,
T = – 2 + 273 = 271 K
Mach angle,
M = 40°
γ = 1.4, R = 287 J/kg K
Velocity of the projectile, V :
Sonic velocity,
Now,
or,
C=
sin α =
V=
γRT =
1.4 × 287 × 271 ~
− 330 m/s
C
340
or sin 40° =
V
V
330
= 513.4 m/s (Ans.)
sin 40°
Example 16.7. A supersonic aircraft flies at an altitude of 1.8 km where temperature is
4°C. Determine the speed of the aircraft if its sound is heard 4 seconds after its passage over the
head of an observer. Take R = 287 J/kg K and γ = 1.4.
Sol. Altitude of the aircraft = 1.8 km = 1800 m
Temperature,
T = 4 + 273 = 277 K
Time,
t=4s
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869
COMPRESSIBLE FLOW
From Fig. 16.3, we have
tan α =
But, Mach number, M =
or,
V=
1800 450
=
4V
V
...(i)
A
1
C
=
sin α
V
C
sin α
tan α =
...(ii)
O
Fig. 16.3
450
450 sin α
=
(C/sin α)
C
sin α
450 sin α
C
=
or cos α =
cos α
C
450
But
B
AB = Vt = 4V
Substituting the value of V in eqn. (i), we get
or,
4V
a
1800 m
Speed of the aircraft, V :
Refer Fig. 16.3. Let O represent the observer and
A the position of the aircraft just vertically over the
observer. After 4 seconds, the aircraft reaches the position represented by the point B. Line AB represents the
wave front and α the Mach angle.
C=
...(iii)
γRT , where C is the sonic velocity.
R = 287 J/kg K and γ = 1.4
...(Given)
∴
C = 1.4 × 287 × 277 = 333.6 m/s
Substituting the value of C in eqn. (ii), we get
cos α =
333.6
= 0.7413
450
1 − cos2 α = 1 − 0.74132 = 0.6712
Substituting the value of sin α in eqn. (ii), we get
∴
sin α =
V=
C
333.6
497 × 3600
=
= 497 m/s =
= 1789.2 km/h (Ans.)
sin α 0.6712
1000
16.6. STAGNATION PROPERTIES
The point on the immersed body where the velocity is zero is called stagnation point. At
this 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 Flow
Consider 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 free
stream 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,
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870
ENGINEERING THERMODYNAMICS
and ps, Vs, ρs and Ts are corresponding values of
pressure, velocity density, and temperature at point
S.
Applying Bernoulli’s equation for adiabatic
(frictionless) flow at points O and S, (given by
eqn. 16.7), we get
Streamlines
Body
FG γ IJ p + V + z = FG γ IJ p + V + z
H γ − 1K ρ g 2 g
H γ − 1K ρ g 2 g
0
0
2
s
0
0
s
S (stagnation point)
O
2
s
s
But z0 = zs ; the above equation reduces to
FG γ IJ p + V = FG γ IJ p + V
H γ − 1K ρ g 2 g H γ − 1K ρ g 2 g
0
0
2
s
s
2
Fig. 16.4. Stagnation properties.
s
0
Cancelling ‘g’ on both the sides, we have
FG γ IJ p + V = FG γ IJ p + V
H γ − 1K ρ 2 H γ − 1K ρ 2
0
0
2
s
0
s
2
s
At point S the velocity is zero, i.e., Vs = 0 ; the above equation becomes
FG γ IJ F p − p I = − V
H γ − 1K GH ρ ρ JK 2
FG γ IJ p F 1 − p × ρ I = – V
H γ − 1K ρ GH ρ p JK
2
FG γ IJ p F 1 − p × ρ I = – V
H γ − 1K ρ GH p ρ JK
2
or,
or,
0
s
0
s
0
s
0
0
s
0
0
s
0
0
0
s
2
0
0
0
2
2
...(i)
ρ0
p
ρ γ
p
p
For adiabatic process : 0γ = sγ or 0 = 0γ or
=
ρs
p
ρ
ρ
ρ0
s
s
s
Substituting the value of
or,
or,
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\M-therm\Th16-1.pm5
0
1
γ
...(ii)
s
ρ0
in eqn. (i), we get
ρs
FG γ IJ p LM1 − p × FG p IJ OP = − V
H γ − 1K ρ MM p H p K PP 2
N
Q
FG γ IJ p R|S1 − FG p IJ U|V = – V
H γ − 1K ρ | H p K | 2
T
W
LM F p I OP
MM1 − GH p JK PP = – V2 FGH γ −γ 1IJK ρp
N
Q
0
s
0
0
0
s
1−
or,
Fp I
GH p JK
0
s
0
0
s
1
γ
0
1
γ
0
γ −1
γ
2
0
2
2
0
V2
1+ 0
2
0
0
FG γ − 1IJ ρ = F p I
H γ K p GH p JK
0
s
0
0
γ −1
γ
...(iii)
871
COMPRESSIBLE FLOW
For adiabatic process, the sonic velocity is given by,
FG3 p = RT IJ
H ρ K
p
ρ
C=
γRT = γ
For point O,
C0 =
γ
Substituting the value of
γp0
= C02 in eqn. (iii), we get
ρ0
p0
p0
or C02 = γ
ρ0
ρ0
F I
GH JK
V
Fp I
1+
(γ – 1) = G J
2C
Hp K
Fp I
M
1+
(γ – 1) = G J
2
Hp K
F p I = LM1 + FG γ − 1IJ M OP
GH p JK
N H 2 K Q
p
L F γ − 1IJ M OP
= M1 + G
p
N H 2 K Q
L F γ − 1IJ M OP
p = p M1 + G
N H 2 K Q
V2
1
ps
1 + 0 (γ – 1) ×
=
2
p0
C02
0
or,
2
0
0
s
2
γ −1
γ
γ −1
γ
0
2
s
F3 V = M I
GH C
JK
γ −1
γ
0
0
0
γ −1
γ
s
or,
0
2
2
0
2
2
0
s
or,
0
0
or,
s
γ
−1
γ
2
0
0
...(iv)
γ
2 γ −1
...(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
OP
LM
Q
N
L γM F 1 + M + 2 − γ M + ...I OP
= p M1 +
JK PQ
MN 2 GH 4 24
I
p γM F
M
2−γ
M + ...J
+
1+
p = p +
G
2
H 4 24
K
ps = p0 1 + γ M02 + γ M04 + γ (2 − γ ) M06
2
8
48
0
0
0
or,
s
But,
M02 =
0
2
0
2
0
0
V02
C0
=
2
V02
FG γp IJ
Hρ K
0
0
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\M-therm\Th16-1.pm5
2
=
V02 ρ0
γp0
4
2
0
4
...(16.18)
F3 C = γp I
GH
J
ρ K
0
2
0
0
872
ENGINEERING THERMODYNAMICS
Substituting the value of M02 in eqn. (16.18), we get
F
GH
I
JK
p0 γ V02ρ0
M2 2 − γ
M04 + ...
×
1+ 0 +
γp0
2
4
24
ps = p0 +
F
GH
I
JK
ρ0V02
M02 2 − γ
+
+
M04 + ...
1
ps = p0 +
2
4
24
or,
...(16.19)
ρ0V0 2
(when compressibility effects are neglected)
...(16.20)
2
The comparison of eqns. (16.19) and (16.20) shows that the effects of compressibility are
isolated in the bracketed quantity and that these effects depend only upon the Mach number. The
Also, ps = p0 +
LM
MN
bracketed quantity i. e.,
F 1 + M + 2 − γ M + ...I OP may thus be considered as a compressibility
GH 4 24
JK PQ
2
0
0
4
correction factor. It is worth noting that :
l 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.
l For larger value of M, as the terms of binomial expansion become significant, the
compressibility effect must be taken into account.
l 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 compressibility
effects.
16.6.2. Expression for Stagnation Density (ρs)
From eqn. (ii), we have
F I or ρ = F p I or ρ = ρ F p I
GH JK
GH p JK
GH p JK
ρ
Fp I
Substituting the value of G J from eqn. (iv), we get
Hp K
OP
L
ρ = ρ MRS1 + FG γ − 1IJ M UV
MMT H 2 K W PP
Q
N
L F γ − 1IJ M OP
ρ = ρ M1 + G
N H 2 K Q
p
ρ0
= 0
ps
ρs
1
γ
s
s
0
0
1
γ
s
0
s
1
γ
0
s
0
s
or,
s
0
0
0
0
1
γ
γ
2 γ −1
1
2 γ −1
16.6.3. Expression for Stagnation Temperature (Ts)
The equation of state is given by :
p
= RT
ρ
For stagnation point, the equation of state may be written as :
ps
1 ps
= RTs or Ts =
ρs
R ρs
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\M-therm\Th16-1.pm5
...(16.21)
873
COMPRESSIBLE FLOW
Substituting the values of ps and ρs from eqns. (16.17) and (16.18), we get
L F γ − 1IJ M OP
p M1 + G
1
N H 2 K Q
T =
R
L F γ − 1IJ M OP
ρ M1 + G
N H 2 K Q
0
0
s
0
0
γ
2 γ −1
1
2 γ −1
F γ − 1 IJ
γ − 1K
LM FG IJ OP
N H K Q
FG
H
1 p L
γ − 1I
O
F
M P
1+ G
=
J
M
H 2 K Q
R ρ N
L F γ − 1IJ M OP
T = T M1 + G
N H 2 K Q
GH γ − 1
1 p0
γ −1
M0 2
1+
=
R ρ0
2
0
0
0
or,
s
0
0
2
γ −1
γ −1
IJ
K
2
...(16.22)
F3 p = RT I
GH ρ
JK
0
0
0
Example 16.8. An aeroplane is flying at 1000 km/h through still air having a pressure of
78.5 kN/m2 (abs.) and temperature – 8°C. Calculate on the stagnation point on the nose of the
plane :
(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
= 277.77 m/s
60 × 60
Pressure of air,
p0 = 78.5 kN/m2
Temperature of air,
T0 = – 8 + 273 = 265 K
For air : R = 287 J/kg K, γ = 1.4
The sonic velocity for adiabatic flow is given by,
C0 =
∴ Mach number,
M0 =
γRT0 =
1.4 × 287 × 265 = 326.31 m/s
V0 277.77
=
= 0.851
C0 326.31
(i) Stagnation pressure, ps :
The stagnation pressure (ps) is given by the relation,
L F γ − 1IJ M OP
p = p M1 + G
N H 2 K Q
L F 1.4 − 1IJ × 0.851 OP
p = 78.5 M1 + G
N H 2 K
Q
s
or,
0
0
γ
2 γ −1
1.4
2 1.4 − 1
s
= 78.5 (1.145)3.5 = 126.1 kN/m2 (Ans.)
dharm
\M-therm\Th16-1.pm5
...[Eqn. (16.17)]
874
ENGINEERING THERMODYNAMICS
(ii) Stagnation temperature, Ts :
The stagnation temperature is given by,
LM FG γ − 1IJ M OP
...[Eqn. (16.22)]
N H 2 K Q
L 1.4 − 1 × 0.851 OP = 303.4 K or 30.4°C (Ans.)
= 265 M1 +
Q
N 2
Ts = T0 1 +
0
2
2
(iii) Stagnation density, ρs :
The stagnation density (ρs) is given by,
ps
= RTs or
ρs
ρs =
or,
ρs =
ps
RTs
126.1 × 10 3
= 1.448 kg/m3 (Ans.)
287 × 303.4
Example 16.9. Air has a velocity of 1000 km/h at a pressure of 9.81 kN/m2 in vacuum
and a temperature of 47°C. Compute its stagnation properties and the local Mach number. Take
atmospheric 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 the
velocity at a Mach number of 0.8.
Sol. Velocity of air,
Temperature of air,
Atmospheric pressure,
Pressure of air (static),
V0 = 1000 km/h =
1000 × 1000
= 277.78 m/s
60 × 60
T0 = 47 + 273 = 320 K
patm = 98.1 kN/m2
p0 = 98.1 – 9.81 = 88.29 kN/m2
R = 287 J/kg K, γ = 1.4
Sonic velocity,
C0 =
γRT0 =
∴ Mach number,
M0 =
V0 277.78
=
= 0.7746
C0
358.6
Stagnation pressure, ps :
The stagnation pressure is given by,
L γ − 1IJ M OP
p = p M1 + FG
N H 2 K Q
L 1.4 − 1 × 0.7746 OP
p = 88.29 M1 +
N 2
Q
s
or,
1.4 × 287 × 320 = 358.6 m/s
0
0
γ
2 γ −1
1.4
2 1.4 − 1
s
= 88.29 (1.12)3.5 = 131.27 kN/m2 (Ans.)
Stagnation temperature, Ts :
LM FG γ − 1IJ M OP
...[Eqn. (16.22)]
N H 2 K Q
L 1.4 − 1 × 0.7746 OP = 358.4 K or 85.4°C (Ans.)
T = 320 M1 +
Q
N 2
Ts = T0 1 +
or,
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\M-therm\Th16-1.pm5
...[Eqn. (16.17)]
0
2
2
s
875
COMPRESSIBLE FLOW
Stagnation density, ρs :
ps
131.27 × 103
=
= 1.276 kg/m3 (Ans.)
RTs
287 × 358.4
Compressibility factor at M = 0.8 :
ρs =
Compressibility factor
=1+
M0 2 2 − γ
+
M04 + ...
4
24
=1+
0.82 2 − 1.4
+
× 0.84 = 1.1702
4
24
(Ans.)
Example 16.10. Air at a pressure of 220 kN/m2 and temperature 27°C is moving at a
velocity 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/m2
Temperature of air, T0 = 27 + 233 = 300 K
Velocity of air, V0 = 200 m/s
Stagnation pressure, ps :
(i) Compressibility is neglected :
ps = p0 +
where ρ0 =
ρ0V0 2
2
p0
220 × 103
=
= 2.555 kg/m3
RT0 287 × 300
2.555 × 2002
× 10–3 (kN/m2) = 271.1 kN/m2. Ans.
2
(ii) Compressibility is accounted for :
The stagnation pressure, when compressibility is accounted for, is given by,
∴
ps = 220 +
ps = p0 +
Mach number,
Whence,
M0 =
V0
=
C0
F
GH
I
JK
ρ0V02
M2 2 − γ
M04 + ...
1+ 0 +
2
4
24
200
=
γRT0
200
1.4 × 287 × 300
F
GH
...[Eqn. (16.19)]
= 0.576
2.555 × 2002
0.5762 2 − 1.4
−3
×
10
1
+
+
× 0.5764
ps = 220 +
2
4
24
I
JK
ps = 220 + 51.1 (1 + 0.0829 + 0.00275) = 275.47 kN/m2 (Ans.)
or,
Example 16.11. In aircraft flying at an altitude where the pressure was 35 kPa and
temperature – 38°C, stagnation pressure measured was 65.4 kPa. Calculate the speed of the
aircraft. 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 K
Stagnation pressure, ps = 65.4 kPa = 65.4 × 103 N/m2
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\M-therm\Th16-1.pm5
876
ENGINEERING THERMODYNAMICS
Speed of the aircraft, Va :
p0V = mRT0 = m ×
where
FG R IJ T or ρ = m = p M
H MK
V RT
0
0
0
0
0 0
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 × 103 ) × 28
= 0.5 kg/m3
8314 × 235
Now, using the relation : ps = p0 +
Va =
or,
ρ0Va2
2
...[Eqn. (16.20)]
2( ps − p)
=
ρ0
2(65.4 × 103 − 35 × 103 )
= 348.7 m/s (Ans.)
0.5
16.7. AREA-VELOCITY RELATIONSHIP AND EFFECT OF VARIATION OF AREA FOR
SUBSONIC, SONIC AND SUPERSONIC FLOWS
For an incompressible flow the continuity equation may be expressed as :
AV = constant, which when differentiated gives
dA
dV
=−
A
V
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 = 0
ρAdV + ρVdA + AVdρ = 0
Dividing both sides by ρAV, we get
AdV + VdA = 0 or
or,
dV dA dρ
+
+
=0
V
A
ρ
or,
...(16.23)
...(16.24)
dρ
dA
dV
=−
–
ρ
A
V
The Euler’s equation for compressible fluid is given by,
...[16.24 (a)]
dp
+ VdV + gdz = 0
ρ
Neglecting the z terms the above equation reduces to,
dp
+ VdV = 0
ρ
This equation can also be expressed as :
dρ
dp
×
+ VdV = 0 or
dρ
ρ
dp
= C2
dρ
But
∴
dharm
\M-therm\Th16-1.pm5
C2 ×
dρ
+ VdV = 0
ρ
or C2
dp dρ
×
+ VdV = 0
dρ
ρ
dρ
= – VdV or
ρ
...[Eqn. (16.10)]
dρ
VdV
=–
ρ
C2
877
COMPRESSIBLE FLOW
Substituting the value of
dρ
in eqn. (16.24), we get
ρ
VdV
dV
dA
+
–
=0
V
A
C2
VdV
dA
dV
dV
=
–
=
A
V
V
C2
or,
F V − 1I
GH C JK
dA
dV
=
(M2 – 1)
A
V
This important equation is due to Hugoniot.
∴
2
2
FG3 M = V IJ
H
CK
...(16.25)
FG dA IJ for the flow of incompressible and comH AK
F dA IJ and FG dV IJ are respectively fractional variations in
pressible fluids respectively. The ratios GH
HVK
AK
Eqns. (16.23) and (16.25) give variation of
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 expression similar to eqn. (16.25), as given below, can be obtained.
dp = ρV2
F 1 I dA
GH 1 − M JK A
...(16.26)
2
From eqns. (16.25) and (16.26), it is possible to formulate the following conclusions of practical significance.
(i) For subsonic flow (M < 1) :
dV
dA
>0;
< 0 ; dp < 0 (convergent nozzle)
V
A
dV
dA
<0;
> 0 ; dp > 0 (divergent diffuser)
V
A
V1
V2 > V1
p2 < p1
r2 < r1
T2 < T1
V2
(a) Convergent nozzle.
V1
V2
V2 < V1
p2 > p1
r2 > r1
T2 > T1
(b) Divergent diffuser.
Fig. 16.5. Subsonic flow (M < 1).
(ii) For supersonic flow (M > 1) :
dV
dA
>0;
> 0 ; dp < 0 (divergent nozzle)
V
A
dV
dA
<0;
< 0 ; dp > 0 (convergent diffuser)
V
A
dharm
\M-therm\Th16-1.pm5
878
ENGINEERING THERMODYNAMICS
V1
V2 > V1
p2 < p1
r2 < r1
T2 < T1
V2
V1
(a) Divergent nozzle.
V2 < V1
p2 > p1
r2 > r1
T2 > T1
V2
(b) Convergent nozzle.
Fig. 16.6. Supersonic flow (M > 1).
(iii) For sonic flow (M = 1) :
Throat
A2 = A1
dA
= 0 (straight flow passage
A
since dA must be zero)
and
dp = (zero/zero) i.e., indeterminate, but
when evaluated, the change of pressure
p = 0, since dA = 0 and the flow is frictionless.
Fig. 16.7. Sonic flow (M = 1).
16.8. FLOW OF COMPRESSIBLE FLUID THROUGH A CONVERGENT NOZZLE
Fig. 16.8 shows a large tank/vessel fitted with a
short convergent nozzle and containing a compressible
fluid. Consider two points 1 and 2 inside the tank and
exit of the nozzle respectively.
Large tank
V1 = 0
p1
Convergent
nozzle
Let p1 = pressure of fluid at the point 1,
V1 = velocity of fluid in the tank (= 0),
r1
T1 = temperature of fluid at point 1,
1
2
V2
ρ1 = density of fluid at point 1, and p2, V2, T2 and
ρ2 are corresponding values of pressure, velocity, temperature and density at point 2.
T1
Assuming the flow to take place adiabatically,
then by using Bernoulli’s equation (for adiabatic flow),
we have
Fig. 16.8. Flow of fluid through a
convergent nozzle.
p2, r2, T2
FG γ IJ p + V + z = FG γ IJ p + V + z
H γ − 1K ρ g 2 g
H γ − 1K ρ g 2 g
1
2
1
1
1
2
2
2
2
2
But z1 = z2 and V1 = 0
∴
or,
FG γ IJ p + V
H γ − 1K ρ g 2 g
FG γ IJ L p − p O = V or γ LM p − p OP V
ρ Q 2
H γ − 1K MN ρ g ρ g PQ 2 g
γ − 1 Nρ
dharm
\M-therm\Th16-1.pm5
γ
p1
=
γ − 1 ρ1 g
1
2
1
2
2
2
2
2
2
2
1
2
1
2
2
2
[Eqn. (16.7)]
879
COMPRESSIBLE FLOW
or
V2 =
FG
H
IJ
K
2γ
p1 p2
−
(γ − 1) ρ1 ρ2
F
I
GH
JK
p
Fρ I ρ = F p I
p
p
= G J or
=
or
ρ
p
H ρ K ρ GH p JK
ρ
or
For adiabatic flow :
V2 =
p1
p
ρ
2γ
1− 2 − 1
ρ2
p1
(γ − 1) ρ1
1
γ
1
2
γ
2
1
1
2
2
γ
1
1
2
2
...(1)
1
γ
...(i)
ρ1
in eqn. (1), we get
ρ2
Substituting the value of
L Fp I
LM
OP
2γ
p M
F
I
V =
GH JK P = (γ − 1) ρ MM1 − GH p JK
MM
PQ
N
N
L F p I OP
2γ p M
PP
V = (γ − 1) ρ M1 − GH p JK
MN
Q
2
or
1
γ
2γ
p1
p
p1
1− 2 ×
p1
p2
(γ − 1) ρ1
2
1
2
1
1
1
2
1
1
1−
1
γ
OP
PP
Q
γ −1
γ
The mass rate of flow of the compressible fluid,
m = ρ2A2V2, A2 being the area of the nozzle at the exit
...(16.27)
LM F I OP
=ρA
MM GH JK PP , [substituting V from eqn. (16.27)]
N
Q
LM F p I OP
2γ
p
× ρ M1 − G J
m= A
(γ − 1) ρ
MN H ρ K PPQ
2 2
or
2γ
p1
p2
1−
p1
(γ − 1) ρ1
1
2
2
1
γ −1
γ
2
2
1
F I
GH JK
From eqn. (i), we have
ρ1
p
= ρ1 2
ρ2 =
p1
( p1/ p2 )1/ γ
∴
Fp I
ρ = ρ G J
Hp K
2
2
2
1
2
γ −1
γ
2
2
γ
1
γ
1
Substituting this value in the above equation, we get
L F p I OP
F
p I M
m=A
GH p JK MM1 − GH p JK PP
N
Q
LMF p I F p I
OP
2γ
=A
p ρ MG J − G J
PP
γ−1
MNH p K H p K
Q
2
2
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\M-therm\Th16-1.pm5
2γ p1
× ρ12
γ − 1 ρ1
1 1
2
2 γ
γ −1
γ
2
1
1
2/ γ
γ −1 2
+
γ
γ
2
2
1
1
880
ENGINEERING THERMODYNAMICS
m = A2
LMF p I
MMGH p JK
N
2γ
p1ρ1
γ−1
Fp I
−G J
Hp K
2/ γ
2
2
1
γ +1
γ
1
The mass rate of flow (m) depends on the value of
point 1).
OP
PP
Q
...(16.28)
p2
(for the given values of p1 and ρ1 at
p1
p2
for maximum value of mass rate of flow :
p1
d
(m) = 0
For maximum value of m, we have
p
d 2
p1
Value of
FG IJ
H K
p2
are constant
p1
As other quantities except the ratio
LMF p I F p I OP
d
PP = 0
(m) = MGH p JK − GH p JK
p I
F
MN
Q
dG J
Hp K
F I
2Fp I
=0
G J – FGH γ +γ 1IJK GH pp JK
γ Hp K
F p I = FG γ + 1IJ F p I or F p I = γ + 1 F p I
GH p JK
GH p JK
G J
H 2 K GH p JK
2 Hp K
FG p IJ = FG γ + 1IJ F p I
H 2 K GH p JK
Hp K
FG p IJ = FG γ + 1IJ or FG p IJ = FG γ + 1IJ
H 2 K Hp K H 2 K
Hp K
F p I = FG 2 IJ
GH p JK
H γ + 1K
Fp I = F 2 I
GH p JK GH γ + 1JK
2/γ
∴
2
2
2
1
1
γ +1
γ
1
or,
2
−1
γ
2
1
or,
2
1
2
−1
γ
2
2−γ
γ
2
1
or,
2
2−γ
γ
2
1
γ
1
1
2
1
2 − γ −1
or,
1
γ
1
1
or,
γ +1
−1
γ
2
γ
2
1
1− γ
γ
2
1
γ −1
γ
2
γ
γ −1
2
1
or,
1
...(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 discharge pressure is less than the critical pressure.
For air with γ = 1.4, the critical pressure ratio,
p2
=
p1
dharm
\M-therm\Th16-1.pm5
FG 2 IJ
H 1.4 + 1K
1.4
1.4 − 1
= 0.528
...(16.30)
881
COMPRESSIBLE FLOW
Relevant relations for critical density and temperature are :
ρ2
=
ρ1
FG 2 IJ
H γ + 1K
1
γ −1
...[16.30 (a)]
T2
2
=
T1
γ +1
Value of V2 for maximum rate of flow of fluid :
Substituting the value of
...[16.30 (b)]
p2
from eqn. (16.29) in eqn. (16.27), we get
p1
LM F I
OP 2γ p F 2 I
V =
MM GH JK
PP = γ − 1 ρ GH 1 − γ + 1JK
N
Q
2γ p F γ + 1 − 2 I
2γ p F γ − 1I
=
=
G
G J
J
γ −1 ρ H γ +1 K
γ − 1 ρ H γ + 1K
γ
2γ p1
2 γ −1
1−
γ − 1 ρ1
γ +1
2
or
V2 =
×
γ −1
γ
1
1
1
1
1
1
2γ p1
(= C2)
γ + 1 ρ1
...(16.31)
Maximum rate of flow of fluid through nozzle, mmax :
Substituting the value of
mmax
p2
from eqn. (16.30) in eqn. (16.28), we get
p1
LMF 2 I
F
2γ I
= A
GH γ − 1JK p ρ MGH γ + 1JK
MN
LMF 2 I
F
2γ I
=A G
H γ − 1JK p ρ MMGH γ + 1JK
N
1 1
2
∴
mmax = A2
= A2
or
2
γ −1
1 1
2
For air, γ = 1.4,
2
γ
×
γ +1 γ
2 × 1.4
p1ρ1
(1.4 − 1)
LMF 2 I
MMGH 1.4 + 1JK
N
F 2 IJ
−G
H γ + 1K
F 2 IJ
−G
H γ + 1K
2
1.4 − 1
γ
γ +1
×
γ −1
γ
γ +1
γ −1
OP
PP
Q
F 2 IJ
−G
H 1.4 + 1K
1.4 + 1
1.4 − 1
OP
PP
Q
OP
PP
Q
7 p1ρ1 (0.4018 − 0.3348)
mmax = 0.685 A2
p1ρ1
Variation of mass flow rate of compressible fluid with pressure ratio
...(16.32)
FG p IJ :
Hp K
2
1
A passage in which the sonic velocity has been reached and thus in which the mass flow
rate 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.
dharm
\M-therm\Th16-2.pm5
882
ENGINEERING THERMODYNAMICS
Fig. 16.9. depicts the variation of actual and
theoretical mass flow rate versus
p2
. Following
p1
mmax
A2
Actual
m/A2
points are worthnoting :
Choked
Subsonic
(i) The flow rate increases with a decrease in
flow
flow
p2
the pressure ratio
and attains the maximum
p1
p
value of the critical pressure ratio 2 = 0.528 for
Theoretical
p1
air.
(ii) With further decrease in exit pressure
below the critical value, the theoretical mass flow
rate decreases. This is contrary to the actual re0.528
sults where the mass flow rate remains constant
p2
Pressure
ratio p
after attaining the maximum value. This may be
1
explained as follows :
Fig. 16.9. Mass flow rate through
At critical pressure ratio, the velocity V2 at
a convergent nozzle.
the 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 within
the nozzle is subsonic with a maximum velocity equal to the sonic velocity at the throat : Thus
once 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
FG p IJ less than critical pressure ratio, since the flow
Hp K
2
1
in the nozzle is being continuously accelerated with the reduction in the throat pressure below the
critical values and hence the velocity cannot reduce. Thus, the mass flow rate for all values of
FG p IJ less than critical pressure ratio remains constant at the maximum value (indicated by the
Hp K
2
1
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,
FG 2γ IJ p
H γ + 1K ρ
F 2 IJ
p
= G
p
H γ + 1K
1
V2 =
Now pressure ratio,
γ
γ −1
2
1
∴
p1 =
For adiabatic flow :
p1
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\M-therm\Th16-2.pm5
...[Eqn. (16.31)]
1
ρ1γ
=
p2
FG 2 IJ
H γ + 1K
p2
ρ2 γ
γ
γ −1
or
F 2 IJ
=p G
H γ + 1K
γ
γ −1
2
FG IJ or ρ = F p I = F p I
H K ρ GH p JK GH p JK
ρ1
p1
=
ρ2
p2
γ
1
1
2
2
1
γ
2
1
−
1
γ
883
COMPRESSIBLE FLOW
∴
Fp I
ρ =ρ G J
Hp K
1
2
−
2
1
γ
F 2 IJ
or ρ G
H γ + 1K
F 2 IJ
or ρ G
H γ + 1K
1
γ
×
γ −1 γ
2
1
1
1− γ
2
Substituting the values of p1 and ρ1 in the above eqn. (16.31), we get
F 2γ I × p FG 2 IJ
V = GH γ + 1JK
H γ + 1K
2
2
γ
1− γ
R| 1 F 2 I
×S ×G
|T ρ H γ + 1JK
1
γ −1
2
F 2γ IJ × p FG 2 IJ
F 2γ IJ × p × FG 2 IJ
= G
= G
H γ + 1K ρ H γ + 1K
H γ + 1K ρ H γ + 1K
F 2γ IJ × p FG γ + 1IJ = γp = C
V = G
H γ + 1K ρ H 2 K ρ
1
γ
+
1− γ γ −1
2
or
2
−1
2
2
2
i.e.,
U|
V|
W
2
2
2
2
2
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 and
density in terms of the Mach number use is made of the continuity, perfect gas, isentropic flow and
energy equations.
For continuity equation, we have
ρAV = constant
Differentiating the above equation, we get
ρ[AdV + VdA] + AVdρ = 0
Dividing throughout by ρAV, we have
dV dA dρ
=0
+
+
V
A
ρ
From isentropic flow, we have
p
dp
dp
= constant or
= γ
γ
p
p
ρ
For perfect gas, we have p = ρRT or
dρ
dT
dp
=
+
T
ρ
p
From energy equation, we have cpT +
V2
+ constant
2
Differentiating throughout, we get
cpdT + VdV = 0 or
FG γR IJ dT + VdV = 0
H γ − 1K
γR dT
dV
+
=0
2
V
γ −1 V
or,
Also, sonic velocity,
dharm
\M-therm\Th16-2.pm5
C=
FG3 c = γR IJ
γ − 1K
H
p
...(i)
γRT
∴ γR =
C2
T
884
ENGINEERING THERMODYNAMICS
C2
in eqn. (i), we get
T
Substituting the value of γR =
C2
dT dV
×
+
=0
V
(γ − 1)T V 2
or,
FG3 M = V IJ
H
CK
1
dT dV
×
+
=0
2
T
V
(γ − 1) M
From the Mach number relationship
M=
V
γRT
(where
γRT = C)
dM
dV
1 dT
=
–
M
V
2 T
Substituting the value of
...(16.35)
dT
from eqns. (16.34) in eqn. (16.35), we get
T
LM
N
dV
dM
dV
1
−
× (γ − 1) M 2
=
–
V
M
V
2
=
or,
...(16.34)
dV
1 dV
+
× (γ – 1) M2
V
2 V
dM
dV
=
M
V
OP
Q
1
LM1 + γ − 1 M OP or dV =
dM
N 2 Q V LM1 + FG γ − 1IJ M OP M
N H 2 K Q
2
2
...(16.36)
Since the quantity within the bracket is always positive, the trend of variation of velocity
and Mach number is similar. For temperature variation, one can write
LM
MM F
MN GH
2
OP
PP
PQ
− (γ − 1) M
dT
dM
=
...(16.37)
γ −1
T
M
2
1+
M
2
Since the right hand side is negative the temperature changes follow an opposite trend to
that of Mach number. Similarly for pressure and density, we have
IJ
K
OP dM
LM
− γM
MM γ − 1 PP M
N1+ 2 M Q
LM
OP
−M
dM
dρ
P
= M
and,
γ
−
1
MM 1 + FG IJ M PP M
ρ
N H 2 K Q
For changes in area, we have
O
LM
− (1 − M ) P dM
dA
= M
A
MN 1 + γ 2− 1 M PPQ M
dp
=
p
2
2
...(16.38)
2
2
...(16.39)
2
2
dharm
\M-therm\Th16-2.pm5
...(16.40)
885
COMPRESSIBLE FLOW
The quantity within the brackets may be positive or negative depending upon the magnitude of Mach number. By integrating eqn. (16.40), we can obtain a relationship between the
critical throat area Ac, where Mach number is unity and the area A at any section where M >
< 1
LM
MN
OP
PQ
γ +1
A
1 2 + (γ − 1) M 2 2( γ − 1)
=
Ac
γ +1
M
...(16.41)
Example 16.12. The pressure leads from Pitot-static tube mounted on an aircraft were
connected to a pressure gauge in the cockpit. The dial of the pressure gauge is calibrated to read
the aircraft speed in m/s. The calibration is done on the ground by applying a known pressure
across the gauge and calculating the equivalent velocity using incompressible Bernoulli’s equation 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 the
density is 0.454 kg/m3 and ambient pressure is 30 kN/m2. The gauge indicates a velocity of
152 m/s. What is the true speed of the aircraft ?
(UPSC)
Sol. Bernoulli’s equation for an incompressible flow is given by,
ρV 2
= constant
2
The stagnation pressure (ps) created at Pitot-static tube,
p+
ρ0V0 2
(neglecting compressibility effects)
2
p0 = 30 kN/m2, V0 = 152 m/s, ρ0 = 1.224 kg/m3
ps = p0 +
Here
...(i)
...(Given)
2
1.224 × 152
×10–3 = 44.139 kN/m2
2
Neglecting compressibility effect, the speed of the aircraft when
ρ0 = 0.454 kg/m3 is given by [using eqn. (i)],
∴
ps = 30 +
44.139 × 103 = 30 × 103 +
0.454 × V02
2
∴
(44.139 − 30) × 103 × 2
= 62286.34
0.454
V0 = 249.57 m/s
Sonic velocity,
C0 =
γRT0 =
Mach number,
M=
V0 249.57
=
= 0.82
C0 304.16
V02 =
or
∴ True speed of aircraft
Hence true speed of aircraft
dharm
\M-therm\Th16-2.pm5
p0
=
ρ0
1.4 ×
30 × 103
= 304.16 m/s
0.454
F M I , neglecting the terms containing higher powers
GH 4 JK
F 0.82 IJ = 1.168
= G1 +
H 4K
Compressibility correction factor = 1 +
of M0 (from eqn 16.19).
γ
=
0
249.57
2
= 230.9 m/s
1.168
= 230.9 m/s (Ans.)
886
ENGINEERING THERMODYNAMICS
Example 16.13. (a) In case of isentropic flow of a compressible fluid through a variable
duct, show that
L + 1 γ − 1) M OP
1 M1 2 (
A
= M M
A
MN 12 ( γ + 1) PPQ
2
γ +1
2( γ − 1)
c
where γ is the ratio of specific heats, M is the Mach number at a section whose area is A and Ac is
the critical area of flow.
(b) A supersonic nozzle is to be designed for air flow with Mach number 3 at the exit
section which is 200 mm in diameter. The pressure and temperature of air at the nozzle exit are
to be 7.85 kN/m2 and 200 K respectively. Determine the reservoir pressure, temperature and the
throat area. Take : γ = 1.4.
(U.P.S.C. Exam.)
Sol. (a) Please Ref. to Art. 16.9.
(b) Mach number,
M=3
Area 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 K
Reservoir pressure, (p)res. :
From eqn. (16.17),
or,
2
F γ IJ
γ − 1K
2
1.4
1.4 − 1
LM1 + FG γ − 1IJ M OPGH
(p)
= (p)
N H 2 K Q
FG
H
L
O
1.4 − 1I
F
(p)
= 7.85 M1 + G
×3 P
J
N H 2 K Q
res.
nozzle
res.
IJ
K = 288.35 kN/m2 (Ans.)
Reservoir temperature, (T)res. :
From eqn. (16.22),
or,
(T)res.
Throat area (critical), Ac :
From eqn. (16.41),
or,
or,
LM FG γ − 1IJ M OP
N H 2 K Q
L F 1.4 − 1IJ × 3 OP = 560 K (Ans.)
= 200 M1 + G
N H 2 K Q
2
(T)res. = (T)nozzle 1 +
2
LM
OP
MN
PQ
0.0314
. − 1) 3 O
1 L 2 + (14
=
M
PPQ
A
3 MN
1.4 + 1
γ +1
A
1 2 + (γ − 1) M 2 2( γ − 1)
=
Ac
M
γ +1
1.4 + 1
2 2 (1.4 − 1)
c
Ac =
or
0.0314 1
= (2.333)3 = 4.23
Ac
3
0.0314
= 0.00742 m2 (Ans.)
4.23
16.10. FLOW THROUGH LAVAL NOZZLE (CONVERGENT-DIVERGENT NOZZLE)
Laval nozzle is a convergent-divergent nozzle (named after de Laval, the swedish scientist
who invented it) in which subsonic flow prevails in the converging section, critical or transonic
conditions in the throat and supersonic flow in the diverging section.
dharm
\M-therm\Th16-2.pm5
887
COMPRESSIBLE FLOW
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).
Throat
Inlet
(p1) 1
Flow
Exit
Receiver
2
3
(a)
a
b
c
d
p1
pc
e
p3
f
Critical
pc = 0.528 p1
Shock wave
fronts
Inlet
Throat
j
Exit
(b)
Fig. 16.10. (a) Laval nozzle (convergent-divergent nozzle) ; (b) Pressure distribution through
a convergent-divergent nozzle with flow of compressible fluid.
— When the receiver pressure is reduced, flow will occur through the nozzle. As long as the
value 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 is
shown by line ‘b’.
— With further reduction in p3, a stage is reached when p2 is equal to critical pressure
pc = 0.528 p1, at this line M = 1 in the throat. This condition is shown by line ‘c’. Flow is
subsonic 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 in
throat is sonic, downstream it is supersonic. Somewhere in the diverging section a
shock wave occurs and flow changes to subsonic (curve d). The flow across the shock is
non-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°C
temperature. The air flows from the tank to the atmosphere through a convergent nozzle. If the
diameter 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)
dharm
\M-therm\Th16-2.pm5
888
ENGINEERING THERMODYNAMICS
p1 = 284 kN/m2 (gauge)
= 284 + 100 = 384 kN/m2 (absolute)
Temperature in the tank, T1 = 24 + 273 = 297 K
Diameter at the outlet of the nozzle, D = 20 mm = 0.02 m
Sol. Pressure in the tank,
π
× 0.022 = 0.0003141 m2
4
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 :
∴
Area, A2 =
Equation of state is given by p = ρRT or ρ =
p
RT
p1
384 × 103
=
= 4.5 kg/m3
RT1
287 × 297
The fluid parameters in the tank correspond to the stagnation values, and maximum flow
rate of air is given by,
∴
ρ1 =
mmax = 0.685 A2
...[Eqn. (16.32)]
p1ρ1
= 0.685 × 0.0003141
384 × 103 × 4.5 = 0.283 kg/s
Hence 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/m2 (abs.)
Temperature inside the vessel, T1 = 20 + 273 = 293 K
Pressure at the outlet of the nozzle, p2 = 1750 kN/m2 (abs.)
R = 287 J/kg K, γ = 1.4
Velocity of air, V2 :
L F p I OP
F
2γ I p M
V = G
H γ − 1JK ρ MM1 − GH p JK PP
N
Q
I
p F
p
From equation of state : = RT J
ρ =
G
RT H
ρ
K
2
where,
1
2
1
1
γ −1
γ
...[Eqn. (16.27)]
1
1
1
=
2500 × 103
= 29.73 kg/m3
287 × 293
Substituting the values in the above equation, we get
F 2 × 1.4 IJ × 2500 × 10 LM1 − FG 1750 IJ
V = G
H 1.4 − 1K 29.73 MMN H 2500 K
3
2
dharm
\M-therm\Th16-2.pm5
1.4 − 1
1.4
OP
PP
Q
889
COMPRESSIBLE FLOW
=
7 × 84090 (1 − 0.903) = 238.9 m/s
V2 = 238.9 m/s (Ans.)
i.e.,
Example 16.16. A tank fitted with a convergent nozzle contains air at a temperature of
20°C. The diameter at the outlet of the nozzle is 25 mm. Assuming adiabatic flow, find the mass
rate 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 K
Diameter at the outlet of the nozzle, D2 = 25 mm = 0.025 m
Area,
A2 = π/4 × 0.0252 = 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.) :
p1
140 × 103
=
= 1.665 kg/m3
RT1
287 × 293
p1 = 140 kN/m2 (abs.)
p2 = atmospheric pressure = 100 kN/m2
ρ1 =
Pressure at the nozzle,
p2
100
=
= 0.7143
p1
140
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
∴ Pressure ratio,
m = A2
2γ
p1ρ1
γ −1
LMF p I F p I
MMGH p JK − GH p JK
N
2
2
γ
1
2
γ +1
γ
1
OP
PP
Q
LM
F 2 × 14. I
.
= 0.0004908 G 1.4 − 1J × 140 × 10 × 1665
H
K
MN(0.7143)
3
= 0.0004908
or
2
1.4 + 1
1.4 − (0.7143) 1.4
OP
PQ
1631700 (0.7143)1.4285 − (0.7143)1.7142
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/m2 (abs.)
p2 = pressure at the nozzle = atmospheric pressure = 100 kN/m2
p2
100
=
= 0.33.
p1
300
The pressure ratio being less than the critical ratio 0.528, the flow in the nozzle will be
sonic, the flow rate is maximum and is given by eqn. (16.32), as
∴ Pressure ratio,
mmax = 0.685 A2 p1ρ1
where,
∴
dharm
\M-therm\Th16-2.pm5
ρ1 =
p1
300 × 103
=
= 3.567 kg/m3
287 × 293
RT1
mmax = 0.685 × 0.0004908
300 × 103 × 3.567 = 0.3477 kg/s (Ans.)
890
ENGINEERING THERMODYNAMICS
Example 16.17. At some section in the convergent-divergent nozzle, in which air is flowing, pressure, velocity, temperature and cross-sectional area are 200 kN/m2, 170 m/s, 200°C and
1000 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/m2
Velocity of air,
V1 = 170 m/s
Temperature,
T1 = 200 + 273 = 473 K
Cross-sectional area,
A1 = 1000 mm2 = 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) :
V12
2 × cp
Ts = T1 +
Stagnation temperature,
= 473 +
ps
=
p1
Also,
FT I
GH T JK
s
170 2
= 487.45 K (or 214.45°C) (Ans.)
2 × (1.0 × 1000)
γ
γ −1
1
F 487.45 IJ
=G
H 473 K
1.4
1.4 − 1
= 1.111
∴ Stagnation pressure,
ps = 200 × 1.111 = 222.2 kN/m2 (Ans.)
(ii) Sonic velocity and Mach number at this section :
C1 =
Sonic velocity,
γRT1 =
1.4 × 287 × 473 = 435.9 m/s (Ans.)
V1
170
=
= 0.39 (Ans.)
C1 435.9
(iii) Velocity, Mach number and flow area at outlet section where pressure is
110 kN/m2 :
...(Given)
Pressure at outlet section, p2 = 110 kN/m2
Mach number,
M1 =
LM FG IJ OP
From eqn (16.17),
N H K Q
222.2
L F 1.4 − 1IJ M OP = (1 + 0.2 M )
= M1 + G
110
N H 2 K Q
F 222.2 IJ = 1.222
(1 + 0.2 M ) = G
H 110 K
.
− 1I
F 1222
M = G
H 0.2 JK = 1.05 (Ans.)
γ
ps
γ −1
γ −1
M22
= 1+
p1
2
0
or,
2
2
1
3.5
1/ 2
or,
dharm
\M-therm\Th16-2.pm5
2
1.4
2 1.4 − 1
2 3.5
0
891
COMPRESSIBLE FLOW
T2
=
Ts
Also,
or,
Fp I
GH p JK
2
γ −1
γ
s
F 110 IJ
= G
H 222.2 K
1.4 − 1
1.4
= 0.818
T2 = 0.818 × 487.45 = 398.7 K
Sonic velocity at outlet section,
C2 =
γRT2 =
1.4 × 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
ρ1A1V1 = ρ2A2V2 or
i.e.,
p2
p1
A1V1 =
A2V2
RT
RT1
2
∴ Flow area at the outlet section,
A2 =
p1 A1V1T2 200 × 0.001 × 170 × 398.7
=
= 6.199 × 10–4 m2
T1 p2V2
473 × 110 × 420.26
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 of
the nozzle :
At throat, critical conditions prevail, i.e. the flow velocity becomes equal to the sonic velocity and Mach number attains a unit value.
From eqn. (16.22),
2
or,
t
t
Hence
Also,
or,
LM FG IJ OP
N H K Q
487.45
L F 1.4 − 1IJ × 1 OP = 1.2 or T = 406.2 K
= M1 + GH
T
2 K
N
Q
Ts
γ −1
Mt 2
= 1+
Tt
2
Tt = 406.2 K (or 133.2°C). Ans.
F I
GH JK
pt
Tt
=
Ts
Ts
γ
γ −1
pt
or
=
222.2
FG 406.2 IJ
H 487.45 K
1.4
1.4 − 1
= 0.528
pt = 222.2 × 0.528 = 117.32 kN/m2. Ans.
Sonic velocity (corresponding to throat conditions),
Ct =
γRTt =
1.4 × 287 × 406.2 = 404 m/s
∴ Flow velocity,
Vt = Mt × Ct = 1 × 404 = 404 m/s
By continuity equation, we have : ρ1A1V1 = ρt AtVt
pt
p1
A1V1 =
AtVt
RT
RT1
t
or,
∴ Flow area at throat, At =
Hence,
dharm
\M-therm\Th16-2.pm5
p1 A1V1Tt 200 × 0.001 × 170 × 406.2
=
= 6.16 × 10–4 m2
T1 pt Vt
473 × 117.32 × 404
At = 6.16 × 10–4 m2 or 616 mm2 (Ans.)
892
ENGINEERING THERMODYNAMICS
16.11. SHOCK WAVES
Whenever a supersonic flow (compressible) abruptly changes to subsonic flow, a shock
wave (analogous to hydraulic jump in an open channel) is produced, resulting in a sudden rise in
pressure, density, temperature and entropy. This occurs due to pressure differentials and when
the Mach number of the approaching flow M1 > 1. A shock wave is a pressure wave of finite
thickness, of the order of 10–2 to 10–4 mm in the atmospheric pressure. A shock wave takes place in
the diverging section of a nozzle, in a diffuser, throat of a supersonic wind tunnel, in front of
sharp 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 Wave
Consider 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 after
a shock wave takes place (M2 < 1).
Normal shock wave
p1, r1, T1
p2, r2, T2
V1 > C1
V2 < C2
M>1
M<1
Fig. 16.11. Normal shock wave.
For analysing a normal shock wave, use will be made of the continuity, momentum and
energy equations.
Assume unit area cross-section, A1 = A2 = 1.
...(i)
Continuity equation :
m = ρ1V1 = ρ2V2
Momentum equation :
ΣFx = p1A1 – p2A2 = m (V2 – V1) = ρ2A2V22 – ρ1A1V12
for A1 = A2 = 1, the pressure drop across the shock wave,
p1 – p2 = ρ2V22 – ρ1V12
...(ii)
p1 + ρ1V12 = p2 + ρ2V22
Consider the flow across the shock wave as adiabatic.
Energy equation :
FG γ IJ p + V = FG γ IJ p + V
H γ − 1K ρ 2 H γ − 1K ρ 2
1
2
1
1
2
2
2
...[Eqn. (16.7)]
2
(z1 = z2, duct being in horizontal position)
or,
FG
H
IJ
K
γ
p2 p1
V 2 − V22
−
= 1
γ − 1 ρ2 ρ1
2
Combining continuity and momentum equations [refer eqns. (i) and (ii)], we get
p1 +
dharm
\M-therm\Th16-2.pm5
(ρ1V1)2
(ρ V )2
= p2 + 2 2
ρ1
ρ2
...(iii)
...(16.42)
893
COMPRESSIBLE FLOW
This equation is ‘known as Rankine Line Equation.
Now combining continuity and energy equations [refer eqns. (i) and (iii)], we get
γ
γ −1
F p I + (ρ V ) = γ F p I + (ρ V )
GH ρ JK 2ρ
G J 2ρ
γ −1 Hρ K
1 1
2
1
1
1
2
2
2
2 2
2
2
2
...(16.43)
This equation is called Fanno Line Equation.
Further combining eqns. (i), (ii) and (iii) and solving for
p2
=
p1
FG γ + 1IJ ρ − 1
H γ − 1K ρ
FG γ + 1IJ − ρ
H γ − 1K ρ
p2
, we get
p1
2
1
2
1
Solving for density ratio
...(16.44)
ρ2
, the same equations yield
ρ1
FG γ + 1IJ p
V
H γ − 1K p
ρ
=
=
V
FG γ + 1IJ + p
ρ
H γ − 1K p
2
1
1
2
1+
2
1
...(16.45)
2
1
The eqns. (16.44) and (16.45) are called Ranking-Hugoniot equations.
One can also express
T2
p2 V2 ρ2
,
,
and
in terms of Mach number as follows :
T1
p1 V1 ρ1
2 γ M12 − (γ − 1)
p2
=
(γ + 1)
p1
...(16.46)
V1
(γ + 1) M12
ρ
= 2 =
V2
(γ − 1) M12 + 2
ρ1
...(16.47)
T2
[(γ − 1) M12 + 2] [2γ M12 − (γ − 1)]
=
T1
(γ + 1)2 M12
...(16.48)
By algebraic manipulation the following equation between M1 and M2 can be obtained.
M22 =
(γ − 1) M12 + 2
2γM12 − (γ − 1)
...(16.49)
Example 16.18. For a normal shock wave in air Mach number is 2. If the atmospheric
pressure and air density are 26.5 kN/m2 and 0.413 kg/m3 respectively, determine the flow conditions 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,
Atmospheric pressure,
Air density,
dharm
\M-therm\Th16-2.pm5
M1 = 2
p1 = 26.5 kN/m2
ρ1 = 0.413 kg/m3
894
ENGINEERING THERMODYNAMICS
Mach number, M2 :
M22 =
=
∴
Pressure, p2 :
(γ − 1) M12 + 2
2γM12 − (γ − 1)
...[Eqn. (16.49)]
(1.4 − 1) × 2 2 + 2
3.6
=
= 0.333
2 × 1.4 × 22 − (1.4 − 1) 11.2 − 0.4
M2 = 0.577 (Ans.)
p2
2 γM12 − (γ − 1)
=
p1
(γ + 1)
=
...[Eqn. (16.46)]
2 × 1.4 × 2 2 − (1.4 − 1) 11.2 − 0.4
=
= 4.5
(1.4 + 1)
2.4
p2 = 26.5 × 4.5 = 119.25 kN/m2 (Ans.)
∴
Density, ρ2 :
ρ2
(γ + 1) M12
=
ρ1
(γ − 1) M12 + 2
=
(1.4 + 1)22
9.6
= 2.667
=
2
+2
1
.
6
(1.4 − 1)2 + 2
ρ2 = 0.413 × 2.667 = 1.101 kg/m3 (Ans.)
∴
Temperature, T1 :
Since p1 = ρ1 RT1,
...[Eqn. (16.47)]
∴ T1 =
p1
26.5 × 103
=
= 223.6 K or – 49.4°C (Ans.)
ρ1 R
0.413 × 287
Temperature, T2 :
T2
[(γ − 1) M12 + 2] [2γM12 − (γ − 1)] [(1.4 − 1)22 + 2] [2 × 1.4 × 22 − (1.4 − 1)]
=
=
T1
(γ + 1)2 M12
(1.4 + 1)2 22
(16
. + 2)(11.2 − 0.4)
= 1.6875
23.04
∴
T2 = 223.6 × 1.6875 = 377.3 K or 104.3°C (Ans.)
Velocity, V1 :
=
C1 =
Since
γRT1 =
1.4 × 287 × 223.6 = 299.7 m/s
V1
= M1 = 2 ∴ V1 = 299.7 × 2 = 599.4 m/s (Ans.)
C1
Velocity, V2 :
C2 =
Since
dharm
\M-therm\Th16-2.pm5
γRT2 =
1.4 × 287 × 377.3 = 389.35 m/s
V2
= M2 = 0.577 ∴ V2 = 389.35 × 0.577 = 224.6 m/s (Ans.)
C2
895
COMPRESSIBLE FLOW
16.11.2. Oblique Shock Wave
As shown in Fig. 16.12, when a supersonic flow
undergoes a sudden turn through a small angle α (positive), an oblique wave is established at the corner. In
comparison with normal shock waves, the oblique shock
waves, being weaker, are preferred.
Shock
Supersonic
flow
The shock waves should be avoided or made as
weak as possible, since during a shock wave conversion
of mechanical energy into heat energy takes place.
a
Sharp corner
Fig. 16.12. Oblique shock wave.
16.11.3. Shock Strength
The strength of shock is defined as the ratio of pressure rise across the shock to the upstream
pressure.
i.e.
Strength of shock
=
p2 − p1 p2
=
–1
p1
p1
=
2 γM12 − (γ − 1)
2 γM12 − (γ − 1) − (γ + 1)
− 1=
γ+1
γ+1
=
2 γM12 − γ + 1 − γ − 1 2 γM12 − 2 γ
2γ
=
=
(M12 – 1)
γ+1
γ+1
γ +1
Hence, strength of shock =
2γ
(M12 – 1)
γ +1
...(16.50)
Example 16.19. In a duct in which air is flowing, a normal shock wave occurs at a Mach
number 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 = 1.5
Upstream pressure,
p1 = 170 kN/m2
Upstream temperature, T1 = 23 + 273 = 296 K
γ = 1.4
(i) Pressure, temperature and Mach number downstream of the shock :
p2
2 γM12 − (γ − 1)
=
p1
γ +1
=
∴
2 × 1.4 × 1.52 − (1.4 − 1) 6.3 − 0.4
=
= 2.458
1.4 + 1
2.4
p2 = 170 × 2.458 = 417.86 kN/m2. Ans.
T2
[(γ − 1) M12 + 2] [2γM12 − (γ − 1)]
=
T1
(γ + 1)2 M12
dharm
\M-therm\Th16-2.pm5
...[Eqn. (16.46)]
...[Eqn. (16.48)]
896
ENGINEERING THERMODYNAMICS
=
∴
[(1.4 − 1) × 1.52 + 2][2 × 1.4 × 1.52 − (1.4 − 1)]
2
(1.4 + 1) × 1.5
2
=
2.9 × 5.9
= 1.32
12.96
T2 = 296 × 1.32 = 390.72 K or 117.72°C. Ans.
M22 =
=
(γ − 1) M12 + 2
2γM12 − (γ − 1)
...[Eqn. (16.49)]
(1.4 − 1) × 1.52 + 2
2 × 1.4 × 1.52 − (1.4 − 1)
=
2.9
= 0.49
5.9
∴
M2 = 0.7. Ans.
(ii) Strength of shock :
Strength of shock
=
p2
– 1 = 2.458 – 1 = 1.458. Ans.
p1
HIGHLIGHTS
1.
2.
A compressible flow is that flow in which the density of the fluid changes during flow.
The characteristic equation of state is given by :
p
= RT
ρ
3.
where p = absolute pressure, N/m2,
ρ = density of gas, kg/m3,
R = characteristic gas constant, J/kg K, and
R = absolute temperature (= t°C + 273).
The pressure and density of a gas are related as :
For isothermal process :
p
For adiabatic process :
4.
p
= constant
ρ
ργ
= constant.
The continuity equation for compressible flow is given as :
ρAV = constant
dA dV
dρ
+
+
=0
A
V
ρ
5.
6.
... in differential form.
For compressible fluids Bernoulli’s equation is given as :
p
V2
loge p +
+ z = constant
ρg
2g
... for isothermal process
FG γ IJ p + V 2 + z = constant
H γ − 1K ρg 2 g
... for adiabatic process.
Sonic velocity is given by :
C=
dharm
\M-therm\Th16-2.pm5
dp
=
dρ
K
ρ
... in terms of bulk modulus
897
COMPRESSIBLE FLOW
7.
Mach number,
(i) Subsonic flow :
(ii) Sonic flow :
(iii) Supersonic flow :
C=
p
=
ρ
RT
... for isothermal process
C=
γp
= γRT
ρ
... for adiabatic process.
V
C
M < 1, V < C... disturbance always moves ahead of the projectile
M = 1, V = C... disturbance moves along the projectile
M > 1, V > C... The projectile always moves ahead of the disturbance.
M=
C
1
=
.
V M
The pressure, temperature and density at a point where velocity is zero are called stagnation pressure (ps),
temperature, (Ts) and stagnation density ρs. Their values are given as :
Mach angle is given by : sin α =
8.
γ
L F γ − 1IJ M02 OP γ − 1
p = p M1 + G
N H 2 K Q
s
o
1
L F γ − 1IJ M02 OP γ − 1
ρ = ρ M1 + G
N H 2 K Q
L F γ − 1IJ M02 OP
T = T M1 + GH
N 2 K Q
9.
s
o
s
o
where p0, ρ0 and To are the pressure, density and temperature at any point O in the flow.
Area-velocity relationship for compressible fluid is given as :
(i) Subsonic flow (M < 1) :
(ii) Supersonic flow (M > 1) :
(iii) Sonic flow (M = 1) :
dA
dV
=
(M2 – 1)
V
A
dV
dA
>0;
< O ; dp < 0
V
A
dV
dA
<0;
> 0 ; dp > 0
V
A
(convergent nozzle)
dV
dA
>0;
> 0 ; dp < 0
V
A
(divergent nozzle)
dV
dA
<0;
< 0 ; dp > 0
(convergent diffuser)
V
A
dA
=0
(straight flow passage since dA must be zero)
A
zero
i.e. indeterminate, but when evaluated,
zero
the change of pressure dp = 0, since dA = 0 and the flow is frictionless.
Flow of compressible fluid through a convergent nozzle :
(i) Velocity through a nozzle or orifice fitted to a large tank :
dp =
10.
(divergent diffuser)
L F I γ γ− 1 OP
F
p
2 γ I p1 M
1− G 2J
V = G
M
J
−
γ
ρ
1
H K 1 M H p1 K PP
N
Q
2
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898
ENGINEERING THERMODYNAMICS
(ii) The mass rate of flow is given by :
LMF I 2γ F I γ γ+ 1 OP
F
I
γ
2
p2
p2
m=A G
H γ − 1JK p1ρ1 MMGH p1 JK − GH p1 JK PP
N
Q
2
(iii) Value of
F p2 I for maximum value of mass rate of flow is given by :
GH p1 JK
F p2 I = F 2 I γ γ− 1 = 0.528
GH p1 JK GH γ + 1JK
(when γ = 1.4)
(iv) Value of V2 for maximum rate of flow of liquid is given as,
FG 2γ IJ p1
H γ + 1K ρ (= C )
V2 =
1
2
(v) Maximum rate of flow of fluid through nozzle,
mmax
2
γ +1
FG 2γ IJ p1ρ1 LMMFG 2γ IJ γ − 1 − FG 2 IJ γ − 1 OPP
= A H γ − 1K
H γ + 1K P
MNH γ + 1K
Q
2
For air, substituting γ = 1.4, we get
mmax = 0.685 A2
11.
p1ρ1
If the pressure ratio is less than 0.528, the mass rate of flow of the fluid is always corresponding to the
pressure ratio of 0.528. But if the pressure ratio is more than 0.528, the mass rate of flow of fluid is
corresponding to the given pressure ratio.
Whenever a supersonic flow (compressible) changes to subsonic flow, a shock wave (analogous to hydraulic jump in an open channel) is produced, resulting in a sudden rise in pressure, density, temperature and
entropy.
p1 +
γ
γ −1
One can also express
(ρ V )2
(ρ1V1)2
= p2 + 2 2
ρ2
ρ1
F p1 I + (ρ1V1)2 = γ F p2 I + (ρ2V2 )2
GH ρ1 JK 2ρ21 γ − 1 GH ρ2 JK 2ρ22
FG γ + 1IJ ρ2 − 1
U|
p2
γ − 1K ρ 1
H
||
=
p1
FG γ + 1IJ − ρ2
||
H γ − 1K ρ1
|V
F
γ + 1I p2 |
1+ G
V1
ρ2
H γ − 1JK p1 ||
=
=
V2
ρ1
FG γ + 1IJ + p2 ||
H γ − 1K p1 |W
... Fanno line Equation
...Rankinge-Hugoniot Equations
p2 V2 ρ2
T
,
,
and 2 in terms of Mach number as follows :
p1 V1 ρ1
T1
2γM12 − (γ − 1)
p2
=
γ +1
p1
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... Ranking Line Equation
...(i)
899
COMPRESSIBLE FLOW
V1
ρ2
(γ + 1) M12
V2 = ρ = (γ − 1) M 2 + 2
1
...(ii)
1
[(γ − 1) M12 + 2] [2 γM12 − (γ − 1)]
T2
=
(γ + 1) 2 M12
T1
M22 =
Also
(γ − 1) M12 + 2
2γM12 − ( γ − 1)
...(iii)
.
OBJECTIVE TYPE QUESTIONS
1.
2.
3.
4.
Choose the Correct Answer :
All real fluids are
(a) incompressible
(b) compressible to some extent
(c) compressible to any extent
(d) none of the above.
A change in the state of a system at constant volume is called
(a) isobaric process
(b) isochoric process
(c) isothermal process
(d) adiabatic process.
A process during which no heat is transferred to or from the gas is called an
(a) isochoric process
(b) isobaric process
(c) adiabatic process
(d) isothermal process.
An adiabatic process is one which follows the relation
(a)
(c)
5.
6.
7.
8.
9.
10.
p
= constant
ρ
p
ρn
= constant (n ≠ γ)
(b)
p
ργ
= constant
(d) v = constant.
An isentropic flow is one which is
(a) isothermal
(b) adiabatic
(c) adiabatic and irreversible
(d) adiabatic and reversible.
Indica upto what Mach number can a fluid flow be considered incompressible ?
(a) 0.1
(b) 0.3
(c) 0.8
(d) 1.0.
Which of the following is the basic equation of compressible fluid flow ?
(a) Continuity equation
(b) Momentum equation
(c) Energy equation
(d) Equation of state
(e) All of the above.
The velocity of disturbance in case of fluids is ...... the velocity of the disturbance in solids.
(a) less than
(b) equal to
(c) more than
(d) none of the above.
Sonic velocity (C) for adiabatic process is given as
(a) C =
γRT 3
(b) C =
(c) C =
γ 2 RT
(d) C = γRT.
γRT
where γ = ratio of specific heats, R = gas constant, T = temperature.
The flow is said to be subsonic when Mach number is
(a) equal to unity
(b) less than unity
(c) greater than unity
(d) none of the above.
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900
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
ENGINEERING THERMODYNAMICS
The region outside the Mach cone is called
(a) zone of action
(b) zone of silence
(c) control volume
(d) none of the above.
A stagnation point is the point on the immersed body where the magnitude of velocity is
(a) small
(b) large
(c) zero
(d) none of the above.
A convergent-divergent nozzle is used when the discharge pressure is
(a) less than the critical pressure
(b) equal to the critical pressure
(c) more than the critical pressure
(d) none of the above.
At critical pressure ratio, the velocity at the throat of a nozzle is
(a) equal to the sonic speed
(b) less than the sonic speed
(c) more than the sonic speed
(d) none of the above.
Laval nozzle is a
(a) convergent nozzle
(b) divergent nozzle
(c) convergent-divergent nozzle
(d) any of the above.
A shock wave is produced when
(a) a subsonic flow changes to sonic flow
(b) a sonic flow changes to supersonic flow
(c) a supersonic flow changes to subsonic flow
(d) none of the above.
The sonic velocity in a fluid medium is directly proportional to
(a) Mach number
(b) pressure
(c) square root of temperature
(d) none of the above.
The stagnation pressure (ps) and temperature (Ts) are
(a) less than their ambient counterparts
(b) more than their ambient counterparts
(c) the same as in ambient flow
(d) none of the above.
Across a normal shock
(a) the entropy remains constant
(b) the pressure and temperature rise
(c) the velocity and pressure decrease
(d) the density and temperature decrease.
A normal shock wave
(a) is reversible
(b) is irreversible
(c) is isentropic
(d) occurs when approaching flow is supersonic.
The sonic speed in an ideal gas varies
(a) inversely as bulk modulus
(b) directly as the absolute pressure
(c) inversely as the absolute temperature
(d) none of the above.
In a supersonic flow, a diffuser is a conduit having
(a) gradually decreasing area
(b) converging-diverging passage
(c) constant area throughout its length
(d) none of the above.
Choking of a nozzle fitted to a pressure tank containing gas implies
(a) sonic velocity at the throat
(b) increase of the mass flow rate
(c) obstruction of flow
(d) all of the above.
A shock wave which occurs in a supersonic flow represents a region in which
(a) a zone of silence exists
(b) there is no change in pressure, temperature and density
(c) there is sudden change in pressure, temperature and density
(d) velocity is zero.
Which of the following statements regarding a normal shock is correct ?
(a) It occurs when an abrupt change takes place from supersonic into subsonic flow condition
(b) It causes a disruption and reversal of flow pattern
(c) It may occur in sonic or supersonic flow
(d) None of the above.
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901
COMPRESSIBLE FLOW
26.
For compressible fluid flow the area-velocity relationship is
(a)
dV
dA
=
(1 – M2)
V
A
(b)
dV
dA
=
(C2 – 1)
V
A
dV
dV
dA
dA
=
(M2 – 1)
(d)
=
(1 – V2).
V
V
A
A
The sonic velocity is largest in which of the following ?
(a) Water
(b) Steel
(c) Kerosene
(d) Air.
Which of the following expressions does not represent the speed of sound in a medium ?
(c)
27.
28.
29.
(a)
K
ρ
(c)
K
p
ρ
(b)
γRT
(d)
dp
.
dρ
The differential equation for energy in isentropic flow is of the form
(a)
dp
dV
dA
+
+
=0
ρ
V
A
(c) 2VdV +
(b) VdV +
dp
=0
ρ
dp
=0
ρ
(d) dp + d (ρV2) = 0.
30.
Which of the following statements is incorrect ?
(a) A shock wave occurs in divergent section of a nozzle when the compressible flow changes abruptly
from supersonic to subsonic state
(b) A plane moving at supersonic state is not heard by the stationary observer on the ground until it passes
him because zone of disturbance in Mach cone trails behind the plane
(c) A divergent section is added to a convergent nozzle to obtain supersonic velocity at the throat
(d) none of the above.
1.
8.
15.
22.
29.
(b)
(a)
(c)
(a)
(b)
ANSWERS
2.
9.
16.
23.
30.
(b)
(b)
(c)
(d)
(c).
3.
10.
17.
24.
(c)
(b)
(c)
(c)
4.
11.
18.
25.
(b)
(b)
(b)
(a)
5.
12.
19.
26.
(d)
(c)
(b)
(c)
6.
13.
20.
27.
(b)
(a)
(d)
(b)
7.
14.
21.
28.
(e)
(a)
(d)
(c)
THEORETICAL QUESTIONS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Differentiate between compressible and incompressible flows.
Give the examples when liquid is treated as a compressible fluid.
When is the compressibility of fluid important ?
What is the difference between isentropic and adiabatic flows ?
What is the relation between pressure and density of a compressible fluid for (a) isothermal process
(b) adiabatic process ?
Obtain an expression in differential form for continuity equation for one-dimensional compressible flow.
Derive an expression for Bernoulli’s equation when the process is adiabatic.
How are the disturbances in compressible fluid propagated ?
What is sonic velocity ? On what factors does it depend ?
What is Mach number ? Why is this parameter so important for the study of flow of compressible fluids ?
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\M-therm\Th16-2.pm5
902
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
ENGINEERING THERMODYNAMICS
Prove that velocity of sound wave in a compressible fluid is given by : C = k/ρ , where k and ρ are the bulk
modulus and density of fluid respectively.
Define the following terms :
(i) Subsonic flow (ii) Sonic flow (iii) Supersonic flow, (iv) Mach cone.
What is silence zone during the disturbance which propagates when an object moves in still air ?
What is stagnation point of an object immersed in fluid ?
What is stagnation pressure ?
What are static and stagnation temperatures ?
Derive an expression for mass flow rate of compressible fluid through an orifice or nozzle fitted to a large
tank. What is the condition for maximum rate of flow ?
What is the critical pressure ratio for a compressible flow through a nozzle ? On what factors does it
depend ?
Describe compressible flow through a convergent-divergent nozzle. How and where does the shock wave
occur in the nozzle ?
What do you mean by compressibility correction factor ?
How is a shock wave produced in a compressible fluid ? What do you mean by the term “Shock strength” ?
UNSOLVED EXAMPLES
1.
2.
3.
4.
5.
6.
7.
8.
9.
A 100 mm diameter pipe reduces to 50 mm diameter through a sudden contraction. When it carries air at
20.16°C under isothermal conditions, the absolute pressures observed in the two pipes just before and after
the contraction are 400 kN/m2 and 320 kN/m2 respectively. Determine the densities and velocities at the
two sections. Take R = 290 J/kg K.
[Ans. 4.7 kg/m3 ; 3.76 kg/m3 ; 39.7 m/s ; 198.5 m/s]
A gas with a velocity of 300 m/s is flowing through a horizontal pipe at a section where pressure is
60 kN/m2 (abs.) and temperature 40°C. The pipe changes in diameter and at this section the pressure is
90 kN/m2. If the flow of gas is adiabatic find the velocity of gas at this section.
Take : R = 287 J/kg K and γ = 1.4.
[Ans. 113 m/s]
An aeroplane is flying at 21.5 m/s at a low altitude where the velocity of sound is 325 m/s. At a certain point
just outside the boundary layer of the wings, the velocity of air relative to the plane is 305 m/s. If the flow
is frictionless adiabatic determine the pressure drop on the wing surface near this position.
Assume γ = 1.4, pressure of ambient air = 102 kN/m2.
[Ans. 28.46 kN/m2]
A jet propelled aircraft is flying at 1100 km/h. at sea level. Calculate the Mach number at a point on the
aircraft where air temperature is 20°C.
Take : R = 287 J/kg K and γ = 1.4.
[Ans. 0.89]
An aeroplane is flying at an height of 20 km where the temperature is – 40°C. The speed of the plane is
corresponding to M = 1.8. Find the speed of the plane.
Take : R = 287 J/kg K, γ = 1.4.
[Ans. 1982.6 km/h]
Find the velocity of bullet fired in standard air if its Mach angle is 30°.
[Ans. 680.4 m/s]
Air, thermodynamic state of which is given by pressure p = 230 kN/m2 and temperature = 300 K is moving
at a velocity V = 250 m/s. Calculate the stagnation pressure if (i) compressibility is neglected and (ii)
compressibility is accounted for.
Take γ = 1.4 and R = 287 J/kg K.
[Ans. 313 kN/m2, 323 kN/m2]
2
A large vessel, fitted with a nozzle, contains air at a pressure of 2943 kN/m (abs.) and at a temperature of
20°C. If the pressure at the outlet of the nozzle is 2060 kN/m2 (abs.) find the velocity of air flowing at the
outlet of the nozzle.
Take : R = 287 J/kg K and γ = 1.4
[Ans. 239.2 m/s]
Nitrogen gas (γ = 1.4) is released through a 10 mm orifice on the side of a large tank in which the gas is at
a pressure of 10 bar and temperature 20°C. Determine the mass flow rate if (i) the gas escapes to atmosphere
(1 bar) ; (ii) the gas is released to another tank at (a) 5 bar, (b) 6 bar.
[Ans. (i) 0.183 kg/s ; (ii) 0.183 kg/s ; 0.167 kg/s]
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\M-therm\Th16-2.pm5
COMPRESSIBLE FLOW
10.
11.
12.
13.
14.
903
Air is released from one tank to another through a convergent-divergent nozzle at the rate of 12 N/s. The
supply tank is at a pressure of 400 kN/m2 and temperature 110°C, and the pressure in the receiving tank
is 100 kN/m2. Determine : (i) The pressure, temperature, and Mach number in the constriction, (ii) The
required diameter of constriction, (iii) The diameter of the nozzle at the exit for full expansion, and the
Mach number.
[Ans. (i) 210 kN/m2 ; 319 K, (ii) 43.5 mm ; (iii) 48 mm ; 1.56]
Oxygen flows in a conduit at an absolute pressure of 170 kN/m2. If the absolute pressure and temperature
at the nose of small object in the stream are 200 kN/m2 and 70.16°C respectively, determine the velocity in
the conduit. Take γ = 1.4 and R = 281.43 J/kg K.
[Ans. 175.3 m/s]
Air at a velocity of 1400 km/h has a pressure of 10 kN/m2 vacuum and temperature of 50.16°C. Calculate
local Mach number and stagnation pressure, density and temperature. Take γ = 1.4, R = 281.43 J/kg K and
[Ans. 1.089 ; 192.358 kN/m2 ; 1.708 kg/m3 ; 399.8 K]
barometric pressure = 101.325 kN/m2.
A normal shock wave occurs in a diverging section when air is flowing at a velocity of 420 m/s, pressure
100 kN/m2, and temperature 10°C. Determine : (i) The Mach number before and after the shock, (ii) The
pressure rise, and (iii) The velocity and temperature after the shock.
[Ans. (i) 1.25 ; 0.91 ; (ii) 66 kN/m2, (iii) 292 m/s ; 54°C]
A normal shock wave occurs in air flowing at a Mach number of 1.5. The static pressure and temperature
of the air upstream of the shock wave are 100 kN/m2 and 300 K. Determine the Mach number, pressure
and temperature downstream of the shockwave. Also estimate the shock strength.
[Ans. 0.7 ; 246 kN/m2 ; 396.17 K ; 1.46]
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\M-therm\Th16-2.pm5