Lectures of Week 1
Introduction to Compressible Flows
Course Title: Compressible Flow and Propulsion System
Course Code: (ME-417)
Class Rule
• On Monday, student will be marked absent if come after 8:40 AM, no tolerance
will be provided. (Section C)
• On Wednesday, student will be marked absent if come after 9:25 AM. (Section C)
• On Monday, student will be marked absent if come after 11:35 AM, no tolerance
will be provided. (Section D)
• On Friday, student will be marked absent if come after 12:15 PM, no tolerance
will be provided. (Section D)
• Each and every student is itself responsible for maintaining his 75% attendance.
• I will not mark attendance of any student who is busy either in workshops,
seminars, internships, industrial visits, society activities, FYP work, any type of
illness etc.
• No compromise in these rules.
Course content
• Governing equations for compressible fluid flow: conservation of mass, momentum and energy.
• Sonic velocity and Mach number, difference between incompressible, subsonic and supersonic flow, propagation of
sound waves, equations for perfect gases in terms of Mach number, optical methods of investigation.
• Isentropic flow of a perfect gas, limiting conditions (choking), effect of area change on flow properties, flow in
convergent and convergent-divergent nozzles, Hugoniot equation, applications of isentropic flow.
• Formation of shock waves, Weak and Strong waves, stationary and moving shock waves, working equations for
perfect gases, operating characteristics of converging diverging nozzle, supersonic diffusers and pitot tube.
• Governing equations for oblique shock waves and Prandtl-Meyer flow, Shock Polar, variation of properties across
an oblique shock wave, expansion of supersonic flow over successive corners and convex surfaces.
• Fannoline, friction parameter for a constant area duct, limiting conditions, isothermal flow in long ducts.
• Flow in ducts with heating or cooling, thermal choking due to heating, correlation with shocks.
• Propulsion applications including rocket nozzles, rocket engine staging, supersonic inlets, and exhaust nozzles for
air breathing propulsion systems. Parametric cycle analysis for ramjet, turbojet, turbofan, and turboprop engines.
Course Learning Outcomes
No.
CLO
PLO
Taxonomy
Level
1
Explain different terms of compressible and isentropic
PLO – 1
flows
C2
2
Solve cases of different non-isentropic flows such as
normal / oblique shock and flows with friction or heat PLO – 2
transfer
C3
3
Analyze various shaft power and aircraft gas turbine
PLO – 3
engines
C4
Test book: Gas Dynamics by M. Haluk Aksel
Introduction
Compressible Flow and Propulsion System
Fluid flow with significant
density change
A machine that produces thrust
to push an object forward
Gas dynamics
Gas turbines
Compressible flow
• The course of compressible flow/gas dynamics is concerned with
the causes and the effects arising from the motion of
compressible fluids particularly gases.
• It is branch of more general subject of fluid dynamics.
• Compressible flow involves significant changes in density. It is
encountered in devices that involve the flow of gases at very high
speeds.
• Compressible
flow
combines
fluid
dynamics
and
thermodynamics in that both are necessary to the development of
the required theoretical background.
Compressible flow
• The analysis of flow problems is based on the fundamental principles given
below:
1. Conservation of mass
2. Newton’s second law of motion
3. Conservation of energy
Continuity Equation
• The continuity equation for a control volume is
• For steady flow, any partial derivative with respect to time is zero and the
equation becomes:
• For one-dimensional flow any fluid property will be constant over an entire cross
section.
• Thus both the density and the velocity can be brought out from under the integral
sign.
• If the surface is chosen perpendicular to V, the integral is very simple to
evaluate.
Continuity Equation
• For steady, one-dimensional flow, the continuity equation for a control volume
becomes
• If there is only one section where fluid enters and one section where fluid leaves
the control volume, continuity equation becomes
• An alternative form of the continuity equation can be obtained by differentiating
equation. For steady one-dimensional flow this means that
• Dividing by ρAV yields
Momentum Equation
• The time rate of change of momentum of a fluid mass equals the net force
exerted on it.
• The integral form of equation is
• If there is only one section where fluid enters and one section where fluid leaves
the control volume steady one-dimensional flow, the momentum equation for a
control volume becomes:
Energy Equation
• The first law of thermodynamics is a statement of conservation of energy. For a
system composed of a given quantity of mass that undergoes a process, we say
that
• The transformed equation that is applicable to a control volume is
• With enthalpy, the one-dimensional energy equation for steady-in-the- mean flow
is
• where q and ws represent quantities of heat and shaft work crossing the control
surface per unit mass of fluid flowing.
Sonic Velocity
• A disturbance at a given point creates a region of
compressed molecules that is passed along to its
neighboring molecules and in so doing creates a
traveling wave.
• The speed at which this disturbance is propagated
through the medium is called the wave speed.
• This speed not only depends on the type of medium
and its thermodynamic state but is also a function
of the strength of the wave.
• The speed of waves of very small amplitude is
characteristic only of the medium and its state.
• Sound waves are infinitesimal waves (or weak
pressure pulses) which propagate at the
characteristic sonic velocity.
Sonic Velocity
• Consider a long constant-area tube filled with fluid and having a piston at one
end.
• The fluid is initially at rest. At a certain instant the piston is given an incremental
velocity dV to the left.
• The fluid particles immediately next to the piston are compressed a very small
amount as they acquire the velocity of the piston.
• As the piston (and these compressed particles) continue to move, the next group
of fluid particles is compressed.
• The wave front is observed to propagate through the fluid at the characteristic
sonic velocity of magnitude a.
Sonic Velocity
• All particles between the wave front and the piston are moving with velocity dV
to the left and have been compressed from ρ to ρ + dρ and have increased their
pressure from p to p + dp.
• For the analysis we choose the wave region as a control volume and assume the
wave front as a stationary wave.
Sonic Velocity
• For an observer moving with this control volume, the fluid appears to enter the
control volume through surface area A with speed ‘a’ at pressure p and density ρ.
• The fluid leaves the control volume through surface area A with speed a –dV,
pressure p + dp and density ρ + dρ.
• When the continuity equation is applied to the flow through this control volume,
the result is
(1)
Sonic Velocity
• Since the control volume has infinitesimal thickness, the shear stresses along the
walls can be neglected.
• We shall write the x-component of the momentum equation, taking forces and
velocity as positive if to the right.
• For steady one-dimensional flow:
(2)
• Equations (1) and (2) are now be combined to eliminate dV,
Sonic Velocity
• The derivative dp/dρ is not unique. It depends entirely on the process.
• Thus it should really be written as a partial derivative with the appropriate
subscript.
• Since we are analyzing an infinitesimal disturbance we assume negligible losses
and heat transfer as the wave passes through the fluid.
• Thus the process is both reversible and adiabatic, which means that it is
isentropic. Therefore, equation of sound can properly be written as
• Sound velocity can be expressed in terms of bulk or volume modulus of
elasticity Ev.
Sonic Velocity
• Since air is more easily compressed than water, the speed of sound in air is much
less than it is in water.
• From Equation, we can conclude that if a fluid is truly incompressible, its bulk
modulus would be large and sonic velocity would be high.
• Equation can be simplified for the case of a gas that obeys the perfect
• gas law:
• For perfect gases, sonic velocity is a function of the individual gas and
temperature only. Sonic velocity is a property of the fluid and varies with the
state of the fluid.
Mach Number
• We define the Mach number as
• If the velocity is less than the local speed of sound, M is less than 1 and the flow is
called subsonic.
• If the velocity is greater than the local speed of sound, M is greater than 1 and the
flow is called supersonic.
Wave Propagation
• Consider a point disturbance that is at rest in a fluid. Infinitesimal pressure pulses
are continually being emitted and they travel through the medium at sonic velocity
in the form of spherical wave fronts.
• To simplify matters we keep track of only those pulses that are emitted every
second.
Wave Propagation
• Now consider a similar problem in which the
disturbance is no longer stationary.
• Assume that it is moving at a speed less than
sonic velocity, say a/2.
• Figure shows such a situation at the end of 3
seconds.
• Note that the wave fronts are no longer
concentric. Furthermore, the wave that was
emitted at t = 0 is always in front of the
disturbance itself.
• Therefore, any person, object, or fluid
particle located ahead will feel the wave
fronts pass by and know that the disturbance
is coming.
Wave Propagation
• Next, let the disturbance move at exactly sonic velocity. Figure shows this case in
which all wave fronts coalesce on the left side and move along with the
disturbance.
• In this case, no region upstream is forewarned of the disturbance as the
disturbance arrives at the same time as the wave front
Wave Propagation
• Now suppose the disturbance is moving
at velocity V > a. The wave fronts
coalesce to form a cone with the
disturbance at the apex.
• This is called a Mach cone. The region
inside the cone is called the zone of
action since it feels the presence of the
waves.
• The outer region is called the zone of
silence, as this entire region is unaware
of the disturbance.
• The half-angle at the apex is called the
Mach angle and is given the symbol μ. It
should be easy to see that
Wave Propagation
• In the subsonic case the fluid can “sense” the presence of an object and smoothly
adjust its flow around the object.
• In supersonic flow this is not possible, and thus flow adjustments occur rather
abruptly in the form of shock or expansion waves.
• Since the supersonic and subsonic flows have different characteristics, it is
suitable to use Mach number as a parameter in our basic equations.
Flow Regimes
• It is useful to illustrate different
regimes of compressible flow by
considering an aerodynamic body
in a flowing gas.
• Far upstream of the body, the flow
is uniform with a free stream
velocity of V∞
• Now consider an arbitrary point in
the flow field, where p, T, ρ, and V
are the local pressure, temperature,
density, and velocity at that point.
Flow Regimes
• All of these quantities are point
properties and vary from one point
to another in the flow. The speed of
sound ‘a’ is a thermodynamic
property of the gas and varies from
point to point in the flow.
• If a∞ is the speed of sound in the
uniform free stream, then the ratio
V∞/a∞ defines the free-stream Mach
number M∞.
• Similarly, the local Mach number,
M is defined as M = V/a, and varies
from point to point in the flow
field.
Flow Regimes
• All of these quantities are point
properties and vary from one point
to another in the flow. The speed of
sound ‘a’ is a thermodynamic
property of the gas and varies from
point to point in the flow.
• If a∞ is the speed of sound in the
uniform free stream, then the ratio
V∞/a∞ defines the free-stream Mach
number M∞.
• Similarly, the local Mach number,
M is defined as M = V/a, and varies
from point to point in the flow
field.
Flow Regimes
• Consider the flow over an airfoil
section as sketched in Figure. Here,
the local Mach number is
everywhere less than unity.
• Such a flow where M < I at every
point, and hence the flow velocity
is everywhere less than the speed of
sound is defined as subsonic flow.
• This flow is characterized by
smooth
streamlines
and
continuously varying properties.
Flow Regimes
• Note that the initially straight and
parallel streamlines in the free
stream begin to deflect far upstream
of the body i.e. the flow is
forewarned of the presence of the
body.
• Also, as the flow passes over the
airfoil, the local velocity and Mach
number on the top surface increase
above their free-stream values.
• However, if M is sufficiently less
than 1, the local Mach number
everywhere will remain subsonic.
Flow Regimes
• For airfoils in common use, if M∞ <
0.8, the flow field is generally
completely subsonic.
• Therefore
to
the
airplane
aerodynamicist,
the
subsonic
regime is loosely identified with a
free stream where M∞ < 0.8.
• If M∞ is subsonic, but is
sufficiently near 1, the flow
expansion over the top surface of
the airfoil may result in locally
supersonic regions, as sketched in
Figure.
• Such a mixed region flow is
defined as transonic.
Flow Regimes
• M∞ is less than 1 but high enough
to produce a pocket of locally
supersonic flow.
• In most cases, this pocket
terminates with a shock wave
across
which
there
is
a
discontinuous and sometimes rather
severe change in flow properties.
• If M∞ is increased to slightly above
unity, this shock pattern will move
to the trailing edge of the airfoil,
and a second shock wave appears
upstream of the leading edge.
• This second shock wave is called
the bow shock, and is sketched in
Figure.
Flow Regimes
• In passing through that part of the bow
shock that is nearly normal to the free
stream, the flow becomes subsonic.
• However, an extensive supersonic
region again forms as the flow
expands over the airfoil surface, and
again terminates with a trailing-edge
shock.
• Both flow patterns sketched in Fig. b
and c are characterized by mixed
regions of locally subsonic and
supersonic flow.
• Such mixed flows are defined as
transonic flows, and 0.8 < M∞ < 1.2 is
defined as the transonic regime.
Flow Regimes
• A flow field where M∞ > 1
everywhere
is
defined
as
supersonic. Consider the supersonic
flow over the wedge-shaped body
in Fig. 1.
• A straight, oblique shock wave is
attached to the sharp nose of the
wedge. Across this shock wave, the
streamline
direction
changes
discontinuously.
• Ahead of the shock, the streamlines
are
straight,
parallel,
and
horizontal; behind the shock they
remain straight and parallel but in
the direction of the wedge surface.
Flow Regimes
• Unlike the subsonic flow in Fig. a,
the supersonic uniform free stream
is not forewarned of the presence of
the body until the shock wave is
encountered.
• The flow is supersonic both
upstream and (usually, but not
always) downstream of the oblique
shock wave.
• The temperature, pressure, and
density of the flow increase almost
explosively across the shock wave
shown in Fig. d.
Flow Regimes
• As M∞, is increased to higher
supersonic speeds, these increases
become more severe. At the same
time, the oblique shock wave
moves closer to the surface, as
sketched in Fig. e.
• The incompressible flow is a
special case of subsonic flow;
namely, it is the limiting case where
M∞→ 0.
• Since M∞ = V∞/a∞ we have two
possibilities:
The former corresponds to no flow and is
trivial. The latter states that the speed of
sound in a truly incompressible flow
would have to be infinitely large.
Flow Regimes
• M∞ is less than 1 but high enough
to produce a pocket of locally
supersonic flow.
• In most cases, this pocket
terminates with a shock wave
across
which
there
is
a
discontinuous and sometimes rather
severe change in flow properties.
• If M∞ is increased to slightly above
unity, this shock pattern will move
to the trailing edge of the airfoil,
and a second shock wave appears
upstream of the leading edge.
• This second shock wave is called
the bow shock, and is sketched in
Figure.
Flow Regimes
• M∞ is less than 1 but high enough
to produce a pocket of locally
supersonic flow.
• In most cases, this pocket
terminates with a shock wave
across
which
there
is
a
discontinuous and sometimes rather
severe change in flow properties.
• If M∞ is increased to slightly above
unity, this shock pattern will move
to the trailing edge of the airfoil,
and a second shock wave appears
upstream of the leading edge.
• This second shock wave is called
the bow shock, and is sketched in
Figure.
Use of Mach Number in governing equations
• Since supersonic and subsonic flows have different characteristics, it would be
instructive to use Mach number as a parameter in our basic equations.
• This can be done easily for the flow of a perfect gas as in this case we have a
simple equation of state
After solving above equation x = 1
After solving above equation x = 16
M=4
Chapter 4
Isentropic Flows
Course Title: Compressible Flow and Propulsion System
Course Code: (ME-417)
Governing Equation
Continuity Equation
𝒎ሶ = 𝝆𝟏 𝑨𝟏 𝑽𝟏 = 𝝆𝟐 𝑨𝟐 𝑽𝟐
Momentum Equation
𝑭𝒑 = 𝒑𝟏 𝑨𝟏 − 𝒑𝟐 𝑨𝟐 = 𝝆𝟐 𝑨𝟐 𝑽𝟐𝟐 − 𝝆𝟏 𝑨𝟏 𝑽𝟐𝟏
Energy Equation
𝑽𝟐𝟏
𝑽𝟐𝟐
𝒉𝟏 +
= 𝒉𝟐 +
𝟐
𝟐
Second Law of thermodynamics
𝒔𝟏 = 𝒔𝟐
Stagnation Relations
• The internal energy and the flow energy of a fluid are frequently combined into a
single term, enthalpy.
• Whenever the kinetic and potential energies of the fluid are negligible, as is often
the case, the enthalpy represents the total energy of a fluid.
• For high-speed flows, such as those encountered in jet engines, the potential
energy of the fluid is still negligible, but the kinetic energy is not.
• In such cases, it is convenient to combine the enthalpy and the kinetic energy of
the fluid into a single term called stagnation (or total) enthalpy.
• It is defined per unit mass as
𝑽𝟐
𝒉𝒕 = 𝒉 +
𝟐
(1)
Stagnation Relations
• When the potential energy of the fluid is negligible, the stagnation enthalpy
represents the total energy of a flowing fluid stream per unit mass.
• Thus the stagnation enthalpy indicates the enthalpy of a fluid when it is brought to
rest adiabatically.
• The properties of a fluid at the stagnation state are called stagnation properties.
• During stagnation process, since the kinetic energy of a fluid is converted to
enthalpy (internal energy or flow energy), the fluid temperature and pressure is
increased.
Stagnation Relations
• Stagnation (or total) temperature is the temperature the gas attains when it is brought
to rest adiabatically.
𝑽𝟐
𝑻𝒕 = 𝑻 +
𝟐𝑪𝒑
(2)
𝑽𝟐
• The term
corresponds to the temperature rise during such a process and is called the
𝟐𝑪𝒑
dynamic temperature.
• For high-speed flows, the stagnation temperature is higher than the static (or ordinary)
temperature.
• For example, the dynamic temperature of air flowing at 100 m/s is about 5 K.
• Therefore, when air at 300 K and 100 m/s is brought to rest adiabatically (at the tip of a
temperature probe, for example), its temperature rises to the stagnation value of 305 K.
• The pressure a fluid attains when brought to rest isentropically is called the stagnation
pressure.
Stagnation Relations
Stagnation Relations
• The stagnation and static properties can be related in terms of Mach numbers.
𝑽𝟐 = 𝑴𝟐 𝒂𝟐
The velocity of sound:
𝒂𝟐 = 𝜸𝑹𝑻
Now equation 1 can be written as:
𝑴𝟐 𝜸𝑹𝑻
𝒉𝒕 = 𝒉 +
𝟐
But
𝜸𝑹
𝑪𝒑 =
𝜸−𝟏
𝑴𝟐 (𝜸 − 𝟏)𝑪𝒑 𝑻
𝒉𝒕 = 𝒉 +
𝟐
𝑴𝟐 𝜸 − 𝟏
𝒉𝒕 = 𝒉 𝟏 +
𝟐
𝑴𝟐 𝜸 − 𝟏
𝑻𝒕 = 𝑻 𝟏 +
𝟐
Stagnation Relations
• The stagnation process is considered isentropic and following relations can be
used for pressure.
𝜸Τ(𝜸−𝟏)
𝜸Τ(𝜸−𝟏)
𝟐
𝒑𝒕
𝑻𝒕
𝑴 𝜸−𝟏
=
= 𝟏+
𝒑
𝑻
𝟐
𝑴𝟐 𝜸 − 𝟏
𝟏Τ(𝜸−𝟏)
𝝆𝒕
= 𝟏+
𝝆
𝟐
• The stagnation quantities Tt, pt etc. can be calculated from the actual conditions of
M, V, T, p, and ρ at a given point in a general flow field.
• The actual flow field itself may not have to be adiabatic or isentropic from one
point to the next.
• The isentropic process is only for definition of total conditions at a point.
Maximum Speed
• A gas attains its maximum speed when it is hypothetically expanded to zero
pressure.
• The static pressure at this state is also zero.
𝑽𝟐
𝑻𝒕 = 𝑻 +
𝟐𝑪𝒑
0
𝑽𝒎𝒂𝒙 =
𝟐𝑪𝒑 𝑻𝒐
𝑽𝒎𝒂𝒙 =
𝜸
𝟐
𝑹𝑻𝒐
𝜸−𝟏
(3)
Critical Speed of Sound
• This is the speed of sound at the sonic state of a perfect gas, where M=1 and can
be given as
𝑉 ∗ = 𝑎∗ = 𝛾𝑅𝑇 ∗
• From equation 2
2𝛾
𝑅(𝑇𝑜 − 𝑇 ∗ )
𝛾−1
𝑉∗ =
• Eliminating T*
∗
𝑉 =
2𝛾
𝑅𝑇𝑜
𝛾+1
• The relation between these three reference speeds can be obtained via equations
Critical Speed of Sound
𝑎∗
=
𝑎𝑜
2
𝛾+1
(4)
𝑉𝑚𝑎𝑥
=
𝑎𝑜
2
𝛾−1
(5)
𝑉𝑚𝑎𝑥
=
∗
𝑎
𝛾+1
𝛾−1
(6)
Critical Speed of Sound
• From eq. 2
𝑽𝟐
𝑻𝒐 = 𝑻 +
𝟐𝑪𝒑
(2)
2𝐶𝑝 𝑇𝑜 = 2𝐶𝑝 𝑇 + 𝑉 2
From eq. 5
2
𝑉𝑚𝑎𝑥
=
𝛾
𝛾
2
𝑅𝑇𝑜 = 2
𝑅𝑇 + 𝑉 2
𝛾−1
𝛾−1
2
𝑎
2
𝑉𝑚𝑎𝑥
=2
+ 𝑉2
𝛾−1
2
𝑎𝑜2
𝛾−1
(8)
2
• Dividing eq 7 by 𝑉𝑚𝑎𝑥
(7)
• Using eq. 4, 5 and 6
𝑉2 +
Equation A is the kinematic form of the
steady, adiabatic energy equation
2
𝑎2
𝑉2
1=
+ 2
2
𝛾 − 1 𝑉𝑚𝑎𝑥 𝑉𝑚𝑎𝑥
• Substituting eq. 8
2
2
𝛾 + 1 ∗2
2
2
2
𝑎 = 𝑉𝑚𝑎𝑥 =
𝑎 =
𝑎
𝛾−1
𝛾−1 𝑜 𝛾−1
Equation A
𝑉2
𝑎2
+ 2=1
2
𝑉𝑚𝑎𝑥 𝑎𝑜
Equation B
Critical Speed of Sound
• Equation B is the equation of ellipse and
is known as the steady flow adiabatic
ellipse.
• All the points on the ellipse have the
same energy.
• Each point differs from others owing to
the relative proportions of the thermal
and kinetic energy and thus corresponds
to different Mach Numbers.
• Mach Number can be obtained by
differentiating eq. B
𝟐 𝒅𝒂
𝑴=−
𝜸 − 𝟏 𝒅𝑽
Critical Speed of Sound
• Thus the change in slope from point to point indicates how the changes in the
Mach Number are related to the changes in speed of sound and velocity.
• Therefore, it is direct comparison of the relative magnitudes of thermal and kinetic
energies.
• As observed from figure, for low Mach Number flows. The changes in the Mach
number are mainly due to the changes in velocity.
• However, at high Mach numbers flows, they are due to the changes in the speed of
sound and the compressibility effects ate dominant.
• One more important point to note in the figure is the case corresponding to 𝑀 ≤ 3,
where the changes in the speed of sound are negligibly small. These flows are
considered to be incompressible.
At a point in a flow passage, the velocity and temperature of air are 802 m/s and
400 K respectively,
a. Calculate the stagnation speed of sound.
b. Calculate the critical speed of sound.
c. Calculate the Mach number and the Mach number referred to critical conditions
at the given state.
d. If the air is accelerated in a suitable flow passage, find the maximum possible
velocity that can be attained. Determined the Mach number referred to critical
conditions corresponding to the state of maximum velocity.
e. Obtain the kinematic form of the adiabatic energy equation and sketch the
steady flow adiabatic ellipse.
Effects of Area Variation on Flow Properties in
Isentropic Flow
• Consider a one dimensional Compressible flow in an infinitesimal duct with
variable area as shown in figure.
• The infinitesimal variation of the cross-sectional area of the duct causes
infinitesimal changes in the flow properties.
• The changes may be related by using basic equations
Effects of Area Variation on Flow Properties in
Isentropic Flow
Continuity Equation
𝑚ሶ = 𝜌𝐴𝑉 = (𝜌 + 𝑑𝜌)(𝐴 + 𝑑𝐴)(𝑉 + 𝑑𝑉)
By simplifying
Momentum Equation
𝑑𝑉 𝑑𝐴 𝑑𝜌
+
+
=0
𝑉
𝐴
𝜌
(1)
𝑝𝐴 − 𝑝 + 𝑑𝑃 𝐴 + 𝑑𝑃 + 𝑑𝐹𝑝 = 𝑚ሶ 𝑉 + 𝑑𝑉 − 𝑚𝑉
ሶ
• Where 𝑑𝐹𝑝 is the infinitesimal pressure forces acting in the x-direction on the side walls of the
control volume.
𝑑𝑝
• On the side walls, the average pressure is 𝑝 + and the cross-sectional area of the side walls
2
perpendicular the flow direction is dA
1
𝑝𝐴 − 𝑝 + 𝑑𝑃 𝐴 + 𝑑𝑃 + (𝑝 + 𝑑𝑝)𝐴 = 𝜌𝐴𝑉 𝑉 + 𝑑𝑉 − 𝜌𝐴𝑉 2
2
Effects of Area Variation on Flow Properties in
Isentropic Flow
By simplifying
𝑑𝑝 = −𝜌𝑉𝑑𝑉
𝑑𝑝
+𝑑
𝜌
𝑉2
2
(2)
=0
From equation 2
𝑑𝑝
−
= 𝑑𝑉
𝜌𝑉
Put in equation 1
𝑑𝐴 𝑑𝑝 𝑑𝜌
− 2+
=0
𝐴 𝜌𝑉
𝜌
𝑑𝐴
𝑑𝑝 𝑑𝜌
=
−
2
𝐴
𝜌𝑉
𝜌
𝑑𝐴
𝑑𝑝
𝑑𝜌 2
=
1−
𝑉
2
𝐴
𝜌𝑉
𝑑𝑝
𝑑𝐴
𝑑𝑝
𝑉2
=
1− 2
2
𝐴
𝜌𝑉
𝑎
𝒅𝑨
𝒅𝒑
𝟐
=
𝟏
−
𝑴
𝑨
𝝆𝑽𝟐
(A)
𝒅𝑨
𝒅𝑽
=−
𝟏 − 𝑴𝟐
𝑨
𝑽
(B)
Effects of Area Variation on Flow Properties in
Isentropic Flow
Effects of Area Variation on Flow Properties in
Isentropic Flow
• Equation describes the variation of pressure with flow area. For subsonic flow (M
< 1), dA and dp must have the same sign.
• That is, the pressure of the fluid must increase as the flow area of the duct
increases and must decrease as the flow area of the duct decreases.
• Thus, at subsonic velocities, the pressure decreases in converging ducts (subsonic
nozzles) and increases in diverging duct (subsonic diffusers).
• In supersonic flow (Ma >1), dA and dp have opposite signs.
• Two common devices involving area change are nozzle and diffuser.
• The same piece of equipment can operate as either a nozzle or a diffuser,
depending on the flow regime.
• Thus a device is called a nozzle or a diffuser because of what it does, not what it
looks like.
One dimensional isentropic flow
• A nozzle is a device that converts
enthalpy (or pressure energy for the
case of an incompressible fluid) into
kinetic energy.
• From Figure we see that an increase
in velocity is accompanied by either
an increase or decrease in area,
depending on the Mach number.
One dimensional isentropic flow
• A diffuser is a device that converts kinetic energy into enthalpy (or pressure
energy for the case of incompressible fluids).
• The highest velocity we can achieve by a converging nozzle is the sonic velocity,
which occurs at the exit of the nozzle.
• To accelerate a fluid, we must use a converging nozzle at subsonic velocities and a
diverging nozzle at supersonic velocities.
• Based on Equation B, which is an expression of the conservation of mass and
energy principles, a diverging section must be added to a converging nozzle to
accelerate a fluid to supersonic velocities.
• The result is a converging– diverging nozzle.
• The fluid continues to accelerate as it passes through a supersonic (diverging)
section. A large decrease in density makes acceleration in the diverging section
possible.
One dimensional isentropic flow
• If the mass flow rate is fixed, there is a minimum cross-sectional area required to
pass this flow and this phenomenon is known as chocking.
• For a given reduction in area, there is a maximum initial Mach number which can
be maintained in a subsonic flow, while there is a minimum initial Mach number
which can be maintained steadily in a supersonic flow.
Throat
One dimensional isentropic flow
Effect of Back Pressure
• Consider the subsonic flow through a converging
nozzle as shown in Figure.
• The nozzle inlet is attached to a reservoir at pressure
pr and temperature Tr.
• The reservoir is sufficiently large so that the nozzle
inlet velocity is negligible.
• The fluid velocity in the reservoir is zero and the
flow through the nozzle is approximated as
isentropic,
• The stagnation pressure and stagnation temperature
of the fluid at any cross section through the nozzle
are equal to the reservoir pressure and temperature,
respectively.
One dimensional isentropic flow
• Now we begin to reduce the back pressure and observe the resulting effects on
the pressure distribution along the length of the nozzle, as shown.
• If the back pressure pb is equal to pt, which is equal to pr, there is no flow and the
pressure distribution is uniform along the nozzle.
• When the back pressure is reduced to p2, the exit plane pressure pe also drops to
p2. This causes the pressure along the nozzle to decrease in the flow direction.
• Now suppose the back pressure is reduced to p3 (= p*), which is the pressure
required to increase the fluid velocity to the speed of sound at the exit plane or
throat).
• The mass flow reaches a maximum value and the flow is said to be choked.
• Further reduction of the back pressure to level p4 or below does not result in
additional changes in the pressure distribution, or anything else along the nozzle
length.
One dimensional isentropic flow
• The effect of back pressure on the nozzle exit pressure pe is:
A converging nozzle with an exit cross sectional area of 0.001 m2 is
operated with air at a back pressure of 69.5 kPa, as shown. The nozzle is
fed from a large reservoir where the stagnation pressure and temperature
are 100 kPa and 60o C respectively. Determine the Mach number and
temperature at the nozzle exit. Also, find the mass flow rate through the
nozzle. Assume one dimensional steady isentropic flows.
One dimensional isentropic flow
Flow in converging diverging nozzle
• Now repeat the previous assumptions for
converging-diverging nozzle and study
the effects of back pressure.
No Flow conditions
• Consider the valve at the exit of the
nozzle is closed, then the pressure is
constant throughout the nozzle. Such that
pe=pb=pt=po.
• As represented by horizontal line
One dimensional isentropic flow
• When the back pressure in the discharge
reservoir is slightly decreased by opening
the valve then the flow is subsonic
throughout the converging-diverging
nozzle. As shown as (i) in fig.
• The static pressure of the fluid decreases
from the entrance to the throat, where it
reaches its minimum value and then
increases in the diverging section.
• Therefore, the converging part of the
nozzle acts as a sub sonic nozzle while the
diverging part acts as a subsonic diffuser
and the passage behaves like a
conventional venturi-tube.
One dimensional isentropic flow
Chocking conditions
• The flow is chocked at the throat where
the Mach number is unity. As shown by
(ii) in figure.
• The throat pressure is equal to the critical
pressure at the exit and is equal to the
back pressure.
• The flow is subsonic at every point except
at the throat so that the diverging part
again acts as a subsonic diffuser.
One dimensional isentropic flow
Non Isentropic Flow Regime
• Flow pattern as indicated by (iii) is a
typical representation of this flow regime,
which is not isentropic.
• A non-isentropic phenomenon, known as
shock transforms the supersonic flow in
the diverging section into a subsonic one.
• This will be covered in chapter 5.
One dimensional isentropic flow
Exit Plane shock conditions
• In this case, the shock phenomenon
moves to the exit plane of the converging
diverging nozzle as represented by (iv) in
figure.
One dimensional isentropic flow
Overexpansion Flow regime
• A typical representation of this flow is given
pattern (v).
• The flow is sonic at the throat where the
Mach number is unity, and supersonic in the
entire diverging section of the nozzle.
• The mass flow rate is invariant with respect
to the back pressure since the flow is choked
at the throat.
• The gas is overexpanded at the exit plane,
because the pressure at the exit plane is lower
than the back pressure.
• The compression, which occurs outside the
nozzle, involves non-isentropic oblique
compression waves which cannot be treated
with one-dimensional, while the flow
throughout the nozzle is isentropic.
One dimensional isentropic flow
Design conditions
• Flow pattern (vi) in figure represents the
condition for which the convergingdiverging nozzle is actually designed.
• The flow is entirely isentropic within and
outside the nozzle such that the exit plane
pressure is identical with back pressure.
• Owing to the chocking at the throat, the
flow is supersonic in the entire diverging
section of the nozzle.
• As long as pe=pb, the shape of the jet
leaving the nozzle is cylindrical.
A converging-diverging nozzle with a throat area of 0.0035 m2 is
attached to a very large tank of air in which the pressure is 125
kPa and the temperature is 47o C as shown in figure. The nozzle
exhausts to the atmosphere with a pressure of 100 kPa. If the
mass flow rate is 0.9 kg/s, determine the exit area and the Mach
number at the throat. The flow in the nozzle is isentropic.
At a point upstream of the throat in a converging-diverging
nozzle, the velocity, temperature and pressure are 172 m/s, 22o C
and 200 kPa, respectively as shown in figure. If the nozzle,
operating at its design condition has an exit area of 0.01 m2 and
discharges to the atmosphere with a pressure of 100 kPa,
determine the mass flow rate and the nozzle throat area.
At a point upstream of throat in a converging diverging nozzle, the velocity,
temperature and pressure are 172m/s, 22o C and 200 kPa respectively as shown in
figure. If the nozzle is operating at its design conditions and has an exit area of 0.01
m2 and discharges to atmosphere with a pressure of 100 kPa. Determine the mass
flow rate and nozzle throat area.
Chapter 5
Normal Shock waves
Course Title: Compressible Flow and Propulsion System
Course Code: (ME-417)
Introduction
• A shock wave represents an abrupt change in fluid properties, in which finite
variations in pressure, temperature, and density occur over a shock thickness
comparable with the mean free path of the gas molecules involved.
• Also, it will be shown that shock wave can only be a compression type, and a
rarefaction shock which requires a decrease in entropy, in an adiabatic process is
not possible.
• It is known that supersonic flows adjusts to the presence of a body by means of
the shock process, whereas subsonic flow can adjust by gradual changes in flow
properties.
• The compression of a gas across a shock wave is instantaneous and it can’t be a
reversible process. Because of irreversibility, the kinetic gas of the gas leaving the
shock wave is smaller than that of the isentropic compression between the same
pressure points.
Introduction
• The reduction in kinetic energy due to the shock appears as heating of the gas to a
static temperature, above that corresponding to the isentropic compression value.
• Consequently, as a gas flows through the shock wave, it experiences decrease in
available energy and increase in its entropy.
Development of a compression wave
• When a piston in a tube is given a steady
velocity, to the right of magnitude dV, a sound
wave travel ahead of a piston, through the
medium in the tube.
• Suppose that the piston is now given a second
increment of velocity, dV, causing a second wave
to move into the compressed gas behind the first
wave.
• The location of these waves are shown in figure
5.1 (b).
• The first and second waves, created by the
acceleration of the piston, travel respectively at
speed of sound a1 and a2, with respect to the gas
into which they are moving.
Development of a compression wave
• Since the second wave is moving into a
compressed gas with a slightly elevated
temperature, then a2 is greater than a1.
• The first wave is moving in stationary gas so the
absolute velocity of this wave is a1. However,
the second wave is moving into a gas which is
already moving to the right at a velocity dV, and
therefore the absolute velocity of the second
wave is a2+dV.
• Hence, the second wave moves faster than the
first and gradually gets closer at time t=t2 as
shown in figure 5.1 (c).
Development of a compression wave
• Now suppose that a piston is accelerated from rest to a finite velocity increment of
magnitude ∆V to the right.
• The finite velocity increment can be thought to consists of a large number of
infinitesimal increments each of magnitude dV.
• Figure 5.2 shows the velocity of piston versus time with the incremental velocity
superimposed.
Development of a compression wave
• The corresponding location of the waves and
pressure distribution in tube, at t=t1 are shown
in figure 5.3 (a).
• The first wave of the series of waves is called
head while the last is known as tail.
• As was demonstrated that the waves next to
the piston tend to overtake those further down
the tube as shown in figure 5.3 (b) and (c) at
time t=t2 and t=t3 respectively.
• The present analysis ceases to be valid,
because the viscous and heat transfer effects
are no longer negligible with such
extraordinary gradients in the velocity and
temperature along the wave.
Development of a compression wave
• Indeed, if the isentropic analysis were
continued up to t=t4, the wave would topple
over to the form shown by the dashed lines in
figure 5.3 (d).
• This is physically absurd, since it means that
at the same time and at the same location, the
fluid has simultaneously three different values
of pressure, density and velocity.
• Therefore viscous and heat conduction effects
intervene soon after t=t3 and produce a normal
shock wave at t=t4.
• The thickness of the shock wave is
approximately 2.5 x10-7 m , so the pressure
and temperature along the wave are very large,
yet not truly infinite.
General Equations for the flow across a normal
shock wave
• The flow through a stationary normal shock wave may be analyzed by considering
one-dimensional flow.
• In the analysis following assumptions are made:
1. The normal shock wave is perpendicular to the streamlines.
2. The normal shock take place at a constant cross-sectional area, since the shock
thickness is very small, compared with the dimensions of the duct.
3. Although the shear stress is present within the boundary layer, it acts on a very
small area since the shock wave is very thin. For this reason, frictional effects
are negligible.
4. The flow process, including the shock waves, can be assumed adiabatic with no
external work, and the effects of the body forces are negligible, since the shock
wave is very thin.
General Equations for the flow across a normal
shock wave
Continuity Equation
𝒎ሶ = 𝝆𝒙 𝑨𝒙 𝑽𝒙 = 𝝆𝒚 𝑨𝒚 𝑽𝒚
Since 𝐴𝑥 = 𝐴𝑦 , the mass flow rate per unit area or the mass flux is given by
ሶ
𝑮 = 𝒎/𝑨
= 𝝆𝒙 𝑽𝒙 = 𝝆𝒚 𝑽𝒚
General Equations for the flow across a normal
shock wave
Momentum Equation
Since the flow across the normal shock wave is assumed to be frictionless, the pressure forces
acting on the control volume of figure 5.5 must be balanced by the rate of change of momentum
across the control volume.
𝒑𝒙 𝑨𝒙 − 𝒑𝒚 𝑨𝒚 = 𝝆𝒚 𝑨𝒚 𝑽𝟐𝒚 − 𝝆𝒙 𝑨𝒙 𝑽𝟐𝒙
𝒑𝒙 − 𝒑𝒚 = 𝑮(𝑽𝒚 − 𝑽𝒙 )
General Equations for the flow across a normal
shock wave
Energy Equation
For an adiabatic process across a normal shock wave with no external work the first law of
thermodynamics becomes
𝑽𝟐𝒚
𝑽𝟐𝒙
𝒉𝒐𝒙 = 𝒉𝒙 +
= 𝒉𝒚 +
= 𝒉𝒐𝒚 = 𝒉𝒐 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝟐
𝟐
General Equations for the flow across a normal
shock wave
The second law of thermodynamics
The flow through a normal shock wave is irreversible, because of the almost discontinuous
property changes across the shock wave, Thus,
𝒔𝒚 > 𝒔𝒙
Equation of state
For any fluid, the equation of state may be given as
𝒉 = 𝒉(𝒔, 𝒑)
and
𝒔 = 𝒔(𝒑, 𝝆)
General Equations for the flow across a normal
shock wave
• Typically the fluid / flow properties before the shock are known and the conditions
that exist after the shock need to be predicted.
• The unknown parameters are four in number (ρ2, p2, h2, V2).
• The four unknowns can be found through the use of the three governing equations
(from continuity, energy, and momentum concepts) if property relations are
known.
• For the case of a perfect gas, equation of state is known. If specific heats are
assumed constant, working equations in terms of Mach numbers and the specific
heat ratio can be obtained.
• For a perfect gas enthalpy is a function of temperature only. The energy and
momentum equations can be written as:
General Equations for the flow across a normal
shock wave
𝑻𝒕𝟏 = 𝑻𝒕𝟐
𝑻𝟏
𝜸−𝟏 𝟐
𝜸−𝟏 𝟐
𝟏+
𝑴𝟏 = 𝑻𝟐 𝟏 +
𝑴𝟐
𝟐
𝟐
𝒑𝟏
𝒑𝟏 +
𝑹𝑻𝟏
(1)
𝑴𝟐𝟏 𝜸𝑹𝑻𝟏
𝒑𝟐
= 𝒑𝟐 +
𝑹𝑻𝟐
𝑴𝟐𝟐 𝜸𝑹𝑻𝟐
𝟐
𝒑𝟏 𝟏 + 𝜸𝑴𝟏
𝟐
= 𝒑𝟐 𝟏 + 𝜸𝑴𝟐
(2)
General Equations for the flow across a normal
shock wave
• Using definition of Mach number, velocity of sound and equation of state, the
continuity equation for perfect gas is
𝒑𝟏 𝑴𝟏
𝑻𝟏
=
𝒑𝟐 𝑴𝟐
(3)
𝑻𝟐
• Once the gas is identified, γ is known, and a state preceding the shock is given in
terms of p1, M1, and T1, equations 1-3 are sufficient to solve for the unknowns
after the shock: p2, M2, and T2.
General Equations for the flow across a normal
shock wave
• The equation can be simplified as:
(4)
• In the previous section, it was shown that
• From above two equations we can obtain
• This relation is known as Prandtl’s relation. This can be written in terms of
velocities a*, V1 and V2.
General Equations for the flow across a normal
shock wave
(5)
• Since a1∗= a2∗. The proof is given below. The total temperature is
• For the critical case the above equation becomes
• Where 𝑎∗ =
𝛾𝑅𝑇 ∗ , Eliminating T*
General Equations for the flow across a normal
shock wave
• Since total temperature after shock does not change then a1∗ = a2∗. This proves
equation (5)
• By solving (4) and (1), (2) and eliminating M2 we get
(6)
(7)
General Equations for the flow across a normal
shock wave
• From continuity equation and total pressure relation along with (4) gives
(8)
(9)
• The end points 1 and 2 (before and after the shock) are well-defined states, but the
changes that occur within the shock do not follow an equilibrium process in the
usual thermodynamic sense.
• For this reason the shock process is usually shown by a dashed or wiggly line.
General Equations for the flow across a normal
shock wave
• Note that when points 1 and 2 are located on the T–s diagram, it can immediately
be seen that an entropy change is involved in the shock process i.e., s2 > s1.
• The reason is that the flow through the shock is adiabatic but irreversible.
General Equations for the flow across a normal
shock wave
*
• For isentropic flow, the area at which Mach number is unity is defined as A ,
which is used as a reference. Since shock is non-isentropic, areas downstream of
the shock cannot be referenced to the area upstream of the shock i.e. A1∗ ≠ A2∗.
• For steady flow however mass flow rates are equal. m1 = m2
• It was shown in previous chapter
• Or
• Since flow is adiabatic across a shock wave Tt1 = Tt2.
General Equations for the flow across a normal
shock wave
• For entropy changes some important relations that come from combinations of the
first and second laws can be used.
(10)
• Differentiating the enthalpy h = u + pv
(11)
• Combining equations (10) and (11) we get
(12)
General Equations for the flow across a normal
shock wave
• Although the assumption of a reversible process is made to derive equations (10)
and (12), the results are equations that contain only properties and thus are valid
relations to use between any end states, whether reached reversibly or not.
• For ideal gas h = CpT and v = RT/P. We get
• Using following relations
General Equations for the flow across a normal
shock wave
• After simplifications
(13)
• The changes in various properties due to shock as given in equations (4), (6-9),
and (13) are shown in graphical form.
• The conservation of mass and energy relations can be combined into a single
equation and plotted on an h-s diagram, using property relations.
• The resultant curve is called the Fanno line, and it is the locus of states that have
the same value of stagnation enthalpy and mass flux (mass flow per unit flow
area).
• Likewise, combining the conservation of mass and momentum equations into a
single equation and plotting it on the h-s diagram yield a curve called the Rayleigh
line.
General Equations for the flow across a normal
shock wave
• Both these lines are shown on the h-s diagram in Figure. As will be proved later,
the points of maximum entropy on these lines (points a and b) correspond to M =
1.
• The state on the upper part of each curve is subsonic and on the lower part
supersonic.
General Equations for the flow across a normal
shock wave
• The Fanno and Rayleigh lines intersect at two points (points 1 and 2), which
represent the two states at which all three conservation equations are satisfied.
• One of these (state 1) corresponds to the state before the shock, and the other
(state 2) corresponds to the state after the shock.
• Note that the flow is supersonic before the shock and subsonic afterward.
Therefore the flow must change from supersonic to subsonic if a shock is to occur.
• The larger the Mach number before the shock, the stronger the shock will be.
From equation (9) and (13)
General Equations for the flow across a normal
shock wave
(14)
• The above equation shows that for 1 > γ >1.66, the entropy change is always
positive when M1 > 1. Therefore for a perfect gas, shock proceeds from supersonic
to subsonic.
Rankine-Hugoniot relation
• The equation relating the pressure and density ratios (for the normal shock) is
called Rankine-Hugoniot relation.
• Hugoniot equation provides a relation in terms of thermodynamic quantities across
the shock without inclusion of velocity / Mach number.
• From equation (7) and (8), eliminating M1 we get
General Equations for the flow across a normal
shock wave
(15)
• The isentropic relation between pressure and
density is
(16)
• Figure shows the pressure-density variation for
isentropic process (Eq. 16) and for normal shock
(Eq. 15).
General Equations for the flow across a normal
shock wave
• The figure shows that
(i) for a given change in specific volume / density, a shock wave results in higher
pressure than an isentropic compression
(ii) weak shocks are nearly isentropic.
• The shock wave, however increases entropy and consequent total pressure loss,
i.e., the shock compression is less efficient than the isentropic compression
An airstream with a velocity of 500 m/s, a static pressure of 70 kPa and a static
temperature of 300 K undergoes a normal shock, as shown in figure. Determine
a. The Mach number and velocity after the normal shock wave,
b. The static conditions after the normal shock wave,
c. The stagnation conditions after the normal shock wave, and
d. The entropy change across the normal shock wave
Operation of supersonic diffusers
• When the flow at the entrance to a converging and diverging duct is supersonic,
the fluid may be slowed down to Mach number of unity or greater at the throat.
• If sonic conditions are achieved, the fluid may either expand again to supersonic
speed or can be compressed at subsonic speeds. In the latter case the device is
called supersonic diffuser.
• It is used in supersonic wind tunnels and in engine intake.
• A simple supersonic wind tunnel can be constructed using a converging diverging
nozzle and a test section. If back pressure is sufficiently reduced the flow will be
supersonic in the test section.
Operation of supersonic diffusers
• Even though the construction of such type of unit is simple but power
consumption will be high due to large pressure variation.
• For this reason supersonic diffuser is used to decelerate the fluid to subsonic
speed.
• The air is drawn from a large reservoir where the stagnation temperature and
pressure are Tt and Pt.
• The wind tunnel discharges into another reservoir where the back pressure is pb.
The back pressure is controllable by means of a fan.
Operation of supersonic diffusers
Operation of supersonic diffusers
• Several pressure distributions are possible as the wind tunnel is started and back pressure is
reduced.
(1) No flow condition (Pb = Pti) e.g. when the fan is not operating
(2) Subsonic flow in test sections and both converging-diverging ducts. Fluid accelerates in
converging sections and decelerates in diverging sections.
(3) Choking - This occurs when fan speed is further increased. The flow becomes choked in the
duct which has smaller diameter throat. If second throat is smaller then flow is choked in second
one and the flow is subsonic in test section.
(4) To obtain supersonic flow in test section, the first throat should be smaller. When back
pressure is reduced further, normal shock occurs in the diverging portion of the first nozzle
(5-7)Unfavorable condition, the shock occurs in the test section (beginning i.e. 5 or end i.e. 7) at
the highest possible Mach number and thus the losses are greatest
(8-10) With further decrease of back pressure, shock location proceeds to second converging
diverging duct which now acts like a supersonic diffuser.
Shock stabilizes in the diverging section diffuser. Flow pattern 10 is most favorable which is
obtained by increasing the back pressure to adjust the location of shock slightly downstream of
the throat.
An airstream with a Mach Number of 2.0, a pressure of 150 kPa and a temperature
of 350 K enters a diverging channel. If the ratio of the exit cross-sectional area to
the inlet cross-sectional is 3.0, determine the back pressure which is necessary to
produce a normal shock wave in the channel with a cross-sectional area equal to
twice the inlet cross-sectional area. Assume steady one dimensional isentropic flow
except through the normal shock wave.
x y
Pi=150 kPa
Ti=350 K
Mi=2.0
Flow
Ax=Ay =2A
Ae=3Ai
Moving Normal Shock
• In many situations a normal shock wave is moving. When an explosion occurs,
shock wave propagates through the atmosphere from the point of explosion.
• Consider a normal shock wave moving in a stationary gas at a constant velocity Vs
with respect to a fixed observer. The conditions before and after shock are ‘b’ and
‘a’ respectively.
• Va is the velocity of gas behind the shock wave with respect to fixed observer.
Moving Normal Shock
• Now suppose the observer is moving with shock wave velocity. Now the shock
will be stationary with respect to observer.
• Static properties are defined as those measured by a device which is moving with
absolute flow velocity. Therefore static properties are independent of observer
velocity. This means
Moving Normal Shock
• Velocity and Mach number before and after the normal shock will be Vx = Vs and
Vy = Vs – Va
• Stagnation properties are measured by bringing the flow to rest with respect to
observer.
• These properties thus depend on observer velocity. The stagnation temperature of
fluid can change across a moving normal shock wave.
Moving Normal Shock
• The difference is:
• But Vb = 0
• For stationary normal shock (which is obtained by moving the observer),
stagnation temperature will be same.
• Solving (A), (B) and (C)
A normal shock wave moves at a constant velocity of 500 m/s into still air with a
temperature and pressure of 300 K and 100 kPa respectively (Figure). Determine
the static and stagnation conditions present in the air after the passage of the wave,
as well as the gas velocity behind the wave.
Chapter 6
Frictional Flow in constant-area ducts
Course Title: Compressible Flow and Propulsion System
Course Code: (ME-417)
Introduction
• In chapter 4, compressible flow in ducts was investigated for the case in which the
changes in flow properties are brought about only by changes in area.
• However, In a real flow situation, frictional forces are present and may have an
important effects in the flow.
• Naturally, the inclusion of frictional terms in the equations of motion makes the
resultant analysis far more complex.
• For this reason, in order to study the effects of friction on the compressible in the
ducts certain restrictions are placed on the flow.
• The first part of this chapter is concerned with compressible flow in constant-area
insulated ducts which eliminates the effects of the change in area and the heat
transfer.
• In the second part of this chapter, the other extreme case, when the constant-area
ducts are extremely long will be treated. This case approximates the flow of a
natural gas through a long uninsulated pipeline where there is sufficient area foe
heat transfer to make the flow non-adiabatic and approximately isothermal.
Governing equations for adiabatic one dimensional flow
of a perfect gas with friction in constant area ducts
Continuity Equation
𝒎ሶ = 𝝆𝟏 𝑨𝟏 𝑽𝟏 = 𝝆𝟐 𝑨𝟐 𝑽𝟐 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
• As long as the area of the duct is constant then
A1=A2=A, then mass flow rate per unit area or
the mass flux, G, is given by
ሶ
𝑮 = 𝒎/𝑨
= 𝝆𝟏 𝑽𝟏 = 𝝆𝟐 𝑽𝟐 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
Momentum Equation
−𝑭𝒇 + 𝒑𝟏 𝑨𝟏 − 𝒑𝟐 𝑨𝟐 = 𝝆𝟐 𝑨𝟐 𝑽𝟐𝟐 − 𝝆𝟏 𝑨𝟏 𝑽𝟐𝟏
Governing equations for adiabatic one dimensional flow
of a perfect gas with friction in constant area ducts
The first law of thermodynamics
𝑽𝟐𝟏
𝑽𝟐𝟐
𝑽𝟐
𝒉𝒐𝟏 = 𝒉𝟏 +
= 𝒉𝟐 +
= 𝒉𝒐𝟐 = 𝒉 +
𝟐
𝟐
𝟐
Equation of state
𝒑 = 𝝆𝐑𝐓
The second law of thermodynamics
𝒔𝟐 > 𝒔𝟏
Governing equations for adiabatic one dimensional flow
of a perfect gas with friction in constant area ducts
To calculate the entropy change,
thermodynamics relation will be used
the
If the above expression is integrated for constant
specific heats, then the entropy change is
An insulated tube with a cross-sectional area of 40 mm2 is fed with air by an
isentropic converging nozzle from a large tank where the temperature and pressure
are 300K and 100 kPa, respectively as shown in figure. At section 1, where the
nozzle joins to the constant-area tube, the pressure is 95 kPa. At section 2, which is
located some distance downstream in the tube, the air temperature is 290 K.
Determine,
a. The mass flow rate
b. The stagnation pressure at section 2, and
c. The frictional force on the duct wall between section 1 and 2.
The Fanno Line
• The Fanno line was first introduced during the discussion of normal shock waves.
• In this section, the equation of the Fanno line for a perfect gas will be derived
• The entropy change is
• Using continuity equation
• From energy equation, for a perfect gas with ℎ = 𝐶𝑝 𝑇, 𝑉 =
𝑉1 = 2𝐶𝑝 (𝑇𝑜 − 𝑇1 )
2𝐶𝑝 (𝑇𝑜 − 𝑇) and
The Fanno Line
• But as
• Which is the equation of the Fanno line for a perfect gas\
• But, from the energy equation,
• Also, for a perfect gas,
• The Mach number is unity at the point of maximum entropy.
The Fanno Line
The Fanno Line
• The upper branch shows the subsonic flows while the lower branch shows the
supersonic flows.
• Since the flow is adiabatic, according to the Second law of thermodynamics, the
entropy must increases.
• Thus, the path of states along any one of Fanno line must be towards right.
• Consequently, if the flow at the some point in the duct is subsonic, the effects of
friction will tend to increase the velocity and the Mach number to decrease the
temperature and pressure of the stream.
• If, on the other hand, the flow is initially supersonic, the effects of friction will
decrease the velocity and Mach number and to increase the pressure and
temperature of the stream.
• A supersonic flow may therefore never become subsonic, and a subsonic flow may
never become supersonic, unless a discontinuity is present.
Effect of Friction on the flow properties along the
Fanno line
• Consider one-dimensional adiabatic flow of a
compressible fluid in a constant-area duct with friction,
as shown in figure.
Continuity Equation
• Neglecting the products of differentials and dividing
𝝆𝑨𝑽.
(1)
Effect of Friction on the flow properties along the
Fanno line
Momentum Equation
• The sum of frictional forces and the pressure forces acting on the
control volume must be balanced by the rate of change of
momentum across the control volume, that is
(2)
• Where, 𝜏𝑤 , is the wall shear stress on the control volume by the
duct and dAw is the wetted wall area over which shear stress is
acting.
• At this point, the coefficient of friction, f, which is defined as the
ratio of the wall shear stress to the dynamic pressure of the
stream, may be introduced.
(3)
Effect of Friction on the flow properties along the
Fanno line
• Ducts of other than circular shape can be included in this
analysis by introducing the hydraulic diameter, D as
• Now, the momentum equation takes the following form
(4)
Effect of Friction on the flow properties along the
Fanno line
The first law of thermodynamics
• For adiabatic flow with no external work, the first law of
thermodynamics for the control volume is
• Neglecting second-order differentials, it is possible to
obtain,
• But, for a perfect gas,
and
, then
(5)
Effect of Friction on the flow properties along the
Fanno line
Equation of state
• The equation of state for a perfect gas,
logarithmically differentiated to yield
, may be
(6)
The second law of thermodynamics
• For adiabatic and frictional flows
(7)
Variation of flow properties with friction
Velocity
(8)
Variation of flow properties with friction
Velocity
(9)
Variation of flow properties with friction
Velocity
• Eliminating
to obtain
from eq. 1, 5 and 6, it is possible
(10)
• The equation 9 will become
(11)
Variation of flow properties with friction
Pressure
• Combining equation 10 and 11
Temperature
Mach number
Density
Variation of flow properties with friction
Stagnation pressure
Impulse function
Variation of flow properties with friction
Stagnation pressure
Non dimensional friction factor
A
• The above equation is a differential equation which
relates the changes in the Mach number to the distance
along the duct, x.
• A general problem for the Fanno line is shown in
figure.
• For a given duct length, L, the integration of above
equation between states 1 and 2 would produce a
complicated function of M1 and M2.
• This function have to be evaluated numerically for
each combination of M1 and M2 encountered in a
problem.
Non dimensional friction factor
• It is known that, for all subsonic and super sonic Fanno line flows, the Mach
number tends towards unity.
• Therefore, for a given value of the inlet Mach number in the duct, there always
exist a hypothetical maximum duct length, Lmax, for which the flow is sonic at the
exit of this hypothetical extension as indicated in figure.
• For this reason, reference conditions may be chosen as critical conditions
corresponding to a Mach number of unity.
• However, it should be noted that these critical conditions are not the same as in
isentropic flow because of the different nature of these flows.
• The task is to integrate above equation between any given state and the critical
state as
B
Non dimensional friction factor
• The right hand side of this equation may be evaluated using integration by parts.
• However, on the left-hand side, the friction factor, f is a function of Reynolds
number,
• For flows in constant area duct the product of density and velocity is constant
along the duct as indicated by the continuity equation and therefore the variation
in the Reynolds number is caused only by the variation in the fluid viscosity.
• Then, a mean friction factor, over the duct length may be defined as
C
• And integration of equation B
Non dimensional friction factor
• This equation gives the maximum value of
corresponds to any given initial
Mach number.
• Since the non-dimension friction factor,
is a function of the Mach
number, then the duct length, L, required for the flow Mach number to change
from an initial value of M1 to a final value of M2 may be found from
Air enters a constant-area insulated duct with a Mach number of 0.60, a pressure of
150 kPa and a temperature of 300 K, as shown in figure. Assuming a duct of length
of 0.45 m, a duct diameter of 0.03 m and an average friction factor of 0.005,
determine
a. The Mach number, pressure and temperature at the duct exit, and
b. The frictional force on the duct
Chocking due to friction
• For a specified value of the Mach number at the inlet of the duct, there is always a maximum
duct length for which flow is sonic at the duct exit.
• If the duct length is increased beyond the maximum length the flow characteristics change
within the duct. The variation in properties / characteristics depends on whether the flow is
subsonic or supersonic.
• Consider a duct in which flow is subsonic at the inlet. The duct is long enough such that Mach
number is unity at the exit.
• The process is shown by curve A on T-s diagram. Point A is located at the point of maximum
entropy.
• If the length is increased above its maximum value up to point B the resistance to flow will
increase which will increase entropy.
• Since Mach number cannot exceed unity as it will lead to decrease in entropy, the inlet
conditions will change (i.e. inlet Mach number, mass flow rate will reduce).
• Hence additional duct length results in reduction of mass flow rate and flow jumps to another
Fanno line. This choking is due to friction.
Chocking due to friction in subsonic flows
Chocking due to friction
• Now consider a duct in which supersonic flow exists at the inlet. The duct is sufficiently long
such that flow decelerates and Mach number is unity (and maximum entropy) at the exit.
• If the duct length is extended, flow resistance is increased. Since entropy cannot decrease to
make flow subsonic, shock appears. As the duct length is increased the shock moves upstream
towards the duct inlet.
Adiabatic ducts fed by converging nozzles
Adiabatic ducts fed by converging nozzles
1. No-flow conditions (Pb=Poi): In this case flow does not take place.
2. Subcritical flow regime: p* < pb < p0i i.e. subcritical flow regime in which back
pressure is slightly less than stagnation pressure. The stream leaving is subsonic
so that the exit pressure is equal to back pressure. Further reduction in back
pressure increases mass flow rate and flow shifts to another Fanno line.
3. Critical flow: Mach number is unity at the exit and the flow is choked. Again,
the exit pressure is equal to back pressure.
4. Supercritical flow regime: Further reduction in back pressure cannot increase
mass flow rate. The flow pattern and Mach number is same within duct.
However after the exit (oblique) expansion waves occur.
Air at a stagnation temperature of 400 K and a stagnation pressure of 300 kPa is
supplied to a constant area insulated duct through an isentropic converging nozzle
as shown in figure. The duct has a diameter of 0.04 m and a length of 8.5 m, and it
discharges to the atmosphere with a pressure of 100 kPa. Assume a friction factor of
0.001.
a. Determine the inlet and exit Mach numbers
b. Determine the pressure and temperature at the exit of the duct.
c. Determine the mass flow rate.
d. Show the process on a T-s diagram
Chapter 8
Steady and two-dimensional supersonic flows
Course Title: Compressible Flow and Propulsion System
Course Code: (ME-417)
Introduction
• Consider a supersonic stream of gas flowing in
both sides of a wall with zero thickness which
has a curved section as shown in figure.
• At the point where the wall start to deflect to
turn, the stream lines adjust to the wall begin to
deflect.
• As a result, a pressure disturbance is created
whose head represents a Mach wave.
• The angle of inclination of this wave to the
streamlines is the Mach angle 𝜇1 .
• As the gas follows the curved part of the wall, an
infinite number of Mach number are generated
owing to the continuous change in the direction
of streamlines.
Introduction
• The gas properties and the flow velocity adjust
to the corresponding variation at the walls
through the Mach waves.
• Therefore, two continuous waves-one on the
upper and another on the lower side of the wall –
are formed.
• The flow is decelerated by continuous
compression waves on the lower side of the wall,
while it is accelerated by continuous expansion
waves on the upper side.
• The curvature of the thin waves begins at the
head of these waves and ends at the tails.
• The state of the gas does not change along the
Mach lines.
Introduction
• As the Mach number of the flow increases, the
angle of inclination of the Mach waves to the
streamlines, that is the Mach angle, decreases.
• As long as the Mach number increases across an
expansion wave, then the Mach angle decrease
from 𝜇1 to 𝜇2 causing the Mach waves to
diverge.
• However, across a compression wave, the Mach
number decreases and the Mach angle increases
from 𝜇1 to 𝜇3 causing the Mach waves to
converge and form an oblique shock waves
which is inclined to the flow direction.
Oblique Shock Waves
Governing Equation for the Oblique Shock wave
Figure 8.4
Governing Equation for the Oblique Shock wave
Continuity Equation
𝒎ሶ = 𝝆𝒙 𝑨𝒙 𝑽𝒙𝒏 = 𝝆𝒚 𝑨𝒚 𝑽𝒚𝒏 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝒎ሶ
𝑮 = = 𝝆𝒙 𝑽𝒙𝒏 = 𝝆𝒚 𝑽𝒚𝒏 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑨
Momentum Equation
𝑷𝒙 𝑨𝒙 − 𝑷𝒙 𝑨𝒙 = 𝝆𝒚 𝑨𝒚 𝑽𝟐𝒚𝒏 − 𝝆𝒙 𝑨𝒙 𝑽𝟐𝒙𝒏
Since the area is constant across the shock wave. Also mass flow rate is constant.
𝑷𝒙 − 𝑷𝒚 = 𝝆𝒚 𝑽𝟐𝒚𝒏 − 𝝆𝒙 𝑽𝟐𝒙𝒏
In the tangential direction there is no change in pressure
𝝆𝒚 𝑨𝒚 𝑽𝟐𝒚𝒕 − 𝝆𝒙 𝑨𝒙 𝑽𝟐𝒙𝒕 = 𝟎
𝑽𝒙𝒕 = 𝑽𝒚𝒕
Governing Equation for the Oblique Shock wave
The first law of thermodynamics
Expanding
However, 𝑉𝑥𝑡 = 𝑉𝑦𝑡
The second law of thermodynamics
Relation for the flow of a perfect gas across an
oblique shock wave
Recall equations from chapter 5 (Normal Shock wave)
Relation for the flow of a perfect gas across an
oblique shock wave
a. Relation between the upstream and the downstream Mach number
b. Relation between the shock wave angle and pressure angle
From figure 8.4
Relation for the flow of a perfect gas across an
oblique shock wave
c. Pressure ratio across an oblique shock wave
d. Temperature ratio across an oblique shock wave
e. Density Ratio across an oblique shock wave
Relation for the flow of a perfect gas across an
oblique shock wave
f. Velocity ratio across an oblique shock wave
g. Stagnation Pressure ratio across an oblique shock wave
e. Entropy change across an oblique shock wave
Chapter 7
Flow in constant-area ducts with heat
transfer
Course Title: Compressible Flow and Propulsion System
Course Code: (ME-417)
Governing Equations
Continuity Equation
𝒎ሶ = 𝝆𝟏 𝑨𝟏 𝑽𝟏 = 𝝆𝟐 𝑨𝟐 𝑽𝟐 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝒎ሶ
𝑮 = = 𝝆𝟏 𝑽𝟏 = 𝝆𝟐 𝑽𝟐 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑨
P1,
h1,
𝝆
T1, 𝟏
V1,
P2,
h2,
𝝆
T2, 𝟐
V2
2
1
Q
Momentum Equation
𝑷𝟏 𝑨𝟏 − 𝑷𝟐 𝑨𝟐 = 𝝆𝟐 𝑨𝟐 𝑽𝟐𝟐 − 𝝆𝟏 𝑨𝟏 𝑽𝟐𝟏
𝑰 = 𝑨𝟏 𝑷𝟏 + 𝝆𝟏 𝑽𝟐𝟏 = 𝑨𝟐 𝑷𝟐 + 𝝆𝟐 𝑽𝟐𝟐 = 𝐀 𝐏 + 𝛒𝑽𝟐 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
Governing Equations
The first law of thermodynamics
ሶ = 𝒎(𝒉
ሶ 𝒐𝟐 − 𝒉𝒐𝟏 )
𝑸 = 𝒎𝒒
P1,
h1,
𝝆
T1, 𝟏
V1,
P2,
h2,
𝝆
T2, 𝟐
V2
2
1
𝑽𝟐𝟐
𝑽𝟐𝟏
ሶ = 𝒎ሶ 𝒉𝟐 +
𝑸 = 𝒎𝒒
− 𝒉𝟏 −
𝟐
𝟐
Equation of state
𝑷 = 𝝆𝐑𝐓
Q
Governing Equations
P1,
h1,
𝝆
T1, 𝟏
V1,
Second law of thermodynamics
𝟐
𝒒 = න 𝑻 𝒅𝒔
𝟏
2
1
From thermodynamics relation,
𝒅𝑷
𝑻𝒅𝒔 = 𝒅𝒉 −
𝝆
P2,
h2,
𝝆
T2, 𝟐
V2
𝒅𝝆
𝑻𝒅𝒔 = 𝒅𝒖 − 𝑹𝑻
𝝆
Q
May be used.. For perfect gas
𝒅𝒉 = 𝑪𝑷 𝒅𝑻
𝒅𝒖 = 𝑪𝒗 𝒅𝑻
𝑷 = 𝝆𝑹𝒅𝑻
𝒅𝑻
𝒅𝑷
𝒅𝑻
𝒅𝝆
𝒅𝒔 = 𝑪𝑷
−𝑹
= 𝑪𝒗
−𝑹
𝑻
𝑷
𝑻
𝝆
𝑻𝟐
𝑷𝟐
𝑻𝟐
𝝆𝟐
𝒔𝟐 − 𝒔𝟏 = 𝑪𝑷 𝐥𝐧
− 𝐑 𝐥𝐧
= 𝑪𝒗 𝐥𝐧
− 𝐑 𝐥𝐧
𝑻𝟏
𝑷𝟏
𝑻𝟏
𝝆𝟏
Air flows with negligible friction through a duct with a constant cross-sectional area
of 0.025 m2, as shown in figure. At section 1, the pressure, temperature, and
velocity are 150 kPa, 350 K and 100 m/s respectively. At section 2, the pressure is
75 kPa. If the air is heated between section 1 and 2, determine,
a. The properties at section 2.
b. The amount of heat transfer, and
c. The entropy change
150 kPa
350 K
100 m/s
75 kPa
2
1
Q
The Rayleigh Line
• The Rayleigh line was introduced in chapter 5 of normal shock wave.
• If conditions upstream of the control volume (state 1) are fixed, and the conditions
are to be found at the down stream section (state 2), then for a particular value of
V2: 𝝆𝟐 may be calculated from the continuity equation, (ii) P2 may be evaluated
from the momentum equation, (iii) T2 may be found from equation of state and
finally s2 may be obtained from the equation
• therefore the continuity equation, the momentum equation and the equation of
state defined a locus of state passing through state 1 which is known as the
Rayleigh line.
• On the T-s diagram, since the energy equation hasn’t yet introduced the Rayleigh
line for a perfect gas represents the state with same mass flux, and the same
impulse function but not necessarily the value of the stagnation temperature.
• Therefore, states on Rayleigh line are reachable from each other in presence of
heat transfer effects.
The Rayleigh Line
The Rayleigh Line
Continuity Equation
𝒎ሶ = 𝝆𝑽𝑨 = 𝝆 + 𝒅𝝆 𝑽 + 𝒅𝑽 𝑨
≈0
𝝆𝑽𝑨 = 𝝆𝑽𝑨 + 𝝆𝑨𝒅𝑽 + 𝑽𝑨𝒅𝝆 + 𝐀𝒅𝝆𝒅𝑽
Neglecting the higher order terms
P
h
T
V
𝝆
P+dP
h+dh
T+dT
V+dV
𝝆 + 𝒅𝝆
dq
2
1
dx
𝒅𝑽 𝒅𝝆 𝟏 𝒅 𝑽𝟐
𝒅𝝆
+
=
+
=𝟎
𝑽
𝝆
𝟐 𝑽𝟐
𝝆
Momentum Equation
ሶ
𝑷𝑨 − 𝑷 + 𝒅𝑷 𝑨 = 𝒎ሶ 𝑽 + 𝒅𝑽 − 𝒎𝑽
𝒅𝑷 = −𝝆𝑽𝒅𝑽
Now using the equation of state 𝑷 = 𝝆𝐑𝐓 and the definition of Mach number 𝑉 2 = 𝑀2 𝛾𝑅𝑇it is possible to obtain
𝒅𝑷
𝜸𝑴𝟐 𝒅(𝑽𝟐 )
=−
𝑷
𝟐
𝑽𝟐
The Rayleigh Line
First law of thermodynamics
𝟐
𝑽
(𝑽 + 𝒅𝑽)
𝑪𝑷 𝑻 +
+ 𝒅𝒒 = 𝑪𝑷 𝑻 + 𝒅𝑻 +
𝟐
𝟐
𝟐
P
h
T
V
𝝆
dq
2
1
≈0
𝑽𝟐
𝑽𝟐
(𝒅𝑽)𝟐
𝑪𝑷 𝑻 +
+ 𝒅𝒒 = 𝑪𝑷 𝑻 + 𝑪𝑷 𝒅𝑻 +
+ 𝑽𝒅𝑽 +
𝟐
𝟐
𝟐
Neglecting the second-order differential, it is possible to obtain
𝑪𝑷 𝒅𝑻 + 𝑽𝒅𝑽 = 𝒅𝒒
𝜸𝐑
But for a perfect gas 𝑪𝑷 = 𝜸−𝟏
and 𝑴 =
P+dP
h+dh
T+dT
V+dV
𝝆 + 𝒅𝝆
𝑽
𝜸𝑹𝑻
𝒅𝑻 𝜸 − 𝟏 𝟐 𝒅 𝑽𝟐
𝜸 − 𝟏 𝒅𝒒
+
𝑴
=
𝟐
𝑻
𝟐
𝑽
𝜸𝑹 𝑻
dx
The Rayleigh Line
Equation of state
𝒅𝑷 𝒅𝝆 𝒅𝑻
=
+
𝑷
𝝆
𝑻
The second law of thermodynamics
𝒅𝒒
𝒅𝒔 =
𝑻
P
h
T
V
𝝆
P+dP
h+dh
T+dT
V+dV
𝝆 + 𝒅𝝆
dq
2
1
dx
Variation of flow properties with heat transfer
Velocity
Density
Temperature
Stagnation Pressure
Pressure
Stagnation Temperature
Mach Number
Entropy
Consider a 12 cm diameter tubular combustion chamber. Air
enters the tube at 400 kPa, 500 K and 70 m/s. Fuel with a heating
value of 39000 kJ/kg is burned by spraying in air. The exit Mach
number is 0.8. Determine the rate at which the fuel is burned and
exit temperature.
Thermal Choking
• In Fanno flow, recall that once sufficient duct was added, or the receiver pressure
was lowered far enough, we reached a Mach number of unity at the end of the
duct.
• Further reduction of the receiver pressure could not affect conditions in the flow
system.
• The addition of any more duct caused the flow to move along a new Fanno line at
a reduced flow rate.
• Subsonic Rayleigh flow is quite similar. Figure shows a given duct fed by a large
tank and converging nozzle. Once sufficient heat has been added, M = 1 at the end
of the duct.
• The T–s diagram for this is shown as path 1–2–3. This is called thermal choking.
• It is assumed that the receiver pressure is at p3 or below. Reduction of the receiver
pressure below p3 would not affect the flow conditions inside the system.
Thermal Choking
Thermal Choking
• However, the addition of more heat will change these conditions.
• Now suppose that we add more heat to the system.
• This would probably be done by increasing the heat transfer rate through the walls
of the original duct.
• However, it is more convenient to indicate the additional heat transfer at the
original rate in an extra piece of duct, as shown in Figure.
• The only way that the system can reflect the required additional entropy change is
to move to a new Rayleigh line at a decreased flow rate (due to high pressure
ratio).
• This is shown as path 1–2’–3’– 4 on the T–s diagram.
• Whether or not the exit velocity remains sonic depends on how much extra heat is
added and on the receiver pressure imposed on the system.
Thermal Choking
• The important thing to remember is that once a subsonic flow is thermally choked,
the addition of more heat causes the flow rate to decrease.
• Just how much it decreases and whether or not the exit remains sonic depends on
the pressure that exists after the exit.
• The parallel between choked Rayleigh and Fanno flow does not quite extend into
the supersonic regime. In Fanno flow extension in length of duct in case of
supersonic flow results in shock.
• This does not happen in supersonic flows.
• Figure shows a M = 3.53 flow that has Tt/Tt* = 0.6139.
• For a given total temperature at this section, the value of Tt/Tt* is a direct
indication of the amount of heat that can be added to the choke point.
Thermal Choking
• If a normal shock were to occur at this point, the Mach number after the shock
would be 0.450, which also has Tt/Tt* = 0.6139.
• Thus the heat added after the shock is exactly the same as it would be without the
shock.
• The situation above is not surprising since heat transfer is a function of stagnation
temperature, and this does not change across a shock.
• The shock may occur at some location preceding the Rayleigh flow such as in a
converging–diverging nozzle which produces the supersonic flow
Thermal Choking
Shaft power cycles
Course Title: Compressible Flow and Propulsion System
Course Code: (ME-417)
Assumptions for Ideal Cycle
The assumption of ideal conditions will be taken to imply the following:
a. Compression and expansion processes are reversible and adiabatic, i.e., isentropic.
b. The change of kinetic energy of the working fluid between inlet and outlet of each component
is negligible.
c. There are no pressure losses in the inlet ducting, combustion chambers, heat-exchangers,
intercoolers, exhaust ducting, and ducts connecting the components.
d. The working fluid has the same composition throughout the cycle and is a perfect gas with
constant specific heats.
e. The mass flow of gas. is constant throughout the cycle.
f. Heat transfer in a heat-exchanger (assumed counterflow) is 'complete', so that in conjunction
with (d) and (e) the temperature rise on the cold side is the maximin possible and exactly
equal to the temperature drop on the hot side.
Simple gas turbine cycle
Simple gas turbine cycle
• The relevant steady flow energy equation is
• where Q and Ware the heat and work transfers per unit mass flow. Applying this to each
component, bearing in mind assumption (b), we have
• The cycle efficiency is
Simple gas turbine cycle
• Making use of the isentropic p - T relation, we have
• where r is the pressure ratio
• The cycle efficiency is then
Simple gas turbine cycle
• The specific work output W, upon which the size of plant for a given power depends, is found to
be a function not only of pressure ratio but also of maximum cycle temperature T3. Thus
• which can be expressed as
• where t = T3/ T1
Heat-exchange cycle
• Using the nomenclature of Fig., the cycle efficiency
now becomes
• With ideal heat-exchange T5 = T4, and on substituting
the isentropic p- T relations the expression reduces to
• Thus the efficiency of the heat-exchange cycle is not
independent of the maximum cycle temperature, and
clearly it increases as tis increased.
• Furthermore it is evident that, for a given value ,of t,
the efficiency increases with decrease in pressure
ratio and not with increase in pressure ratio as for the
simple cycle.
Heat-exchange cycle
• This is to be expected because in this limiting case the
Carnot requirement of complete external heat reception
and rejection at the upper and lower cycle temperature is
satisfied.
Reheat cycle
Cycle with reheat and heat-exchange
Cycles with intercooled compression
Methods of accounting for component losses
The performance of real cycles differs from that of ideal cycles for the following reasons
a. Because fluid velocities are high in turbomachinery the change in kinetic energy between inlet and
outlet of each component cannot necessarily be ignored. A further consequence is that the compression
and expansion processes are irreversible adiabatic and therefore involve an increase in entropy.
b. Fluid friction results in pressure losses in combustion chambers and heat exchangers, and also in the
inlet and exhaust ducts. (Losses in the ducts connecting components are usually included in the
associated component losses.)
c. If a heat-exchanger is to be of economic size, terminal temperature differences are inevitable; i.e. the
compressed air cannot be heated to the temperature of the gas leaving the turbine.
d. If a heat-exchanger is to be of economic size, terminal temperature differences are inevitable; i.e. the
compressed air cannot be heated to the temperature of the gas leaving the turbine.
e. The values of CP and 𝛾 of the working fluid vary throughout the cycle due to changes of temperature
and, with internal combustion, due to changes in chemical composition.
f.
With internal combustion, the mass flow through the turbine might be thought to be greater than that
through the compressor by virtue of the fuel added. In practice, about 1-2 per cent of the compressed
air is bled off for cooling turbine discs and blade roots, and we shall see later that the fuel/air ratio
employed is in the region 0·01-0·02.times
Stagnation properties
• Applying the concept of stagnation properties to an adiabatic compression, the energy equation
becomes
• Similarly for a heating process without work transfer,
• The stagnation pressure is thus defined by
Stagnation properties
Compressor isentropic efficiency
Turbine isentropic efficiency
Compressor and turbine efficiencies
Isentropic efficiency
• Because turbomachines are essentially adiabatic, the ideal process is isentropic and the
efficiency is called an isentropic efficiency.
• Making use of the concept of stagnation enthalpy or temperature to take account of any change
in kinetic energy of the fluid between inlet and outlet we have, for the compressor.
• For a perfect gas 3
• Similarly the turbine isentropic efficiency is defined as
Pressure losses
Heat-exchanger effectiveness
Mechanical losses
Polytropic Efficiency
• Consider an axial flow compressor consisting of a number of successive stages.
• If the blade design is similar in successive blade rows it is reasonable to assume
that the isentropic efficiency of a single stage, 𝜂𝑠 , remains the same through the
compressor.
•
Determine the specific work output, specific fuel consumption, and cycle
efficiency for a heat exchange cycle, with following specifications.
Compressor Pressure ratio : 4.0
6
Turbine Inlet Temperature 1100 K
Isentropic efficiency of compressor : 0.85
2
Isentropic efficiency of turbine : 0.87
1
Mechanical Transmission efficiency : 0.99
Combustion efficiency : 0.98
Heat Exchanger Effectiveness : 0.80
Pressure losses
Combustion Chamber : 2% comp. delivery pressure
Heat exchange air side: 3% comp. delivery pressure
Heat exchange gas side: 0.04 bar
Ambient conditions : 1 bar, 285 K
5
3
4
Gas Turbine Cycles for Aircraft Propulsion
Dr. Syed Muhammad Asad Akhter
Assistant Professor
Mechanical Engineering Department
Course Title: Compressible Flow and Propulsion System
Course Code: (ME-417)
Propulsion Systems
• Jet Propulsion is similar to the release of an inflated balloon.
The inflated balloon has compressed air pressing equally against
all the sides.
The air rushes out the open hole at the bottom.
The action is that the air is pushed out in one direction.
The reaction is that the balloon flies in the other direction.
Gas turbine cycles for propulsion
Gas turbine cycles for propulsion
• Jet engines work by taking in air, compressing it, mixing it with fuel and igniting
it to produce hot exhaust gases.
• These gases are then channeled through a turbine which powers the compressor.
• The fast moving exhaust gases exit through a nozzle to produce thrust that propels
aircraft.
Gas turbine cycles for propulsion
• Aircraft gas turbine cycles differ from shaft power cycles in that the useful power
output is in the form of thrust:
i. the whole of the thrust of the turbojet and turbofan is generated in propelling
nozzles,
ii. with the turboprop most is produced by a propeller with only a small
contribution from the exhaust nozzle.
• A second distinguishing feature is the need to consider the effect of forward speed
and altitude on the performance.
• The differing requirements for take-off, climb, cruise and maneuvering are also
recognized.
Gas turbine cycles for propulsion
• High velocity air first flow through the diffuser where it is decelerated, increasing
its pressure and then air is compressed in the compressor.
Gas turbine cycles for propulsion
• The air is mixed with fuel in the combustion chamber, where the mixture is burned
at constant pressure.
Gas turbine cycles for propulsion
• The high pressure, high temperature combustion gases partially expand in the
turbine producing enough power to drive the compressor and auxiliary equipment.
• Finally, the exhaust gases expand in the nozzle to the ambient pressure and leave
the aircraft at the high velocity.
Gas turbine cycles for propulsion
Gas turbine cycles for propulsion
• Consider the schematic diagram of a propulsive duct shown in Figure.
• Relative to the engine, the air enters the intake with a velocity Ca equal and
opposite to the forward speed of the aircraft.
• The power unit accelerates the air so that it leaves with the jet velocity Cj.
• For simplicity the mass flow m is assumed constant (i.e. the fuel flow is
negligible), and thus the net thrust F due to the rate of change of momentum is
(1)
• When the exhaust gases are not expanded completely to Pa in the propulsive duct,
the pressure Pj in the plane of the exit will be greater than Pa
• In this case there will be an additional pressure thrust exerted over the jet exit area
and will be added with the momentum thrust i.e.
(2)
Gas turbine cycles for propulsion
• When the aircraft is flying at a uniform speed Ca in level flight, the thrust is equal
and opposite to the drag of the aircraft at that speed.
• If complete expansion to Pa is assumed in the propelling nozzle and then equation
(1) will be applicable.
• From the equation it is clear that the required thrust can be obtained by designing
the engine to produce either a high-velocity jet of small mass flow or a lowvelocity jet of high mass flow.
Various Types of Jet propulsion Units
• Various propulsion units in the order of decreasing mass flow and increasing jet
velocity are shown.
Various Types of Jet propulsion Units
Performance Parameters of Jet Propulsions
• The propulsion efficiency 𝜂𝑃 can be defined as the ratio of the useful propulsive
energy or thrust power (FCa) to the available kinetic energy.
• The available kinetic energy is sum of thrust energy and unused kinetic energy of
the jet.
• 𝜂𝑃 is not an overall power plant efficiency, because the unused enthalpy in the jet
is ignored. The equation shows that
a. F is a maximum when Ca = 0, i.e. under static conditions, but 𝜂𝑃 is then zero;
b. 𝜂𝑃 is a maximum when Cj/Ca = 1 but then the thrust is zero.
• We may conclude that although Cj is greater than Ca the difference is not too
great.
Performance Parameters of Jet Propulsions
• The thrust and fuel consumption of a jet propulsion unit vary with both cruise
speed and altitude. The flight regimes found to be suitable for the various types of
gas turbine engine are shown in figure.
• The propulsion efficiency is a measure of the effectiveness with which the
propulsive duct is being used for propelling the aircraft.
• The efficiency of energy conversion within the powerplant itself is symbolize by
𝜂𝑒 .
• The rate of energy supplied by the fuel is mfQnet where mf is the fuel mass flow.
This is converted into potentially useful kinetic energy for propulsion
𝑚 𝐶𝑗2 −𝐶𝑎2
2
together with unusable enthalpy in the jet i.e., 𝑚𝐶𝑃 (𝑇𝑗 − 𝑇𝑎 )
Performance Parameters of Jet Propulsions
• The overall efficiency 𝜂𝑜 is the ratio of the useful work done in overcoming drag
to the energy in the fuel supplied
Performance Parameters of Jet Propulsions
• A crude comparison of different engines can be made if the engine performance is
quoted at two operating conditions:
i. the sea-level static performance at maximum power that must meet the· aircraft
take-off requirements and
ii. the cruise performance at the optimum cruise speed and altitude of the aircraft
for which it is intended. However more useful parameter is specific fuel
consumption (SFC) which, for aircraft engines, is defined as the fuel
consumption per unit thrust. 𝜂𝑜 and SFC are related as:
Performance Parameters of Jet Propulsions
• With a given fuel, the value of Qnet will be constant and it can be seen that the
overall efficiency is proportional to Ca/SFC.
• Another important performance parameter is the specific thrust Fs, namely the
thrust per unit mass flow of air (e.g., Ns/kg).
• This provides an indication of the relative size of engines producing the same
thrust because the dimensions of the engine are primarily determined by the
airflow.
• Size is important because of its association not only with weight but also with
frontal area and the consequent drag. SFC and specific thrust are related by
• When estimating the cycle performance at altitude we shall need to know the way
in which ambient pressure and temperature vary with height above sea level.
Performance Parameters of Jet Propulsions
• The variation depends to some extent upon the season and latitude, but it is usual
to work with an average or 'standard' atmosphere.
• For a given speed in m/s the Mach number rises with altitude up to 11000 m
because the temperature is falling
Intake and propelling nozzle efficiencies
Intake
Intake and propelling nozzle efficiencies
• Because of the significant effect of forward speed, the intake must be considered
as a separate component:
• The intake is a critical part of an aircraft engine installation, having a significant
effect on both engine efficiency and aircraft safety.
• The prime requirement is to minimize the pressure loss up to the compressor face
while ensuring that the flow enters the compressor with a uniform pressure and
velocity, at all flight conditions.
• Non-uniform, or distorted, flow may cause compressor surge which can result in
either engine flame-out or severe mechanical damage due to blade vibration
induced by unsteady aerodynamic effects.
• Even with a well-designed intake, it is difficult to avoid some flow distortion
during rapid maneuvering.
Intake and propelling nozzle efficiencies
• Current designs of compressor require the flow to enter the first stage at an axial
Mach number in the region of 0.4 - 0.5.
• Since there is no heat or work transfer, the stagnation temperature is constant
although there is a loss of stagnation pressure due to friction and due to shock
waves at supersonic flight speeds.
Intake and propelling nozzle efficiencies
• Figure (a) shows acceleration of the fluid external to the inlet which occurs when
the inlet operates at a velocity lower than the design value.
• Figure (c) shows deceleration of the fluid external to the inlet which occurs at a
velocity higher than design.
• Under static conditions or at very low forward speeds the intake acts as a nozzle in
which the air accelerates from zero velocity or low Ca to C1 at the compressor
inlet.
• At normal forward speeds, however, the intake performs as a diffuser with the air
decelerating from Ca to C1 and the static pressure rising from pa to p1.
• The total pressure at compressor inlet can be calculated from total temperature
relation given by T01 = Ta + Ca2/2Cp (since T0 does not change in inlet).
• The pressure rise in inlet (p01 - pa) is referred to as the ram pressure.
Intake and propelling nozzle efficiencies
• At supersonic speeds, the pressure rise is due to shock at inlet followed by that due
to subsonic diffusion in the remainder of the duct.
• The intake efficiency can be expressed in a variety of ways, but the two most
commonly used are
i. the isentropic efficiency 𝜂𝑖 (defined in terms of temperature rise) and
ii. the ram efficiency 𝜂𝑟 (defined in terms of pressure rise).
• The total properties and isentropic relations for inlet (case c) are given below. For
inlet and exit static properties are used in relations.
Intake and propelling nozzle efficiencies
Intake and propelling nozzle efficiencies
Intake and propelling nozzle efficiencies
• 𝜂𝑟 can be shown to be almost identical in magnitude to i and the two quantities are
interchangeable, however 𝜂𝑟 is easier to measure experimentally.
• 𝜂𝑖 would be less than this for supersonic intakes, the value decreasing with
increase in inlet Mach number.
• For supersonic intakes, it is more usual to quote values of stagnation pressure ratio
p01/p0a (due to high velocity) as a function of Mach number.
• p01/p0a is called the pressure recovery factor of the intake.
• The supersonic inlet is also required to provide the proper quantity and uniformity
of air to the engine over a wider range of flight conditions.
• The inlet design for supersonic case is more challenging due to flow deceleration
through shock waves.
• The approaching air should be subsonic before entering the compressor.
• This is done by CD duct for a supersonic flow.
Intake and propelling nozzle efficiencies
• For the subsonic inlets, near-isentropic internal diffusion can be relatively easily
achieved, and the inlet flow rate adjusts to the demand.
• The internal aerodynamic performance of a supersonic inlet is a major design
problem, sometimes it requires variable geometry to minimize inlet loss and drag
and provide stable operation.
• An approximate relation for the pressure recovery factor relating to the shock
system itself is
• which is valid when 1 <Ma< 5
Propelling nozzles
• 'Propelling nozzle' refers to the component in which the working fluid is expanded
to give a high-velocity jet.
• With a simple jet engine, usually there is a single nozzle downstream of the
turbine.
• The turbofan may have two separate nozzles for the hot and cold streams, or the
flows may be allowed to mix and leave from a single nozzle.
Propelling nozzles
• When thrust boosting is required, an afterburner may be incorporated in the jet
pipe.
• Between the turbine exit and propelling nozzle there is a jet pipe of a length
determined by the location of the engine in the aircraft.
Propelling nozzles
• In the transition from the turbine annulus to circular jet pipe some increase in area
is provided to reduce the velocity, and hence friction loss, in the jet pipe.
• Depending on the location of the engine in the aircraft, and on whether reheat is to
be incorporated for thrust boosting, the 'propelling nozzle’ comprises some or all
of the items.
• The use of a convergent-divergent nozzle can result in significant increase in
engine weight, length and diameter which would in turn result in major
installation difficulties and a penalty in aircraft weight.
• At operating pressure ratios less than the design value, a convergent-divergent
nozzle of fixed proportions can be less efficient because of the loss incurred by the
formation of shock wave in the divergent portion.
Propelling nozzles
• For these reasons aircraft gas turbines normally employ a convergent propelling
nozzle. A secondary advantage of this type is the relative ease with which the
following desirable features can be incorporated:
i. Variable area which is often required when an afterburner is incorporated
ii. Thrust reverser to reduce the length of runway required for landing, used almost
universally in civil transport aircraft.
Propelling nozzles
• Most of the jet noise is due to mixing of the high-velocity hot stream with the cold
atmosphere, and the intensity decreases as the jet velocity is reduced.
• For this reason the jet noise of the turbofan is less than that of the simple turbojet.
• The-noise level can be reduced by accelerating the mixing process At high
supersonic speed the large ram pressure rise in the intake results in a very high
nozzle pressure ratio.
• The value of p04/pa is then many times larger than the critical pressure ratio and
may be as high as 10-20 for flight Mach numbers in the range 2-3. In that case CD
nozzle can be utilized.
• Two approaches are commonly used: one via an isentropic efficiency 𝜂𝑗 and the
other via a specific thrust coefficient KF.
Propelling nozzles
• KF is defined as the ratio of the actual specific gross thrust, namely
[mC5 + A5(p5 -pa)]/m, to that which would have resulted from isentropic flow.
• 𝜂𝑗 is defined as
• It follows that for given inlet conditions (p04, T04) and an assumed value of 𝜂𝑗 , T5
is given by
• Putting M5 = 1 we have the familiar expression
Propelling nozzles
• Having found Tc from equation, we see from Fig. 3.10(b) that the value of 𝜂𝑗
yields Tc’ which is the temperature reached after an isentropic expansion to the
real critical pressure Pc, namely
Propelling nozzles
• The remaining quantity necessary for evaluating the pressure thrust A5(Pc - Pa), is
the nozzle area A5. For a given mass flow m, this is given approximately by
• Determination of the specific thrust and SFC for a simple turbojet engine,
having the following component performance at the design point at which the
cruise speed and altitude are M= 0.8 and 10,000 m.
• From the I.S.A. table, at 10,000 m
The turbofan engine
• The turbofan engine was originally conceived as a method of improving the
propulsion efficiency of the jet engine by reducing the mean jet velocity,
particularly for operation at high subsonic speeds.
• In the turbofan a portion of the total flow by-passes part of the compressor,
combustion chamber, turbine and nozzle before being ejected through a separate
nozzle as shown in Fig. 3.15.
• Figure 3.15 shows an engine with separate exhausts, but it is sometimes desirable
to mix the two streams and eject them as a single jet of reduced velocity.
The turbofan engine
• Turbofan engines are usually described in terms of bypass ratio, defined as the
ratio of the flow through the bypass duct (cold stream) to the flow at entry to the
high-pressure compressor (hot stream).
• It immediately follows that
• For the particular case where both streams are expanded to atmospheric pressure
in the propelling nozzles, the net thrust is given by
The turbofan engine
• The design .point calculations for the turbofan are similar to those for the turbojet;
in view of this, only the differences calculation will be outlined.
a. Overall pressure ratio and turbine inlet temperature are specified as before, but it
is also necessary to specify the bypass ratio (B) and the fan pressure ratio (FPR).
b. From the inlet conditions and FPR, the pressure and temperature of the flow
leaving the fan and entering the bypass duct can be calculated. The mass flow
down the bypass duct can be established from the total flow and the bypass
ratio. The cold stream thrust can then be calculated as for the jet engine, noting
that air is the working fluid. It is necessary to check whether the fan nozzle is
choked or unchoked; if choked the pressure thrust must be calculated.
c. In the two-spool configuration shown in Fig. 3.15 the fan is driven by the LP
turbine. Calculations for the HP compressor and turbine are quite standard, and
conditions at entry to the LP turbine can then be found. Considering the work
requirement of the low pressure rotor,
The turbofan engine
• The value of B may range from 0·3 to 8 or more, and its value has a major effect
on the temperature drop and pressure ratio required from the LP turbine.
• If the two streams are mixed it is necessary to find the conditions after mixing by
means of an enthalpy and momentum balance; this will be considered following
an example on the performance of an engine similar to that shown in Fig. 3.15.