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GasDynamics 1
Isentropic Flow
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GasDynamics
Agenda
Introduction
Derivation
Stagnation properties IF in a converging and converging-diverging
nozzle
Application
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GasDynamics
Introduction
q=0
P T
Consider a gas in horizontal sealed cylinder with apiston at one end. The gas expands outwards
moving the piston and performing work. The walls
of the piston are insulated and no heat transfer takes
place. Is this an isentropic process?
The mathematical relationships between
pressure, density, and temperature are
known as the isentropic flow relations.
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GasDynamics
Isentropiccore flow
Introduction
Examples of isentropic flows: Jet or rocket nozzles,
diffusers Airfoils
But in reality there is no real flow is entirely
isentropic!!
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GasDynamics
Derivation
For isentropic flow:
And:
So:
Applying energy eqn to get relation between T & M:
1
1
2
1
1
2
1
2
=
=
P
P
T
T
RTa =
1
1
2
1
1
2
2
1
2
2
1
2
=
==
P
P
a
a
T
T
22
2
22
2
11
VTc
VTc pp +=+
( )( )2
2
2
1
2
1
1
2
21
21
TcV
TcV
T
T
p
p
+
+=
RT
V
a
VM
==
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GasDynamics
Derivation cont.
But:
And:
So:
1
1
2
1
1
2
2
1
2
2
1
2
=
==
PP
aa
TT
222
2
1
22M
c
R
RT
V
Tc
V
pp
=
=
2
2
2
1
1
2
2
11
2
1
1
M
M
T
T
+
+
=
1
2
1
2
2
1
2
2
11
2
11
+
+
=
M
M
P
P
1
1
2
1
2
2
1
2
2
11
2
11
+
+
=
M
M
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GasDynamics
Derivation cont.
To find relation between A & M:
Using relation -M and T-M
Where:
222111 VAVA =
=
2
1
2
1
1
2
V
V
A
A
=
22
11
2
1
1
2
RTM
RTM
A
A
( )
( )121
2
1
2
2
2
112
1
1
2
2
12
1
1
21
1
1
2
2
1
1
2
2
11
2
11
+
+
+
+
=
=
=
M
M
M
M
K
K
M
M
K
K
K
K
M
M
A
A
2
2
11 MK
+=
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GasDynamics
Basic Equations for
One-Dimensional Compressible Flow
Control Volume
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GasDynamics
Basic Equations for
One-Dimensional Compressible Flow
Continuity
Momentum
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GasDynamics
Basic Equations for
One-Dimensional Compressible Flow
Second Law of Thermodynamics
First Law of Thermodynamics
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GasDynamics
Basic Equations for
One-Dimensional Compressible Flow
Property Relations
Equation of State
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GasDynamics
Isentropic Flow of an Ideal Gas
Area Variation
Basic Equations for Isentropic Flow
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GasDynamics
Isentropic Flow of an Ideal Gas
Area Variation
Isentropic Flow
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GasDynamics
Stagnation Conditions
Total (Stagnation) conditions :
A point (or points) in the flow where V = 0.
Fluid element adiabatically slow down
A flow impinges on a solid object
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GasDynamics
Stagnation Conditions (cont.)
From Energy Equation and the first law of thermodynamics
Total enthalpy = Static enthalpy + Kinetic energy (per unit mass)
Steady and adiabatic flow h0 = const (h01 = h02)
Steady, inviscid, adiabatic flow T0 = const
Isentropic flow P0 = constand 0 = const
(Slow down adiabatically and reversibly)
For a calorically perfect gas , h0 = CPT0 or h = CP T
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GasDynamics
Stagnation condition is a condition that would
exist if the flow at any point was isentropically
brought to or come from rest (V =M= 0).
Stagnation values:
0
0
0
0
=
=
=
==
TT
PP
MV
+=
20
2
11 M
T
T
120
2
11
+=
M
P
P
1
1
20
2
11
+=
M
Stagnation Properties
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GasDynamics
Stagnation Properties cont.
Examples
1
0
2
Stagnation point is point 0.
0
0
0
0
=
=
=
==
TT
PP
MV
Stagnation point is inside the chamber.
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GasDynamics
Example:
M T0/T P0/P 0/ a0/a A/A*
0.50 -
-
2.40
Isentropic Relations in Tabular Form
+=
20
2
11 M
T
T
120
2
11
+=
M
P
P
1
1
20
2
11
+=
M
( )121
2
2
*1
1
1
21 +
+
+
+=
M
MA
A
2
1
20
2
11
+= M
a
a
0.50 1.05000 1.18621 1.12973 1.02470 1.33984
2.40 2.15200 17.08589 7.59373 1.50000 2.63671 36.74650
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GasDynamics
Pitot Probe Measurementfor Compressible Flow:
V = 0
P0
PIncompressible flow (Bernoulli eqn):
Compressible flow:
20
21 VPP = ( ) PPV =
02
120
2
11
+=
M
P
P
=
11
21
0
P
PM
+
=
111
21
0
P
PP
a
V
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GasDynamics
Example (Compressible pitot tube)
20
Given: Air at u = 750 fts , Mercury manometer which reads a
change in height of 8 inches.
Find: Static pressure of air in psia
Assume: Ideal gas behavior for air
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GasDynamics
Analysis: First consider the manometer which is governed by fluid statics. In fluid
statics, there is no motion, thus there are no viscous forces or fluid inertia; one
thus has a balance between surface and body forces. Consider the linear
momentum equation:
21
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GasDynamics 22
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GasDynamics 23
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GasDynamics
Critical condition is a conditionthat would exist if the flow was
isentropically accelerated
or decelerated until M = 1.
M*=1 T*, P*,*, A*, a* =
Critical Conditions:
+
+
+=
2
1
1
1
2*M
T
T
12
1
1
1
2*
+
+
+=
M
P
P
1
1
2
1
1
1
2*
+
+
+=
M
2
1
2
1
1
1
2*
+
+
+= M
a
a
1
2
3
4
( )( )12
1
2
1
1
1
2* +
+
+
+=
MMA
A
A=A*
M=1
M1
0
1
2
3
4
5
6
7
0 1 2 3
Mach number
A/A*
5
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GasDynamics
Homeworks
1. Calculate the Mach number of two aircraft both travelling with anairspeed of 300m/s. One is traveling at sea level (T=250C); the
other at an altitude of 11km (T=-160C.)
2. A perfect gas with is traveling at Mach 3 with a static
temperature of 250K, a static pressure of 101kPa, and a static
density of 1.4077kg/m3
. Determine the stagnation temperature,pressure, and density values.
3. An aircraft is flying at 80m/s at sea level where the temperature
is 200C, density is 1.225kg/m3 and pressure is 1030.1mbar.
Assuming R=287 J/kgK what Mach number is the aircraft flying?
Air stagnates near the leading edge. Assuming isentropic
compressible flow calculate the stagnation pressure. Assuming
incompressible flow, use Bernoullis equation to calculate the
stagnation pressure. What is the error in assuming
incompressible flow at this Mach number?
4.1=