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Gas Dynamics 1
Isentropic Flow
Gas Dynamics
Agenda
• Introduction• Derivation• Stagnation properties• IF in a converging and converging-diverging
nozzle • Application
Gas Dynamics
Introduction
q=0
P ρ T
Consider a gas in horizontal sealed cylinder with a piston 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.
Gas Dynamics
Isentropiccore flow
Introduction Examples of isentropic flows: Jet or rocket nozzles,
diffusers Airfoils
But in reality there is no real flow is entirely isentropic!!
Gas Dynamics
Derivation
For isentropic flow:
And:
So:
Applying energy eqn to get relation between T & M:
1
1
2
1
1
2
1
2−−
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟⎠
⎞⎜⎜⎝
⎛=
γγγ
ρρ
PP
TT
RTa γ=
1
1
2
1
1
221
22
1
2−−
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟⎠
⎞⎜⎜⎝
⎛==
γγγ
ρρ
PP
aa
TT
22
22
2
21
1VTcVTc pp +=+
( )( )2
22
12
1
1
2
2121
TcVTcV
TT
p
p
++
=
RTV
aVM
γ==
Gas Dynamics
Derivation cont.
But:
And:
So:
1
1
2
1
1
221
22
1
2−−
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟⎠
⎞⎜⎜⎝
⎛==
γγγ
ρρ
PP
aa
TT
222
21
22M
cR
RTV
TcV
pp
−=⎥⎥⎦
⎤
⎢⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡= γγ
γ
22
21
1
2
211
211
M
M
TT
−+
−+= γ
γ
1
21
22
1
2
211
211
−−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−+
−+=
γγ
γ
γ
M
M
PP
11
21
22
1
2
211
211
−−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−+
−+=
γ
γ
γ
ρρ
M
M
Gas Dynamics
Derivation cont.
To find relation between A & M:
Using relation ρ-M and T-M
Where:
222111 VAVA ρρ = ⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
2
1
2
1
1
2
VV
AA
ρρ
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
22
11
2
1
1
2
RTMRTM
AA
γγ
ρρ
( )( )12
1
21
22
2
112
1
1
2
2
121
1
21
1
1
2
2
1
1
2
211
211
−+
−+
−
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−+
−+=⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
γγ
γγ
γ
γ
γ
M
M
MM
KK
MM
KK
KK
MM
AA
2
211 MK −+= γ
Gas Dynamics
Basic Equations forOne-Dimensional Compressible Flow• Control Volume
Gas Dynamics
Basic Equations forOne-Dimensional Compressible Flow
• Continuity
Momentum
Gas Dynamics
Basic Equations forOne-Dimensional Compressible Flow
Second Law of Thermodynamics
• First Law of Thermodynamics
Gas Dynamics
Basic Equations forOne-Dimensional Compressible Flow
Property Relations
• Equation of State
Gas Dynamics
Isentropic Flow of an Ideal Gas– Area Variation
• Basic Equations for Isentropic Flow
Gas Dynamics
Isentropic Flow of an Ideal Gas– Area Variation
• Isentropic Flow
Gas Dynamics
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
Gas Dynamics
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 = const and 0 = const
(Slow down adiabatically and reversibly)
• For a calorically perfect gas , h0 = CPT0 or h = CP T
Gas Dynamics
Stagnation condition is a condition that wouldexist if the flow at any point was isentropicallybrought to or come from rest (V =M= 0).
Stagnation values:
0
0
0
0
ρρ ===
==
TTPPMV⎥⎦
⎤⎢⎣⎡ −+= 20
211 M
TT γ
120
211
−
⎥⎦⎤
⎢⎣⎡ −+=
γγ
γ MPP
11
20
211
−
⎥⎦⎤
⎢⎣⎡ −+=
γγρρ M
Stagnation Properties
Gas Dynamics
Stagnation Properties cont.Examples
10
2
Stagnation point is point 0.
0
0
0
0
ρρ ===
==
TTPPMV
Stagnation point is inside the chamber.
Gas Dynamics
Example:
M T0/T P0/P ρ0/ρ a0/a A/A* θ
0.50 --
2.40
Isentropic Relations in Tabular Form
⎥⎦⎤
⎢⎣⎡ −+= 20
211 M
TT γ
120
211
−
⎥⎦⎤
⎢⎣⎡ −+=
γγ
γ MPP
11
20
211
−
⎥⎦⎤
⎢⎣⎡ −+=
γγρρ M
( )121
2
2* 1
11
21 −+
⎟⎟⎠
⎞⎜⎜⎝
⎛+−+
+=
γγ
γγ
γM
MAA
21
20
211 ⎥⎦
⎤⎢⎣⎡ −+= M
aa γ
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
Gas Dynamics
Pitot Probe Measurementfor Compressible Flow:
V = 0
P0
PIncompressible flow (Bernoulli eqn):
Compressible flow:
20 2
1 VPP ρ=− ( )ρ
PPV −= 02
120
211
−
⎥⎦⎤
⎢⎣⎡ −+=
γγ
γ MPP
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡−⎟
⎠⎞⎜
⎝⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛−
=−
11
21
0γ
γ
γ PPM
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡−⎟
⎠⎞⎜
⎝⎛ +−
⎟⎟⎠
⎞⎜⎜⎝
⎛−
=−
111
21
0γγ
γ PPP
aV
Gas Dynamics
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 psiaAssume: Ideal gas behavior for air
Gas Dynamics
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; onethus has a balance between surface and body forces. Consider the linearmomentum equation:
21
Gas Dynamics 22
Gas Dynamics 23
Gas Dynamics
Critical condition is a conditionthat would exist if the flow wasisentropically acceleratedor decelerated until M = 1.
M*=1 T*, P*, ρ*, A*, a* =
Critical Conditions:
⎥⎦
⎤⎢⎣
⎡+−+
+= 2
11
12* M
TT
γγ
γ
12
11
12* −
⎥⎦
⎤⎢⎣
⎡+−+
+=
γγ
γγ
γM
PP
11
2
11
12* −
⎥⎦
⎤⎢⎣
⎡+−+
+=
γ
γγ
γρρ M
21
2
11
12*
⎥⎦
⎤⎢⎣
⎡+−+
+= M
aa
γγ
γ
1
2
3
4
( )( )12
1
2
11
12* −
+−
⎥⎦
⎤⎢⎣
⎡+−+
+=
γγ
γγ
γMM
AA
A=A*M=1
M<1 M>1
0
1
2
3
4
5
6
7
0 1 2 3
Mach number
A/A*
5
Gas Dynamics
Homeworks1. Calculate the Mach number of two aircraft both travelling with anairspeed of 300m/s. One is traveling at sea level (T=250C); theother at an altitude of 11km (T=-160C.)
2. A perfect gas with is traveling at Mach 3 with a statictemperature of 250K, a static pressure of 101kPa, and a staticdensity 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 temperatureis 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 isentropiccompressible flow calculate the stagnation pressure. Assumingincompressible flow, use Bernoulli’s equation to calculate thestagnation pressure. What is the error in assumingincompressible flow at this Mach number?
4.1=γ