analysis of a disturbance in a gas flow p m v subbarao associate professor mechanical engineering...
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Analysis of A Disturbance in A Gas Flow
P M V SubbaraoAssociate Professor
Mechanical Engineering DepartmentI I T Delhi
Search for More Physics through Mathematics .…
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Analysis of Plane Disturbance
• A control volume for this analysis is shown, and the gas flows from left to right.
• The conditions to the right of the disturbance are uniform, but different from the left side and vice versa.
•The thickness of disturbance is very small.•No chemical reactions.
•There is no friction or heat loss at the disturbance.
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Conservation of Mass Applied to 1 D Steady Flow
0.
Vt
Conservation of Mass:
Conservation of Mass for 1DSF:
0. V
Integrate from inlet to exit :
onstant. CVdVV
x
x
x
x
u
T
u
y
y
y
y
u
T
u
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Gauss Divergence Theorem
Constantˆ.. AV
dAnVVdV
If the velocity is normal to the area :
ConstantA
udA
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yyyxxx AuAu
The area of the disturbance is constant.
yyxx uu
Conservation of mass:
Conservation of momentum: The momentum is the quantity that remains constant because there are no external forces.
x
x
x
x
u
T
u
y
y
y
y
u
T
u
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Conservation of Momentum Applied to 1 D Steady Flow
pVV
.
VdpVdVVVV
.
Using gauss divergence theorem:
AA
dAnpVdAnV ˆ.ˆ.
x
x
x
x
u
T
u
y
y
y
y
u
T
u
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If the velocity is normal to the area :
AA
pdAdAu2
Steady, Inviscid 1-D Flow, Body Forces negligible
xxxyyyyyxx AuAuApAp 22
22xxyyyx uupp
The area of the disturbance is constant.
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Conservation of Energy Applied to 1 D Steady Flow
wqVet
e
.
Steady flow with negligible Body Forces and no heat transfer is an adiabatic flow
wVe
.
For a blissful fluid the rate of work transfer is only due to pressure.
VdAnpVe
.ˆ.
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For a total change from inlet to exit :
AV
VdAnpVdVe
.ˆ.
Using gauss divergence theorem:
AA
VdAnpdAVe
.ˆ
One dimensional flow normal to the area of cross section
0 xxxxyyyyxxxyyy AueAueuApuAp
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0 xyxxxyyy ememuApuAp
Using conservation of mass
With negligible body forces:
2
2uie
022
22
x
xy
yxxxyyy
uim
uimuApuAp
022
22
x
xy
yxxx
xxx
yyy
yyy ui
ui
uA
uAp
uA
uApm
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022
22
x
xy
yx
x
y
y ui
ui
ppm
022
22
x
x
xx
y
y
yy
upi
upim
022
22
x
xy
y
uh
uhm
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The process is adiabatic, or nearly adiabatic, and therefore the energy equation can be written as:
22
22y
yx
x
uh
uh
For calorically perfect gas:
0
22
22TC
uTC
uTC p
yyp
xxp
The equation of state for perfect gas reads
yyyxxx RTpRTp :
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Solution of Simultaneous Equations• If the conditions upstream are known, then there are four
unknown conditions downstream. • A system of four unknowns and four equations is solvable.• There exist multiple solutions because of the quadratic
form of equations. • Out of these multiple solutions, some are physically
possible and some are not.• These Physically possible solutions refer to the universal
law of direction of happening.• Different Physically possible solutions will lead to
development of different products or processes. • The only tool that brings us to the right direction of
happening is the second law of thermodynamics.• This law dictates the direction of happening : Across the
disturbance the entropy can increase or remain constant.
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•In mathematical terms, it can be written as follows:
0 xyxy ssss
For an ideal gas :
0lnln
x
y
x
yp p
pR
T
TC
0ln1
ln
x
y
x
y
p
p
T
T
• We will not use isentropic conditions.• Use more algebra to reduce the number of variables.
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Summary of Equations
yyxx uu Conservation of mass:
22xxyyyx uupp Conservation of momentum:
Conservation of Energy:
22
22y
yx
x
uh
uh
The equation of state for perfect gas
yyyxxx RTpRTp :
0ln1
ln
x
y
x
y
p
p
T
T
Constraint:
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Change in Mach Number between points x & y
yyxx uu Conservation of mass:
Dividing this equation by cx
x
yy
x
xx c
u
c
u x
y
y
yy
x
xx c
c
c
u
c
u
x
yyyxx c
cMM
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y
x
y
x
x
y
c
c
M
M
22xxyyyx uupp Conservation of momentum:
2
2 cp
pc
22yyyxxx upup
22
22
yyyy
xxxx u
cu
c
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22
22
yyyy
xxxx u
cu
c
Dividing this through by cx2/
2
2
2
2
x
yyy
x
yyxx c
cM
c
cM
2
2
2
1
1
y
x
y
x
x
y
c
c
M
M
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2
2
2
1
1
x
y
y
x
x
y
c
c
M
M
y
x
y
x
x
y
c
c
M
M
&
2
2
2
1
1
y
x
y
x
y
x
y
x
c
c
M
M
c
c
M
M
x
y
y
x
x
y
M
M
M
M
c
c2
2
1
1
Momentum Equation : Continuity Equation :
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2121
22y
y
yx
x
xupup
Energy equation in terms for pressure and velocity for a perfect gas
2
22
2
1
2
1
2y
yx
x uc
uc
Dividing this by
1
2 2
xc
1
2
11
2
1 2
2
2y
x
yx M
c
cM
22
22y
ypx
xp
uTc
uTc
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12
1
12
1
2
22
y
x
x
y
M
M
c
c
x
y
y
x
x
y
M
M
M
M
c
c2
2
1
1
22
2
2
2
2
1
1
12
1
12
1
x
y
y
x
y
x
M
M
M
M
M
M
Energy Equation :
Combined Mass, Momentum and Energy Conservation :
Combined Mass & Momentum Equation :
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Combined Mass, Momentum and Energy Conservation
x
x
x
x
u
T
u
y
y
y
y
u
T
u
22
2
2
2
2
1
1
12
1
12
1
x
y
y
x
y
x
M
M
M
M
M
M
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22222222 112
11
2
11 yxyxy MMMMMM
x
0221 222222 xyxyxy MMMMMM
0221 22222244 xyxyxyxy MMMMMMMM
Nothing Happening :
2222 0 xyxy MMMM
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12
2)1(2
22
x
xy
M
MM
With a disturbance between x & y,
This equation relates the downstream Mach number to the upstream.
It can be used to derive pressure ratio, the temperature ratio, and density ratio across the disturbance.
0221 2222 xyxy MMMM
2)1(21 222 xxy MMM
If there is something happening between x & y
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x
y
y
x
x
y
T
T
M
M
c
c
12
1
12
1
2
22
12
2)1(2
22
x
xy
M
MM
Substitute value of My
112
2)1(
21
12
1
2
2
2
x
x
x
x
y
M
M
M
T
T
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1222)1(1
122122
22
xx
xx
x
y
MM
MM
T
T
22
22
1
1221
x
xx
x
y
M
MM
T
T
1
12 2
x
xx
yy
x
y M
T
T
p
p
21
12
2
x
x
x
y
M
M
y
x
y
x
x
y
c
c
M
M
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Change in Entropy Across the disturbance
11
ln
x
y
x
yxy
p
p
T
T
R
ss
x
y
x
ypxy p
pR
T
TCss lnln
x
y
x
ypxy
p
p
T
T
R
C
R
sslnln
x
y
x
yxy
p
p
T
T
R
sslnln
1
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121
22
22
1
12
1
1221ln
x
x
xxxy M
M
MM
R
ss
R
ss xy
Mx
Infeasible
Physically possible solution 2
Solution - 1
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The Nature of Irreversible Phenomenon
Mx
My constant=1.4
This Strong Irreversibility is called as Normal Shock.
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Nature of Normal Shock
• The flow across the shock is adiabatic and the stagnation temperature is constant across a shock.
• The effect of increase in entropy across a shock will result in change of supersonic to subsonic flow.
• The severity of a shock is proportional to upstream Mach Number.
• Normal Shock is A severe irreversible Diffuser.• No capital investment.• Can we promote it ?
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Normal Shock Past F-18
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Reentry Interface Gas Dynamics
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Occurrence of Normal Shock Reentry Vehicles
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Weakening of Normal Shock
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Brayton Cycle for Jet Propulsion
A BC
D
1'-11" 10" 1'-2" 1 2 3 4Compressor
Turbine +Nozzle
Turbofan
Burner
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Gas Dynamic Analysis of Turbojet Engine
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Jet Engine Inlet Duct
• All jet engines have an inlet to bring free stream air into the engine.
• The inlet sits upstream of the compressor and, while the inlet does no work on the flow.
• Inlet performance has a strong influence on engine net thrust.
• Inlets come in a variety of shapes and sizes with the specifics usually dictated by the speed of the aircraft.
• The inlet duct has two engine functions and one aircraft function .
• First : it must be able recover as much of the total pressure of the free air stream as possible and deliver this pressure to the front of the engine compressor .
• Second : the duct must deliver air to the compressor under all flight conditions with a little turbulence .
• Third : the aircraft is concerned , the duct must hold to a minimum of the drag.
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• The duct also usually has a diffusion section just ahead of the compressor to change the ram air velocity into higher static pressure at the face of the engine .
• This is called ram recovery .
• SUBSONIC INLETS• A simple, straight, short inlet works quite well. • On a typical subsonic inlet, the surface of the inlet
from outside to inside is a continuous smooth curve with some thickness from inside to outside.
• The most upstream portion of the inlet is called the highlight, or the inlet lip.
• A subsonic aircraft has an inlet with a relatively thick lip.
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Subsonic Inlet Duct
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Axi-Symmetric Inlets
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Rectangular Inlets
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High Supersonic Flight
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Supersoinc Intakes