M a = 1.5
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Transcript of M a = 1.5
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Ma = 1.5
supersonic subsonic
Tape along ceilingand wall of super-sonic nozzle.
NORMAL
SHOCK
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Ma = 1.5
supersonic subsonic
NORMAL
SHOCK
• can occur if converging, diverging and constant area channels• shock wave involves supersonic to subsonic flow• associated with rapid deceleration, pressure and entropy rise• only To is constant across shock
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Can occur in internal and external flow, must be between supersonic to subsonic, irreversible, “abrupt discontinuity” (0.2 microns ~ 10-5 in)
Ma = 1.7
Ma
=1.5
normal shock(irreversiblediscontinuity)
Interested in changes across shock rather than what’s happening in shock
Important for design of inlets for high performance aircraft and supersonic wind tunnels
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Thickness is about 0.2 microns (10-5 inches)“treat” as abrupt discontinuity
p, , T can be very largeDecelerations may be of the order
of tens of millions of gs
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Look at fundamental equations: cons. of mass, momentum and energy, 2nd Law of thermodynamics, property relations
for an ideal gas with constant specific heats
Because shock so thin, A1 = A 2 and Rx = 0;because control volume boundaries are far from shock, no gradients so no heat transfer
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Quasi-One-Dimensional, Steady, FBx = 0, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0,
effects of gravity = 0, ideal gas, cv, cp is constant
Propertyrelationsfor idealgas withcv and cp
constant
Cons. Of Mass
Cons. of Momentum
Cons. of Energy
2nd Law of Thermodynamics
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Quasi-One-Dimensional, Steady, FBx = 0, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0,
effects of gravity = 0, ideal gas, cv, cp is constant
Conservation of Mass
Shock width is extraordinarily thin so A1 = A2
1V1 = 2V2 = (dm/dt)/A
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Quasi-One-Dimensional, Steady, FBx = 0, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0,
effects of gravity = 0, ideal gas, cv, cp is constant
Momentum Equation
Shock width is extraordinarily thin so Rx is negligible;
dm/dt = VAp1 + 1V1
2 = p2 + 2V22
0
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Quasi-One-Dimensional, Steady, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0,
effects of gravity = 0, ideal gas, cv, cp is constant
First Law of ThermodynamicsConservation of Energy
No temperature gradients at stations 1 and 2 so adiabatic
h1 + V12 = h2 + V2
2 ; h01 = h02
0
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h1 + V12 = h2 + V2
2 h01 = h02
h = cp T h01 = c0T01
h02 = c0T02
T01 = T02
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Quasi-One-Dimensional, Steady, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0,
effects of gravity = 0, ideal gas, cv, cp is constant
Second Law of ThermodynamicsConservation of Energy
No temperature gradients at stations 1 and 2 so adiabatic
s22V2A - s11V1A 0; s2 > s1
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Second Law of ThermodynamicsConservation of Energy
s22V2A - s11V1A 0; s2 > s1
2nd Law by itself is little help in calculating entropy.To calculate entropy use 1st and 2nd Laws, ideal gas,constant cp to get:
s2 – s1 = cpln(T2/T1) – Rln(p2/p1)
Quasi-One-Dimensional, Steady, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0,
effects of gravity = 0, ideal gas, cv, cp is constant
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Quasi-One-Dimensional, Steady, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0,
effects of gravity = 0, ideal gas, cv, cp is constant
Equation of State ~ Ideal Gas
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(1)
(2)
(2)
(3)
(4)(5)
(6)
To recap:Major simplifying features ~Because so thin
A1 = A 2; Rx = 0Because boundaries of control volume are far from shockNo gradients of V, T, , p there
Q/dm = 0
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breath
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(1)
(2)
(2)
(3)
(4)(5)
(6)
6 unknowns: p1, 1, T1, s1, h1, and V1
6 equations:
and one constraint:
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(1)
(2)
(2)
(3)
(4)(5)
(6)
Fanno Line friction
Rayleigh Line heat tranfer
Normal shock must satisfy eqs: 1-6, so must lie on intersection of Fanno and Rayleigh lines.
IGNORE
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BREATH
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s1
h1
1
s2
h2
2
s2 > s1
Property Changes Across Shock
“explanation”
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To = constant
Entropy increases forirreversible process
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p02 < p01
T2 > T1
T = To/[1 + (k-1)M2/2]T2/T1 = [1 + (k-1)M1
2/2]/[1 + (k-1)M2
2/2]
p02 < p01 (prob. 11.2)
T2 > T1
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Since T2 > T1, then h2 > h1
..so V2< V1
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Since V decreases across shock,then must increase for V to
remain constant.
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Since V is constant across shockand V decreases across shock, then
p must increase across shock.Can also see from Ts diagram.
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Since V decreases across shock and T increases across shock
then Ma = V / (kRT)1/2 must decrease supersonic – to – sonic across shock.
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?Make sure you
know what goes in here
GIVEN
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Want: p02/p01 = f(M1); T2/T1 = f(M1); p2/p1 = f(M1); V2/V1 = f(M1); M2 = f(M1); 2/ 1 = f (M1)
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s1
h1
1
s2
h2
2
(1)
(2)
(2)
(3)
(4)(5)
(6)
6 equations 6 unknowns
Hard to solve, much easier if had ratios (e.g. p1/p2) in terms of M1
Ma1 Ma2
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Strategy ~
Step #1: Obtain property ratios, p1/p2, T1/T2, etc. in terms of M1 and M2.
Step #2: Develop relationship between M1 and M2.
Step #3: Recast property ratios, p1/p2, T1/T2, etc. in terms of M1.
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Want: p02/p01 = f(M1, M2); T2/T1 = f(M1,M2); p2/p1 = f(M1,M2); V2/V1 = f(M1,M2);
M2 = f(M1); 2/ 1 = f (M1, M2)
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To/T = 1 + {(k – 1)/2}M2
T01 = T02
1T2/T1 = f(M1, M2)
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Want: p02/p01 = f(M1, M2); T2/T1 = f(M1,M2); p2/p1 = f(M1,M2); V2/V1 = f(M1,M2);
M2 = f(M1); 2/ 1 = f (M1, M2)
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V2/V1 = f(M1, M2)
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Want: p02/p01 = f(M1, M2); T2/T1 = f(M1,M2); p2/p1 = f(M1,M2); V2/V1 = f(M1,M2);
M2 = f(M1); 2/ 1 = f (M1, M2)
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2/1 = f(M1, M2)
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Want: p02/p01 = f(M1, M2); T2/T1 = f(M1,M2); p2/p1 = f(M1,M2); V2/V1 = f(M1,M2);
M2 = f(M1); 2/ 1 = f (M1, M2)
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p2/p1 = f(M1, M2)
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T2/T1 = f(M1, M2)
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M2 = f(M1)
=
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M2 = f(M1)
=
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M2 = f(M1)
Let: L = M12(1 + M1
2 (k-1)/2) / (1 +k M12)2
after much algebra(M2
2)2 ((1/2)(k-1) – k2L) + M22(1-2kL) – L = 0
Gas Dynamics - John
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ds > 0M1> 0
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Want: p02/p01 = f(M1, M2); T2/T1 = f(M1,M2); p2/p1 = f(M1,M2); V2/V1 = f(M1,M2);
M2 = f(M1); 2/ 1 = f (M1, M2)
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Have: T2/T1, V2/V1, 2/1, p2 /p1 = f(M1,M2); M2= f(M1)Still need: p02/p01 = ?
po/p = [1 + M2{(k – 1)/2}]k/(k-1)
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Want: p02/p01 = f(M1, M2); T2/T1 = f(M1,M2); p2/p1 = f(M1,M2); V2/V1 = f(M1,M2);
M2 = f(M1); 2/ 1 = f (M1, M2)
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Have: T2/T1, V2/V1, 2/1, p2 /p1 = f(M1,M2); M2= f(M1)Still need: p02/p01 = ?
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And similarly can substitute for M2 = f(M1) for other properties to get:
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EXAMPLE
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T1 = 0oCp1 = 60 kPa (abs)V1 = 497 m/s
T2 = 87oC
Find: M2, V2, P02
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T1 = 0oCp1 = 60 kPa (abs)V1 = 497 m/s
T2 = 87oC
Find: M2, V2, P02
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T1= 273o K , p1 = 60 kPa,V1 = 497 m/s, T2 = 360oKFind: M2,V2, p02,
M1 = V1/(kRT1) = 1.5M2 = 0.70
p02 = p010.93po1 /p1 = [1 + M1
2{(k – 1)/2}]k/(k-1)
p01= p13.67 = 220.2p02 = p010.93p02 = 205 kPa (abs)
V2 = V1/1.24 = 267 m/sOther ways ~ M1 = V1/(kRT1); T01= T1(1 + M1
2(k-1)/2); T01 = T02; M2 =[{2/(k-1)}{(T02/T2) – 1}]1/2
V2 = M2/(kRT2)p2 = p1 + 1V1(V1 – V2); 1 = p1/(RT1); p02 = p2[1 + M2
2(k-1)/2]k/(k-1)
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THE
END
REALLY