Upload
semah
View
43
Download
1
Tags:
Embed Size (px)
DESCRIPTION
supersonic. subsonic. N O R M A L. S H O C K. Tape along ceiling and wall of super- sonic nozzle. M a = 1.5. supersonic. subsonic. N O R M A L. S H O C K. M a = 1.5. can occur if converging, diverging and constant area channels - PowerPoint PPT Presentation
Citation preview
Ma = 1.5
supersonic subsonic
Tape along ceilingand wall of super-sonic nozzle.
NORMAL
SHOCK
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
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
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
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
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
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
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
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
h1 + V12 = h2 + V2
2 h01 = h02
h = cp T h01 = c0T01
h02 = c0T02
T01 = T02
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
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
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
(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
breath
(1)
(2)
(2)
(3)
(4)(5)
(6)
6 unknowns: p1, 1, T1, s1, h1, and V1
6 equations:
and one constraint:
(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
BREATH
s1
h1
1
s2
h2
2
s2 > s1
Property Changes Across Shock
“explanation”
To = constant
Entropy increases forirreversible process
1
23
4
56
7
8
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
1
23
4
56
7
8
1
23
4
56
7
8
Since T2 > T1, then h2 > h1
..so V2< V1
1
23
4
56
7
8
Since V decreases across shock,then must increase for V to
remain constant.
1
23
4
56
7
8
Since V is constant across shockand V decreases across shock, then
p must increase across shock.Can also see from Ts diagram.
1
23
4
56
7
8
Since V decreases across shock and T increases across shock
then Ma = V / (kRT)1/2 must decrease supersonic – to – sonic across shock.
?Make sure you
know what goes in here
GIVEN
Want: p02/p01 = f(M1); T2/T1 = f(M1); p2/p1 = f(M1); V2/V1 = f(M1); M2 = f(M1); 2/ 1 = f (M1)
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
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.
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)
To/T = 1 + {(k – 1)/2}M2
T01 = T02
1T2/T1 = f(M1, M2)
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)
V2/V1 = f(M1, M2)
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)
2/1 = f(M1, M2)
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)
p2/p1 = f(M1, M2)
T2/T1 = f(M1, M2)
M2 = f(M1)
=
M2 = f(M1)
=
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
ds > 0M1> 0
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)
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)
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)
Have: T2/T1, V2/V1, 2/1, p2 /p1 = f(M1,M2); M2= f(M1)Still need: p02/p01 = ?
And similarly can substitute for M2 = f(M1) for other properties to get:
EXAMPLE
T1 = 0oCp1 = 60 kPa (abs)V1 = 497 m/s
T2 = 87oC
Find: M2, V2, P02
T1 = 0oCp1 = 60 kPa (abs)V1 = 497 m/s
T2 = 87oC
Find: M2, V2, P02
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)
THE
END
REALLY