Isentropic Flow In A Converging Nozzle

1
Isentropic Flow In A Converging Nozzle
2
Converging Nozzle
M 0
x 0
M can equal 1 only where dA 0 Can one find M gt
1 upstream of exit?
3
M can equal 1 only where dA 0 Can one find M gt
1 upstream of exit?
Converging Nozzle
M 0
x 0
No, since M 0 at x 0, can not increase to gt
1 without at some x 1 which is not possible
because dA ? 0 anywhere but at exit.
4
Now that we have isentropic equations can explore
the following problems (assuming they are
isentropic)
Converging Nozzle
pb may or may not equal pe
How the mass flow changes with decreasing back
pressure in a converging nozzle?
5
pb pe if Melt1
pbpo V0 pb0 Vmax
pb
pe
Converging nozzle operating at various back
pressures. Plotting mass flow as a function of
pb/po as pb decreases from vacuum (pb/po 0)
to when the nozzle is closed (pb/po 1 then po
everywhere and flow 0).
6
Me 1 Me lt 1
?tVtAt
0 0.528po
p0 Mass Flow Rate vs
Back Pressure
po/p 1 (k-1)/2M2k/(k-1) pe/po
2/(k1)k/(k-1) k1.4, pe/po 0.528
7
po/pe 1 (k-1)/2M2k/(k-1) po/pe 1
(k-1)/2k/(k-1) po/pe (k
1)/2k/(k-1) pe/po 2/(k1) k/(k-1)
k1.4, pe/po 0.528
pb pe
p pe
8
What is the maximum mass flux through
nozzle? dm/dtmax dm/dtchoked ?VA f(Ae,
To, po, k, R) A Ae V c kRT1/2
2k/(k1)RTo1/2 Eq. 11.22 ? p/(RT) p
po2/(k1)k/(k-1) Eq. 11.21a T
To2/(k1) Eq. 11.21b
9
dm/dtmax dm/dtchoked ?VA ?VA
p/(RT)2k/(k1)RTo1/2Ae ?VA
po2/(k1)k/(k-1)/(RTo2/(k1))2k/(k1)R
To1/2Ae ?VA Aepo(RTo)(-11/2) k1/2
2/(k1)(k/(k-1) 1 1/2) (k/(k-1) 1 1/2)
2k-2(k-1)(k-1)/(2k-1) k1/(2k-1) ?VA
Aepo (RTo)-1/2k1/22/(k1)k1/(2k-1) ?VA
Aepo (k/RTo-1/2) k1/22/(k1)k1/(2k-1)
10
dm/dtmax dm/dtchoked ?VA ?VA
p/(RT)2k/(k1)RTo1/2Ae ?VA
po2/(k1)k/(k-1)/(RTo2/(k1))2k/(k1)R
To1/2Ae ?VA Aepo(RTo)(-11/2) k1/2
2/(k1)(k/(k-1) 1 1/2) (k/(k-1) 1 1/2)
2k-2(k-1)(k-1)/(2k-1) k1/(2k-1) ?VA
Aepo (RTo)-1/2k1/22/(k1)k1/(2k-1) ?VA
Aepo (k/RTo-1/2) k1/22/(k1)k1/(2k-1)
11
dm/dtmax dm/dtchoked ?VA ?VA
p/(RT)2k/(k1)RTo1/2Ae ?VA
po2/(k1)k/(k-1)/(RTo2/(k1))2k/(k1)R
To1/2Ae ?VA Aepo(RTo)(-11/2) k1/2
2/(k1)(k/(k-1) 1 1/2) (k/(k-1) 1 1/2)
2k-2(k-1)(k-1)/(2k-1) k1/(2k-1) ?VA
Aepo (RTo)-1/2k1/22/(k1)k1/(2k-1) ?VA
Aepo (k/RTo-1/2) k1/22/(k1)k1/(2k-1)
12
k 1.4 R 287 mks 0.04Aepo/To1/2
Note! Maximum mass flow rate will depend on the
exit area Ae, properties of the gas, k and R,
conditions in the reservoir, po and To but not
the pressure at the exit, pe.
13
(Air so k 1)
As long as pb/po gt 0.528 then pe pb and M at
throat is lt 1. pe If pb/po is 0.528 then pb
and M at throat is 1. If pb/po is lt 0.528 then
pe lt pb and M at throat is 1.
14
pb po Me lt 1 pb lt po Me lt 1
?s 0
pb lt po Me lt 1
Me 1
Me 1
?s gt 0
15
Once pep, even if pe is continually lowered
nothing happens upstream of the
throat. Disturbances traveling at the speed of
sound can not pass throat and propagate
upstream.
16
True or false?
If pe gt p then pe pb but if pe p then as
pb decreases pe p and pblt pe.
17
True or false?
If pe gt p then pe pb but if pe p then as
pb decreases pe p and pblt pe.
18
True or false?
If pe gt p then pe pb but if pe p then as
pb decreases pe p and pblt pe.
TRUE
19
Should be able to be able to explain why pe can
never be lower than p.
20
Regime II pb/po lt p/po Isentropic to throat
Me1 pe p gt pb nonisentropic expansion occurs
after leaving throat
Regime I 1? pb/po ? p/po isentropic and pepb
21
(No Transcript)
22
Label a, b, c, d, e from above on this graph.
23
(No Transcript)
24
Given Ae 6 cm2 po 120 kPa To 400K pb
90kPa Find pe dm/dt
25
Check for choking p/po 0.528 p 0.528 po
0.528 (120 kPa) 63.4 kPa If back pressure is
less than this than flow is choked. pb 90 kPa gt
p so flow is not choked and
pe pb 90 kPa
Given Ae 6 cm2 po 120 kPa To 400K pb
90kPa Find pe dm/dt
26
dm/dt ?VA ?e Ve Ae
Isentropic relations po/pe 1(k-1)Me2/2k/(k-1
) To/Te 1 (k-1)Me2/2 ?o/?e 1
(k-1)Me2/21/(k-1) pe ?eRTe Me Ve/ce Ve/
(kRTe)1/2 Ae/A (1/Me)(1(k-1)Me2/(k1)/2(k
1)/2(k-1)
Given Ae 6 cm2 po 120 kPa To 400K pb
pe 90kPa Find pe dm/dt
27
dm/dt ?VA ?e Ve Ae
po/pe 1(k-1)Me2/2k/(k-1) Me2 5(po/pe)2/7
1 0.428 Me 0.654 To/Te 1
(k-1)Me2/2 Te To / 1 0.2 Me2 400/1
0.2(0.654)2 368K
Given Ae 6 cm2 po 120 kPa To 400K pb
pe 90kPa Find pe dm/dt
28
dm/dt ?VA ?e Ve Ae
Me 0.654 kPa Te 368K Me Ve/ce Ve/
(kRTe)1/2 Ve Me (kRTe)1/2 0.654(1.4)(287)(26
8)1/2 Ve 251 m/s ?e pe/(RTe)
90,000/(287)(368) 0.851 kg/m3
?e Ve Ae 0.128 kg/s
Given Ae 6 cm2 po 120 kPa To 400K pb
pe 90kPa Find pe dm/dt
29
Given Ae 6 cm2 po 120 kPa To 400K pb
90kPa 45 kPa Find pe dm/dt
30
Check for choking p/po 0.528 p 0.528 po
0.528 (120 kPa) 63.4 kPa If back pressure is
less than this than flow is choked. pb 45 kPa lt
p so flow is choked and
pe ? pb pe p 63.4 kPa
Given Ae 6 cm2 po 120 kPa To 400K pb
45kPa Find pe dm/dt
31
dm/dtchoked 0.04Aepo/To1/2
k 1.4 R 287 mks 0.04Aepo/To1/2
?e Ve Ae 0.04(0.0006)120000/4001/2 0.144
kg/s
Given Ae 6 cm2 po 120 kPa To 400K pb
pe 90kPa Find pe dm/dt
32
THE END
until next time
33
(No Transcript)
34
(No Transcript)
35
(No Transcript)
36
(No Transcript)
37
(A/A)(1/M2)(2/k1)(1(k-1)M2/2(k1)/(k-1)
38
(No Transcript)
39
(No Transcript)
40
(No Transcript)
41
(No Transcript)
42
CONVERGING
CONVERGING-DIVERGING
I S E N T R O P I C
43
i if flow is slow enough, Vlt0.3M, then ?
incompressible so B. Eq. holds. ii still
subsonic but compressibility effects more
apparent, B.Eq. Not good. iii highest pb where
flow is choked Mt1 i, ii and iii are all
isentropic flows
But replace Ae by At.
44
Note diverging section decelerates subsonic
flow, but accelerates supersonic flow. What is
does for sonic flow depends on downstream
pressure, pb.
45
Note diverging section decelerates subsonic
flow, but accelerates supersonic flow. What is
does for sonic flow depends on downstream
pressure, pb. There are two Mach numbers, one lt
1, one gt1 for a given C-D nozzle which still
supports isentropic flow.
46
Flow can not expand isentropically to pb so
expand through a shock. Flows are referred to as
being overexpanded because
pressure p in nozzle ltpb.
When Mgt1 and isentropic then flow is said to be
at design conditions.
Lowering pb further will have no effect
upstream, where flow remains isentropic. Flow
will go through 3-D irreversible expansion. Flow
is called underexpanded, since
additional expansion takes place outside the
nozzle.
47
(No Transcript)
48
(No Transcript)
49
(No Transcript)
50
(No Transcript)
51
(No Transcript)
52
(No Transcript)
53
(No Transcript)
54
(No Transcript)
55
(No Transcript)