Transcript Document

Isentropic Flow In A Converging Nozzle
Converging Nozzle
M=0
x=0
M can equal 1 only where dA = 0
Can one find M > 1 upstream of exit?
M can equal 1 only where dA = 0
Can one find M > 1 upstream of exit?
Converging Nozzle
M=0
x=0
No, since M = 0 at x = 0, can not increase to > 1
without at some x =1 which is not possible because
dA  0 anywhere but at exit.
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?
pb = pe if Me<1
pb=po; V=0
pb=0; V=max
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).
Me = 1
Me < 1
=tVtAt
0
0.528po
p0
Mass Flow Rate vs Back Pressure
po/p = [1 + {(k-1)/2}M2]k/(k-1)
pe*/po = {2/(k+1)}k/(k-1); k=1.4, pe*/po =0.528
po/pe = [1 + {(k-1)/2}M2]k/(k-1)
po/pe* = [1 + {(k-1)/2}]k/(k-1)
po/pe* = [(k + 1)/2}]k/(k-1)
pe*/po = [2/(k+1)] k/(k-1)
k=1.4, pe*/po =0.528
p* = pe
pb = pe
What is the maximum mass flux through nozzle?
dm/dtmax = dm/dtchoked = *V*A* = f(Ae, To, po, k, R)
A* = Ae
V* = c* = [kRT*]1/2 = [{2k/(k+1)}RTo]1/2 Eq. 11.22
 * = p*/(RT*)
p* = po[2/(k+1)]k/(k-1)
T* = To[2/(k+1)]
Eq. 11.21a
Eq. 11.21b
dm/dtmax = dm/dtchoked = *V*A*
*V*A* = {p*/(RT*)}{[{2k/(k+1)}RTo]1/2}Ae
*V*A* =
{po[2/(k+1)]k/(k-1)}/(R{To[2/(k+1)]){[{2k/(k+1)}RTo]1/2}Ae
*V*A* = Aepo(RTo)(-1+1/2) [k1/2] [2/(k+1)](k/(k-1) –1 +1/2)
(k/(k-1) –1 +1/2) = [2k-2(k-1)+(k-1)]/(2[k-1]) = [k+1]/(2[k-1])
*V*A* = Aepo (RTo)-1/2k1/2[2/(k+1)][k+1]/(2[k-1])
*V*A* = Aepo (k/[RTo]-1/2) k1/2[2/(k+1)][k+1]/(2[k-1])
dm/dtmax = dm/dtchoked = *V*A*
*V*A* = {p*/(RT*)}{[{2k/(k+1)}RTo]1/2}Ae
*V*A* =
{po[2/(k+1)]k/(k-1)}/(R{To[2/(k+1)]){[{2k/(k+1)}RTo]1/2}Ae
*V*A* = Aepo(RTo)(-1+1/2) [k1/2] [2/(k+1)](k/(k-1) –1 +1/2)
(k/(k-1) –1 +1/2) = [2k-2(k-1)+(k-1)]/(2[k-1]) = [k+1]/(2[k-1])
*V*A* = Aepo (RTo)-1/2k1/2[2/(k+1)][k+1]/(2[k-1])
*V*A* = Aepo (k/[RTo]-1/2) k1/2[2/(k+1)][k+1]/(2[k-1])
dm/dtmax = dm/dtchoked = *V*A*
*V*A* = {p*/(RT*)}{[{2k/(k+1)}RTo]1/2}Ae
*V*A* =
{po[2/(k+1)]k/(k-1)}/(R{To[2/(k+1)]){[{2k/(k+1)}RTo]1/2}Ae
*V*A* = Aepo(RTo)(-1+1/2) [k1/2] [2/(k+1)](k/(k-1) –1 +1/2)
(k/(k-1) –1 +1/2) = [2k-2(k-1)+(k-1)]/(2[k-1]) = [k+1]/(2[k-1])
*V*A* = Aepo (RTo)-1/2k1/2[2/(k+1)][k+1]/(2[k-1])
*V*A* = Aepo (k/[RTo]-1/2) k1/2[2/(k+1)][k+1]/(2[k-1])
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.
(Air so k = 1)
As long as pb/po > 0.528 then pe = pb and M at throat is < 1.
pe If pb/po is = 0.528 then = pb and M at throat is = 1.
If pb/po is < 0.528 then pe < pb and M at throat is = 1.
pb = po; Me < 1
pb < po ; Me < 1
s = 0
pb < po ; Me < 1
Me = 1
Me = 1
s > 0
Once pe=p*, 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.
True or false?
If pe > p* then pe = pb; but if pe =
p* then as pb decreases pe = p*
and pb< pe.
True or false?
If pe > p* then pe =
pb; but if pe = p*
then as pb decreases
pe = p* and pb< pe.
True or false?
If pe > p* then pe =
pb; but if pe = p*
then as pb decreases
pe = p* and pb< pe.
TRUE
Should be able to be able to
explain why pe can never be
lower than p*.
Regime I:
1 pb/po  p*/po
isentropic
and pe=pb
Regime II:
pb/po < p*/po
Isentropic to throat
Me=1
pe = p* > pb
nonisentropic
expansion
occurs after
leaving throat
Label a, b, c, d, e from above on this graph.
Given:
Ae = 6 cm2; po = 120 kPa;
To = 400K; pb = 90kPa
Find:
pe; dm/dt
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 > 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
dm/dt = VA = e Ve Ae
Isentropic relations:
po/pe = [1+(k-1)Me2/2]k/(k-1);
To/Te = 1 + (k-1)Me2/2;
o/e = [1 + (k-1)Me2/2]1/(k-1)
pe = eRTe
Me = Ve/ce = Ve/ (kRTe)1/2
Ae/A* = (1/Me){[(1+(k-1)Me2]/(k+1)/2}(k+1)/2(k-1)
Given: Ae = 6 cm2; po = 120 kPa;
To = 400K; pb = pe = 90kPa
Find: pe; dm/dt
dm/dt = VA = e Ve Ae
po/pe = [1+(k-1)Me2/2}k/(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
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)(268)]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
Given:
Ae = 6 cm2; po = 120 kPa;
To = 400K; pb = 90kPa = 45 kPa
Find:
pe; dm/dt
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 < 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
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
THE END
… until next time
(A/A*)=(1/M2){(2/[k+1])(1+(k-1)M2/2}(k+1)/(k-1)
CONVERGING-DIVERGING
CONVERGING
I
S
E
N
T
R
O
P
I
C
i – if flow is slow enough, V<0.3M, 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; Mt=1
i, ii and iii are all isentropic flows
But replace Ae by At.
Note: diverging section decelerates subsonic flow, but accelerates
supersonic flow. What is does for sonic flow depends on downstream
pressure, pb.
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 < 1, one >1 for a
given C-D nozzle which still supports isentropic flow.
Flow can not expand isentropically
to pb so expand through a shock.
Flows are referred to as being
overexpanded
because pressure p in nozzle <pb.
When M>1 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.