Exit Conditions of a Vapor Stream though a Nozzle 1

[radaes]

Hi,

I'm trying to calculate the exit conditions of a high-pressure LPG nozzle. I have the dimensions (interior @ state 1 and exit) of the nozzle and the initial conditions of the vapor (P1, T1), as well as the mass flow rate (mdot).
P1 & T1 will give me density (rho)
since mdot=rho*A*V, i can solve for velocity, thereby defining state 1.
i need to find pressure & velocity at the exit of the nozzle.
i'm treating it as an adiabatic process and ignoring frictional effects (for now).
I know that Pexit is (supposed to be) below ambient atmospheric. I also know I can't just use Q1=Q2, since this is a compressible process and rho is a function of the pressure, which is a function of velocity, which is a function of rho, which is a function of pressure....
I'm stuck in a loop. any help, suggestions, or ideas?

cheers,

rad

 

[Latexman]

Sailoday,

After some study, you were right.

Sailoday and aviat,

I believe I know where my thinking went astray now. I kept thinking this problem was like condition 3 in Fig. 5.21 (a) in Vol. 1 of Shapiro and Fig. 3c in aviat’s link – a normal shock in the nozzle. Why? Because when the ambient pressure is between the exit pressure that results when the flow just chokes (condition 2 in Shapiro and Fig. 3b in aviat’s link) and the exit pressure that results when the “design condition” is reached (condition 6 in Shapiro and Fig. 3f in aviat’s link), I don’t see how to determine which flow condition and regime you have. After playing with the on-line nozzle, I see Sailoday was right. How do you determine which flow condition and regime you have in this case?


Good luck,
Latexman

 

[sailoday28]

Latexman (Chemical)
How do you determine which flow condition and regime you have in this case?
The conditions within the nozzle and to the exit plane will be the same for regimes III and IV (Shapiro)based on isentropic flow. [Path (4), the shock occurs at the exit.]

The path of the flow downstream of nozzle has to be determined from measurements outside the nozzle.

 

[Latexman]

Sailoday,

But I was asking about regimes II and III. At least I think I was asking about regimes II and III. I'll ask another way. Refer to Shapiro, Vol. 1, p. 140, Fig. 5.21 (a):

As P[sub]B[/sub] is reduced from condition 2 to condition 4 a normal shock wave will move from the throat (condition 2) to the exit plane (conditoion 4). This is regime II.

As P[sub]B[/sub] is further reduced from condition 4 to condition 6 (the design condition) a compression occurs beyond the exit plane with oblique shock waves. This is regime III.

Also see "Experimental Results" on pages 141 to 143 for pictures of this.

My question is as follows. Let's say I know P[sub]exit subsonic[/sub] (condition 2) and P[sub]exit supersonic[/sub] (condition 6) from solving for M[sub]subsonic[/sub] (condition 2) and M[sub]supersonic[/sub] (condition 6) by using the area ratio equation (use A[sub]exit[/sub]/A* in Eq. 4.19) and using Eqs. 4.14b. If P[sub]B[/sub] is somewhere between P[sub]exit subsonic[/sub] and P[sub]exit supersonic[/sub] like this problem, how do I tell whether I have regime II or III? I don't know how to determine the P[sub]B[/sub] for condition 4.

Good luck,
Latexman

 

[sailoday28]

Latexman
"As PB is reduced from condition 2 to condition 4 a normal shock wave will move from the throat (condition 2) to the exit plane (conditoion 4)".---THIS IS THE REGION BOUNDED BY REGION II.

"As PB is further reduced from condition 4 to condition 6 (the design condition) a compression occurs beyond the exit plane with oblique shock waves. This is regime III."
WE ARE IN AGREEMENT
A/A* should yield 2 solutions, one for subsonic and one for supersonic.
Using the isentropic relations eq.(4-14b)sould then yield the pressure at the exit with no shock.
With no shock compare the calculated pressure at the exit from the Mach no with the back pressure. This will determine whether external flow is with regions III or IV.

With a shock at the exit, Pdownsteam at exit=pback pressure.
c=sound speed
Mass flux W/A= pexit*gamm*(Mdownsteam of shock)/Cdownsteam of shock.
W/A and P at exit are known. M and C downsteam both unknown
BUT
From energy equation Co^2=c^2 +(gamma-1)/2*vel^2 down stream of shock. or (Co/c)^2=(gamm-1)/2*M^2downsteam of shock. Again M and C downstream of shock are unknown.
2 equations, 2 unknowns. If Mdownsteam <1, Then path is (4)
I hope I haven't rambled on too much.
Please check my formulation and comment.
Regards

 

[sailoday28]

LATEXMAN REVISE MY LAST POSTING OF

"c=sound speed
Mass flux W/A= pexit*gamm*(Mdownsteam of shock)/Cdownsteam of shock.
W/A and P at exit are known. M and C downsteam both unknown
BUT
From energy equation Co^2=c^2 +(gamma-1)/2*vel^2 down stream of shock. or (Co/c)^2=(gamm-1)/2*M^2downsteam of shock. Again M and C downstream of shock are unknown.
2 equations, 2 unknowns. If Mdownsteam <1, Then path is (4)"
I hope I haven't rambled on too much.
Please check my formulation and comment.
Regards

WITH Mx=MACH NO. UPSTREAM OF SHOCK AT EXIT, My MACH NO. DOWNSTREAM OF SHOCK ----Mx AND Px KNOWN
PG 118 SHAPIRO EQ.5.16B YIELDS Mx
EQ 5.14 THEN YIELDS Py THE DOWNSTREAM PRESSURE. IF Py MATCHES BACK PRESSURE, THEN THAT IS THE PATH.

Again, I appologize to those without access to "The dynamics and thermodynamics of compressible fluid flow"

 

[aviat]

I found a textbook in the library titled “Introduction to Fluid Mechanics” by James E.A.John and William L. Haberman, which has a complete chapter on this subject.
They use the tables from NACA, which start on page 21 in the site below. The tables are for air with specific heats of 1.4, so you may want to adjust for propane, which has a specific heat of 1.13. NACA assumed that for a one-dimensional flow in a channel the flow was isentropic (adiabatic and reversible) and frictionless. With isentropic flow stagnation or total temperature and pressure remain constant while static temperature and pressure varies though system.

P1 = 240 kPa
T1 = 330 K
Area 1 = 2.267 cm^2
Area throat = 0.3739 cm^2
Area exit = 0.45 cm^2
Pb (back pressure) = 101.3 kPa
G = 1.4
All pressures are in kPa.
Pb/P1 = 0.4221 which < (Pb/P1)crit (or 0.5283) therefore, nozzle is chocked for all back pressure below 0.5283(240)= 126.8. M1 for chocked nozzle (from tables) P = 0.5283(240)= 126.8, T = 0.8333(330) = 275K.
Mdot =( Pe/(R*T)*At*Ve = (Pe/R*T)*At* sqrt(G*R*T)
= (126.8kPa/m^2)*(0.3739x10-4m^2)/((0.2870kN/kg.K)*275K) * sqrt(1.4*287N.m/kg*275K) = 0.024kg/s.
Exit V = A/A* = 0.45/0.3739 = 1.2. From supersonic tables for A/A* of 1.2 Me=1.54
Ve =Me* sqrt (G*R*T) T=0.6897(330) = 223.8, which = 1.54*sqrt((1.4*(287.0 N.m/kg.K)*223.8K) = 461.8 m/s.
Again in order to get more accurate numbers you will have to adjust for propane.

From these tables you can also determine pressures across normal shocks. From the above we know that for back pressures 0< Pb < 126.8 nozzle is chocked. For a shock forming just downstream of throat, with a A/A* of 1.2 in subsonic table: M = 0.59, p/pt = 0.7901 Pb = 0.7901(240), therefore shock will form at 189.6. For shock at nozzle exit, with isentropic flow in nozzle up to the exit plane, the Mach number M1 just before the shock, it can be found in the supersonic table for, M1 = 1.54 P1/Pt = 0.2570 or P1 = 61.68. From right side of table, after M2, (called Normal Shock Table in text) we find P2/P1 = 2.6 so P2 = 160.4. Therefore a normal shock will appear in the nozzle for 160.4 < Pb < 189.6. Oblique shocks will appear in the nozzle at the exit for 61.68 < Pb < 160.4 and expansion waves for Pb < 61.68.
I hope some of this makes sense! I had to leave a lot of details out.

http://naca.larc.nasa.gov/reports/1953/naca-report-1135/

 

[Latexman]

Sailoday and aviat,

Thanks for pointing me in the right direction. I had it figured out mentally from Sailoday's explanation, and then aviat graciously provided a working example (the icing on the cake). I understand whats going on now. Thanks again!


Good luck,
Latexman