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Exit Conditions of a Vapor Stream though a Nozzle

Exit Conditions of a Vapor Stream though a Nozzle

Exit Conditions of a Vapor Stream though a Nozzle

(OP)
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

RE: Exit Conditions of a Vapor Stream though a Nozzle

"High-pressure" is a relative term.  Tell us all that you know, like the value and units of the knowns (T1, P1, and mdot) and the calculated value (rho) and the dimensions of the nozzle and piping/tubing.  That way we can narrow down our responses.

Good luck,
Latexman

RE: Exit Conditions of a Vapor Stream though a Nozzle

(OP)
You're right, that may have been (aka "was) a stupid statement... i was thinking "high velocity" and somehow that morphed into "high pressure" when i wrote the post...
anyway: propane vapor is my fluid
p1=240 kPa
t1=330 K
my handy property calculator spits out rho=3.9667.. call it 4.0 kg/m^3
mdot=0.0408 kg/s
area1=1.267 cm^2 (upstream of converging section)
areaThroat=0.3739 cm^2
areaExit= 0.45 cm^2

that gets me a velocity @ point1 of 80.5 m/s (V=mdot/rho*A)
taking rho as a constant, i can get velocities at my other points.... except rho is not a constant.
I tried Bernoulli for compressible gas flow, but that requires either knowing the velocities or the pressures at both points... and just guesstimating the numbers kinda blew up in my face... so now i'm stuck and outta ideas.  

anyone want to take a stab at explaining it to me?

thanks and cheers,

rad

RE: Exit Conditions of a Vapor Stream though a Nozzle

Are the distances from "state 1" to the throat and from the throat to the exit short?  Say, 1-2 cm each?

Good luck,
Latexman

RE: Exit Conditions of a Vapor Stream though a Nozzle

(OP)
point 1 to throat ~2.4cm, throat to exit ~.5cm

are you going to tell me that the distances are too short for fully developed flow?

cheers,
rad

RE: Exit Conditions of a Vapor Stream though a Nozzle

You need to calculate c* (the sonic velocity of gas at the throat).  Assume choked flow (Mach no. =1) at the throat.

RE: Exit Conditions of a Vapor Stream though a Nozzle

It is stated that Pexit is supposed to be below ambient atmospheric.   Does this mean the exit pressure is below the back pressure?  
A good example of a converging-diverging nozzle with varying back pressure conditions is illustrated on pg 140, Vol 1 -of Shapiro--The Dynamics and Thermodynamics of Compressible Fluid Flow.

For the conditions you stated, supersonic flow will exist from the nozzle throat to exit---NO shocks.

RE: Exit Conditions of a Vapor Stream though a Nozzle

(OP)
well, c=sqrt(k*R*T), for propane k~1.1, R~189, T~330 -->
c~262 m/s, which gives me a M=1.04... taking into account that this entire model is "more or less", most of my numbers are "close enough", which pretty much means chocked flow @ the throat already

The vapor is supposed to exit the nozzle below atmospheric pressure, which sets up a vacuum that entrains air @ ambient into the vapor stream... since i know that the mixing takes place, i know that the exit pressure has to be < 1atm.  i just want to know what the pressure/velocity of the vapor stream is at exit.  the numbers i'm coming up with are too low (velocity) or too high (pressure).

any other ideas?

cheers,
rad

RE: Exit Conditions of a Vapor Stream though a Nozzle

My Crane TP410 said k (Cp/Cv) = 1.15 for propane.  The critical pressure ratio = 0.5744.  p* = 138 kPa.  T* = 307 K.  rho* = 2.45 kg/m^3.  With the area ratio (Aexit/Athroat), I get Msubsonic = 0.5983 and Msupersonic = 1.4880.

If subsonic, I get Treceiver = 321 K and Preceiver = 196 kPa.

If supersonic, I get Treceiver = 283 K and Preceiver = 74 kPa.

You said the receiver pressure is below atmospheric (101 kPa), so whether it'll be sub or supersonic depends on what that really is and if it's controlling or not.

Btw, is your mdot is too high?  I got 0.022 kg/sec in the throat.

Good luck,
Latexman

RE: Exit Conditions of a Vapor Stream though a Nozzle

I posted previously without reading your last post.  Sorry.  Since it exhausts into air at ambient T and P (101 kPa), the flow in the diverging section of the nozzle will be supersonic part of the way to the exit, then there will be a normal shock wave where the flow changes irreversibly from supersonic to subsonic.  To keep it simple, I'd use the exit conditions in my previous post for subsonic flow.  Actual conditions will vary due to the entropy change associated with the shock wave.

Good luck,
Latexman

RE: Exit Conditions of a Vapor Stream though a Nozzle

Latexman (Chemical)If a shock occurs, then the exit pressure will match the back pressure.  I believe the exit pressure will correspond to that of the exit Mach no which will be supersonic.

RE: Exit Conditions of a Vapor Stream though a Nozzle

Sailoday,

I'm glad you rejoined.

I agree.  The exit pressure = the back pressure, which is ambient pressure of about 101 kPa.

I disagree the exit flow will be supersonic.

On page 140 and using Fig. 5.21 (a), the exit pressure is between the subsonic design curve (condition 2) and the supersonic design curve (condition 4).  It is as condition 3.  Therefore the following applies, "As regime II is entered, a normal shock wave appears downstream of the throat and the process aft of the shock comprises subsonic deceleration."

So, the flow is supersonic from the throat to the shock wave, but after the shock wave the flow is subsonic and remains subsonic to the exit and outside to the air.

That's how I see it.

I wimped out and did not calculate the changes in properties across the shock wave.  I hope using the subsonic solutions' properties are close enough.  Can you comment on that?  Thanks!

Good luck,
Latexman

RE: Exit Conditions of a Vapor Stream though a Nozzle

Latexman
As I see it, refering again to Shapiro, with path 4, exit and back pressure are equal only if the shock occurs at the exit plane.  
The original statement of the problem is that exit pressure is less than the back pressure. Doesn't this indicate that the flow conditions are either in region III or IV just below path (4)? and that oblique shock waves occur downstream of the exit??

Sorry for those without the bible of Shapiro?

Regards

RE: Exit Conditions of a Vapor Stream though a Nozzle

Sailoday,

Since radAES said "I know that Pexit is (supposed to be) below ambient atmospheric."  I assumed by the "(supposed to be)" he was not so sure, so I did not put much faith in that and used my own judgement.  If it is true, you are right.  If it is not true, then I am right.

radAES, can you clarify?

Good luck,
Latexman

RE: Exit Conditions of a Vapor Stream though a Nozzle

(OP)
I know for a fact that the mixing takes place, and I know that the explanation for the mixing is that Pexit is lower than Pair, which sets up a vacuum that "pulls" the air into the vapor stream; therefore, I am very comfortable with saying that Pexit is less than 1atm.

On a different note: I'm very sure of my mass flowrate - that's just based on the rating of the nozzle (in BTU/hr) and the energy content of LP (BTU/kg) - yes, I know I'm mixing unit systems, but I despise standard - so sue me :)

cheers,
rad
"Remember, if you leave it to the last minute, it'll only take a minute"

RE: Exit Conditions of a Vapor Stream though a Nozzle

radAES,

When the propane leaves the nozzle, it is a "free jet" if it enters an area whose cross-sectional area is > 5 times the area of the nozzle exit.  I bet this is your case.  When a "free jet" leaves its outlet and goes into the surrounding fluid, the jet will entrain the surrounding fluid, expand, and decelerate.  The momentum of the jet is transferred to the surrounding fluid being entrained.  While there may be some fluid movement caused by static pressure differences, the resulting mixing due to just this component will be minimal.  In your case, the vast majority of the “jet mixing” will be due to momentum transfer.

The 2X difference in flow could be due to many, many things.

Good luck,
Latexman

RE: Exit Conditions of a Vapor Stream though a Nozzle

Sailoday,

Do you have access to Perry’s Chemical Engineers’ Handbook (7th Edition)?  It speaks to our little debate on this case very well.  On page 6-25 under Convergent/Divergent Nozzles (De Laval Nozzles) it discusses the A/A* equation, “With A set equal to the nozzle exit area, the exit Mach number, pressure and temperature can be calculated.  Only if the exit pressure equals the ambient discharge pressure is the ultimate expansion velocity reached in the nozzle.  Expansion will be incomplete if the exit pressure exceeds the ambient discharge pressure; shocks will occur outside the nozzle.  If the calculated exit pressure is less than the ambient discharge pressure, the nozzle is overexpanded and compression shocks within the expanding portion will result.”

I thought this was well said.

Good luck,
Latexman

RE: Exit Conditions of a Vapor Stream though a Nozzle

Latexman
I don't have Perry's Handbook.
"Only if the exit pressure equals the ambient discharge pressure is the ultimate expansion velocity reached in the nozzle."
Refering again to Shapiro pg 140 Vol 1.  If path is on (4),(5),or (6)  of Filg 5.21a----Then I believe the exit plane Mach nos for each path will be the same.

 "If the calculated exit pressure is less than the ambient discharge pressure, the nozzle is overexpanded and compression shocks within the expanding portion will result.”--I have to digest this last statement.
Regards

RE: Exit Conditions of a Vapor Stream though a Nozzle

In supersonic flow the relationship between pressure and velocity is opposite to subsonic flow. If the exit pressure is less than ambient there, there should be an oblique wave beyond the nozzle exit, and nozzle is overexpanded. This would occur if the the diverging section is supersonic. M1 occurs at the throat of a converging-diverging nozzle. See site below.
http://www.engapplets.vt.edu/fluids/CDnozzle/cdinfo.html
As far as determining pressure and velocity, it involves expansion waves and obilique shock waves, which are two and three dimensional in nature.

RE: Exit Conditions of a Vapor Stream though a Nozzle

aviat (Aerospace) states"As far as determining pressure and velocity, it involves expansion waves and obilique shock waves, which are two and three dimensional in nature"

I don't disagree.--However, as Aviat further states this is downstream of the exit plane----
This will not affect mass flow rate in the nozzle.

RE: Exit Conditions of a Vapor Stream though a Nozzle

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

RE: Exit Conditions of a Vapor Stream though a Nozzle

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.

RE: Exit Conditions of a Vapor Stream though a Nozzle

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 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 regime 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.

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

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

Good luck,
Latexman

RE: Exit Conditions of a Vapor Stream though a Nozzle

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


RE: Exit Conditions of a Vapor Stream though a Nozzle

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"

RE: Exit Conditions of a Vapor Stream though a Nozzle

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/

RE: Exit Conditions of a Vapor Stream though a Nozzle

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

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