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
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
Good luck,
Latexman
RE: Exit Conditions of a Vapor Stream though a Nozzle
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
Good luck,
Latexman
RE: Exit Conditions of a Vapor Stream though a Nozzle
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
RE: Exit Conditions of a Vapor Stream though a Nozzle
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
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
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
Good luck,
Latexman
RE: Exit Conditions of a Vapor Stream though a Nozzle
RE: Exit Conditions of a Vapor Stream though a Nozzle
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
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
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
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
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
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
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
http:/
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
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
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
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
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
"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
"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
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:/
RE: Exit Conditions of a Vapor Stream though a Nozzle
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