Velocity of air coming out of a nozzle

[lrthomas]

Hello, first time poster here!

I need to find the speed of air directly coming out of a nozzle connected to a compressed air tank. Inside the tank I have 30 psi gauge pressure (~207 kPa) and the diameter of the nozzle is 1.5 mm. Atmospheric pressure outside.

Isn't there a simple formula for this...? I've been searching for quite a while now, I haven't come across anything somewhat straight-forward. I need to ask real people.

I appreciate it, will give thanks!

 

[Simon Bridge]

Probably: $p+\frac{1}{2}\rho v^2 + \rho gh = constant$
the v is velocity.

There is also:
http://www.engineersedge.com/wwwboard/posts/15354.html (compressed air escaping a tank)
http://www.tlv.com/global/TI/calculator/air-flow-rate-through-orifice.html (calculator)

 

[lrthomas]

Perfect. Thank you very much!

 

[boneh3ad]

So, the equation given by Simon Bridge is wrong in this case. You can't use it. The equation he cited,
p+\frac{1}{2}\rho v^2 + \rho gh = constant,
is the Bernoulli equation, which has no meaning for compressible flow, and this situation is definitely one where compressibility is an issue. Luckily, this is still an easy problem.

The question you have to ask is whether or not the flow is choked, meaning does the velocity reach Mach 1 at the orifice exit. To check that you have to look at the actual expansion process of the air in your particular configuration. Now, assuming your orifice is that same exit area all the way through or else is narrowest on the outer skin of your tank (converging), then there is a very simple number you have to compare you situation to:
\dfrac{p}{p_0} \leq 0.528.
In this case, p is your atmospheric pressure (14.7 psia in this case) and p_0 is the absolute pressure in the tank (44.7 psia in this case). For a straight or converging duct/orifice, if this pressure ratio is below this critical value, the flow is choked and your flow is moving at exactly Mach 1 at the exit of the hole. In your case, the flow is choked so the exit velocity is going to be Mach 1. What this translates to in terms of actual velocity is going to depend on the temperature at the exit, which is also relatively easy to calculate.

Perhaps more interesting, given a few conditions, you can directly calculate the mass flow you are getting through your system (make sure units are consistent). For any choked flow, the mass flow through the nozzle or duct is going to be
\dot{m} = \dfrac{p_{01}A^*}{\sqrt{T_{01}}}\sqrt{\dfrac{\gamma}{R}}\left( \dfrac{2}{\gamma + 1} \right)^{\frac{\gamma + 1}{2(\gamma-1)}}.
The variables here are the mass flow, \dot{m}; the total pressure in the tank, p_{01}; the smallest cross-sectional area of your orifice (throat area), A^*; the total temperature if your tank, T_{01}; the ratio of specific heats (=1.4 for air), \gamma; and the specific gas constant for your gas (about 287 J kg^-1 K^-1 for air), R.

Keep in mind that as your tank vents, unless you have a powerful compressor refilling the tank, your tank pressure is going to go down. As long as that pressure ratio is less than the critical value the exit Mach number will still be 1, but your rate of mass flow will change, as will your velocity slightly since the gas is going to cool off in the tank as it loses pressure. It will continue in that fashion until it reaches that critical pressure ratio, at which point it will take on different characteristics, also based on isentropic flow assumptions. If it continues to get slower as the air leaves the tank to the point where the Mach number falls below about 0.3, then and only then would Bernoulli's equation stand a reasonable chance of giving you a reasonable answer.

 

[Simon Bridge]

So, the equation given by Simon Bridge is wrong in this case. You can't use it. The equation he cited,
p+\frac{1}{2}\rho v^2 + \rho gh = constant,
is the Bernoulli equation, which has no meaning for compressible flow, and this situation is definitely one where compressibility is an issue.

... which is why I also gave a couple of links ;)
You got here with the redirect before I did - thanks.

 

[rppearso]

What if the flow is subsonic?

 

[SteamKing]

If the air flow has a velocity below about 0.3 Mach (approx. 100 m/s at sea level), it can be considered incompressible for all practical purposes. But for velocities above this figure, even though the flow is subsonic, compressibility effects must be considered.

 

[rppearso]

So do you have any references or equations for subsonic velocity above mach 0.3? The mass flow equation above is for sonic flow. I found a reference in Shapiro (1953)

\V = /[2k/k-1 * R(T_o -T) * (1-(P_o/P)^((k-1)/k)]^(0.5).

Ugh still learning to use latext.

 

[SteamKing]

Shapiro is the standard reference for compressible flow; in fact, it is the model for most subsequent texts on the subject.

Compressible flow encompasses several different regimes: isothermal, adiabatic, or some combination of the two. Flow thru ducts with friction introduces an added twist.

The OP's nozzle diameter (1.5mm) is a bit on the small side. We don't know anything about the hose connecting the nozzle to his reservoir, the size of the reservoir, or any other pertinent details.

 

[boneh3ad]

But you do know the pressure ratio, and barring substantial pressure loss through a hose, the flow will be choked and sonic (at least until the tank pressure falls).

 

[aksyzfr1]

What if the tank has propane gas and temperature falls inside as the gas vents out. How do take the temperature then?

 

[aksyzfr1]

I want to calculate the velocity of gas Escaping a propane tank to the atmosphere. Tank pressure is 7kg/cm^2 at equilibrium. Area of orifice is 2.5mm.

 

[rppearso]

I have used a software that was based on https://www.amazon.com/gp/product/0471066915/?tag=pfamazon01-20

I am still going through this book but the software is very good about predicting sonic flow through a pipe, mass flow and temperature drop. Even though this text (from MIT) does not have Navier stokes it does have Eulers equation which just neglects sheer forces at the wall which is a reasonable assumption for compressible gas flow. Once you create the fanno line calculator then its just a matter of calculating the equivalent length. Its a super handy tool if your dealing with single gas phase flow at high mach numbers up to mach 1.

For the full Navier stokes equations and CFD modeling I really like this book - https://www.amazon.com/gp/product/0070016852/?tag=pfamazon01-20

 

[Anachronist]

I'm more interested in un-choked flow below the critical pressure. Suppose I have an air tank pressurized to 1.5 atm of absolute pressure (0.5 atm above ambient) and outside it's 1 atm. I know $\gamma=1.4$ for air, and the density of air at 1 atm is 1.225 kg/m². The tank has a hole in it. What is the velocity of air through the hole?

According to https://en.wikipedia.org/wiki/Bernoulli's_principle#Compressible_flow_in_fluid_dynamics
$$\frac {v^2}{2}+\left( \frac {\gamma}{\gamma-1}\right)\frac {p}{\rho} = \left(\frac {\gamma}{\gamma-1}\right)\frac {p_0}{\rho_0}$$
where the part on the right side of the equals symbol is a constant, and:

$p$ is the pressure (I assume it's the internal total pressure [measured gauge pressure plus ambient pressure])
$\rho$ is the density (I assume this is the density of the pressurized air inside the tank)
$v$ is the velocity
$p_0$ is the total pressure (I assume this means ambient pressure)
$\rho_0$ is the total density (I assume this means ambient density)​

Then
$$v = \sqrt{\frac{ 2\gamma}{\gamma-1} \left ( \frac{p_0}{\rho_0} - \frac{p}{\rho} \right )}$$
This looks to me like it would evaluate to zero always, since the two $p/\rho$ terms cancel out.

 

[rppearso]

I would buy a copy of Crane Technical paper 410 for quick equations on compressible flow - https://store.flowoffluids.com/shop/publications/TP410.html
Also we have a macro software at my work that was written based on equations from a 1953 MIT textbook - compressible fluid flow -
https://www.amazon.com/gp/product/0471066915/?tag=pfamazon01-20

Compressible fluid flow is much more complicated than liquid flow.

 

[Anachronist]

Well, here's what I ended up doing.

From the Bernoulli equation above, with the right side being constant, I have:
$$v=C_v\sqrt{\frac{2 \gamma}{\gamma-1}\frac{p}{\rho}}$$
In this case $p$ is excess pressure, the difference between the pressure inside the pressure vessel and ambient pressure outside, and $\rho$ is the density of air inside the vessel. So I need to solve for that velocity coefficient $C_v$.

The critical pressure at which the flow is choked is when
$$\frac{P_a}{P_c} = \left ( \frac{2}{\gamma+1} \right ) ^\frac{\gamma}{\gamma-1} = 0.528\text{ for }\gamma=1.4$$
where $P_a=101325\text{ Pa}$ is atmospheric (ambient) pressure, and $P_c=191801\text{ Pa}$ is the critical pressure at which the flow is choked at the speed of sound $c$:
$$c=\sqrt{\gamma R T}=345.5\text{ m/s at }297\text{ K (20°C)}$$
where $R$ is the ideal gas constant divided by gas molecular weight (8.3145 / 0.02896 for air), and $T$ is the air temperature in Kelvin.

Further, the density of the air at critical pressure is
$$\rho_c = \rho_a \frac{P_c}{P_a}=2.314\text{ kg/m^3}$$
where $\rho_a$ is the density of ambient air, 1.225 kg/m³.

So now I have enough to solve for $C_v$:
$$C_v = \frac{c}{\sqrt{\frac{2 \gamma}{\gamma-1}\frac{P_c-P_a}{\rho_c}}} = 0.661$$

Then, for any internal absolute pressure $P$ less than $P_c$, we can calculate the internal air density $\rho$ and plug it into the first equation above:
$$v=C_v\sqrt{\frac{2 \gamma}{\gamma-1}\frac{P-P_a}{\rho}}$$

If the pressure on one side of the orifice is constant and less than the critical choke pressure, then that should be the velocity.

If the pressure is coming from an air tank of finite volume, this velocity is just a snapshot in time. The pressure, and therefore velocity, will decrease over time. We can approximate the pressure remaining after a tiny time interval by multiplying the mass flow rate by the time interval to get the air mass lost during the time interval, determine the new density as mass divided by volume, and thereby determine a new pressure at the end of the time interval, and repeat the calculation.

The mass flow rate is given by
$$m' = C_c \rho A v$$
where $C_c$ is the contraction coefficient (values I've seen are 0.97 for a rounded edge and 0.62 for a sharp edge), and $A$ is the area of the orifice.

If it isn't a slow leak, but a rapid decompression starting from way above the critical choke pressure, we could even get fancy and use adiabatic expansion to determine temperature drop as pressure goes down in the tank, which in turn would affect the pressure as well as the speed of sound.

 

[MikeSpike]

Hello, first time poster here!

I need to find the speed of air directly coming out of a nozzle connected to a compressed air tank. Inside the tank I have 30 psi gauge pressure (~207 kPa) and the diameter of the nozzle is 1.5 mm. Atmospheric pressure outside.

Isn't there a simple formula for this...? I've been searching for quite a while now, I haven't come across anything somewhat straight-forward. I need to ask real people.

I appreciate it, will give thanks!

I'd like to have a small scale (desktop?) CO2 blaster I could "feed" CO2 snow, accelerate the particles with pressure regulated gas from an air compressor/tank or a CGA-320 conventional carbon dioxide cylinder and blow it out a nozzle. I was hoping to introduce the "snow" into the airstream by coupling a Venturi device in the tubing. Makers of dry ice blasting equipment claim their particle velocity exiting the nozzle to be supersonic. I began my search for a formula to calculate the nozzle velocity and came across your thread. I joined the forum today (5/30/2019) to engage those of you who've demonstrated interest in the OP topic. Should I start a new thread? Or, is this an appropriate place to raise this related issue?

 

[MikeSpike]

Also, can someone tell me what PUSH notifications are? It's slightly OT to ask about it here, I realize that. I enabled them, thinking that doing so would enable eMail notices of replies to this thread? Now I'm thinkin' that might not be the function of push notifications.

 

[anorlunda]

what is your application? High speeds for a toy could be dangerous, but necessary for other purposes.

 

[MikeSpike]

dry ice blaster

 

[ChethanKodase]

So, the equation given by Simon Bridge is wrong in this case. You can't use it. The equation he cited,
p+\frac{1}{2}\rho v^2 + \rho gh = constant,
is the Bernoulli equation, which has no meaning for compressible flow, and this situation is definitely one where compressibility is an issue. Luckily, this is still an easy problem.

The question you have to ask is whether or not the flow is choked, meaning does the velocity reach Mach 1 at the orifice exit. To check that you have to look at the actual expansion process of the air in your particular configuration. Now, assuming your orifice is that same exit area all the way through or else is narrowest on the outer skin of your tank (converging), then there is a very simple number you have to compare you situation to:
\dfrac{p}{p_0} \leq 0.528.
In this case, p is your atmospheric pressure (14.7 psia in this case) and p_0 is the absolute pressure in the tank (44.7 psia in this case). For a straight or converging duct/orifice, if this pressure ratio is below this critical value, the flow is choked and your flow is moving at exactly Mach 1 at the exit of the hole. In your case, the flow is choked so the exit velocity is going to be Mach 1. What this translates to in terms of actual velocity is going to depend on the temperature at the exit, which is also relatively easy to calculate.

Perhaps more interesting, given a few conditions, you can directly calculate the mass flow you are getting through your system (make sure units are consistent). For any choked flow, the mass flow through the nozzle or duct is going to be
\dot{m} = \dfrac{p_{01}A^*}{\sqrt{T_{01}}}\sqrt{\dfrac{\gamma}{R}}\left( \dfrac{2}{\gamma + 1} \right)^{\frac{\gamma + 1}{2(\gamma-1)}}.
The variables here are the mass flow, \dot{m}; the total pressure in the tank, p_{01}; the smallest cross-sectional area of your orifice (throat area), A^*; the total temperature if your tank, T_{01}; the ratio of specific heats (=1.4 for air), \gamma; and the specific gas constant for your gas (about 287 J kg^-1 K^-1 for air), R.

Keep in mind that as your tank vents, unless you have a powerful compressor refilling the tank, your tank pressure is going to go down. As long as that pressure ratio is less than the critical value the exit Mach number will still be 1, but your rate of mass flow will change, as will your velocity slightly since the gas is going to cool off in the tank as it loses pressure. It will continue in that fashion until it reaches that critical pressure ratio, at which point it will take on different characteristics, also based on isentropic flow assumptions. If it continues to get slower as the air leaves the tank to the point where the Mach number falls below about 0.3, then and only then would Bernoulli's equation stand a reasonable chance of giving you a reasonable answer.

So the mass flow rate at Mach 1 doesn't depend on pressure on the other side? I mean does the pressure on the other side matters only to determine whether the flow is compressible or incompressible? Isn't the mass flow rate still higher if there is near-vacuum on the other side?

 

[boneh3ad]

So the mass flow rate at Mach 1 doesn't depend on pressure on the other side? I mean does the pressure on the other side matters only to determine whether the flow is compressible or incompressible? Isn't the mass flow rate still higher if there is near-vacuum on the other side?

Once the flow is choked, no further lowering of the downstream pressure can affect the mass flow rate. Only the upstream conditions matters at that point.