martinholmesuk
Computer
Hi,
Does compressed (15psi) air pick up speed when it goes around a 90 degree bend?
Thank you
Martin
Does compressed (15psi) air pick up speed when it goes around a 90 degree bend?
Thank you
Martin
Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations JStephen on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
pressure / speed
- Thread starter martinholmesuk
- Start date
- Status
Martin,
The mass flow at the inlet and outlet of the tube is the same but the pressure (and density) at the outlet is slightly lower due to the pressure drop accross the elbow, therefore air does pick up speed if it goes around a 90 degree bend.
The velocity increase is proportional to the ratio of pressures upstream and downstream of the tube.
Vd = Pu x Vu / Pd
where
Vd = downstream velocity
Pu = upstream absolute pressure
Vu = upstream velocity
Pd = downstream absolute pressure
The mass flow at the inlet and outlet of the tube is the same but the pressure (and density) at the outlet is slightly lower due to the pressure drop accross the elbow, therefore air does pick up speed if it goes around a 90 degree bend.
The velocity increase is proportional to the ratio of pressures upstream and downstream of the tube.
Vd = Pu x Vu / Pd
where
Vd = downstream velocity
Pu = upstream absolute pressure
Vu = upstream velocity
Pd = downstream absolute pressure
jbmr,
The it looks like you're trying to use Bernouli's equation which can be stated (ignoring the elevation term, in U.S. units you need to be really careful how you divide a pressure in lbf/in^2 by a density in lbm/ft^3):
vup2/(2*g)+Pup/rho =vdown2/(2*g)+Pdown/rho
if density is constant (a requirement for this equation) then:
vdown=(2*g*(Pup- Pdown)/rho+vup2)0.5
If you try to do the relationship as a first-order proportionality you get good units, but wrong answers.
David Simpson, PE
MuleShoe Engineering
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
The harder I work, the luckier I seem
The it looks like you're trying to use Bernouli's equation which can be stated (ignoring the elevation term, in U.S. units you need to be really careful how you divide a pressure in lbf/in^2 by a density in lbm/ft^3):
vup2/(2*g)+Pup/rho =vdown2/(2*g)+Pdown/rho
if density is constant (a requirement for this equation) then:
vdown=(2*g*(Pup- Pdown)/rho+vup2)0.5
If you try to do the relationship as a first-order proportionality you get good units, but wrong answers.
David Simpson, PE
MuleShoe Engineering
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
The harder I work, the luckier I seem
For incompressible fluids of constant density, without heat exchange, flowing in a tube of constant diameter, friction would translate into [Δ]T, a rise in temperature (or internal energy), of the fluid, with both the inlet and outlet velocities remaining unchanged.
The friction work would then be equal to C[Δ]T, where C is the specific heat.
Compressibility effects will be small when the ratio of fluid velocity to the velocity of sound is small compared to unity.
zdas04,
The Bernoulli’s equation also includes a frictional pressure drop term. The rho v squared term is negligible in comparison with this term in the case of a fluid flowing in a pipe of constant section and for this reason you can assume that the pressure difference accross the elbow is equivalent to the frictional pressure drop.
The equation quoted in my previous post is just an application of the ideal gas law assuming isothermal compressible flow.
I agree with 25362, if the velocity is low the flow can be considered as incompressible and the pressure drop through the elbow is negligible. In this case the velocity change is zero. On the other hand, if the velocity is high the pressure drop and density variation accross the elbow (and any other piping accident) will induce a velocity increase.
I think you will agree for instance that the gas velocity is higher downstream of a control valve than upstream.
Regarding temperature variation, gas pressure drop is generally considered as an isenthalpic depressurisation. The temperature variation for such a depressurisation is zero in the case of an ideal gas.
For a non ideal gas the temperature drops a little bit due to the Joule Thomson effect but I would think this is negligible for air at 15psi.
The Bernoulli’s equation also includes a frictional pressure drop term. The rho v squared term is negligible in comparison with this term in the case of a fluid flowing in a pipe of constant section and for this reason you can assume that the pressure difference accross the elbow is equivalent to the frictional pressure drop.
The equation quoted in my previous post is just an application of the ideal gas law assuming isothermal compressible flow.
I agree with 25362, if the velocity is low the flow can be considered as incompressible and the pressure drop through the elbow is negligible. In this case the velocity change is zero. On the other hand, if the velocity is high the pressure drop and density variation accross the elbow (and any other piping accident) will induce a velocity increase.
I think you will agree for instance that the gas velocity is higher downstream of a control valve than upstream.
Regarding temperature variation, gas pressure drop is generally considered as an isenthalpic depressurisation. The temperature variation for such a depressurisation is zero in the case of an ideal gas.
For a non ideal gas the temperature drops a little bit due to the Joule Thomson effect but I would think this is negligible for air at 15psi.
- Thread starter
- #7
martinholmesuk
Computer
Thanks everyone, Very much a tech answer.
jbmr,
"Bernoulli's equation includes a frictional pressure drop term"???? I just pulled out my Grad School notes on the derivation and his FIRST assumption in going from Euler's equation is that there is no friction (Euler's first assumption was no change in density). There is a velocity term, a pressure term, and an elevation term. That is it. People move the constants around to fit a specific situation (i.e., sometimes it is P*g and Z*g instead of v^2/g), but as long as the three terms end up with the same units it is valid. The version I used above results in everything resolving to a length term. You can also move constants around to end up with a length squared over a time squared term. Either way Bernoulli's equation does not have a friction term.
In 25 years of messing with fluid flow I've never had a case where the velocity^2/rho term was negligible in a real-world fluid-flow calculation except in creep flow (hydraulics). Even in Laminar flow at high pressures it is far from negligible.
There isn't a velocity term in any version of the ideal gas law I've ever seen. You can easily include a mass-flow term, but that only lets you infer a velocity from the characteristics of the container.
Back to the original poster's question. "Does gas speed up as it goes around a 90 degree bend?". The change bulk in velocity is related to the pressure drop. Crane 410 says that a standard elbow is the same pressure drop as 14 ft of pipe so unless you have a very high specific pressure drop, the velocity won't change in any way that is measureable. On the other hand the gas has to accelerate around the 90, so you end up with skewed velocity profiles and any particular streamline will show a velocity change.
David Simpson, PE
MuleShoe Engineering
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
The harder I work, the luckier I seem
"Bernoulli's equation includes a frictional pressure drop term"???? I just pulled out my Grad School notes on the derivation and his FIRST assumption in going from Euler's equation is that there is no friction (Euler's first assumption was no change in density). There is a velocity term, a pressure term, and an elevation term. That is it. People move the constants around to fit a specific situation (i.e., sometimes it is P*g and Z*g instead of v^2/g), but as long as the three terms end up with the same units it is valid. The version I used above results in everything resolving to a length term. You can also move constants around to end up with a length squared over a time squared term. Either way Bernoulli's equation does not have a friction term.
In 25 years of messing with fluid flow I've never had a case where the velocity^2/rho term was negligible in a real-world fluid-flow calculation except in creep flow (hydraulics). Even in Laminar flow at high pressures it is far from negligible.
There isn't a velocity term in any version of the ideal gas law I've ever seen. You can easily include a mass-flow term, but that only lets you infer a velocity from the characteristics of the container.
Back to the original poster's question. "Does gas speed up as it goes around a 90 degree bend?". The change bulk in velocity is related to the pressure drop. Crane 410 says that a standard elbow is the same pressure drop as 14 ft of pipe so unless you have a very high specific pressure drop, the velocity won't change in any way that is measureable. On the other hand the gas has to accelerate around the 90, so you end up with skewed velocity profiles and any particular streamline will show a velocity change.
David Simpson, PE
MuleShoe Engineering
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
The harder I work, the luckier I seem
- Thread starter
- #9
martinholmesuk
Computer
Might I just add:-
It's from a turbo charger which is heating up the air (until the intercooler does it job) What other ways can I reduce the air temp?
Cheers
It's from a turbo charger which is heating up the air (until the intercooler does it job) What other ways can I reduce the air temp?
Cheers
- Status
Similar threads
- Question
- Locked
- Question
- Question