My dumb question of the day...what is “pressure recovery”?

[travis]

I keep seeing this pop up from time to time, especially in regards to chamber deshrouding work and bowl porting. What is it, what does it do, how do you measure it, etc?

[GARY C]

In simple terms it's the airs ability to follow the chamber wall around the valve without loosing velocity, you need a chamber wall really close to the valve to be able to tailor it as lift increases.
At very low lift the air exits around the valve in a more unified method but as lift increases it transitions to the far side and the chamber wall in a since becomes part of the port but if the chamber wall is to far then the pressure does not recover as it dumps into the cylinder due to the change in size. Here is an image of a well designed port and chamber and a link to an image of a bad chamber exit that I could not get to post here. https://www.bing.com/images/search?view ... ajaxhist=0

[JoePorting]

Might be better explained through an old Stanford University video of Flow Visualization. Focus on minute 13 through 22 for the diffuser part since the air exiting the valve seat area and going into the chamber is a lot like air going through a diffuser. The point here is that you don't want the exit walls at such a large angle so as to cause the flow to stall. The video doesn't give us an ideal wall angle, but suggests it's much closer than people think. Hence, in developing flow past the intake valve angle, the top angle and the chamber wall should be right up close to the actual flow path for max intake flow.

[hoffman900]

Pressure is the inverse of velocity. It’s about recovering as much pressure as possible from a given amount of velocity. Remember,the column slows down drastically as it passes through the valve seat and into a large cylinder. Obviously, we want pressure in the cylinder. Look at a rocket nozzle, and it’s the same thing, just that the minimum cross section is changing as the valve moves.

[GARY C]

Pressure is the inverse of velocity. It’s about recovering as much pressure as possible from a given amount of velocity. Remember,the column slows down drastically as it passes through the valve seat and into a large cylinder. Obviously, we want pressure in the cylinder. Look at a rocket nozzle, and it’s the same thing, just that the minimum cross section is changing as the valve moves.

So your trying to use the velocity of the port to create pressure in the cylinder?

[hoffman900]

Pressure is the inverse of velocity. It’s about recovering as much pressure as possible from a given amount of velocity. Remember,the column slows down drastically as it passes through the valve seat and into a large cylinder. Obviously, we want pressure in the cylinder. Look at a rocket nozzle, and it’s the same thing, just that the minimum cross section is changing as the valve moves.

So your trying to use the velocity of the port to create pressure in the cylinder?

Yes. High velocity = low pressure, as velocity decays, pressure rises. The better you handle slowing down the air column, the more pressure can be recovered.

This is jet engines, rockets, our hearts, etc all work.

[GARY C]

Pressure is the inverse of velocity. It’s about recovering as much pressure as possible from a given amount of velocity. Remember,the column slows down drastically as it passes through the valve seat and into a large cylinder. Obviously, we want pressure in the cylinder. Look at a rocket nozzle, and it’s the same thing, just that the minimum cross section is changing as the valve moves.

So your trying to use the velocity of the port to create pressure in the cylinder?

Yes. High velocity = low pressure, as velocity decays, pressure rises. The better you handle slowing down the air column, the more pressure can be recovered.

This is jet engines, rockets, our hearts, etc all work.

So if I understand correctly and can word this correctly... as the piston drops in the bore and creates a negative pressure you are trying to match or exceed that with the positive pressure from the port?

Just trying to get a visual per say to figure out how one would go about measuring the changes, it does not seem like it would be a simple test on a flowbench due to the bench continually vacuuming out the bore.

[Stan Weiss]

While these numbers are not measured but calculated. Look at velocity and cfm per sq @ the curtain area and how the number go in the oppiste of the port.

Stan

Code: Select all

_____Intake________------_Curtain_-----___Effective__---_Throat_--____----_Valve_---____----_MCSA_----
_Lift______CFM_____fps_______DC_____Area____Area______fps_______DC______fps_______DC______fps_______DC
_.2000__166.000__284.338___0.8117___1.401___1.137__126.769___0.3619__102.005___0.2912__131.357___0.3750
_.3000__227.000__259.216___0.7400___2.102___1.555__173.353___0.4949__139.488___0.3982__179.627___0.5128
_.4000__294.000__251.793___0.7188___2.802___2.014__224.519___0.6409__180.659___0.5157__232.645___0.6641
_.5000__357.000__244.599___0.6983___3.503___2.446__272.630___0.7783__219.371___0.6262__282.497___0.8064
_.6000__400.000__228.384___0.6520___4.203___2.741__305.468___0.8720__245.794___0.7017__316.524___0.9036
_.7000__426.000__208.482___0.5952___4.904___2.919__325.323___0.9287__261.771___0.7473__337.098___0.9623
_.8000__432.000__184.991___0.5281___5.605___2.960__329.905___0.9418__265.458___0.7578__341.845___0.9759
_.9000__434.000__165.198___0.4716___6.305___2.973__331.433___0.9461__266.687___0.7613__343.428___0.9804
1.0000__433.000__148.335___0.4235___7.006___2.967__330.669___0.9440__266.072___0.7596__342.637___0.9781
Avg_____352.111__219.482___0.6266___4.203___2.412__268.897___0.7676__216.367___0.6177__278.629___0.7954


_____Intake_______%_Step___--------_CFM_per_Sq._In._-------____L/D___Lift___________Sq_In
_Lift______CFM__Increase___Throat____Valve__Curtain_____MCSA__Ratio___mm____M^3/s____Area____fps
_.2000__166.000____________52.820___42.502__118.474___54.732__.090___5.08__0.0783___1.053__378.430
_.3000__227.000____36.75___72.230___58.120__108.006___74.845__.135___7.62__0.1071___1.706__319.260
_.4000__294.000____29.52___93.550___75.274__104.914___96.935__.179__10.16__0.1388___2.377__296.900
_.5000__357.000____21.43__113.596___91.405__101.916__117.707__.224__12.70__0.1685___3.052__280.706
_.6000__400.000____12.04__127.278__102.414___95.160__131.885__.269__15.24__0.1888___3.143__305.468
_.7000__426.000_____6.50__135.551__109.071___86.867__140.457__.314__17.78__0.2010___3.143__325.323
_.8000__432.000_____1.41__137.461__110.607___77.080__142.436__.359__20.32__0.2039___3.143__329.905
_.9000__434.000______.46__138.097__111.119___68.832__143.095__.404__22.86__0.2048___3.143__331.433
1.0000__433.000_____-.23__137.779__110.863___61.806__142.765__.448__25.40__0.2044___3.143__330.669
Avg_____352.111___________112.040___90.153___91.451__116.095

CARFOR - Registered to:   Stan Weiss

Version: 4.1.0 -- May 30, 2019

[jed]

Go to Advanced Engine Tech, look for thread "2 questions 4 the experts" it dosents deal with pressure recovery
Specifically but parts of it do discuss pressure recovery remotely.

Modok posted this and it helped me to understand pressure recovery in the chamber and further into the cylinder.
It is on page 4 at 12:50


"How about this. A pipe closed at one end. how does that resonate? it's called a "quarter wave resonator", and it's called that because it's length is one quarter of the total wave length.
A wave, or cycle, is a full cycle, meaning ends up back where it started. I had a terrible time trying to talk to guys that think of harmonics in musical terms, as they don't use the same terminology, so I have to state clearly that a cycle is a cycle.

So the way such a thing resonates is it cycles between velocity and pressure. Lets say one end of the pipe is closed, and the air in it HAS velocity flowing toward the closed end....what will happen?
The velocity will turn to pressure in the closed end.. Once it has built pressure in the closed end, then that pressure will make the air flow the other way. Now it has velocity going the other way, and will continue until stopped. Then there is LOW pressure in the closed end, and that stops it. Then the air flows toward the low pressure. And we are back where we started. One cycle."

The velocity in the port has very little pressure. High velocity equal low pressure. How that port velocity exits the valve seat apateure and is transitioning to pressure can cause flow to stall.
Please don't repeat what I have written

[jed]

viewtopic.php?f=15&t=23564
Look at this thread

[jed]

viewtopic.php?f=15&t=34978&p=420534&hil ... sh#p420334
More on pressure recovery

[Carnut1]

Resized_20181002_142719_290.jpeg

This may help explain what pressure recovery is. It may not. In the picture the main metering venturi for my water plant. 36" main in and out center is either 20" or 16" this meter has been in constant service since 1957. I have seen 24 mgd flow with 1500 hp pushing water though it. We usually feed between 77 to 84 psi to the system. How efficient is it? I am not sure, it is dead quiet while operating which I thought was strange. Flow is from bottom to top in picture. Notice the d.p. nipples for pressure gauge. Gauge calculates flow due to differential pressure according to Bernoulli's equation. The shape accelerates the flow to reduce pressure at the center. The pressure recovery occurs from small back to large at a slower rate that is where the pressure recovery is happening. I hope this helps and is not confusing. Thanks, Charlie

[mk e]

So if I understand correctly and can word this correctly... as the piston drops in the bore and creates a negative pressure you are trying to match or exceed that with the positive pressure from the port?

Just trying to get a visual per say to figure out how one would go about measuring the changes, it does not seem like it would be a simple test on a flowbench due to the bench continually vacuuming out the bore.

As I understand it the port velocity and conversion to cylinder pressure tries to match the cylinder but
due to system losses and lag it can't until the piston has slowed near the bottom. By then the velicoty is near max and easily over fills the cylinder yielding VE numbers above 100%. The exhaust kicks the flow moving during overlap, the piston continues to accelerate the flow, the the flow velocity it turned back into pressure in the cylinder as the intake closes....ideally. This is the simple version, and were all the wave dynamics discussion begin. Again, as I understand it.

[gruntguru]

Flow in a pipe with varying CSA.
For air you can neglect the gravitational term in Bernoulli's equation leaving P + 1/2pv^2 = const at any cross section along the pipe. P is static pressure, p is density, v is velocity. The second term 1/2pv^2 is referred to as "dynamic pressure".

What the equation is saying is - you can convert pressure into velocity and back again. If there are no losses (due to friction, excessive turbulence etc), the final pressure in the cylinder (assumed stagnant) will be atmospheric because the air was initially at that pressure and zero velocity before entering the intake. In fact (again assuming no losses) the "total" pressure (static + dynamic) should equal atmospheric at any cross section along the intake tract.

"Pressure recovery" is simply describing how much of the total pressure is retained in the transition from maximum velocity near the valve seat and "zero" velocity in the cylinder. If there were no losses (perfect design), the total pressure would be the same at both points.

[GARY C]

So if I understand correctly and can word this correctly... as the piston drops in the bore and creates a negative pressure you are trying to match or exceed that with the positive pressure from the port?

Just trying to get a visual per say to figure out how one would go about measuring the changes, it does not seem like it would be a simple test on a flowbench due to the bench continually vacuuming out the bore.

As I understand it the port velocity and conversion to cylinder pressure tries to match the cylinder but
due to system losses and lag it can't until the piston has slowed near the bottom. By then the velicoty is near max and easily over fills the cylinder yielding VE numbers above 100%. The exhaust kicks the flow moving during overlap, the piston continues to accelerate the flow, the the flow velocity it turned back into pressure in the cylinder as the intake closes....ideally. This is the simple version, and were all the wave dynamics discussion begin. Again, as I understand it.

It just got me to wondering if there would be any good info to gather If I had a pressure gauge in place of the spark plug and could instantaneously stop the airflow between the motors and the plenum on my Flowbench and record the pressure spike... May require a timed intake valve closing in conjunction with it... I think a similar idea was raised sometime back by randy331.