What happens at the “choke” point?

General engine tech -- Drag Racing to Circle Track

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digger
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Re: What happens at the “choke” point?

Post by digger »

the SAE paper says a mean inlet mach number (averaged over time period IVO to IVC)) of 0.55 or greater shows the VE to fall off the proverbial cliff based on the inlet valve mean effective area (curtain area)
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modok
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Re: What happens at the “choke” point?

Post by modok »

Yeah. That's the classic.
it may be found to be higher than .55, in moments where, due to putting the brakes on momentum in the runner, you are developing higher than atmospheric pressure, and thus able to push more through the restriction, because the inlet pressure IS actually higher. The mach limit calculates as if atmospheric pressure is acting on the choke point, over a time period, but really.......... the instantaneous pressure at the choke point can be higher or lower at any given moment. Perhaps if an engine had no intake manifold, then .55 or .5 would prove more uniformly correct.
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Re: What happens at the “choke” point?

Post by digger »

The test had no intake manifold or exhaust. The area used for calculation is not the minimum cross section area of the port. The Mach index will be higher than the mean inlet Mach number.
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Re: What happens at the “choke” point?

Post by modok »

Maybe that's true, I don't know. I think the SAE article you are referencing was different from the one I thought it was, two Japanese guys?, but it's been 15 years since I read it so, I don't know if I could even find it now.
but, ah......whats your point?
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Re: What happens at the “choke” point?

Post by digger »

that at the choke point it might not be actually choking but might be simply be too fast. if its too fast such then the velocity gradients may be not conducive to fill the cylinder with ever decreasing period of time as rpm rise . this implies the shape will play a role in how fast flow can be. a somewhat uniform gradient will tolerate relatively high average speed that might be governed by friction and pumping losses but in a complicated section of the port the VE might drop because the effective flow area actually reduces as the air is unable to follow the shape leading to high flow seperation.
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Re: What happens at the “choke” point?

Post by DrillDawg »

Does anyone figure the mach number at the valve/throat to see what it's relationship is with the port mach number? And what piston speed/position and valve lift is that mach number achieved?
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Re: What happens at the “choke” point?

Post by user-30563 »

DrillDawg wrote: Mon Jun 25, 2018 9:26 am Does anyone figure the mach number at the valve/throat to see what it's relationship is with the port mach number? And what piston speed/position and valve lift is that mach number achieved?
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Re: What happens at the “choke” point?

Post by GerryP »

As with everything, it depends upon how the air mass approaches and leaves the restriction. Do some low-level research on the de Laval nozzle. It can help explain why a port that shouldn't be making power, does.
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Re: What happens at the “choke” point?

Post by groberts101 »

digger wrote: Mon Jun 25, 2018 4:31 am that at the choke point it might not be actually choking but might be simply be too fast. if its too fast such then the velocity gradients may be not conducive to fill the cylinder with ever decreasing period of time as rpm rise . this implies the shape will play a role in how fast flow can be. a somewhat uniform gradient will tolerate relatively high average speed that might be governed by friction and pumping losses but in a complicated section of the port the VE might drop because the effective flow area actually reduces as the air is unable to follow the shape leading to high flow seperation.
Right on. And to your next post.. where is a port most tolerant of air speed.. before or after the turn into the cylinder?
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