A testing little question on flow rates

[FlowSpecialist]

See how you do. We should all be fairly familiar with flowbenches and in particular the fact that there is no one standard for the pressure drop to test at. People use 10" of water, 20", 25", 28", and even some ratings based on inches of mercury. Happily though we can convert the flow at one pressure drop to any other by applying the square root law.

Flow rate is proportional to the square root of pressure drop. So if we have 100 CFM @ 10" of water and want to know how many CFM that would be at 20" we multiply by root 20/10 = 1.41.

Now ignoring pulse tuning, normally aspirated engines operate at atmospheric pressure of 14.7 psi maximum pressure drop across the head and inlet manifold. If we want to squeeze more air into the cylinder we can increase that pressure drop by adding a turbocharger and raising the inlet manifold pressure and hence the pressure drop into the cylinder. So here's the question.

If we double the pressure drop on a flowbench we get 1.41 times as much airflow. If we take an engine with 100 bhp and turbocharge it to twice atmospheric pressure i.e. one atm of boost on the boost gauge then how much more air should it now flow and therefore what should be the new power output? (ignore parasitic losses and assume 100% turbocharger efficiency)

If the answer is not the same as for a flowbench then why not?

Dave

[MAP]

Hi Flowspecialist,

I hope it's OK for me to take a stab at this question, although I'm not at all familiar with flowbenches. I've never even seen one.

But if power is directly proportional to volume velocity of air (and presumably fuel in a fixed proportion,) then according to the "square root" law you cited:


Airflow volume velocity: w/o turbo - U, w/ turbo - 1.41*U
BHP: w/o turbo - 100hp, w/ turbo - 141hp

Nevertheless, I have a question:

I've always heard then when flow is laminar, that indeed
U = P^1/n, where n=2

But as the Reynolds number climbs, and flow becomes turbulent, that n>2, and is nondeterministic, due to the stochastic nature of turbulent flow. More often than not, however, n is claimed to be about 3 for turbulent flow.

Since airflow inside an engine is usually largely turbulent, then how can it be said so universally that if we double the pressure drop, that volume velocity increases by a factor SQRT(2)?

(Honest question - I'm not trying to be argumentive.)

As for "If the answer is not the same as for a flowbench then why not?", since you asked two questions, there are two answers, of which only one is applicable to a flowbench (namely, volume velocity.) Again, with my limited knowledge, provided the air is entering the test assembly with the same flow pattern as would be created by having a turbo present, then the relative volume velocities ought to be the same. But if the flow patterns are different for any reason (exit turbulence from the turbo being an obvious example,) then they would diverge.

Best,
Mark

[FlowSpecialist]

When flow is turbulent flow rate is proportional to the square root of pressure drop. When flow is laminar at very low Reynolds numbers flow rate becomes proportional to pressure drop. Flow in an engine or on a flow bench at any reasonable pressure drop will not be laminar.

I'll wait until people have had a chance to reply before commenting on my original question. Not many takers though it seems. Come on Dusty, if you're doing a degree that covers fluid mechanics how about having a pop at this one?

[MAP]

Flow Specialist,

As for the value of "n," I was recalling something incorrectly from related hydrodynamic losses for ship motion. Thank you. Nevertheless, n=2 for turbulent flow is still only approximate, since turbulent flow is by definition stochastic, with its uncertainty being related to Reynolds number as well.

But be that as it may, might I suggest dropping the implied "I'm smarter than you" tone, and simply enlighten us with the answer?

Thanks,
Mark

[rookie]

Originally Posted by MAP

Flow Specialist,

But be that as it may, might I suggest dropping the implied "I'm smarter than you" tone, and simply enlighten us with the answer?

Thanks,
Mark

Based on the post and coments made by flow specialist, I get the feeling that is the purpose of his posts.

Mark, someone tells me you are in to car stereo systems, is this so??

[MAP]

Hi Rookie,

Speaking about Flow Specialist in the third person - he seems smart and I look forward to learning from him. On the other hand, I work with brilliant people with seemingly quadruple-digit IQ's all the time, so when I see egos start to come into play, I exit stage left really fast.

Car stereos: I've designed speakers for cars, but truth be told, my heart's in very high performance indoors audio. Thanks for asking, however!

Best,
Mark

[FlowSpecialist]

Originally Posted by MAP

Flow Specialist,

As for the value of "n," I was recalling something incorrectly from related hydrodynamic losses for ship motion. Thank you. Nevertheless, n=2 for turbulent flow is still only approximate, since turbulent flow is by definition stochastic, with its uncertainty being related to Reynolds number as well.

But be that as it may, might I suggest dropping the implied "I'm smarter than you" tone, and simply enlighten us with the answer?

Thanks,
Mark

I've always thought the most enlightening way to consider a subject is by examining questions which make you rethink what you thought you knew. That's what I spend a large part of my life doing to try and understand engines better. I like being set puzzles to solve and I like posing them. However if doing that on here is going to be taken as some sort of ego trip rather than a way of provoking a stimulating debate which might uncover something new then not a problem. Maybe we don't like to learn in the same way.

I'm just a bit disappointed that with people like DV prepared to share their knowledge on here that there isn't a more spirited debate on each of the topics and that it seems only a small number of people really contribute at all. Perhaps it needs more time for people to find their way here but it's a shame that such good articles are apparently being written for the benefit of very few.

As to n=2 yes that's only an approximation because factors like the Reynolds number, discharge coefficient and expansibility factor vary slightly with the conditions but it's a very close approximation for most purposes if the conditions are not changing by a huge amount.

Dave

[rookie]

Originally Posted by FlowSpecialist

I'm just a bit disappointed that with people like DV prepared to share their knowledge on here that there isn't a more spirited debate on each of the topics and that it seems only a small number of people really contribute at all. Perhaps it needs more time for people to find their way here but it's a shame that such good articles are apparently being written for the benefit of very few.

Dave

I may be totally wrong but I feel this is a poke at the intelligent of the members of GFN and that you feel it needs time for more people like you.

I’m sorry, but I am not politically correct….

Anyhow most people hear do not debate David’s articles because they know his attention to detail (may be you should read his books they apply to more than Chevy’s) and it is somewhat senseless to argue with a dyno pull.

Instead, most of us build and test engines using the info learned from his writings and try different things along with the facts at hand to try to expand on the results already obtained.

I've always thought the most enlightening way to consider a subject is by examining the facts.

Then as you said, maybe we don't like to learn in the same way.

[MAP]

Dear Flow Specialist,

Apologies if I misunderstood you - your clarification brought the discussion back to its proper course. Your taunt to Dusty however (not trying to defend him; I'm sure he's more than capable of defending himself,) came across a bit strong.

Well, now that I think this is behind us, let's jump in. Where are the truths, and where are the fallacies with respect to the perceived theories on which your original questions were based?

Best,
Mark

[Lasse]

Originally Posted by FlowSpecialist

If we double the pressure drop on a flowbench we get 1.41 times as much airflow. If we take an engine with 100 bhp and turbocharge it to twice atmospheric pressure i.e. one atm of boost on the boost gauge then how much more air should it now flow and therefore what should be the new power output? (ignore parasitic losses and assume 100% turbocharger efficiency)

If the answer is not the same as for a flowbench then why not?

Dave

Answer in not the same as for a flowbench, because flowbench measures volumetric flow, but engine likes mass flow. Charging air to twice atmospheric pressure doubles massflow.