A testing little question on flow rates

[FlowSpecialist]

To answer the original question it's easiest to start with incompressible flow. That's more intuitive to visualise and you can measure it with a bucket which is far simpler than ploughing through the flow equations.

With incompressible flow the density of the medium can't alter and therefore the fluid condition upstream and downstream of the restriction is the same. Flow rate is always proportional to the square root of the pressure drop across the restriction. Most importantly it doesn't matter what the pressures upstream and downstream are, only their relative magnitude. It could be 20 psi upstream and 10 psi downstream or 3000 psi upstream and 2990 downstream. The difference is still 10 psi.

With compressible flow things get more complex. The square root law still applies but ONLY when the upstream conditions are held constant. In other words to increase the pressure drop without changing anything else we have to change the downstream conditions i.e suck harder. If we change the upstream conditions then the density of the fluid changes and so does the mass flow rate.

For an airflow bench testing inlet ports the upstream condition is atmospheric pressure. The bench sucks air into the inlet port, through the head and into the bench - or more properly atmospheric pressure pushes air into the depression created by the vacuum pump but let's stick to the word 'suck' for now even though physicists hate it.

If we change the pressure drop the upstream conditions still stay the same which is vital if the square root law is to apply. If the pressure drop doubles then the mass flow rate goes up by root 2.

For a turbocharged engine we are not actually applying the same conditions. We are increasing the pressure drop by changing the upstream conditions not the downstream ones. That brings a second factor into play which is air density.

The first impact of doubling inlet manifold pressure is that inlet manifold air density doubles. That changes the resistance to motion of the fluid. It has twice the mass per unit volume but it's also harder to get moving. Luckily a second square root law comes into play. In fact there are more square roots in flow theory than under square trees.

The total flow into the engine is now made up of two factors.

1) With twice the pressure drop the mass flow rate increases by root 2 for a given upstream air density.

2) For a given pressure drop the mass flow rate increases by root 2 if the air density doubles.

The total flow into the engine is now root 2 x root 2 = 2

In other words it stays exactly proportional to manifold pressure so with 1 atm of boost the 100 bhp engine becomes a 200 bhp one - not a 141 bhp one.

However, now go back to the bit about flow benches testing inlet flow. What happens when you try to test exhaust flow? It might be the reason why your very expensive flow bench doesn't actually work at all if it hasn't been properly designed.

Dave

[rookie]

Originally Posted by FlowSpecialist

There, we've found something we can actually agree on. That wasn't so hard after all



When David asked me to come and have a look on here at what he was getting up to these days I hoped to be able to answer a few questions, pose a few of my own and also learn something from the experience and knowledge of others on here.

I wasn't expecting it to turn into a pissing contest quite so soon or I perhaps wouldn't have bothered.

I have read plenty of David's work thank you, albeit not the Chevy books, and he doesn't need you to tell me what sort of regard I ought to hold his expertise in. He's known that well enough for 30 years.

BTW, the word 'debate' doesn't mean to contradict. It means to discuss, ask further questions.

They do say that England and America are two countries divided by a common language so I'll put this down to teething problems of that nature for now and say no more about it.

Dave

I’m not at liberty to discuss why I was asked to post hear.

There is a puzzle for ya. See if you can find out who asked me hear and why.

My comment about you reading David’s books were based on what you said here.
I'm not familiar with these formulae but then I haven't read DV's book or books on Chevy engines. Not engines we deal with much here in Blighty. I did by chance read Bill Jenkins excellent Chevy book many years ago but that didn't go into this sort of thing anyway.
high velocity porting on a 2v per cyl pushrod motor

How long does it usually take your discussions to turn in to pissing contests and how does one judge those?

BTW, the definition of debate - To discuss or argue opposing points.

You can look the other two up your self.

In addition, who are They?

[DavidVizard-GFN]

After having a great trip to Grenada then to VIR I find myself on this thread and a little perturbed that things may be getting out of hand. I am reluctant to step in but can I please ask that we all excercise a little restraint here as I am feeling a little like the piggy in the middle.

That said let me give you an easy answer to the original question. First there is not a 14.7 psi pressure drop into the cylinder. For a typical high perf engine it averages (I emphasize average) about 50 inches. So if an intake manifold has one atmospher pressure in it (403 inches H2O) then the cylinder will have 353 inch it at about BDC. If we double the intake pressure to 806 inches and the induction cylcle now has a 100 inch pressure drop across the valve (it wont be quite that but for the sake of an easy argument lets say it is as a worst case situation) then the cylinder pressure at about BDC will be 706 inches.
Thats twice what it was before- ignoring heat thats twice the density and thus twice the power.

DV

[MAP]

Greetings David,

Thank you, David, and I gladly take responsibility for my share in this and promise not to create any further strife.

Back to Flow Specialist's explanation: that was a good, and mostly qualitative, explanation for your claim that with double the pressure in, we get twice the massflow and twice the power out (even in my very first post, when I followed your logic to deduce that 100hp would become 141hp, I knew that something was missing since we know that power goes nearly in direct proportion to inlet pressure.)

Having said all this, however, you've broached a complex topic and offered a minimalistic explanation. I'd like to see this proven by breaking each point down to first principles, and citing the appropriate constitutive equations. (My training is basically to question claims until and unless taken to this level of rigor.) To clarify, I'm convinced of the overall result, but I remain unconvinced about the validity of the individual steps used to arrive at that result. For example: you will probably point me to Bernoulli's equation for flow going as square root of pressure gradient, but in one of the first posts, you claimed that airflow was directly proportional to pressure for laminar flow with a low Reynolds number. Yet, Bernoulli's equation is predicated on laminar flow with time-invariant streamlines.

If you feel this would be too ponderous to post here, could you kindly point me to a good source (or sources) to probe matters further?

Thank you,
Mark

[FlowSpecialist]

Originally Posted by MAP

Greetings David,

Thank you, David, and I gladly take responsibility for my share in this and promise not to create any further strife.

Back to Flow Specialist's explanation: that was a good, and mostly qualitative, explanation for your claim that with double the pressure in, we get twice the massflow and twice the power out (even in my very first post, when I followed your logic to deduce that 100hp would become 141hp, I knew that something was missing since we know that power goes nearly in direct proportion to inlet pressure.)

Having said all this, however, you've broached a complex topic and offered a minimalistic explanation. I'd like to see this proven by breaking each point down to first principles, and citing the appropriate constitutive equations. (My training is basically to question claims until and unless taken to this level of rigor.) To clarify, I'm convinced of the overall result, but I remain unconvinced about the validity of the individual steps used to arrive at that result. For example: you will probably point me to Bernoulli's equation for flow going as square root of pressure gradient, but in one of the first posts, you claimed that airflow was directly proportional to pressure for laminar flow with a low Reynolds number. Yet, Bernoulli's equation is predicated on laminar flow with time-invariant streamlines.

If you feel this would be too ponderous to post here, could you kindly point me to a good source (or sources) to probe matters further?

Thank you,
Mark

If you accept the premise as being correct but want to follow the maths further then I think I have to leave that up to you but I can point you here more easily than trying to photocopy BS 1042 on fluid flow in closed conduits.

Large Diameter Orifice for Gas Flow

You can see from the equations part way down the page that mass flow is proportional to the square root of (density x pressure drop).

Dave

[MAP]

Greetings Flow Specialist,

Alright; that's a satisfactory answer. Thank you. Neither would I want to post a first-principles discussion of the Schroedinger wave equation outside of an audience of physicists, I suppose.

Best,
Mark

[FlowSpecialist]

Originally Posted by MAP

Greetings Flow Specialist,

Alright; that's a satisfactory answer. Thank you. Neither would I want to post a first-principles discussion of the Schroedinger wave equation outside of an audience of physicists, I suppose.

Best,
Mark

I'd be happy for you to do that. Me and my cat (Schroedinger in joke) are both alive and well despite numerous observations. I keep telling her that if I don't observe her to be alive she may well not be but she resolutely stays alive despite this. I don't have the heart to put her in a box and test it more thoroughly.

Dave

[rrrobert]

um bumbling around this q , i thought of this , if the flow goes up then thats fair enough 1,41 times bhp ,but the density has gone up as well ,so i would have jumped in with both feet and said 1.41 times the density increase ,and wiat to be embarrised for not getting it right or thinking about it for longer to preserve my standing in the comunity lol

regards robert

pleae bear in mind iv spent the day at the scrapyard 15 ft in the air in a scorching british sun taking off another fiat head for flow testing !


just realized , nothing has changed flow eff wize on the bench ,just the depression , so the head would still make the same bhp ,but with double the density from the turbo ,then double the power all things being equal .(thats a great phrase)

[MAP]

Greetings,

The comment I made about posting concerning the SWE was something I rather regret. My purpose is not to suggest that I can do it, as if I somehow have a need to prove whatever measure of an intellect I may have before the people reading here, but rather that certain topics are simply too deep and complex to be handled efficiently in an internet forum. As many have said, internet fora are a mile wide, but usually only about an inch deep. Thankfully, GFN has succeeded remarkably in increasing that depth, but the nature of the beast remains daunting.

This thread, I think, may have pushed this limit a bit too far, I fear. And FS, your qualitative explanation was good, but fundamentally unsatisfying from a physical/mathematical point of view. Nevertheless, sincere thanks to you for trying to offer a reasonable explanation, although P,V,T (and therefore rho) constraints for the turbo "output," which is seen as an "input" to the motor, need to be addressed before the entire argument can be called comprehensive - for example, if we double P, rho is not necessarily doubled. We need to look at at least one more equation of state before we can make that claim.

Thanks again,
Mark

[GordonShumway]

I'm otta here.