Q1=Q2(P2/P1)^.5 for predicting injector flow rate. From where does this come? 2

[carwhisperer]

I have seen the above formula in several places around the world wide web. Q2=Q1(P2/P1)^.5, where Q2 is new mass flow rate, Q1 is original mass flow rate, P2 is new pressure and P1 is original pressure. The formula is supposed to predict flow rate of an injector using different fuel pressures, usually increased pressure to add more fuel for forced induction.

Setting aside whether the formula gives any useful results, or other concerns such as slowing pintle movement, etc., do you know how this formula is derived? I tried deriving it with Bernoulli's equation and I couldn't quite get there. My assumption is that you need to use Bernoulli's twice, as in treat it as two different systems with 2 different left hand pressures but the same right hand pressure (intake manifold pressure). But this didn't seem to work.

Any insight would be greatly appreciated!
Brian

 

[BrianPetersen]

"Velocity pressure" = 1/2 x density x velocity squared (Bernoulli)

Now make a ratio of one to the other. The 1/2 and the density cancel out.

So P2/P1 = (V2/V1)^2
So (P2/P1)^.5 = V2/V1

Then it is a simple understanding that Q = V x A. The A (cross-sectional area of the nozzle) cancels when making a ratio of one to the other since it stays the same.

So (P2/P1)^.5 = Q2/Q1
And then Q2 = Q1 x (P2/P1)^.5

There is an implicit assumption that we are dealing with incompressible turbulent flow. For low-viscosity liquids distant from their boiling temperature under the conditions seen in a gasoline injector nozzle, this is a good assumption.

 

[carwhisperer]

Well, cool, it sounds like I was on the right track. Thank you for the response. I believe that the following version of Bernoulli's will work: P1 + .5rv1^2 = P2 + .5rv2^2, where r = mass density and assuming change in head is not applicable.

It seems to me, however, that you need another statement along the lines of: P3 +.5rv3^2 = P4 + .5rv4^2, where P1 = baseline pressure, P3 = new pressure and P2=P4 = intake manifold pressure. Then the new velocity (which is proportional to mass flow rate?) would be v3= v1((P2-P1+.5rv2^2)/(P2-P3+.5rv4^2))^.5. If I made a mistake in my Algebra let me know. Or, if there are things that should be neglected, let me know.

Brian

 

[GregLocock]

No, it is more of an orifice rule than Bernouilli. Bernouilli only applies along a stream line.



Cheers

Greg Locock


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[carwhisperer]

No, it is more of an orifice rule than Bernouilli. Bernouilli only applies along a stream line.

OK, do you know how the original equation is derived? What is the basis for it?
Brian

 

[GregLocock]

The volumetric flow through an orifice area A is driven by a pressure drop P and is equal to P^.5*A*discharge coefficient.

http://www.efunda.com/formulae/fluids/calc_orifice_flowmeter.cfm

He uses Bernouilli like you did, so i was wrong. sorry.

Cheers

Greg Locock


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[carwhisperer]

That's interesting but I don't think it is addressing the issue in which I am interested. It seems to address only a single system. When you change the upstream pressure in a system, in my mind, that creates a new system. I only see that page referring to one system at a time. Furthermore, I don't see the equation of interest, Q2=Q1(P2/P1)^.5, stated on that page. What am I missing?
Brian

 

[BrianPetersen]

The pressure that we are talking about in these formulas is the differential pressure across the injector nozzle. This is the difference between the regulated fuel rail pressure, and whatever the intake manifold pressure is.

If you are trying to derive the full equation for flow through the injector then yes, there is going to be a whole lot more going on.

If that's what you are trying to do, then remember that the intake manifold pressure isn't constant through the engine's cycle, which means the instantaneous flow rate is going to be different depending on your injection timing. Are you injecting it during a low point in the manifold pressure through the engine's cycle or during a high point?

At a certain point you just say the heck with it and calibrate a look-up table rather than trying to derive formulas, and just give the engine what it wants based on empirical testing. That's how real ECUs do it.

You asked a simple question ... and now you are overcomplexifying it.

 

[carwhisperer]

OK thanks and I'm sorry to be tiresome. I think I get your point about differential pressure, and I think I see how that helps.

But I think all that does is change my formula to this: v3= v1((Pd1+.5rv2^2)/(Pd2+.5rv4^2))^.5, where Pd is differential pressure and Pd1 is P2-P1 and Pd2 is P2-P3.

I'm sure you are right about the empirical approach being more practical. However, I still want to know how the often quoted rule of thumb equation is derived. If you can show me how to start with Bernoulli's full equation and end up with Q1=Q2(P2/P1)^.5, I'm happy.

But when you start with this, "Velocity pressure" = 1/2 x density x velocity squared (Bernoulli), I don't know what you're doing.

If you don't know the answer, it's OK!
Brian

 

[BrianPetersen]

I already went through the derivation.

In any turbulent flow stream, the difference in pressure between upstream and downstream conditions is related to the velocity pressure multiplied by some factor that is dependent on a ton of other factors but which is constant for given geometry and fluid conditions. Whether it's the drag on an external body flowing through stationary air (e.g. the drag on a car, airplane, etc moving through the air) or the amount of fluid flowing through a duct or nozzle, the same concept applies. Open up your fluid dynamics textbook, it's all in there ...

 

[carwhisperer]

Thanks for the advice! I think to further the discussion between just you and I would be unproductive.

But, if anyone else is still reading this, do you know the math/theory in getting from P1 + .5rv1^2 = P2 + .5rv2^2 (Bernoulli's neglecting the head pressure term) to Q2=Q1(P2/P1)^.5 (the popular formula for predicting injector flow rate with a change in rail pressure)?

Or, if that's not the proper way to get there, what is?

To work with this equation, I think one needs to define the terms. I think this makes sense: P1 is pump pressure, P2 is intake manifold pressure. r is mass density. v1 is upstream velocity. v2 is down stream velocity.

Brian

 

[NormPeterson]

That Q2=Q1(P2/P1)^.5 formula is evidently for normally aspirated engines, where it can be assumed that the absolute pressure downstream of the injector can be neglected (conveniently making all those terms disappear). For a first cut determination that your current injectors will still be adequate at some higher power level, it'd certainly be reasonable to ignore 1 or 2 manifold psia at WOT compared to 43 or so fuel psig. P1 and P2 being the before and after fuel line pressures (to the left in the orifice picture in the link).

Pressurizing the manifold is a different situation, but it seems a first cut there to get the new flow Q2 would have pressure P2 now include the manifold absolute pressure, which for 10 psi "boost" would be around 25 psia. Not negligible any longer.

Either way, it's still one system, just with different pressures on both sides of the injector creating different differential pressures P1 and P2. It's these differential pressures that (roughly) determine the flows Q1 and Q2.


Norm

 

[Lou Scannon]

Since the beginning of PFI, the rail fuel pressure has been referenced to the intake manifold, so boosted applications are not a special case.

"Schiefgehen wird, was schiefgehen kann" - das Murphygesetz

 

[BrianPetersen]

^ Aye, but nowadays, a lot of EFI systems are returnless - they have an in-tank pump and regulator which is not referenced to intake pressure and there is only one fuel hose going out to the engine! The fuel pressure is not referenced to intake manifold pressure because the in-tank fuel regulator has no way of knowing what the intake pressure is.

My fuel injected motorcycles are all like that.

 

[dgallup]

It does not matter if the vehicle is naturally aspirated or boosted or whether the fuel rail has a return or is returnless. The static flow rate through a port fuel injector will be proportional to the square root of the pressure drop across it. I have made this measurement on virtually every brand and model of injector in existence. The correlation is extremely good.

The old systems kept a constant pressure differential between the fuel rail and the intake manifold as this makes life much simpler for the ECU to calculate the pulse width to send to the injector. Newer ECU's have so much compute power it is not a problem for them to calculate the static flow rate and look up the injector dead time for a given pressure drop and figure the correct pulse width.

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[Lou Scannon]

the rail fuel pressure has been referenced to the intake manifold

Let me rephrase that...: "the fuel metering calculation has accounted for the delta P available to the fuel injector".

"Schiefgehen wird, was schiefgehen kann" - das Murphygesetz