Bernoulli and 350 ft/s, 146 cfm/in²

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Earlie A

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I hate watching technical Youtube videos when the guy on the video really doesn't know what he's talking about. Today, I may be that guy. The Bernoulli equation as it relates to intake ports is a complicated subject and I do not pretend to understand much of it. With the help of some guys on Speed-talk, I have managed to absorb some things. That is what I want to share.

Everything I'm talking about here has to do with theoretical dry air at 60 degrees F moving through an intake port on a flow bench with the flow bench depression (the difference between inlet and outlet pressure) set to 28 inches of water. These numbers do not represent what is happening in a running engine.

If you read much about flow bench testing you will soon see references to a maximum average port velocity of 350 ft/s and a maximum potential flow rate of 146 cfm/sq in. It's difficult to understand what these numbers mean and where they come from. The 350 ft/s is especially confusing because parts of a port CAN exceed 350 ft/s average velocity. The key word there is PART of the port, not the whole port. The 146 cfm/sq in is confusing because of the sq in part. Square inches of what? I don't think I can explain all of those questions, but maybe I can help some of you see the origin of these numbers. Discussions like this usually lead to other people jumping in and sharing knowledge that I do not have, so I look forward to learning something on this subject myself.

I'm not going to attempt to explain much of the Bernoulli principle. I do not fully understand it myself. If you are interested do the Google search and read up on it. I'll only attempt to explain the parts that relate to our discussion. Referencing the attached drawing of a port with a wide mouth at point 1 and a small discharge at point 2, the Bernoulli principle states that the sum of all fluid energies at point 1 must equal the sum of all energies at point two. So the internal energy + kinetic energy + potential energy at point 1 must equal the internal energy + kinetic energy + potential energy at point 2. Potential energy has to do with the height (elevation) of the fluid. In this case, the height at point 1 equals the height at point 2. There is no change in height from point 1 to point 2, so there is no change in potential energy. So we can delete potential energy from our equation. It now becomes internal energy + kinetic energy at point 1 = internal energy + kinetic energy at point 2. For our purposes, internal energy is represented by static pressure of the fluid (air) and the kinetic energy is a product of the density of the air multiplied by the velocity squared. It's like the energy in a moving bullet. The mass of the bullet can do damage, but the velocity does more damage.

Now, let's assume that the mouth of this intake port is so large that the velocity of the air at point 1 is zero. Our equation now becomes internal energy at point 1 = internal energy + kinetic energy at point 2. Or stated another way, static pressure 1 = static pressure 2 + kinetic energy 2. Or, just rearranging the equation static pressure 1 - static pressure 2 = kinetic pressure 2. We know that static pressure 1 - static pressure 2 is the pressure differential between the two points, which is the 28 inch depression across the flow bench. So now we have 28 inches of H2O = kinetic energy 2. So what this is saying is that 28 inches of water (which is 1.012 psi which is also 145.7 lbf/sq ft) is the amount of the static pressure at point 1 that is converted entirely to kinetic energy at point 2. Originally at point 1 there is an absolute pressure of 14.969 psi (this is standard atmospheric pressure) available to push the air through the port. Since we know that there is 1.012 psi less at point 2, the absolute pressure at point 2 is 13.675 psi. This is all the energy that is available to push air through the port. That's it. That's the limitation. If all of that 1.012 psi gets 100% converted from pressure energy to kinetic (velocity) energy, that is the limit of velocity at point 2.

In equation form this is P1-P2=ρv²/2 where P=pressure, ρ=density at point 2 and v=velocity at point 2

Here's one area where the Bernoulli principle is a little errant for air flow in a port. The Bernoulli equation used in this discussion is for incompressible fluids. Air is compressible. However, if air is flowing at low speeds relative to the speed of sound, and if there is no turbulence then Bernoulli can be used for air. Obviously this is a stretch as well because air flows quite fast in the port and does have turbulence. Nonetheless, these errors are somewhat small so we will continue.

If we assume the air is incompressible, then the density at 2 = density at point 1. So now we can write:

P1-P2=ρv²/2 where P=pressure, ρ=density at point 1 and v=velocity at point 2. So now reference the paper below where the numbers are plugged in and the equation is solved for v2.

We see that v2=350.3 ft/s. So what this is saying is that if air at point 1 loses 1.012 psi of pressure in traveling to point 2, it must gain 350.3 ft/s in average velocity. This is the maximum gain. There is no more gain possible. Can't happen.............unless the density at point 2 is lower than the density at point 1 (which it is) or unless there is a venturi effect. That's how velocities in certain areas of the port can exceed 350.3 average. These concepts are even more complicated than what we have discussed so far.

One more point to make. 146 cfm per square inch is really the same thing as 350.3 ft/s. See the attached paper for the unit conversions. If you convert seconds to minutes and multiply by 1 ft²/144 in² (which is really multiplying by 1), you end up with 146 cfm/in². For the purposes of this discussion, the 146 cfm/in² applies to the discharge area at point 2. We'll leave it there for now other than to say this. The 146 cfm/in² is an important number in calculating port efficiencies and discharge coefficients. The challenge is accurately defining the discharge area. There does not seem to be a universal standard.

If I've made errors, please kindly point them out so I can edit this post. I really tried to get it right, but I'm sure I've butchered something.

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The average guy or gal on boob tube is basically trying to make as long of a video as possible for more possible commercials and drive their market on the leader board up. Look at UTG- such long wind that it gives me anxiety watching him. The only guy I know that knows everything about Mopar’s but gets so much…. Wrong?!?

Anyhow- I’m not sure what this formula has to do with an average guy but nice work and I hope somebody chimes in. in regards to this question.
Again EarlyA you have taken it to the next level!
 
I admire your deep thought that goes into all this. I’m going to have to delve into this post later tonight at the end of my day and try to bring my level of deep thinking to your level(wish me good luck). Vastly interested subject for me. Looking forward to opening the door and “break on through to the other side”.
 
Here's an equivalent without the math. Let's say you're racing a 3200 lb car with a 400 hp engine. You can get about 116 mph in the quarter mile, but the theoretical limit is 130 mph. The 1.012 psi on the flow bench is like the 400 hp, it's all the power there is. The 3200 lb car is like the density of the air in the port. It's the mass that's being pushed. There is a limit to how fast 400 hp can push a 3200 lb car just like there's a limit to how fast 1.012 psi can push air with a density of 0.0764 lbm/ft3. Lighten the car and it can go faster on 400 hp. Reduce the air density and it can go faster than 350 ft/s with 1.012 psi of pressure drop. The 3200 lb car will never get to the 130 mph theoretical limit just like the 0.0764 air will never get to the 350 ft/s theoretical limit.
 
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This will be way over my IQ and pay grade level but I’ll try to follow along. I sticka da bur in da thing that go round and round and make aluminum chips. Lol
 
This will be way over my IQ and pay grade level but I’ll try to follow along. I sticka da bur in da thing that go round and round and make aluminum chips. Lol
There's scientific formulas for everything but common sense can prevail and make it work. A friend worked at Deere in engine development. An engineer designed the cam for their new diesel they were developing. The lobes were damn near square. Denny said it wouldn't work and it didn't. He contacted Harvey Crane and he designed a cam for the engine that worked well.
 
Here is the source of some of my confusion on this subject. In book after book and in video after video, the theoretical limits of 350 ft/s and 146 cfm/in² are used. Several months ago, in pitot tube probing the combustion chamber side of the intake valve (see picture below), I was getting local velocities of well over 350 ft/s. I thought, well, that's OK as long as the average velocity is below the limit of 350ft/s. Look at the sixteen pitot tube readings in pink. The average is 370 ft/s.

The answer is twofold. One reason is because of the venturi effect. Air is speeding up to get through the venturi created by the valve and valve seat. The other answer is because of the 28 in H2O pressure drop across the intact port. The 350 ft/s velocity limit is the limit at the density of air at the inlet of the port (14.696 psi). Because the air in the chamber is at a lower pressure (13.675 psi) than the air at the inlet to the port, the density in the chamber is lower. So, as air travels down the intake port, it's pressure is reducing and therefore its density is also reducing. Since the mass flow rate of air through the port cannot change, the velocity has to increase as the air gets closer to the chamber. So, even if the area of the intake port is constant, the velocity will be increasing as the air travels down the port.

One other confusing part of this pitot testing in the chamber was the average cfm/in² readings. These can be seen in blue. Again, the average number of 154 cfm/in² seems to violate the maximum theoretical value of 146. But again, the reason it's doing so is because of the venturi effect and the reduction in density associated with high air speeds. The 154 number would yield a discharge coefficient of 154/146=1.05 if the 146 is used as the theoretical maximum. It should not be possible to get discharge coefficients higher than 1, but it is common at low lifts. I believe the explanation has to do with scfm vs acfm. The 154 number is cfm at a lower density than the 146 number. If both numbers were converted to standard conditions (scfm), I believe that a discharge coefficient of greater than 1 could never be achieved.

Discharge coefficients are about as intimidating as shaping a short side radius. I think the reason is because of the complexities of scfm vs acfm and the difficulty of measuring the valve curtain area. More on that later.

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