Mach .55 flow HP loss

[Rick360]

I have read about the HP loss that occurs when intake velocity exceeds mach .55 and have found this interesting, yet I am concerned about being able to prevent this in my next engine.

Questions:
Is the mach .55 only seen in a running engine and is there a way to measure it at high pressures?

Does it show up on a flowbench any other way at reasonable pressures (<40"wc)?

What velocity when measured at 28" would do this?

Why does it reduce HP?

Is the overall flow thru the engine reduced when this happens?

Thanks,
Rick

[Walline]

I am no expert on this, but This is what I understand. Someone please correct me if I am wrong. Sound would travel at 1200 fps in cylinder head conditions. So .55-.60 times the speed of sound is 660-720fps. Air moving into the engine stops and starts with valve closing. When the valve opens the air must move. It takes energy to move the air,so at higher speeds(fps)it takes more energy to move it. I beleive it is called inertia block. If you want to calculate it on your engine use. . .

LPV= 0.00353*RPM*S*B2 divided by CA

LPV= limiting port velocity
S = stroke
B = bore
CA= min port cross section

406 Chevy .00353*6500RPM*3.75 Stroke*4.155squared divided by 2.35CA = 632fps
Hope this helps

[ozrace]

I am no expert on this, but This is what I understand. Someone please correct me if I am wrong. Sound would travel at 1200 fps in cylinder head conditions. So .55-.60 times the speed of sound is 660-720fps. Air moving into the engine stops and starts with valve closing. When the valve opens the air must move. It takes energy to move the air,so at higher speeds(fps)it takes more energy to move it. I beleive it is called inertia block. If you want to calculate it on your engine use. . .

LPV= 0.00353*RPM*S*B2 divided by CA

LPV= limiting port velocity
S = stroke
B = bore
CA= min port cross section

406 Chevy .00353*6500RPM*3.75 Stroke*4.155squared divided by 2.35CA = 632fps
Hope this helps

I think you pretty much covered it Walline.
It takes power to draw the mixture into the cylinder. The higher the velocity is - the more power is required.
High velocity air provides more power throughout the range, but there comes a point where the extra power from increased air velocity is less than the power lost to the work of drawing air in. That point is around .55 Mach on most conventional 2 valve designs.
It can vary, depending on port design and other factors, and is generally higher for 4 valves.
The formula is a useful guide, but you have to remember that many ports only flow significant air through a portion of the port, whilst the best designed ports flow well through almost the entire port (even velocity gradient). A poor port design will probably need to be larger because you aren't using the entire esp. port across the short turn.
Larry from maxracesoftware has said that when the pitot diff. pressure exceeds the test pressure, the velocity is too high, and power will suffer.
When someone with as much experience as Larry - on Flow Bench AND Dyno - shares something like that with us, I take it as fact....

[ozrace]

Sorry - that should have read:-
A poor port design will probably need to be larger because you aren't using the entire port, especially across the short turn.

[Darin Morgan]

I am no expert on this, but This is what I understand. Someone please correct me if I am wrong. Sound would travel at 1200 fps in cylinder head conditions. So .55-.60 times the speed of sound is 660-720fps. Air moving into the engine stops and starts with valve closing. When the valve opens the air must move. It takes energy to move the air,so at higher speeds(fps)it takes more energy to move it. I beleive it is called inertia block. If you want to calculate it on your engine use. . .

LPV= 0.00353*RPM*S*B2 divided by CA

LPV= limiting port velocity
S = stroke
B = bore
CA= min port cross section

406 Chevy .00353*6500RPM*3.75 Stroke*4.155squared divided by 2.35CA = 632fps
Hope this helps

I think you pretty much covered it Walline.
It takes power to draw the mixture into the cylinder. The higher the velocity is - the more power is required.
High velocity air provides more power throughout the range, but there comes a point where the extra power from increased air velocity is less than the power lost to the work of drawing air in. That point is around .55 Mach on most conventional 2 valve designs.
It can vary, depending on port design and other factors, and is generally higher for 4 valves.
The formula is a useful guide, but you have to remember that many ports only flow significant air through a portion of the port, whilst the best designed ports flow well through almost the entire port (even velocity gradient). A poor port design will probably need to be larger because you aren't using the entire esp. port across the short turn.
Larry from maxracesoftware has said that when the pitot diff. pressure exceeds the test pressure, the velocity is too high, and power will suffer.
When someone with as much experience as Larry - on Flow Bench AND Dyno - shares something like that with us, I take it as fact....

I have often been perplexed as to why the actual VE and overlap flows are not considered in this equation? Would your LPV not be higher in your 406 Chevy if the measured VE was 112%?
?

[Walline]

Darin I get what you are saying about VE numbers. I think it would be higher with anything over 100 right. Are you saying this equation errors on LPV? Also what do you mean by overlap flows?

[Darin Morgan]

Darin I get what you are saying about VE numbers. I think it would be higher with anything over 100 right. Are you saying this equation errors on LPV? Also what do you mean by overlap flows?

If I take the equation and use the Mean CA and multiply it by the Observed VE ( total flow through engine including overlap flow) the answer seems a lot closer to reality. I know that multiplying the constant in the equation along with all the other variables cant be correct but the error is so low over the range it works in it seems of very little consequence. Anyone know where, how or who derived the constant of .00353?



Overlap flow and VE are the observed ( not trapped ) air flow numbers you see on the dyno screen. All engines waste some air fuel mixture out the ex during overlap somewhere throughout the power band so that 112% VE number you see on the dyno screen is trapped + the wasted flow out the exhaust. You may only have a true VE of about 108 to 109%. It depends on chamber design. camming ect ect. This topic has been discussed on another thread by max race software but I cant remember where.

[maxracesoftware]

Anyone know where, how or who derived the constant of .00353?

(Pi/4) = (3.141592654 / 4) = .785398163

360 * .785398163 = 282.7433388 or .003536777


the .00353 is just the 1/x reciprocal of 283.286119

its a combination of conversion constants into 1 constant = 282.286119
it converts inches into area and inches into feet

it should really be = 282.7433388 or .003536777

How can you use 282.7433388 ??

you can use this constant along with Air Velocity FPS to solve for what is the required Intake Valve diameter needed for a certain "Peak HP RPM"

Intake_Valve = (( RPM * CID ) / ( Cylinders * 314.5 * 282.7433388 )) ^.5

where;
RPM = the point you want Peak HP to occur
CID = total engine size in Cubic Inches
Cylinders= the number of engine cylinders
314.5 = Air velocity in Feet per Second
282.7433388 = Units Constant
^ .5 = Square Root of a Number

2.515 = (( 9000 * 500 ) / ( 8 * 314.5 * 282.7433388 )) ^ .5


Lets say you have a NHRA SuperStock Chevy SBC Engine
4.065 Bore X 3.493 Stroke = 362.661 CID with a 1.940" OD Intake Valve

1.940 = (( 7200 * 362.661 ) / ( 8 * 306.7 * 282.7433388 )) ^ .5

7200 RPM for NHRA SS 350 with 041x heads 1.940/1.500 valves
is very close to the Norm average


Why the FPS difference between ProStock and SuperStock ??
314.5 -vs- 306.7

the ProStock port can handle higher air velocity more efficient because of
less port centerline axis -to- valve axis ..and a better overall shaped port
with more constant cross-sectional area

goes back to some of same reasons SuperStock old style Cylinder Heads like 23 degree can only effectively use .37 to .39 L/D Ratios
whereas, modern Cyl Heads can use .41 to .42+ L/D Ratios
without Loss in HP or Torque

Lets say your ProStock head design is as inefficient as a SuperStock Head
then the required Intake Valve diameter would be larger =>

2.547" Int OD = (( 9000 * 500 ) / ( 8 * 306.7 * 282.7433388 )) ^ .5

[maxracesoftware]

Larry from maxracesoftware has said that when the pitot diff. pressure exceeds the test pressure, the velocity is too high, and power will suffer.
When someone with as much experience as Larry - on Flow Bench AND Dyno - shares something like that with us, I take it as fact....-ozrace

ozrace....thanks for the vote of confidence

from a SpreadSheet =>

VALVE-------------------------------Prime Choke
LIFT--------28 in------- Area------- Velocity
0.2--------- 145-------- 1.587------ 83.755
0.3--------- 230-------- 2.380------ 132.852
0.4--------- 335-------- 3.173------ 193.502
0.5--------- 410-------- 3.966------ 236.823
0.6--------- 475-------- 4.760------ 274.368
0.7--------- 525-------- 5.553------ 303.249
0.8--------- 550-------- 6.346------ 317.690
0.9--------- 555-------- 7.139------ 320.578
1----------- 560-------- 7.933------ 323.466

23.87" Pitot Probe Velocity Pressure..at 1.000" Lift ..depending upon Port's Corner Radius

23.87 " when Flow Testing at 28" inches H2O

the 23.87" in the smallest "Port" Cross-sectional area doesn't equal or exceed the Flow Test Pressure in a ProStock Head ..even at 1.000" Lift

very hard to see evidences "Sonic Choke" inside modern ProStock Intake port (not including Short Turn Apex area)...just barely at early stages of Choke.....if you were to epoxy up that smallest cross-sectional area to the point where you would measure 28" or higher on a Pitot Probe...all the while not seeing any loss in Flow...and if the Heads would still Flow = 560 cfm @ 1.000"...you would see HP/Torque Losses on the Dyno, especially after the RPM point of Peak HP....after that point, HP would drop like a rock with a Choke problem

you would need a cam with 1.149" lift to see effective 1.000" Lift
or epoxy that area a little smaller ...and Choke would occur around .900 to 1.000 of lift...the Head would stop flowing and pumping losses would increase until either side of choke velocity or max lift point

the head will not flow anymore at Choke, but pumping losses will increase

very , very easy to have a Sonic Choke problem in an old SuperStock SBC head if you aren't careful


if the ProStock's smallest cross-sectional area is reduced to 3.840 sq.inches area and flow 560 CFM at 28" of Test Pressure you will be at Sonic Choke in live engine...if you increase Flow , or Cam Lift, or reduce this area even further, you will be into severe Sonic Choke
HP will drop 75+ HP easily

3.840 area = ( 560_CFM / 350_FPS ) * 2.4

350_FPS = velocity @ 28" Test Pressure


LPV= 0.00353*RPM*S*B2 divided by CA

LPV= limiting port velocity
S = stroke
B = bore
CA= min port cross section

657.94 fps = (.00353 * 9000 * 3.62 * 4.687 * 4.687 ) / 3.840 area

98.8 " inches in live engine
.5896 Mach at 59 F

Trends=>
if you have a Port on the edge of Choke...increasing things like E/I Ratio
can cause you to loose HP/Torque ..it will make you spread centers and reduce E/I Ratio

[SBC]

Great post Larry! Thanks

[SBC]

Great post Larry! Thanks

[Guest]

Great post Larry! Thanks

I agree.
There are several people on this Forum that make it very special.
Thanks to you all, and thanks to Don for making a place for it to come together.

[ozrace]

Great post Larry! Thanks

(sorry - didn't log in properly)

I agree.
There are several people on this Forum that make it very special.
Thanks to you all, and thanks to Don for making a place for it to come together.[/quote]

[bill jones]

-Larry,,,where did the 4.687 come from? what does it represent?

[shawn]

Bill,
that's the bore on his example. Also happens to be the bore on a prostock type motor.In his example-

(.00353 * 9000 * 3.62 * 4.687 * 4.687 ) / 3.840 area

9000=rpm
3.62=stroke
4.687=bore

shawn