lets take a one liter cylinder with no valves, camshaft etc

just a 1 liter cylinder with a piston and a stopcock...the stopcock is open so the pressure in the cylinder is the same as the barometric pressure

i believe that barometric pressure is generally about 14.5 psi

now lets close the stopcock & move the piston until it compresses the air to 1/2 liter..

what would the pressure in the cylinder actually be..
since a pressure guage reads 0 at atmosheric pressure would the guage then read 14.5 or 29 or something different

now compress the air to 100 cc
what is the pressure now???

does anyone know / remember how to calculate the correct answers??

------------------
Rich
Cocoa Beach, Fla
Team Chevelle #380
46 but feeling like 20 when i'm in my 70 SS 454
wa3men@aol.com
www.chevelles.com/showroom/70_SS_454.jpg (http://www.chevelles.com/showroom/70_SS_454.jpg)

Be Back

[This message has been edited by Greybeard (edited 08-28-2000).]

Rich,
Geeez, had to dig out the old college books to be sure, but you are correct. The pressure is determined by Boyle's law. The formula says:
P2 = P1 x (V1 / V2)

P2=new pressure
P1=old pressure
V1=old volume
V2=new volume

In your case P2 = 14.5 x ( 1 / .5 ) = 29

Now that you have me totally confused http://www.chevelles.com/forum/smile.gif http://www.chevelles.com/forum/smile.gif

Hope this helped,

------------------
Bill Koustenis
Advanced Automotive Machine
Waldorf Md

1971 Heavy Chevy - original owner
Team Chevelle #100

What about

P*v = n*R*T

BillK was on it. Your idea of "appearent psi" can be confused do to what kind of gauge you have. Is it "atmospheric" or the other one... that is if you are not at sea level your "atms" are different.

So most use a common "zero" scale.

If I say any more, I'll have to go back and demand an A instead of the B I got in Thermodynamics. http://www.chevelles.com/forum/mad.gif

------------------
used to be "72chevelle350TH350" the fever got me too
BB414, TH400 (POS!), 3.73 posi,
13.634sec @ 102.5 MPH

Picture of me roasting the tires (http://www.auburnextremeracing.org/drivers/mike/)
Video of me staging (smoke of course) (http://www.notabusinessracing.org/videos/mike_chevelle_burnout.avi)

One comlication if you compress gas quickly and by large factor (like ~10 in engine) is that the temperature of the gas goes up, so the pressure ratio and volume ratio will different.

Thermodynamics??? I did this my sophomore year in chemistry class - no kidding!

------------------
1971 Chevelle Malibu - 406 sbc TH350

pics at www.adam_nova.homestead.com/Untitled1.html (http://www.adam_nova.homestead.com/Untitled1.html)

One thing that you have not factored in is the heat transfer from the cylinder. This will influence everything. If there is no heat transfer (perfectly insulated) it is called adiabatic compression. If there is perfect heat transfer to the outside, then it is called isothermal compression. What real engines and compressors have is called polytropic compression, which is a blend of the two. The polytropic constant is determined by the rate of heat transfer from the cylinder which is influenced by coolings system stuff and by the rate of compression. Sorry to rain on your parade . You can guess what my major was. However, if you would like to play with this, then assume no heat transfer and use the equation PV = nRT where p is Pressure, V is Volume, n is the mass of the gas inside (blow this off and assume it is constant) , R is the Universal Gas constant and T is the absolute temperature (add 459 to the Fahrenheit value).You can blow off the gas constant for yoiur study. Now you can get back the earier post about Boyle's law and get going. What I am trying to show here is that there is actually a lot going on in what seems like a simple process. OK, now I will cut to the Chase; this math shows why you cannot calculate the compression pressure by knowing the mechanical compression ratio, in addition to not knowing what the heck the valve timing is doing to the mechanical ratio.

Since it's a hypothetical question, assume the temp remains the same. There's a whole other set of physics laws to deal with that. The simplest answer is, if you reduce the volume by half, you double the pressure. Like Bill said, Boyle's Law.

To calculate the compression ratio.

Displacement + chamber volume divided by chamber volume. [1000cc + 100cc]/100cc = 11-1

At the halfway point; [500cc + 600cc]/ 600cc or 1.83-1.

Okay now I'm confused. We will all spen days kicking each other in the shins about polotics or street racing or why ford sucks.

Why is it that when the real technical questions get asked the sky falls and the college texts come out?

Perhaps we should have a whole new forum for the heavy questions.

------------------
Francis Taracido Gold# 201
sniper0666@aol.com
History shows again and again how nature points out the folly of man.
Oh No! You Know She's Got To Go!
Go Go Quadzilla! <oo=+=oo>
http://members.aol.com/sniper0666/page/intro.html

70 SS 454 if i understand your thinking correctly, then what you really want to know is if building a motor for high cylinder pressures is really worth it. Jwagner nail this one on the head with explaining the complications you can get into when working at real world situations. But what you should really want to shoot for (in my opinion) for a polytropic process such as your engine (now keep in mind that this is a real world situation), then you should shoot for high cylinder pressures to get the most amount of work out of the system possible. Use the equation of W(work) going from start to finish (in other words start = 1 to finish =2) = (P2V2-P1V1)/(1-n). Now with your situation of having the ideal gas law equation used for your perfect situation, PV=nRT and you're still looking for the pressures at P1 and P2 along with your volumes at V1 and V2 then you would want to set this up for high cylinder temps getting into the end of the system for the most amount of work to be done. Now granted in a real world situation super high cylinder temps can lead to the heat sticking around and bad things settling in such as overheating, pre-ignition and such, but with your "perfect situation" I would recommend shooting for the maximum amount of work with W=(nR(T2-T1))/(l-n). Just my two cents worth guys http://www.chevelles.com/forum/smile.gif

------------------
I need GM style rear 1/4's for 70-72!! e-mail me.

My head hurts...

------------------
SOLD IT NOW IM STYLIN, IN my 90 VETTE (91 updated) soon to trade for an SS
WANNA TRADE????
My Chev-Hell Page (http://hometown.aol.com/jnkb2cool/home.html)

New Member #783
Ft.Worth, Tx
LOOKING FOR 70* SS

ok guys heres what im getting at...

if the cranking compression is say 147lbs...then does that mean that the CR must be AT LEAST 10:1 if the cylinder was ideal, with no valve timing or heat transfer to skew the math??

if you take a cylinder that is 1 liter at atmosheric pressure of 14.7 and compress it to 100 cc would the pressure not be 10 X = 147 the original pressure??

so does that mean that any engine that cranks at 150 psi must be AT LEAST 10:1 CR

and one that cranks at 180 must be AT LEAST 12.25 :1 CR
we all agree you cant calculate CR based on cold crnaking pressures but is it fair to say that we can calculate the MINIMUM CR and that the actual could be NO LESS than that figure??


------------------
Rich
Cocoa Beach, Fla
Team Chevelle #380
46 but feeling like 20 when i'm in my 70 SS 454
wa3men@aol.com
www.chevelles.com/showroom/70_SS_454.jpg (http://www.chevelles.com/showroom/70_SS_454.jpg)

[This message has been edited by 70 SS 454 (edited 08-29-2000).]

Compression ratios deal with cylinder volumes rather than pressure. The compression ratio of any engine is the total cylinder volume (including combustion chamber) with the piston at the bottom of the cylinder vs the volume of the cylinder with the piston at the top of the cylinder.

[This message has been edited by Randy Mosier (edited 08-29-2000).]

Rich, I'm sorry but it won't work. The assumption of "no heat transfer" aka adiabatic will wind up skewing the math in the wrong direction, because the temperature of the gas will rise when it is compressed. The rise in temperature will cause the pressure will go up by MORE than the ratio of volumes. The law for adiabatic changes in volume is

temperature * volume ^ (gamma-1) = constant

where temperature scale is from absolute zero, and gamma is about 1.4 for air. You can see that the tempature will go up considerably when volume of gas is reduced by a factor of 10. You could use this equation along with PV=NRT to see what the pressure does. The real-world situation is more complicated, somewhere between adiabatic and constant tempature.

A small point: Also don't forget that our pressure gauges read the pressure above atmospheric, so a reading of 150 psi on the gauge is more like 165 psi absolute pressure.

70 SS 454
You are off on the wrong page. "Cranking compression" is really cranking pressure. The 147 lbs is really 147 lbs per sq. inch (psi). Cranking pressure is really a meaningless number except it is nice when all 8 cylinders are nearly equal. For example, a big cam will greatly reduce cranking pressure.

Craig Berland

i think jwagner & carl hit it..

cranking pressure is higher that one would calculate directly from the CR due to temp rise which also directly increases pressure..the effect of the cam would be opposite ie would tend to reduce cranking pressure