How does displacment affect compression...
Guest
Joined: May 2002
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This is a basic geometry problem I could answer exactly for you if I had a chance to spend about 5 minutes working out the equations on a piece of paper.
My equation will assume that the piston crown has not been modified by the piston manufacturer to make up for the extra cylinder volume
If not the easy answer is that it will cause a slight increase in compression.
Unfortunately I am too busy at the moment trying to get work done and catch up on TR posts, but if no one gets around to it by later tonight I will let ya know.
My equation will assume that the piston crown has not been modified by the piston manufacturer to make up for the extra cylinder volume
If not the easy answer is that it will cause a slight increase in compression.
Unfortunately I am too busy at the moment trying to get work done and catch up on TR posts, but if no one gets around to it by later tonight I will let ya know.
Last edited by 0HP930; Mar 20, 2003 at 07:00 AM.
Guest
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Im gonna do a little research... I still don't think it would raise compression, but Im neither a renound tuner, nor a math major....
The way Im thinking... is that your not changing the rate of compression IE stroke of the motor. So your not actually compressing it any further... your just compressing more mixture at the same rate...
The way Im thinking... is that your not changing the rate of compression IE stroke of the motor. So your not actually compressing it any further... your just compressing more mixture at the same rate...
Guest
Posts: n/a
Here is what my boss says about this....
---- SNIP ----
PV=nRT : Ideal Gas Law
I'm using volume for the discussion, even though it is radius that is changing. The volume is proportional to the square of the radius.
(1)Assuming your engine stuffs moles of fuel into the combunstion chamber until a certain pressure is reached then the Pressure(P)
before your stroke is constant. (2)Also, say you increase the Volume(V) of your combustion chamber by 10%, and therefore the moles of air/fuel mix are also increased by 10%.(n)
(3)R is a constant, and doesn't change.
(4)T is temperature, and lets assume this doesn't change.(Reasonable)
So we have before the stroke:
Unmodified chamber: P(V) = (n)RT
Modified chamber: P(V*1.1) = (n*1.1)RT
Lets say your stroke compresses the chamber by a 2:1 ratio. So after the stroke we have:
Unmodified chamber: P(V*0.5) = nRT -> P = (nRT)/(V/0.5) -> P = 2 * [ (nRT) / (V) ] (Basically 2x the pressure before the stroke... Big surprise, half the volume double the pressure...)
Modified chamber: P(V*1.1*0.5) = (n*1.1)RT
P = (1.1*n*RT) / (0.55 * PV)
= 2 * [ (nRT)/(V) ]
(This is the same as the unmodified chamber.)
So, if you proportionally increase BOTH the *VOLUME* and the amount of fuel air mixture then the pressure
should remain the same. I know very little about cars, but if their fuel system only puts out a certain amount of fuel
and won't put out 10% more fuel & air if you increase the available volume by 10% then something different will happen.
---- SNIP ----
---- SNIP ----
PV=nRT : Ideal Gas Law
I'm using volume for the discussion, even though it is radius that is changing. The volume is proportional to the square of the radius.
(1)Assuming your engine stuffs moles of fuel into the combunstion chamber until a certain pressure is reached then the Pressure(P)
before your stroke is constant. (2)Also, say you increase the Volume(V) of your combustion chamber by 10%, and therefore the moles of air/fuel mix are also increased by 10%.(n)
(3)R is a constant, and doesn't change.
(4)T is temperature, and lets assume this doesn't change.(Reasonable)
So we have before the stroke:
Unmodified chamber: P(V) = (n)RT
Modified chamber: P(V*1.1) = (n*1.1)RT
Lets say your stroke compresses the chamber by a 2:1 ratio. So after the stroke we have:
Unmodified chamber: P(V*0.5) = nRT -> P = (nRT)/(V/0.5) -> P = 2 * [ (nRT) / (V) ] (Basically 2x the pressure before the stroke... Big surprise, half the volume double the pressure...)
Modified chamber: P(V*1.1*0.5) = (n*1.1)RT
P = (1.1*n*RT) / (0.55 * PV)
= 2 * [ (nRT)/(V) ]
(This is the same as the unmodified chamber.)
So, if you proportionally increase BOTH the *VOLUME* and the amount of fuel air mixture then the pressure
should remain the same. I know very little about cars, but if their fuel system only puts out a certain amount of fuel
and won't put out 10% more fuel & air if you increase the available volume by 10% then something different will happen.
---- SNIP ----
Guest
Posts: n/a
but you're not increasing the volume uniformly. You're only increasing the cylinder, the combustion chamber is the same. So you're starting with MORE air fuel (the bigger cyl) and pushing it into a smaller (relitively, but actually the same) space. Compression should increase, but probably not enough to matter.
Livin' in Seattle
Joined: Nov 2000
Posts: 2,813
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Originally posted by Cronic
Here is what my boss says about this....
---- SNIP ----
PV=nRT : Ideal Gas Law
I'm using volume for the discussion, even though it is radius that is changing. The volume is proportional to the square of the radius.
(1)Assuming your engine stuffs moles of fuel into the combunstion chamber until a certain pressure is reached then the Pressure(P)
before your stroke is constant. (2)Also, say you increase the Volume(V) of your combustion chamber by 10%, and therefore the moles of air/fuel mix are also increased by 10%.(n)
(3)R is a constant, and doesn't change.
(4)T is temperature, and lets assume this doesn't change.(Reasonable)
So we have before the stroke:
Unmodified chamber: P(V) = (n)RT
Modified chamber: P(V*1.1) = (n*1.1)RT
Lets say your stroke compresses the chamber by a 2:1 ratio. So after the stroke we have:
Unmodified chamber: P(V*0.5) = nRT -> P = (nRT)/(V/0.5) -> P = 2 * [ (nRT) / (V) ] (Basically 2x the pressure before the stroke... Big surprise, half the volume double the pressure...)
Modified chamber: P(V*1.1*0.5) = (n*1.1)RT
P = (1.1*n*RT) / (0.55 * PV)
= 2 * [ (nRT)/(V) ]
(This is the same as the unmodified chamber.)
So, if you proportionally increase BOTH the *VOLUME* and the amount of fuel air mixture then the pressure
should remain the same. I know very little about cars, but if their fuel system only puts out a certain amount of fuel
and won't put out 10% more fuel & air if you increase the available volume by 10% then something different will happen.
---- SNIP ----
Here is what my boss says about this....
---- SNIP ----
PV=nRT : Ideal Gas Law
I'm using volume for the discussion, even though it is radius that is changing. The volume is proportional to the square of the radius.
(1)Assuming your engine stuffs moles of fuel into the combunstion chamber until a certain pressure is reached then the Pressure(P)
before your stroke is constant. (2)Also, say you increase the Volume(V) of your combustion chamber by 10%, and therefore the moles of air/fuel mix are also increased by 10%.(n)
(3)R is a constant, and doesn't change.
(4)T is temperature, and lets assume this doesn't change.(Reasonable)
So we have before the stroke:
Unmodified chamber: P(V) = (n)RT
Modified chamber: P(V*1.1) = (n*1.1)RT
Lets say your stroke compresses the chamber by a 2:1 ratio. So after the stroke we have:
Unmodified chamber: P(V*0.5) = nRT -> P = (nRT)/(V/0.5) -> P = 2 * [ (nRT) / (V) ] (Basically 2x the pressure before the stroke... Big surprise, half the volume double the pressure...)
Modified chamber: P(V*1.1*0.5) = (n*1.1)RT
P = (1.1*n*RT) / (0.55 * PV)
= 2 * [ (nRT)/(V) ]
(This is the same as the unmodified chamber.)
So, if you proportionally increase BOTH the *VOLUME* and the amount of fuel air mixture then the pressure
should remain the same. I know very little about cars, but if their fuel system only puts out a certain amount of fuel
and won't put out 10% more fuel & air if you increase the available volume by 10% then something different will happen.
---- SNIP ----
A larger bore will require a set of aftermarket pistions and therefore your compression ratio will be completely dependent on your selection.
Guest
Posts: n/a
Originally posted by Cronic
Im gonna do a little research... I still don't think it would raise compression, but Im neither a renound tuner, nor a math major....
The way Im thinking... is that your not changing the rate of compression IE stroke of the motor. So your not actually compressing it any further... your just compressing more mixture at the same rate...
Im gonna do a little research... I still don't think it would raise compression, but Im neither a renound tuner, nor a math major....
The way Im thinking... is that your not changing the rate of compression IE stroke of the motor. So your not actually compressing it any further... your just compressing more mixture at the same rate...
I used to think the way you do, but I was wrong, and so are you
Seriously though, while it seems it wouldn't change, it does. I can't explain exactly why, but its just one of those things you have to trust.
sooo...where's the turbo?
Joined: Oct 2002
Posts: 1,624
Likes: 1
it's just like this...you increase the size of the cylinder, so more air gets sucked in to compensate for the larger size...that is, if you use the exact same pistons, just one larger to compensate for the increased bore...liikike for instance..my subaru is 10:1 and 2.5 liters, yet the pistons are dished....while 10:1 pistons in a 1.6 liter engine are waaay more than just dished
sooo...where's the turbo?
Joined: Oct 2002
Posts: 1,624
Likes: 1
oh yah...good call....that even makes more sense, because subaru stroke is like half of a honda stroke, which means if they were both the same stroke the honda would have massively domed pistons
