diabolical1

iTrader: (3)

Hey All,

I have a few questions, and I figured I'd like to try to turn this into a one stop reference resource while I'm at it. I'm going to try to get as much info on compression for the different engine models throughout the years so that I can post it here. If it's good enough maybe we can have a stick or archive ...

Anyway here are my questions:
1. Is there a formula that I can use to turn numbers that I get from a compression test in the numbers you get for static compression? For example, if I were to get a 110 PSI reading from both rotors, is there a way to estimate what compression rotors are giving me that number (9.0:1, 9.4:1, 9.7:1, etc.)?

I'm asking because I'm in the process of polishing my rotors and considering the fact that I am removing material from the recesses, I want to know if there's a way to figure out how much deviation from the stock static compression I have.

I'm thinking that there may be too many variables (seals, springs, housing surface, etc.), but I'm not too informed on this topic so I figured I'd ask.

2. Does anyone have any ideas on how to measure the volume of a rotor recess and just how significant would that be in estimating compression (assuming that it is significant at all)?

I was thinking of placing a piece of packaging tape over the face and pipetting water into the recess, but I'm not sure how effective that will be.

3. Do changes in port size (stock vs. street or bridge, etc.) or port size disparity (intake size vs. exhaust size) affect static compression?

Thanks for any answers and like I said, I'll be trying to gather some info to post as well.

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peejay

No.

At least, none that I am aware of. You can make one, if you knew ALL of the variables and accounted for them... and there are SO MANY variables, many of which are rough guesstimates, that there isn't that much point to doing so.

Riddle me this: You have a piston engine with 9:1 compression that makes 200psi on a compression test. How is that possible, given that 9 times atmospheric pressure is less than 200psi? Compressing the air makes it hotter, and inefficient compression (no compression is 100% efficient) adds heat, which adds to the pressure... now how much heat is lost to the engine block, and how quickly is that heat lost (so we can take cranking SPEED into account)... and this means we have to take ambient atmospheric conditions into account as well as the temperature of the engine assembly... and we haven't even gone into cam (port) timing yet and how it affects compression test numbers...

No wonder they say "Compression test numbers are just a guide to the state of the engine's health and should not be taken as absolutes"!

diabolical1

iTrader: (3)

figured as much. i also figured you'd probably be the only one to try to answer this, peejay.
thanks much.

i guess i'll just have to see how power i make with the rotors when they're done and try not to lose sleep over it too much. i'll just hope that either the compression change is really minute or that the new seals and springs will compensate for whatever i lose.

Wargasm

In order to estimate rotor recess volume, your water idea might work... or you could try using a fine sand with a ruler to "level it" off. Then you pour the sand onto a sensitive scale and weigh the grains to estimate the chamber volume.

diabolical1

iTrader: (3)

well, i'm going to keep plugging away at it. i've done the practice rotor and i've started the ones i plan to build the motor with. so there's no turning back now. i'll have to wait until i get home (about 2 more weeks) so i can break out the references and make up that compression table to post.

just want to say thanks to Peejay and Wargasm for the input, i really appreciate it - and anyone else that wants to add (or subtract) a technical tidbit.

badfish229

There is a formula for compression ratio in rotary engines. Let me express the formula, then I will explain what each term means .

For starters,

e=eccentricity
R=rotor center-to-tip distance
a=angle

The main thing to understand is the diagrams. This is an ideal rotary engine with triangular flanks. The top image represents the clearance volume, and the bottom image represents the displacement volume.

Observing Figure 7.7, suppose a the maximum differential area, which is

dA[max]=2ydx=2[e*sin(3a)+R*sin(a)]*d{e*cos(3a)+R*sin(a)}

You can divide dA[max] by R^2, and then differentiate with respect to a, to get

A[max]/(R^2)=-2(Integrand){[(e/R)*sin(3a)+sin(a)]*[3*(e/R)*sin(3a)+sin(a)],a,0,60}

Now, you have a dimensionless variable for the maximum differential area over the rotor center-to-tip distance squared. Note that angle a only varies from 0 to 60 degrees, the limits of the integration.

After some plugging and chugging,

A[max]/(R^2)=pi*[(e/R)+(1/3)]-3^(1/2)*[1-6(e/R)] (1)

You can also get the clearance area over the rotor center-to-tip distance squared.

A[min]/(R^2)=pi*[(e/R)+(1/3)]-3^(1/2)*[1+6(e/R)] (2)

Phew. Okay, now that you have these two variables, divide equation (1) by equation (2) to get the compression ratio, or

CR={A[max]/(R^2)}/{A[min]/(R^2)}

From observation, the compression ratio is related to the eccentricity ratio (e/R). (e/R) has many other uses, such as detmermining displacement, as well as power.

Compression ratio does not increase with size but increases by changing (e/R).

Since a real rotary engine's rotors are rounded, you must take in account for the added volume per flank.

You can calculate the segment volume as

A[s]/(R^2)={(3/8)[theta-sin(theta)]}/{[sin^2(theta)]/2]}

Now, the compression ratio goes to

CR={A[max]/(R^2)-A[s]/(R^2)}/{A[min]/(R^2)-A[s]/(R^2)}

In case you guys didn't already know , the rotor recess does not affect the displacement, only the compression ratio.

As for defining the rotor recess volume, just follow what Wargasm explained .

badfish229

As for your 3rd question, I believe porting an engine has little to do with compression ratio. Porting an engine only increases the flow area, not the clearance and displacement volumes.

Dokta

just check this out, its a lot less confusing.

http://www.personal.utulsa.edu/~kenn...n/chapter7.pdf

dont even bother reading it unless you have taken calc II

badfish229

Where do you think I got that information from?

diabolical1

iTrader: (3)

well thanks for the formulae. sheesh! thanks for also making me realize that Calc II was about 9 years ago - and i'm old! i have not gone over it with any of my old texts yet (and i've had no need to apply this stuff to keep it fresh in mind), but at a glance, if i'm understanding correctly, this only calculates static compression, no? it may be me being obtuse, but i figured it would be best to work backwards from an actual compression reading simply to account for inconsistencies in my home-polishing skills and paint a more accurate picture. however, i'll try not to let your labor go to waste, if i can remember how to do this stuff, i'll calculate the compressions based on what i have - but my guess is i'm going to get 9.4, like Mazda did regardless of the changes i made.

well, i have no reason to doubt you on this. again, maybe my understanding is lacking or i'm being too simplistic with all of this, but with larger ports allowing more air in to be compressed, how would it not affect the CR (assuming no changes in rotor recess volume)?

again thanks for the responses.

badfish229

No problem .

Well, I tend to look at porting rotaries like changing camshaft profiles on piston engines. Does changing a cam profile result in change in the compression ratio? IMO, you're actually changing the "duration" at which the ports will stay open. More air is going to come in, but it's not going to change the clearance and displacement volumes.

Mathematically speaking, look at either the equations I wrote or the PDF file Dokta left. Note that compression ratio has nothing to do with mass flow rates. It only has to do with the clearance and displacement volumes.

There is a way to relate compression ratio and mass flow rate to get power.

I hope this helps .

Dokta

Actually dynamic compression ratio is determined by camshaft profiles and rod ratio. I'm still not experienced enough with rotaries to know how the relationship between dynamic and static compression works. But in my experience with piston engines, dynamic compression ratio (the actual compression that the engine "sees") has always been lower than the static compression ratio. I have read some papers that there are ways of increasing the dynamic compression ratios over the static compression ratios by using just the right amount of overlap, changing cam timing, tweaking variable valve timing engines, using aluminum rods, and utilizing porting techniques. I suppose with proper porting on a rotary, you could increase the dynamic compression ratio within a small rpm range. But again, my experience in this area is limited. Perhaps there is someone else on this board who knows more about the dynamic compression ratios in a rotary?

badfish229

Well, we were talking about static compression ratio, which is where the equations came from.

You are right. Changing the length of connecting rods on piston engines will have an adverse affect on the compression ratio. That is similar, but not analagous, to changing the rotor center-to-tip distance, I think. Dynamic compression ratio is different from static compression ratio in that it uses the position of the piston with respect to the intake valve closing rather than BDC of the crank stroke to find the clearance volume. Due to this fact, dynamic compression ratio is always lower than static compression ratio. Also, compression ratio is determined by the engine's design, therefore it never changes throught its operation. You could flow a million CFM of working fluid, but the compression ratio is constant. Remember compression ratio has nothing to do with volumetric, let alone mass flow rate.

In rotary engines, you could find the dynamic compression ratio by determining the position of the rotor with respect to the port closing.

diabolical1

iTrader: (3)

i was thinking along these lines, but i guess i wasn't making a distinction between dynamic and static compressions. i just thought that more air being compressed by the same unchanged rotor would increase compression (at least in theory). obviously you guys are more informed on the technical specifics, so i can't argue yea or nay on the matter, but it's still nice to learn.

badfish229

I have never heard of being able to increase the compression ratio over a small rpm range. Dokta's hypothesis may hold true in a sense, but overall, compression ratio is constant throughout the powerband. Even if this were true, I wouldn't worry about maximizing or minimizing the CR over some arbitrary engine speed. What would you have to gain? What if you lose power over that range? Would it be worth it? I want to illustrate how mass flow rate has little to do with compression ratio .

Remember the Ideal Gas Law for mass flow rates? Just in case anyone forgot, the equation is

p(AV)=mR[G]T

p=pressure

(AV)=Area x Velocity

m=mass flow rate

R[G]=Ideal Gas Constant, which I believe is (R/M); M describes the molar mass of the working fluid, which in this case for air, is 28.97 (I can't remember the units; I hope someone can add to this ).

T=absolute temperature, in either Rankine or Kelvin.

Rearranging slightly,

m=p(AV)/{R[G]T}

As you can see, the mass flow rate is directly proportional to the area.

As I said earlier, compression ratio has little to do with mass flow rate, but it does influence it in the working chamber.

Suppose you replace variable A with A[min] from the compression ratio equation. I don't want to get into the messy math (you can experiment for yourself ), but your result will ultimately lead to putting the compression ratio into the mass flow formula.

The mass flow rate I am talking about is always changing; changing A is going to adversely affect the mass flow rate. But that's not what we are here to find.

The mass flow rate for the intake port is constant, and will not influence the compression ratio.

The inlet mass flow rate is calculated from the port area; this area cannot be derived from the compression ratio.

What does all this mean? The mass flow rates at the intake port, as well as the exhaust port, will never influence the compression ratio. The area of the ports has nothing to do with the clearance or displacement volume.

Dokta

You guys need to remember that the static compression ratio is almost never the actual compression ratio in the engine. I am most familiar with piston engines, so I'll talk about them first.

The compression ratio that is calculated from the static cr, piston rod length and the intake valve closing point is called the trapped compression ratio. Sometimes people call it the dynamic compression ratio, but this is wrong. It does not take into account many of the factors of an engine that is acually running.

The actual dynamic compression ratio changes with RPM, volumetric efficiency, rod ratio, resonance of the engine, pulsing pressure waves from the opening and closing of the intake valves, port size, port surface finish, valve geometry, valve seat angle, valve overlap, intake manifold design, and many other factors too numerous to mention here. Some race engines (such as nascar) that have limitations on their static cr, have been able to increase thier dynamic compression ratio to be higher than static (as much was 115% higher). These engines have higher volumetric efficiencies (i.e. a 3.0 liter engine that ingests 3.3 liters of air per cycle). Dynamic cr increases with torque, and for the most part follows the torque curve along the rpm range.

For example, me and friend were recently able to increase the volumetric efficiency of a race engine to 105%. Most of the work was done by my friend on the intake manifold. He took into account my timing of the intake cam, and was able to take advantage of the pressure waves created when the intake valve slams shut. The intake manifold was designed so that at a certain RPM range (11,000-13,000) that pressure wave would travel up the runner, pressurize the main chamber, and "bounce" back down the runners just in time to catch the intake valve on an adjacent cylinder as it is opening. This localized area of higher pressure is able to ram an intake charge into the cylinder and hold it there until the intake valve closed. A similar design was used with the headers so that there is an area of low pressure at the exhaust valves during the entire time they are open. (Actually it was just keeping away the localized high pressure until after they closed) Then with lots of dyno time , i was finally able to get the exhaust valves to stay open just the right amount into the intake stroke. The low pressure at the exhaust valves helped to "suck" the spent exhaust gasse out of the cylinder, and provide a greater differential pressure between the intake and exhaust, further helping charge the cylinder with more air than it was supposed to contain.

That is just one way to increase the dynamic compression ratio of an engine for a certain rpm range. It really helps increase the torques numbers right in the power band where we need them. I'm sure the formula 1 guys have dozens of other tricks.

As I said, im not as familiar with rotary engines, but I am sure there are ways of increasing dynamic compression ratio in these applications as well.

badfish229

You're correct. Dynamic compression ratio changes with time, or in this case, angular velocity. One could assume the Wankel rotary engine exhibits Otto Cycle characteristics. The equations I used describe static compression ratio. This varies far from what dynamic compression ratio is. I figured dynamic compression ratio changed with time; it is after all dynamic .

So, how does one account for dynamic compression ratio?

diabolical1

iTrader: (3)

well, as promised ... i know that a few months have passed since i started this thing, but i never forgot. it's just really hard to find time when you don't have the internet in your apartment and you're in nursing school.

diabolical1

iTrader: (3)

i know that i'm missing a few things here, and i think that the 10A years may be a little off, but for the most part, i think i got everything else right. so for anyone interested, i hope this chart is helpful.

cheers
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by the way, it's a Word document. when i tried to cut and paste, it was not in neat columns, so sorry for any inconvenience.