I need to figure out the distance 1 rotor tip passes from TDC to BDC. I've got tons of formulas that I'm having a hard time figuring out.

 

Why? Tell us what, exactly, you are trying to find, and perhaps we can find another way to find it.

If you are looking for the actual shift in center of mass of the rotor, it doesn't matter where the apexes are. It is simply the distance travelled by the eccentric shaft during 180 degrees of movement. There is no amount of rotation of the rotor that will change its center of mass.

So, unlike a piston engine, the eccentric shaft does not induce strictly vertical motion like a crankshaft does to a piston via the connecting rods.

There is energy expended in causing the rotor to accelerate and decelerate in its rotation, via the gears attached to the eccentric shaft, as it is moved around the rotor housing, which will cause torsional vibration in the engine.

But, there is nothing in the rotary that will cause vibration in a strictly linear fashion, like the pistons, and to a lesser extent, the valvetrain does in a piston engine.

 

rotor stroke is 3 x e

defined as total normal dsistance traveled by the face of the rotor, in one stroke.

search articles by Mike C

 

Yea theres no actual stroke. It's a non reciprecating engine.

 

my bad i mean reciprocating engine

 

Just as with a recip engine piston face, the wankel stroke is the vector normal distance traveled by the face of the rotor.

stroke = 2 x e for piston engine, in 180 deg crank rotation
stroke = 3 x e for wankel, in 270 deg crank rotation

 

there isnt a stroke! just like there isnt a bore! stop trying to force it in...

 

no Santa? gee wiz

here's a good derivation of swept volume (and effective stroke) at the end of this link.

http://www.mikesdriveway.com/misc/rotor.doc

 

Ok, I know that it's not a piston. I know it's not a reciprocating engine. I'm doing an assignment for my Thermal Dynamics class. We're supposed to figure out the Bore/Stroke ratio, Mean piston (rotor) speed, Brake HP, Brake Mean Effective Pressure, and specific Power (HP/CID) for an engine of our choosing. I took the challenge of applying these measurements to a Rotary Engine.

There is a stroke in a rotary if it is defined as: the distance traveled from BDC to TDC, or Swept Volume. Would you agree that a rotor travels a certain distance, and sweeps a certain volume? That's all Stroke and Bore are, and I'm trying to accurately compare a reciprocating engine to a rotary engine.

KevinK2 thanks for the formulas and links. What does 'e' stand for in your equation?

 

Have fun transferring what are largely piston-engine equations to the rotary.

Especially the one about BMEP. LOL.

Swept volume? What counts for swept volume on a rotary? The volume swept by the sum of both rotors' one lobe during the entire intake phase?

 

This technical paper by Yamamoto should help. Section 2.4.3 calculates "stroke volume".

http://www.rotaryrefs.net/rotaryrefs...-KYamamoto.pdf

e is the crank pin or e-shaft offset. .59" for a 13B. The stroke (3e) is the sum of all the differential normal displacements of the rotor face (first link). It is not a continuous line/curve you can draw.

each of 6 .65L active chambers will complete a full 4 stroke cycle in 3 crank revs, so a complete engine cycle takes 3 revs, and will displace 3.9L of air. 13B rotaries are labeled as 1.3L, based on 1 rev needed to fire all the plugs for each housing (2 of the 6 chambers). Piston engine cycles require 2 revs, and will pump their rated displacement in 2 revs.

the "bore" area is the piston projected area for recips. It's the single rotor face projected area for wankels, ie the rectangle defined by 2 apex seal contact lines.

There are some sae papers on wankles, which is your best place to get real info. You will get a lot of barbaric mis-info on this forum.

I suggest you not try to learn the themodynamics of an otto cycle engine, at the same time you are trying to learn how a rotary otto works. Pick a piston engine to learn on. As was said, BMEP and hard-to-define "rotor" speeds are tough to do. There may be another (simpler) way to bring wankle analysis into the class.

 

well, BMEP should be easy.

formula is BMEP = Tq x 2.475 / L

Tq is rated torque, ft-lbs
L is displacement in 2 revs, = 2.6L

So for RX8 at 160 ft-lbs, BMEP = 152 psi ( a bit on the low side... )

 

I looked this up on Google.

http://www.engineersedge.com/wwwboard/messages/798.html

Go to the page to see any follow-ups. I cut 'n' pasted his original question here.

Posted by Carmon Colvin (12.166.72.98) on November 18, 2002 at 17:12:57:

First of all I am NOT an engineer so please forgive me if I use the wrong terminology or similar laymen mistake.

I am trying to calculate the BMEP (Brake Mean Effective Pressure) of a Wankel rotary engine. Searching the net I have found one formula that is slightly different for a 4-stroke and 2-stroke type of engine.

2-stroke BMEP = (HP * 6500) / (L * RPM)
4-stroke BMEP = (HP * 13000) / (L * RPM)
source: http://engineersedge.com/engine_formula_automotive.htm

The only difference between the two formulas are the numbers 6500 and 13000. The only correlation with a 2x difference between the two types of motors are the 2-stroke motor uses 100% of is displacement for combustion during one revolution where a 4-stroke motor uses 50% of its displacement for one revolution.

The only way I can make this work between those formulas is to look at it this way.


BMEP = (HP * (6500 / DU%) ) / (L * RPM)
Using the Displacement % used as DU%

OR

2-stroke BMEP = (HP * (6500 / 1.0)) / (L * RPM)
4-stroke BMEP = (HP * (6500 / .5)) / (L * RPM)

A Wankel Rotary engine uses 33% of its displacement for combustion for one revolution output from the motor.

Wankel Rotary BMEP (HP * (6500 / .333)) / (L * RPM)

IS THIS A CORRECT ASSUMPTION?

Data from an actual dyno run (measured at the wheels) of a slightly modified Mazda 13b (1308cc) non turbo motor put out 168 HP at 7000 RPM and 147 ft lbs of torque @ 4000 RPM.

BMEP = (168 * (6500 / .333)) / (1.308 * 7000)
BMEP = (168 * 19500) / 9156
BMEP = 3276000 / 9156
BMEP = 357.8
This number seems really high compared to other BMEP readings I have seen for 4 and 2 stroke engines.
And am I correct in assuming this is in pounds per liter?

If this is completely wrong please help me with a formula that will accurately give me BMEP of a Rotary Engine.

Thanks.


Carmon Colvin

 

I aslo found that link.

Responder asks "IS THIS A CORRECT ASSUMPTION?"

answer, not quite.

The L represents total displacement per engine cycle, not Mazda's 1 rev rating. He actually said it, but did not follow with it:

"A Wankel Rotary engine uses 33% of its displacement for combustion for one revolution output from the motor."

So for this example were L only represented 1 rev, BMEP = 357.8/3 = 119 psi

Using my torque formula:

BMEP = 147*2.475 / 2.6 = 140 psi (higher than 119 at the torque peak, as expected).

 

To clarify my ancient reply ...

You asked: "Stroke" of a 13b?"I need to figure out the distance 1 rotor tip passes from TDC to BDC. I've got tons of formulas that I'm having a hard time figuring out."

As far as total motion, I've seen it derived for a rotor face using Calculus, and it was 3 x e. You get the same answer if you back calculate it, based on the piston engine formula:

Bore area (rotor face projected area) x stroke = displacement (of rotor face). The rotary stroke is close to a half circle of the rotor at radius e, = (pi) x e.

Also, contrary to another reply, there is no accel'n or decel'n forces at constant speed. The rpm of the rotor is 1/3 shaft speed. There is a primary unbalance due to the offset "e" of the rotor mass center. Counter weights are used.

 

See next post

 

I've seen stroke derived for a rotor face using Calculus, and it was 3 x e. You get the same answer if you back calculate it, based on the piston engine formula:
Bore area (rotor face projected area) x stroke = displacement (at one rotor face).

For the few still interested, I ran the numbers:

R = 4.134" = rotor effective radius, from spin center to apex seal contact line
e = .59" = offset of rotor center from crankshaft centerline
w = 3.15" = width of rotor
D = 40 cu in = Displacement per chamber (655 cc's, .655 L )

For a piston engine, the bore area is [(pi)/4] d ^2. For a wankel, it's the distance between apex seals times the rotor width (as measured at the housing bore):
2*R*cos(30)*w, = (5.58)(3.15) = 22.555 sq in = face area

The displacement "D" of one of the 6 rotor faces is 40 cu in
Stroke = D / (face area) = 1.77"

Stroke / e = 1.77/.59 = 3.000, so stroke = 3 x e, same as derived using calculus by Mike C.

-----------------------------------------------------------------------

Correction to my post number 6:
stroke = 2 x e for piston engine, in 180 deg crank rotation
stroke = 3 x e for wankel, in 540 deg crank rotation (was 270) , (180 deg rotor rotation, from TDC to BDC)

The stroke (3e) is the sum of all the differential normal displacements of the rotor face, relative to the housing inner surface. During the power stroke the volume expansion is based on the rotor face moving toward the centerline of the shaft, while the housing inner surface gets further from that centerline. It is not a continuous line/curve you can draw.

New link to K Yamamoto's book and volume calc, at post 11: From Yamamoto's Book

Great video from EE:

EE Wankel Model

 

My post number 6 was mostly correct. For wankel, 270 deg crank rotation means 90 deg @ rotor, going from TDC to BDC on the power and intake strokes. But in addition to rotor face motion of 2 x e, the housing inner surface gets further away from the crankshaft center as rotation occurs. This is the additional "e" distance to get 3 x e for the total stroke.

"e" is the offset of the rotor center, relative to the crankshaft center, same as with piston engine crankshaft and rod journal.

The video shows the stroke as you watch 90 deg of rotor rotation, from min to max chamber volumes.

.