# Swiroset.com

Powering future

## Explanation of the volumetric efficiency of a rotary engine

According to Auto-Ware.com, the volumetric efficiency is:

… used to describe the amount of fuel / air in the cylinder relative to normal atmospheric air. If the cylinder is filled with fuel / air at atmospheric pressure, the engine is said to have a volumetric efficiency of 100%. On the other hand, superchargers and turbochargers increase the pressure entering the cylinder, giving the engine a volumetric efficiency greater than 100%. However, if the cylinder is vacuuming, then the engine is less than 100% volumetric efficient. Normally aspirated engines normally operate between 80% and 100% VE. So now when you read that a certain combination of manifold and cams turned out to be 95% VE, you know that the higher the number, the more power the engine can produce.

Characteristics of a rotary engine compared to a 4-stroke piston engine:

The rotor of a rotary engine completes one stroke for every 270 ° of crankshaft rotation:

• 270º suction of crank rotation.
• Compression 540º of crank rotation.
• Combustion 810º of crank rotation.
• 1080º crank rotation escapement.

In other words, a rotary engine with 1080 crankshaft rotation is needed to complete an intake, compression, combustion, and exhaust cycle. Or 3 crankshaft rotations per cycle.

A piston completes one stroke every 180º of crank rotation:

• Aspiration 180º of crank rotation.
• 360º compression of crank rotation.
• Combustion 540º of crank rotation.
• Exhaust 720º of crank rotation.

A piston engine requires 720º of crankshaft rotation to complete one cycle. That is, 2 complete revolutions of the crankshaft.

A rotor turns at 1/3 the speed of the crankshaft. In other words, for every 1 revolution of a rotor, the crankshaft has made 3 revolutions. For example, when a vehicle’s tachometer reads 9000 rpm, a rotor turns at 3000 rpm.

In two-rotor engines, the front and rear rotors are 180º apart from each other. A 360º rotation of the crankshaft will make the 2 rotors go through the combustion stroke. Since each combustion chamber is -in the case of a 13B- 654cc, each 360º of crankshaft rotation will displace a total of 1308cc.

To interpolate the cycles and volume displaced by a rotary engine vs. a 4 piston engine, we can use the following logic:

• A piston engine needs 720º of crankshaft rotation to complete one cycle.
• In a rotary engine, 720º will produce 4 times of combustion:
1. 360º of crankshaft rotation => 2 combustion times.
2. 720º = 360º x 2
3. 720º of crankshaft rotation => 4 combustion times.
4. 4-stroke = 654cc x 4
5. 4 stroke = 2616cc

For the sake of simplicity, we can stipulate that a 1.3L two-rotor rotary engine is similar to a 2.6L 4-piston, 4-piston engine. It may not be academically correct, but it is a relatively simple way to visualize how the rest of this writing and the formulas normally applied to piston engines can be applied to a rotary engine.

Also, applying the same calculations used to determine volumetric efficiency (VE) in a piston engine, but for a rotary engine, will yield optimistic results. If we were to consider the rotary engine, a 1.3L displacement 4-stroke engine, the results would give a VE of over 100% in more than one instance, which is highly unrealistic.

Did it make any sense? Hmm, maybe not, but try to go through the following steps as I try to make sense of what I have collected so far.

A little experiment …

Well, today I finally gave up and decided to do a little experiment that I came across while looking for an effective method to calculate volumetric efficiency in a vehicle without having to start the car’s engine. I came across the following experiment: Calculate the volumetric efficiency of your car

I will assume that you are too lazy or tired to follow the link, so I will explain a bit what the experiment entails.

The experience requires the following: (1) motor vehicle; (1) OBD-II scan tool; (1) Stock air intake with a Mass Air Flow (MAF) sensor in the stock setting – according to the author, a slight variation from the factory stock, such as removing the screen or repositioning the sensor will give the experiment little value- (1) section of private, safe and desert road.

Once you have acquired all of those items, the procedure is quite painless. Attach the scan tool to the vehicle and make sure it can report the following: Engine RPM, Intake Air Temperature, and Airflow. Using the deserted stretch of the private road, run the vehicle from a low rpm engine (2500 rpm) @ WOT to the red line (or as far as you want the sample to go) while recording the intake air temperature (IAT) , engine speed (RPM) and intake air flow (IMAF).

Once you’ve recorded your data, re-read the experiment from the link provided and start calculating numbers! Its principle seems straightforward: based on the theoretical volumetric airflow calculated for your engine (a Renesis in this case), and the data you recorded, you can approximate the actual VE of your particular engine. I will provide the formulas that I used at the end of this article. For now, let’s take a look at this graph. [http://www.myrotarycar.com/mazdarx8/images/13B.MSP.Volumetric.Efficiency.020205.a.gif].

Theoretical volumetric airflow was calculated assuming that a 13B MSP rotary engine has a similar displacement in 720 ° of crankshaft rotation as a 2.6 liter 4-stroke piston engine. Watch how VE rises as the engine speed increases, until it reaches 5500 rpm. This is where the engine is rated to produce maximum torque, therefore it is safe to assume that the VE will peak at or near 5500 rpm. Also, you can safely assume that the volumetric efficiency plotted against engine speed will mimic the shape and characteristics of the torque curve produced by the engine.

Note that the EV plotted is somewhat linear: it starts at 80% and rises slightly above 100%. If the results of this experiment could be validated and the parameters I used were accurate, it would mean that the Renesis engine, at least in my car, is indeed very efficient for a normally aspirated internal combustion engine, definition of VE above.

Calculation of the volumetric efficiency (VE) for the Renesis rotary engine (13B MSP):

We will use the following values ​​obtained during our data recording:

Data:

Intake Air Temperature (IAT) = 82ºF

Engine speed (RPM) = 8561 rpm

Air Flow (MAF) = 27.3 pounds / minute

THEORETICAL CALCULATION OF AIR FLOW:

Formula:

[(ED) x (rpm) x (VE)] / [(ES) x (C)] = TAF

Variables:

ED = Engine displacement [in³]

rpm = motor speed [RPMs]

VE = volumetric efficiency [%]

ES = Engine stroke coefficient [#]

C = Conversion coefficient from in³ to ft³

TAF = theoretical air flow [ft³]

Solving:

[(159.64in³) x (8561rpm) x (1)] / [(2) x (1728 in³/ft³)] = TAF

TAF = 395.42 ft³

Values:

ED = 2.6 liters (1308 cc x 2) >> 159.64 in³

rpm = I think 8561rpm arbitrarily.

VE = Since this corresponds to theoretical VE, we assume VE = 100% (1)

ES = Since we simplify a 13B engine to a 4-stroke piston engine, thus 2.6L, we use a coefficient of 2.

C = 1728 in³ / ft³

CALCULATION OF AIR DENSITY AND TEMPERATURE:

Formula:

[(t1) / (t2)] = [(d2) / (d1)]

Variables:

t1 = air temperature for a known density [ºR]

t2 = Intake air temperature measured by the IAT sensor [ºR]

d1 = Density of air for a known temperature [lb/ft³]

d2 = Density of intake air [lb/ft³]

Solving for [d2]:

[(t1) / (t2)] x (d1) = (d2)

[(491.67ºR) / (541.67ºR)] x (0.0808 lb / ft³) = d2

d2 = 0.073341 lb / ft³

Values:

t1 = 32ºF >> 491.67ºR

t2 = 82ºF >> 541.67ºR

d1 = 0.0808 lb / ft³

CALCULATION OF THE VOLUMETRIC FLOW RATE:

Formula:

[(MF) / (d2)] = AVF

Variables:

MF = Mass flow rate taken from CANScan [lb/minute]

d2 = Density of intake air [lb/ft³]

AVF = actual volumetric flow rate [ft³/minute]

Solving:

[(27.3lb/minute) / (0.073341lb/ft³)] = AVF

AVF = 372,233 ft³ / minute

Values:

MF = 27.3 pounds / minute

d2 = 0.073341 lb / ft³

CALCULATION OF THE VOLUMETRIC EFFICIENCY:

Formula:

[(AVF) / (TAF)] = VE

Variables:

AVF = actual volumetric flow rate [ft³/minute]

TAF = theoretical air flow [ft³/minute]

VE = volumetric efficiency [%]

Solving:

[(372.233ft³/minute) / (395.42ft³/minute)] = AVF

AVF = 0.94 >> 94%

Values:

AVF = 372,233 ft³ / minute

TAF = 395.42 ft³ / minute

Is this remotely close to being accurate? I really do not know! I just took the time to do some research through different channels and gather information. If you have any comments, or would like to make a suggestion or correct something, please contact me!