Table of Contents
Definition
For a thermal engine, the combustion process depends on the air-fuel ratio inside the cylinder. The more air we can get inside the combustion chamber, the more fuel we can burn, the higher the output engine torque and power.
Since air has mass, it has inertia. Also, the intake manifold, the valves and the throttle are acting as restrictions for the air flow into the cylinders. By volumetric efficiency we measure the capacity of the engine to fill the available geometric volume of the engine with air. It can be seen as a ratio between the volume of air drawn the cylinder (real) and the geometric volume of the cylinder (theoretical).
Formula
Most of the internal combustion engines used nowadays on road vehicles, have a fixed volumetric capacity (displacement), defined by the geometry of the cylinder and the crank mechanism. Strictly speaking, the total volume of an engine Vt [m3] is calculated function of the total number of cylinders nc [-] and the volume of one cylinder Vcyl [m3].
The total volume of the cylinder is the sum between the displaced (swept) volume Vd [m3] and the clearance volume Vc [m3].
The clearance volume is very small in comparison with the displacement volume (e.g. ratio 1:12) so it can be neglected when calculating the volumetric efficiency of the engine.
Image: Basic piston and cylinder geometry parameters of internal combustion engines
where:
IV – intake valve
EV – exhaust valve
TDC – top dead center
BDC – bottom dead center
B – cylinder bore
S – piston stroke
r – connecting rod length
a – crank radius (offset)
x – distance between the crank axis and the piston pin axis
θ – crank angle
Vd – displaced (swept) volume
Vc – clearance volume
The volumetric efficiency ηv [-] is defined as the ratio between the actual (measured) volume of intake air Va [m3] drawn into the cylinder/engine and the theoretical volume of the engine/cylinder Vd [m3], during the intake engine cycle.
The volumetric efficiency can be regarded also as the efficiency of the internal combustion engine to fill the cylinders with intake air. The higher the volumetric efficiency the higher the volume of intake air in the engine.
In case of indirect fuel injection engines (mainly gasoline) the intake air is mixed with fuel. Since the amount of fuel is relatively small (ratio 1:14.7), compared with the amount of air, we can neglect the fuel mass for volumetric efficiency calculation.
The actual intake air volume can be calculated function of air mass ma [kg] and air density ρa [kg/m3]:
Replacing (4) in (3) gives the volumetric efficiency equal to:
Usually, on the engine dynamometer, intake air mass flow rate is measured [kg/s] instead of air mass [kg]. Therefore, we need to use air mass flow rate for volumetric efficiency calculation.
where:
Ne [rot/s] – engine speed
nr [-] – number of crankshaft rotations for a complete engine cycle (for 4-stroke engine nr = 2)
From equation (6), we can write the intake air mass as:
Replacing (7) in (5) gives the volumetric efficiency equal with:
The volumetric efficiency is maximum 1.00 (or 100%). At this value, the engine is capable of drawing all of the theoretical volume of air available into the engine. There are special cases in which the engine is specifically designed for one operating point, for which the volumetric efficiency can be slightly higher than 100 %.
If intake air pressure pa [Pa] and temperature Ta [K] are measured in the intake manifold, the intake air density can be calculated as:
where:
ρa [kg/m3] – intake air density
pa [Pa] – intake air pressure
Ta [K] – intake air temperature
Ra [J/kgK] – gas constant for dry air (equal to 286.9 J/kgK)
Example
Let’s consider a compression ignition (diesel) engine with the following parameters:
nr = 2
pa = 1.5 bar
Ta = 40 °C
Ra = 286.9 J/kgK
Ne = 1000 rpm
maf = 0.0375 kg/s
For the above engine parameters, calculate the volumetric efficiency.
Step 1. Calculate the intake air density using equation (9). Make sure that all measurement units match.
The air intake pressure was converted from bar to Pa and the temperature from °C to K.
Step 2. Calculate the volumetric efficiency of the engine using equation (8).
The engine displacement was converted from L to m3 and the engine speed from rpm to rps.
Image: Volumetric efficiency function of intake air pressure and engine speed
The volumetric efficiency of an internal combustion engine depends on several factors like:
- the geometry of the intake manifold
- the intake air pressure
- the intake air temperature
- the intake air mass flow rate (which depends on engine speed)
Usually, engines are designed to have the maximum volumetric efficiency at medium/high engine speed and load.
You can also check your results using the volumetric efficiency calculator below.
Calculator
| Vd [L] | nr [-] | pa [bar] | Ta [˚C] | Ra [J/kgK] | Ne [rpm] | ma [kg/s] |
| Air density, ρa [kg/m3] = | ||||||
| Volumetric efficiency, ηv [%] = | ||||||
For any questions or observations regarding this tutorial please use the comment form below.
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RODNY
CAN YOU HELP CALCULATE MY 152.2CC MOTORCYCLE FOR FUEL AND AIR MIXTURE PLUS SPART TIMING..THANK YOU
jani129
It seems your Va as intake air volume is understood at intake manifold pressure. It is a theoretical volume, the amount of air passed through, would take at intake manifold pressure. I get the impression not everybody does it this way, as in some do it at outside air pressure. I do prefer intake manifold pressure method. Now the volumetric efficiency charts that some cars might use add another twist. the chart has two axles, one for rpm and the other for intake manifold pressure. but the pressure sensor is not measuring absolute intake manifold pressure, rather it is measuring it relative to outside air pressure. this adds more ambiguity.
i want to design an ecu. Is the kind of manifold air pressure sensor that is relative to outside pressure, cheaper or more reliable than absolute sensor? absolute sensor could improve accuracy in case of high altitude or weird weather. is this very theoretical consideration, not important?
tomas
so as you gain rpm the volumetric efficiency goes down?
Art
Sorry to be picky, but equation (9) says
ρa = pa / (Ra + Ta)
Shouldn’t this be
ρa = pa / (Ra x Ta) (multiply, not add)
Alberto
It looks like the calculator is not using the RPM value, it’s always 1000 RPM so the VE got high numbers, I’m working with a calculator and used your formulas, this post totally makes sense but I was validating my function with the calculator and got different values.
Anthony Stark
There was a bug in the script. Fixed it, should be ok now. Thanks!
amir muhammad
design a six-liter race car engine that operates on a four-stroke cycle. decide what the design speed will be, and then give the number of cylinders, bore, stroke, piston rod length, average piston speed, imep, brake torque, fuel used, af, and brake power, all at design speed. all parameter values should be within typical, reasonable values and should be consistent with the other values. state what assumptions you make (e.g., mechanical efficiency, volumetric efficiency, etc.)
Mojtaba
Hi, how are you, i am mojtaba
Excusme, I need to know how much air flow needed to ignite a diesel engine with the following specifications, please help me, if possible tell me the formula, thank you very much.
Bore: 128mm
Stroke: 144mm
Displacment: 21.93 cm^
Rpm: 2100
Rate: 662kw, 900psi @2100rpm
Number of sylinder: 12
Coprssion ratio: 16:1
Plase hel me…
Muammer
A SI engine with 1.6 liter displacement volume operating with excess air ratio λ=1.1, volumetric efficiency of 90%, engine speed of 5000 rpm, and indicated thermal efficiency of 30%. Exhaust gas temperature at exhaust valve is 600 ⁰C. Determine,
a. Indicated engine power
b. Exhaust power
c. Cooling power
Mantas
I dont understand why i am getting efficiency in hundreds of percent…
My case:
2,8L
Boost pressure (at the MAP sesnosr) 0,9 bar (boosted engine)
Intake temp is arround 20 celsius
5600 RPM
Calculated air flow is 317,3 g/s
Where do i make a mistake?
mason'sracin
double check that your giving measurements in the proper format. It looks like you should be inputting kg/s not g/s. also, your volumetric efficiency will likely be higher than 100% with boost, as your flow rate will be higher.
Thomas
Hi Anthony
You said ” What is important is oxygen, not the actual air. The reason is that oxygen makes combustion possible. For example air contains also nitrogen, which is not needed because leads to the formation of nitrogen oxides (NOx). The more air you have in the cylinder, the more oxygen, the more fuel you can burn”,Thus, the volume of air ( mainly nitrogen) has no effect on engine’s power output? I mean this volume (apart from O2 that causes combustion to occur) when compressed can be lead to bigger explosion. Sorry to use explosion term.
Junaid
Hi:
I need to calculate these shown below:
Brake Power
◦ Brake Mean Effective Pressure
◦ Volumetric Efficiency
◦ Thermal Efficiency
◦ Air / fuel ratio
To calculate these, i be given these data such as: engine speed, mass airflow, fuel volume flow, pumping torque, brake torque, intake port gas temperature, exhaust gas temperature, indicated torque, friction torque, heat transfer rate, intake port pressure, reference pressure, exhaust port pressure, fuel, type, fuel density, displacement, low heating value, ambient reference temperature, bore, stroke, inlet valve diameter, exhaust valve diameter, inlet valve maximum lift and exhaust valve maximum lift
RPD
I’m curious about a calculation I’m trying to run. I have an IC engine, that used 4.7 gallons of LPG, over 12.5 hours. The engine is 220cc displacement. Pressure was 1Bar. Temp was 22C, RPM is 3600 and its a 4 stroke engine.
How would you compute this?
For density I get 1.17 kg/m^3
For Ma I get ~ 0.00019788 kg/s * 2 / 3600 = 0.000000110. That cant be right… I’m feeling like I am mixing up liquid vs gas density or forgetting some basic principal, when calculating mass flow:
4.7 Gal LPG / 12.5 hrs
0.376 gal/hr
0.000104444 gal/sec
0.39536523 ml/sec
0.197880298 g/sec
0.00019788 kg/sec
I’m using 0.5005 g/ml for density of propane liquid. I see some 0.495 and some 0.510 ratings.
John
Do these equations change in a boosted engine? If so, how? Would love to see an example using a turbocharged engine.
Anthony Stark
The equations are the same for a boosted engine because it takes into account the intake air pressure. For the data in the calculator you can see that the intake air pressure is 1.5 bar (boosted engine).
Carly P.
The Volumetric Efficiency Calculator looks not OK! The VE formula is Ok
It calculate the data in this article OK, but not with my data.
I have the Toyota NASCAR engine:
5.866liter
0.0503 kg/sec
8700rpm
0.981 Baro intake
31 degr Celsius isue temp
1.242 Dentisy
The VE% must be 104%. The calculator is making 91%
Could you explain why it’s Good calculation with your formula and not with the calculator??
Do I make a mistake?
Gr. Carly
Anthony Stark
Hi Carly,
I’ve run the formula with your parameters and got the same result as the calculator:
Vd = 5.866;
nr = 2;
pa = 0.981;
Ta = 31;
Ra = 286.9;
Ne = 8700;
ma = 0.0503;
roa = (pa*10^5)/(Ra*(Ta+273.15))=1.1242183
nv = (ma*nr)/(roa*Vd*1e-3*(1000/60))=91.528540
If I put the air density of 1 kg/m3 then I get around 103% VE but this is wrong because the air density needs to be recalculated function of pressure and temperature.
So the formulas and the calculator are outputting the right values.
Thomas
Can anyone explain how much power we can gain from: 2.0L car, volume efficiency 110%, compression ratio 12:1, fuel ratio 13:1? Pls qoute the power produced from above and let me know the loses as routine (whatever it is)
Also is there any formula to do for furure use?
G. Amba Prasad Rao
I found this website more useful right from basics to advanced systems explained.
Since much discussion is going on Alternate fuels and emissions.
It would be better if some calculations with respect to emissions from engines or vehicles is provided.
Javier Albarracin
Is there a way to theoretically calculate voleff? without knowing mass air flow? Mass air flow depends of Voleff and vice versa so this is useless if you wanted to simulate it.
Anthony Stark
It’s a bit tricky. You have a couple of options in order to determine the volumetric efficiency for simulation purposes.
1. Measure the intake air mass flow rate, or pressure and temperature, on a similar real engine and then calculate the volumetric efficiency.
2. If the engine is a new design, use computational fluid dynamics (CFD) simulations to get the air mass flow rate and then calculate the volumetric efficiency.
3. Use the volumetric efficiency data from a similar engine/simulation and try to calibrate it until you have a match in the exhaust output (p, T).
If the engine design is really new, you have no other option but CFD, unfortunately.
Alan
I have a 2011 kia forte koup ex 2.0 6 speed manual transmission car what is the volumetric efficiency for it
Thomas
Another question is, we know that perfect combustion for fuel like gasoline is having 14.7kg air against 1kg gasoline , on the other hand, a 2 L car @4000rpm will need 4000 litre of air per minute (2 x (4000/2) and 540/000 l ( about 600kg) per hour. So, a 2l car which consumes 4.5 L of gasoline, it requir around 67kg air, while we see air consumption is different, so engine runs lean, keep in mind that , our calculation is based perfect cylinder filling, while in reality does not happen, so runs more lean. If all these are correct, why tons of pollution we face everyday????pls help me to get out of this.Thanks.
Anthony Stark
When you say “lean” is just looking at the numbers. In reality, inside the cylinders, you don’t know what’s happening. There are a number of causes which determines incomplete combustion and hence air pollution. Have a look at this article:
https://x-engineer.org/automotive-engineering/internal-combustion-engines/performance/effects-of-vehicle-pollution-on-human-health/
It gives an overview of the factors which contribute to incomplete combustion.
Thomas
Thanks Anthony
air volume is important for what? Air volume is useful and important Per se or its stored oxygen? As we know, the more air, the more power, because more fuel can burn, so our intention of absorbing more air, is reaching more oxygen? Or no, the volume itself is required and can provide us power “indirectly” by its mass?
Thanks
Anthony Stark
Hi Thomas,
What is important is oxygen, not the actual air. The reason is that oxygen makes combustion possible. For example air contains also nitrogen, which is not needed because leads to the formation of nitrogen oxides (NOx). The more air you have in the cylinder, the more oxygen, the more fuel you can burn, the higher the pressure, the higher the torque, the higher the power.
Thomas
Good explanation.
But Have some questions which may not be sooo relevant, As we know 1 litre of gasoline has about 9kw of energy, now questions are:
1- as SI engines have 25~30% efficiency, it means what it gains is 1.8~2 kw of that 9kw?? Apart from losses.
2- how theoretically is possible to get 100% efficiency or better to say , gaining that 9kw!!
3- fulll burning means obtaining that 9kw??
Regards
Anthony Stark
Hi Thomas,
1. 25% efficiency means that if the theoretical power within the fuel is 100 kW, what you get at the crankshaft is 25 kW.
2. This is practical impossible since you’ll need to have 100% of the fuel burned and absolutely no heat transfer from the burning mixture to the engine block.
3. only if there is no heat transfer to the engine block.
Alok
it is a wonderful explanation- free from any jargon.
I wish to know if it is for any reason that equation 3 in denominator uses Vd instead of Vd+Vc as the air volume sucked in the engine.
jeremie
I agree with flavio, air density decrease with low air temperature, strange !
Anthony Stark
Air density increases with lower temperature. The reason is that the molecules have lower kinetic energy and they stay close to each other hence more molecules in the same volume => higher density.
wilson
thanks a lot for the simplicity.
Flávio Paoliello
We know that a higher air density is beneficial to volumetric efficiency of an engine, hence the use of intercooler. So I did not undersstand why air density appear in the denominetor of the volumetric eff formula. Thanks for explaining.
Anthony Stark
The article explains step by step the final equation. Air density appears in the denominator because it’s used to calculate the intake air volume, see equation (4).
Jordi
Thank god!
Tim Edes
Excellent explanation…thank you !
Ramniwas patidar
Very good sysmatic explanation in simple way. Thanks
Fajrul
How to know the value of volumetric efficiency theoretically depends on engine speed (rpm)?