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).

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 *V _{t} [m^{3}]* is calculated function of the total number of cylinders

*n*and the volume of one cylinder

_{c}[-]*V*.

_{cyl}[m^{3}]The total volume of the cylinder is the sum between the displaced (swept) volume *V _{d} [m^{3}]* and the clearance volume

*V*.

_{c}[m^{3}]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.

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** *η _{v} [-]* is defined as the ratio between the actual (measured) volume of intake air

*V*drawn into the cylinder/engine and the theoretical volume of the engine/cylinder V

_{a}[m^{3}]_{d}[m

^{3}], 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 *m _{a} [kg]* and air density

*ρ*:

_{a}[kg/m^{3}]Replacing (4) in (3) gives the volumetric efficiency equal to:

\[\eta_v = \frac{m_a}{\rho_a \cdot V_d} \tag{5}\]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:

*N _{e} [rot/s]* – engine speed

*n*– number of crankshaft rotations for a complete engine cycle (for 4-stroke engine

_{r}[-]*n*)

_{r}= 2From equation (6), we can write the intake air mass as:

\[m_a=\frac{\dot{m}_a \cdot n_r}{N_e} \tag{7}\]Replacing (7) in (5) gives the volumetric efficiency equal with:

\[\bbox[#FFFF9D]{\eta_v = \frac{\dot{m}_a \cdot n_r}{\rho_a \cdot V_d \cdot N_e}} \tag{8}\]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 *p _{a} [Pa]* and temperature

*T*are measured in the intake manifold, the

_{a}[K]**intake air density**can be calculated as:

where:

*ρ _{a} [kg/m^{3}]* – intake air density

*p*– intake air pressure

_{a}[Pa]*T*– intake air temperature

_{a}[K]*R*– gas constant for dry air (equal to

_{a}[J/kgK]*286.9 J/kgK*)

### Example – How to calculate the volumetric efficiency

Let’s consider a compression ignition (diesel) engine with the following parameters:

\[ \begin{split}V_d &= 3.8 \text{ L} \\

n_r &= 2 \\

p_a &= 1.5 \text{ bar} \\

T_a &= 40 \text{ } ^\circ \text{C} \\

R_a &= 286.9 \text{ J/kgK} \\

N_e &= 1000 \text{ rpm} \\

\dot{m}_a &= 0.0375 \text{ kg/s}

\end{split} \]

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 *m ^{3}* and the engine speed from

*rpm*to

*rps*.

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 calculator below.

### Volumetric Efficiency Calculator

V_{d} [L] | n_{r} [-] | p_{a} [bar] | T_{a} [˚C] | R_{a} [J/kgK] | N_{e} [rpm] | m_{a} [kg/s] |

Air density, ρ_{a} [kg/m^{3}] = | ||||||

Volumetric efficiency, η_{v} [%] = |

For any questions or observations regarding this tutorial please use the comment form below.

Don’t forget to Like, Share and Subscribe!

## 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)?