# How can I estimate the rate of consumption of oxygen in a car engine?

I need some approximate figures for how many litres of oxygen per minute a typical car engine requires. How would I estimate this?

• How is just doing the math not the obvious and simple answer? – Olin Lathrop Jun 21 '16 at 11:09
• @OlinLathrop you're assuming the OP has knowledge of chemical reactions, chemical equilibrium equations, and how to apply that knowledge to a real world system. While these topics may be obvious and simple to many engineers and car/engine enthusiasts they certainly aren't to most laypeople. – BarbalatsDilemma Jun 21 '16 at 13:54
• @Barb: That would then be the first question to ask. – Olin Lathrop Jun 21 '16 at 14:28
• @Olin: Good point, hopefully this question will inspire more questions on the site. I was trying to say that sometimes people are unaware of how much there is to learn about a topic, so it may take a few questions to figure out what the 'right' questions are. – BarbalatsDilemma Jun 21 '16 at 15:57

Do the math.

Pick a reference car and get its fuel milage at some speed. From that you can compute the amount of gasoline used in one minute. Now look at the chemistry of gasoline and figure out how much oxygen your one minute worth of gasoline requires to completely burn.

There are more details going on in a internal combustion engine, like nitrogen reacting with oxygen, not all carbon fully reacting with oxygen, etc. However, the above should be a reasonable starting point.

## Example

My car gets 50 miles/gallon at 55 miles/hour (it's a Honda Civic Hybrid, actually gets a bit more than that on average). That's 1.1 gallons/hour = 0.0183 gal/minute = 69.4 ml/min. The density of gasoline is about 750 g/l, so we have about 52 g of gasoline used every minute.

We'll approximate gasoline as being pure octane, which is C8H18. Since carbon has a atomic weight of 12 and hydrogen of 1, that of octane is 114.

In ideal combustion, each carbon atom combines with two oxygen, and every two hydrogen with one oxygen. One octane molecule will therefore combine with 25 oxygen atoms. The atomic weight of oxygen is 16, so the total atomic weight of the oxygen required for one octane molecule is 400.

This means that 400 units of oxygen will be used for each 114 units of octane by mass. That means our 52 g of gasoline require 182 g of oxygen.

The question now comes down to what the volume of 182 g of oxygen is. If you just wanted to know the quantity of oxygen, then mass is more useful anyway. I'll therefore determine the volume of typical "air" that is required to get that much oxygen.

Oxygen is diatomic, so has a molecular weight of 32 in air. That means our 182 g of oxygen is 5.7 moles. Let's say air is 21% oxygen, so that means we need 27.2 moles of air. At 20 °C and 1 atm, a mole of ideal gas has a volume of 24 liters. Therefore, the final answer within all the specific conditions picked above is that the engine will suck in about 650 liters of air per minute. Since some of these conditions can vary quite widely, and the fuel efficiency used is well off to the efficient end for current cars, I'll round up and say it takes 1 cubic meter of air per minute.

Note that there wasn't any engineering here. This was just plugging in numbers that are all easily looked up.

• Olin is this 650 per hour or minute? thanks! – ergon Jun 21 '16 at 19:11
• @ergo: Read the whole answer. This was clearly stated. Nonetheless, I added more of a summary to the end. – Olin Lathrop Jun 21 '16 at 20:01

You need to do a "mass balance". Here's roughly how:

How many litres of fuel per hour does the car use?

Work out the mass of this volume of fuel (I assume we can look up the density somewhere.)

From this, and assuming the fuel is gasoline C8H18. Convert the known mass to moles. (This is the tricky step that a chemist would know.)

From the equation of combustion of gasoline, work out how many moles of O2 you need. Then convert that to a mass of oxygen.

Air happens to be 79% nitrogen and 21% oxygen by volume and weighs a little over a kilogram per cubic metre (we'd have to look it up). So work out how many kilos of air you need and then its volume in cubic metres.

(I haven't done this in over twenty years so I'll leave you to work out how to find the mass and volume of air from the oxygen mass. It'll require juggling a few numbers but it shouldn't be too hard.)

In the above we're assuming the engine mix uses the ideal ("stoichiometric") quantity of air and there's full combustion with no side reactions like carbon monoxide, oxides of nitrogen etc. So it's an approximate figure.