# Homework Help: Thermodynamics Otto Cycle engine

1. Apr 6, 2007

### Gordon Arnaut

I'm trying to estimate the temperature of combustion in an Otto Cycle engine, using a formula based on the First Law:

T4 = T3 + fQ /cv

Where T4 is the combustion temperature, T3 is the air temperature after compression, f is the fuel/air ratio, Q is the heat energy of the fuel, and cv is the specific heat at constant volume.

Assuming T3 is 750 degrees Kelvin, the fuel is gasoline which has a heat content of 43,500 kj/kg, the fuel air ratio is 0.07 (close to stoichiometric), and cv at this temperature is 0.8 kj-kg/K, I get a very high number: 4556 K.

I think the temperature should be much lower. Is there another approach to this problem?

Regards,

Gordon.

2. Apr 7, 2007

### Andrew Mason

Show us how you worked it out. What are you using for Cv of the post-combustion gases?

AM

3. Apr 7, 2007

### lpfr

At 4500 K an important fraction of molecules dissociate. As an example, at 2400 K, 5% of water molecules are dissociated in O2 and H2. This means that a significant fraction of the combustion heat is used to dissociate the molecules. Then you formula is simply not applicable. You do not know how much heat is used to heat the gas and how much to maintain the molecules dissociated.
Of course, as temperature lowers, molecules recombine and give back the "missing" heat.

Last edited: Apr 7, 2007
4. Apr 7, 2007

### Gordon Arnaut

Andrew, it's just simple artihmetic using hte values I provided.

We don't know the cv post-combustion because we don't know the temperature.

Regards,

Gordon.

5. Apr 7, 2007

### Andrew Mason

Your answer may not be all that wrong. Gasoline puts out a lot of energy.

The Cv for air would be 800 J/Kg K but for H2O it is about 2000 - 3000 J/Kg K. H2O would form part of the post-combustion gases and it is the Cv of the post-combustion gases that you have to use.

Also, in an internal combustion engine, the piston starts moving down before combustion is complete so there is a cooling due to expansion.

AM

6. Apr 14, 2007

### Gordon Arnaut

Thanks, Andrew.

You're right. The answer I got is actually correct.

What threw me was that in Otto cycle mathematical models, the temperature of combustion mentioned is always a lot lower.

However, if you take the high temperature number I calculated above and multiuply it by the thermal efficiency of the engine which is about 0.3, you get a combustion temperature of about 1500 K, which is inline with the mathematical models.

Regards,

Gordon.

7. Apr 15, 2007

### lpfr

To multiply the temperature by the efficiency is just a devoid of sense as to multiply by $$\sqrt{10}$$ or $$\log_{10}2$$ or divide by $$\pi$$.
But if it is all you need to be happy, it is OK.

But, if you are not absolutely happy, I suggest that you post the same question in a chemistry forum. You'll be surprised on how much the chemists know about combustion.
P.S. I'm physicist not chemist.

8. Apr 15, 2007

### Andrew Mason

That sounds like the temperature of the exhaust, not the temperature of combustion.

The temperature at the bottom of the stroke (exhaust) is [itex]T_{bottom} = T_{top}r^{1-\gamma}[/tex] where r is the compression ratio. So for a high compression ratio engine, (max. r = 13) the temperature of the exhaust is:

$$T_{bottom} = 4500*13^{-.4} = 1613 K.$$

As I said before, in a real Otto cycle, the fuel does not burn instantaneously. Some burns during the downstroke. So the temperature never reaches the theoretical top temperature.

AM