# Thermodynamics Otto Cycle engine

• Gordon Arnaut
In summary, the temperature of combustion in an Otto cycle engine is much lower than the temperature mentioned in the mathematical models.
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.

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

AM

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:
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.

Gordon Arnaut said:
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.
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

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.

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.

Gordon Arnaut said:
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.
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

## 1. What is a thermodynamics Otto Cycle engine?

A thermodynamics Otto Cycle engine is a type of internal combustion engine that uses a four-stroke cycle to convert heat energy into mechanical work. It is commonly used in automobiles and other vehicles.

## 2. How does a thermodynamics Otto Cycle engine work?

The engine operates by intaking a fuel-air mixture into the combustion chamber, compressing it, igniting it, and then releasing the exhaust gases. This process repeats in a constant cycle, with the energy from the combustion of the fuel-air mixture being converted into mechanical work to power the vehicle.

## 3. What is the efficiency of a thermodynamics Otto Cycle engine?

The efficiency of an Otto Cycle engine is typically around 20-30%, meaning that only a small percentage of the energy from the fuel is actually converted into mechanical work. This is due to factors such as heat loss and friction within the engine.

## 4. What are the advantages of a thermodynamics Otto Cycle engine?

The Otto Cycle engine is relatively simple and efficient, making it a popular choice for vehicles. It also has a higher power-to-weight ratio compared to other types of engines, making it well-suited for use in smaller vehicles.

## 5. What are the limitations of a thermodynamics Otto Cycle engine?

One limitation of the Otto Cycle engine is its reliance on fossil fuels, which contribute to pollution and climate change. It is also not as efficient as other types of engines, such as diesel engines, which can achieve higher fuel efficiency. Additionally, the four-stroke cycle can only operate at a limited range of speeds, which can result in decreased efficiency at lower speeds.

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