Energy transformations in an IC engine cylinder

[vcsharp2003]

Homework Statement

What are the exact transformations occurring in a cylinder of an automobile car that uses internal combustion engine?

Relevant Equations

None

A drop of fuel is ignited in an engine cylinder, that produces heat, light and sound energies from the chemical energy stored in the drop of oil.
What I am not clear about is how heat energy gets transformed into mechanical work? I think the heat energy produced from ignition flows from burnt fuel to the surrounding air inside the cylinder, which causes the air temperature to rise; as the temperature rises, the molecules in air start moving with higher kinetic energy, and these high kinetic energy molecules push the piston of the cylinder to produce mechanical work. Is this correct?

 

[hmmm27]

Pretty much. There's a slight increase in the number of gaseous molecules in the chamber, which also contributes to the pressure.

 

[Lnewqban]

 

[Chestermiller]

Homework Statement:: What are the exact transformations occurring in a cylinder of an automobile car that uses internal combustion engine?
Relevant Equations:: None

A drop of fuel is ignited in an engine cylinder, that produces heat, light and sound energies from the chemical energy stored in the drop of oil.
What I am not clear about is how heat energy gets transformed into mechanical work? I think the heat energy produced from ignition flows from burnt fuel to the surrounding air inside the cylinder, which causes the air temperature to rise; as the temperature rises, the molecules in air start moving with higher kinetic energy, and these high kinetic energy molecules push the piston of the cylinder to produce mechanical work. Is this correct?

There is no heat involved. In the framework of thermodynamics, heat would involve flow of thermal energy across the boundary of the cylinder to the engine block.

What happens thermodynamically is that the energy released by the chemical reaction goes into raising the temperature of the combustion gases. For an ideal gas mixture, this would cause the pressure of the gas to substantially rise (at constant cylinder volume). The increased gas pressure forces the piston to move and drive the crankshaft.

 

[vcsharp2003]

There is no heat involved. In the framework of thermodynamics, heat would involve flow of thermal energy across the boundary of the cylinder to the engine block.

Heat is not flowing into or out of the cylinder i.e. it's an adiabatic process, but won't heat energy be generated as a result of the drop of fuel burning?

 

[jack action]

There is no heat involved. In the framework of thermodynamics, heat would involve flow of thermal energy across the boundary of the cylinder to the engine block.

So there is no heat involved in the Otto or Diesel cycles?

the heat energy produced from ignition flows from burnt fuel to the surrounding air inside the cylinder,

The heat is produced by a combination of the fuel AND the air inside the cylinder. For the combustion to happen, the fuel must be well mixed with the air, and each molecule set heats up locally as it ignites.

as the temperature rises, the molecules in air start moving with higher kinetic energy, and these high kinetic energy molecules push the piston of the cylinder to produce mechanical work.

Yes.

 

[Chestermiller]

So there is no heat involved in the Otto or Diesel cycles?

Not during the ignition and power stroke, which is what the OP is referring to.

The heat is produced by a combination of the fuel AND the air inside the cylinder. For the combustion to happen, the fuel must be well mixed with the air, and each molecule set heats up locally as it ignites.

Yes.

This energy does not enter as heat through the walls of the cylinder. It is released by breaking old- and making new chemical bonds.

 

[Chestermiller]

Heat is not flowing into or out of the cylinder i.e. it's an adiabatic process, but won't heat energy be generated as a result of the drop of fuel burning?

This is not heat energy in thermodynamic parlance. If we are talking about a thermodynamic process, we should use proper thermodynamics terminology. It is internal energy change as a result of making and breaking chemical bonds.

 

[Lnewqban]

For gasoline, the air/fuel ratio is about 15; therefore, the liquid to gas expansion of the burning fuel also contributes to the pressure increase inside the chamber.

Please, see:
https://en.m.wikipedia.org/wiki/Air–fuel_ratio#Engine_management_systems

In real life, there are several factors that affect combustion and resulting peak pressure:
* Compression ratio is 9.0 to 11.0 for regular gasoline engines, the higher the ratio the better regarding efficiency.
Turbo charged engines deliver more power, but need special fuel.
Please, see:
https://en.m.wikipedia.org/wiki/Compression_ratio

* Ignition and valves timing: Because the piston can move very fast, the time at which the flame and peak pressure reach the top of the piston is very important. Because combustion gases of previous cycle remain in the cylinder and get mixed with fresh mix, proper ventilation is key, which is controlled by valve’s timing.
Please, see:
https://en.m.wikipedia.org/wiki/Ignition_timing

https://en.m.wikipedia.org/wiki/Valve_timing

* Not all the generated heat can be used for gas expansion, since the metals of the engine’s parts (especially exhaust valves) can only tolerate so much increase of their own temperature to keep dimensions and good lubrication. Hence, the cooling system needed.
Please, see:
https://en.m.wikipedia.org/wiki/Internal_combustion_engine_cooling

:)

 

[jack action]

This energy does not enter as heat through the walls of the cylinder. It is released by breaking old- and making new chemical bonds.

In every Otto cycle analysis I've seen, it is always referring to a "heat transfer" and represented with a $Q$.

And I have never heard anyone stating that the Otto cycle doesn't apply to an internal combustion engine because there is no heat transfer with the exterior.

 

[Chestermiller]

In every Otto cycle analysis I've seen, it is always referring to a "heat transfer" and represented with a $Q$.

And I have never heard anyone stating that the Otto cycle doesn't apply to an internal combustion engine because there is no heat transfer with the exterior.

Even during the ignition and power stroke? If it is spoken of that way, it is playing it fast and loose with the proper terminology. Basically what it boils down to is that the net effect of internal release of chemical energy within the gas is the same as if an equal amount of heat energy were transferred to the working fluid through the wall of the cylinder.

 

[vcsharp2003]

This is not heat energy in thermodynamic parlance.

I was under the impression that IC engine was an example of a heat engine as mentioned in Second Law of Thermodynamics. But from your explanation, it seems that it's not an example of a heat engine since heat energy is not being converted to work. Heat engine, by definition, is something that converts heat energy to work ( in a partial manner).

 

[Chestermiller]

I was under the impression that IC engine was an example of a heat engine as mentioned in Second Law of Thermodynamics. But from your explanation, it seems that it's not an example of a heat engine since heat energy is not being converted to work. Heat engine, by definition, is something that converts heat energy to work ( in a partial manner).

Well, the working fluid certainly doesn't pass through a closed cycle in the sense that the final state is identical to the initial state. So it certainly isn't a heat engine in the standard sense.

 

[vcsharp2003]

Well, the working fluid certainly doesn't pass through a closed cycle in the sense that the final state is identical to the initial state. So it certainly isn't a heat engine in the standard sense.

What would an example of a heat engine that we use nowadays?

 

[Chestermiller]

What are your thoughts on this?

 

[256bits]

I was under the impression that IC engine was an example of a heat engine as mentioned in Second Law of Thermodynamics. But from your explanation, it seems that it's not an example of a heat engine since heat energy is not being converted to work. Heat engine, by definition, is something that converts heat energy to work ( in a partial manner).

Why would it not be an example of a heat engine.

The heat is produced internally by the combustion process, an exothermal reaction, and transferred to the gases raising their temperature.
With external combustion engines, the heat is transferred to the working fluid through a barrier. The working fluid then goes through its processes producing work and rejecting heat. Closed examples would be the Carnot and the Stirling engines.
A steam engine rejects the heat by rejecting the working fluid. The next cycle comprises a new mass of working fluid. As does most of the other heat engines such as that used for turning the turbine in a jet engine. The working fluid does not return to its starting point with the confines of a particular volume.

 

[Chestermiller]

Why would it not be an example of a heat engine.

The heat is produced internally by the combustion process, an exothermal reaction, and transferred to the gases raising their temperature.
With external combustion engines, the heat is transferred to the working fluid through a barrier. The working fluid then goes through its processes producing work and rejecting heat. Closed examples would be the Carnot and the Stirling engines.
A steam engine rejects the heat by rejecting the working fluid. The next cycle comprises a new mass of working fluid. As does most of the other heat engines such as that used for turning the turbine in a jet engine. The working fluid does not return to its starting point with the confines of a particular volume.

Look guys, all I'm saying is that, in thermodynamics, energy produced internally by the combustion process, an exothermic reaction, raising the temperature of the gases is not considered "heat;" it is considered internal energy change due to reaction $\Delta U_R$. The term "heat" is reserved for exchange of thermal energy with the surroundings, which is not what is happening here.

Now, if you want to call it heat, knock yourself out. But it is not proper thermodynamic terminology.

 

[256bits]

Now, if you want to call it heat, knock yourself out. But it is not proper thermodynamic terminology.

There is no disagreement with what you have said. There is no Q crossing the boundary.
The work comes from the change in internal energy.

For anyone looking at it on sites on the internet, ( I didn't come across an exception ) they will find a heat engine pretty much being described as having a cold reservoir and a hot reservoir, As you have described, the Otto cycle does not have a hot reservoir with which to transfer heat, Yet, the process is stated as having a heat addition in an isochoric process ( or isobaric case may be or Otto or Diesel ), with ΔU = Q - W.
All these sites seems to be lacking the proper treatment in that regard.

 

[vcsharp2003]

What are your thoughts on this?

I came across the following at
https://en.m.wikipedia.org/wiki/Hea...s of heat engines,work as the desired product.
"Everyday examples of heat engines include the thermal power station, internal combustion engine, firearms, refrigerators and heat pumps. Power stations are examples of heat engines run in a forward direction in which heat flows from a hot reservoir and flows into a cool reservoir to produce work as the desired product. Refrigerators, air conditioners and heat pumps are examples of heat engines that are run in reverse, i.e. they use work to take heat energy at a low temperature and raise its temperature in a more efficient way than the simple conversion of work into heat (either through friction or electrical resistance)"

 

[vcsharp2003]

Well, the working fluid certainly doesn't pass through a closed cycle in the sense that the final state is identical to the initial state. So it certainly isn't a heat engine in the standard sense.

Below is the diagram that I came across in the Physics textbook by Sears & Zemansky. Isn't the IC engine going through a cyclic process according to this diagram as it returns to its original pressure-volume state at point a?

 

[Chestermiller]

Below is the diagram that I came across in the Physics textbook by Sears & Zemansky. Isn't the IC engine going through a cyclic process according to this diagram as it returns to its original pressure-volume state at point a?

View attachment 290043

Let's test your understanding of a some thermodynamics fundamentals.

True of false:

1. In a thermodynamic cycle, the thermodynamic state of the working fluid is the same at the beginning and end of thermodynamic cycle.

2. In a thermodynamic cycle, the internal energy of the working fluid is the same at the beginning and end of the cycle.

3. At point b in the diagram, the thermodynamic state of the gas mixture before step bc is the same as the thermodynamic state of the gas mixture after step ab.

4. In step bc of the Otto cycle, heat $Q_H$ is actually transferred across the boundary of the cylinder from the engine block to the gas mixture.

5. In thermodynamics, the heat Q in the first law equation represents both heat in transit across the boundary of the system and heat generated by chemical reaction within the system.

 

[Lnewqban]

Isn't the IC engine going through a cyclic process according to this diagram as it returns to its original pressure-volume state at point a?

Please, see:
https://www.nuclear-power.com/nucle...ycle-otto-engine/actual-and-ideal-otto-cycle/

https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node26.html

You could imagine that the ideal cooling process in actuality occurs in the atmosphere, assuming it as part of the engine.

 

[Chestermiller]

There is no disagreement with what you have said. There is no Q crossing the boundary.
The work comes from the change in internal energy.

For anyone looking at it on sites on the internet, ( I didn't come across an exception ) they will find a heat engine pretty much being described as having a cold reservoir and a hot reservoir, As you have described, the Otto cycle does not have a hot reservoir with which to transfer heat, Yet, the process is stated as having a heat addition in an isochoric process ( or isobaric case may be or Otto or Diesel ), with ΔU = Q - W.
All these sites seems to be lacking the proper treatment in that regard.

Yes, for the initial Isochoric temperature rise step, both Q and W are zero. So we have for that step:
$$\Delta U=Q-W=0$$with, per unit mass, $$\Delta U=\Delta U_R+C_v\Delta T=0$$where $\Delta U_R$ is the change in internal energy per unit mass due to chemical reaction and Cv is the heat capacity per unit mass of products. So the temperature rise in this step is $$\Delta T=\frac{(-\Delta U_R)}{C_v}$$which is positive for an exothermic reaction.

 

[jack action]

@Chestermiller :

Nobody is saying you are technically wrong, but it's not only what is found on the Internet, it's also written in every book and taught at every University in every thermodynamics class:

An internal combustion engine (ICE) is a heat engine in which the combustion of a fuel occurs with an oxidizer (usually air) in a combustion chamber that is an integral part of the working fluid flow circuit.

The Otto, Diesel, or Brayton cycles don't specify that the engine must have an internal combustion (heck, it doesn't even assume a combustion). In theory, you can build an engine based on these cycles with an external combustion.

Imagine we build 3 imaginary engines, build on the same cycle, with the same thermodynamics specs:

1. One works with an external combustion, transfers heat through the cylinder;
2. One has a combustible mixture injected into the combustion chamber that stays near the walls and heats up air in the middle. There is an imaginary boundary between the combustible mixture and the air because the combustible mixture never mixes with the air;
3. One has the same combustible mixture mixed in with the air (pure internal combustion as we know it).

What you are saying is that engines 1 and 2 are heat engines and the third one isn't. But I never heard of an "internal energy" engine. And the end results are the same for every engine, impossible to differentiate thermodynamically speaking. So if engine 3 is not a heat engine, what is it? And I would appreciate a source. (Let me also recall you that we are in the Introductory Physics Homework Help section)

 

[vcsharp2003]

Let's test your understanding of a some thermodynamics fundamentals.

True of false:

1. In a thermodynamic cycle, the thermodynamic state of the working fluid is the same at the beginning and end of thermodynamic cycle.

2. In a thermodynamic cycle, the internal energy of the working fluid is the same at the beginning and end of the cycle.

3. At point b in the diagram, the thermodynamic state of the gas mixture before step bc is the same as the thermodynamic state of the gas mixture after step ab.

4. In step bc of the Otto cycle, heat $Q_H$ is actually transferred across the boundary of the cylinder from the engine block to the gas mixture.

5. In thermodynamics, the heat Q in the first law equation represents both heat in transit across the boundary of the system and heat generated by chemical reaction within the system.

1. True
2. Not sure, but I do know that state variables P,V,T return to their original values
3. True (P,V values at b are same whether we consider ab or bc)
4. False
5. True ( Q is heat added which is the sum of all heat energies involved including inflow, outflow)

 

[bob012345]

Heat is not flowing into or out of the cylinder i.e. it's an adiabatic process, but won't heat energy be generated as a result of the drop of fuel burning?

Of course. You are using the word heat in a common sense way but some answers above are using specific technical terms that use different words like bookeeping terms depending on what is happening.

 

[vcsharp2003]

Call it merely an internal energy change but the reaction products contain a great amount of energy in the form of heat and do a lot of work. It amounts to a release of internal chemical energy as heat. And most of it does pass as heat out of the cylinder into the engine block to dissipate anyway.

Could it be that the ignited fuel becomes a heat reservoir as soon as it's ignited, which then causes the temperature of entire volume of gases in the cylinder to rise? So, effectively heat flows from ignited fuel i.e. heat reservoir to surrounding gases within the same cylinder.

 

[Chestermiller]

1. True
2. Not sure, but I do know that state variables P,V,T return to their original values

So are you saying that the initial state before combustion is the same as the final state after combustion or not? Are you aware that internal energy of a gas mixture depends not only on P, V, and T, but also the chemical composition of the mixture, and if that changes due to chemical reaction, U changes?

3. True (P,V values at b are same whether we consider ab or bc)

Wrong. What you are saying is that there is no difference between a mixture of gasoline and air before combustion and the corresponding mixture of combustion products after combustion. The internal energies, entropies, and enthalpies of these mixtures, even at the same P and V are different.

4. False

Correct.

5. True ( Q is heat added which is the sum of all heat energies involved including inflow, outflow)

Incorrect. In thermodynamics, the heat Q in the first law equation represents only heat in transit across the boundary of the system. Energy released by chemical reaction is not included in Q.

Fundamentals of Engineering Thermodynamics, Moran et al: "The symbol Q denotes an amount of energy transferred across the boundary of a system in a heat interaction with the system's surroundings."

 

[Chestermiller]

@Chestermiller :

Nobody is saying you are technically wrong, but it's not only what is found on the Internet, it's also written in every book and taught at every University in every thermodynamics class:

I would never teach it that way, especially if the student is supposed to be learning the subject correctly.

The Otto, Diesel, or Brayton cycles don't specify that the engine must have an internal combustion (heck, it doesn't even assume a combustion). In theory, you can build an engine based on these cycles with an external combustion.

Imagine we build 3 imaginary engines, build on the same cycle, with the same thermodynamics specs:

1. One works with an external combustion, transfers heat through the cylinder;
2. One has a combustible mixture injected into the combustion chamber that stays near the walls and heats up air in the middle. There is an imaginary boundary between the combustible mixture and the air because the combustible mixture never mixes with the air;
3. One has the same combustible mixture mixed in with the air (pure internal combustion as we know it).

What you are saying is that engines 1 and 2 are heat engines and the third one isn't.

Maybe I went too far when I said the the IC engine is not a heat engine. What I meant was that it is not a standard heat engine operating in a closed cycle. Moran et al, Fundamental of Engineering Thermodynamics discuss "standard-air" Otto and diesel cycles, in which the gas is inert, and in which heat actually is transferred to the working fluid in the Isochoric temperature rise step. However, this differs from an IC engine "cycle" where no heat exchange with the surroundings occurs in this step.

But I never heard of an "internal energy" engine. And the end results are the same for every engine, impossible to differentiate thermodynamically speaking. So if engine 3 is not a heat engine, what is it? And I would appreciate a source. (Let me also recall you that we are in the Introductory Physics Homework Help section)

Even though the end results are the same, I disagree with the philosophy of dumbing down course material to make it more palatable to the novice student, even if it is "a little incorrect fundamentally (with the intention of later correcting it as the student covers more advanced material). This always causes confusion for the student (as, for example, in the situation where heat capacity is defined in terms of heat rather than in terms of internal energy and enthalpy).

 

[Chestermiller]

Call it merely an internal energy change but the reaction products contain a great amount of energy in the form of heat and do a lot of work. It amounts to a release of internal chemical energy as heat. And most of it does pass as heat out of the cylinder into the engine block to dissipate anyway.

It does not release internal chemical energy as heat. The release of internal chemical energy goes directly into raising the temperature with no intermediate amount of heat leaving or entering the gas. And I'm talking only about the combustion step in the process. Of course, later in the cycle, actual heat does pass out of the cylinder, in transit across the cylinder boundary.

 

[Chestermiller]

Could it be that the ignited fuel becomes a heat reservoir as soon as it's ignited, which then causes the temperature of entire volume of gases in the cylinder to rise? So, effectively heat flows from ignited fuel i.e. heat reservoir to surrounding gases within the same cylinder.

Your system is the entirety of the gas, and no heat flows into or out of this system from the surroundings (the engine block) during the combustion step in the process.