# Diesel Cycle - Heat Engines - Thermo.

1. Feb 17, 2005

Hello,

If anybody could answer this question for me that would be great. I have been wondering about it all morning. In the Diesel cycle for heat engines, on the second stage of operation, the Volume increases while pressure stays the same. I understand how this can be accomplished by raising the temperature, but is it done by the injection of highly compressed gas in the piston at the top (combustion position). I understand the Otto cycle and how it raises the pressure while volume is constant but I don’t see how this can be accomplished the other way around. Are my assumptions correct? If somebody could fill me in on this information or give me a good site to look it up on, that would be great.

Regards,

2. Feb 18, 2005

### minger

I wondered this myself throughout Thermo I and II. It wasn't until earlier this semester in the class in taking now (Internal Combustion Engines) that our prof "tried" to explain this assumption. From what I understand, it is due to the fact that the pressure from the gas coming in is at such high pressure that it does work on the piston. You obviously need a change in a specific volume for work to be done, so it is constant pressure with work being done as opposed to constant volume.

When the piston is at TDC, there is a certain amount of material in the combustion chamber. When you inject the fuel into the cylinder, you are essentially adding material, and thus increasing the volume (I guess you say you increase the volume because the pressure is already so high, compression ratios ~18), and since the pressure is so high, it stays constant.

This is the best I got for an explanation. This is kind of what would say based on what my prof has said and what my just logical interpretations of the cycle are.

3. Feb 18, 2005

### Clausius2

I assume that Minger and you are talking about the combustion stage, where pressure remains constant and volume increases, aren't you?.

Well, whereas the Otto cycle represents better the behavior of gasoline engines, the Diesel cycle is representative of Diesel combustion type (obvious, isn't it?). The reason for the pressure remaining the same is due to the proper combustion mechanism, although I encourage you to know that in the real world, all engines (diesel or gasoline) are a mix of Otto+Diesel engines. The cycle you are referring to, is an "ideal" representation of a real Diesel engine, where there is a stage of pressure rising with the volume remaining constant at the TDC, and a stage of volume increasing with pressure constant, and it has to do with the two main phases of combustion inside a diesel engine: the premixed and the diffusion combustion. In your cycle, the premixed combustion has been deleted, and only the diffusion stage is represented.

The diffusion stage of combustion is characterized by a deflagrative flame. Once the fuel is injected at the top dead center, the pressure is so high that fuel and air burn like a premixed flame in a detonation mode (auto-ignition). That phase is represented by the segment of volume constant and pressure rising which I have said it has been erased in your "ideal" cycle. Once this occurs, the rest of the fuel burns in a diffusion flame mode, sustained by the former ignition. You should know that compressible effects in deflagrative flames are usually neglected, because the flame velocity is much smaller than the sound velocity. So that, as the flame advances inside the combustion chamber, there is no effect on the pressure. Therefore, in this stage the pressure remains constant and volume increases due to the crank movement.

As I have said before, real engines are a compound between pure Diesel and pure Otto engines.

4. Feb 22, 2005

that answeres my question. Thanks Clausius2.

5. Feb 22, 2005

### Andrew Mason

The only way that I can see a volume of burning air/fuel mixture remaining at constant pressure is for the heat being added by the burning fuel to occur gradually over the whole downstroke so that it can be converted entirely to work (which would also seem to be thermodynamically impossible).

So, I can barely understand how one could have constant pressure during ignition if the volume is rapidly expanding, but according to this animation:

http://www.rawbw.com/~xmwang/myGUI/DieselG.html

the burning of fuel occurs at the top of the stroke and the volume changes very little when the fuel ignites.

Are you suggesting that the burning is more gradual and takes place uniformly over the entire downstroke?

AM

6. Feb 23, 2005

### Clausius2

Hi Andrew, by the way the first thing I have to say to you is you're doing a great work at Homework section, congratulations.

Open the figure attached. That's a SEMI-REAL diesel engine, proposed by J. Heywood.

In fact, the animation is a bit rough and it doesn't help too much. The combustion process in a Diesel Engine is started by the ignition of fuel some grades before the TDC. Then there is an ignition delay till the detonation occurs. This detonation is a premixed combustion which provokes a high peak of pressure in a short volume variation. Usually, this stage is not represented in non advanced internal combustion engines books. This process would be modelled by the segment a-b. In this stage, the process of combustion is so rapid than the crank shaft roughly moves.

On the other hand, the combustion keeps on by means of that ignition, which promotes the propagation of diffusion flames. The diffusion stage is a SLOW speed combustion stage. So that, while the crank returns to the BDC, there is two counteracting effects:

-An increasing of pressure in the combustion chamber. Although we know that a deflagration doesn't provoke much change in the pressure COMPARED with the pressure itself, due to the closed environment and heating the pressure will raise (not as larger as in the detonation stage).

-An increasing of volume and so a decreasing in pressure because of gas expansion.

If we apply a (simplified) Energy equation to the combustion chamber, we will come to:

$$\frac{dP}{d\alpha}=\frac{\gamma-1}{V}\frac{dQ_{released}}{d\alpha}-\frac{\gamma P}{V}\frac{dV}{d\alpha}$$ where alpha is the crank angle.

The two effects (expansion and heat release are being counterated each other as the expansion stroke runs). You could check that the MAXIMUM (dP=0) pressure is reached in some point of the expansion stroke.

Don't take my words literally and watch to the animation, because both things doesn't match. Also, the figure I posted is a simplification. The real cycle is something smoothed around the lines I have drawed. But the physics of the event is what I have just said you.

#### Attached Files:

• ###### Stroke I.bmp
File size:
27.9 KB
Views:
131
Last edited: Feb 23, 2005
7. Feb 23, 2005

### Andrew Mason

Clausius2:

Thanks very much. Your drawing is very helpful. I take it from your explanation that ignition occurs in a to b and continues from b to c. I can see how the burning of some of the fuel during part of the downstroke will maintain a relatively constant pressure for a while. It is the 'a to b' part that is missing from most PV diagrams of the diesel cycle.

AM

8. Feb 23, 2005

### Clausius2

Yes it is. You're right.

The fact is that drawing represents most of the nowadays engines (either diesel or spark ignition). I remember me when taking the classic thermodynamic classes of cycles, and now I realise that real cycles are not as an "ideal" as it seemed to be.

Regards.