# Calculating heat of compression?

1. May 25, 2010

### ISX

I am very interested in how hot a diesels compression chamber gets right before it shoots the fuel in and ignites. I want to know a formula that I can use to get the peak temperature. I know there are several factors like time and piston/engine heat absorbing, but if the engine is 200F, I think we can use that for something? I am not too good at this stuff but want to learn how to figure this. Let me know what other factors you need to know. Compression ratio is 17.0:1. Bore is 4.02" stroke is 4.72". Thanks for any help you can provide.

2. May 25, 2010

### Gordianus

The Diesel compression cycle is usually considered to be adiabatic (i.e. no heat exchange).
Clearly this is a bit too much but it gives an approximate answer.
Having said this, you can use the following relationship that links the initial (i) and final (f) states of such compression:
Pi*Vi^K=Pf*Vf^K

Where K=1.4 for air. You can assume Pi= 1 atmosphere and the validity of the ideal gas equation

3. May 26, 2010

### ISX

Care to simplify that down a little? I'm not that good at physics so your going to have to explain what the letters mean. Thanks.

4. May 26, 2010

### xxChrisxx

It's basically

Pressure * volume^k = constant.

However if you use

Temperature * volume(K-1) = constant.

K is 1.4 for air.

So you start with the inlet temperature and volume. Then as the volume reduces the temperaute increases.

However you can't use this heat for anything, it's just not worth the effort. The only way you can use it usefully is to expand it, which defeats the point of compressing it to increase efficiency. The temperature also helps the combustion process.

5. May 26, 2010

### ISX

Alright I am still lost. I understand it but what is the K-1 for? I understand airs compression constant is 1.4 but why do you subtract one? I also do not in the calculations where I get temperature based on how much it is compressed.

Hmm, if the temperature going in is 100F and volume is 60 cubic inches to the (1.4-1=0.4) power, I get 514 as the constant. I am thinking I just figured pressure rather than temperature it increased to.

I know why it gets hot and what purpose it serves I am just curious to how hot it gets.

6. May 26, 2010

### xxChrisxx

K is just the isentropic constant for an adiabatic process. The fact its constant just means that you can use a ratio to find the new temperature.

Temperature of inlet * volume at inlet ^ 0.4 = Temperature final ^ volume final ^0.4

So we can reduce that to:
T1* (17/1)^0.4 = T2
This is the inlet temperature * by the compression ratio ^ 0.4.

NOTE: You must use the Kelvin scale for thermodynamic processes.

Inlet T=297K

297^17^0.4 = T2
922K = 648 degrees C = 1199F

7. May 26, 2010

### ISX

Ah, makes perfect sense now. How do I figure in boost pressure, and why are you subtracting one from the 1.4 thing?

Last edited: May 26, 2010
8. May 26, 2010

### xxChrisxx

This gives the relationship for adiabatic processes.

You figure in boost pressure by using PV^K = constant. Using the inlet pressure as boost + atmospheric.

So for example on an N/A car the inlet pressure can be assumed as 1 bar/atm (they are close enough that I dont make the distinction. Say you have .5 bar boost. Then you set the inlet P as 1.5bar

If you'll hang fire i'll post a pic. EDIT: I don't have office on the new pc so I can't upload a screenshot of what I've done thats similar to this.

9. May 26, 2010

### ISX

I just can't understand why you subtract one from 1.4.

So if I use PV^K to find the constant, I use 1.4 or 0.4? Would that be 1.5 bar times V to the K power? What is V?

10. May 26, 2010

### xxChrisxx

K is gamma on the wiki page. (I don't know how to put a gamma on here). Gamma for air is 1.4.

You dont need to konw how to derive it to use the equations. The relationship is shown in the derivation on the wiki page. They derive the equation

PV^gamma = constant.

However we konw from PV=nRT (ideal gas law) that P V and T and connected. When you substitute in different variables, the powers are different.

In short, don't worry yourrself about why they are like that as you can just the the derived equations blindly no problem.

11. May 26, 2010

### ISX

Alright so let me see if I got it. To find temp of 1.5bar air in 17 compression:

1.5(60 cubic inches^0.4)= 7.715 constant. Now where in the T1*(17/1)^0.4 = T2 equation do I use that constant?

12. May 26, 2010

### rcgldr

Don't forget that the glow plug retains heat from previous cycles (or it's preheated at engine startup), and is hotter than the compressed air inside the cylinder. The ignition point starts at the glow plug, then a flame front travels away from the glow plug towards the cylinder, somewhat smoothly (otherwise you'd get pings or knocking due to multiple or premature ignition points).

13. May 26, 2010

### ISX

I have a grid heater.

14. May 27, 2010

### xxChrisxx

You don't use the constant in that way. You are also using gamma - 1 when you should be using gamma.

When we say two things are the same we can say they are equal. However when we have more things that are all equal it's easier to mathematically say the 'result' is constnat than to explicity state that all relationships are equal.

So for example

P(inlet)*V(inlet)^gamma = constant
P2*V2^gamma = constant.
P3*V3^gamma = const.

etc etc etc

What is is mathematically saying is that you can put inlet and p2v2 = to one another.
P1V1^1.4 = P2V2^1.4

So rearranging this to find P2.
P1 * (V1^1.4/V2^1.4) = P2

This can be tidied up to make:
P1 * (V1/V2)^1.4 = P2.

This relationship works for all values of V1 and V2. so you could use it to make a plot of how pressure increases with respect to volume decrease. However beucase we are considering a special case, A full compression. V1/V2 is the comrpession ratio:1.
So V1=17/V2=1 = 17

So:
P1*(CR^1.4) = P2

1.5*17^1.4= P2
This gives the result of a final pressure of: 79.2 bar.

If there is any point above you didn't get just let me know and i'll have a go at explaining in more detail.

15. May 27, 2010

### ISX

I get all that now but it's like your going from one thing to another without showing how they intertwine. The original formula just had T1*CR^0.4 which I don't see how you figure the boost pressure into. I assume it has something to do with the 0.4 part but am not sure. How about I give you a compression ratio of 17, a boost pressure of 10psi, and a starting temperature of 100F and you write down all your work in order, showing how you go from those 3 factors, to an end result temperature in Fahrenheit for when the piston is fully compressed. I am starting to grasp everything your saying just not seeing where it all mixes together, so if you could just do that example so that I could see exactly how it all works together, that would be a lot easier for me to understand. Thanks for all the help so far!

16. Jan 1, 2011

### ISX

I got the calculator to work based on this http://en.wikipedia.org/wiki/Adiabatic_process where they did the engine example same as I am doing and I got the temp and everything. I have everything on an excel spreadsheet so I can play with it but the way they did it, atmospheric pressure and boost pressure have no effect on the temperature of compression. I am leaving the initial temp the same but I would have thought more pressure in there would mean higher temp when compressed. What do I need to do to get it to factor in the initial pressures of the combustion chamber in order to get different readings when I add boost pressure?

Or does the air always end up the same temp when I leave the initial temp the same on everything but vary the pressures? Seems like it would make a difference but I am beginning to wonder.

17. Jan 2, 2011

### Gordianus

Assuming an adiabatic compression, the temperature ratio is solely given by the compression ratio. Old Diesel engines (non-turbocharged) usually had 20:1 to 22:1 compression ratios. This ensured a temperature increse well above the flash point of diesel fuel.
What do you mean by "boost pressure"?

18. Jan 2, 2011

### ISX

Boost pressure as in pressure from a turbo.

So no matter how much boost there is, if the intake air is the same temp, the compressed air will also be the same high temp it compresses to?

Last edited: Jan 2, 2011
19. Jan 2, 2011

### Gordianus

The turbocharger takes air at atmospheric pressure and room temperature and releases air at higher pressure and temperature. Then, the air goes through a heat exchanger (intercooler) that lowers the air temperature almost back to room temperature (this isn't strictly true). This way, in every admission cycle, the engines receives more air and can deliver more power. However, since the entered the engine at almost the same temperature, and the compression ratio remains the same, the temperature increase is almost the same.

20. Jan 2, 2011

### ISX

Yeah I knew how the turbo worked I was just trying to get the facts straight that intake air at any pressure doesn't matter. As in, an engine with 100psi being forced in but at 0F, will have the same heat at the end of compression as an engine with 10psi at 0F being forced in.. Pressure changes, but the temp doesn't in the end.