Otto cycle efficiency

[NBO]

The efficiency of the Otto cycle was descibed in many papers and could be expressed as 1-1/CR^(k-1). Could somebody help me with explanation what k is? We have 2 gases: before and after combustion and k is different.

 

[ione]

Specific heat ratio: cp/cv

 

[LostHippie]

I believe in this case, k is the ratio of specific heats for whatever working fluid you are working with.

k = Cp/Cv

Cp = specific heat at constant pressure
Cv = specific heat at constant volume

k is used in your isentropic compression and expansion processes in an ideal otto cycle calc. If you are doing calculations for a real otto cycle, then you have to take into account that these processes are no longer isentropic and have efficiencies associated with them.

So to use an example, we will consider the compression process, in an ideal otto cycle, the initial and final pressure and temperature can be easily found through the isentropic equations that utilize k. In a non-ideal, real otto cycle, there is an isentropic efficiency associated with the compression cycle due to friction in the moving parts, heat transfer from the cylinder walls, etc. This means that the actual temperatures and pressures will differ, sometimes significantly, from the ideal calculated temperatures and pressures.

I hope I haven't rambled too much, I have been classified as scatter-brained at times ^_^ and I hope this has somewhat helped.

Peace,
LostHippie

 

[NBO]

Thanks Hippie,
Combustion engine has 2 working fluids.
Before combustion: k1=Cp1/Cv1
After combustion: k2=Cp2/Cv2
and k1 is not equal to k2.

Which one should I use for efficiency calculation?
Thanks.

 

[ione]

Take the specific heat ratio of air (slight changes with temeprature).

 

[LostHippie]

I agree with ione...usually for theoretical calculations you use the air-standard otto cycle setup, where k is the specific heat ratio of air.

peace,
LostHippie