- #1
Kozy
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I am attempting to improve a spreadsheet I created back in uni some 5-6 years ago, which models all the velocities and accelerations of the pistons and rods, as well calculating pressure, temperature and resulting torque. It is only a simple model so far, originally it started simply by calculating the temperature and pressure prior to ignition, then had an input for pressure at ignition (P4) from which the PV through power stroke was then calculated.
I have dug this out of my hard drive and would like to update it, replacing the P4 input with a calculated value taking into account AFR and compression ratio. So far I am having some difficulty with the 'Qin' heat added portion of this.
I have two methods for calculating this and I am unsure which, if any is correct.
Version 1:
Qin = V*1/(R*FAR)
Where:
V= V1+V2
R = 8.314
AFR = 1/AFR
This gives me 45951 of undisclosed units, I suspect joules.
Ignition temperature is then:
T4 = T3+(Qin/Cv)
Where:
T3 = 770°K
Qin = 45951
Cv = 3R/2 =12.471
Which gives me 4455°K
This method results in a torque figure with a BMEP of 291psi, however a leaner AFR does result in a hotter ignition temperature which would appear to be correct.
The second method was a bit more simple:
T4 = T3 + (HHVfuel * Mfuel)
Where:
T3 = 770°K
HHVfuel = 47.3Mj/kg
Mfuel = Mair * FAR = 4.6e-5
This resulted in a T4 of 2935°K which produced a better torque figure (BMEP of 170psi), but which become hotter with a richer AFR, which does not appear to be correct.
Can someone point me in the right direction with this? I've been chasing my tail for a while and in the absence of any thermodynamics textbooks I've not been able to find a decent answer to my problem.
I have dug this out of my hard drive and would like to update it, replacing the P4 input with a calculated value taking into account AFR and compression ratio. So far I am having some difficulty with the 'Qin' heat added portion of this.
I have two methods for calculating this and I am unsure which, if any is correct.
Version 1:
Qin = V*1/(R*FAR)
Where:
V= V1+V2
R = 8.314
AFR = 1/AFR
This gives me 45951 of undisclosed units, I suspect joules.
Ignition temperature is then:
T4 = T3+(Qin/Cv)
Where:
T3 = 770°K
Qin = 45951
Cv = 3R/2 =12.471
Which gives me 4455°K
This method results in a torque figure with a BMEP of 291psi, however a leaner AFR does result in a hotter ignition temperature which would appear to be correct.
The second method was a bit more simple:
T4 = T3 + (HHVfuel * Mfuel)
Where:
T3 = 770°K
HHVfuel = 47.3Mj/kg
Mfuel = Mair * FAR = 4.6e-5
This resulted in a T4 of 2935°K which produced a better torque figure (BMEP of 170psi), but which become hotter with a richer AFR, which does not appear to be correct.
Can someone point me in the right direction with this? I've been chasing my tail for a while and in the absence of any thermodynamics textbooks I've not been able to find a decent answer to my problem.
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