Re: Help me for design calculations
This took a little time to remember, but here's what I did:
Total volume is given as 75cc. Compressible volume is the calculated cylinder:
PI*(46/2)^2*42=69800 cubic mm=69.8cc
Compressed volume would be 75-69.8=5.2cc
Determine compression ratio:
75/5.2=14.423:1
At intake cycle, assume air pressure in cylinder is 1 atmosphere (atm). Thus, compressed pressure is 14.423 atm. Also assume that the air/fuel mix is compressed adiabatically...the mix doesn't heat up due to compression.
Let's also assume for simplicity that gasoline is composed of octane C8H18 and in gaseous form and combustion is complete. The chemical equation for it's combustion is:
2•C8H18 + 25•O2 --> 16•CO2 + 18•H2O
The ratio change in gas pressure will be 34:27 after combustion. Remember though that only the 21% oxygen in the air is consumed. So, only that partial pressure will be affected. The partial pressure before combustion is:
21% * 14.423 = 3.029 atm
and after combustion is:
3.029 * 34/27 = 3.814 atm
Assuming adiabatic combustion, total chamber pressure is:
14.423*0.79 + 3.814 = 15.208 atm
Maximum force on the piston is just after TDC. Area of piston head is
PI*(46/2)^2= 1661 mm^2
Using conversion from atm to pascals (Pa or N/mm^2), force would be:
15.208*(.101325 N/mm^2)*1661= 2559.6 N
Now assume that maximum torque occurs half way through the combustion cycle. Chamber volume is:
5.2 + 34.9 = 40.1 cc
Assuming adiabatic expansion, use gas law to determine new pressure:
P1•V1 = P2•V2
15.208•5.2 = P2•40.1
P2 = 1.972 atm
Torque arm length is equal to the crank throw or 1/2 the stroke length, 21mm or 0.021 m. Torque is calculated to be:
1.972*.101325*1661*0.021= 6.97 N•m
Calculating the torque at just past TDC would require some force vector calculation, but this is a good start I think. Hope this helps!
Man, what a way to celebrate my 2,000th post!
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Last edited by jdmccright; 03-19-2011 at 11:19 AM.
Reason: Added stuff
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