timberwolf
Addicted to ArboristSite
I did a little experiment today working on trying to figure out just how much saw was cooled by the evaporative effects of the fuel/oil. But as I see the results I think this may be worthy of a tread of it's own.
Hypothesis:
Given that the stocheometric mixture for air/gasoline is 14.5:1 it should be possible to calculate how much fuel would be used knowing the effective compressed volume and the RPM.
Math
44mm piston 33mm stroke and with my exhaust timing = 24.9 CC effective volume compressed by the piston.
X 13,500 RPM = .336 m3 of air
X air density 1.286 kg/m3
= .432 kg of air
divide by 14.5:1 air/fuel ratio
=27.9 grams of fuel or 34.9 ml to run 1 min at 13,500 (just over 2 tablespoons)
Here is what I did to test the theory:
-Saw MS 260, pipe was on it, so pipe stayed on for the test ( I think this was best anyway because it should have reduced fuel losses in the scavenging cycle)
-16"bar .325 square filed 8 pin sprocket
-added 200 ml 0f 40:1 93 octane gas mixed with synthetic/castor based oil
-ran the saw for 1 min in the wood (13,500 average RPM) (16 10" cookies)
-Dumped remaining fuel out into a container and refilled with a graduated syringe to determine how much fuel was used
I found that the saw used 50 ml (40.15 grams) for the 1 min test. (just over 3 tablespoons)
Conciliation:
By taking the theoretical fuel consumption, divided buy the actual I am getting 69.8%. I guess this makes sense as some fuel is unburned, some is lost back out the carb, and also there is all the physics involved with scavenging. If the pipe is working more charge should be packed into the cylinder before being compressed by the piston and hence upping the real life fuel consumption.
Now divided by 13,500 RPM to determine that the saw uses 0.0037 ml (.00297 grams) per stroke. To put this in tangible terms that a little less than 1/10 of a drop per stroke. 1/10 of a drop is not a lot of coolant.
Strange though 1/10th of a drop 13,500 times a min = 3/4 of a gallon per hour in the cut!
Think I will try this again muffler, open header, and with a different saw and see what comes out of it..
Hypothesis:
Given that the stocheometric mixture for air/gasoline is 14.5:1 it should be possible to calculate how much fuel would be used knowing the effective compressed volume and the RPM.
Math
44mm piston 33mm stroke and with my exhaust timing = 24.9 CC effective volume compressed by the piston.
X 13,500 RPM = .336 m3 of air
X air density 1.286 kg/m3
= .432 kg of air
divide by 14.5:1 air/fuel ratio
=27.9 grams of fuel or 34.9 ml to run 1 min at 13,500 (just over 2 tablespoons)
Here is what I did to test the theory:
-Saw MS 260, pipe was on it, so pipe stayed on for the test ( I think this was best anyway because it should have reduced fuel losses in the scavenging cycle)
-16"bar .325 square filed 8 pin sprocket
-added 200 ml 0f 40:1 93 octane gas mixed with synthetic/castor based oil
-ran the saw for 1 min in the wood (13,500 average RPM) (16 10" cookies)
-Dumped remaining fuel out into a container and refilled with a graduated syringe to determine how much fuel was used
I found that the saw used 50 ml (40.15 grams) for the 1 min test. (just over 3 tablespoons)
Conciliation:
By taking the theoretical fuel consumption, divided buy the actual I am getting 69.8%. I guess this makes sense as some fuel is unburned, some is lost back out the carb, and also there is all the physics involved with scavenging. If the pipe is working more charge should be packed into the cylinder before being compressed by the piston and hence upping the real life fuel consumption.
Now divided by 13,500 RPM to determine that the saw uses 0.0037 ml (.00297 grams) per stroke. To put this in tangible terms that a little less than 1/10 of a drop per stroke. 1/10 of a drop is not a lot of coolant.
Strange though 1/10th of a drop 13,500 times a min = 3/4 of a gallon per hour in the cut!
Think I will try this again muffler, open header, and with a different saw and see what comes out of it..
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