Freeing a seized engine with compressed air

[DJ54]

Years ago when a few of us were restoring old hit & miss engines, & tractors, if one was really stuck bad, we'd, as mentioned braze a pipe fitting to an old spark plug, with the porcelain knocked out. Reduce down to female 1/8" pipe, then screw a 1/8" grease zerk in it, and pump the cylinder full of grease. You'll need to remove the rocker arm, so all valves are closed. Once the cylinder is full, keep pumping. A normal hand grease gun will produce approx. 12,000 psi. Multiply that times the diameter of the piston, and you can see that creates quite a force. It's a mess to clean up, but never had one we couldn't get out.

On tractors, they were usually so bad, we needed to replace the sleeves anyway, so not a big problem. Grease was salvaged to put on plows & such.

 

[quicksandfarmer]

Let's run some numbers as I have nothing to do this morning.....it just rained on my "down" hay so I can't bale it .

Thinking about force applied, like with a torque wrench in ft-lbs. Force = pressure x area. Pressure (considering the valves are closed which as stated above may not be the case.....tough luck) is say 125 psi; best you can get from a single stage compressor.

For area let's take a single cylinder of a Ford and a Johnny Popper....guessing at the diameters but close enough to see where we are. Ford take 3". Popper take 7".

Area of the piston is 22/7....aka 3.14 x (ｽ the diameter x ｽ the diameter).

Ford: 3.14 x (3"/2 x 3"/2) = 3.14 x 2.25 = 7" surface area.

Force applied to piston is 125 psi x 7" area = 883 ft-lbs.......Yepper that's a sizeable force pushing down on that piston. No need to go to the popper but let's do it anyway.

Popper: 3.14 x (7"/2 x 7"/2) = 3.14 x 12.25 = 38.5" surface area.

125 psi x 38.5" area = 4810 ft-lbs. Yepper that really IS a sizeable force.

Add that to a long stroke for good torque and you see why John decided that a 2 jug, syrup bucket engine would be to a farmer's advantage when needing something to do some serious work.

Course what's missing from the above is the added whammy when the injector squirts a mist of diesel in the compressed air generated by the compression stroke.

So, compare that to you on your pipe wrench on the crank shaft trying to rock the engine back and forth after you soaked it good with some sort of release agent.

Yes sir. Compressed air is a tool. Course with a multi cylinder engine luck would have it that you could have at least one cylinder with the valves closed.

Sorry to hear about your hay. I hope it can be saved. Around here this late in the year I don't think you'd have a hope of drying it.

I am going to correct your math a bit. If you multiply PSI -- pounds per square inch -- by bore area in square inches you get pounds of force, not foot-pounds. To get foot-pounds you have to multiply by the length of the lever arm in feet. That depends on the position of the crank, at TDC it's zero, at 90 degrees it's the length of the crank, and at 180 it's zero again. You can figure out the length of the crank by visualizing that the piston goes through a full stroke when the crank goes from TDC to 180, so the crank length is half the stroke. So let's say your stroke is four inches, the crank length would be two inches, or one sixth of a foot, and you would divide those force numbers by six to get foot-pounds. They're still pretty big numbers.

Pulling it all together the maximum torque is force times crank length, which is equal to PSI times bore area times half the stroke divided by twelve (to convert inches to feet). Now bore area times stroke has a name, it's the engine displacement (per cylinder). So you could also say that torque equals PSI times engine displacement divided by 24. Which is kind of an interesting result, that the torque generated doesn't depend on the engine geometry, just its total displacement.

Also, 125 PSI is about 9 atmospheres of pressure. It would be the pressure that an engine with a compression ration of 9:1 would generate on the compression stroke. So while the forces generated by compressed air are considerable, they're not much compared to the forced in the engine while running.

 

[09112]

What a great feeling when that engine finally comes free and oil/water flies out the plug holes. PB Blaster did the job and with just a little nudge with a pipe wrench and six foot extension pipe. Thanks for all the good advise I got from the guys/gals on this site I saved an engine that was doomed to the scrap heap. Now I'll get on with the restoration and ask more questions later. Love that Allis C which I hope to make into a CA.

 

[GLyford]

Thank you for circling back and letting us know the results.

Have a soft spot for Allis, Dad had a '57 B.

 

[RedNeckRacin]

Glad to hear it worked out! :thumbsup:

 

[Texasmark]

Yank the rockers and pushrods and you can have all of the valves closed. Then you can try the pistons that are about halfway through the stroke.

That solves that problem.....great idea and takes just a few minutes, especially if you have the old style rail rocker system like in my 4 cyl Fords.

 

[Texasmark]

Sorry to hear about your hay. I hope it can be saved. Around here this late in the year I don't think you'd have a hope of drying it.

I am going to correct your math a bit. If you multiply PSI -- pounds per square inch -- by bore area in square inches you get pounds of force, not foot-pounds. To get foot-pounds you have to multiply by the length of the lever arm in feet. That depends on the position of the crank, at TDC it's zero, at 90 degrees it's the length of the crank, and at 180 it's zero again. You can figure out the length of the crank by visualizing that the piston goes through a full stroke when the crank goes from TDC to 180, so the crank length is half the stroke. So let's say your stroke is four inches, the crank length would be two inches, or one sixth of a foot, and you would divide those force numbers by six to get foot-pounds. They're still pretty big numbers.

Pulling it all together the maximum torque is force times crank length, which is equal to PSI times bore area times half the stroke divided by twelve (to convert inches to feet). Now bore area times stroke has a name, it's the engine displacement (per cylinder). So you could also say that torque equals PSI times engine displacement divided by 24. Which is kind of an interesting result, that the torque generated doesn't depend on the engine geometry, just its total displacement.

Also, 125 PSI is about 9 atmospheres of pressure. It would be the pressure that an engine with a compression ration of 9:1 would generate on the compression stroke. So while the forces generated by compressed air are considerable, they're not much compared to the forced in the engine while running.

Yes on the ft-lbs being an error. I caught that just now. As you said we are just talking about force applied. Great job of bringing in more pertinent information as to what's going on, things I didn't know of to add. And yes the squirt if diesel does make a difference.

I just tossed out some numbers while having my morning coffee and yes didn't have all the facts but it answered my question as to is it feasible.

I only had to free up one engine and it was a 4 cyl 300 International. Forget what I put down the throat but did the rock the crank with the wrench thing and got it out.

On the hay yes, it's ok. Thanks for caring. What's not ok is the hay I did this spring. We had lots of rain and the hay was sudan-sorghum with large stems that don't seal well in a round bale sitting out in the weather (water logged soil, not flooded, but lots of water) like grasses can do. I'm having to scrap 2 cuttings.

But all isn't lost as I will put it in my poop slinger and dump it back on the field, run over it with the bush hog and disc it in after I get the winter crop off next spring.

 

[buickanddeere]

What is the definition of Stuck ? Sitting inside a building and won稚 roll over after sitting a few months ? Or dragging something out of a fence row and trying to turn it over ?