Will the Real Compression Test Please Stand Up!

[Baconaman]

1. The gauge MUST have a Schrader type valve at the very tip of the hose where it screws into the cylinder. No adapter can be used.
2. The engine must be cold. Testing hot will result in approximately 10 PSI less.
3. Pull the engine over until the gauge won't go any higher. This will vary depending on displacement of the saw and length of the hose. There is no set number.
4. The throttle does not need to be held open on a piston ported chainsaw engine.
5. Don't waste your time testing an engine that has just been assembled. It should be run, then allowed to cool off completely.

I appreciate your concise list. Thanks

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[Baconaman]

All of it depends, are you selling it to me on ebay? If so, add three shots of gear oil to the cylinder, then connect a hose from a 175psi compressor to the carburetor opening and plug the exhaust (one crank is all that is necessary turning slow) take reading and add 30psi for each 100 feet over sea level then put the results in the sales ad.

For the 3 shots of gear oil, do I have to measure it with a shot glass or can I just free pour like in my earlier years as a bartender?

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[kevin j]

here is more than you want to know about Schrader valves but understanding the concept also helps understand check valves and relief valves

The Schrader valve doesn’t pop open and stay open after it reaches the 10 psi. There’s always is a differential pressure. It’s like any other check valve, or relief valve, which is basically just a check valve with a high spring pressure. The upstream engine side pressure is balanced by the downstream pressure plus the spring force. So the downstream gauge side will always read lower.

There are hydraulic or pneumatic valves that do what you are envisioning they’re called sequence valves but they reference atmosphere and have a third port. Those would be used in scenarios where you want a certain function to not operate until the line pressure reaches a certain pressure but then you want to valve fully open and not create a pressure drop. typically used in a clamp and drill circuit on machine tools
where the drill does not start until the clamp pressure is reached, but after that it don’t want the pressure drop reducing the drill pressure

 

[HarleyT]

I appreciate your concise list. Thanks

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My list was a whole lot more conciser!!!

 

[lone wolf]

1. The gauge MUST have a Schrader type valve at the very tip of the hose where it screws into the cylinder. No adapter can be used.
2. The engine must be cold. Testing hot will result in approximately 10 PSI less.
3. Pull the engine over until the gauge won't go any higher. This will vary depending on displacement of the saw and length of the hose. There is no set number.
4. The throttle does not need to be held open on a piston ported chainsaw engine.
5. Don't waste your time testing an engine that has just been assembled. It should be run, then allowed to cool off completely.

Nailed it!

 

[dswensen]

here is more than you want to know about Schrader valves but understanding the concept also helps understand check valves and relief valves

The Schrader valve doesn’t pop open and stay open after it reaches the 10 psi. There’s always is a differential pressure. It’s like any other check valve, or relief valve, which is basically just a check valve with a high spring pressure. The upstream engine side pressure is balanced by the downstream pressure plus the spring force. So the downstream gauge side will always read lower.

There are hydraulic or pneumatic valves that do what you are envisioning they’re called sequence valves but they reference atmosphere and have a third port. Those would be used in scenarios where you want a certain function to not operate until the line pressure reaches a certain pressure but then you want to valve fully open and not create a pressure drop. typically used in a clamp and drill circuit on machine tools
where the drill does not start until the clamp pressure is reached, but after that it don’t want the pressure drop reducing the drill pressure

I very much appreciate this discussion and the education - thank you.

So, If I have read you properly, assuming that the spring holds the Schrader valve closed up to 10PSI of pressure (which is hypothetical for the purposes of this discussion), and you pull the starter cord until the gauge reads 120 PSI, the compression is really 110 PSI (120 - 10)?

I still am thinking about this and having trouble wrapping my mind around the idea that somehow, 10 PSI is eaten up holding the spring compressed and the valve open. It seems like once the valve opens, the pressure on both sides of the valve would equalize.

 

[blsnelling]

No. 120

 

[rwoods]

As a fool I will wade in - a reading of 120 with a 10# valve is actually 130 (spring pressure plus stored pressure on the gauge side) but you use the gauge reading for your measurement.

Like a seesaw with a 10# weigh on one side, you have to add 10# to the other side to balance it.

Ron

 

[lone wolf]

As a fool I will wade in - a reading of 120 with a 10# valve is actually 130 (spring pressure plus stored pressure on the gauge side) but you use the gauge reading for your measurement.

Like a seesaw with a 10# weigh on one side, you have to add 10# to the other side to balance it.

Ron

Never heard that before.

 

[kevin j]

No, cylinder = 130 - 10 psi drop across valve = 120 gauge.

the actual pressure drop if probably a psi or two on the special schraders, vs. 10 to 30 on the tire valves. But lets keep using 10 psi spring as the example, it is much easier to visualize. Off topic, the spring force is probably fractions of an ounce, but the area of the valve seat is so small it takes the 10 psi to lift the valve slightly off its seat.

If the cylinder would be 130, it loses 10 psi across the cracking of the valve, and gauge would read 120.

If the cyl. was 110,you can’t get an increase to 120 on the gauge.

Think of it as a spring loaded loaded check valve (which it is), in a hole in the wall of a tank (the combustion chamber of the cylinder). At just over 10 psi, it will crack and relieve some flow out of the tank. However, it doesn’t just open up and stay open like you are visualizing.

The sum of the forces trying to open the valve are the pressure in the tank, the 130 psi used in this example, times the seat area of the valve. Tiny little hole.
The sum of the forces trying to close the valve are the outside pressure (the gauge side) times the valve seat area plus the spring force on that valve area, which we say equates to 10 psi.

So 130 is trying to open, and x + 10 is trying to close. The valve will leak some air into the gauge until x is 120 and the valve is seated closed again. But that 10 psi diffefential is always there.


Another bunny trail, most of you will ignore this, but the opening (cracking) area of the poppet or valve is NOT truly the same as the effective area once it slightly opens. There are fluid forces on a larger area, so the pressure has to reduce slightly below the set point before the valve closes. Example: a direct acting relief valve set to crack open at 1500 psi, will rise slightly with flow across it to 1500-1600-1700 psi etc., but then if flow is reduced, it has to drop down below 1500, maybe 1450 or 1475 before it closes fully. This is hysterisis. Doesn’’t matter on a log splitter, but the rising pressure with flow, and the hysterisis for reset can matter a lot in some circuits (like oil cooler control, filter bypass, or hydrostatic charge pressure control). The flow rise is handled by using a ‘pilot operated’ relief valve where a tiny pilot orifice controls a larger spool for the actual main flow. It also reduces the hysterisis.
The real cheap and simple relief valves are just a spring pushing a ball against a seat. Cheap, rugged, tolerate dirt, but a lot of pressure rise and hysterisis.
Next step up is a poppet style valve and spring, and the better valve is the pilot operated spool valve. the pilot is built into the inside of the spool as part of a cartridge valve usually.

 

[dswensen]

As a fool I will wade in - a reading of 120 with a 10# valve is actually 130 (spring pressure plus stored pressure on the gauge side) but you use the gauge reading for your measurement.

Like a seesaw with a 10# weigh on one side, you have to add 10# to the other side to balance it.

Ron

Thanks for this. I'm still not seeing it, but it's likely my small mind that can't picture it.

Without being able to prove this one way or the other with my equipment, I'm going with gauge readings. Once the valve is held open, I can't picture why the pressure on both sides of the valve would not be equal. Certainly the pressure on the cylinder side is higher until the valve opens, but not after.

Thanks guys.

 

[dswensen]

No, cylinder = 130 - 10 psi drop across valve = 120 gauge.

the actual pressure drop if probably a psi or two on the special schraders, vs. 10 to 30 on the tire valves. But lets keep using 10 psi spring as the example, it is much easier to visualize. Off topic, the spring force is probably fractions of an ounce, but the area of the valve seat is so small it takes the 10 psi to lift the valve slightly off its seat.

If the cylinder would be 130, it loses 10 psi across the cracking of the valve, and gauge would read 120.

If the cyl. was 110,you can’t get an increase to 120 on the gauge.

Think of it as a spring loaded loaded check valve (which it is), in a hole in the wall of a tank (the combustion chamber of the cylinder). At just over 10 psi, it will crack and relieve some flow out of the tank. However, it doesn’t just open up and stay open like you are visualizing.

The sum of the forces trying to open the valve are the pressure in the tank, the 130 psi used in this example, times the seat area of the valve. Tiny little hole.
The sum of the forces trying to close the valve are the outside pressure (the gauge side) times the valve seat area plus the spring force on that valve area, which we say equates to 10 psi.

So 130 is trying to open, and x + 10 is trying to close. The valve will leak some air into the gauge until x is 120 and the valve is seated closed again. But that 10 psi diffefential is always there.


Another bunny trail, most of you will ignore this, but the opening (cracking) area of the poppet or valve is NOT truly the same as the effective area once it slightly opens. There are fluid forces on a larger area, so the pressure has to reduce slightly below the set point before the valve closes. Example: a direct acting relief valve set to crack open at 1500 psi, will rise slightly with flow across it to 1500-1600-1700 psi etc., but then if flow is reduced, it has to drop down below 1500, maybe 1450 or 1475 before it closes fully. This is hysterisis. Doesn’’t matter on a log splitter, but the rising pressure with flow, and the hysterisis for reset can matter a lot in some circuits (like oil cooler control, filter bypass, or hydrostatic charge pressure control). The flow rise is handled by using a ‘pilot operated’ relief valve where a tiny pilot orifice controls a larger spool for the actual main flow. It also reduces the hysterisis.
The real cheap and simple relief valves are just a spring pushing a ball against a seat. Cheap, rugged, tolerate dirt, but a lot of pressure rise and hysterisis.
Next step up is a poppet style valve and spring, and the better valve is the pilot operated spool valve. the pilot is built into the inside of the spool as part of a cartridge valve usually.

Using this logic, and your thoughts that the "pressure drop" across a tire gauge Schrader valve is 10-30 PSI, you are saying that when my tire gauge is reading 32 PSI, the pressure in the tire is actually 42 to 62 PSI?

Hmmmm ......

 

[rwoods]

Not to worry, I can't prove it either. It was explained to me that the valve has the spring pressure plus the air pressure pushing against it. To open it the outside pressure has to overcome both. When the outside pressure can no longer hold the valve open, it closes trapping the compressed air between the valve and the gauge. The spring pressure exerts no force on the air between the two so the gauge is actually reacting to the trapped compressed air. Not an engineer and am easily fooled so don't count on me or my explanation. I do know that for performance and measuring purposes the gauge reading is what you use not some calculation off it unless it has a known deviation.

Ron

 

[rwoods]

Using this logic, and your thoughts that the "pressure drop" across a tire gauge Schrader valve is 10-30 PSI, you are saying that when my tire gauge is reading 32 PSI, the pressure in the tire is actually 42 to 62 PSI?

Hmmmm ......

Apples and oranges. You mechanically crack the valve when you check the air pressure in a tire. Thus the gauge is reading the actual pressure in the tire. Ron

 

[TheTone]

Excellent question. But there's a big difference here. Assume you're filling from a pump with 120 psi. The pump pressure overpowers the tire valve. You don't just hold the nozzle on the valve or you would overinflate the tire. You let off and check the pressure with your gauge. There is a rod in the tire pressure gauge that opens the tire valve by mechanical force. It doesn't matter what the valve spring pressure is in the tire valve, it is overcome by the force of your pushing on the gauge. So now you have an open valve letting air at the tire pressure flow into the gauge. The tire gauge will now read the tire pressure minus the opening pressure of the tire gauge Schrader valve, hopefully small.

 

[HarleyT]

 

[rwoods]

Excellent question. But there's a big difference here. Assume you're filling from a pump with 120 psi. The pump pressure overpowers the tire valve. You don't just hold the nozzle on the valve or you would overinflate the tire. You let off and check the pressure with your gauge. There is a rod in the tire pressure gauge that opens the tire valve by mechanical force. It doesn't matter what the valve spring pressure is in the tire valve, it is overcome by the force of your pushing on the gauge. So now you have an open valve letting air at the tire pressure flow into the gauge. The tire gauge will now read the tire pressure minus the opening pressure of the tire gauge Schrader valve, hopefully small.

Right on. I forgot that some tire gauges have a valve as well; those that don't are reading the actual tire pressure. Same principles. Ron

 

[Baconaman]

Apples and oranges. You mechanically crack the valve when you check the air pressure in a tire. Thus the gauge is reading the actual pressure in the tire. Ron

Of course we do! Thanks for pointing that out because I was stuck there as well.

 

[Baconaman]

No, cylinder = 130 - 10 psi drop across valve = 120 gauge.

the actual pressure drop if probably a psi or two on the special schraders, vs. 10 to 30 on the tire valves. But lets keep using 10 psi spring as the example, it is much easier to visualize. Off topic, the spring force is probably fractions of an ounce, but the area of the valve seat is so small it takes the 10 psi to lift the valve slightly off its seat.

If the cylinder would be 130, it loses 10 psi across the cracking of the valve, and gauge would read 120.

If the cyl. was 110,you can’t get an increase to 120 on the gauge.

Think of it as a spring loaded loaded check valve (which it is), in a hole in the wall of a tank (the combustion chamber of the cylinder). At just over 10 psi, it will crack and relieve some flow out of the tank. However, it doesn’t just open up and stay open like you are visualizing.

The sum of the forces trying to open the valve are the pressure in the tank, the 130 psi used in this example, times the seat area of the valve. Tiny little hole.
The sum of the forces trying to close the valve are the outside pressure (the gauge side) times the valve seat area plus the spring force on that valve area, which we say equates to 10 psi.

So 130 is trying to open, and x + 10 is trying to close. The valve will leak some air into the gauge until x is 120 and the valve is seated closed again. But that 10 psi diffefential is always there.


Another bunny trail, most of you will ignore this, but the opening (cracking) area of the poppet or valve is NOT truly the same as the effective area once it slightly opens. There are fluid forces on a larger area, so the pressure has to reduce slightly below the set point before the valve closes. Example: a direct acting relief valve set to crack open at 1500 psi, will rise slightly with flow across it to 1500-1600-1700 psi etc., but then if flow is reduced, it has to drop down below 1500, maybe 1450 or 1475 before it closes fully. This is hysterisis. Doesn’’t matter on a log splitter, but the rising pressure with flow, and the hysterisis for reset can matter a lot in some circuits (like oil cooler control, filter bypass, or hydrostatic charge pressure control). The flow rise is handled by using a ‘pilot operated’ relief valve where a tiny pilot orifice controls a larger spool for the actual main flow. It also reduces the hysterisis.
The real cheap and simple relief valves are just a spring pushing a ball against a seat. Cheap, rugged, tolerate dirt, but a lot of pressure rise and hysterisis.
Next step up is a poppet style valve and spring, and the better valve is the pilot operated spool valve. the pilot is built into the inside of the spool as part of a cartridge valve usually.

What he said. I was just about to send an almost identical post but KJ beat me to it. Mine was going to be a lot more technical with a lot of formulas and really hard math stuff but I think he covered it well. As we say in the valve biz, don't bolt your bonnet too tight or you might blow your butt weld end!

 

[blsnelling]

Once the valve opens the gauge is going to see the full pressure.