Ford: High altitude requires 0W oil. Why?

ZeeOSix

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Engine horsepower will be reduced because of lower oxygen content; so, they recommend thinner oil (less oil-viscosity drag) to make up for some of the lost horsepower at high altitudes.
Ford is only recommending an oil with a lower W rating at 7500+ ft while at -4F or below. So they are focusing on cold start-up conditions.
 
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... Look at an example where there was just enough gravity to keep the oil at the bottom of the sump and enough air pressure on it to keep it from vaporizing. Then vary the air pressure above the oil from say 5 PSIA to 100 PSIA. If there was essentially no gravity and at 5 PSIA there wouldn't be much of a driving force to push oil into the pump inlet. If the air pressure was increased to 100 PSIA the driving force would be much higher. ...
Imagine a basketball made out of much stronger material than rubber. We fill 1/4 of the ball's volume with motor oil, the rest is air and inflate it so the pressure gauge reads 5.3 PSI. That's 5.3 PSI over ambient, or a 5.3 PSI differential between inside the ball and outside the ball, so the absolute pressure inside the basketball is 14.7 + 5.3 = 20 PSI.

Now we take this ball into space where there is no gravity. The ball is strong enough to hold the pressure; it doesn't explode. It is still 20 PSI inside the ball. Because there's no gravity, the oil is floating around in the ball. The air pressure inside the ball doesn't force the oil to one side or another, the oil floats around inside the ball freely intermixing with the air, sticking to the inside walls of the ball due to surface tension, forming little oil balls whose size depends on the oil's surface tension and the air pressure.

Now we increase the air pressure in the ball: double it, triple it, whatever. This pressure doesn't force the oil to once side or area. It continues floating around inside the ball. The pressure might make the little oil balls floating around, smaller as the increased pressure acts against the surface tension.

By analogy, my point is that air pressure doesn't force the oil down into the pan or into the oil pump intake. Only gravity does that.
 

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Imagine a basketball made out of much stronger material than rubber. We fill 1/4 of the ball's volume with motor oil, the rest is air and inflate it so the pressure gauge reads 5.3 PSI. That's 5.3 PSI over ambient, or a 5.3 PSI differential between inside the ball and outside the ball, so the absolute pressure inside the basketball is 14.7 + 5.3 = 20 PSI.

Now we take this ball into space where there is no gravity. The ball is strong enough to hold the pressure; it doesn't explode. It is still 20 PSI inside the ball. Because there's no gravity, the oil is floating around in the ball. The air pressure inside the ball doesn't force the oil to one side or another, the oil floats around inside the ball freely intermixing with the air, sticking to the inside walls of the ball due to surface tension, forming little oil balls whose size depends on the oil's surface tension and the air pressure.

Now we increase the air pressure in the ball: double it, triple it, whatever. This pressure doesn't force the oil to once side or area. It continues floating around inside the ball. The pressure might make the little oil balls floating around, smaller as the increased pressure acts against the surface tension.

By analogy, my point is that air pressure doesn't force the oil down into the pan or into the oil pump intake. Only gravity does that.
Assuming that you are familiar with airplane engines, I have a question: is the oil pump on an airplane engine mounted above the oil tank/pan, making it dependent on atmospheric pressure, or is it located below or inside of the oil tank, making it independent of atmospheric pressure?
 
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Ford is only recommending an oil with a lower W rating at 7500+ ft while at -4F or below. So they are focusing on cold start-up conditions.
It is both the cold start and cold-engine drivability. High altitudes and low temperatures both cause the engine to run poorly, and you don't want to aggravate the problem by using a high-viscosity oil that increases the engine friction.

What people don't realize is that since a SAE 0W-30 has a higher viscosity index (VI) than a SAE 5W-30, it will have a lower operational viscosity throughout the warm-up period as a result, greatly improving the drivability during warm-up.

As an example, HPL PCMO 0W-30 has an operational viscosity of 2,198 cSt at −20 C, whereas HPL PCMO 5W-30 has an operational viscosity of 3,076 cSt at the same temperature.

 

ZeeOSix

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It is both the cold start and cold-engine drivability. High altitudes and low temperatures both cause the engine to run poorly, and you don't want to aggravate the problem by using a high-viscosity oil that increases the engine friction.What people don't realize is that since a SAE 0W-30 has a higher viscosity index (VI) than a SAE 5W-30, it will have a lower operational viscosity throughout the warm-up period as a result, greatly improving the drivability during warm-up.
I highly doubt that is the reason. Computer controlled engines take into account the ambient pressure and temperature and tune the engine accordingly. So it's not going to "run poorly" with 5W-30 vs 0W-30 when at 7500 ft and -4F. Still sticking with the pumpability issue at those conditions.
 
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Assuming that you are familiar with airplane engines, I have a question: is the oil pump on an airplane engine mounted above the oil tank/pan, making it dependent on atmospheric pressure, or is it located below or inside of the oil tank, making it independent of atmospheric pressure?
On typical Lycoming engines, the oil pump is in the accessory housing just outside the sump. It draws oil from the sump, through a hole/passageway. That oil intake that feeds the pump, from which it draws, is just above the bottom of the sump. The pump is at about the same level horizontally as the sump.

... Now we take this ball into space where there is no gravity. ... The air pressure inside the ball doesn't force the oil to one side or another, the oil floats around inside the ball freely intermixing with the air, sticking to the inside walls of the ball due to surface tension, forming little oil balls whose size depends on the oil's surface tension and the air pressure.
...
In the space basketball thought experiment: suppose there is gravity to push the oil inside the ball to once side. We orient the ball so its inflation valve is at the bottom of the oil collected by the gravitational field. Now the air pressure inside the ball tries to push/squirt that oil out the valve. Increasing air pressure in the ball increases the force at which the oil squirts out.

However, the only reason this happens is because of the pressure differential inside the ball (20 PSI) and outside (0 PSI or vacuum). In the engine, ambient atmospheric pressure is the same everywhere. I don't see where reducing ambient pressure creates a differential.
 
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ZeeOSix

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Imagine a basketball made out of much stronger material than rubber. We fill 1/4 of the ball's volume with motor oil, the rest is air and inflate it so the pressure gauge reads 5.3 PSI. That's 5.3 PSI over ambient, or a 5.3 PSI differential between inside the ball and outside the ball, so the absolute pressure inside the basketball is 14.7 + 5.3 = 20 PSI.

Now we take this ball into space where there is no gravity. The ball is strong enough to hold the pressure; it doesn't explode. It is still 20 PSI inside the ball. Because there's no gravity, the oil is floating around in the ball. The air pressure inside the ball doesn't force the oil to one side or another, the oil floats around inside the ball freely intermixing with the air, sticking to the inside walls of the ball due to surface tension, forming little oil balls whose size depends on the oil's surface tension and the air pressure.

Now we increase the air pressure in the ball: double it, triple it, whatever. This pressure doesn't force the oil to once side or area. It continues floating around inside the ball. The pressure might make the little oil balls floating around, smaller as the increased pressure acts against the surface tension.
But that analogy doesn't really apply to the conditions being discussed with oil under gravity and ATM pressure in an engine sump.

By analogy, my point is that air pressure doesn't force the oil down into the pan or into the oil pump intake. Only gravity does that.
Not a very good analogy, so let's look at how a PD actually works. Assume the PD is built so tight that if you put air pressure above ATM on the inlet side that it would be sealed enough to basically not let any air escape past the rotors. In real life, that's not the case, but if a PD pumps are designed and built correctly they are pretty close to sealing the inlet from the outlet for all practical purposes, so for this discussion let's assume that's the case.

When the PD rotates it moves the entire volume from the inlet side to the outlet side, leaving the inlet side chamber empty and under some level of vacuum. Any forces acting on the oil in the sump will try to push the oil towards that void vacuum pocket in the pump. Like mentioned before, if you could change the gravity force and the atmospheric force on the oil, it will change the total force on the oil that is trying to push it into the pump inlet. Therefore, if the engine is at a higher elevation with less ATM pressure, there will be less total pressure trying to move the oil to the pump inlet. If you went high enough in elevation were the air pressure was essentially zero, then you would only have the gravitational force on the oil, and it would be less total force than if there was gravity + ATM pressure on the oil. With a well fuctioning PD pump, it really doesn't matter what the outlet side is seeing ... only that the inlet side volume empties and becomes a void under vacuum for the fluid to flow into. With a non-PD pump, it would be a different behavior.
 

ZeeOSix

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In the space basketball thought experiment: suppose there is gravity to push the oil inside the ball to once side. We orient the ball so its inflation valve is at the bottom of the oil collected by the gravitational field. Now the air pressure inside the ball tries to push/squirt that oil out the valve. Increasing air pressure in the ball increases the force at which the oil squirts out.

However, the only reason this happens is because of the pressure differential inside the ball (20 PSI) and outside (0 PSI or vacuum). In the engine, ambient atmospheric pressure is the same everywhere. I don't see where reducing ambient pressure creates a differential.
On a well sealed PD pump, the outlet conditions on the pump don't really effect what's happening on the inlet side of the pump - see your bold comments in your quote. Therefore, your space basketball analogy is basically similar to how a PD pump would react with varying gravitational and air pressure forces trying to push the fluid into the inlet side of the pump.
 
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It seems that the outlet conditions on the pump do affect the inlet side. Why? Because the pump has a single rotor that simultaneously drives input and output. As one of the pump's rotor faces draws from the intake, another rotor face is pushing its previously drawn oil out the other side. Any force or differential pressure you apply at one side of the pump, exerts a force on the rotor which affects the other side.

Thus ambient pressure which affects both sides of the pump equally should have no effect on the pump. If it makes intake harder, it makes output easier, and vice versa. Net effect zero.
 

ZeeOSix

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It seems that the outlet conditions on the pump do affect the inlet side. Why? Because the pump has a single rotor that simultaneously drives input and output. As one of the pump's rotor faces draws from the intake, another rotor face is pushing its previously drawn oil out the other side. Any force or differential pressure you apply at one side of the pump, exerts a force on the rotor which affects the other side.

Thus ambient pressure which affects both sides of the pump equally should have no effect on the pump. If it makes intake harder, it makes output easier, and vice versa. Net effect zero.
Disagree ... why can an unregulated PD pump put 100s of PSI on the outlet side and still create a vacuum on the inlet side. Without a pressure regulating valve on the outlet side, a PD pump will literally blow-up whatever it's feeding or blow itself up. It couldn't do that if the inlet and outlet chambers were not separated and sealed by the rotors/impellers.
 
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Disagree ... why can an unregulated PD pump put 100s of PSI on the outlet side and still create a vacuum on the inlet side. Without a pressure regulating valve on the outlet side, a PD pump will literally blow-up whatever it's feeding or blow itself up. It couldn't do that if the inlet and outlet chambers were not separated and sealed by the rotors/impellers.
Of course. The inlet and outlet chambers are separated and sealed. But that doesn't prevent forces on the output side from applying at the intake side. The pump rotor is essentially a multi-faced wheel or set of gears that spans both sides (inlet to outlet). If you push on the rotor at any point, that force is applied to the entire rotor thus transmitted to all points - it moves as a unit. Pressure on the output side is a force applied to the rotor face approaching the outlet, and thus to the entire rotor, including the rotor face on the inlet side. If the pump motor isn't strong enough, the output pressure causes the rotor to slow down which reduces the oil drawn in on the intake side. Thus outlet conditions directly affect intake conditions.

The pump does work on the fluid it is pumping. That work comes from the power source driving the pump. The pressure the pump creates on the output side exerts a resistance to the pump, which is a torque against the pump rotor. The vacuum the pump creates on the input side exerts another resistance to the pump, which is another torque against the pump rotor. The power source driving the pump applies the opposing torque to keep the rotor spinning. The power needed to drive the pump to create that pressure differential is that torque multiplied by the rotational speed of the pump.
 

ZeeOSix

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Of course. The inlet and outlet chambers are separated and sealed. But that doesn't prevent forces on the output side from applying at the intake side.
Just how does the pressure at the pump outlet effect the pressure on the pump inlet when they are separated and sealed from each other. If the outlet pressure effected the inlet side, it wouldn't be possible for a healthy PD pump to produce a vacuum on the inlet side while at the same time producing 100s of PSI on the outlet side ... regardless of what forces are acting wherever in the pump.

The pump rotor is essentially a multi-faced wheel or set of gears that spans both sides (inlet to outlet). If you push on the rotor at any point, that force is applied to the entire rotor thus transmitted to all points - it moves as a unit. Pressure on the output side is a force applied to the rotor face approaching the outlet, and thus to the entire rotor, including the rotor face on the inlet side. If the pump motor isn't strong enough, the output pressure causes the rotor to slow down which reduces the oil drawn in on the intake side. Thus outlet conditions directly affect intake conditions.
See previous response ... the outlet conditions don't effect the inlet conditions on a properly designed PD pump.

As long as the pump is turning, it's moving all the volume that enters the pump inlet and pushes that same volume by force out the outlet. The inlet runs at a vacuum which allows fluid to enter (based in the driving force on the fluid), and the pump output will pressurize to whatever it takes to move the inlet volume through the resistance to flow. It will literally blow-up something due to excessive pressure in the process if the pump is not regulated. And while that's going on, the inlet just keeps running at a vacuum and taking in fluid. Whatever forces on the rotor you're fixated on has no bearing on how the pump performs. As long as you have enough power to turn it, it will performed as described.

The pump does work on the fluid it is pumping. That work comes from the power source driving the pump. The pressure the pump creates on the output side exerts a resistance to the pump, which is a torque against the pump rotor. The vacuum the pump creates on the input side exerts another resistance to the pump, which is another torque against the pump rotor. The power source driving the pump applies the opposing torque to keep the rotor spinning. The power needed to drive the pump to create that pressure differential is that torque multiplied by the rotational speed of the pump.
Of course, but again ... as long as you can turn the PD pump, it will operate and perform as described. Since the outlet really has no effect on the inlet, then the forces (gravity and atmospheric pressure) acting on the fluid (in the sump) will have an effect on how the fluid/oil flows to the pump inlet chamber. Once the fluid makes it there, it will be pumped to whatever is connected to the outlet. If the fluid/oil can't fully make it to the inlet chamber, or doesn't make it at all, then there is a pumpability issue. That's why SAE came up with a defined "pumpability viscosity" standard in SAE J300. At some point the forces on the oil in the sump can not get it to the pump inlet. If you reduce some of that driving force by going to a high elevation, then at some elevation & temperature combination it may effect pumpability.
 
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... Whatever forces on the rotor you're fixated on has no bearing on how the pump performs. As long as you have enough power to turn it, it will performed as described.
Forces on the rotor clearly do impact the pump's performance: as those forces increase (greater pressure differential), it requires more power to operate the pump. Newton's 3rd law: when the pump rotor exerts a force on the fluid, that fluid exerts and equal & opposite force on the rotor.
... the forces (gravity and atmospheric pressure) acting on the fluid (in the sump) will have an effect on how the fluid/oil flows to the pump inlet chamber. ... If you reduce some of that driving force by going to a high elevation, then at some elevation & temperature combination it may effect pumpability.
This is the nut of the issue. I think we agree that if the air pressure in the oil sump was below ambient (for example you sealed the oil sump from atmosphere and used a vacuum pump to reduce its air pressure), this would make it harder for the pump to draw oil into its intake. By sealing the sump you've created a pressure differential between the intake and output sides of the pump. The more difficult question is what happens when there is no pressure differential, but ambient pressure -- everywhere, including both sides of the pump -- is reduced.
 

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Here is a theory: When an engine is started, the oil pump almost immediately creates a vacuum in the oil pickup tube, which is filled quickly by the oil due to the atmospheric pressure on the oil in the pan. At -10F, it is filled much more slowly, because the oil is much thicker at that temperature. But still quickly enough with a 5W, based on CCS and MRV testing which is done at low altitudes. When the altitude increases, the atmospheric pressure drops, and at some point, slows the filling of the pickup tube down sufficiently to allow for an unacceptable number of dry cranking cycles leading to engine wear. So at a certain altitude (say above 7500') a 0W oil is preferable at cold temperatures (say below -4F).
 
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Here is a theory: When an engine is started, the oil pump almost immediately creates a vacuum in the oil pickup tube, which is filled quickly by the oil due to the atmospheric pressure on the oil in the pan. At -10F, it is filled much more slowly, because the oil is much thicker at that temperature. But still quickly enough with a 5W, based on CCS and MRV testing which is done at low altitudes. When the altitude increases, the atmospheric pressure drops, and at some point, slows the filling of the pickup tube down sufficiently to allow for an unacceptable number of dry cranking cycles leading to engine wear. So at a certain altitude (say above 7500') a 0W oil is preferable at cold temperatures (say below -4F).
Yes, that's the theory. However, the part I bold faced doesn't seem right. It's gravity, not atmospheric pressure, that causes the oil in the pan to fill in and replace the oil sucked into the oil pump intake. It can't be atmospheric pressure, because that pressure is ambient -- there is no atmospheric pressure differential across the intake & output sides of the oil pump.

I emphasize atmospheric because of course, the pump itself creates a pressure differential across the intake & output sides! But atmospheric pressure does not contribute to this because it's ambient - equal on all sides.
 

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Yes, that's the theory. However, the part I bold faced doesn't seem right. It's gravity, not atmospheric pressure, that causes the oil in the pan to fill in and replace the oil sucked into the oil pump intake. It can't be atmospheric pressure, because that pressure is ambient -- there is no atmospheric pressure differential across the intake & output sides of the oil pump.

I emphasize atmospheric because of course, the pump itself creates a pressure differential across the intake & output sides! But atmospheric pressure does not contribute to this because it's ambient - equal on all sides.
Please re-read. The pump creates a vacuum. The atmospheric pressure causes the oil to fill the vacuum. Gravity cannot make the oil go up the tube. You must be misunderstanding something.
 
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Here's a simple diagram:
1663344287755.jpg

The oil pump reduces pressure at point I and increases pressure at point O. This is a pressure differential maintained by the power delivered to the pump. Suppose ambient pressure is Pa. The pressure we measure with a gauge at point O is Po, and at point I is Pi. Because we measure it with a gauge, that is relative to ambient pressure Pa. So the absolute pressure at point O is Po + Pa, and at point I is Pi + Pa.

For example suppose Po is 30 PSI and Pi is -10 PSI. The pump delivers 30 PSI of oil to the engine, creates 10 PSI of vacuum to draw fresh oil from the intake, and maintains a pressure differential of 40 PSI.

The question is, how does Pa affect this? The pressure difference across the pump is Po - Pi. It is independent of Pa. Any change in Pa changes Po and Pi equally. Put differently: if Pa drops, the pump has more resistance on the intake side and less on the output side. And vice versa. All forces in balance, no net change in the pressure differential across the pump, or the power needed to drive it.
 

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Here's a simple diagram:
View attachment 117208
The oil pump reduces pressure at point I and increases pressure at point O. This is a pressure differential maintained by the power delivered to the pump. Suppose ambient pressure is Pa. The pressure we measure with a gauge at point O is Po, and at point I is Pi. Because we measure it with a gauge, that is relative to ambient pressure Pa. So the absolute pressure at point O is Po + Pa, and at point I is Pi + Pa.

For example suppose Po is 30 PSI and Pi is -10 PSI. The pump delivers 30 PSI of oil to the engine, creates 10 PSI of vacuum to draw fresh oil from the intake, and maintains a pressure differential of 40 PSI.

The question is, how does Pa affect this? The pressure difference across the pump is Po - Pi. It is independent of Pa. Any change in Pa changes Po and Pi equally. Put differently: if Pa drops, the pump has more resistance on the intake side and less on the output side. And vice versa. All forces in balance, no net change in the pressure differential across the pump, or the power needed to drive it.
This is all fine and good, but it has nothing to do with my theory. My theory is all about the pickup tube, and your diagram does not even have one. Also, no car engines that I am aware of work this way. You show a partially submerged oil pump. Don't cars generally have oil pumps attached to the engine block with the oil pan below it and a pickup tube in between?
 
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This is all fine and good, but it has nothing to do with my theory. My theory is all about the pickup tube, and your diagram does not even have one. Also, no car engines that I am aware of work this way. You show a partially submerged oil pump. Don't cars generally have oil pumps attached to the engine block with the oil pan below it and a pickup tube in between?
True, the diagram is simplified. Draw the pump outside the sump with a pickup tube. I don't see how it materially changes anything.

BTW, I would love to understand why you and others are so convinced that atmospheric/ambient pressure makes a difference. I'm glad this thread is evolving as a discussion rather than an argument, appreciate everyone's patience.
 
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Alien

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True, the diagram is simplified. Draw the pump outside the sump with a pickup tube. I don't see how it materially changes anything.

BTW, I would love to understand why you and others are so convinced that atmospheric/ambient pressure makes a difference. I'm glad this thread is evolving as a discussion rather than an argument, appreciate everyone's patience.
Think about the pickup tube issue, and how it approximates how a mercury barometer works, which after all is an instrument used to measure atmospheric pressure. Forget the pressure side of the pump for the moment. Then re-read my theory and some of @ZeeOSix 's explanations. Maybe it will make sense then.
 
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