Noak and wear
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Originally Posted By: Shannow
You must be overlooking a lot of websites to be cherry picking the data that you are showing.
No, If I'm wrong I'll admit that. I don't care. I just want to learn something.
Do you want to take a shot at calculating it? Haven't you done this before, being in the business and all. Give a poor electrical engineer a hand.
You must be overlooking a lot of websites to be cherry picking the data that you are showing.
No, If I'm wrong I'll admit that. I don't care. I just want to learn something.
Do you want to take a shot at calculating it? Haven't you done this before, being in the business and all. Give a poor electrical engineer a hand.
Q=29.3*10E-9(L+0.0043*(W/D))*m*D^2*N
m=1,000*(2C/D)
Q= flow in gallons per minute
C=bearing clearance in inches lets assume 0.001
D=journal dia in inches
L=bearing length in inches bearing width
N=rotational speed rpm?
W=steady load to be supported
let's do a 2" journal
L about 0.5"
N=6000rpm
W= max torque 400ft*lbs/ number of bearings (4) = 100
This is going to be a low number because of the 10E-9 factor in there.
m=1,000*(2C/D)
Q= flow in gallons per minute
C=bearing clearance in inches lets assume 0.001
D=journal dia in inches
L=bearing length in inches bearing width
N=rotational speed rpm?
W=steady load to be supported
let's do a 2" journal
L about 0.5"
N=6000rpm
W= max torque 400ft*lbs/ number of bearings (4) = 100
This is going to be a low number because of the 10E-9 factor in there.
OK, I'll take a shot at the equation you posted in drops per minute.
D=2.0
L= 1.0
m= 1.5 (.003 diametrical clearance)
W=10
N= 1000, 2000
Q1= 3.32x0.01(1+0.0043)10/2))1.5x4x1000
Q1=1000.2828 drops per minute
Q2= 3.32x0.01(1+0.0043)10/2))1.5x4x2000
Q2=2000.5656 drops per minute
Looky there! Double the speed = double the volume.
Q=~RPM was pure coincidence. I pulled D,L, and W from my behind.
Ed
D=2.0
L= 1.0
m= 1.5 (.003 diametrical clearance)
W=10
N= 1000, 2000
Q1= 3.32x0.01(1+0.0043)10/2))1.5x4x1000
Q1=1000.2828 drops per minute
Q2= 3.32x0.01(1+0.0043)10/2))1.5x4x2000
Q2=2000.5656 drops per minute
Looky there! Double the speed = double the volume.
Q=~RPM was pure coincidence. I pulled D,L, and W from my behind.
Ed
I'm getting something like 0.000352356 gallons per minute
doubling of that is insignificant. The oil pump is putting out like 50 gals per minute at 6000rpm.
doubling of that is insignificant. The oil pump is putting out like 50 gals per minute at 6000rpm.
W= max torque 400ft*lbs/ number of bearings (4) = 100
No, W= weight of the rotating element in pounds; weight of crankshaft in lbs, not torque.
Ed
No, W= weight of the rotating element in pounds; weight of crankshaft in lbs, not torque.
Ed
Originally Posted By: edhackett
W= max torque 400ft*lbs/ number of bearings (4) = 100
No, W= weight of the rotating element in pounds; weight of crankshaft in lbs, not torque.
Ed
I think its the total load, not just the weight of the crankshaft. It's meant to be the pressure against the bearing.
W= max torque 400ft*lbs/ number of bearings (4) = 100
No, W= weight of the rotating element in pounds; weight of crankshaft in lbs, not torque.
Ed
I think its the total load, not just the weight of the crankshaft. It's meant to be the pressure against the bearing.
Last edited:
Directly from your source:
Quote:
the only value left to find is the load (W). This is simply a matter of determining the weight of the rotating element divided by the number of bearings.
This equation is for steady state. You are correct in thinking that total load is important. We don't have the dynamic load on the bearings due to the reciprocating mass or from combustion events. That's another set of equations that have to be factored in. That will make W larger, but since that is a constant, the Q/RPM remains the same.
What did you use for m?
Ed
Quote:
the only value left to find is the load (W). This is simply a matter of determining the weight of the rotating element divided by the number of bearings.
This equation is for steady state. You are correct in thinking that total load is important. We don't have the dynamic load on the bearings due to the reciprocating mass or from combustion events. That's another set of equations that have to be factored in. That will make W larger, but since that is a constant, the Q/RPM remains the same.
What did you use for m?
Ed
Corrected, I was off by a factor of 10.
OK, I'll take a shot at the equation you posted in drops per minute.
D=2.0
L= 1.0
m= 1.5 (.003 diametrical clearance)
W=10
N= 1000, 2000
Q1= 3.32x0.001(1+0.0043)10/2))1.5x4x1000
Q1=100.0283drops per minute
Q2= 3.32x0.001(1+0.0043)10/2))1.5x4x2000
Q2=200.0566drops per minute
Looky there! Double the speed = double the volume.
Ed
OK, I'll take a shot at the equation you posted in drops per minute.
D=2.0
L= 1.0
m= 1.5 (.003 diametrical clearance)
W=10
N= 1000, 2000
Q1= 3.32x0.001(1+0.0043)10/2))1.5x4x1000
Q1=100.0283drops per minute
Q2= 3.32x0.001(1+0.0043)10/2))1.5x4x2000
Q2=200.0566drops per minute
Looky there! Double the speed = double the volume.
Ed
ow30 is a little bit heavier than what you idealised, but
http://www.opieoils.co.uk/p-716-castrol-edge-0w-30-fst-fully-synthetic-engine-oil.aspx
aint too far off and meets your necessary approvals..
http://www.opieoils.co.uk/p-716-castrol-edge-0w-30-fst-fully-synthetic-engine-oil.aspx
aint too far off and meets your necessary approvals..
I found some calculations for force on the rods. It looks like 6200 lbs for a stock Chevy 350.
W=6200
D=2.0
L=0.5
m=1.05 (.0021)
Q@1000 RPM =.1924 gal/min x 8 rods =1.54 gal/min
Q@2000 RPM = 3.08 gal/min
Q@4000 RPM = 6.16 gal/min
Q@6000 RPM = 9.24 gal/min
That's just for the rods. I'd guess W for the mains is less, plus cam bearings. Those values seem to be at least in the ball park, given ~20 gpm for a HV pump. Anyway, that's what I come up with. It's likely much, much more complicated than that. It always is.
Ed
W=6200
D=2.0
L=0.5
m=1.05 (.0021)
Q@1000 RPM =.1924 gal/min x 8 rods =1.54 gal/min
Q@2000 RPM = 3.08 gal/min
Q@4000 RPM = 6.16 gal/min
Q@6000 RPM = 9.24 gal/min
That's just for the rods. I'd guess W for the mains is less, plus cam bearings. Those values seem to be at least in the ball park, given ~20 gpm for a HV pump. Anyway, that's what I come up with. It's likely much, much more complicated than that. It always is.
Ed
ed,
good work, and you are right, going back to first principals for bearings is an iterative process, described here
http://nptel.ac.in/courses/IIT-MADRAS/Machine_Design_II/pdf/5_4.pdf
Turtle,
Originally Posted By: turtlevette
T = Film thickness…generally .001 inch… but it may vary based on oil type and application[/i]
This doesn't have a speed component.
you can't assume a film thickness, you have to calculate it from the above link I gave...film thickness is dependent on viscosity, load, and surface speed difference (there's that pesky "N" again)
Don't need to even do any worked calcs when you see your quoted formula, dimensional analysis says flow is proportional to speed...if you look around Bobistheoilguy, I've done the full design calcs a couple of times for a couple of different scenarios to show how stuff works...eg...
http://www.bobistheoilguy.com/forums/ubbthreads.php/topics/1212318/4
But if you need to keep playing, have a look at this simple online calculator
http://www.tribology-abc.com/calculators/c9_3.htm
It won't let you change everything, but will help develop your intuition for what parameters affect what, and in what direction.
good work, and you are right, going back to first principals for bearings is an iterative process, described here
http://nptel.ac.in/courses/IIT-MADRAS/Machine_Design_II/pdf/5_4.pdf
Turtle,
Originally Posted By: turtlevette
T = Film thickness…generally .001 inch… but it may vary based on oil type and application[/i]
This doesn't have a speed component.
you can't assume a film thickness, you have to calculate it from the above link I gave...film thickness is dependent on viscosity, load, and surface speed difference (there's that pesky "N" again)
Don't need to even do any worked calcs when you see your quoted formula, dimensional analysis says flow is proportional to speed...if you look around Bobistheoilguy, I've done the full design calcs a couple of times for a couple of different scenarios to show how stuff works...eg...
http://www.bobistheoilguy.com/forums/ubbthreads.php/topics/1212318/4
But if you need to keep playing, have a look at this simple online calculator
http://www.tribology-abc.com/calculators/c9_3.htm
It won't let you change everything, but will help develop your intuition for what parameters affect what, and in what direction.
Originally Posted By: turtlevette
Originally Posted By: edhackett
W= max torque 400ft*lbs/ number of bearings (4) = 100
No, W= weight of the rotating element in pounds; weight of crankshaft in lbs, not torque.
Ed
I think its the total load, not just the weight of the crankshaft. It's meant to be the pressure against the bearing.
Yes, the radial load on the bearing will be a sum of the cylinder pressure and dynamic loads due to engine speed. Weight of the crankshaft is so small compared to operating loads, that it is nearly insignificant. The magnitude and direction of the radial load on a bearing changes continuously through the engine cycle. Think more of a peak cylinder pressure of ~1100 psi for a modern naturally aspirated engine acting on whatever the piston area is. Then you need to know operating speed, rotating and reciprocating masses, crank stroke and rod length to calculate the dynamic force.
Originally Posted By: edhackett
W= max torque 400ft*lbs/ number of bearings (4) = 100
No, W= weight of the rotating element in pounds; weight of crankshaft in lbs, not torque.
Ed
I think its the total load, not just the weight of the crankshaft. It's meant to be the pressure against the bearing.
Yes, the radial load on the bearing will be a sum of the cylinder pressure and dynamic loads due to engine speed. Weight of the crankshaft is so small compared to operating loads, that it is nearly insignificant. The magnitude and direction of the radial load on a bearing changes continuously through the engine cycle. Think more of a peak cylinder pressure of ~1100 psi for a modern naturally aspirated engine acting on whatever the piston area is. Then you need to know operating speed, rotating and reciprocating masses, crank stroke and rod length to calculate the dynamic force.
Originally Posted By: edhackett
I found some calculations for force on the rods. It looks like 6200 lbs for a stock Chevy 350.
W=6200
D=2.0
L=0.5
m=1.05 (.0021)
Q@1000 RPM =.1924 gal/min x 8 rods =1.54 gal/min
Q@2000 RPM = 3.08 gal/min
Q@4000 RPM = 6.16 gal/min
Q@6000 RPM = 9.24 gal/min
That's just for the rods. I'd guess W for the mains is less, plus cam bearings. Those values seem to be at least in the ball park, given ~20 gpm for a HV pump. Anyway, that's what I come up with. It's likely much, much more complicated than that. It always is.
Ed
Total oil flow for a modern LS-type engine is ~10 gpm at max power with the oil running ~100C at the inlet.
I found some calculations for force on the rods. It looks like 6200 lbs for a stock Chevy 350.
W=6200
D=2.0
L=0.5
m=1.05 (.0021)
Q@1000 RPM =.1924 gal/min x 8 rods =1.54 gal/min
Q@2000 RPM = 3.08 gal/min
Q@4000 RPM = 6.16 gal/min
Q@6000 RPM = 9.24 gal/min
That's just for the rods. I'd guess W for the mains is less, plus cam bearings. Those values seem to be at least in the ball park, given ~20 gpm for a HV pump. Anyway, that's what I come up with. It's likely much, much more complicated than that. It always is.
Ed
Total oil flow for a modern LS-type engine is ~10 gpm at max power with the oil running ~100C at the inlet.
OVERKILL
$100 Site Donor 2021
Originally Posted By: turtlevette
You also are in disagreement with the motor oil university articles posted on this site.
Yes.
Quote:
There's bound to be more stuff around.
Which you seem to have found and appear to be at least trying to have a reasonable discussion about now.
Quote:
Am I going to have to gather more info before you believe what I'm saying?
Yes. You have to provide information that proves your position. Quoting 101 doesn't do that as you've already gathered. But the thread has progressed significantly since this post and you seem to be starting to understand "the Shannow group's" position now, no?
Quote:
It's important for everyone to understand how an oil pump works to also understand viscosity and flow in an engine. Something we discuss daily.
It's science and engineering and even if you had 100k posts on BITOG you can't change that.
I agree wholeheartedly, which means you need to understand it too, not assume that everybody else here is wrong and that you are the only one who knows what he's talking about.
You also are in disagreement with the motor oil university articles posted on this site.
Yes.
Quote:
There's bound to be more stuff around.
Which you seem to have found and appear to be at least trying to have a reasonable discussion about now.
Quote:
Am I going to have to gather more info before you believe what I'm saying?
Yes. You have to provide information that proves your position. Quoting 101 doesn't do that as you've already gathered. But the thread has progressed significantly since this post and you seem to be starting to understand "the Shannow group's" position now, no?
Quote:
It's important for everyone to understand how an oil pump works to also understand viscosity and flow in an engine. Something we discuss daily.
It's science and engineering and even if you had 100k posts on BITOG you can't change that.
I agree wholeheartedly, which means you need to understand it too, not assume that everybody else here is wrong and that you are the only one who knows what he's talking about.
OVERKILL
$100 Site Donor 2021
Originally Posted By: turtlevette
It's semantics whether you call the spring a relief spring or pressure regulating spring.
Sure, but that wasn't your initial position:
Originally Posted By: turtlevette
It's not a relief. It's a pressure regulator which is partially open at anything above fast idle.
So at least we've got that out of the way
Originally Posted By: turtlevette
The crux of the disagreement is whether the pump bypasses most of the time (my assertion) or almost never (shannow group).
If you have 40psi hot idle at 600 rpm what is the approx pressure at 1200? 80?
then what's the pressure at 2400? 160?
what's the pressure at 4800?
and on to redline. It doesn't take many rpm to get into bypass.
Have you never owned an engine with an oil pressure gauge? I've posted numerous times on this that observed pressure does not double with RPM due to the reasons Shannow has now touched on, which are the other components in the engine, particularly the bearings, consuming more oil downstream.
My old SBF was ~32PSI hot idle (850RPM), ~45psi at 3,500 and onto the relief (65psi) by around 5,000-5,500 IIRC.
Originally Posted By: turtlevette
We can argue about where it occurs
Easy enough to determine on any vehicle with an oil pressure gauge really.
Originally Posted By: turtlevette
and that a thinner oil will bypass at a higher rpm.
I don't think there's anything to argue about there. If observed oil pressure is lower with a thinner oil (which, in my experience, it is, but only by a few PSI), then it will take more RPM to create enough pressure to open the relief than it will with a heavier lubricant.
Quote:
Why do you suppose there is so much emphasis on developing variable displacement pumps vs the constant displacement most vehicles have now. Efficiency. An oil pump bypassing most of the time wastes energy.
Approached from another angle: Oil that is cooler, and subsequently thicker, will lead to unnecessarily high oil pressure, as the engine doesn't need (can't consume) the volume of oil being supplied. Taking an SBC for example, not on the relief, rolling down the road at 60Mph and 2,500RPM, sump temps are lowered by airflow over the pan so operating viscosity is up. Oil pressure is 40psi. Does the engine need 40psi of oil pressure to be properly lubricated in that state or is the engine just wasting power? Note the relief is still closed in this scenario.
Variable displacement pumps (which there is a great article on from the ASME here )
Originally Posted By: ASME
Fixed-displacement oil pumps currently circulate oil in most automobiles. Designers typically oversize the pumps to handle the harshest engine operating conditions. Most of the time, they consume more power and deliver significantly higher oil pressure than needed. They contain pressure-relief valves as a crude, cost-effective, and reliable way to avoid excessively high oil pressures. But these designs are inefficient, losing significant amounts of energy at high oil flows typical in internal-combustion engines.
as noted, work to "solve" that problem, by providing the volume and pressure the engine needs at a given point rather than the scenario depicted above with the conventional pump.
Quote:
If bypass events were rare why bother to develop a more efficient pump?
See above.
Originally Posted By: turtlevette
Think about it.
We have been and are.
Originally Posted By: turtlevette
And remember the pressure monitoring point on most engines is after the filter where there is a bit of pressure drop. You may see 40psi on the gauge when there is really 50psi at the pump.
Let's get this right.
Yes, the filter can provide some pressure drop. That's going to depend on the viscosity of the lubricant of course and the volume moving through the media. Hot oil is going to provide much less of a pressure drop than cold oil and the same goes with higher volume and of course a combination of the two. Note that the total pressure drop is controlled by the bypass in the filter itself, as it will only allow a set amount of differential pressure before it starts bypassing the media.
Gary Allen on here quite some time ago (RIP) had a rather cool setup on his jeep 4.0L where he put oil pressure gauges before and after the filter to monitor bypass events FWIW.
Yes, let's get this right. Which means not attacking anyone's credentials because you don't agree with them and instead sticking to civil discussion.
It's semantics whether you call the spring a relief spring or pressure regulating spring.
Sure, but that wasn't your initial position:
Originally Posted By: turtlevette
It's not a relief. It's a pressure regulator which is partially open at anything above fast idle.
So at least we've got that out of the way
Originally Posted By: turtlevette
The crux of the disagreement is whether the pump bypasses most of the time (my assertion) or almost never (shannow group).
If you have 40psi hot idle at 600 rpm what is the approx pressure at 1200? 80?
then what's the pressure at 2400? 160?
what's the pressure at 4800?
and on to redline. It doesn't take many rpm to get into bypass.
Have you never owned an engine with an oil pressure gauge? I've posted numerous times on this that observed pressure does not double with RPM due to the reasons Shannow has now touched on, which are the other components in the engine, particularly the bearings, consuming more oil downstream.
My old SBF was ~32PSI hot idle (850RPM), ~45psi at 3,500 and onto the relief (65psi) by around 5,000-5,500 IIRC.
Originally Posted By: turtlevette
We can argue about where it occurs
Easy enough to determine on any vehicle with an oil pressure gauge really.
Originally Posted By: turtlevette
and that a thinner oil will bypass at a higher rpm.
I don't think there's anything to argue about there. If observed oil pressure is lower with a thinner oil (which, in my experience, it is, but only by a few PSI), then it will take more RPM to create enough pressure to open the relief than it will with a heavier lubricant.
Quote:
Why do you suppose there is so much emphasis on developing variable displacement pumps vs the constant displacement most vehicles have now. Efficiency. An oil pump bypassing most of the time wastes energy.
Approached from another angle: Oil that is cooler, and subsequently thicker, will lead to unnecessarily high oil pressure, as the engine doesn't need (can't consume) the volume of oil being supplied. Taking an SBC for example, not on the relief, rolling down the road at 60Mph and 2,500RPM, sump temps are lowered by airflow over the pan so operating viscosity is up. Oil pressure is 40psi. Does the engine need 40psi of oil pressure to be properly lubricated in that state or is the engine just wasting power? Note the relief is still closed in this scenario.
Variable displacement pumps (which there is a great article on from the ASME here )
Originally Posted By: ASME
Fixed-displacement oil pumps currently circulate oil in most automobiles. Designers typically oversize the pumps to handle the harshest engine operating conditions. Most of the time, they consume more power and deliver significantly higher oil pressure than needed. They contain pressure-relief valves as a crude, cost-effective, and reliable way to avoid excessively high oil pressures. But these designs are inefficient, losing significant amounts of energy at high oil flows typical in internal-combustion engines.
as noted, work to "solve" that problem, by providing the volume and pressure the engine needs at a given point rather than the scenario depicted above with the conventional pump.
Quote:
If bypass events were rare why bother to develop a more efficient pump?
See above.
Originally Posted By: turtlevette
Think about it.
We have been and are.
Originally Posted By: turtlevette
And remember the pressure monitoring point on most engines is after the filter where there is a bit of pressure drop. You may see 40psi on the gauge when there is really 50psi at the pump.
Let's get this right.
Yes, the filter can provide some pressure drop. That's going to depend on the viscosity of the lubricant of course and the volume moving through the media. Hot oil is going to provide much less of a pressure drop than cold oil and the same goes with higher volume and of course a combination of the two. Note that the total pressure drop is controlled by the bypass in the filter itself, as it will only allow a set amount of differential pressure before it starts bypassing the media.
Gary Allen on here quite some time ago (RIP) had a rather cool setup on his jeep 4.0L where he put oil pressure gauges before and after the filter to monitor bypass events FWIW.
Yes, let's get this right. Which means not attacking anyone's credentials because you don't agree with them and instead sticking to civil discussion.
Originally Posted By: edhackett
I found some calculations for force on the rods. It looks like 6200 lbs for a stock Chevy 350.
W=6200
D=2.0
L=0.5
m=1.05 (.0021)
Q@1000 RPM =.1924 gal/min x 8 rods =1.54 gal/min
Q@2000 RPM = 3.08 gal/min
Q@4000 RPM = 6.16 gal/min
Q@6000 RPM = 9.24 gal/min
That's just for the rods. I'd guess W for the mains is less, plus cam bearings. Those values seem to be at least in the ball park, given ~20 gpm for a HV pump. Anyway, that's what I come up with. It's likely much, much more complicated than that. It always is.
Ed
6200lbs is an impulse load that decays quickly during the power stroke. Also understand not every rotation is loaded. I think we need to average the load which is what I was trying to do by just taking the peak torque and using that. Using the peak impulse load inflates the numbers.
It sure isn't a simple calc. I've made mention to Shannow before that a turbine is a much simpler solve because the loads are smooth and steady in one direction.
You also have to wonder if the impulse load would break down the hydrodynamic film for a short time so you have a mixed or boundary condition for milliseconds on each power stroke.
I think the only way to find out how much oil an engine will "take" vs rpm is to set up an experiment.
I wonder if anyone had done actual testing and measurement of this?
I found some calculations for force on the rods. It looks like 6200 lbs for a stock Chevy 350.
W=6200
D=2.0
L=0.5
m=1.05 (.0021)
Q@1000 RPM =.1924 gal/min x 8 rods =1.54 gal/min
Q@2000 RPM = 3.08 gal/min
Q@4000 RPM = 6.16 gal/min
Q@6000 RPM = 9.24 gal/min
That's just for the rods. I'd guess W for the mains is less, plus cam bearings. Those values seem to be at least in the ball park, given ~20 gpm for a HV pump. Anyway, that's what I come up with. It's likely much, much more complicated than that. It always is.
Ed
6200lbs is an impulse load that decays quickly during the power stroke. Also understand not every rotation is loaded. I think we need to average the load which is what I was trying to do by just taking the peak torque and using that. Using the peak impulse load inflates the numbers.
It sure isn't a simple calc. I've made mention to Shannow before that a turbine is a much simpler solve because the loads are smooth and steady in one direction.
You also have to wonder if the impulse load would break down the hydrodynamic film for a short time so you have a mixed or boundary condition for milliseconds on each power stroke.
I think the only way to find out how much oil an engine will "take" vs rpm is to set up an experiment.
I wonder if anyone had done actual testing and measurement of this?
Originally Posted By: Shannow
It's nearly constant current in analogy, but delivery volume decreases some with thinner lubricants, as more leaks around the gears and back to the suction.
If that's your explanation for your assertion that heavier oil flows faster through an engine than lower viscosity oil it doesn't explain the lower system back pressure that lower viscosity allows.
As is so often the case on BITOG the discussion has gotten sidetracked with a lot of extraneous points that don't address that critical bottom line. For a given rpm and oil temperature a lower viscosity oil will allow lower system back-pressure which is what an oil pressure gauge actually measures.
The only explanation that makes sense to me is the lower back- pressure is a result of the lower viscosity oil leaking out of the bearings and journals in addition to flowing through them less restricted and therefore with greater flow. That is after all the definition of viscosity; a measure of resistance to flow.
And the difference in back-pressure between a light 20 grade and say a 50 grade is substantial, as much as a 50%. Does that mean the 20 grade oil will flow 50% more oil? I don't know, there could be other leaks within the oil pump itself but I'm of the opinion that a good portion of the reduced back-pressure is a result of less restricted oil flow afforded by the lower viscosity oil.
It's nearly constant current in analogy, but delivery volume decreases some with thinner lubricants, as more leaks around the gears and back to the suction.
If that's your explanation for your assertion that heavier oil flows faster through an engine than lower viscosity oil it doesn't explain the lower system back pressure that lower viscosity allows.
As is so often the case on BITOG the discussion has gotten sidetracked with a lot of extraneous points that don't address that critical bottom line. For a given rpm and oil temperature a lower viscosity oil will allow lower system back-pressure which is what an oil pressure gauge actually measures.
The only explanation that makes sense to me is the lower back- pressure is a result of the lower viscosity oil leaking out of the bearings and journals in addition to flowing through them less restricted and therefore with greater flow. That is after all the definition of viscosity; a measure of resistance to flow.
And the difference in back-pressure between a light 20 grade and say a 50 grade is substantial, as much as a 50%. Does that mean the 20 grade oil will flow 50% more oil? I don't know, there could be other leaks within the oil pump itself but I'm of the opinion that a good portion of the reduced back-pressure is a result of less restricted oil flow afforded by the lower viscosity oil.
Originally Posted By: CATERHAM
and therefore with greater flow.
And the difference in back-pressure between a light 20 grade and say a 50 grade is substantial, as much as a 50%. Does that mean the 20 grade oil will flow 50% more oil? I don't know, there could be other leaks within the oil pump itself but I'm of the opinion that a good portion of the reduced back-pressure is a result of less restricted oil flow afforded by the lower viscosity oil.
If the bypass is closed throughout the rpm range at operating temp, how can that be?
I'm on your side on this, and have been trying to prove the bypass is open through most of the rpm range. Shannow thinks the pump flow almost exactly matches the dynamic flow requirements of hydrodynamic bearings which I don't buy. I don't know for sure who is right at this point but I'm not as sure of my stance now.
That's the reason for all the extraneous discussion. It all hinges on that. Maybe a new thread needs to be opened on that subject.
Noak vs wear has gone by the wayside.
and therefore with greater flow.
And the difference in back-pressure between a light 20 grade and say a 50 grade is substantial, as much as a 50%. Does that mean the 20 grade oil will flow 50% more oil? I don't know, there could be other leaks within the oil pump itself but I'm of the opinion that a good portion of the reduced back-pressure is a result of less restricted oil flow afforded by the lower viscosity oil.
If the bypass is closed throughout the rpm range at operating temp, how can that be?
I'm on your side on this, and have been trying to prove the bypass is open through most of the rpm range. Shannow thinks the pump flow almost exactly matches the dynamic flow requirements of hydrodynamic bearings which I don't buy. I don't know for sure who is right at this point but I'm not as sure of my stance now.
That's the reason for all the extraneous discussion. It all hinges on that. Maybe a new thread needs to be opened on that subject.
Noak vs wear has gone by the wayside.
For the sake of simplicity I'm assuming the relief or by-pass valve is closed or rather the engine is operating below the maximum system allowed back-pressure limit.
For example, say the maximum system back-pressure limit is 95 psi and a 5W-50 at 100C oil temp's and 5,000 rpm generates 90 psi and a 0W-20 generates 60 psi at the same oil temp' and rpm.
There's your 50% higher oil pressure with the 50 grade oil.
At a somewhat lower oil temp' the oil pump will start to enter by-pass mode with the 5W-50 which will further reduce the oil flow but the example I've given is to take the capped pressure relief out of the equation.
For example, say the maximum system back-pressure limit is 95 psi and a 5W-50 at 100C oil temp's and 5,000 rpm generates 90 psi and a 0W-20 generates 60 psi at the same oil temp' and rpm.
There's your 50% higher oil pressure with the 50 grade oil.
At a somewhat lower oil temp' the oil pump will start to enter by-pass mode with the 5W-50 which will further reduce the oil flow but the example I've given is to take the capped pressure relief out of the equation.
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