Coolant pressure at M30 engines

Honolulu said:
1. Coolant system pressure is always the same throughout the cooling system. It is a continuous system filled with an incompressible fluid thus equally pressurized throughout.
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If the pressure is the same throughout the system, how is flow induced?
 
Beaudave said:
Hi Erik. I don't totally agree with you. Without falling into rocket scientist calculations, I am assuming (I may be wrong of course), that I am loosing 2-3 cm of vaporized fluid thrugh the cap, simply because I don't lose any on small distances, even on a daily base.
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I agree with you Beaudave. 2-3 cm in the reservoir is not that much fluid overall. Spillage or leakage of fluid would, as others have noted, mean the leak is below the level of fluid in the system, and would likely result in a rapid loss of coolant.

It is important to remember your chemistry (what a few have called "rocket science"). Matter exists in three basic phases (not withstanding actual rocket science and nuclear physics!). Solid, liquid, and vapor. It is not super obvious, but these phases can all exist at the same time. For example, we see ice cubes floating in a glass of water, or icebergs floating in the ocean. What is not as easy to see is that vapor also exists even when the fluid is not actually boiling (think clouds in the sky).

So, with those basics out of the way, the chemistry (or is it physics?) of the situation is this. A liquid will always have some amount of vapor above the surface. This is called the "vapor pressure". For now, let's just consider a system where there is nothing other than the liquid and vapor of that liquid (in a real system there may be other gases, like air, but we can ignore them here). In a fixed volume, the pressure, temperature and number of gas molecules follow a well-known relationship.

PV=nRT,

where P is the pressure, V is the non-liquid volume of the system, R is a constant (called the natural gas constant), and T is the temperature in degrees Kelvin. Importantly, n is the number of moles of the gas (a mole is a way of measuring the number of molecules, so you don't have to work with gigantic numbers). That n is important because it represents some fraction of that "2-3 cm of liquid" . From this relation (called the "natural gas law"), you can see that for a given volume, the pressure is directly related to the temperature. Note, if the volume is not constrained (causing the pressure to rise) a liquid will evaporate, even below the boiling point. This is because with an unlimited volume, the number of molecules of the vapor will be unconstrained, so the liquid will constantly turn to vapor until there is no liquid left (think puddles in the sun, or clothes drying on the line), even without any liquid "boiling".

What is not as obvious is that in a closed (fixed volume) system, this liquid-gas system is stable over all temperatures. As the temperature rises, the vapor pressure rises, and as the vapor pressure rises the gas molecules are condensed back into liquid. This means that, in this closed system, the temperature can theoretically rise without limit.

So, by now you are probably asking yourself, why is all of this important? Well, the relation shows that there is coolant vapor present in the reservoir at all temperatures where the coolant is a liquid, even if the liquid is not boiling. And, because of the (rearranged) relation P=nRT/V we can see that the smaller the volume, the higher the pressure at any given temperature. If there is ANY vapor leakage from the system (the n gets smaller), then those vapor molecules will be replaced from the liquid to keep the pressure constant at whatever pressure the leaking system will support (the available volume of coolant molecules in the liquid is comparatively infinite). Remember the volume of "empty space" above the coolant in the coolant reservoir, is actually fairly small, so pressure can build up fairly quickly

So that was a long foray into science as a way of saying if the seal on the cap is leaky, such that the cap will allow vapors to leave even below the cap's spring temperature, you will lose vapor. You may also be losing vapor at other joints in the system (i.e where hoses are clamped), and even through the hoses themselves. I'd follow your friend's "rocket science" advice, change the cap, and tighten all of the hose clamps.

Sorry for the science lecture!!

Scott
 
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ScottAndrews said:
There will be pressure variations, for example one side of a pump vane vs the other, but in an incompressible fluid, the DENSITY of the fluid does not change.
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That's not what he said. The pressure drops in a cooling system can be quite high. As I recall from an old Stewart Components tech tip, 20# is not unusual. The highest pressure point is the outlet of the water pump, and the lowest point is the suction side of the pump.
The old-style GM Radiators with the pressure cap on the high-pressure side of the radiator had a serious problem of losing coolant because the pump pressure caused the relief valve to lift. Moving the pressure relief to the low-pressure side of the radiator reduced this effect significantly.
 
ScottAndrews said:
... Matter exists in three basic phases (not withstanding actual rocket science and nuclear physics!). Solid, liquid, and vapor. It is not super obvious, but these phases can all exist at the same time.
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Yes, and what about cavitation?
 
Mike Goble said:
That's not what he said. The pressure drops in a cooling system can be quite high. As I recall from an old Stewart Components tech tip, 20# is not unusual. The highest pressure point is the outlet of the water pump, and the lowest point is the suction side of the pump.
The old-style GM Radiators with the pressure cap on the high-pressure side of the radiator had a serious problem of losing coolant because the pump pressure caused the relief valve to lift. Moving the pressure relief to the low-pressure side of the radiator reduced this effect significantly.
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First, let's assume a system that has no excess air. So the entire cooling system is filled with coolant except the reservoir. I this case the only pressure variations in the system, other than in the vapor section of the reservoir, will be due to the pump. Both sides of the pump are exposed to the vapor pressure in the reservoir, so, if the pump is not pumping, the pressure throughout the system will be nominally the same. The only variations would then be a result of head pressure, with coolant in the lower parts of the system at a very slightly higher pressure than those at the top of the system.

Now introduce a pump. A pump works by definition by creating a pressure differential. This is because the pump vaned are pushing the fluid molecules in one direction, so those molecules on one side of the vane will be pressed against the vane, while those on the other side will have the vane moving away from them. Since there is no vapor in the system, this pressure difference will propagate and equalize across the entire fluid mass in much the same way a column of fluid will have higher pressure at the bottom than at the top. Imagine a simple system with a pump connected to a circular tube that links one side of the pump to the other. in steady state, as the pump turns, the fluid will move, and the pressure on the high side of the pump will be greater than the low side, and in the tube, the pressure will linearly change from the high point to the low point, exactly filling inthe pressures required together from the high side of the pump vane to the low side. While the density of the fluid doesn't (can't) change there are more molecules of fluid pressing in one direction than in the other simply because the pump vane is creating pushing them in that direction creating what I suppose might be considered as an artificial pressure differential at the vane (artificial because it is the motion of the vane that creates the "pressure", not the density of the fluid.

With a vapor reservoir in the midst of this setup, the reservoir will create a static pressure (at a given temperature) that will propagate throughout the fluid. SO, at any point tithe fluid the pressure will be that static pressure of the vapor plus the artificial pressure the pump (which will vary as one goes from the high side of the pump to the low side,.
 
Arde said:
Yes, and what about cavitation?
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Hey, I'm a laminar kinda guy. I don't know nuthin' about no cavitation!

I think cavitation would simply mean that the pump vane is moving faster than the fluid can react, and this presumably causes the fluid molecules on the back side of the pump vane to fail to fill in as the vane moves. the result is a negative pressure that causes the fluid to briefly vaporize (which is why you see bubbles in a cavitating pump or prop). The result is turbulent flow that is very inefficient.

It might be easiest to think of the pump like a propeller or an Archimedes screw. As the screw turns it pushes the fluid molecules along, and because some are pushed forward those behind immediately fill in. Now if the screw/prop turns too fast, the molecules in front get pushed along because they have nowhere else to go, but those behind the screw/prop blade can't keep up, so we get a low pressure area that causes the fluid to vaporize. The vapor doesn't move like the fluid, and so the vane cannot push it as effectively, and the amount of flow decreases.

Here's an explanation of the above.
 
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curious
asked about a 2cm or 3cm lose,
the reservoir has a total of 8cm fluid heigh plus some air
in a very inaccurate calculation 3cm of that reservoir is 800cm3, almost one liter
we say that it is a lot,
but is only long distances
i think small distances add up,
and you end up with longer distances in the addition

curious again,
refill instead of finding and solving the leak

well, good
 
Is there a UV dye that you can add to the coolant, like they use in AC systems? That way you could see where the coolant is escaping.

I still think it is just a hard cap seal that is allowing the vapor to escape out the vent without actually boiling.
 
deQuincey said:
curious
asked about a 2cm or 3cm lose,
the reservoir has a total of 8cm fluid heigh plus some air
in a very inaccurate calculation 3cm of that reservoir is 800cm3, almost one liter
we say that it is a lot,
but is only long distances
i think small distances add up,
and you end up with longer distances in the addition

curious again,
refill instead of finding and solving the leak

well, good
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800 cm3 times 280 Million cars in the US, if it is evaporation it is a big deal.

Now Scott's guidance on vapor pressure may provide the solution, the fluid is a mixture of distilled water and glycol which have different vapor pressures at a given temperature, so if the leak is evaporative it stands to reason that it would dilute the water more than the glycol, no?

So start with a 50-50% mixture and measure the ratio after coolant loss. If the ratio is preserved at 50-50 you have a coolant fluid loss (leak, coolant into the cylinder, radiator, hoses, clamps), if the ratio is glycol dominated, then it is evaporative/vapor loss.
 
Arde said:
800 cm3 times 280 Million cars in the US, if it is evaporation it is a big deal.

Now Scott's guidance on vapor pressure may provide the solution, the fluid is a mixture of distilled water and glycol which have different vapor pressures at a given temperature, so if the leak is evaporative it stands to reason that it would dilute the water more than the glycol, no?

So start with a 50-50% mixture and measure the ratio after coolant loss. If the ratio is preserved at 50-50 you have a coolant fluid loss (leak, coolant into the cylinder, radiator, hoses, clamps), if the ratio is glycol dominated, then it is evaporative/vapor loss.
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Arde, you are BRILLIANT!!!
 
I would want to know where the loss is occurring.

Google “radiator pressure tester”. Apply the device with appropriate adapter to the reservoir and pressurize to observe if there is a pressure loss and corresponding leak. Also test the radiator cap.

EDIT:Sorry guys, I confused this thread with the one about losing 2-3cms of coolant on the reservoir…
 
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