I did that once. I had a leak, a hose spraying iirc, just as we arrived outside some Federal building, not a museum, just south of the Mall in DC. My friend's girfriend worked there and we were picking her up at the end of the day. She took me down to the engineer's room and he gave me a piece of pipe 2 or 3" long. Now I probably couldn't get into the building even if I needed blood.
I guess you need to know how a centrifugal pump works. Pressure rise across the pump is function of the square of its speed. Double the pump speed and the delta P across the pump increases 4X. Expansion due to heat will increase system pressure if it is in a closed system. If a fluid can expand without being constrained---no significant change in pressure. MLD
Hi, It would also matter how hot the water already is. How old is the car? Is the rad fan. electric? Thermo. clutch driven with belt? Or real old car with straight belt driven? No water temp. gauge on the dash?(very good idea to have one)
Well, I give each of you half credit. LOL! At a certain RPM, the pump head (outlet-inlet pressure) and flow rate are both are fairly indept of the coolant temperature as long as the coolant stays a liquid (normal condition) when going through the pump. The engine heat raises the coolant temperature and the pressure (remember this from skool... PV=nRT), but (and this is a big butt) as long as the pump turns at the same RPM, the head should stay about the same.
FWIW, here's a decent drawing of a water pump cross-section
Well, I give each of you half credit. LOL! At a certain RPM, the pump head (outlet-inlet pressure) and flow rate are both are fairly indept of the coolant temperature as long as the coolant stays a liquid (normal condition) when going through the pump. The engine heat raises the coolant temperature and the pressure (remember this from skool... PV=nRT), but (and this is a big butt) as long as the pump turns at the same RPM, the head should stay about the same.
FWIW, here's a decent drawing of a water pump cross-section
A couple of comments--PV=nRT is an equation used in gas flow calculations not when the fluid is a liquid. Don't understand where you are coming from. Just for the record, at constant speed the pressure rise across a centrifugal pump does not remain constant. Typically, there is a droop (loss of delta P) as the pump flow demand increases. Relatively insignificant at first but if the flow demand gets large enough, then the Pump Delta P can drop significantly. Flow demand is dictated by the characteristics of the the system in question--that is, how the delta P vs Flow of the system (line losses etc.) matches up with the delta P vs flow of the pump. Where the two intersect will be the operating point of the System. The idea is to match them so that the intersection takes place where the droop in pump deta P is relatively insensitive to flow demand. MLD
Riddle me this. In a CLOSED system FILLED with INCOMPRESSIBLE liquid, the pressure is whatever the pressure is. The pump can't put more pressure on the output side than is on the input side because it's a CLOSED system FILLED with INCOMPRESSIBLE liquid.
Do pump principles applied to OPEN systems really apply?
Great Question. My original question remains unanswered .... Does the water pump ADD another 10 to 30 psi internal pressure (as the sources I originally was looking at claimed) and if it does, how come that extra pressure, on top of the already existing 12 psi heat pressure, not cause the cap to blow off from the excess pressure now totaling from 20 to 40 psi? All I can think is that the pump only adds perhaps
2 to 5 psi and that by the time it "gets" to the cap area, flow resistance has dissipated it down to 1 or 2 psi and it has ceased to be a problem. Or maybe the pump only "adds" negative pressure, i.e. suction at the inlet, or some combo of all that.
Don't understand either of your comments. Could you expand your explanations so they make some sense? Do you know what happens if you dead end a centrifugal pump (Zero flow) while its running?? The discharge pressure might not change but you better be prepared to see the fluid temperature skyrocket. You seem to to make a big point out of INCOMPRESSIBLE fluid. Got some news for you, Ready---fluids, aka liquids, are not INCOMPRESSIBLE!!! Have you ever heard of "Bulk Modulus", entrained air or compressible flow as they apply to liquids? Do you know what would happen to the pressure in a closed system if the fluid temperature (say water) was increased but the fluid was not able to expand due to the closed (or fixed) system volume? Hint: Delta P=(BM) x (Delta V)/V Clue: Pressure can increase up to the thousands! Want to try and conduct your own experiment? Close the water inlet shut off valve in your house. Keep all faucets closed and of course, lock the hot water tank relief valve so that it doesn't open. Now just crank up the temperature of your water heater. This ought to seek out your system's weak link. MLD
It's my contention that the pump can add no pressure unless it has somewhere to get the water to pump. If it's coming from the closed system, there ain't none available unless you vaporize some on the source side.
Can't argue with your theory. Problem is that it doesn't apply here in any significant amount. If the pressure exceeds the cap pressure, it vents. The pump didn't add the pressure. In your scenario, the pump added heat. So, get back to the topic. In a closed car cooling system under normal operation can the pressure on the output side of the pump significantly exceed the pressure on the input side?
I think we can exclude any vaporization of the liquid leading to excess pressure. If it did, it would vent and, eventually, there'd be no more liquid to pump.
But thank you for the clue and the hint and the nitpicking.
The water pump can develop mabee a couple of pounds pressure on the outlet side into a plugged rad with the cap off. With the cap on, with no air in the system, significantly less. But at best, only a pound or two. Will running the water pump dead headed cause a temp increase? Sure, a very small amount. Inconsequential compared to the heat output of the engine. It will NOT cause a pressure increase in the closed system. The increased pressure increases the boiling point of the coolant. It also helps get and keep entrained air out of the cooling fluid. The only thing that causes the pressure rize is temperature. The pressure reduces back to atmospheric when the temperature drops back to room temperature. Current production vehicles have a "catch tank" that holds excess coolant if any is forced out to regulate the pressure, and it is drawn back in on cooldown.
Yes, and it looks like a lot of you have no idea how much it can, and does.
I'd like to hear the rationale in a closed system.
If the pump can't put more pressure on the output side, then what exactly makes the water flow? Why does the water flow faster the faster the pump runs? Even in an open system, like a pool, the system is under pressure, atmospheric pressure. In an engine cooling system, the overall pressure of the whole system will rise with temp, but there is still a pressure delta across the pump.
Yes, of course they do.
It's exactly the same rational as in an open system. If there is no pressure difference in a closed system, why would the pump move water at all?
It does have somewhere to get the water to pump, it's through the engine cooling system. And without a pressure difference, by what physics do you explain the movement of the coolant through that system?
Nonsense, the pump generates pressure just like any pump would.
It will cause a pressure increase on the output side of the pump versus in the input side.
The increased pressure increases the boiling point of the
I don't believe that's true. If you measured the system pressure with respect to the atmosphere right at the pump output, it should be higher than the pressure measured similarly at the pump input. Both those points would rise or fall with the temperature of the coolant.
Simply put, it's a pump and there is a restriction on the outlet side. An engine cooling system isn't a swimming pool with a pump in the middle of it.
On 5/1/2014 7:05 PM, mike wrote: ...
Fluid flowing or static? (Yes, it makes a difference.) The static pressure of a stagnant system would be uniform and steady, yes.
Nonense. The pump performs work on the fluid and while one doesn't lose fluid in a well-maintained automotive cooling system, it is not a constant volume system, it does have expansion volume thru the overflow and even water isn't totally incompressible and pressure does go up.
Pump principles are principles... There's this thing about "principles"--if they don't apply, then they aren't. :)
Nonsense, while running (and the thermostat open) it's circulating the water. It performs mechanical work on the fluid raising the output pressure in the process which it _must_ do in order to overcome the flow restrictions and pressure drops along the way before it gets back to the pump inlet again. Of course the physics of it is that the pump doesn't "create" pressure--it imparts kinetic energy from the impellers to the fluid to create velocity therein. It is the resistance to flow that creates the measured pressure even though it is commonly referred to "increasing (pressure) head".
The pressure is a variable value from a maximum at the pump outlet to a minimum at the pump inlet. The pressure at the radiator cap will be something under the 12 psi or so of the cap release spring or else it would lift.
So, while your statement above about the closed system and no water isn't right, the idea that a pump doesn't "create" pressure is correct; but the pressure is higher at the pump outlet as compared to the inlet because the input kinetic energy and output velocity thus imparted are reflected in a higher pressure owing to the design of the pump volutes. If didn't have a higher pressure at that point, then there'd be no driving force sufficient to get through the other downstream flow restrictions. It's analogous to needing voltage to "push" a current thru a load.
So what? Pumps are made to create pressure and force fluids past resistance in the system. A "restriction" is just like any other resistance in a piping system, eg elbows, the length of the pipe, etc.
An
Simply put, a pump is always a pump and it can only move fluid via a pressure difference between the intake and output side. Otherwise, explain the physics whereby a pump moves fluid without a pressure differential. You can't, not on this planet.
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