Pumps and peak velocity

myv65 · 10-17-2002, 12:26 PM

There's been a few discussions lately about maximizing velocity within a water cooling system. I'd like to point out that it is a simple matter to calculate the absolute maximum velocity that a given pump can produce. This information is really of limited use, as we'll see shortly, but may prove interesting to some.

Pumps all have a dead-head pressure limit. This is the static pressure they'll generate when zero flow occurs. Static pressure and velocity are simply different forms of energy with the former representing potential and the latter representing dynamic.

Ignoring losses, you can equate head with velocity using the same equations for free-fall of an object. Namely that distance equals 1/2 * acceleration * time^2. If you have a pump that dead-heads at 1 meter, then you use the equation to solve for time. Time equals (distance * 2 / acceleration)^(1/2), so time equals (1 * 2 / 9.81)^(1/2) = 0.45 seconds. Velocity equals time * acceleration, so maximum theoretical velocity equals 0.45 * 9.81 = 4.42 meters/second.

Using dead head levels of 1.5 and 2 meters yields 5.42 and 6.26 meters/second.

OK, so what does this really mean? If you dead-head the pump and pop a hole at the same elevation as the pump outlet, you'll never see a velocity higher than the equations show. The actual velocity you would see is a function of the hole. Standard engineering charts define head loss for various types of orifice and entrace conditions. In general, you'll never get more than about 70% of the value shown by the equations above. More realistically, a number around 50% would be more accurate.

I said earlier this information is of limited use. These conditions exist only when a pump is dead-headed. As soon as you allow some flow, the maximum velocity you'll ever see begins to drop. For a given pump, you could run tests to determine the maximum obtainable velocity versus flowrate. It will vary with flowrate and obviously vary for different pumps. The only thing you can easily state with confidence is that a given pump will never exceed X.XX meters/second.

Just something to keep in mind as you try to balance flow restrictions/velocity increases with your particular pumps and an easy way to peg absolute limits that you can't physically attain.

Alchemy · 11-03-2002, 01:50 AM

Might be helpful to show the excerpt of the Bernoulli equation that you probably got this from initially:

V = SQRT( 2 * g * TSH)

Where SQRT is "Square Root of"
g = acceleration of gravity, 9.8 m/sec^2 or 32.2 ft/sec^2
TSH = total static head, aka pump head

Note also that this only applies for accelerating fluid that is initially stationary, such as when the pump is drawing water from a reservoir. You can get faster velocities with an in-line system since the water doesn't have to be accelerated, it just has to overcome friction.

Where did you get the "50% to 70%" value?

Alchemy

myv65 · 11-03-2002, 09:03 AM

The 50-70% is just a rough ballpark for "typical" entrance/exit coefficients. From the content of your previous posts, I'm certain you already know that if you have a volume of water under a certain pressure and "pop a hole" in the container, the velocity of the water leaving the container will be less than 100% of the potential due to pressure.

10-17-2002, 12:26 PM	#1
myv65 Cooling Savant Join Date: May 2002 Location: home Posts: 365	Pumps and peak velocity There's been a few discussions lately about maximizing velocity within a water cooling system. I'd like to point out that it is a simple matter to calculate the absolute maximum velocity that a given pump can produce. This information is really of limited use, as we'll see shortly, but may prove interesting to some. Pumps all have a dead-head pressure limit. This is the static pressure they'll generate when zero flow occurs. Static pressure and velocity are simply different forms of energy with the former representing potential and the latter representing dynamic. Ignoring losses, you can equate head with velocity using the same equations for free-fall of an object. Namely that distance equals 1/2 * acceleration * time^2. If you have a pump that dead-heads at 1 meter, then you use the equation to solve for time. Time equals (distance * 2 / acceleration)^(1/2), so time equals (1 * 2 / 9.81)^(1/2) = 0.45 seconds. Velocity equals time * acceleration, so maximum theoretical velocity equals 0.45 * 9.81 = 4.42 meters/second. Using dead head levels of 1.5 and 2 meters yields 5.42 and 6.26 meters/second. OK, so what does this really mean? If you dead-head the pump and pop a hole at the same elevation as the pump outlet, you'll never see a velocity higher than the equations show. The actual velocity you would see is a function of the hole. Standard engineering charts define head loss for various types of orifice and entrace conditions. In general, you'll never get more than about 70% of the value shown by the equations above. More realistically, a number around 50% would be more accurate. I said earlier this information is of limited use. These conditions exist only when a pump is dead-headed. As soon as you allow some flow, the maximum velocity you'll ever see begins to drop. For a given pump, you could run tests to determine the maximum obtainable velocity versus flowrate. It will vary with flowrate and obviously vary for different pumps. The only thing you can easily state with confidence is that a given pump will never exceed X.XX meters/second. Just something to keep in mind as you try to balance flow restrictions/velocity increases with your particular pumps and an easy way to peg absolute limits that you can't physically attain.

11-03-2002, 01:50 AM	#2
Alchemy Cooling Savant Join Date: Oct 2002 Location: Boston Posts: 238	Might be helpful to show the excerpt of the Bernoulli equation that you probably got this from initially: V = SQRT( 2 * g * TSH) Where SQRT is "Square Root of" g = acceleration of gravity, 9.8 m/sec^2 or 32.2 ft/sec^2 TSH = total static head, aka pump head Note also that this only applies for accelerating fluid that is initially stationary, such as when the pump is drawing water from a reservoir. You can get faster velocities with an in-line system since the water doesn't have to be accelerated, it just has to overcome friction. Where did you get the "50% to 70%" value? Alchemy

11-03-2002, 09:03 AM	#3
myv65 Cooling Savant Join Date: May 2002 Location: home Posts: 365	The 50-70% is just a rough ballpark for "typical" entrance/exit coefficients. From the content of your previous posts, I'm certain you already know that if you have a volume of water under a certain pressure and "pop a hole" in the container, the velocity of the water leaving the container will be less than 100% of the potential due to pressure.

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