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post #11 of 21
You are talking about this bit of information they are giving?

"DDC+ w/ Petra's Top: 12.00V -- 8.1psig (18.68 feet) | 13.80V -- 10.1psig (23.3 feet)"

I believe they are just converting the psi to feet of head....
So at 13.8V it has 23.3 feet of head, or 10.1psi. (23.3 x 0.433 = 10.1)

Another thing to note probably, is that the psi reading you are getting on your rig is a flowing condition, meaning it is the pressure of flowing water. The max head reading they have in that test is a non flowing condition. (You can see in the picture where they are pinching the tube so it doesn't flow. Thats why its the "max". 0 flow, all pressure, in the business we call that pressure "churn pressure", the pressure a pump produces at a zero flow condition.
Pinch your tubing right after our gauge and see if the gauge gives a different reading.

edit:
Quote:
Originally Posted by ira-k View Post
EDIT:[/B] Never mind I see now....Thats just his measurement of the Psi at max head....I was looking at it wrong...
Not the max psi AT max head.
They are the same reading, just expressed differently. Like saying Psi, or Bar. ie. 1 Bar (14.5 psi) same reading different unit of expression...
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post #12 of 21
I know now... I was looking at it wrong ....Thanks..I was confusing loop pressure with head pressure....That cleared it up for me......Thats why I put a gauge in it...If it starts getting plugged up the Psi will rise..
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post #13 of 21
Ira-K is right. Head is not pressure. Pressure has an area component (as in Pounds per Square Inch.) Head has no area component.

Head is a measure of the pump’s Specific Energy. If you raise a pound of lead one foot off of a table, you have 1 foot-lbs of energy. That is Specific Energy. If you swap the pound of lead for a pound of feathers, you still have one foot-lbs of energy. That is why the head rating of a pump doesn’t change with the fluid.

You can apply the head rating of a pump to a particular fluid to get a pressure rating for that fluid. Of course, we are concerned with water. But if you apply the same head rating to oil or to a very light fluid, you’ll get different PSI ratings.

An easy way to think about the relationship is to think of a shaft spinning at, say 1000 RPM. You can’t answer the question, “how far do I travel per minute†without adding a wheel to the shaft. The wheel provides the missing information (diameter) needed to answer the question. Of course, the answer will change depending on the size of the wheel. Head would be the “1000 RPMâ€, water would be the “wheelâ€, and Specific Gravity of the fluid would be “diameter of the wheel.â€

Mastacator, I believe you misread the chart. Power consumption goes down as head increases and flow decreases. That’s because power consumption is based on how much water is moved…not the head. The pump *always* produces the same amount of head (energy) regardless of flow. The only question is how much of that head goes to moving water and how much goes to overcoming friction. Adding the two for any given loop will always equal the total head of the pump.
post #14 of 21
GrayStar is there any way to figure the amount of head I have used with my in-line pressure gauge? I'm using the 18W DDC W/Petra's top and 5 'of 7/16 tubing..My pressure gauge is reading 5 Psi...Thanks
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post #15 of 21
Quote:
Originally Posted by ira-k View Post
Thats probably what it is, plus I left my tubing a little over long on my rad also....The first test there for max head....Am I reading it wrong or what? Isn't that showing the Psi it takes to get to max head? Thanks EDIT: Never mind I see now....Thats just his measurement of the Psi at max head....I was looking at it wrong...
You don’t use an amount of head. The pump produces head (remember head is energy) and all that energy is consumed in the loop.

What’s really important to know is your flow rate...which you should be able to determine from a pressure measurement.

Pressure is a force applied upon an area. In a typical water-cooling loop the pressure will start at a positive level at the pump outlet (pushing against the tubing,) and then drop as the water travels, and end at a negative level at the inlet (sucking the tubing in.) For any pump, the difference between the inlet pressure and outlet pressure should always equal the head rating converted to pressure.

At zero flow, the pressure at the outlet is the max and the pressure at the inlet is zero. At free flow it’s reversed...the outlet reads zero and the inlet is –max (suction.) Therefore, the pressure reading of the outlet, converted to head, will correlate directly to a flow rate on the P-Q curve.

5 PSI of water equals 11.53 ft of head. Petra hasn’t provided a P-Q chart for his top but I can guesstimate that you’re somewhere in the 1.4 GPM range…probably about 0.1 GPM more than a standard DDC+. Since you’re well above the 1 GPM mark that I personally recommend for a CPU-only loop, you can definitely consider pump flow to be a non-issue in your loop.

You can perform your own conversion of head to pressure at http://www.onlineconversion.com/ under the Pressure section.

If you want to know what the pressure drop is across the components, that's simply 9.971 - 5. You can convert that to a head rating if you like, which is 11.46. That would be the friction head of the loop.
post #16 of 21
Thanks a lot....I only have a cpu in it and I'm getting 1.4 flow....Great!!! I put the gauge in so I could tell if the loop was plugging up and to tell when it's bleed, the needle will stop bouncing when it is...I have that book-marked...Thanks a lot for the formula....Now if I can get other people to use a gauge we can compare our flow through different loops, all different of course but it could give us a general idea any way....Thanks Again I really appreciate you taking the time to explain this.....
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post #17 of 21
Quote:
Originally Posted by ira-k View Post
Thanks a lot....I only have a cpu in it and I'm getting 1.4 flow....Great!!! I put the gauge in so I could tell if the loop was plugging up and to tell when it's bleed, the needle will stop bouncing when it is...I have that book-marked...Thanks a lot for the formula....Now if I can get other people to use a gauge we can compare our flow through different loops, all different of course but it could give us a general idea any way....Thanks Again I really appreciate you taking the time to explain this.....
No prob! Remember that the guage has to be right at the outlet for best accuracy.
post #18 of 21
Quote:
Originally Posted by Graystar View Post
No prob! Remember that the guage has to be right at the outlet for best accuracy.
I have it right after the pump and before the rad....Thanks
LL
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post #19 of 21
Quote:
Originally Posted by Graystar View Post
Ira-K is right. Head is not pressure. Pressure has an area component (as in Pounds per Square Inch.) Head has no area component.

Head is a measure of the pump’s Specific Energy. If you raise a pound of lead one foot off of a table, you have 1 foot-lbs of energy. That is Specific Energy. If you swap the pound of lead for a pound of feathers, you still have one foot-lbs of energy. That is why the head rating of a pump doesn’t change with the fluid.

You can apply the head rating of a pump to a particular fluid to get a pressure rating for that fluid. Of course, we are concerned with water. But if you apply the same head rating to oil or to a very light fluid, you’ll get different PSI ratings.

An easy way to think about the relationship is to think of a shaft spinning at, say 1000 RPM. You can’t answer the question, “how far do I travel per minute†without adding a wheel to the shaft. The wheel provides the missing information (diameter) needed to answer the question. Of course, the answer will change depending on the size of the wheel. Head would be the “1000 RPMâ€, water would be the “wheelâ€, and Specific Gravity of the fluid would be “diameter of the wheel.â€

Mastacator, I believe you misread the chart. Power consumption goes down as head increases and flow decreases. That’s because power consumption is based on how much water is moved…not the head. The pump *always* produces the same amount of head (energy) regardless of flow. The only question is how much of that head goes to moving water and how much goes to overcoming friction. Adding the two for any given loop will always equal the total head of the pump.
Head is a term that indicates the measure of pressure in excess of atmospheric pressure at a point in a column of fluid. Head is pressure, regardless of an area component. Head is normally measured in feet of height (or meters), because a column of fluid would be thought of as vertical, and therefore is measured in feet(meters, etc) of height, but there would be pressure exerted at the base of that column and therefore can get converted to a unit having an “area componentâ€.
You are correct sir, in that, converting head to Psi for a fluid will be dependant on the fluid itself. As stated, the resultant Psi you will get is dependant on the specific gravity of the fluid in question. In my previous posts for converting head to Psi I used 0.433, as that is the specific gravity of fresh water, since water is what we were referring to.
You are also correct in your analogy of using “foot-lbs†for specific energy.
Comparing a 2†diameter pipe 10’ high vs a 4†diameter pipe 10’ high, may have been a little more relavent….. The Feet of Head of each column would be same also because of their height. The Psi at the base would be the same, given both columns of fluid were the same fluid, due to the fluids specific gravity, regardless of the “area component†or diameter of the pipe.
Still your specific energy, or total energy, would be best represented by: pressure head + elevation head + velocity head, not simply “headâ€.
I’m with you on whether it’s a pound of lead or a pound of lead, its still a pound…But, the head actually produced by a pump will change with the fluid being pumped. The rating of a pump is based on the fluid the pump is designed to move. A water pump rated at 20 feet of head will have less/more head if used to pump a fluid with a different specific gravity.
Your analogy of the spinning shaft is a good one….. Problem is, you can still travel the same distance in the same amount of time with a different RPM (head, as stated by you) and a different diameter of wheel (specific gravity). And so IMO not really good for making a relationship between the components.

I may have read the chart wrong for the power sonsumption, if the "dashed line" is the curve for that measurement, there's no key to that line. I see it correctly, thanks.

Its clear we both understand feet of head. And its good we could help out Ira-k. But its probably a good idea to not high jack this thread any longer.
Well done Graystar!
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post #20 of 21
If you open a fluid engineering textbook and look up the definition of head, you’ll find something like this:

“Specific energy or energy per unit weight of fluidâ€

The same textbook may define pressure as:

“The application of external or internal forces to a bodyâ€

Head is energy; pressure is force. Energy is the capacity of a system to do work, and work is defined as the application of a force. These two physical properties are closely related, but are not considered the same. Atmospheric pressure is not a factor in calculating the head of a pump.

Quote:
Originally Posted by Mastacator View Post
Still your specific energy, or total energy, would be best represented by: pressure head + elevation head + velocity head, not simply “headâ€.
This statement doesn’t make sense. What you described is the head of a system. That does not best represent the head of a pump, which is what we’re concerned with. You can calculate the head of a pump by subtracting the total energy at the inlet from the outlet. Maybe that’s what you were referring to? In any case, in closed systems such as our PC water-cooling loops, 100% of the resistance to flow is friction head. All other forms of head have a value of zero. System head will always equal the pump head.

Quote:
Originally Posted by Mastacator View Post
Your analogy of the spinning shaft is a good one….. Problem is, you can still travel the same distance in the same amount of time with a different RPM (head, as stated by you) and a different diameter of wheel (specific gravity).
Right. But don’t forget that the analogy equated distance with pressure. A low head/heavy fluid will give the same pressure as a high head/light fluid...seems just fine to me.
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