Sump pump setup
- Thread starter awais98
- Start date
bigpuffer1
New member
i did not notice any type of check valve on the pressure side of the pump. i also noticed your return "lockline" goes fairly deep in the display tank. a good 10 gallons could drain back to the sump in a pump or power failure. have you tested to make sure you don't overflow?
Kawi9_cf
Member
awais98
SPS
Hi Bigpuffer,
Thank you for your observation.
I did not put any check valve in the return line, you think that is a neccesity? I know it would help with decreasing the backflow in case of pump failure, but I also think it would decrease the flow?
Now after you pointed out, yes I see the lockline failrly deep, I will just adjust it, so it isnt too deep. Thank you.
Also, I have a safety drain that I drilled on the left side of the sump, which is 1.5 inch and that should drain the water in case it ever overflows. See the attached picture.
Thank you for your post and please keep up the good work of helping us newbies!
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awais98
SPS
Thats a very interesting article. Makes things so much clearer.
Thank you for sharing it with me
JonnyQuest
New member
The vapor pressure of the water will not noticeably change even if 100% saturated with O2, C02, N2 etc. and their effects can be ignored when determining the vapor pressure of salt water. If it had any significant effect on the vapor pressure there would be curves of vapor pressure vs. amount of dissolved gases in water for river/lake/sea water. The salt will only very slightly decrease the vapor pressure [increase NPSH(a)] from pure water and again can be ignored. If you want an example to properly calculate the NPSH(a) for aquarium water see the following posts.
http://www.reefcentral.com/forums/showthread.php?t=2239620
There is no energy penalty of using a pump to "lift" the water assuming the friction losses are the same.
High spots in a pump suction lines are avoided in industrial applications since vapor/air can get trapped in them. For example, your two 90° elbows going over the tank top. For aquariums it should not be an issue but some would place a small valve at the high point to remove any trapped air (pull a vacuum) and then close. It seems that you drilled the sump for your safety drain, why did you not drill it for you pump suction line? If you drill a new hole for your pump suction line then you don't ever have to worry about the siphon to your pump inlet.
I think most create a siphon break in the pump outlet plumbing so a check valve does not need to be used which would add unnecessary pressure losses and a potential failure point.
http://www.reefcentral.com/forums/showthread.php?t=2239620
There is no energy penalty of using a pump to "lift" the water assuming the friction losses are the same.
High spots in a pump suction lines are avoided in industrial applications since vapor/air can get trapped in them. For example, your two 90° elbows going over the tank top. For aquariums it should not be an issue but some would place a small valve at the high point to remove any trapped air (pull a vacuum) and then close. It seems that you drilled the sump for your safety drain, why did you not drill it for you pump suction line? If you drill a new hole for your pump suction line then you don't ever have to worry about the siphon to your pump inlet.
I think most create a siphon break in the pump outlet plumbing so a check valve does not need to be used which would add unnecessary pressure losses and a potential failure point.
awais98
SPS
That was very informative. Thank you Johnny quest.
Well you motivated me to drill a hole for the pump suction line, just finished it! Did not break anything! Phew.
Now I never have to worry abt siphon or priming!! I Feel better.
I did not drill initially coz I did not think that height of the Pump would be an issue with priming. And later I was afraid/ loathe to drain the water and drill, lest I would crack the glass.
I will upload a picture as soon as I'm on a laptop. Well here's t pic! I didn't know I could upload from iPad!!
Thank you
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sleepydoc
Team RC
I've never installed/used a centrifugal pump, so pardon my ignorance, but couldn't the sump just be placed below the water level of the sump? It should stay primed without a check valve then.
uncleof6
Active member
With the pump outlet above the water line in the sump, the NPSHa is a negative number, and a foot valve is required to maintain the prime on the pump. This is called "suction lift." Suction head is when the NPSHa is kept positive, e.g. the pump inlet is below the water line in the sump, keeping the pump inlet flooded, and no special considerations are needed to insure the pump stays primed.
With the pumps we use, we really don't want to install them in a suction lift scenario.
With the pumps we use, we really don't want to install them in a suction lift scenario.
awais98
SPS
New to the hobby, I 'assumed' it meant the same.....
JonnyQuest
New member
NPSH(a) is not negative number for 80°F water if the pump inlet is above the sump water line. Total Suction Head (Hs or Hl) is not the same as NPSH(a)! If NPSH(a) was always a negative number if the pump inlet was above the fluid level, a pump could never be placed higher than the level it was pumping from since NPSH(r) is always a positive number. This is obviously not the case.
JonnyQuest
New member
Obvious??
The post was completely inaccurate and needed to be corrected for the benefit of others.
It is also not the first time that NPSH(a) has been stated incorrectly.
The post was completely inaccurate and needed to be corrected for the benefit of others.
It is also not the first time that NPSH(a) has been stated incorrectly.
uncleof6
Active member
Most probably know that a pump can operate with a suction pressure that is less than atmospheric pressure (14.7psia.) It is referred to as a vacuum. On a pressure gauge it would be a negative number. HUH? That is totally incorrect--is it? 14.7psia = 0psig. (pounds per square inch absolute vs pounds per square inch gauge.) Well, a pressure qauge measures pressure--not vacuum, so stating so would be "stating the obvious." However, the pressure at the pump intake, expressed in psig, (which in fact it is,) can be negative. (a negative plus a negative is a positive, so it is added in the formula rather than subtracted) I am still to tired to deal with this.
Keeping the NPSHa positive rather than negative. It is a play on words. In this hobby there may be one or two people really interested in this, and they can look it up if they want to pursue it--IF they want, the rest want to know what they can get away with. Put plainly, keep the pump below the water line in the sump, unless you want large plumbing, and a bigger pump to do the same job.
Keeping the NPSHa positive rather than negative. It is a play on words. In this hobby there may be one or two people really interested in this, and they can look it up if they want to pursue it--IF they want, the rest want to know what they can get away with. Put plainly, keep the pump below the water line in the sump, unless you want large plumbing, and a bigger pump to do the same job.
JonnyQuest
New member
To the 1 or 2 people that may be interested:
Based on the rant of absolute and gauge pressures, the point of my post was obviously missed. I guess bolding was insufficient.
Total Suction Head (Hs or Hl) is not the same as NPSH(a)!
It is also clear that the units of NPSH(a) are not understood as well.
Pump Suction Pressure (Total Suction Head) is a pressure measurement and is not interchangeable with NPSH(a). Net Positive Suction Head Available [NPSH(a)], translation, the available pressure drop from the pump inlet before the fluid will vaporize, expressed in terms of fluid head. Pressure drop, NPSH(a) and NPSH(r) are never expressed by using the terms absolute or gauge as they do not apply. NPSH(a) can't be directly measured with a pressure gauge, while the pump suction pressure can.
For example, two identical systems utilizing water will have the same pump suction pressure and can have very different NPSH(a) values if the water temperatures are different between the two systems. The higher the water temperature, the higher the vapor pressure is, the closer the water is to the boiling point, the less pressure drop is available before it will vaporize and therefore the lower the NPSH(a) value. Clearly the Pump Suction Pressure and NPSH(a) values are very different and with different units.
The NPSH(a) for 80°F water when the pump is elevated (to a limit) above the water line is positive, not some imaginary negative gauge value. The pump suction pressure will be a negative gauge pressure but the NPSH(a) will not be a negative number.
I am curious on an explanation why a pump that is elevated above the water line will need to be bigger.
Based on the rant of absolute and gauge pressures, the point of my post was obviously missed. I guess bolding was insufficient.
Total Suction Head (Hs or Hl) is not the same as NPSH(a)!
It is also clear that the units of NPSH(a) are not understood as well.
Pump Suction Pressure (Total Suction Head) is a pressure measurement and is not interchangeable with NPSH(a). Net Positive Suction Head Available [NPSH(a)], translation, the available pressure drop from the pump inlet before the fluid will vaporize, expressed in terms of fluid head. Pressure drop, NPSH(a) and NPSH(r) are never expressed by using the terms absolute or gauge as they do not apply. NPSH(a) can't be directly measured with a pressure gauge, while the pump suction pressure can.
For example, two identical systems utilizing water will have the same pump suction pressure and can have very different NPSH(a) values if the water temperatures are different between the two systems. The higher the water temperature, the higher the vapor pressure is, the closer the water is to the boiling point, the less pressure drop is available before it will vaporize and therefore the lower the NPSH(a) value. Clearly the Pump Suction Pressure and NPSH(a) values are very different and with different units.
The NPSH(a) for 80°F water when the pump is elevated (to a limit) above the water line is positive, not some imaginary negative gauge value. The pump suction pressure will be a negative gauge pressure but the NPSH(a) will not be a negative number.
I am curious on an explanation why a pump that is elevated above the water line will need to be bigger.
uncleof6
Active member
So looking it up on wiki is not needed: You need to flow x gallons to the tank. The pump will flow x gallons at y total dynamic head. (from the flow curve of the pump.) The total dynamic head on the system is also = y. So the system is well matched. Due to a brain cloud you figure the same pump will flow the same if you drop the suction vessel below the pump suction. Wrong. Suction head is head pressure - vapor pressure yes? Worst case would seem to be 14.7 psia (0psig) at the pump suction. But wait, we have introduced a negative number, because the suction vessel water level is below the pump suction. A negative and a negative is a positive so it is added to the vapor pressure. A vacuum is needed to bring the water up. (<14.7psia) Anyway who cares.
What happens is you use part of the pumps capability to raise the water up to the suction intake. Therefore you have increased the total dynamic head on the system. So rather than flowing x gph at y' total dynamic head the pump will lose "oomph" on the flow curve or flow w gph at z' of total dynamic head. So you need a bigger pump that will flow x gph at z' of total dynamic head. The same pump is not going to do it, no matter how many times you read wiki. Practical application.
Another consideration is we are not dealing with water. We are dealing with salt water (sea water) with a high concentration of dissolved solids and entrained gases. The vapor pressure of salt water is higher (lower boiling temperature.)
Happy reefing.
What happens is you use part of the pumps capability to raise the water up to the suction intake. Therefore you have increased the total dynamic head on the system. So rather than flowing x gph at y' total dynamic head the pump will lose "oomph" on the flow curve or flow w gph at z' of total dynamic head. So you need a bigger pump that will flow x gph at z' of total dynamic head. The same pump is not going to do it, no matter how many times you read wiki. Practical application.
Another consideration is we are not dealing with water. We are dealing with salt water (sea water) with a high concentration of dissolved solids and entrained gases. The vapor pressure of salt water is higher (lower boiling temperature.)
Happy reefing.
JonnyQuest
New member
Um, No. Pump suction head = pressure at the liquid surface (atmospheric pressure) + static liquid head at pump inlet - dynamic head losses to the pump inlet.
There is just one fatal flaw of your "logic". For both cases the head generated by the pump is y'. For the case where the pump is above the sump water level, there indeed is work being done "lifting" the water to the inlet but it is exactly offset by the reduction in work the pump does on the outlet side to get the water to the tank level since it already "lifted" it part of the way on the inlet side.
Let's look at two cases pumping X gpm where the dynamic head losses (inlet + outlet) of the pump are the same between the cases. The sump water level is at the reference level (value 0) and the tank water level is 10 ft. above the sump water level. Both the sump and tank are exposed to atmospheric pressure.
Case 1 - The pump inlet is below the sump water line by 2ft.
The static head at the inlet is therefore +2ft. and the outlet is +12ft. Therefore the static portion of the head generated by the pump is 10ft. [12 - 2].
Case 2 - The pump is 2ft above the sump water line.
The static head at the inlet is -2ft. and the outlet is +8ft. Therefore the static portion of the head generated by the pump is exactly the same 10 ft. [8 - (-2)].
Since the dynamic head loses, change in static head and flow rates are the same for the two cases, the pump is performing the exact same amount of work.
Let's look at the same problem based on the potential energy of the water. The water changed energy equivalent to raising it 10ft. for both cases. If the pump required more work to change the water potential energy by the same amount, where did this imaginary extra work go if both systems have the same dynamic head losses?
Indeed the exact same pump will work. The laws of physics don't change just because the pump is located above the sump water level.
However, there will be a difference if the pump that is placed above the water level requires a foot valve verses a pump that is placed below the water level that does not. Then there is difference in dynamic head because of the extra pressure drop of the foot valve, not simply because the pump is above the sump water level.
I was just curious of someone's "logic" that breaks every law that deals with the conservation energy.
This was already addressed.
Wrong again. The vapor pressure of salt water is slightly lower which has a slightly higher boiling point than pure water.
As a lead process (chemical) engineer that designs (parts of) multi-billion dollar refineries and petrochemical plants I only need to use my brain for these relatively simple concepts and have plenty of practical applications using very large industrial pumps (~50hp).
