Placing a temperature probe

Jump to latest Follow Reply
G

Gepool

Active member
May 3, 2022
34
-
Hello all! Is it OK to put my thermal probe (which is measuring my swimming pool's temperature) AFTER the pump and AFTER the sand filter?

I am asking because if I need to place it between the pump and and the sand filter I will need a plumber, otherwise is is easier for me.
 
Gepool said:
Hello all! Is it OK to put my thermal probe (which is measuring my swimming pool's temperature) AFTER the pump and AFTER the sand filter?

I am asking because if I need to place it between the pump and and the sand filter I will need a plumber, otherwise is is easier for me.
Click to expand...
Depends. Do you have a heater? Do you have solar heating?
 
Gepool said:
Hello all! Is it OK to put my thermal probe (which is measuring my swimming pool's temperature) AFTER the pump and AFTER the sand filter?

I am asking because if I need to place it between the pump and and the sand filter I will need a plumber, otherwise is is easier for me.
Click to expand...
It's best practice to install before the filter. Why does this require a plumber in your case?

Many probes simply require a drill bit and a hose clamp and can go anywhere including in a fitting (though not ideal).

You may get inaccurate reads. Sand filters particularly may have a thermal impact on your water due to friction.
 
bradgray said:
It's best practice to install before the filter. Why does this require a plumber in your case?

Many probes simply require a drill bit and a hose clamp and can go anywhere including in a fitting (though not ideal).

You may get inaccurate reads. Sand filters particularly may have a thermal impact on your water due to friction.
Click to expand...
Unfortunately I dont have a photo right now, but the piping before the filter is very-very short and the only option is either to extend it or drill on fitting, which I dont like to do... Do you think the impact will be above 1-2C? I can compensate that loss in the sensor reading...
 
If your pump uses 500 watts at the impeller to move 50 gpm, the total heat input is 0.5 kWh in one hour, which is 1,706 BTU.

50 gpm is 3,000 gph, which is 25,000 lbs of water.

So, the maximum temp rise from skimmer to return from friction through the entire system would be 1,706/25,000 = 0.06824 degrees Fahrenheit.
 
  • Like
Reactions: Gepool and HermanTX
Here's an example of 3,000 watts, which is 10,236 btu per hr.

120 gpm is 7,200 gph, which is 60,000 lbs. per hour.

So that is 10,236 btu/60,000 lb = 0.1702 degrees Fahrenheit maximum temperature rise possible due to friction from intake to return.

1674156320286.png
 
  • Like
Reactions: Gepool

Enjoying this content?

Support TFP with a donation.

Give Support
The head loss through a TR-140 sand filter at 120 gpm is 14.7 feet.

Note: This seems too low and I suspect that they might have meant PSI instead of feet on the graph.

That equates to a gravitational potential energy = 1.196 MJ (megajoules) = 332.2 W h (watt hours) = 0.3322 kW h (kilowatt hours) = 1,133.6 BTU_IT (IT British thermal units)

1,133.6/60,000 lb = 0.01889 degrees Fahrenheit.



1674157762508.png
 
Last edited:
JamesW said:
The head loss through a TR-140 sand filter at 120 gpm is 14.7 feet.
Click to expand...
That's only 6.36 psi and that is low for a sand filter at 120 gpm.

A more realistic number is 30 psi, which is 69.3 feet, which increases the gravitational potential energy to 5,344 BTU, which could add 0.089 degrees Fahrenheit.
 
  • Like
Reactions: BC78
JamesW said:
Here's an example of 3,000 watts, which is 10,236 btu per hr.

120 gpm is 7,200 gph, which is 60,000 lbs. per hour.

So that is 10,236 btu/60,000 lb = 0.1702 degrees Fahrenheit maximum temperature rise possible due to friction from intake to return.

View attachment 469794
Click to expand...
Geeze @JamesW , excellent response as always.

Thanks for clarifying an anecdotal mistruth I've always believed. I'll send this to the Hayward rep that convinced me of it!
 
  • Like
Reactions: JamesW
The TDH (Total Dynamic Head) of a system is related to amount of power required to move the water through a system at a specific flow rate.

For example, a flow of 120 gpm at 67 feet of head requires about 1.514 kW (kilowatts).

In the graph, we can see that the pump actually uses 3.0 KW.

So, the pump uses 3,000 watts (4.023 HP), to deliver 1,514 watts of power (2.0303 hp) to actually move the water.

The efficiency is 1.514 kW/3 kW = 0.505 = 50.5%.

The 1,514 watts is the actual power required to move water through the system due to friction.

The rest (1,486 watts) is due to inefficiency and energy loss from things like motor heat etc.

1,514 watts is 5,166 btu/hr, which could add 0.0861 degrees Fahrenheit to the water at 120 gpm if it all turned into heat.

3,000 watts is 10236 btu/hr, which could add 0.1706 degrees Fahrenheit to the water at 120 gpm if it all turned into heat.

Below is for the IntelliFlo, which shows a THP (Total Horsepower) of 3.95, but a HHP (Hydraulic Horsepower) of 1.88, which is 1.88/3.95 = 0.47595 = 47.595% efficiency.

The full load amps is 16 at 230 volts, which is 3,680 watts (3.680 KW) compared to the 3.2 KW shown.

3,680 watts is 4.92 HP compared to the 3.95 shown, so they have some leeway in how they report their metrics.

1674225634026.png
1674226706286.png
 
Last edited:
  • Like
Reactions: BC78
Total Head is a combination of static head and dynamic head.

Static head is the difference from the water surface where you pull from (vacuum/suction) to the water surface where you return to.

The elevation of the suction and return ports is not relevant.

For most systems, there is no net static head because the water comes from the pool and returns to the pool.

If there is an elevation change like pulling from a trough and returning to a pool for an infinity edge, then there is a static head loss.

Dynamic head is the resistance due to friction as the water gets pushed through the system.

The Total head is usually just one number as if the elevation change was the only difference with no pipe friction.

So, a Total head of 67 feet is the static and dynamic.

For example if you have two bodies of water and 67 feet of elevation difference in surface level, you are lifting 60,000 lbs of water 67 feet every hour.

The amount of energy it takes is based on the difference in potential energy of 60,000 lbs of water at a height difference of 67 feet.

Note: Assumes a constant height difference like for a spillover infinity edge or if both bodies of water were sufficiently large so as not to change levels.


1674244390921.png
1674246116997.png

http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html

1674246332580.png

https://www.calculatorsoup.com/calculators/physics/gravitational-potential.php

You also have to account for the energy needed to get the water moving from being at rest in the pool to moving at some speed in the pipes (kinetic energy).

For example, at 7 ft/sec, the kinetic energy is 58.67 BTU.

5,165.95 BTU + 58.67 BTU = 5,224.62 btu per hour at 120 gpm.

1674249418662.png

https://www.calculatorsoup.com/calculators/physics/kinetic.php
 
Last edited:
  • Like
Reactions: BC78
You must log in or register to reply here.
Thread Status
Hello , This thread has been inactive for over 60 days. New postings here are unlikely to be seen or responded to by other members. For better visibility, consider Starting A New Thread.
Top Bottom
Back
Jump to latest Follow Reply