# Washington Hp Home DIY Geothermal

Discussion in 'Geothermal Heat Pump Testimonials' started by Hp Home, Dec 30, 2015.

1. ### docjenserWell-Known MemberIndustry ProfessionalForum Leader

Mark, you told someone here asking for help that a 5th grader can do the math, and that you have 35 years of experience. Waiting for your answer on the math question!

2. ### Mark CustisNot soon.Industry ProfessionalForum Leader

Enjoy your wait. You are not my boss.

3. ### Hp HomeMember

I'm confused why unbalanced circuits and balancing valves is a bad thing?

I did room by room heat loss calculations and estimated radiant walls output at 20 btu/sf. The radiant tubing lengths are proportional to the individual room loads and surface area of the heat emitter.

I think for the pump sizing I need to be looking at 9 gpm at 15 ft of head. A single Grundfos Alpha is not even close. So I am open to suggestions on the most efficient solution.

4. ### Mark CustisNot soon.Industry ProfessionalForum Leader

Well with actuators on each zone we do not need the balancing valve as much. Some people think they cause too much flow restriction.

5. ### Hp HomeMember

I guess I had the vocabulary confused. No balancing valve then, just the actuators.

I am beginning to think what this needs is 2 pumps for the radiant. One for the slab downstairs (~6 gpm @ 10') and one for the walls upstairs (~3 gpm @ 15').

6. ### docjenserWell-Known MemberIndustry ProfessionalForum Leader

Again,

balancing valves are creating a flow restriction, even if they are wide open. Thus you need more pumping power, bigger pumps. The flow circuit with the highest pressure drop will determine the pressure you need to achieve adequate flow in your circuits. Why would you put a balance valve on it to obstruct it even further? Specifically with a variable speed circulator which you now have to set higher.
Check the Wilo Stratos pumps as an alternative.

HP: 20 btus/sf at what temperature?

No need for the second pump. Just a constant pressure variable speed will do.

Not sure where you get the the 15 ft/hd from? Keep one thing in mind, even if one circuit sees lesser flow, the water flows through it slower, thus loosing more heat to the space.
Also don't you have 2 zones upstairs, one with a 20' header, the other one with a 100' header? One way, correct?

Mark, how is the 5th grader math coming? Any chance you get this done? Sounded very easy when you described it......
I know I am not your boss. I actually know that for a fact.

7. ### docjenserWell-Known MemberIndustry ProfessionalForum Leader

HP, what is the CV of the 3 zones valves you'll be using?

8. ### Hp HomeMember

20 btu/sf at a delta T of 10F. This is where I got frustrated with the formula. Seems to me like 160-150 delta is going to be moving more heat than 90-80 delta but everything I read says it doesn't matter. Anyways I got as much tubing in each room as I could fit in the interior walls and it worked out pretty proportional to the room loads.

15' of head is my attempt at math. But it's only as good as the numbers input to the equation.

Upstairs is 6 zones on two 3-circuit manifolds. The 100' and 20' is round trip total effective length. Not sure the Cv of the valves, they came as part of the generic preassembled manifolds.

9. ### docjenserWell-Known MemberIndustry ProfessionalForum Leader

The BTU/sf are at a certain temperature of entering water, and assume a certain flow.

The delta T is the temperature the water looses when going through the radiant system, either one circuit or the whole system. So water going in at 90F and leaving at 80F has a delta t of 10F. Now water going in at 160F and going out at 150F also looses 10F, and at the same flow rate delivers the same amount of BTUs.

Now, if the room is 70F and the temperature difference to the water at 90F is much less than when the water has 160F going in, that is capacity what the hotter water enables. In other words, given the same flow, a radiator at 160F will loose more heat (delta T) because the temp difference between the radiator and the room is higher.

So higher temperature matters, because the hotter water can loose more temperature (delta T) than the closer water.

Now flow comes into the game. If the water looses 10F at 3 gpm it delivers 15,000 BTUs. If the water looses 3F at 10 gpm it also delivers 15,000 BTUs. (deltaT x gpm x 500).

So the 20btu/sf is the output at a certain entering water temperature assuming that the temperature will remain stable, that why it also called the average water temp (within the radiator).

So you have already the manifolds for the circuits? With balancing valves? Now, did the manifolds also come with motorized zone valves? Need their CV. Do you plan to have 2 zone valves opening up at the same time for the upstairs zone?

10. ### Hp HomeMember

Thank you I understand now why the heat emitted by 10f delta T is the same at different temp levels. Room temp and flow.

I do not have motorized zone valves. The manifolds look like this-

And each circuit has actuators like this-

11. ### Hp HomeMember

Doing some math I now come up with a very different answer for head loss.

With so many variables to pin down its like trying to hit a moving target.

Here is what I came up with.

To calculate for the longest loop there is 3 segments- 1. 250' of 1/2" pex, 2. 100' of 1" pex, 3. 20' of 1.25" copper

So I calculated for each segment using H=(acL)(f^1.75) and added the three results.

a= fluid loss factor, I used .055 for 110F water
c= pipe size coefficient
L= total effective length of pipe
f= flow rate in gpm

1. 250' of 1/2" pex

f=1,604 btu/5000 = 0.3208 gpm

H= 0.055 x 0.71213 x 250 x (0.3208 ^ 1.75) = 1.338

2. 100' of 1" pex

f= 3,000 btu/ 5000 = 0.6 gpm

H=0.055 x 0.04318 x 100 x (0.6 ^ 1.75) = 0.0971

3. 20' of 1.25" copper

f= 9 gpm

H= 0.055 x 0.0068082 x 20 x (9 ^ 1.75) = 0.3502

1.338 + 0.0971 + 0.3502 = 1.7853 feet of head.

Flow rate is the tricky variable. For example if I use the f= btu/(500 x delta T) formula for the 1.25" copper-

28,000 btu/5000 = 5.6 gpm

But I know the three ton system should flow 9 gpm at full load and I know the heat pump puts out 28,000 btuh and 500 is a constant so I solve for delta T-

Delta T = 28,000 btuh/ (500 x 9) = 6.22F

Do I now start over, recalculating flow rates based on a 6.22F delta T?

12. ### docjenserWell-Known MemberIndustry ProfessionalForum Leader

You are getting there. The 6.22 F delta t is the maximum delta at incoming water temp of 32F for your whole radiant system. Now, sometimes the heat pump can produce more BTUs, when the loop outside is warmer because you are not in the middle of the winter yet. But in general, pushing more water through your radiant will only result in a lesser delta T (and therefore wastes pumping power) since the amount of heat you can transfer is pretty much limited by how much heat the heat pump can make.
So I would say that 6-7F delta is your magic number you should try to shoot for (for your radiant as a whole).

Now your problem is balancing, since you have different flow due to different pipe length and different circuits. You don't want to have 7 out of 9 gallons go to the upstairs space, and the slab only gets 2 gallons (not the case here, I just give an example).

Your flow rates for each circuit and sub circuit are all over, so your delta t for each circuit will be all over, unless you apply flow restrictions, which then also increase your pumping power. But most of the time it is not necessary if you understand who things can be self balancing. Sometimes you don't have a choice, and using the right pumps, it might not cost you much more to operate to increase the flow.

But it is much more than 5th grader math to do this efficiently, which some might not fully understand even after 35 years in the field.

Some, even after 35 years, might increase the water temp of the system, rev up the circulation (and get a really big one) and then use balance valves on each circuit to slow down the flow again, like in older boiler systems. But that would result in a much less efficient system.

I know I mentioned it before, but you should read this thread too. Equally important. https://www.geoexchange.org/forum/threads/the-great-buffer-bypass-tank-tee-discussion.6650/

What is the brand and model of your actuators?

13. ### Hp HomeMember

Learning stuff is fun. Thank you for the help, I am getting there.

The great buffer tank bypass thread was spawned from this thread so yep I have read it and I intend to pipe that way even if the sketches I posted don't show it.
I might even like to pipe it both ways and use logging to collect data on the difference, but that is future tinkering.

The six actuators I have are made by Stellantrieb. They are 24V, normally closed, 250mA max switch on current.
I have six Braeburn 2020 programmable thermostats to control them.

14. ### Hp HomeMember

I went through all the math again using the 6.22F delta T to calculate flow and came up with 3.65 feet of head.

15. ### docjenserWell-Known MemberIndustry ProfessionalForum Leader

You are getting there. just run your math by me quickly.

16. ### Hp HomeMember

At 6.22F delta T times the 500 constant gives 3110 for figuring flow.

1) 250' of 1/2" pex

f= 1,604 btuh/ 3110= 0.5157 gpm

H= 0.055 x 0.71213 x 250 x (0.5157 ^ 1.75) = 3.073

2) 100' of 1" pex

f= 3,000 btuh/ 3,110= 0.9646

H= 0.055 x 0.04318 x 100 x (0.9646 ^ 1.75) = 0.223

3) 20' of 1.25" copper

f= 9 gpm

H= 0.055 x 0.0068082 x 20 x (9 ^ 1.75) = 0.350

Total of the three-
3.073 + 0.223 + 0.350 = 3.646 feet of head

17. ### Hp HomeMember

Moving targets are frustrating this reminds me of chasing chickens.

When I use this equation for the slab (where most of the load goes)-

300' of 1/2" pex

f= 22,000 btuh/ 3,110= 7.074 gpm

Divided by 6 circuits on the manifold- 7.074 gpm/ 6= 1.179 gpm

H= 0.055 x 0.71213 x 300 x (1.179 ^ 1.75)= 15.675 feet of head

This is why I had been thinking it was around 15'.

But if I go back to a 10F delta T-

f= 22,000 btuh/ 5,000= 4.4 gpm

4.4 gpm/ 6= 0.733 gpm

H= 0.055 x 0.71213 x 300 x (0.733 ^ 1.75)= 6.815 feet of head

18. ### docjenserWell-Known MemberIndustry ProfessionalForum Leader

What about when the other zones open up, and some water flows that way?

Calculating flow has nothing to do with BTUs. The water does not know that it needs to deliver BTUs. It just follows the path of least resistance. But with more water flow the resistance goes up exponentially. So a pathway with lesser flow resistance will allow more water to flow through, which now will increase this specific flow resistance. So it will balance out, but some circuits will see more flow than others, thus their delta T will be less (for each foot), since the water is flowing through faster and cannot loose as much heat (for each foot of pipe). You have 12 total pathways,

You have a target of 9 gpm at full capacity, how of that will flow into the slab (hint: not 7.094 gpm, because that comes off the GPM requirement for the zone) and how much into the first floor?
What happens if the first floor shuts down and all the water goes into the slab? And what happens if the first floor opens up and the slab shuts down? What happens if both are running? Those are your 3 scenarios. The one for the flow and pump selection is when both are running?

In other words when all zones are open (9 gpm flow), what is the total pressure needed to make sure that each of the 12 circuits is having the same pressure drop. That is your magic number, than they are balanced. That gives you the ft/hd the pump needs to make to pump 9 gpm through your system. Lets ignore water density at different temperatures for now, and other other minor factors like pipe roughness etc.

Still waiting for Mark here to help out......he might be looking for a 5th grader.

19. ### Mark CustisNot soon.Industry ProfessionalForum Leader

Doc:

You are too kind.

20. ### Hp HomeMember

The plan was for the slab to be the main heat source for the house. With 90F water temp it can put out 32,000 btu's. Outdoor reset would set water temp based on how the slab is heating the main floor.

I thought that when the radiant walls upstairs called for heat the actuator would open and a delta P pump would just speed up to compensate. I'm seeing now how it's not entirely that simple.

Upstairs is lofted with ventilation to help move slab heat up there so the heating load for those 6 radiant wall rooms is small. So at least we won't be too cold if I get it wrong

I think this is a little over the head of most fifth graders but I am glad this forum has input from different experts both well known and not, it makes for a well rounded learning experience to see different points of view.