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HRV balancing ?
Last Post 14 Nov 2010 12:24 PM by Daler. 11 Replies. |
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Daler
New Member
Posts:10
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| 04 Nov 2010 12:38 PM |
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Not sure this is the right place to ask this question, but appreciate the help if you can offer it --
We have a well-maintained 10 year old HRV. It is as a compliment to our baseboard heat so the HRV is independantly run via the attic and (10) ceiling vents total. I have a hunch that the system is unbalanced.
In place of a Magnehelic guage, I rigged up a simple DIY water manometer (plastic tubing loop with colored water in it) and inserted each of its ends into the supply-to-house air duct and the stale air return duct -- essentially stratling the ducts. A couple of simple single pitot tubes were fashioned to probe the air streams in each duct (manomter tubes fitted over them).
For some reason this method of checking balance between the 2 air streams doesn't seem to work. The probe on the supply stream slightly depresses the liquid, which under pressure it should, but the return stream seems to only apply a slight amount of suction at the pitot tube thereby enhancing the effect of the other tube. There appears however good air flow in the return stream.
Is there something about this process that I'm missing here?
Here's a drawing of a Lifebreath device that is very similar to ours, including fan locations.
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jonr
Senior Member
Posts:5341
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| 04 Nov 2010 12:52 PM |
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My understanding is that balance is most correct when the indoor and outdoor pressures are equal (or slightly over/under if that is what you want). So those are the pressures I would measure and monitor (ie, ignore the pressure in the ducts). |
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Daler
New Member
Posts:10
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| 04 Nov 2010 01:22 PM |
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So, leaving the HRV aside, and without a blower door, how can this measurement be done with the equipment I have? |
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jonr
Senior Member
Posts:5341
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| 06 Nov 2010 09:08 AM |
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The problem is that the pressure difference between indoors and outdoors is so small that you won't be able to see it on a water manometer. I'm not aware of a very low cost way to measure this.
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Daler
New Member
Posts:10
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| 06 Nov 2010 10:47 AM |
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Posted By jonr on 06 Nov 2010 09:08 AM
The problem is that the pressure difference between inside and outside is so small that you won't be able to see it on a water manometer. I'm not aware of a very low cost way to measure this.
Hmm. Perhaps this is where a Magnehelic guage really comes in? I kinda figured my water manometer might be a problem as the volume of air appears ok but the pressure to activate the manometer properly is not sensitive enough. Factoring in a Mag guage, whould a simple pitot tube then work? -- or would it have to be a manufactured probe specifically designed for this procedure? |
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jonr
Senior Member
Posts:5341
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| 12 Nov 2010 12:44 PM |
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I'm going to be testing one of these (a RS-232 smart pressure sensor good for .002 pascals):
http://www.sensirion.com/en/pdf/product_information/Datasheet_differential_pressure-sensor_ASP1400_E.pdf
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Daler
New Member
Posts:10
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| 12 Nov 2010 01:57 PM |
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That device looks like the cat's meow ..but I doubt it's $350+ tag would be worth it to me for just casual use. I do use similar (laptop) interfaces for troubleshooting auto mechanics and think these devices are the wave of the future for most anything technically analytical. |
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Lee Dodge
Advanced Member
Posts:714
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| 12 Nov 2010 05:10 PM |
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Daler-
You have missed the idea of what you need to be measuring. From the drawing, they want you to measure the velocity in the ducting by measuring the differential pressure between the static pressure and the total pressure, where,
P + 1/2 rho v^2 = P0
where:
P is the static pressure
rho is the air density
v is the velocity (and the ^2 means velocity squared
P0 is the total or stagnation pressure
P0 is what you measure with the pitot tube that is facing upstream. P is measured with a probe at right angles to the flow, which you could make by simply making a pressure connection at the side of the duct. If you see the same differential pressure, then you have matched the air velocities in the ducting, assuming that the air densities are similar, which they should be. You don't need to do the calculations to solve for the velocities -- you just need to match them.
You can use your manometer setup, but one side needs to go to the total pressure and the other to a static pressure. Without doing the calculations, I don't know if your manometer is sensitive enough. What you are measuring currently is not giving you a clue about matching the velocities as far as I can tell.
Lee Dodge
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Lee Dodge,
<a href="http://www.ResidentialEnergyLaboratory.com">Residential Energy Laboratory,</a>
in a net-zero source energy modified production house
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jonr
Senior Member
Posts:5341
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| 12 Nov 2010 07:19 PM |
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I'll guess about 1 mm/water (if you believe in matching duct velocities vs indoor/outdoor pressures). But I don't know the device or the velocities that they expect.
http://www.engineeringtoolbox.com/pitot-tubes-d_612.html |
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Daler
New Member
Posts:10
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| 13 Nov 2010 01:10 PM |
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Lots to think about here. Thanks for clarifying the process, guys. I did make up a couple of simple pitot tubes out of 5/16" tube-stock. Bent the tips out 90°, and chamfered the nozzle. Interestingly, most of the commercial sales literature I've come across indicate a diameter of just 1/4" -- will my dimensions make any difference in application? Correct me if I'm off base here ..what I will do is is stretch one end of the manometer over my pitot tube and place the tube facing up-stream in the duct, and then followup by just plugging the other end of the monameter into the side of the duct (ahead or behind the pitot?). Hopefully, the manometer can be read accurately enough. Then I'll repeat the process on the other side of the HRV -- then compare the 2 readings. Does this make sense? As for proving the math, I'm not an engineer and math is not my strong attribute  ** And speaking of math : I have another query relating to our HRV. It has to do with the fresh air supply ceiling vents. They are the 4" standard round spin-out type. The manufacturer has suggestions on the amounts of cfm air movement per room requirements. To make it "simple", they provide a table that compares the amount of disk turns with the resulting cfm. For instance, "62 cfm = 16 screw turns (fully open)" But they fail to suggest the number of screw turns at half open (@50 cfm). Logic would dictate that would be 8 turns but it would also seem to equate to 31 cfm instead. I suppose the shape of the vent opening and it's adjustable disk has a bearing on the actual amount. But I suspect it may have more to do with the actual shape of the diffuser, etc. My question here is : for any given air flow, how can it be determined what each turn of the disk would equal how much cfm -- how did they come up with their 50 cfm @ half-open figure?  |
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Lee Dodge
Advanced Member
Posts:714
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| 13 Nov 2010 02:05 PM |
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Daler-
You are correct in your interpretation, and the dimensions of the pitot tube make no difference. You are using the pitot tube to measure the pressure exerted by the moving air, but there is no flow in the tube (no flow through pressure gauge), so the tube dimensions don't matter. It does matter that the pitot tube face upstream (does not have to be perfect), and that it is NOT located near the edge of the duct. The static pressure measurement is just providing a measure of the "reference pressure" that does not include the extra pressure from the moving air flow, and it is measured by a tube that is inserted into the duct, or just ends at the edge of the duct, in the same approximate area as the pitot tube, but with the open end at right angles to the air flow. The pitot tubes that are shown in your diagram appear to be a tube within a tube, where the inner tube is the upstream facing tube to measure the total pressure, and the outer tube has a hole in it or is terminated such that it measures the static pressure (details unclear from figure). On my Venmar AVS EKO 1.5, there are built-in test points, and I assume that they must be measuring the static and total pressures, but it is unclear from the diagram provided. I assume that your unit does not have built in pressure taps?
Typical values that are shown for my Venmar unit are a pressure drop of 0.67 inches (17 mm) of water on the fresh air side corresponds to a flow of 125 CFM, while on the exhaust side the pressure drop is 0.70 inches (18 mm). Again I am assuming that the Venmar unit is measuring the differential pressure between a static and total pressure, but it is unclear from their diagram.
jonr-
I was not suggesting that balancing the flow rates was the best way to balance the HRV, but that what is recommended by the manufacturers. Without much experience in actually balancing HRVs, that is certainly the first approach that I would take also.
Lee Dodge
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Lee Dodge,
<a href="http://www.ResidentialEnergyLaboratory.com">Residential Energy Laboratory,</a>
in a net-zero source energy modified production house
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Daler
New Member
Posts:10
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| 14 Nov 2010 12:24 PM |
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I'll try your suggestion when I get a chance. Our Eneready HRV Model 2000 [ http://www.enereadyproducts.com/prod01.htm ] has no taps in it for retrieving balancing information. Unlike some other competitive brands, the installation manual is very limited in setup details :
 -- nomimal 200 cfm (vents : 5 intake, 5 exhaust). "whisper" grilles are like those in the picture in the previous post.
That's all they have to offer when it comes to balancing.
It seems their balancing instructs are for doing it locally through the adjustment of individual airflow grilles. Except for the hints at the bottom of the page, there is not much reference to actual HRV measuring.
These vent adjustments are what I was trying to determine through my last post. As you may see, the co-relation between fully open and half open appear non-reflective to the number of screw turns, especially for cfm values in between. I suspect there is some sort of formula to determine this factor??
I suppose that once the vents are adjusted correctly, then I can proceed with a rudimentary global adjustment (via dampers if necessary) at the HRV --
With all of our attic-run ducting being insulated 6" flex, it is difficult in inserting and positioning the pitot tube in the center of the airstream and pointing it correctly -- doesn't have a ridged body of a metal duct to maintain the position without the device wiggling around a lot. But, if I all I need is to attach the static end of the manometer to the inside edge of the duct with the pitot tube through an incision a few inches away it should be easier to both determine the duct center and upstream pointing direction to obtain some sort of reading ..that is of course, providing that the water manometer is sensitive enough to register the small comparative changes between pressure and static (PO & P) / exhaust and fresh air ducts  |
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