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  1. #1
    Join Date
    Feb 2012
    Posts
    42

    another silly psych question

    No, this isn't about my "macguyvered" windowshaker-turned-dehum; that's
    been put away for the moment because the real system is doing so much
    better at water removal. So I've been studying that a little -- yep, actually
    weighing buckets. Four to five pounds every hour, what with this bit of
    a heat wave we've had in the northeast.

    The question is this: if I know the db/wb on both sides of my coil, and
    know exactly how much water was removed per hour, is that enough to
    take and stare at a psych chart and equations and determine everything else?
    This is an inverter system running right steadily near setpoint so it's doing
    very little sensible cooling, i.e. just pluggin' away on the humidity and never
    cycling. I know the rough power input but don't know the running COP at
    the time, and am not sure what the CFM is with a loaded coil and the fan on
    low. Can I determine the CFM and the total enthalpy change from just coil
    deltas and pints, or is that too many unknowns?

    No actual system issues, just trying to sling a little theory around. Sorry
    if this isn't the right place for it... I suppose that after a few more pints I
    wouldn't *care* what the CFM was anymore...

    _H*

  2. #2
    Join Date
    Jan 2004
    Location
    Lancaster PA
    Posts
    67,755
    Would need to know actual CFM to plot/calc anything of value.
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  3. #3
    Join Date
    May 2012
    Location
    San Diego, CA
    Posts
    352
    Quote Originally Posted by hobbit View Post
    No, this isn't about my "macguyvered" windowshaker-turned-dehum; that's
    been put away for the moment because the real system is doing so much
    better at water removal. So I've been studying that a little -- yep, actually
    weighing buckets. Four to five pounds every hour, what with this bit of
    a heat wave we've had in the northeast.

    The question is this: if I know the db/wb on both sides of my coil, and
    know exactly how much water was removed per hour, is that enough to
    take and stare at a psych chart and equations and determine everything else?
    This is an inverter system running right steadily near setpoint so it's doing
    very little sensible cooling, i.e. just pluggin' away on the humidity and never
    cycling. I know the rough power input but don't know the running COP at
    the time, and am not sure what the CFM is with a loaded coil and the fan on
    low. Can I determine the CFM and the total enthalpy change from just coil
    deltas and pints, or is that too many unknowns?

    No actual system issues, just trying to sling a little theory around. Sorry
    if this isn't the right place for it... I suppose that after a few more pints I
    wouldn't *care* what the CFM was anymore...

    _H*
    With your dry and wet bulb temperatures you can find enthalpy, for CFM you need to know your cooling capacity of your coils to get your CFM or if you have your CFM you can verify your cooling capacity. Otherwise you will have 2 unknowns and 1 equation...
    You can call me Sam

    It should be a crime to be a mechanical engineer in San Diego
    Summer Design Temperature: 83 F Dry Bulb ~ 69 F Wet Bulb (California Climate Zone 7)

  4. #4
    Join Date
    Nov 2004
    Location
    SW FL
    Posts
    6,244

    Hmm

    Quote Originally Posted by hobbit View Post
    This is an inverter system running right steadily near setpoint so it's doing
    very little sensible cooling, i.e. just pluggin' away on the humidity and never
    cycling.

    No actual system issues, just trying to sling a little theory around.
    It is not at all justified to say " very little sensible cooling" when the system is well matched to the actual cooling requirements ( heat gain).

    Sensible cooling is likely > 70% of the total cooling !.!!
    Designer Dan
    It's Not Rocket Science, But It is SCIENCE with "Some Art". ___ ___ K EEP I T S IMPLE & S INCERE

    Define the Building Envelope and Perform a Detailed Load Calc: It's ALL About Windows and Make-up Air Requirements. Know Your Equipment Capabilities

  5. #5
    Join Date
    Jul 2004
    Location
    Massachusetts
    Posts
    6,829
    You're striking right at the heart of the DOE and their difficulty in applying actual efficiency ratings to inverter technology products/AC/HP. Since they operate at an infinite number of CFM's, they operate at an almost infinite number of efficiency's. And depending on the indoor WB, the amount of latent and sensible is always changing.
    If YOU want change, YOU have to first change.

    If you are waiting for the 'other guy' to change first, just remember, you're the 'other guy's' other guy. To continue to expect real change when you keep acting the same way as always, is folly. Won't happen. Real change will only happen when a majority of the people change the way they vote!

  6. #6
    Join Date
    Feb 2012
    Posts
    42
    That's about what I was thinking, re: COP. The outdoor unit was
    drawing all of 3 or 4 amps at 240V, less than a kilowatt for several
    hours. This particular experiment went on fairly far into the
    evening after any solar load was gone and sensible temps were
    fairly equalized; the room temp changed *very* slowly as the
    unit continued to drag more water out of the air which is why
    I'm assuming the sensible load was pretty small. The house
    isn't very large; a traditional system probably would have been
    delivering 10-minute bursts of arctic blast and shutting down.

    I haven't been able to measure CFM with a loaded coil yet;
    clearly that's the next step... that and a more exact figure on
    power input should give me ballpark COP.

    _H*

  7. #7
    Join Date
    Jul 2005
    Posts
    5,367
    Yes, you could. Shoot me some actual numbers and I'll run the calc and post it here.

  8. #8
    Join Date
    Nov 2004
    Location
    SW FL
    Posts
    6,244
    DB/ WB In 75.0 / 63.0
    DB/ WB Out 52.0/ 51.2
    CFM 600
    Q total 1.68 tons ( seems to correspond to ~1,000 watts / 1.34 HP)
    Q sensible 1.24 tons
    Q latent 0.44 tons
    Condensate = 3.68 lbs/ hr
    Specific Heat Ratio 73.6%

    Need the XLS file, e-mail me.


    1 yyy xxx zzz
    2 yyy xxx zzz
    3 yyy Return Air Coil xxx Difference zzz
    4 DB (' F) yyy 75.0 54.0 xxx 21 51.0 24.0 zzz 52.0 23.0
    5 WB (' F) yyy 63.0 53.0 xxx 10 50.2 12.8 zzz 51.2
    6 Relative Humidity ( %) yyy 51.58 93.85 xxx 94.79 zzz 94.89
    7 Dew Point ('F) yyy 56.038 52.333 xxx 3.705 49.624 6.414 zzz 50.645
    8 Humidity Ratio ( lb /lb) yyy 0.0095335 0.008309 xxx 0.00750571 zzz 0.0077999
    9 Specific Volume (ft^3 lb) yyy 13.681 13.118 xxx 13.025 zzz 13.05616
    10 Enthalpy ( BTU/ lb) yyy 28.4326 21.9752 xxx 6.4574 20.3735 8.0591 zzz 20.9358 7.497
    11 Grains ( 7000 * Humidity Ratio) yyy 66.7345 58.163 xxx 52.540 zzz 54.599
    12 yyy xxx zzz
    13 W - grains / lb : CHART yyy 67.2 58.6 xxx zzz
    14 COMPARISON yyy 1.0070 1.0075 xxx zzz
    15 yyy xxx zzz
    16 Flow /Rate CFM yyy 1000 1000 xxx 1000 600 600 zzz 600
    17 Flow /Rate CF Hr yyy 60000 60000 xxx 36000 zzz 36000
    18 yyy xxx zzz
    19 Lb /Hr (CF /Hr * Humidity Ratio) yyy 4385.80 4573.94 xxx 2631.48 2764.01 2641.48 zzz 2757.32
    20 Humidity / Hr yyy 41.812 38.005 xxx 25.087 20.746 25.18254958 zzz 21.507
    21 --- Condensate ( lbs /hr) ---- yyy 3.807 xxx 4.341 zzz 3.676
    22 Q Total = 4.5 *CFM * Enthalpy Diff. yyy xxx 29,058.3 2.422 21,759.6 1.813 zzz 20,241.4 1.687
    23 Q sensible = 1.08 * CFM * Temp.Diff. yyy xxx 22,680.0 1.890 15,552.0 1.296 zzz 14,904.0 1.242
    24 Q latent = Q Total - Q sensible yyy xxx 6,378.3 0.532 6,207.6 0.517 zzz 5,337.41 0.445
    25 Sensible Heat Ratio (S.H.R.) yyy xxx 0.780 0.715 zzz 0.736
    26 yyy xxx zzz
    Designer Dan
    It's Not Rocket Science, But It is SCIENCE with "Some Art". ___ ___ K EEP I T S IMPLE & S INCERE

    Define the Building Envelope and Perform a Detailed Load Calc: It's ALL About Windows and Make-up Air Requirements. Know Your Equipment Capabilities

  9. #9
    Join Date
    Nov 2004
    Location
    SW FL
    Posts
    6,244

    Thumbs up

    Readable ( Understandable ?) PDF with column alignment is attached.
    Attached Images Attached Images
    Designer Dan
    It's Not Rocket Science, But It is SCIENCE with "Some Art". ___ ___ K EEP I T S IMPLE & S INCERE

    Define the Building Envelope and Perform a Detailed Load Calc: It's ALL About Windows and Make-up Air Requirements. Know Your Equipment Capabilities

  10. #10
    Join Date
    Jan 2009
    Location
    Keokuk, IA
    Posts
    5,520
    Quote Originally Posted by hobbit View Post
    That's about what I was thinking, re: COP. The outdoor unit was
    drawing all of 3 or 4 amps at 240V, less than a kilowatt for several
    hours. This particular experiment went on fairly far into the
    evening after any solar load was gone and sensible temps were
    fairly equalized; the room temp changed *very* slowly as the
    unit continued to drag more water out of the air which is why
    I'm assuming the sensible load was pretty small. The house
    isn't very large; a traditional system probably would have been
    delivering 10-minute bursts of arctic blast and shutting down.

    I haven't been able to measure CFM with a loaded coil yet;
    clearly that's the next step... that and a more exact figure on
    power input should give me ballpark COP.

    _H*

    After sunset, the ambient temps are still fairly high and the home holds a lot of heat from being baked in the sun all day. Especially the roof. There's a couple thousand lbs of composite shingles up there along with the on the sides of the house the exterior sheathing material. But it takes some time to heat up the exterior of the home so peak loads are usually around 4-7PM, not 12-4PM as you'd expect based on the angle of the sun and ambient temps. Further, in most climates, the dewpoint will climb abruptly around dusk.

    Long run times have a dramatic impact on system efficiency. I've seen number quotes as high as a 25% difference. This is because the system can take 8-10 minutes to reach a steady state operating efficiency. As mentioned, I don't think this is fully taken into account as well on Inverter driven systems.

  11. #11
    Join Date
    Jul 2005
    Posts
    5,367
    Quote Originally Posted by dan sw fl View Post
    Readable ( Understandable ?) PDF with column alignment is attached.
    Still not readable. If you used the state points and cfm at the top of you post, then your results aren't very accurate either. The simplified formulas that you're using are ok for ballpark estimates of capacity when cfm is known, but not accurate enough to determine cfm from state points and condensate removal rate.

  12. #12
    Join Date
    Feb 2012
    Posts
    42
    It finally got warm enough again that I was able to get the coil loaded
    up and try to take a CFM measurement. The filter slot opening is
    about the same dimension as the main return duct, so I pull the
    filter, block off the real return by wedging something flat across
    the whole opening, and then do the average-anemometer scanning
    game across the 20" x 6" filter housing opening. On fan low, I get
    4.0 MPH which I read as 293 CFM. Which is odd, because the spec
    for this AHU swears up and down it's 420 CFM on low and 600 on hi.

    It seems that the ECM blower tries to maintain some constant flow,
    because it speeds up quite a bit to compensate for static pressure
    [e.g. if I substantially block return flow, it goes nuts]. How does
    the fan driver actually determine that's happening??

    Anyway, dan_sw's figures are pretty close except for the CFM;
    translating from the cheap temp/humidity meters I dropped inside
    the ductwork I get 67 db / 60 wb on return, and 52 db / 51 wb on the
    supply -- which is taken a few feet from the AHU; I don't have direct
    access to the top of the coil but IR-shooting the distro box on top
    [from a piece of black tape stuck on it, or the joint mastic, because
    IR guns don't like bare galvanized] I see the duct temp right around 50
    so the air coming out right there has to be just shy of saturated.

    Each pound of air's little adventure through this craziness is visually
    represented in this chart, where the red and blue dots are various
    start and end points I logged, the pink is something vaguely like the
    average psychrometric path through the coil, and the green is my guess
    as to what happens after it's shoved through the supply. Now what I'm
    having trouble figuring out is what endpoint to use in calculating
    the enthalpy change -- be it at 293 CFM or 420 or whatever, I guess
    the latent+sensible change is represented by my pink line, and the
    latent only is read on the vertical scale as delta grains or pounds
    of water?? I'm still not 100% clear on that part, particularly for
    for figuring out COP. For most of the run in question, the whole
    shebang [outdoor + indoor units] was drawing about 3.2 amps
    at 240V or 768 watts. That's a pretty modest compressor speed.

    I agree about the "flywheel" effect of the house into the evening,
    particularly over last week when the upstairs was quite warm and
    likely re-radiating a bunch of that into the downstairs and the
    air in question here. With the place still in its "leaky" configuration
    I have no clear idea what the sensible load would be -- it's not even
    going to even vaguely match my heat-loss calculation from last winter.

    Anyway, there's a buttload more data to romp around in. Gotta
    go dump another bucket!

    _H*

  13. #13
    Join Date
    Jul 2005
    Posts
    5,367
    With the temperature and condensate data posted your cfm is running between 517 and 646. The wide range is due to the 4 to 5 lb/ hr statement in your first post.

    517 for 4lb/hr and 646 for 5 lb/hr.

    Note that 517/646 is roughly equal to 4/5.

    With this type of calculation the point at which you measure supply wet and dry bulb temps is immaterial, so long as it's actually in the supply stream somewhere, isn't too close to the coil, and isn't mixed air. The measurements must be accurate too.

    Assuming that it's moving the rated 600 cfm, the data stacks up like this:

    Attached Images Attached Images  
    Last edited by hvacrmedic; 07-23-2012 at 12:12 AM.

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