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Discussion Starter #1
Hi,




I have a SSZ140301A with CAPF3636*6A coil with TXV. There is only the TXV at the indoor coil for AC Mode, no TXV in the outdoor unit for the Heat pump Heat Mode.

My variable speed blower is set to 1125 CFM
I measured:
Outdoor Temperature: 75
Indoor Temperature at return: 71.3 (After 20+ minutes of running)
Indoor Temperature at supply: 51.3 (After 20+ minutes of running)
Indoor wet bulb Temp: 65.5

From the Datasheet using a larger coil the expanded cooling data suggests:
At 1181 CFM, Indoor Dry Bulb = 70, Indoor wetbulb = 67, Outdoor Temp = 75
MBh = 31.3
Hi PR =281
Lo PR =137
Amps = 3.2
Delta T = 12
S/T = 0.45


Some Questions:
1. What will my smaller coil (3636 vs 3642) do to the data?
2. My BTU/hr is only 24,300 using the 1125*1.08*20F formula, the table suggests I should be getting 31,300, is this a sign of something?
3. The Delta T, is that the out door or indoor temp rise?
4. What is S/T?
5. What will be the signs if the system is under or over charged?
 

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Hi,




I have a SSZ140301A with CAPF3636*6A coil with TXV. There is only the TXV at the indoor coil for AC Mode, no TXV in the outdoor unit for the Heat pump Heat Mode.

My variable speed blower is set to 1125 CFM
I measured:
Outdoor Temperature: 75
Indoor Temperature at return: 71.3 (After 20+ minutes of running)
Indoor Temperature at supply: 51.3 (After 20+ minutes of running)
Indoor wet bulb Temp: 65.5

From the Datasheet using a larger coil the expanded cooling data suggests:
At 1181 CFM, Indoor Dry Bulb = 70, Indoor wetbulb = 67, Outdoor Temp = 75
MBh = 31.3
Hi PR =281
Lo PR =137
Amps = 3.2
Delta T = 12
S/T = 0.45


Some Questions:
1. What will my smaller coil (3636 vs 3642) do to the data?
2. My BTU/hr is only 24,300 using the 1125*1.08*20F formula, the table suggests I should be getting 31,300, is this a sign of something?
3. The Delta T, is that the out door or indoor temp rise?
4. What is S/T?
5. What will be the signs if the system is under or over charged?
That formula is only for the sensible heat and not the latent heat. Q=m delta h

Delta T is indoor
 

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Hi,




I have a SSZ140301A with CAPF3636*6A coil with TXV. There is only the TXV at the indoor coil for AC Mode, no TXV in the outdoor unit for the Heat pump Heat Mode.

My variable speed blower is set to 1125 CFM
I measured:
Outdoor Temperature: 75
Indoor Temperature at return: 71.3 (After 20+ minutes of running)
Indoor Temperature at supply: 51.3 (After 20+ minutes of running)
Indoor wet bulb Temp: 65.5

From the Datasheet using a larger coil the expanded cooling data suggests:
At 1181 CFM, Indoor Dry Bulb = 70, Indoor wetbulb = 67, Outdoor Temp = 75
MBh = 31.3
Hi PR =281
Lo PR =137
Amps = 3.2
Delta T = 12
S/T = 0.45


Some Questions:
1. What will my smaller coil (3636 vs 3642) do to the data?
2. My BTU/hr is only 24,300 using the 1125*1.08*20F formula, the table suggests I should be getting 31,300, is this a sign of something?
3. The Delta T, is that the out door or indoor temp rise?
4. What is S/T?
5. What will be the signs if the system is under or over charged?
1-decrease the total BTU slightly.
2-that's total BTU
3-indoor
4-sensible/total
5-head pressure and subcool out of range
Cooling BTU's are a mixture of sensible heat removal(temp diff) and latent
heat removal (moisture). CFM*1.08*delta T only gives sensible. As the humidity in the house changes the Delta T changes because the system will convert some of the unused latent capacity to sensible.
 

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Discussion Starter #4
1-decrease the total BTU slightly.
2-that's total BTU
3-indoor
4-sensible/total
5-head pressure and subcool out of range
Cooling BTU's are a mixture of sensible heat removal(temp diff) and latent
heat removal (moisture). CFM*1.08*delta T only gives sensible. As the humidity in the house changes the Delta T changes because the system will convert some of the unused latent capacity to sensible.

So since my Delta T doesn't match the tables Delta T, does that mean my wet-bulb measurement is off and it should have been much lower which corresponds to a wet-bulb temp of 59 and a temp delta 17?

How do you do the wet-bulb measurement in the field, do you put the wetbulb in the return air flow so it gets active convection, or do you leave it out in the open? Does that even matter?

Also, my weather station in my office was saying the RH was 54% when I took these measurements, which suggests my TWet should have been 58.5 based on a 71.3 Tdry using the conversion from wet/dry to RH. So that supports my claim that I didn't take an accurate wetbulb..


So that points me to the first column in the table that has:
a wetbulb of 59
MBTU of 27.6
S/T 0.82 Which equals a sensible MBTU of 22.632 vs my 24300
Delta T of 17
HiPr 247
LoPr 118
Amps 2.9
kW 2.13
 

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Discussion Starter #5 (Edited)
In looking at these cooling tables, they list kW=Total System Power, they also list AMPs as Outdoor unit (compressor+fan)

If Power = Amps times Volts, then you'd think 220v times Amps should equal kW. Right?

My table shows 26.7 MBh, 2.42kW, 4.3 Amps.

I know my system is moving close to the rated output, why does 220v * 4.3 Amps = 946 Watts? Did they do all these tests on 480 volt three phase or something? The system should be drawing 11 amps, or the system voltage should be 562 volts.. What's also interesting is the Heating table goes up to 11.8 amps at a 2400 watt input power to the outdoor unit. This translates to ~200 volts which seems reasonable. Is because of the power factor of the load changes at the compressor and it screws up the amp meter reading?

What gives? I'm looking at the Goodman SSZ14 datasheet.
 

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S/T is Sensible to total percent/ratio.

Did you take those readings in the return plenum and supply plenum? that is where they must be taken to get an accurate reading for capacity measurements.

You need to take wetbulb of both the return and suppl, then convert that to enthalpy, and the drybulb of both the return and supply, and then use the total enthalpy formula to determine total BTUs, and then subtract the sensible from that to determine how much latent capacity you have/had at those conditions.
 

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Hi Jeff,
Your electrical discrepancies arrise from attempting to apply DC electrical theory (ohms law) Watts = volts x amps to an A/C circuit which doesn't work on DC theory. In an AC circuit you have to combine inductive reactance, capacitive reactance and your resistance to come up with overall reistance to current flow in an AC circuit which is called impedance. You can look up formulas to calculate all of the above if you desire to do so.

Your system's sesnible heat removal capacity in Btu's will vary significantly with the wet bulb temp of your indoor air and fpm of air velocity through the evaporator coil. As you make more condensate you are using more of your capacity to remove the latent heat content in the air (approx.970 btu/lb) of water condensed out of airstream, which leaves you less capacity to make sensible heat change. The speed which air is moved through the coil is important in that it changes the contact time with the cold surface of the coil thereby alloowing more/less of the air to reach dew point temp. and condense. The advantage of more condensate is that by lowering the relative RH in the structure to a lower level you will feel comfortable at a higher dry bulb temperature (sensible heat content). (eg. [email protected]% vs [email protected]%) You can access all this information from a psychrometric chart.
 

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Discussion Starter #8
Hi Jeff,
Your electrical discrepancies arrise from attempting to apply DC electrical theory (ohms law) Watts = volts x amps to an A/C circuit which doesn't work on DC theory. In an AC circuit you have to combine inductive reactance, capacitive reactance and your resistance to come up with overall reistance to current flow in an AC circuit which is called impedance. You can look up formulas to calculate all of the above if you desire to do so.

Do you know the units of the Amp values in the table? Is it Amps RMS? Is the table suppose to support a Tech doing a field measurement with a clamp on current meter? Have you ever checked the current on a condenser unit and cross correlated it with the table?
 

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Discussion Starter #9
Hi Jeff,
You can look up formulas to calculate all of the above if you desire to do so.

I'm still having a really hard time explaining this. Power Factor would suggest if we have an inductive load, the system has to draw more current to achieve the actual load, so that would suggest the current goes up, right? Why is the current going down so drastically in the table when the power is only slightly going down? Is it a really distorted waveform coming from the compressor that creates these tiny current values?

Do you know what I'm talking about?

http://www.goodmanmfg.com/Portals/0/pdf/SS/SS-SSZ14.pdf

Look at page 20.

KW at 65 degrees is 2.40 and Amps is 11.8.
But at 35 degrees, KW is 2.14 (90% of the value)
But now Amps is 6.6 (55% of the value)

Is this because the AC current meter can't read a distorted waveform?
 

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Don't worry about power factor. You need to take entering and leaving wetbulb readings(at the furnace, not at a register), along with actual CFM readings.
 
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