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## Proper cfm

How do you verify proper blower speed/cfm in cooling mode? The only way I know to do is static pressure/ manufacturer chart or a anemometer. Is there any other way if the chart isn't available or you don't have a anemometer?

2. Switch it over to heat and use

Btuh/ delta t x 1.08

If you know what kind of motor the unit is using then you can confirm if it's within "range" by using the static pressure. Residential psc motors max out at .5"w.c. and most ECM motors go up to .8"w.c.

3. If it's electric heat be sure all elements are operating before doing the BTU/DeltaT test. I've seen a lot of electric heaters with an element or two not working.

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For electric heat btu/hr = amps x volts x 3.41, so

cfm = amps x volts x 3.41/(1.08 x deltaT)

The wattage can vary considerably from one unit to the next, even when the same size heater package is installed in both. All loads, including the heating elements, the blower motor, relays, circuit board, electric air cleaners, etc.,., must be accounted for, since all of that electricity is turned into heat. Due to differences in supply voltage it's best to actually measure the voltage and amp draw to calculate the wattage rather than entering the heater kit rated watts.

A 5 kw strip at 250 volts will draw considerrably more wattage than the same strip at 230 volts.

Assume an element resistance of 11 ohms when hot. P = V^2 /R

At 250 volts we get P = 62500/11 = 5682 watts

At 230 volts we get P = 52900/11 = 4809 watts

This corresponds to 15% error in estimated airflow, and doesn't even take into account differences in motor wattage. The latter can also vary considerably from one unit to the next, even with the same motor installed, due to differences in duct static and speed tap selection.

Measurement error already provides about a 10% uncertainty in the estimated cfm, so you'll want to be as precise as possible with the btu calculation before using the temp rise method.
Last edited by hvacrmedic; 02-09-2013 at 10:26 AM.

5. Originally Posted by hvacrmedic
For electric heat btu/hr = amps x volts x 3.41, so

cfm = amps x volts x 3.41/(1.08 x deltaT)

The wattage can vary considerably from one unit to the next, even when the same size heater package is installed in both. All loads, including the heating elements, the blower motor, relays, circuit board, electric air cleaners, etc.,., must be accounted for, since all of that electricity is turned into heat. Due to differences in supply voltage it's best to actually measure the voltage and amp draw to calculate the wattage rather than entering the heater kit rated watts.

A 5 kw strip at 250 volts will draw considerrably more wattage than the same strip at 230 volts.

Assume an element resistance of 11 ohms when hot. P = V^2 /R

At 250 volts we get P = 62500/11 = 5682 watts

At 230 volts we get P = 52900/11 = 4809 watts

This corresponds to 15% error in estimated airflow, and doesn't even take into account differences in motor wattage. The latter can also vary considerably from one unit to the next, even with the same motor installed, due to differences in duct static and speed tap selection.

Measurement error already provides about a 10% uncertainty in the estimated cfm, so you'll want to be as precise as possible with the btu calculation before using the temp rise method.
Thanks for the electric heat breakdown hvacrmedic.

I believe this clarifies one thing I was wondering about, we should be reading amps on the feed and not at the element because we aren't taking into account blower amps and boards.

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Originally Posted by mason
Thanks for the electric heat breakdown hvacrmedic.

I believe this clarifies one thing I was wondering about, we should be reading amps on the feed and not at the element because we aren't taking into account blower amps and boards.
Right. If you can't get your uncertainty in cfm measurment down to around 10%, then you stand a chance of adjusting airflow in the wrong direction, doing more harm than good. Blower heat can produce up to 10% of the total wattage with two 5kw strips engaged, or up to 17% with one 5kw strip engaged.

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Another thing to keep in mind is in the cooling mode the coil will be wet and you wont move as much air due to its higher pressure drop.

Measuring airflow in the field will always be a guestimate in most cases. So, do the best you can, but dont get tunnel vision using the airflow you guestimated.

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Good point bigtime. My air handler delivers 1202 cfm (checked at supply registers with Flow hood) with dry coil, 672 cfm with wet coil. Manufacture's air flow data is for dry coil condiitons and without filter on my York equipment.

9. Originally Posted by chill67
Good point bigtime. My air handler delivers 1202 cfm (checked at supply registers with Flow hood) with dry coil, 672 cfm with wet coil. Manufacture's air flow data is for dry coil condiitons and without filter on my York equipment.
1200 down to less than 700 just from the coil being wet? Was the blower set at the same speed for both tests?

10. Originally Posted by 54regcab
1200 down to less than 700 just from the coil being wet? Was the blower set at the same speed for both tests?
Almost couldn't be.

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Originally Posted by beenthere
Almost couldn't be.
Or the coil is wearing a lint blanket.

12. Originally Posted by chill67
Good point bigtime. My air handler delivers 1202 cfm (checked at supply registers with Flow hood) with dry coil, 672 cfm with wet coil. Manufacture's air flow data is for dry coil condiitons and without filter on my York equipment.
That is a pretty serious reduction in airflow, might want to do a pressure drop on that coil and a TESP on the system. Never hurts to confirm the fan speed setup first.

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Same fan speed and entire system is only a year old, no lint blanket on coil. I was shocked as to my findings as well. Wonder why there are no numbers calculated for air flow with wet coil??????

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