@darrelludelhoven6503

6 years ago

No one provides Techs with the formulas for right sizing filter rack areas to achieve proper velocities through them; I will:
HVAC Airflow- Duct, filter’s Ak and other Component Sizing Formulas:
Ak area of the filter: CFM / required or desired fpm velocity
Example: 3-Ton @1200-CFM / 280-fpm= 4.2857142-Ak
Heating only 858-CFM / 220-fpm = 3.9 (Ak) SF open-air-area.
My filter numbers:
I have (3) 12” x 32” filters 384-sq.ins., each x 3-filters is 1152-sq.ins / 144 is 8-sq.ft. | so, 3.9-Ak sq.ft., / 8-sq.ft filter area is 0.4875% or, filter’s Ak %.
**A tested 59°F heating mode temp-rise x 1.1 is 64.9 furnace 55,500-output / 64.9 is 855-CFM 3.9 X 220 = 858-CFM; Had a tested 855-CFM; very close. …
Therefore, a single 16X25 filter is 400-sq.ins. / 144 is 2.7777 400*.4875=195-sq.ins./ 144=1.3541666-open-air Ak of the filter).
Filter’s Ak = CFM / fpm velocity through the filter in fpm
FPM= CFM / SF Ak AREA of filter
CFM = fpm X SF (Ak) AREA of the filter
Required Ak % of the filter’s area= smaller actual Ak area ‘/’ filters actual larger listed Square Foot size area.
**Print & Save all these Formulas & USE them to properly size filter rack areas to achieve proper low filter fpm velocities...!

Also; a 6” duct; so 6" x* 6” = 36 *x .7854 = the 28.2744-sq.ins" / 144 = 0.19635-Ak-sq.ft area.
** I like the Purolator P-312 white Industrial Strength Disposable Synthetic Panel MERV 4 filters - decent pressure drop with good filtering - you have to order them on the Internet; has a .16" PD @300-fpm; .10" PD @200-fpm velocity
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**7 out 10 HVAC systems by exhaustive testing have airflows out of tolerances, this can cause high cost ECM motors to exceed their rpm Rated limits & thus overheat & fail…Also results in very high cost compressor failures!
**There is also a direct correlation between the blowers 0.5” or 0.6”WC and the total equivalent length (TEL) of the duct system regarding the amount of remaining available steady pressure. Jack Rise having done many hundreds of Manual D’s friction rate sizing has stated that you may need to have 0.4 inches if you have a total equivalent length of the ductwork in order to get a friction rate design of 0.08 inches of water column for adequate velocity in the supply runs.
** Okay Manual D; .4” * 100 = 40 / ‘a very high’ 500-TEL is 0.08” WC friction rate to achieve enough velocity. Manual D on the Internet shows an improper example of designing friction rate for the supply side air-flow; you need a much higher rated static pressure oversized blower than .6”. to have .25” to.3” of Available Static Pressure (ASP) remaining.
They show a 0.6 inch water column blower and all pressure drops result in an Available Static Pressure (ASP) of only 0.17” WC; .17” * 100 is 17 / 500-TEL is 0.034-FR; and therefore, supply air ducts would be sized way to big and would result in an extreme low velocity of airflow through the ducts.

**A more reasonable situation is .25” (ASP) Available Static Pressure & 300-TEL, or less; 25/300-TEL is .083” (FR) Friction Rate to achieve proper airflow velocities.

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