CFM Confusion-equations and calculations 1

[DirtySuby]

I am new to the forum, and have little to no experience with designing HVAC systems. I have searched these forums and other websites for help on the subject. It seems that each equation I find for certain calculations, I get a different answer than what another equation for the same topic gave me.
Anyways, my first basic problem is finding the required CFM for a heating system.
The area that needs to be heated is approx. 10ft X 4.25ft X 7 ft. I need to figure out the required CFM for this volume. I used an airflow equation I found on engineeringtoolbox.com, and calculating the BTU with the equation (exposed sq. ft*(max inside temp-min oustide temp) * heat loss value) [aka 1/R-value]=BTU. That equation could be wrong, I found that on a greenhouse website. Using other calculators I came up with smaller values.
In any case, the value I came up with was 2247.7 BTU. Seems low? Using the Airflow equation from engineeringtoolbox.com (

http://www.engineeringtoolbox.com/design-ventilation-systems-d_121.html)

I came up with about 0.036821905 m^3/s or 78 ft^3/min. once again, seems very low.
Finding the necessary CFM is just the beginning of my problem, but if anyone can lead me in the right direction, I will attempt to take it from there. Thanks!

 

[dpoppeli]

Heat-lost = heat-input given steady-state conditions. In US units, heat loss, q = U x A x delta-T, and the heat input required is q = 1.08 x CFM x delta-T. That latter formula can be derived from the standard convective heat equation. delta-T is highest room temperature minus lowest design outside temperature which is about -5 deg-F 99.6% of the time in northern Indiana according to my '97 ASHRAE handbook. Check all your potential heat losses, e.g. perimeter floor losses, windows, doors, roofs, infiltration (often rule-of-thumb number which depends on construction quality, building exposure or number of exits/entries by people), and then add a safety factor and pickup load if required. Pickup load is the non-steady-state part and is the amount of extra heat required after the building/room has cooled down over night (or other) and you want it to heat back up in a reasonable amount of time. This can be as much as 40% extra capacity required in a house furnace. It is why a furnace is often much larger then would appear to be required from a calculated figure. Your space is not large so not a lot of heat may be required.

I notice you're a student at Purdue. Suggest you perhaps join AHSHRRAE and then either pick up a copy of their "Principles of Heating, Ventilating, and Air Conditioning" which is a sort of textbook which summarizes much info in their Fundamentals handbook, or "HVAC Simplified". Books appear to have bargain pricing and can be found at ashrae.org.

 

[dpoppeli]

I should clarify that delta-T for q-input is different. In a reciculating air system this would be the return (room) air-temp minus the heated-air-temp (in human comfort conditions may want to limit to around 95 deg-F so as not to melt your occupants).

 

[UtilityLouie]

I'd also say that heat loss usually is not the driving force behind how much heat you have to bring into the building. The driving force is the minimum ventilation required in the room as described by the International Mechanical Code.

If you're not air conditioning an occupied space, usually the minimum ventilation required is 6 air changes and hour. Heating that air flow up to a comfortable temperature to supply to the space is where your heating load comes from. Q=1.08*CFM*delta T. The discharge temperature can then be set to offset the heat loss of the room.