Planning New Grow; looking for AC feedback

[SoCalInTx]

I am planning a grow room in my garage. Planning to have all the walls framed, insulated and drywalled so save on electricity costs. I live in an area that gets very hot in the summer and know I'll need to keep the new room cooled.

The dimensions of the room I am planning is 5ft deep, 8ft wide and 8ft high. Planning to have a dedicated veggie room and flowering room by splitting the width in 1/2 (4ft & 4ft) by using "panda" plastic.

I am wondering if I'll need an AC unit in both sections or if one AC unit will be enough to keep both sides at the right temp. I worry about one room being a perfect temp, but the other too hot. Planning to run one 400 or 600 watt HPS in the flowering and CFLs in the veggie room.

I'm looking at is a Frigidaire 6,000 btu and the cheapest I've found is ~$177 USD. If I go with only 1 unit how can I be sure both rooms stay within temp? And/or does anyone have another suggestion for keeping the rooms cool for less $$.

Sorry this is so long, but I want to be sure I do it right the first time and will really appreciate the help and suggestions!!

[bigtopsfinn]

This is from 00420 at ICmag:


Calculating The Size of a Air Conditioner
OK guys so this come's up as or in a thread just about every other day....
with this formula a room would be able to get in the 70-72 deg range even when it's 100+ but yet still maintain humidity at night time with proper air movement

Calculating Heat Load
Heat is measured in either BTU or Kilowatts. 1KW is equivalent to 3500BTUs.
The heat load depends on a number of factor's

1.The floor area of the room
2.The heat generated by equipment
3 the heat from lighting
4.The number of room occupants
5.The ambient temp ( your room's starting temp ) this will be added asap im still playing with the #'s

1.Floor Area of Room

Room Area BTU = L x W x 40 ( H = 8foot + 5btu per foot after that)

If you have a wall that is facing the sun add in for the extra heat

Sun facing wall BTU = L x H x 40

2.Equipment
This is trickier to calculate than you might think. The wattage on equipment is the maximum power consumption rating, the actual power consumed may be less. However it is safer to overestimate the wattage than underestimate it.

Equipment BTU = Total wattage for equipment x 3.5

Ballast in room BTU = ballast wattage/2 x 3.5

3.Lighting

Lighting BTU = Total wattage for all lighting x 4

Air cooled hood BTU = wattage x 4 / 2
@
400= 200-250 cfm
600= 250-300 cfm
1k = 300-350 cfm
^ is just a guide cfm per hood will be on your ducting/SP....

4.Occupants
even being that im only there for 2 or so of the 12 hrs i like it to be able to handle the extra sweat when i smoke one.

Total Occupant BTU = Number of occupants x 400

Total Cooling Required

Add all the BTUs together.

Total Heat Load = Room Area BTU + Total Occupant BTU + Equipment BTU + Lighting BTU

If your using a portable a/c

Total Heat Load BTU x 2

^this has been from threads on this site i dont have any real experince with portable a/c's other then one and it was not in a growroom but if you plan on getting one id think about one thats bigger then you thought about :)


sample

so if we have a 3 x 3 room with 1k and 200 watt's in fan's/pump's

for room we need 360 btu
for fan we need 700 btu
1k = 4000btu
ballast in the room = 1500 btu

we would need a/c that is 6560 btu i would round up to the next size 7kbtu


air exchange
aka intake/exhaust
complete air exchange every 4-5 minutes is average for a grow room/greenhouse (co2 control)....
for heat control (no a/c) i like to use 2.5 minutes {1M is best imo}
m=air exchange in min's

l x w x h = cf / m = cfm

lets use are room from above as a sample
3 x 3 x 8 = 72 /2.5 = 28.8 (30)
highend turn over = 72cfm
lowend turn over = 30cfm
now for the light 1k = 300-350cfm

exhaust system and air-cooled hood should be separated but most ppl cant/dont ( it's spendy) so add them together if your using 1 fan for both

the fan size we need is 330-422 remember you have duct loss (SP) in the sample room i would use a 440 cfm fan

SEER & EER ratings NO they are not the same.....

EER, or the Energy Efficient Ratio, is a measure of how efficiently a cooling system will operate when the outdoor temperature is at a specific level - usually 95°F. EER is calculated as a simple ratio of BTU's to the amount of power a unit consumes in watts. Here is an example using an air conditioner with 12,000 BTU's and consuming 1500 watts of power:

EER = BTU's / Watts

12,000 / 1500 = 8

EER = 8


While it is true that the higher the EER and BTU's, the more efficient the cooling system, many make the mistake of purchasing oversized air conditioners and ignoring EER ratings. The following is an example of an air conditioner with 12,000 BTU's and 1200 watts:

12,000 / 1200 = 10

EER =10

This would mean that this second unit can produce the same amount of cooling but more energy efficiently. Therefore, to save money on your monthly electric bill, choose a cooling system by getting an appropriately sized unit with a high EER.


SEER(seasonal energy efficiency ratio) measures how efficiently a residential central cooling system (air conditioner or heat pump) will operate over an entire cooling season, as opposed to a single outdoor temperature. As with EER, a higher SEER reflects a more efficient cooling system. SEER is calculated based on the total amount of cooling (in Btu) the system will provide over the entire season divided by the total number of watt-hours it will consume:

SEER = seasonal Btu of cooling / seasonal watt-hours used