I posted this at ReefCentral but thought I would post here too.
I have been using fans in my hood to exhaust the heat from the metal halides to the outside. Decided that I wanted to upgrade to an external fan on the outside of the wall to pull more air through the hood as well reduce noise in the house. Question is how to size it? So let's apply a little science.
First the change in heat can be defined as
Q = mCp dT
m is the mass of the substance
Cp is the specific heat capacity of the substance
dT is the change in temperature
Since we know how much heat we are adding to the system we need to rearrange the equation to find the mass of air needed and hence the flow. Thus we get:
V = Q / p Cp (T1 - T2)
V is the volume of air
p is the density of air (1.2 Kg / cubic meters)
Cp is 1005 J/Kg K
T1 is the desired temperature of the tank in Celsius
T2 is the supply temperature in Celsius.
This equation can be applied to a hood or to a fish room. For the hood them T2 is the room temperature.
If you are applying this to a fish room then T2 would be the temperature of the air coming out of the AC vents. You would also need to subtract out the air flow of the AC system from the final calculation of the air flow. However since many people would not be able to measure this there is probably no need to really worry about it.
So let's do the calculation for my hood. I have two 250w MH. Of course there are other sources of heat in the system that you could add. But for this I am just going to use the heat from the lights. So
Q= 500 W.
T1 = 80F = 26.7 C.
T2 = 73F = 22.8 C.
V = 500 / (1.2 * 1005 * (26.7-23.9)) = 0.1063 cubic meters/ sec
Next convert to cubic feet per minute since most fans are rated this way.
0.1063 * 2118 = 225 CFM
So I selected a FanTech RVF 4 which is rated at 124 cfm, 19 watts, for 4 inch duct. I decided I could do with a little rise in heat as the next fan up was rated at 193 cfm but at 93 watts. To make sure you remove all the excess heat then you would need a fan that would exceed this number or lower the temperature of the air going in. If you have a long duct run then you would have to take into account the fan rating with pressure loss. However since I am just going through a wall I decided not to worry about it.
Now I just need to install the fan and see how well it works.
Hope this helps someone.
I have been using fans in my hood to exhaust the heat from the metal halides to the outside. Decided that I wanted to upgrade to an external fan on the outside of the wall to pull more air through the hood as well reduce noise in the house. Question is how to size it? So let's apply a little science.
First the change in heat can be defined as
Q = mCp dT
m is the mass of the substance
Cp is the specific heat capacity of the substance
dT is the change in temperature
Since we know how much heat we are adding to the system we need to rearrange the equation to find the mass of air needed and hence the flow. Thus we get:
V = Q / p Cp (T1 - T2)
V is the volume of air
p is the density of air (1.2 Kg / cubic meters)
Cp is 1005 J/Kg K
T1 is the desired temperature of the tank in Celsius
T2 is the supply temperature in Celsius.
This equation can be applied to a hood or to a fish room. For the hood them T2 is the room temperature.
If you are applying this to a fish room then T2 would be the temperature of the air coming out of the AC vents. You would also need to subtract out the air flow of the AC system from the final calculation of the air flow. However since many people would not be able to measure this there is probably no need to really worry about it.
So let's do the calculation for my hood. I have two 250w MH. Of course there are other sources of heat in the system that you could add. But for this I am just going to use the heat from the lights. So
Q= 500 W.
T1 = 80F = 26.7 C.
T2 = 73F = 22.8 C.
V = 500 / (1.2 * 1005 * (26.7-23.9)) = 0.1063 cubic meters/ sec
Next convert to cubic feet per minute since most fans are rated this way.
0.1063 * 2118 = 225 CFM
So I selected a FanTech RVF 4 which is rated at 124 cfm, 19 watts, for 4 inch duct. I decided I could do with a little rise in heat as the next fan up was rated at 193 cfm but at 93 watts. To make sure you remove all the excess heat then you would need a fan that would exceed this number or lower the temperature of the air going in. If you have a long duct run then you would have to take into account the fan rating with pressure loss. However since I am just going through a wall I decided not to worry about it.
Now I just need to install the fan and see how well it works.
Hope this helps someone.
