elpepe2
Computer
- May 31, 2007
- 6
Hi,
I need some advise about the specifications of an air conditioner.
I have a system that dissipates heat through a rectangular flat piece of stainless steel (a panel). Heat rate is 600 W. I need to keep the surface of the panel at some T1 temperature (about 15 Cent degrees). I am evaluating in buying a commercial air conditioner to provide cold air over the surface of the panel. Certainly I thought in a closed loop layout because I need to avoid condensation around the set up, and I am working with temperatures around the dew point of the enviroment(the panel is thermally isolated from the enviroment). Some of the air conditioner I saw also have dehumidification facilities, wich is a plus.
From the specifications I found that some provide 400 CFM at a 6" nozzle, which leads to 0.18 m3/s, and this means air passing over the panel at 10 m/s.
I made some calculations using the Newton's Law of cooling:
Qdot = h*A*(T1-Ta)
where Qdot is the heat rate exchange form the panel to the fluid, h is the heat transfer coefficient, and Ta is the air temperature.
I found for a flat panel and a fluid flowing parallel to it that the Nusselt number can be expressed:
Nu = h*L/k = 0.037*Re^(4/5)*Pr^(1/3)
Where L is the length in the direction of the flow,k is the dynamic conductivity of the air, Pris the Prandtl number = 0.71 (air at 15 Cent degrees), and the Re is the Reynold number = 4.1 e+5 in my case, which corresponds to a turbulent flow.
That leads to a heat transfer coefficient of 39 W/(m2*K).
In my geometry it means I need to send air at 2 cent degrees!!!. This is not what I was thinking about. I mean, at first I was thinking that the air should to be at a lower temperature from the panel but not so low...
Now, I revised my assumptions and maybe some adjustments can be done.
1- First, the configuration should not be a flat panel whith air flowing over it. As I need a closed loop, the panel itself could be thought one of the flat faces of a rectangular duct. I think in this case the Nusselt number should be different, but I could not find any expression for it yet. Could anybody help me in this subject?
2- I have an additional problem. Air conditioners has nozzle of circular section, so I need to adapt the inlet and outlet nozzles to a rectangular section (or maybe a semicircular section)for making my panel a part of the duct. This adaptations will lead to a pressure drop in the system, but I don't know how to estimate it. Manufacturers of air conditioner I saw do not provide the fan curve of their equipment but knowing the pressure drop maybe helps in future information requests. In this point I need some tips or guide for calculating the pressure drop.
Thanks in advance,
I need some advise about the specifications of an air conditioner.
I have a system that dissipates heat through a rectangular flat piece of stainless steel (a panel). Heat rate is 600 W. I need to keep the surface of the panel at some T1 temperature (about 15 Cent degrees). I am evaluating in buying a commercial air conditioner to provide cold air over the surface of the panel. Certainly I thought in a closed loop layout because I need to avoid condensation around the set up, and I am working with temperatures around the dew point of the enviroment(the panel is thermally isolated from the enviroment). Some of the air conditioner I saw also have dehumidification facilities, wich is a plus.
From the specifications I found that some provide 400 CFM at a 6" nozzle, which leads to 0.18 m3/s, and this means air passing over the panel at 10 m/s.
I made some calculations using the Newton's Law of cooling:
Qdot = h*A*(T1-Ta)
where Qdot is the heat rate exchange form the panel to the fluid, h is the heat transfer coefficient, and Ta is the air temperature.
I found for a flat panel and a fluid flowing parallel to it that the Nusselt number can be expressed:
Nu = h*L/k = 0.037*Re^(4/5)*Pr^(1/3)
Where L is the length in the direction of the flow,k is the dynamic conductivity of the air, Pris the Prandtl number = 0.71 (air at 15 Cent degrees), and the Re is the Reynold number = 4.1 e+5 in my case, which corresponds to a turbulent flow.
That leads to a heat transfer coefficient of 39 W/(m2*K).
In my geometry it means I need to send air at 2 cent degrees!!!. This is not what I was thinking about. I mean, at first I was thinking that the air should to be at a lower temperature from the panel but not so low...
Now, I revised my assumptions and maybe some adjustments can be done.
1- First, the configuration should not be a flat panel whith air flowing over it. As I need a closed loop, the panel itself could be thought one of the flat faces of a rectangular duct. I think in this case the Nusselt number should be different, but I could not find any expression for it yet. Could anybody help me in this subject?
2- I have an additional problem. Air conditioners has nozzle of circular section, so I need to adapt the inlet and outlet nozzles to a rectangular section (or maybe a semicircular section)for making my panel a part of the duct. This adaptations will lead to a pressure drop in the system, but I don't know how to estimate it. Manufacturers of air conditioner I saw do not provide the fan curve of their equipment but knowing the pressure drop maybe helps in future information requests. In this point I need some tips or guide for calculating the pressure drop.
Thanks in advance,
