How do I calculate the future cooling demand

Replies

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  Ramani Aswath

  Thee are too many loose ends in the description. If the total area is 7700 m^2, it cannot all be in one plane. The biggest heat input seems to be from solar radiation. At any given time the rays are hitting the area at different angles, which will also be changing continuously with the time of the day. There will be a variation with the season of the year because of the variation in the angle of incidence.

  It is unlikely that the heat input fro lighting is uniform over the 7700 m^2 area.

  Unless the actual 3d geometry of the enclosed space is known it will not be possible to arrive at any meaningful figures.

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  Rasmus Brenøe Ruus

  Hi Ramani 

  What more information would you need to get a figure? It doesn't have to be a meaningful figure to begin with. 

  
  

  
  

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  Ramani Aswath

  What is this total area of 7700 m^2? Floor area? Area of walls plus floor plus roof.

  It is impossible to get any meaningful figure unless one can evaluate the total area under solar radiation. 220.7 W/sqmeter also seems very high. It is better to take the total lighting wattage installed for load calculation.

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  Rasmus Brenøe Ruus

  Hi Ramani

  
  

  The total power is 1700.05 kW from lighting it is this high because it's a greenhouse =) and the total area is floor area.

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  Ramani Aswath

  Just what I suspected. Then the roof is probably glass, maybe tinted, which will reduce the transmitted solar wattage. Greenhouses are used when the outside ambient is too cold for sustaining growth.

  Why not just circulate the outside air itself to control the temperature?

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  Rasmus Brenøe Ruus

  Hi Ramani

  it's an already installed installation, that's why they don't do that, and because in the summer it's too hot to just use the air, and in the winter it's too cold. 

  
  

  I'm glad you're so fast at responding, thanks. Will there be any easy way to calculate the future cooling demand though?

• 

  Ramani Aswath

  Since it is a green house, solar radiation will be the maximum load, added to this will be the 1700.05 kW from lighting. The peak worst case input will be 7700 x 1.2 + 1700 , which works out to a massive about 1.1 MW.

  This is assuming that he sun is directly above, which may not be the case in very north or south latitudes. Far away from the tropical latitudes, the summer sunlight hours will be more than in winter and the angle of incidence more nearly normal. Also the loss of heat through the walls will be more in winter to the much cooler ambience outside, which will reduce the amount of cooling load.

  At least in winter a suitable air-to-air heat exchanger can be used to remove excess heat from inside with appropriate temperature sensors.

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  Rasmus Brenøe Ruus

  Okay thank you very much Ramani.

  
  

  Have a great day, thanks for the help!