Ashrae 62.1-2007 addresses for mechanical rooms as 0.06 CFM/ft2. But nothing for garbage compactor. I have done a project with compactor, where the vendor of garbage compactor himself took all the responsiblity for ventialtion.
If you are in design stage and have some garbage compactor on board, they can tell you their requirements of ventilation.
1. How hot is "too hot" -- what is the temperature you're seeing now? What temperature would you like to keep as an upper limit?
2. What is the design outdoor temperature for cooling in your location? (Also, what is your location?)
3. How many square feet area and cubic feet volume is your mechanical room?
4. What are the magnitudes of the cooling loads in your room?
5. What is your current ventilation rate?
The ASHRAE ventilation rate has nothing to do with cooling, it's for air quality control. If you can provide this forum with the answers to the questions above, you'll get some good advice I bet.
How about sizing an exhaust fan for 6-10 air changes/hr and connecting it to a thermostat. When the temperature gets too high, the fan kicks on and opens a damper to an intake louver.
Not pretty, but moving air feels better than stagnant still air.
Hi All,
I've seen in a book how to solve this kind of problems. In a solution for selection of fan for mechanical control room author determines all equipment, what is power input (in kW/hr), and efficiency. After that you can calculate dissipated heat (based on power input and efficiency), and based on max allowable indoor temperature and air supply temperature you can calculate amount of air removed in order to keep temperature below allowavle temperature.
The main problem is I do not remember what was the book and I also badly needed it because I have to select the fan form my conrol room. Does Anybody know detail of this calculations?
Thank you,
Try Q=M*Cp*(Delat T)
Q = Total heat laod (W)
M = Mass Flow Rate of Air (in your case requried)(M3/s)
Cp = 4.18 for air at standard conditions
Delta T = Change in air temp required (K)
Q=M*Cp*(Delta T)
Q- total heat load (in Watts)
M- Mass flow rate of air (kg/hr)
Cp = 1.005 kJ/kg x K - Specific heat at constant temp.
Delta T temperature difference.
I found 1.005 value from book. Where your 4.18 came from?
Hi everyone,
I'd like to ask another question: What is more practical delta T between inlet and outlet air? I assume to calculate worst operating condition (hot summer, July temperature ~ +30 C) and assuming 10 C ( +40 C) outlet temperature. Am I missing something? I realize that it depends how fast enclosures will be cooled down, though.
Thank you,
