Is it possible to meet the cooling demand with just the use of fan coil units. There will be lake breeze but the breeze pattern is unavailable. The cooling load of each room is below 1 ton.

Subin Where do you live? What is the normal relative humidity in the summer? Where would you get the cold water for the fan coil device, and what temp would it be at? A 1 ton load is 12,000 btu’s per hour. If you assume you can get a 10 degree difference between incoming water and outgoing water across the fan coil (big if) your flow rate would have to be 12000 btu/s per hour divided by(8 btu/s per gallon * 10 degrees difference) or 150 gallons/hour or 2.5 gallons per minute, per room. For there to be any latent heat removal ie humidity the temp of the water in the coils must be quite a bit below the dew point, and condensate must be able to drain out somehow. More questions than answers. Cheers, Eric

Thank you Mr. Anderson for the quick reply, I have done the calculation and its 9800 Btu/hr. So it would require 110 gallons/hr(with 10 degree difference) just to cool the room. The RH is quite high reaching the maximum of 85. I am yet to decide on the water source and its temperature. I am thinking of solar cooling but it doesn't seem cost effective, so in a bit of dilemma here.

Big radiators (like an entire floor, wall or ceiling) work best with moderate temperatures. In a well sealed and properly pressure balanced building, latent load can be so low that a little bit of refrigerant based cooling/dehumidification applied to the ventilation air is sufficient to handle humidity.

IMO, using uncooled ground water for 3/4 of cooling load is possible in some cases. Using nighttime cooled (evaporative, radiant and convective) water should be possible in climates with hot days and cool nights.

Except for dry climates, I would not use a cooling system that had no dehumidification capability.

It's been pretty slow around these parts lately. Likely because everyone is now fully engaged during the work week with actual construction. For those who wonder why weight tonnage is used to rate AC cooling capacity...

One ton of AC cooling capacity is equivalent to the cooling capacity that would be obtained by using one ton (i.e., 2000 pounds) of ice per day to accomplish this cooling, which is equivalent to 12,000 BTU/hour of cooling capacity.

So, how was this 12,000 BTU/hour cooling capacity derived?

When ice is at its melting point of 32° F and absorbs heat, the temperature of the ice does not change. Instead, the ice melts to form liquid water. The amount of heat needed to melt ice into liquid water at 32° F is called the latent heat of fusion which has a value of 144 BTUs per pound. If you have a pound of ice at 32° F, it takes 144 BTUs of heat to completely melt all of it.

If you have one ton of ice, it takes 144 BTUs per pound times 2000 pounds or 288,000 BTUs of heat to melt it completely. Producing this amount of heat at 100% efficiency would require about 84 KWH of electricity, 2 gallons of fuel oil, 282 cubic feet of natural gas, 3 pounds of propane, or 31 pounds of dry Douglas fir. You could completely melt this one ton of ice in one hour, one day, or one year, depending on how quickly you heat it. At some point, it was decided to use one day for this standard. So, if one ton of ice melts uniformly over one day or 24 hours, it absorbs heat at the rate of 288,000 BTUs divided by 24 hours or 12,000 BTU/hour.

A slight educational digression… After the ice has been melted into liquid water, it only takes 1 BTU to raise the temperature of one pound of liquid water by 1° F. This also happens to be the definition of a BTU (British Thermal Unit). So, it only takes an additional 180 BTUs to raise the temperature of this one pound of liquid water from 32° F to 212° F, the boiling point. When water is at its boiling point of 212° F and absorbs heat, the temperature of the water does not change. Instead, the water vaporizes to form steam. The amount of heat needed to vaporize water at 212° F is called the latent heat of vaporization which has a value of 970.4 BTUs per pound. If you have a pound of liquid water at 212° F, it takes 970.4 BTUs of heat to completely vaporize all of it. So, if you are starting out with a pound of ice that you want to melt, heat up to 212° F and completely vaporize into steam, it will take a total of 1294.4 BTUs (i.e., 144 plus 180 plus 970.4) of heat to accomplish this. If you have one ton of ice, it takes 1294.4 BTUs per pound times 2000 pounds or 2,588,800 BTUs of heat to vaporize it completely. Producing this amount of heat at 100% efficiency would require about 758 KWH of electricity, 19 gallons of fuel oil, 2,538 cubic feet of natural gas, 28 pounds of propane, or 280 pounds of dry Douglas fir.