a simple method to figure WPSF

[Mos`Hamid AKzurie]

1) Measure the room's width and length. You can also use the floor plans to determine the room's size.

2) Convert the room length and width into feet if you're using a floor plan that gives these dimensions in yards or meters. If the length and width are measured in yards, multiply the values by three. If the they are given in meters, multiply the numbers by 3.280839895. For example, a room is 5.6 by 3.2 meters converts to 18.37 (5.6 x 3.280839895) by 10.50 (3.2 x 3.280839895) feet.

3) Multiply the length of the room by width to calculate the area in square feet. In the previous example, the area of the room was 18.37-by-10.5--or 192.89 square feet.

4) Obtain energy consumption (in watts) in a room from documentation, specifications or elsewhere. For example, the lighting in the rooms comes from six 75-watt and two 100-watt lamps. That corresponds to 6 x 75 + 2 x 100 = 650 watts.

5) Divide the wattage consumed in the room by its area in square feet to calculate watts per square foot. In our example, 650 watts divided by 192.89 square feet is 3.37 watts per square feet.

Lets say the main service switchboard is rated for 1,600-ampere, 277/480-volt, 3-phase, 4-wire service with room for another identical main service switchboard. The existing configuration will provide the equivalent of approximately six watts per square foot of gross building area for this mostly open storage warehouse building with day lighting. Industrial Bldg. is 183,894 square feet.

Ohm's Law. The potential difference Vab across a resistance R from a to b with a current I from a to b is given by Ohm's Law, an Empirical law, Vab = IR.

My main question... Is there a formula to determine the watts per square foot, including the power factor? It has been too long since I worked with this and cannot make sense of it anymore. I forgot what the PF is as well. Any help is appreciated. Thanks!!

[BAMF]

Well, first of all the fact that the gear is rated for 1600 amps does not mean that it is served by a transformer capable of supplying 1600 amps. But let’s take for granted, for now anyway, that it is.

1600 amps times 480 volts times 1.732 (the square root of three, a term that always finds its way into three phase calculations) gives you a total of 1,330,000 VA, or 1,330 KVA. If all the loads were resistive (i.e., no motors, no inductive lighting ballasts, no welders), you could use this value as though it were the same as watts. Thus, in your building of 183,894 square feet, you have 1,330,000/183,894, or about 7.2 watts per square foot.

If there are inductive loads, and this is likely, you would get watts from VA by multiplying VA times the power factor. Let’s assume a power factor of 0.90, just for discussion.

1,330,000 VA times 0.9 is 1,197,000 watts. Then, 1,197,000/183,894 is about 6.5 watts per square foot.

[BAMF]

If you have a small wind turbine with a blade diameter of 1 m (about 3 ft) and an operating efficiency of 20% at a wind speed of 6 m/sec (about 13.4 mph). Then, to calculate how much power the turbine can generate at this wind speed:

Rotor swept area: Area = Π × (Diameter/2)2 = 3.14 × (1/2)2 = 0.785 m2

Available power in the wind: Pwind= Air Density × Area × v3/2 = 1.2 × 0.785 × 63/2 = 101.7 watt

Then the power that can be extracted from the wind assuming 20% turbine efficiency is:
Pturbine=0.20 × 101.7 = 20.3 watts

If this ran continuously for a year (about 8,750 hours) then it would produce: 20.3 watts × 8,750 hours = 177,625 watt-hours, or about 177 kWh in a year.

[Mos`Hamid AKzurie]

I may use solar as additional means of energy. How would I calculate for energy?

[BAMF]

On average (as a general "rule of thumb") modern photovoltaics (PV) solar panels will produce 8 - 10 watts per square foot of solar panel area. For example, a roof area of 20 feet by 10 feet is 200 square-feet (20 ft x 10 ft). This would produce, roughly, 9 watts per sq-foot, or 200 sq-ft x 9 watts/sq-ft = 1,800 watts (1.8 kW) of electric power.

One kilowatt-hour (1 kWh) means an energy source supplies 1,000 watts (1 kW) of energy for one hour. Generally, a solar energy system will provide output for about 5 hours per day. So, if you have a 1.8 kW system size and it produces for 5 hours a day, 365 days a year: This solar energy system will produce 3,285 kWh in a year (1.8 kW x 5 hours x 365 days).

If the PV panels are shaded for part of the day, the output would be reduced in accordance to the shading percentage. For example, if the PV panels receive 4 hours of direct sun shine a day (versus the standard 5 hours), the panels are shaded 1 divided by 5 = 20% of the time (80% of assumed direct sun shine hours received). In this case, the output of a 200 square-foot PV panel system would be 3,285 kWh per year x 80% = 2,628 kWh per year.

Hope this helps