to calculate kw per hour

Kilowatt is a unit of power, which is energy per unit time. You can't ask how many kW you use in an hour. It's like saying I'm traveling 50 miles per hour for 24 hours, and I want to know how many miles per hour I'm using per minute.

Similarly if the maximum demand on a power station is 2700kW, that is the maximum energy per unit time. I don't see where time comes into it.

Hello, Guest

So is 2700 W the maximum demand of your system? What does this value refer to?

I'm surprised that the diversity factor is less than one. Are you sure it is not the demand factor you've cited?

Consider a conveyor belt made up of six sections, each driven by a 2 kW motor. As material is transported along this belt, it is first carried by section 1, then section 2, and each section after that in turn until the final section is reached.

In this example only one section of conveyor is carrying material at any given time. Therefore five motors are only handling no-load mechanical losses (say 0.1 kW) keeping the belts moving whilst one motor is handling the load (say 1 kW). The demand presented by each motor when it is carrying its load is 1 kW, the sum of the demand loads is 6 kW but the maximum load presented by the system at any given time is only 0.1 kW x 5 + 1 kW = 1.5 kW.

In this example the total energy (not power) used by this system in a 24-hour period is 1.5 kW x 24 hours, or 36 kW-hours. Neither the diversity factor nor the demand factor are needed for this computation.

The diversity factor for this system is:

sum of possible actual demand loads = 6 kW / max actual demand = 1.5 kW

hence

6 kW / 1.5 kW = 4.

Note that the diversity factor is greater than one. Meanwhile the demand factor for this system is:

sum of actual demand loads = 1.5 kW / maximum possible demand = 12 kW

hence

1.5 kW / 12 kW = 0.125

Does this explain things a bit?

Kilowat is power and there is no per hour it is an instantaneous amount.

What we are interested in is KWH (Kilowatt per Hour ) ie the total energy consumed.

It is like integrating the amount

So the energy consumption

= ∫P(t) dt

This is simplified as

= P_av.t = P_max.df.t

where P_max = max Power (say unit KW)

df= diversity demand factor = P_max/P_av (Unitless)

t = time

so if we take the Energy consumption in a day

then

E = P_max*df*24 KWHr : this is the most common unit and is called a unit

= P_max*df*24*60 KWMin

= P_max*df*24*60*60 KWSec

= P_max*df*24*60*60 * 1000 Joules etc.

In your case per day energy consumption

= 2700 KW * 0.85 * 24 Hr

= 55080 KWHr or 55080 Units

Since

P_max = 2700 and df = 0.85

Diversity factor refers to the average percent expected to be in use of the available connected load. Helpful when planning the panel size required. Industrial complexes would have a higher diversity factor than a commercial complex, and a residence even lower.

A diversity factor of .85 would mean that 85% of the connected load would be in use on average. Given 2700 kW as a max load, a diversity factor of .85 results in 2295 kWhr as the average usage each hour. This is average kW X one hour, or kWhr.

A full 24 hour period as stated would result in 55080 kWhr. There is no need to add the confusion of converting to 'per minute' by dividing this result by 60.

Our guest will be well served to become certain of the difference between kW and Kwhr. These terms equate to "how fast" and "how far". I hope this has helped decrease your confusion. Please reply so we may know if we have accomplished that goal.

CJM