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Power and energy

The key idea

Power tells you how fast a system supplies or uses energy. Energy tells you how much it supplies or uses over a period. Plant design needs both.

Two numbers with different jobs

A 10 kW load uses 10 kWh when it operates at full power for one hour. If it operates for six hours, it uses 60 kWh. The equipment must still supply 10 kW during those hours.

Now supply the same 60 kWh in three hours. The load needs 20 kW. The energy total is unchanged, but the generator, inverter and connecting cable face a higher demand while the load operates.

This is why a customer's monthly energy bill is useful input, but cannot by itself determine plant capacity. You also need the timing and peak of demand.

Compare the same energy in two days

Start with 10 kW for six hours. Predict the daily energy before selecting the 20 kW, three-hour case. Then change the operating time while you keep power fixed.

Same energy, different capacity
Daily energy
60 kWh
Peak demand
10 kW
Load factor
25.0%
03581013kW00:0006:0012:0017:0023:00
  • Demand
Read the exact values
Chart values in kW
Time / stepDemand
00:000.00
01:000.00
02:000.00
03:000.00
04:000.00
05:000.00
06:000.00
07:000.00
08:0010.00
09:0010.00
10:0010.00
11:0010.00
12:0010.00
13:0010.00
14:000.00
15:000.00
16:000.00
17:000.00
18:000.00
19:000.00
20:000.00
21:000.00
22:000.00
23:000.00
10 kW
6 h
Compare equal energy

Model note · Constructed teaching load. Each point represents the mean power for one hour; operation starts at 08:00. No field measurements.

The area under the power curve is energy. The highest point is peak demand. The load factor compares average demand with peak demand over the stated period. A low load factor means that the peak is large relative to the average; it does not tell you whether the customer's use is efficient.

Apply the distinction to equipment

PV capacity is commonly stated as a DC power rating. Battery storage has both an energy rating in kWh and a charge/discharge power limit in kW. A generator has a power limit and consumes fuel while it operates. None of these ratings, alone, states how much useful energy the complete plant will deliver in a year.

The math, if you want itOptional — the page reads completely without it

For a load at constant power P over a time t, and for a sampled profile with hourly means Pi and step Δt:

energy from constant power

E = P · t

energy from a sampled profile

E = Σ Pi · Δt

Kilowatts multiplied by hours give kilowatt-hours. The explorer uses hourly mean power and a one-hour step, so the energy is simply the sum of the 24 bars.

load factor over the period T

LF = EPpeak · T

T is the full period, 24 hours here. The 10 kW, six-hour example gives 60 / (10 × 24) = 0.25, or 25%. The 20 kW, three-hour example has the same energy and half the load factor, 12.5%. At zero demand the explorer reports 0% by display convention.

See it in Phasor

Use the load builder to inspect daily energy and the sizing peak together. Keep the time profile attached to the design so the simulation can check when the energy is needed.

Continue the design path

Next, learn how customer timing changes the total load, then separate battery power from stored energy. For AC equipment ratings, read active and reactive power.