From a power reading to an energy bill
Power tells you how quickly energy is transferred. A meter accumulates energy over an interval. For changing output, calculate each interval separately and add the MWh. Active power performs work; reactive power affects voltage and equipment loading. Three-phase formulas require line-to-line RMS voltage and balanced conditions.
How much electricity is in this hour?
Change the height or width of a block. Its area is energy; its height is power.
Second block fills the remaining 40 minutes. Original teaching data—not a real plant record.
Energy = (120 × 20/60) + (30 × 40/60)
Read the working and assumptions
These are constant powers within each block. Add interval energies, then divide by the full elapsed hour. The simple mean, 75 MW, is only reliable here when the durations are equal or both powers are the same. For changing output, use a time series or a justified ramp model—not just the highest reading.
Try 120 MW for 20 minutes and 60 MW for 40 minutes: equal energy despite different powers. This exercise does not change your quiz result.
The lesson
8 chaptersBrowse the chapters
01Power and energy: read the area, not just the height
A generator is described as ‘100 MW’. That tells you a power rating—not how much electricity it produced today. To find the energy, you also need its actual output and how long it operated.
| Quantity | Question it answers | Unit |
|---|---|---|
| Power, P | How quickly is energy being delivered? | MW |
| Energy, E | How much crossed the meter during the period? | MWh |
For one interval
E = P̄ × Δt
- E
- Energy delivered (MWh)
- P̄
- Interval-average power (MW)
- Δt
- Elapsed time (hours)
Use average power for the interval. Convert minutes to hours before multiplying.
Two different outputs. The same energy in each interval.
Work the half-hour total
Convert each duration
20 ÷ 60 = ⅓ h; 10 ÷ 60 = ⅙ h
The MW readings cover unequal lengths of time.
Calculate and add the energy
(60 × ⅓) + (120 × ⅙) = 20 + 20 = 40 MWh
Calculate each block separately; then add quantities with the same unit.
Recover the average power
40 MWh ÷ 0.5 h = 80 MW
Divide the total energy by the whole elapsed period.
Result: 40 MWh delivered; 80 MW average across 30 minutes.
Check your understanding
If 120 MW lasted 20 minutes instead, with the first interval unchanged, what would the totals be?
Show reasoning
60 MWh over 40 minutes; average power 90 MW.
60 × 20/60 + 120 × 20/60 = 60 MWh. The two durations are now equal, so the simple average also works.
02Capacity, output and meter readings are not interchangeable
A plant can have a 200 MW nameplate, offer less than 200 MW and produce something different again. Before using a number, identify which record it came from.
| Record | What it describes | What to check |
|---|---|---|
| Nameplate capacity | An equipment rating under specified conditions | Gross or net rating; stated conditions |
| Available capacity | What can presently be offered or used | Outages, fuel and temperature limits |
| Dispatch target | The requested operating level | Target time and ramping requirements |
| Metered output | What was actually measured | Metering boundary, interval and revisions |
Capacity factor for a stated period
CF = Eactual ÷ (Prated × T) × 100%
This example uses nameplate capacity. A published dataset may use a different capacity basis: record it and keep gross/net boundaries consistent.
A six-hour operating record
Calculate reference energy
200 MW × 6 h = 1,200 MWh
This is the energy corresponding to continuous nameplate output for the period.
Compare measured energy
600 ÷ 1,200 × 100% = 50%
The meter records 600 MWh across those same six hours.
Result: Capacity factor = 50% for these six hours—not necessarily for the year.
A cumulative meter is a running total
If readings rise from 12,340 MWh to 12,380 MWh, the period energy is 40 MWh. Check meter resets, units, revisions and the timestamps before subtracting. The closing reading alone is not that period’s consumption.
Check your understanding
A 200 MW plant exports 900 MWh over the same six hours. What changed?
Show reasoning
Capacity factor rises to 75%; the nameplate remains 200 MW.
900 ÷ (200 × 6) = 0.75. A larger energy total is not evidence of a larger equipment rating.
03Voltage and current: calculate the power, then the loss
| Quantity | Meaning | Unit |
|---|---|---|
| Voltage, V | Electrical potential difference | volt (V) |
| Current, I | Rate of electric charge flow | ampere (A) |
| Resistance, R | Opposition in an ideal resistive element | ohm (Ω) |
Steady DC at the stated terminals
P = V × I
Volts multiplied by amperes gives watts. AC calculations also depend on waveform and phase relationships.
A 48 V device draws 20 A
Calculate the terminal power
48 × 20 = 960 W = 0.96 kW
This is electrical input at the device terminals, not automatically its useful mechanical or computing output.
Add a duration
0.96 kW × 2 h = 1.92 kWh
Assume voltage and current stay constant for the two hours.
Result: Power: 0.96 kW. Two-hour energy: 1.92 kWh.
Heating in a fixed resistance
Ploss = I² × R
Ohm’s law V = I × R applies to the ideal resistive element. Hold R constant when comparing currents.
Double the current; four times the resistive loss
04Active, reactive and apparent power
An 80 kW load does not necessarily require only 80 kVA of equipment capacity. The voltage and current relationship matters as well as useful energy transfer.
| Symbol | Quantity | Typical unit |
|---|---|---|
| P | Active power: average net energy-transfer rate | kW or MW |
| Q | Reactive power: reactive component of AC power | kvar or Mvar |
| S | Apparent power: RMS voltage-current loading | kVA or MVA |
An 80–60–100 power triangle
Under sinusoidal conditions
S² = P² + Q²; power factor = P ÷ S
Use compatible prefixes. With harmonics or unbalance, the simplified triangle may be insufficient.
Read the example
Calculate power factor
80 kW ÷ 100 kVA = 0.8
Power factor is a ratio, not a measure of fuel-conversion efficiency.
Find reactive-power magnitude
√(100² − 80²) = 60 kvar
Do not add 80 + 60 to find apparent power: the components form a right triangle.
Result: S = 100 kVA, P = 80 kW, |Q| = 60 kvar; power factor = 0.8.
Balanced sinusoidal three-phase power
P = √3 × VLL × Iline × power factor
- VLL
- Line-to-line RMS voltage (V)
- Iline
- RMS line current (A)
Use line-to-line RMS voltage and line current. Volts and amperes give watts; divide by 1,000 for kW.
One substitution
At 400 V line-to-line, 100 A and power factor 0.8, P = √3 × 400 × 100 × 0.8 ≈ 55.4 kW. Using 230 V phase-to-neutral in the same expression without conversion would give the wrong result.
05Turn the physical quantities into a commercial calculation
A price in A$/MWh cannot be multiplied by MW alone to produce dollars. First calculate the energy sold or purchased in the interval.
Unadjusted energy value
Value = P̄ × Δt × energy price
Units reconcile as MW × h × A$/MWh = A$. This is dimensional arithmetic, not a complete market settlement calculation.
48 MW exported for five minutes
Find interval energy
48 × (5 ÷ 60) = 4 MWh
Use the average export at the stated metering point.
Apply the illustrative price
4 × A$250/MWh = A$1,000
A synthetic, constant price is used for this interval.
Result: A$1,000 of unadjusted energy value, before losses and other settlement adjustments.
The same distinction explains battery duration
A battery’s MW limit controls the delivery rate; usable MWh limits how long it can sustain that rate. Both numbers need the same delivery boundary.
Ideal constant-output duration
Duration = usable energy ÷ output power
60 deliverable MWh ÷ 50 MW = 1.2 hours = 72 minutes. Assume no other commitments reduce the available energy or power.
06Check the number before using it in a decision
A useful result is a number that someone else can reconstruct. Before sharing it, answer these five questions.
| Check | Question |
|---|---|
| Quantity | Is this power, energy, capacity or an accumulated meter reading? |
| Units | Have minutes become hours, and kW become MW where necessary? |
| Boundary | Is it measured at the generator, site connection, battery or customer meter? |
| Time and sign | Which period is covered? Is export positive, and are intervals equal? |
| Feasibility | Does the result respect power, usable-energy and duration limits? |
Check your understanding
A colleague reports ‘100 MW × A$80/MWh = A$8,000’. What information is missing?
Show reasoning
The duration and the meaning of the 100 MW figure.
If 100 MW is actual average output over one hour, energy is 100 MWh and unadjusted value is A$8,000. A nameplate rating alone cannot establish either amount.
Where to apply this next
Use these checks in dispatch and settlement (Topics 14 and 17), storage (Topics 5 and 37), and data-centre demand (Topic 42). Keep physical calculations separate from contract terms and current market rules.
07A ramp is a slope, not a rectangle
The operator's question
A unit starts at 0 MW and reaches 120 MW after 30 minutes. A colleague multiplies 120 by 0.5 hours and reports 60 MWh. That assumes the unit was at 120 MW for the entire half-hour. The end reading alone cannot establish the delivered energy.
Only for an assumed linear ramp
E = [(Pstart + Pend) / 2] × Δt
Use hours for duration. The relationship is exact for a linear ramp, not for arbitrary output between two samples.
Calculate the original ramp example
State the shape
0 → 120 MW, increasing linearly
A straight-line rise forms a triangle under the power curve.
Find average power
(0 + 120) / 2 = 60 MW
This endpoint mean works because the change is linear.
Integrate over time
60 × 0.5 = 30 MWh
The endpoint-as-rectangle estimate overstates this example by 30 MWh.
Result: 30 MWh for the assumed linear ramp; verify the shape against interval data.
| Evidence available | Useful method | Limitation |
|---|---|---|
| Interval-average MW | Multiply each average by its duration; add | Confirm averaging period and missing intervals |
| Cumulative energy meter | Subtract consecutive readings | Check reset, rollover and boundary |
| Only start and end MW | Use an explicit ramp assumption or request more data | Endpoints do not reveal intermediate trips or plateaus |
Check your understanding
If the plant reaches 120 MW immediately and stays there for the same half-hour, is 30 MWh still correct?
Show reasoning
No. That different profile delivers 60 MWh.
120 MW × 0.5 h = 60 MWh. Same end power and elapsed time do not imply the same energy.
08Turn a calculation into a defensible operating note
From record to decision
- 01
Define
Name the meter, timezone, interval and whether power is gross generation or net export.
- 02
Reconcile
Align durations; identify gaps, duplicates and revisions before multiplying or summing.
- 03
Explain
Report the result, uncertainty and the decision it can actually support.
Why the meter matters
Synthetic example: a generator produces 100 MW while auxiliaries consume 4 MW, both constant for 30 minutes. Gross energy is 50 MWh. Net export at that boundary is 48 MWh. Applying a price to gross output instead of the relevant settled quantity changes the commercial conclusion.
Same time period, stated boundary
Enet = (Pgross − Pauxiliary) × Δt
Only when these are matching average powers at a consistent boundary; other losses and settlement adjustments are excluded here.
Check your understanding
Write the two-sentence note for the 100 MW example.
Show reasoning
The plant exported an estimated 48 MWh over the stated 30-minute period, assuming constant 100 MW gross output and 4 MW auxiliary load. Confirm the net-export meter and any other applicable adjustments before using the estimate for settlement.
The note states quantity, period, boundary, assumptions and the evidence still required. It does not claim that this teaching estimate is an actual settlement amount.
Watch the explanation
Key terms in plain English
- Voltage
- Electrical potential difference: the driving condition for current in a circuit. Voltage level is not itself a quantity of energy.
- Power factor
- Active power divided by apparent power. A lower power factor can mean more current is needed for the same active power; it is not the same as energy efficiency.
- MW
- Megawatt: a measure of power, or how quickly energy is transferred. A plant's maximum MW is not its actual output in every interval.
- MWh
- Megawatt-hour: a quantity of energy. A steady 1 MW for one hour delivers 1 MWh.
Supporting diagrams
From power to energy
Add energy across intervals, not unweighted power readings.
- 01 · First interval60 MW × ⅓ h
20 minutes → 20 MWh
- 02 · Second interval120 MW × ⅙ h
10 minutes → 20 MWh
- 03 · Total40 MWh
Over 30 minutes → 80 MW average