Why cheap energy is not always available energy
A factory needs power through the evening. Compare what different technologies can actually deliver, what it costs to run them, and what an annual cost figure leaves out.
The lesson
8 chaptersBrowse the chapters
01Start with the hour you need to supply
A factory needs 50 MW from 18:00 to 22:00. A developer offers a 100 MW solar farm. Is that enough? Comparing the two nameplates is tempting, but it answers the wrong question. The factory needs 200 MWh during particular hours, not an annual total produced whenever the weather permits.
Three questions before comparing a technology
- 01
Energy
How many MWh, measured at which boundary?
- 02
Timing
During which hours, and for how long?
- 03
Delivery
Can the plant and network supply those hours?
What you will work out
Trace how a plant produces electricity, calculate a thermal unit’s running cost, and distinguish annual production from dependable output. Then test two ways to serve the factory’s four-hour evening load. Topic 1 supplies the MW-to-MWh calculation; this unit adds technology and cost.
Check your understanding
The solar farm produces more than 200 MWh over the day. Does that settle the factory’s question?
Show reasoning
No. Daily energy sufficiency is not hourly matching.
You still need the evening output profile, any storage or alternative supply, and the limits at the connection point. The factory scenario uses invented numbers throughout this unit.
02Follow the energy through the plant
Technology names conceal different conversion chains. Coal and gas plants turn fuel into heat and mechanical rotation before generating electricity. Solar PV converts light into DC electricity and uses an inverter to supply AC. Wind turns a rotor; hydro uses moving water. These differences determine which inputs, losses and operating limits matter.
A thermal plant’s energy boundary
- 01
Fuel in
7.5 GJ of fuel energy
- 02
Conversion
Turbine, generator and auxiliaries
- 03
Net electricity
1 MWh = 3.6 GJ exported
Convert heat rate to net efficiency
η = 3.6 ÷ H
- H
- Fuel energy per net electrical output (GJ/MWh)
- η
- Net electrical efficiency (fraction)
Use net electrical output and the same fuel heating-value basis throughout. Multiply the ratio by 100 for a percentage.
Two plants, the same net electricity
Plant A
3.6 ÷ 7.5 = 0.48 = 48%
One net MWh requires 7.5 GJ of fuel.
Plant B
3.6 ÷ 10 = 0.36 = 36%
One net MWh requires 10 GJ. Higher heat rate means more fuel, not higher efficiency.
Result: Plant A uses 25% less fuel per net MWh than B: (10 − 7.5) ÷ 10. This does not yet tell you which plant is cheaper to build or quicker to start.
03Compare operating roles, not just fuel types
No single ranking covers every job. A plant that is economical over a long run may be unsuitable for a short, unexpected peak. Use the table as a checklist for a particular asset, not as a claim that every plant in a technology family behaves identically.
| Technology | Useful role | Evidence to request |
|---|---|---|
| Coal-fired steam plant | Sustained output when committed | Minimum stable load, start time, fuel supply, outages, cooling and ramp limits. |
| Combined-cycle gas turbine (CCGT) | Recover exhaust heat to produce additional power | Heat rate across loading levels, start costs, gas deliverability and ambient-temperature derating. |
| Open-cycle gas turbine (OCGT) | Meet short peaks where its start and ramp capability suit the task | Start readiness, ramp test, fuel availability and cost for the expected operating hours. |
| Wind and solar PV | Produce energy when the resource is available | Time-series resource and output, clipping, curtailment, connection limits and correlated low-output periods. |
| Reservoir hydro | Schedule output within water and plant limits | Usable water, head, inflows, environmental releases and the value of saving water for later. |
| Battery / pumped storage | Move energy between periods; provide qualified services | Usable MWh, charge state, MW limit, efficiency, recharge opportunity and competing commitments. |
Check your understanding
A 100 MW unit is at 40 MW and can ramp at 3 MW/min. Can it deliver 80 MW five minutes later?
Show reasoning
Not under these assumptions: the limit is 55 MW.
40 + 3 × 5 = 55 MW. Nameplate capacity does not override ramp limits. This simplified calculation also assumes fuel, network and other limits do not bind.
04Build the running cost, one unit at a time
A simplified thermal running cost
c = H × f + v
- f
- Delivered fuel price (A$/GJ)
- v
- Variable operations and maintenance (A$/MWh)
- c
- Cost per additional electrical output (A$/MWh)
Illustrative short-run calculation, excluding starts, no-load costs and any additional applicable obligations.
Gas at A$12/GJ
Fuel
7.5 GJ/MWh × A$12/GJ = A$90/MWh
The GJ units cancel. Use delivered fuel cost at the plant, not an unrelated hub quotation.
Variable operation
A$90 + A$5 = A$95/MWh
The A$5/MWh assumption covers incremental operation and maintenance only.
Fuel-price sensitivity
7.5 × (14 − 12) = A$15/MWh
At A$14/GJ, running cost rises to A$110/MWh, holding heat rate and other costs unchanged.
Result: A fuel-price change passes through in proportion to heat rate. A higher-heat-rate unit is more exposed per MWh.
A positive margin may not justify a start
Suppose price is A$120/MWh and running cost is A$95/MWh. Producing 100 MWh earns A$2,500 before start and other costs. If a required start costs A$4,000, that isolated run loses A$1,500 on this incremental calculation. A unit already online faces a different decision. Do not allocate sunk construction cost to an extra MWh and then call it marginal cost.
Check your understanding
How much output covers that A$4,000 start at a constant A$25/MWh margin?
Show reasoning
160 MWh before any other incremental costs.
4,000 ÷ 25 = 160 MWh. This is a break-even energy calculation, not permission to dispatch: feasible duration, loading, prices and operating limits still matter.
05Annual MWh do not tell you when energy arrives
An annual-energy estimate
E = capacity × 8,760 × capacity factor
Non-leap year. Define the capacity and metering boundary consistently; this is not a dependable-capacity calculation.
A synthetic 100 MW project
Estimate annual production
100 × 8,760 × 0.30 = 262,800 MWh
A 30% capacity factor gives an average of 30 MW across the year.
Identify what is missing
Nothing in this annual total tells you the output during the factory’s four evening hours, a hot still week, or a network outage.
Result: Use interval profiles for matching supply to load. Use multiple weather years and outage assumptions for reliability analysis; one average is insufficient.
The price received depends on the production profile
Consider two one-hour intervals priced at A$20 and A$100/MWh. A plant generates 80 MWh in the first and 20 MWh in the second. Revenue is A$3,600, so its energy-weighted capture price is A$36/MWh—not the unweighted market average of A$60/MWh. The cheapest production hours can also be the lowest-price hours.
Capture price
Capture price = Σ(Eᵢ × pᵢ) ÷ ΣEᵢ
Use matched intervals and a consistent price and output boundary. This simple example excludes losses, contracts and other revenues.
06Choose the cost measure that answers your question
| Question | Useful measure | What it cannot establish alone |
|---|---|---|
| Should another MWh be produced now? | Incremental operating cost, opportunity cost and feasible output | A dispatch outcome or lifetime investment return. |
| How do new generation projects compare? | Levelised cost of electricity (LCOE) with consistent assumptions | The value of output at particular times or a profitable contract price. |
| What portfolio can meet demand? | Time-resolved system modelling and system costs | The cash flow of a particular investor. |
| Will this project earn an adequate return? | Project cash flow, NPV and sensitivities | Reliability of the entire electricity system. |
CSIRO’s GenCost distinguishes individual-technology costs from system-level modelling. Read the methodology and scenario alongside a cost chart. Ask which generation, storage and network costs are inside the boundary, which year the costs represent, and how financing and utilisation are treated. Do not add a second firming allowance to a metric that already includes it.
A bid is not an audited cost statement
In the NEM, AEMO dispatches through a centrally coordinated process. Offers, availability and operating limits matter; your simplified fuel calculation is not itself an observed offer or dispatch target. A claim about an actual interval needs the market records, not just a plausible cost narrative.
Check your understanding
Project A has a lower LCOE than B. Is A necessarily the better supplier for the factory’s evening demand?
Show reasoning
No—compare delivery during those hours and the cost of the remaining supply.
The screening metric does not settle profile matching, network access, contract allocation or investment risk. It is useful within its stated boundary, not a universal ranking.
07Serve the factory’s four-hour evening load
Return to the 50 MW factory. For this synthetic screen, demand is constant for four hours; grid transfer is unconstrained and neither option has an outage. Ignore taxes, network charges, capital recovery and contract settlement. We are comparing the stated operating costs—not choosing an investment.
| Input | Gas option | Storage option |
|---|---|---|
| Required output | 50 MW for 4 hours | 50 MW for 4 hours |
| Plant capability | At least 50 MW; ready for the required start | 50 MW / 200 MWh usable AC-delivered energy, fully charged |
| Operating assumptions | 7.5 GJ/MWh; gas A$12/GJ; variable O&M A$5/MWh | 80% AC round-trip efficiency; charging A$40/MWh; degradation A$8/MWh discharged |
| Additional assumption | A$4,000 start required | Enough charging time and connection capacity; no competing use of stored energy |
Base-case comparison
Energy to deliver
50 MW × 4 h = 200 MWh
Both options must meet the same output boundary.
Gas
200 × (7.5 × 12 + 5) + 4,000 = A$23,000
A$19,000 running cost plus one A$4,000 start.
Storage
(200 ÷ 0.80) × 40 + 200 × 8 = A$11,600
Charging purchases are 250 MWh. Charging costs A$10,000 and assumed degradation costs A$1,600. Do not apply efficiency again to the stated AC-deliverable 200 MWh capacity.
Result: Storage has the lower specified operating cost in this base case, by A$11,400. This is not proof of lower whole-life cost, firm availability or the best use of a scarce charge.
Check your understanding
What happens if charging power costs A$100/MWh?
Show reasoning
Storage rises to A$26,600; gas remains A$23,000 under the fixed assumptions.
250 × 100 + 1,600 = 26,600. The break-even charging price is (23,000 − 1,600) ÷ 250 = A$85.60/MWh. Cost rankings can reverse.
08Turn the calculation into an analyst’s recommendation
A useful brief names the decision, reports a result and states what would change it. It does not announce that one technology is universally best. Your reader should be able to reproduce the arithmetic and see which evidence is still missing.
| Evidence | Why it could change the answer |
|---|---|
| Load profile and critical-load requirements | Defines hours, peaks, flexibility and the consequence of a supply gap. |
| Plant operating envelope and outage history | Tests whether output is physically available when required. |
| Delivered fuel terms / interval charging prices | Replaces assumed input costs and checks transport or charging constraints. |
| Network studies and connection agreement | Tests deliverability rather than assuming a copper-plate system. |
| Capital, maintenance and contract cash flows | Moves from an operating-cost screen to an investment or procurement decision. |
Continue the analysis
Topic 6 follows the gas supply chain; Topics 13–15 explain offers, dispatch and constraints. Topic 23 builds the investment cash flow. Topics 36–38 examine renewable yield, batteries and longer-duration storage. The same physical quantities should remain consistent across all of them.
Watch the explanation
Key terms in plain English
- MWh
- Megawatt-hour: a quantity of energy. A steady 1 MW for one hour delivers 1 MWh.