Plant sizing and network checks
The key idea
A plant can have enough annual energy while its feeder has excessive voltage drop or current. Energy adequacy and electrical delivery need separate checks.
Two different boundaries
Plant sizing compares generation, demand and storage across time. It asks how equipment and dispatch serve the load. Network analysis uses connectivity, impedance, voltage and source controls to calculate what reaches each bus.
An annual surplus is not even proof of hourly energy adequacy. The surplus may occur at midday while demand is unmet at night. After resolving that timing problem, you still need to test the network.
The common mistake is to read a successful energy result as approval of every cable and transformer. Those limits are not established by the annual balance.
Keep the plant fixed and change the feeder
The teaching plant has 60 kW capacity, 120 MWh of annual available energy and 100 MWh of annual demand. Those totals stay fixed. The feeder screen uses a separate 40 kW endpoint operating case.
Increase the feeder length. Then select the larger conductor. Watch voltage drop and losses change even though no plant rating changes.
Annual energy ledger · fixed
- Plant capacity
- 60 kW / 0.06 MW
- Available energy
- 120 MWh
- Demand
- 100 MWh
- Voltage drop
- 6.82%
- Thermal loading
- 116.6%
- Peak line loss
- 3.22 kW
Voltage drop exceeds the teaching 5% limit. Current exceeds the assumed rating. Annual energy totals above stay unchanged.
Read the values and assumptions
| Size | R (Ω/km) | X (Ω/km) | Rating (A) |
|---|---|---|---|
| 35 | 0.87 | 0.08 | 55 |
| 70 | 0.44 | 0.08 | 100 |
Model note · Constructed scenario: 60 kW plant, annual available energy 120 MWh and annual demand 100 MWh. These aggregate totals do not prove hourly adequacy. Feeder screen: 40 kW endpoint load, 400 V, power factor 0.9; fixed-current linear approximation. A 5% drop screen and assumed conductor ratings are teaching limits, not a utility standard. Loss is peak kW, not annual kWh.
The assumed small conductor is also below the required current rating at this operating point. Shortening it reduces voltage drop but does not raise its assumed continuous-current rating. This shows why “fixing voltage” and “fixing thermal loading” are separate decisions.
Read the units at the handoff
A capacity of 60 kW is 0.060 MW. An energy total of 120,000 kWh is 120 MWh. These are unit conversions, not changes in equipment performance.
The Phasor hybrid domain uses kW and kWh. Network elements use their defined power, energy, voltage and impedance units. Inspect the converted element values when materializing or binding a design. Do not divide a value by 1,000 a second time.
The math, if you want itOptional — the page reads completely without it
For a balanced three-phase feeder carrying P at line voltage V and power factor cos φ, the current is:
line current
I = P√3 · V · cos φ
With 40 kW, 400 V and power factor 0.9, I ≈ 64.15 A. Using the total phase resistance R and reactance X of the run:
the approximate line-to-line drop
ΔV ≈ √3 · I · ( R cos φ + X sin φ )
For 300 m of the small conductor, R = 0.87 × 0.3 = 0.261 Ω, and the drop is about 27.3 V, or 6.8%.
three-phase resistive loss
Ploss = 3 · I² · R
About 3.22 kW here. That is a loss at the selected operating point; multiplying by 8,760 would assume this current persists all year, which this lesson does not do.
Use the right level of analysis
This screen estimates current at nominal voltage and uses a linear voltage-drop equation. It does not iterate constant-power load current as voltage changes. It omits unbalance, transformer regulation, neutral effects, controls and fault behaviour.
Use an actual load-flow calculation for the network case, then the appropriate fault and protection studies. EPRI’s power-flow overview distinguishes snapshot and time-varying distribution analysis.
See it in Phasor
Materialize a saved plant version or bind a study to existing elements. Complete electrical data, run the required network study, and compare on the same demand and operating basis. A missing generator reactance cannot be recovered from an annual energy total.
Continue
Learn how distribution networks connect customers, then compare conductor size and cost.