Load Flow Calculator
Calculate power flow, losses, and voltage drop through transmission lines.
About this calculator
This calculator is a simplified single-line power-flow estimator, not a full network load-flow (power-flow) study. Real load-flow analysis solves a nonlinear system of power-balance equations across every bus and branch in a network -- typically with an iterative numerical method such as Newton-Raphson (fast, quadratic convergence, the industry standard for large systems) or Gauss-Seidel (simpler, slower, linear convergence) -- given generator and load injections at each bus, and returns voltage magnitude and angle everywhere plus every branch's flow. This calculator instead takes a single known current and power factor and directly computes what that current does flowing down one line segment of stated length, resistance, and reactance: Real Power and Reactive Power are the two components three-phase apparent power splits into (S = √3·V·I, with Real Power = S·cos φ and Reactive Power = S·sin φ).
Line Losses use the standard 3·I²·R resistive-heating formula, and because that formula has no voltage term, the same current produces identical Line Losses regardless of Line Voltage -- which is exactly why utilities step transmission voltage up: moving the same power at a lower current, for the same line resistance, cuts I²R losses dramatically. Voltage Drop % uses the standard approximate three-phase drop formula, √3·I·(R·cos φ + X·sin φ), expressed as a percentage of Line Voltage. What this calculator does NOT model: multiple interconnected buses and lines, generator dispatch or reactive-power support devices, or convergence to a self-consistent whole-network state -- all of which real load-flow software solves for simultaneously.
Inputs
Results
Real Power
113,535.9 kW
Line Losses
3,750 kW
≈ 250 homes' peak draw
How to Use This Calculator
- Enter Line Voltage, Line Current, and Power Factor.
- Set Line Length, Resistance, and Reactance.
- Review Real Power (kW) and Line Losses (kW).
- Use Reactive Power (kVAR) and Voltage Drop (%) to inform your decision.
How the result changes with Line Voltage
| Line Voltage | Real Power | Line Losses |
|---|---|---|
| 69 | 56,768 kW | 3,750 kW |
| 104 | 85,563.3 kW | 3,750 kW |
| 207 | 170,303.9 kW | 3,750 kW |
| 345 | 283,839.8 kW | 3,750 kW |
What each input means
- Line Voltage
- Line-to-line voltage of the transmission line.
- Line Current
- Current flowing through the line.
- Power Factor
- Load power factor (lagging).
- Line Length
- Length of the transmission line.
- Resistance
- Line resistance per mile.
- Reactance
- Line reactance per mile.
How this is calculated
Worked example, using the default values
- Identify Input Parameters4 parametersLine Voltage = 138, Line Current = 500, Power Factor = 0.95, Line Length = 50 = 6 input(s) provided
- Calculate Real PowerReal Power113535.9 = 113535.9
- Calculate Line LossesLine Losses3750 = 3750
- Calculate Reactive PowerReactive Power37317.5 = 37317.5
- Calculate Voltage DropVoltage Drop7.88 = 7.88
Engine last updated . Checked against 1 independently-derived test — how we verify calculators. Built by Paul Gunder, a software engineer, not a licensed financial, medical, or legal professional.
Frequently Asked Questions
Is this the same thing as a full power-flow (load-flow) study of a network?
No. This calculator estimates power, losses, and voltage drop for one known current flowing down a single transmission line segment. A real load-flow study solves power-balance equations across an entire interconnected network of buses and branches using iterative methods like Newton-Raphson or Gauss-Seidel, typically starting from generator and load injections rather than an already-known line current.
Why do Line Losses stay the same if I raise Line Voltage but keep Current fixed?
Line Losses here follow 3·I²·R, which depends only on Line Current and the line's total resistance, not on Line Voltage -- so raising Line Voltage alone leaves Line Losses unchanged across this calculator's full 1-765 kV range. This is precisely why power systems transmit at high voltage: delivering the same power at a lower current (for a fixed resistance) sharply reduces I²R losses along the line.
Does raising Line Current increase Reactive Power along with Real Power?
Yes, across this calculator's full 1 to 5,000 A range at fixed voltage and power factor: Reactive Power rises as Line Current rises, alongside Real Power, because both come from the same apparent power (S = √3·V·I) split by the fixed power factor into a real component and a reactive component. Neither one falls as current climbs -- a heavier load draws more of both -- except at unity power factor (1.0), where Reactive Power is always zero regardless of Current, since sin(φ) = 0 there.
Why does Voltage Drop % fall as I raise Line Voltage, even though the drop itself doesn't change?
Voltage Drop % is the absolute Voltage Drop divided by Line Voltage, and the absolute drop here depends only on Line Current, resistance, reactance, and Power Factor -- not on Line Voltage itself. So the same absolute drop becomes a smaller percentage on a higher-voltage line, which is why utilities can tolerate longer feeder runs at higher voltages before Voltage Drop % becomes a problem.
What happens to Line Losses if I increase Line Length or Resistance?
Both increase Line Losses across their full declared ranges, holding Line Current fixed: Line Length and Resistance multiply together into the line's total resistance, which feeds directly into the 3·I²·R loss formula, so a longer line or a higher per-mile resistance produces proportionally more resistive heating loss for the same current.
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