Per-Unit And Load Flow For A Three-Bus System In PowerWorld: How To Get The Base Right

You have three buses, two voltage levels, a table of ohms per kilometre and a PowerWorld licence that opens to an empty grid. The usual first mistake is typing ohms into a program that wants per-unit values, or the reverse. Most marks in this kind of task are won or lost on the base, not on the solver.

Quick answer: Pick one system MVA base (100 MVA is the PowerWorld default), pick a voltage base for each voltage zone, convert every line with Zbase = kV² ÷ MVA, enter the pu values, solve the power flow and read voltages, angles and flows. Always state your base and explain any transformer implied between voltage levels.

What a per-unit and load flow task actually asks for

Coursework of this type, such as the three-bus problem in a Coventry University power systems module taught at PSB Academy Singapore, usually has three scored parts. You calculate line parameters by hand, build the network in PowerWorld Simulator, then run the load flow and present a solved diagram. The hand calculation checks that you understand what the software does for you.

For similar engineering briefs, browse the Coventry University question hub.

Step 1: choose the bases before touching a number

The per-unit system divides every quantity by a base of the same dimension. Two bases are free choices: the apparent power and the voltage. Current and impedance bases then follow (Glover, Overbye and Sarma, 2017).

One base MVA is used across the whole network. A different voltage base is used on each side of a transformer, in the ratio of the transformer’s rated voltages. Choose 100 MVA unless the brief tells you otherwise, because it keeps numbers tidy and matches the default case base in PowerWorld (PowerWorld Corporation, n.d.).

Step 2: convert line data, with a worked example

Take a generic 22 kV overhead line, 15 km long, with R = 0.02 Ω/km and X = 0.04 Ω/km, on a 50 MVA base. This is not your brief; it just shows the method.

1. Total R = 0.02 × 15 = 0.30 Ω and total X = 0.04 × 15 = 0.60 Ω.
2. Zbase = 22² ÷ 50 = 9.68 Ω.
3. Rpu = 0.30 ÷ 9.68 = 0.0310 and Xpu = 0.60 ÷ 9.68 = 0.0620.

For a cable with shunt susceptance B in siemens, use Bpu = B × Zbase, because admittance is the inverse of impedance. If the data is given per kilometre, multiply by the length first. If your brief says to ignore the shunt admittance on a line, enter zero and say so.

Step 3: deal with the voltage-level mismatch

A frequent trap is a network where one bus is at a different nominal voltage from the others, yet the brief connects it by a line. A line has no ratio, so an implied transformer sits somewhere between the two voltage levels. Your report should say this in one or two sentences, not ignore it.

State which voltage you use for each line’s impedance base and why. The cable ohms in the brief are valid only at the voltage where they were measured, and moving them to the other base changes the pu value by the square of the voltage ratio. For a 33 kV to 11 kV ratio of three, that factor is nine, which is why an unexplained choice costs marks.

If your module allows it, you may add an ideal transformer with a tap of 1.0 pu in PowerWorld so that the model is consistent. Mention that a real transformer would carry its own impedance, normally a few per cent, and that you have neglected it because the brief gives none.

Changing base is also easy to get wrong. To move a pu impedance from one MVA base to another, multiply by new MVA ÷ old MVA: 0.05 pu on 100 MVA becomes 0.0125 pu on 25 MVA.

Step 4: build and solve in PowerWorld Simulator

The software side is mostly clicking, but the order matters. Create the case in edit mode, add buses with nominal kV, mark one as slack, add generators and loads, then add branches with R, X and B in pu on the case base. Switch to run mode and solve the power flow.

After it solves, label bus voltages in pu and kV, angles in degrees and line flows in MW and MVAr on the one-line diagram. Check them against a quick sanity test: generation must equal load plus losses, and the losses on short, low-impedance lines should be small.

Where marks are lost

Markers of engineering reports tend to be forgiving about software layout and unforgiving about unstated assumptions. In our experience of reviewing student drafts, the same few errors come up each term.

• Never stating the base MVA or the voltage base, so the pu values cannot be checked.
• Mixing 33 kV and 11 kV impedance bases without comment.
• Pasting a screenshot without saying what it shows or whether the case converged.
• Giving pu values to six decimals but no comment on whether they are reasonable.
• Putting the answers in the wrong order, when the brief asks for each question on a new page.

Be honest about limits. A comment such as “the pu resistance is small because the lines are short” is better than a silent table.

Checklist before you export to PDF

• System MVA base and each voltage base are written down and justified.
• Every Zbase calculation is shown once, in full, then summarised in a table.
• Shunt susceptance is either converted or explicitly ignored.
• The PowerWorld diagram shows voltages, angles and line flows, and states that the case converged.
• Each figure is numbered, captioned and referred to in the text.
• All sources are in the Coventry Harvard reference list and cited in the text (Coventry University, n.d.).

Frequently asked questions

Which MVA base should I choose for a per-unit calculation?

Any base is mathematically valid, but 100 MVA is the common choice and matches PowerWorld’s default case base. Use it unless your brief specifies another. State it at the start of the question and keep it the same everywhere.

How do I convert ohms to per-unit?

Divide the ohmic value by the base impedance, where base impedance is the base voltage squared divided by the base MVA. For example, 0.2 Ω on a 33 kV, 100 MVA base is 0.2 ÷ 10.89 = 0.0184 pu.

What do I do when buses have different voltage levels?

Assign a voltage base to each zone, linked by the transformer ratio. Explain that the connection implies a transformer, say what you assume about it, and show the base you used for each line. The explanation is part of the marks.

Why does PowerWorld say the case did not converge?

Common causes are no slack bus, an unrealistic load, wrong units in the branch data, or generator limits that cannot supply the load. Check units first, because ohms typed into a pu field is the most common mistake.

How should I present the load flow results?

Show a solved one-line diagram with bus voltages, angles and line flows, plus a short results table. Then comment on voltage levels, losses and loading rather than just pasting the picture.

Can I use sample answers for my own coursework?

Use them only to compare method and presentation with your own working. Our 6032EE power systems worked sample is for reference only and must not be submitted as your own work.

If you are stuck on the base or the model, our report writing help and coursework support pages explain how we review your own draft, and the student reviews show what others say about the process. Send us your working a few days before the Blackboard deadline, not on the night. For subject-specific support with power systems coursework, see our electrical power engineering assignment help page.

References

Coventry University (n.d.) The Coventry University guide to referencing in Harvard style. Available at: https://libguides.coventry.ac.uk/cuharvard/2b (Accessed: 13 October 2026).

Glover, J.D., Overbye, T.J. and Sarma, M.S. (2017) Power system analysis and design. 6th edn. Boston, MA: Cengage Learning.

PowerWorld Corporation (n.d.) Power flow solution: general. Available at: https://www.powerworld.com/WebHelp/Content/MainDocumentation_HTML/Power_Flow_Solution_General.htm (Accessed: 13 October 2026).