For information and reference only – not to be submitted as your own work. Use this extract to check your method and presentation against your own calculations, then redo every figure and every PowerWorld screenshot from your own model, as your institution requires.
This sample shows how the calculation parts of a 6032EE Power Systems coursework for Coventry University at PSB Academy Singapore can be answered: the solar and load-factor questions (Question 1) and the per-unit line parameters for a three-bus network (Question 3). Question 2, the 3,000-word power plant report, is covered only as a structure outline.
How the brief is built
The coursework is an individual report in Coventry Harvard style, submitted as a PDF through Blackboard, with each question on a new page. Question 1 is pure calculation, Question 2 is a data-led report on a modern plant in Europe, the USA or Australia, and Question 3 asks for a three-bus model in PowerWorld Simulator. Our Coventry University question hub lists related briefs.
Question 1 – Solar energy and load factor (model working)
(i) Solar insolation and peak sun hours
Solar insolation is the solar energy that reaches a unit area of a surface over a stated period, usually quoted in kWh/m² per day. Peak sun hours express that energy as the number of hours of full-strength sunshine at a reference irradiance of 1 kW/m² (Duffie and Beckman, 2013).
Peak sun hours = daily insolation ÷ 1 kW/m² = 5 kWh/m²/day ÷ 1 kW/m² = 5 hours per day. The monthly average is therefore treated as five hours of 1 kW/m² every day.
(ii) Number of 180 W panels
The daily load is 2,500 Wh. One 180 W panel delivers, in the ideal case, 180 W × 5 h = 900 Wh per day. The 12 V rating fixes the system voltage and battery arrangement; it does not change the energy count.
| Case | Working | Panels (exact) | Panels (round up) |
|---|---|---|---|
| Ideal (no losses) | 2,500 ÷ (180 × 5) | 2.78 | 3 |
| With 75% overall derating (illustrative assumption) | 2,500 ÷ (180 × 5 × 0.75) | 3.70 | 4 |
State clearly which case you are answering. The brief gives no losses, so the ideal result of three panels is the direct answer. A derating factor for wiring, temperature, dirt, charge controller and battery efficiency is realistic, and the 75% above is only an illustration; if you use a derated figure, say where the value comes from.
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(iii) Energy per year and reserve capacity
Load factor = average demand ÷ maximum demand, so average demand = 0.5 × 12 MW = 6 MW.
Energy generated per year = 6 MW × 8,760 h = 52,560 MWh (equivalently 12 MW × 0.5 × 8,760).
The utilisation factor is taken here as maximum demand ÷ installed capacity. Installed capacity = 12 MW ÷ 0.4 = 30 MW, so the reserve capacity = 30 − 12 = 18 MW. If your lecture notes define utilisation factor differently (for example energy-based), state your definition and redo the step.
(iv) Installed capacity for a 20% reserve margin
Installed capacity = maximum demand × (1 + reserve margin) = 60 MW × 1.2 = 72 MW, which leaves a reserve of 12 MW above the 60 MW peak.
| Part | Key formula | Result |
|---|---|---|
| (i) | Peak sun hours = insolation ÷ 1 kW/m² | 5 h/day |
| (ii) | N = daily load ÷ (P × PSH) | 2.78, so 3 panels |
| (iii) | E = LF × MD × 8,760 | 52,560 MWh; 30 MW; 18 MW reserve |
| (iv) | Capacity = MD × 1.2 | 72 MW |
Question 2 – Power plant report (structure only)
The report is worth a third of the marks and carries a word count of about 3,000 words, so plan it by the mark split in the brief. This sample does not name a plant or quote figures; pick your own plant and take every number from a source you can cite.
- Introduction and background (about 500 words): plant type, location, commissioning date, owner, why it counts as modern.
- Generation operation (about 1,000 words): installed capacity, maximum demand, reserve margin from reliable sources, generation-mix chart, demand chart.
- Two renewable types (about 1,000 words): contribution of each, a chart of the country’s future renewable plan, and your own comments.
- Summary and conclusions (about 400 words), then the reference list.
Use the International Energy Agency, IEEE, IET and the national electricity authority for data, and cite every figure and chart in Coventry Harvard style (Coventry University, n.d.).
Question 3 – Per-unit line parameters
Choosing the base and the 33/11 kV issue
Bus 1 and Bus 2 are at 33 kV and Bus 3 is at 11 kV, so a line between a 33 kV and an 11 kV bus implies a transformer that the brief does not show. The per-unit method needs one MVA base for the whole system and a voltage base for each voltage zone, linked by the transformer ratio (Glover, Overbye and Sarma, 2017). Here the base is 100 MVA, with 33 kV for Buses 1 and 2 and 11 kV for Bus 3. PowerWorld also uses a 100 MVA case base by default (PowerWorld Corporation, n.d.).
Zbase = kV² ÷ MVA: at 33 kV, 33² ÷ 100 = 10.89 Ω; at 11 kV, 11² ÷ 100 = 1.21 Ω.
Line values
| Line | R (Ω) | X (Ω) | Zbase (Ω) | R (pu) | X (pu) | B (pu) |
|---|---|---|---|---|---|---|
| 1-2 overhead, 20 km | 0.200 | 0.300 | 10.89 | 0.01837 | 0.02755 | ignored |
| 1-3 cable, 10 km (11 kV base) | 0.050 | 0.010 | 1.21 | 0.04132 | 0.00826 | 0.0121 |
| 2-3 cable, 10 km (11 kV base) | 0.050 | 0.010 | 1.21 | 0.04132 | 0.00826 | 0.0121 |
The cable susceptance is 0.001 S/km × 10 km = 0.01 S, and Bpu = B × Zbase = 0.01 × 1.21 = 0.0121.
If you instead treat the cables as operating on the 33 kV base, the same ohmic values give R = 0.00459 pu, X = 0.00092 pu and B = 0.1089 pu. Explain which voltage you assume for the cables and why; the choice changes the answer by a factor of nine.
Commentary
All pu resistances and reactances are small, in the range 0.001 to 0.05, which is realistic for short, low-impedance lines. The overhead line has an X/R ratio of 1.5, while the cables have a ratio of only 0.2, because the brief gives them a very low reactance. Mention that a real 33/11 kV transformer would add its own impedance, usually several per cent on its own rating, and that the brief ignores it.
The load of 100 MW + j50 MVAr becomes 1.0 + j0.5 pu on the 100 MVA base, a total of 111.8 MVA or 1.118 pu.
PowerWorld Simulator – what you must capture
You cannot reuse anyone else’s screenshots, so build the model yourself and capture these items.
- Enter the three buses with their nominal kV (33, 33, 11) and mark Bus 1 as the slack.
- Add the two generators with X = 0.01, 0 to 2000 MW, plus or minus 9900 MVAr limits, and AGC and AVR switched on.
- Add the three branches using your pu R, X and B values on the 100 MVA case base.
- Add the load of 100 MW and 50 MVAr at Bus 3.
- Run the power flow, then capture a one-line diagram showing bus voltages (pu and kV), angles and line MW and MVAr flows.
- Check that the solution converged and that flows balance at each bus, and describe the results in your own words.
Describe what each screenshot shows and refer to the figure number in the text.
Need a second opinion on your own model? Our electrical engineering assignment help page explains what we support; the guide per-unit and load flow in PowerWorld walks through the method in more detail.
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).
Duffie, J.A. and Beckman, W.A. (2013) Solar engineering of thermal processes. 4th edn. Hoboken, NJ: John Wiley & Sons.
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).
