Level 3 Renewable energy

From the resource to the installation: solar irradiation and wind of a site, sizing of generators, batteries, electronics, wiring and solar pumps.
Tutorial Version 2.0.0 November 2026 EN / FR Click for the French version
Level 3 — Renewable energy
NRJXpert 2.0.0 · November 2026

📘Introduction

Objectives

  • Estimate the irradiation received by tilted modules from the horizontal irradiation; choose the tilt.
  • Read a module datasheet and calculate its power in real conditions.
  • Draw up an energy balance and size the PV array, battery, charge controller and inverter of a stand-alone system.
  • Choose cable cross-sections and protection, and cost the wiring.
  • Estimate the yield of a grid-connected plant (performance ratio).
  • Characterise a wind site with the Weibull distribution and calculate the production of a wind turbine.
  • Size a solar pump (TDH, hydraulic energy, peak power, tank) and check the result with an hourly simulation.

Prerequisites: levels 1 and 2 (§ 2.4). Duration: 28 to 34 h.

↑ Back to top

☀️3.1 Solar resource

Solar geometry

The declination δ (angle between the equator and the direction of the sun) varies from −23.45° (21 December) to +23.45° (21 June): δ = 23.45·sin(360·(284 + n)/365), n = day number. The sunset hour angle is ωs = arccos(−tan φ·tan δ) (φ = latitude) and the day length 2ωs/15 hours.

H₀ = (24/π)·Gsc·(1 + 0.033 cos(360n/365))·(cos φ cos δ sin ωs + ωs sin φ sin δ)    Gsc = 1 367 W/m²
Clearness index Kt = H/H₀    diffuse fraction Hd/H = f(Kt) (Erbs correlation)

H₀ is the irradiation that would be received without the atmosphere; Kt measures the transparency of the sky (0.5 to 0.7 in the Sahel; lower in the rainy season and in dusty weather).

Tilted plane (Liu-Jordan, Klein)

Hβ = H·[ (1 − Hd/H)·Rb + (Hd/H)·(1 + cos β)/2 + ρground·(1 − cos β)/2 ]

The beam component is multiplied by Rb (geometric ratio tilted/horizontal), the diffuse by the fraction of sky seen, and the ground reflects a share ρground (albedo, 0.2; 0.3 on light sand).

Example 3.1 — Dakar in December (φ = 14.7° N)

Results (calculated by NRJXpert)

Mean day n = 344: δ = −23.05°; ωs = 83.59°; day length 11.15 h; H₀ = 7.96 kWh/m²/d. With H = 4.28 kWh/m²/d (locations database), Kt = 0.538 and Hd/H = 0.394.

Tilt β0°11°15°30°
Hβ December4.284.754.895.25
Hβ June5.875.545.394.69

A steeper tilt favours winter and penalises summer. For a stand-alone system NRJXpert chooses the tilt that maximises the worst month: 11° (December and August are then equal, 4.75 kWh/m²/d); for a grid-connected system, the one that maximises the annual energy: 14° (2 027 kWh/m²/yr instead of 1 986 flat).

Lab 5 — Weather data of a site

  1. Tools › Site weather data: search a location (e.g. Kédougou, Podor, Ziguinchor), source NASA POWER hourly.
  2. Compare horizontal H, H on the optimal plane, temperature and wind at 10 m with Dakar. Export to CSV.
  3. Use for PV: the Site tab of the PV window receives the irradiation and temperatures of the site.
↑ Back to top

☀️3.2 The photovoltaic module

A cell is a large illuminated diode: I = Iph − I₀(e(V + RsI)/(n·VT) − 1) − (V + RsI)/Rsh. Datasheets give, at standard test conditions (STC: 1 000 W/m², 25 °C, AM1.5): Voc, Isc, Vmp, Imp, Pp = Vmp·Imp (watt-peak) and the temperature coefficients.

Isc ∝ G    P ≈ Pp·(G/1000)·(1 + γ(Tc − 25))    γ ≈ −0.40 %/°C (crystalline silicon)
Tc = Ta + (NOCT − 20)/800 · G    (NOCT ≈ 45 °C)

Example 3.2 — 400 Wp module at noon in Dakar

G = 800 W/m², Ta = 35 °C, NOCT = 45 °C, γ = −0.40 %/°C.

Show the solution

Tc = 35 + 25/800 × 800 = 60 °C; P = 400 × 0.8 × (1 − 0.004 × 35) = 275 W: heat costs 14 %.

Connection. In series voltages add up (string); in parallel currents add up. A shaded module limits its whole string (bypass diodes). PWM charge controller: the modules work at the battery voltage (12, 24, 48 V); 36-cell modules (Vmp ≈ 17 V) are needed per 12 V. MPPT charge controller: a DC-DC converter makes the array work at its maximum power whatever the battery voltage; essential with current modules of 60 to 144 cells (Vmp 30 to 40 V).

↑ Back to top

☀️3.3 Sizing a stand-alone PV system

Method

  1. Load assessment: for each appliance, power × number × hours of operation; daily energy Ed (Wh/d) and maximum simultaneous apparent power Smax (VA, for the inverter).
  2. Needs referred to the battery: Qd = Ed/(U·ηconv·ηcables·ηbat) (Ah/d), with ηconv = 0.9 for AC appliances, 0.98 for the cables, 0.9 for the battery.
  3. Design irradiation: worst month on the plane of the modules (or annual mean if a seasonal deficit is accepted).
  4. PV array: PWM: Np = ⌈Qd/(H·0.9·Imp)⌉ strings of Ns = ⌈1.2·U/(0.9·Vmp)⌉; MPPT: Pp = Qd·U/(H·0.9·fT·0.96) with fT = 1 − 0.004·(Ta + 25 − 25).
  5. Battery: useful capacity C = Qd·autonomy(h)/24; Ns = U/Ubatt, Np = ⌈C/(Cbatt·DoD)⌉.
  6. Charge controller: current ≥ 1.25 × array current (PWM: Np·Isc; MPPT: Pp/U). Inverter: ≥ 1.25 × Smax.
  7. Wiring (§ 3.4), costs (level 4), and month-by-month check: production, needs, coverage rate.
Numbers are rounded up (rounding to the nearest integer under-sizes: this was an error of the original program, corrected in NRJXpert). Among the equipment of the database, the option « choose at minimum life-cycle cost » keeps the cheapest combination over the operating period, replacements included.

Design margin. The “Design margin (%)” field of the Options tab increases the needs (Ah/d) before the array and battery are sized; it covers the uncertainty on the load profile, demand growth and ageing (10 % by default in new projects). The examples of this tutorial are calculated with a 0 % margin so that they can be checked by hand: set 0 to recover their values.

Example 3.3 — Rural health post near Dakar

ApplianceNumber × powerHours/dEnergy (Wh/d)
LED lamps (cos φ 0.9)10 × 15 W5 (18:00 – 23:00)750
Vaccine refrigerator (cos φ 0.8)1 × 150 W12 (one hour out of two)1 800
Computer1 × 60 W4240
Fans2 × 50 W6600
Total3 390

24 V bus, MPPT charge controller, 72 h autonomy, DoD 60 %. Imposed equipment (indicative 2026 prices): 400 Wp module (Vmp 31.0 V, Imp 12.9 A, Isc 13.6 A; 110 kF), 12 V 200 Ah AGM battery (180 kF), 60 A MPPT controller (150 kF), 1.5 kVA inverter (300 kF).

Show the solution (identical to the NRJXpert calculation)

Battery: Qd = 3 390/(24 × 0.9 × 0.98 × 0.9) = 177.9 Ah/d; C = 177.9 × 72/24 = 533.8 Ah useful; 2 batteries in series × ⌈533.8/(200 × 0.6)⌉ = 5 in parallel = 10 batteries.

Array: December, H11° = 4.75 kWh/m²/d, Ta = 22 °C → fT = 1 − 0.004 × 22 = 0.912. Pp = 177.9 × 24/(4.75 × 0.9 × 0.912 × 0.96) = 1 141 Wp. String: ⌈1.3 × 24/(31.0 × 0.82)⌉ = 2 modules; ⌈1 141/800⌉ = 2 strings → 4 modules, 1 600 Wp.

Charge controller: 1 600/24 = 66.7 A → 2 controllers of 60 A. Inverter: Smax = 365 VA → one 1.5 kVA inverter.

Balance: production 2 530 kWh/yr, coverage 100 %, useful energy 1 237 kWh/yr. Total investment 3 185 kF of which wiring 235 kF; levelised cost of the kWh (8 %, inflation 3 %, 20 years) 580 F/kWh. The batteries account for 77 % of the life-cycle cost (5 replacements in 20 years).

Lab 6 — Sizing the health post

  1. Help › Open a tutorial example… › pv_case_de_sante.nxpv. Go through tabs 1 (Site: Dakar, horizontal plane converted), 2 (Needs: the 4 appliances and their schedules), 3 (Options and costs: MPPT, 24 V, imposed equipment, wiring).
  2. F5 (Size) and find the values of example 3.3 in tab 4.
  3. Set the bus to 48 V: the array drops to 3 modules (1 200 Wp), the batteries to 4 × 3 = 12, the controller to a single unit, and the wiring to 129 kF. Explain each change.
  4. Untick MPPT: a warning reports that 31 V modules on a 24 V bus lose a large part of their power with a PWM controller.
  5. Size on the annual mean instead of the worst month: how many months in deficit?
  6. Go back to 24 V MPPT and tick Choose the equipment at minimum life-cycle cost: NRJXpert goes through the database (current equipment, indicative 2026 prices) and keeps 2 TOPCon modules of 620 Wp, 6 LFP lithium batteries 25.6 V 100 Ah, 3 MPPT controllers of 30 A and one 24 V 2 kVA inverter: investment 2 982 kF, levelised cost of the kWh 339 F/kWh instead of 580. Explain: the LFP battery is more expensive to buy, but it lasts 12 years instead of 4, accepts 90 % discharge instead of 50 % and has a better efficiency.
↑ Back to top

🧵3.4 Wiring: cross-sections, protection and costs

Wiring commonly accounts for 5 to 10 % of the investment of a stand-alone PV system, and a wrong cross-section is paid for in energy losses, heating, or even fire. NRJXpert sizes each link according to NF C 15-100 / IEC 60364-5-52 and guide UTE C15-712-1:

  1. Design current Ib = 1.25 × rated current (PV side: 1.25·Isc, since the irradiance can exceed 1 000 W/m²).
  2. Protection: standard rating In ≥ Ib (gPV fuse, DC or AC circuit breaker).
  3. Heating: Iz = Iz0(S)·kθ·kg ≥ In, with kθ = √((90 − θa)/60) (XLPE insulation) and kg the grouping factor. At 40 °C and kg = 0.8: k = 0.913 × 0.8 = 0.730.
  4. Voltage drop at rated current: ΔU = b·(ρL/S·cos φ + λL·sin φ)·I, b = 2 in DC and single-phase; maximum 3 % on the PV array side, 1 % on the battery side, 3 % on the AC side.
  5. Cross-section chosen = the larger of the two, standardised (1.5; 2.5; 4; 6; 10; 16; 25; 35; 50; 70; 95; 120; 150; 185; 240; 300 mm²).
  6. Cost = conductors (fixed part + part × cross-section, per metre of conductor) + installation (per metre of link) + protection; maintenance 1 %/yr; lifetime 25 years.

Example 3.4 — Why the bus voltage changes everything

960 Wp array (PWM, 12 V modules), 15 m between the array and the controller, ΔU ≤ 3 %. Compare a 24 V and a 48 V bus.

Show the solution

24 V: I = 36 A (short-circuit current of the array); heating: In = 50 A, Iz0 ≥ 50/0.730 = 68.5 A → 10 mm². Voltage drop: S ≥ 2 × 0.0225 × 15 × 36/(0.03 × 24) = 33.8 mm² → 35 mm² (ΔU = 2.89 %). The voltage drop decides.

48 V: I = 18 A; S ≥ 8.4 mm² → 10 mm² (ΔU = 2.53 %). A cross-section 3.5 times smaller for the same power: the relative voltage drop varies as 1/U². The complete wiring of the system (4 links) goes from 292 to 147 kF.

Lab 7 — Wiring of the health post

  1. In the example pv_case_de_sante.nxpv, tab 3, Wiring section: lengths 15, 3, 3 and 20 m, temperature 40 °C.
  2. After F5, read the table of links in tab 4. At 24 V: array → controller 2 × 10 mm² (62 V in MPPT, ΔU 2.96 %), controller → battery and battery → inverter 2 × 50 mm² (100 A protection), inverter → board 2 × 1.5 mm² (10 A protection).
  3. Set the array → controller length to 40 m: which cross-section? Switch to aluminium: which cross-section and which cost?
  4. Wiring assumptions and prices…: replace the prices by those of a local supplier and observe the effect on the cost of the kWh.
↑ Back to top

🏠3.5 Grid-connected PV

Without a battery, all the production is fed in (or self-consumed). The reasoning is in specific yield (kWh/kWp/yr):

E = Pp · Hβ,yr · PR    PR (performance ratio) = 0.75 to 0.85: temperature, dust, mismatch, cables, inverter

Example 3.5 — 10 kWp in Dakar

Show the solution

Tilt 14°: Hβ = 2 027 kWh/m²/yr; with PR = 0.78: 1 581 kWh/kWp/yr, i.e. 15.8 MWh/yr. Tab 6 (« Economic viability ») of NRJXpert calculates hour by hour with the I-V model of the module, the temperature (NOCT), inverter clipping, system losses and degradation; it finds ≈ 1 560 kWh/kWp (PR ≈ 0.78) with the typical days of the database.

↑ Back to top

🌬️3.6 Wind resource and the Weibull distribution

The power of the wind crossing an area A is P = ½ρAv³: doubling the speed multiplies the power by 8. The speed fluctuates; its distribution generally follows a Weibull distribution:

f(v) = (k/c)(v/c)k−1 e−(v/c)k    F(v) = 1 − e−(v/c)k    v̄ = c·Γ(1 + 1/k)    power density = ½ρc³Γ(1 + 3/k)

c (m/s) is the scale parameter, k the shape parameter (k = 2: Rayleigh distribution; large k: steady wind, e.g. trade winds). NRJXpert fits c and k by maximum likelihood on the measurements, taking calms into account.

Height extrapolation (logarithmic profile, roughness z₀): v(H) = v(Hmeas)·ln(H/z₀)/ln(Hmeas/z₀). Air density: ρ = 1.225 kg/m³ at 15 °C at sea level; 1.17 at 28 °C.

Example 3.6 — Coastal site: c₁₀ = 5.5 m/s, k = 2.2

Roughness z₀ = 0.03 m (open terrain), hub at 30 m, ρ = 1.17 kg/m³.

Show the solution

At 10 m: v̄ = 5.5 × Γ(1.455) = 4.87 m/s; density = 118 W/m². Height factor ln(1000)/ln(333) = 1.189: c₃₀ = 6.54 m/s, v̄₃₀ = 5.79 m/s, density 199 W/m² (+68 % for +19 % speed). The wind is below 3 m/s (cut-in of a small machine) 16.5 % of the time at 30 m.

↑ Back to top

🌬️3.7 Wind production and sizing

P̄ = ∫ P(v)·f(v) dv    E = 8 760·P̄·(1 − losses)    capacity factor CF = E/(Pn·8 760)
density correction (IEC 61400-12-1): P(v) read on the curve at v·(ρ/1.225)1/3

Design margin. As for PV, the “Design margin (%)” field increases the needs used for the number of machines and the storage (10 % by default in new projects); the example below is calculated with 0 %.

Example 3.7 — 10 kW machine on the site of example 3.6

Power curve (kW): 0 up to 3 m/s; 0.6 at 4; 1.3 at 5; 2.3 at 6; 3.6 at 7; 5.2 at 8; 7.0 at 9; 8.7 at 10; 10 from 11 to 25 m/s. Diameter 7 m, losses 6 %.

Show the solution

Numerical integration: P̄ = 2.79 kW; gross energy 24.4 MWh/yr, net 23.0 MWh/yr; capacity factor 26 %. Mean power coefficient: 2 787/(199 × 38.5) = 0.36. To cover 100 kWh/d (36.5 MWh/yr), ⌈36.5/23.0⌉ = 2 machines are needed on an annual balance — more to cover the least windy month (tab 4 of the wind window).

Lab 8 — « Wind systems » window

  1. Tools › Wind systems, tab 1: load the example record Dakar-Yoff 1975.
  2. Tab 2: read c, k and the power density month by month; compare regression and maximum likelihood.
  3. Tab 3: choose a wind turbine, examine its power curve. Tab 4: monthly production, capacity factor, number of machines for given needs.
  4. Tab 5: grid-connected system, Wiring section (40 m per machine, 50 m to the connection point): read the cross-sections and the wiring cost in the results and the report.
↑ Back to top

💧3.8 Solar water pumping

In a village without a grid, a pump fed by photovoltaic modules fills a tank: the stored water replaces the battery (“direct solar” pumping). A battery can also be added to pump early in the morning or at the end of the day. Sizing starts from the water demand (m³/d) and the total dynamic head (TDH), not from an electrical load profile as in § 3.3.

TDH = HG + PC   (static head + head losses in the pipe)
Energy method: Eel = CH·V·TDH / Rp   with CH = ρg/3 600 = 2.725 Wh/(m³·m) and Rp = ηp·ηm·ηconv
Pc = Eel / (Ens·ηc)   (Ens: peak-sun hours; with a battery, also divide by ηB)
Power method: Qmax = π·V/(2·ΔT); PH = Qmax·TDH/367 (kW, Q in m³/h); Pel = PH/(ηp·ηm); Pc = Pel/(ηconv·ηc)·1 000/G

The pumping energy is proportional to the volume and to the head: doubling the head doubles the module area. The design month is the one where Pc is largest (high demand, little sun). The choice of pump (surface or submersible centrifugal, piston or progressive-cavity positive displacement…) depends on the head and the flow: NRJXpert offers selection advice and a database of 26 pumps (“Pumps” tab of the equipment database).

The textbook methods ignore the starting threshold of the motor-pump and the low-irradiance hours: the hourly simulation checks the share of the demand that is covered and can enlarge the array. The tank (usually 1 to 3 days of consumption) absorbs the variations of the sun.

Example 3.8 — Village of 500 inhabitants: 25 m³/d at 15 m TDH

500 inhabitants × 50 l/d = 25 m³/d; TDH = 15 m (entered); Ens = 5 kWh/m²/d; submersible centrifugal pump directly coupled: ηp = 0.5, ηm = 0.85, ηconv = 1; array losses 10 % (ηc = 0.9); pumping over 9 h; G = 800 W/m²; design margin 0 %. Size the array with both methods, then check by simulation.

Show the solution

Energy: Eel = 2.725 × 25 × 15/0.425 = 2 404 Wh/d; Pc = 2 404/(5 × 0.9) = 534 Wp → 6 modules of 100 Wp (3 in series × 2 in parallel = 600 Wp).

Power: Qmax = π × 25/(2 × 9) = 4.36 m³/h; PH = 4.36 × 15/367 = 178 W; Pel = 178/0.425 = 420 W (motor-pump power); Pc = 420/0.9 × 1 000/800 = 583 Wp.

Hourly simulation: the 600 Wp array covers only 86 % of the demand (late start, low-irradiance hours); 9 modules (3 × 3 = 900 Wp) are needed to reach 100 %. The proposed tank is 50 m³ (2 days). Conclusion: the formulas give an order of magnitude, the simulation fixes the actual array.

With a battery (same data, ηB = 0.85): Pc = 2 404/(5 × 0.9 × 0.85) = 628 Wp. The battery is expensive and has to be replaced: for the same use, the tank is almost always cheaper.

Design margin. In NRJXpert, the “Design margin (%)” field (10 % by default in new projects) increases the needs before the calculation. With 10 %, the example becomes: demand 27.5 m³/d, Eel = 2 645 Wh/d, Pc = 588 Wp. The tutorial examples are calculated with a 0 % margin so that they can be checked by hand.

Lab 8 bis — « Solar water pumping » window

  1. Help › Open a tutorial example… › pompage_village.nxpomp (or Renewables › Solar water pumping…, then Open).
  2. Tab 2: read the demand (25 m³/d). Tab 3: TDH = 15 m. Tab 4: choose the Energies method, then Powers: compare Pc (534 and 583 Wp).
  3. Tab 5: run the simulation; read the coverage (86 %) and the tank level. Choose the Simulation method: the array becomes 900 Wp.
  4. Tab 4: set the margin to 10 % and the configuration to With battery; note the effect on the peak power and on the cost (tab 6, comparison with the diesel motor-pump).
  5. Equipment database › Pumps tab: browse the pumps and pick the right one for “Pump from database”.
↑ Back to top

🎯Exercises

Exercise 1 — Solar home system

A family wants to supply 4 lamps of 7 W (5 h/d), a 60 W television (4 h/d) and charge phones (20 Wh/d), at 12 V DC (no inverter). H = 5 kWh/m²/d, 12 V modules of 100 Wp (Imp = 5.6 A), 12 V 100 Ah batteries, 2 days of autonomy, DoD 50 %. Size the system.

Show the solution

Ed = 140 + 240 + 20 = 400 Wh/d; Qd = 400/(12 × 0.98 × 0.9) = 37.8 Ah/d (no converter). Modules (PWM): 37.8/(5 × 0.9 × 5.6) = 1.5 → 2 modules. Battery: 37.8 × 2/0.5 = 151 Ah → 2 batteries of 100 Ah in parallel. Controller: 1.25 × 2 × Isc (≈ 6 A) = 15 A → 20 A controller.

Exercise 2 — Cross-section of an array cable

An MPPT array of 6 strings delivers Isc = 6 × 13.6 A at 380 V, 60 m from the technical room; ΔU ≤ 1 %, 45 °C, kg = 0.7. Cross-section? (Use the calculator above.)

Show the solution

I = 81.6 A; Ib = 102 A; protection 125 A; k = √(45/60) × 0.7 = 0.606 → Iz0 ≥ 206 A → 50 mm²; voltage drop: S ≥ 2 × 0.0225 × 60 × 81.6/(0.01 × 380) = 58 mm² → 70 mm² (ΔU = 0.83 %).

Exercise 3 — Tower height

On the site of example 3.6, by how much does the power density increase from an 18 m tower to a 30 m tower? And if the roughness is 0.3 m (village, trees)?

Show the solution

z₀ = 0.03: (ln(1 000)/ln(600))³ = (6.908/6.397)³ = 1.26 (+26 %). z₀ = 0.3: (ln(100)/ln(60))³ = (4.605/4.094)³ = 1.42 (+42 %): the rougher the terrain, the more height pays.

Exercise 4 — Pumping from a borehole

A borehole has a static level of 20 m and a drawdown of 4 m and delivers into a water tower whose inlet is at 6 m; lump-sum head losses: 10 % of HG. Demand: 12 m³/d. Ens = 5.5 kWh/m²/d, ηp = 0.45, ηm = 0.85, ηconv = 1, array losses 10 %, pumping over 8 h, G = 800 W/m², 250 Wp modules. Calculate the TDH, the peak power with both methods and the number of modules.

Show the solution

HG = 20 + 4 + 6 = 30 m; PC = 3 m; TDH = 33 m. Rp = 0.45 × 0.85 = 0.3825. Energy: Eel = 2.725 × 12 × 33/0.3825 = 2 821 Wh/d; Pc = 2 821/(5.5 × 0.9) = 570 Wp → 3 modules (750 Wp); with a 10 % margin: 627 Wp, still 3 modules. Power: Qmax = π × 12/16 = 2.36 m³/h; PH = 2.36 × 33/367 = 212 W; Pel = 212/0.3825 = 554 W; Pc = 554/0.9 × 1 000/800 = 769 Wp → 4 modules. The more cautious power method is the one to keep before the hourly simulation.

↑ Back to top

📋Level 3 quiz

1. For a stand-alone PV system, the tilt chosen by NRJXpert maximises:

  1. the annual irradiation
  2. the June irradiation
  3. the irradiation of the worst month
  4. the module temperature
The system must work in the least sunny month; for grid-connected systems the annual energy is maximised.

2. When the cell temperature increases, it is mainly:

  1. the current that decreases
  2. the voltage that decreases
  3. the current that vanishes
  4. the power that increases
Voc decreases by about 0.3 %/°C; Isc increases very slightly.

3. A 60-cell module (Vmp ≈ 31 V) on a 24 V battery requires:

  1. an MPPT charge controller
  2. a PWM charge controller
  3. two modules in parallel
  4. an inverter
With a PWM controller the module would work at ≈ 28 V instead of 31 V: loss of power.

4. The autonomy of a battery is meant to cover:

  1. the peak power
  2. the inverter losses
  3. the night energy only
  4. successive days without sun
Night energy is covered by the daily cycle; autonomy covers the overcast days.

5. For the same power and length, going from 24 to 48 V divides the cross-section imposed by the voltage drop by:

  1. √2
  2. 2
  3. 4
  4. 8
The current is halved and the admissible relative drop applies to a voltage twice as high: S ∝ I/U = P/U².

6. The factor 1.25 applied to the short-circuit current of a PV array accounts for:

  1. cable ageing
  2. an irradiance that may exceed 1 000 W/m²
  3. the voltage drop
  4. the battery temperature
UTE C15-712-1: Ib = 1.25·Isc (cloud reflection, albedo).

7. A performance ratio (PR) of 0.78 means:

  1. 78 % of the energy the array would produce at STC is delivered
  2. the module efficiency is 78 %
  3. the inverter efficiency is 78 %
  4. 78 % of the time is sunny
PR = E / (Pp × Hβ/1 kW/m²).

8. If the mean wind speed increases by 10 %, the power density increases by about:

  1. 10 %
  2. 21 %
  3. 27 %
  4. 33 %
1.1³ = 1.331 (at constant shape k).

9. A high Weibull shape parameter (k = 3 to 4) indicates:

  1. a very variable wind
  2. a steady wind
  3. a strong wind
  4. many calms
The distribution narrows around the mean (trade winds).

10. In the 20-year life-cycle cost of a stand-alone PV system, the largest item is often:

  1. the modules
  2. the wiring
  3. the batteries (replacements included)
  4. the inverter
Example 3.3: 77 % of the life-cycle cost with batteries replaced every 5 years.

11. In a “direct solar” water pumping system, energy storage is provided by:

  1. a lead-acid battery
  2. the water tank
  3. a flywheel
  4. a capacitor
Water is stored, not electricity: a tank of 1 to 3 days of consumption plays the role of the battery.

12. If the TDH doubles, with unchanged flow and efficiencies, the daily electrical pumping energy:

  1. stays the same
  2. doubles
  3. is multiplied by 4
  4. is multiplied by √2
Eel = CH·V·TDH/Rp is proportional to the TDH.

13. In example 3.8, a 10 % design margin brings the peak power from 534 Wp to about:

  1. 540 Wp
  2. 588 Wp
  3. 641 Wp
  4. 1 068 Wp
The margin increases the needs: 534 × 1.1 = 588 Wp.
↑ Back to top
← Level 2 — Power networksLevel 4 — Economic studies →

👤Author

LT
Prof. Lamine THIAW
Department of Electrical Engineering
École Supérieure Polytechnique
Cheikh Anta Diop University, Dakar
lamine.thiaw@ucad.edu.sn
lamine.thiaw@esp.sn

November 2026