Sizing of Solar PV Systems: Load, Load Factor, Days of Autonomy, Battery, Inverter and Array Size

This chapter completes the Solar photovoltaic (PV) systems part of the Engineering Sciences (XE) paper, Section XE8.1 of XE8 Energy Science. It answers the second half of the heading “Solar photovoltaic (PV) systems: … sizing of PV systems (load factor, days of autonomy, battery size, inverter size, PV array size), maximum power point tracking”. It teaches how a daily load is built up, what the load factor means, how peak sun hours and derating give the array size, how days of autonomy, depth of discharge and efficiency give the battery bank, how the inverter and charge controller are rated, and how modules are arranged in strings. The numericals are daily energy and load factor, array peak watts, battery ampere-hours, inverter rating, charge-controller current, battery-bank arrangement and daily yield, all with every factor stated in the question.

1. Load assessment and load factor

Sizing starts with the load. List every appliance, its power rating P (W) and the hours h it runs per day; the daily energy is E = Σ P·h (Wh/day). The peak demand is the largest power drawn at one moment, which is the sum of the loads that can run together, not of all appliances. The load factor is the ratio of average to peak power over the period considered; for a day, LF = E/(Ppeak × 24 h). It lies between 0 and 1. LF = 1 is a constant load; a small LF means a spiky load that is on briefly.

The two quantities size different parts. The energy E (with sunshine data) sizes the array and the battery; the peak sizes the inverter and the wiring. For the same E, a lower load factor means a higher peak and so a larger inverter. Worked example. A home uses 7.2 kWh/day with a 1.5 kW peak: the average power is 7.2/24 = 0.30 kW and LF = 0.30/1.5 = 0.20. The peak equals the average only if the load is flat.

A stand-alone (off-grid) system has array, charge controller, battery and, for AC loads, an inverter; the battery makes up the gap between sun and demand. A grid-connected system has array and grid-tied inverter only, with the grid as the store, so there is no battery to size.

2. Peak sun hours and PV array size

The daily insolation on the tilted array, in kWh/m²/day, equals the number of peak sun hours (PSH): the hours of a hypothetical 1 kW/m² sun that would deliver the same energy. An insolation of 5 kWh/m²/day is 5 PSH; it is not 5 hours of daylight. A module rated Prated (peak watts, at 1 kW/m²) therefore delivers about Prated × PSH per day before losses.

Real output falls short of this because of cell heating, soiling, wiring and mismatch, and the efficiency of the controller or inverter and of battery charging. Combine them into one system derating factor η_sys (given in a problem; typical values are below 1, around 0.6 to 0.8). Then Parray (Wp) = E_dc/(PSH × η_sys), where E_dc is the daily energy the array must supply (the load divided by the inverter efficiency for AC loads). The same formula run backwards gives daily yield = Parray × PSH × η_sys, and the ratio of actual yield to Parray × PSH is the performance ratio.

⚠️ Design for the worst month, not the average
A stand-alone system that must work all year is sized on the month of lowest insolation (or highest load). Using the annual average PSH of 5 when the worst month gives 3.5 undersizes the array by a factor 5/3.5 = 1.43 in that month. Also use insolation on the tilted plane, not on the horizontal.

3. Battery sizing: autonomy, depth of discharge, efficiency

Days of autonomy N is the number of consecutive days the battery alone must carry the load with no useful sun. Depth of discharge (DoD) is the fraction of nominal capacity that may be used; a battery is not drained fully, to protect its life, so the nominal capacity must exceed the energy drawn. With battery efficiency ηb on the energy that passes through it,

Battery energy (Wh) = E_dc × N/(DoD × ηb) and capacity (Ah) = Wh/Vbus. Halving the permitted DoD doubles the bank; doubling N doubles it; the array is unaffected by N because it is sized for one day's energy. Cells are wired in series to reach the bus voltage and in parallel to reach the ampere-hours: twelve-volt, 100 Ah batteries for a 48 V, 400 Ah bank need 48/12 = 4 in series and 400/100 = 4 such strings, 16 in all.

⚠️ Ah and Wh are tied by the bus voltage
Capacity in Ah is meaningless until the bus voltage is stated: the same 8366 Wh is 349 Ah at 24 V but 174 Ah at 48 V. DoD and ηb go in the denominator; putting them in the numerator makes the bank smaller than the load needs.

4. Inverter, charge controller and module arrangement

The inverter converts DC to AC and is rated by the power it must deliver, not by the daily energy. Rule: inverter rating ≥ (the largest AC load that can run simultaneously) × a safety factor, commonly 1.25 (stated in a problem). Loads that surge on starting, such as motors and compressors, can draw several times their running power; either the inverter must tolerate that surge or the load must be started alone. Inverter efficiency enters the energy chain: E_dc = E_ac/ηinv.

The charge controller protects the battery from overcharge and deep discharge. It is rated for the array current: Icontroller ≥ Np × Isc,module × 1.25, since Isc, not Imp, is the largest the array can pass and 1.25 allows for irradiance above 1 kW/m² (the factor is stated in a problem). For the module count, N = Parray/Pmodule, rounded up; the series count per string is set by the bus or the inverter's input window, Ns = Vbus/Vmp,module for a direct-coupled array, and Np = N/Ns. For an MPPT controller a longer string is allowed, subject to the cold-weather Voc limit of the previous chapter.

What sets the size of each component
ComponentGoverned byFormula
Arraydaily energy, PSH, lossesE_dc/(PSH × η_sys)
Batterydaily energy, autonomy, DoDE_dc × N/(DoD × ηb)
Invertersimultaneous peak (and surge)Psim × 1.25
Charge controllerarray short-circuit currentNp × Isc × 1.25

5. Tilt, module count and string layout

A fixed array faces the equator (south in the northern hemisphere). A tilt close to the latitude maximises the yearly energy; a steeper tilt favours winter and a shallower one summer, so a stand-alone system sized on its worst month leans to the steeper side. Shading must be avoided: even a small shadow on one module can cut a whole series string, as the previous chapter explained.

Worked example (layout). A 2.0 kWp array uses 250 Wp modules with Vmp = 30 V, Imp = 8.33 A. The module count is 2000/250 = 8. The MPPT controller accepts strings of four modules (4 × 30 = 120 V at the maximum power point), so Ns = 4 and Np = 8/4 = 2. The array current at the maximum power point is 2 × 8.33 = 16.7 A at 120 V, which is 2000 W. If the controller had to be directly coupled to a 24 V battery bank, modules of a lower Vmp would be needed, with Ns chosen so that Ns × Vmp matches the charging voltage.

🎯 Why a derating factor sits in the formula
Nameplate watts are an STC figure; no installed array sits at STC. The derating factor is the product of the temperature factor, soiling, wiring and mismatch losses and the conversion efficiencies, for example 0.95 × 0.97 × 0.98 × 0.95 × 0.90 ≈ 0.77. It is the most sensitive input after PSH, because the array size varies as its inverse.

6. A full worked design

A cabin has four 10 W lamps for 6 h, two 60 W fans for 8 h and a 100 W TV for 4 h, all AC. E_ac = 4 × 10 × 6 + 2 × 60 × 8 + 100 × 4 = 240 + 960 + 400 = 1600 Wh/day. If all run together the peak is 40 + 120 + 100 = 260 W. The data given: inverter efficiency 0.90, bus 24 V, PSH 5, η_sys 0.75, autonomy 2 days, DoD 0.5, battery efficiency 0.85, inverter safety factor 1.25.

  • DC energy: E_dc = 1600/0.90 = 1777.8 Wh/day.
  • Array: 1777.8/(5 × 0.75) = 474 Wp; 474/250 = 1.9, so with 250 Wp modules 2 modules (500 Wp) are needed.
  • Battery: 1777.8 × 2/(0.5 × 0.85) = 8366 Wh, so 8366/24 = 349 Ah at 24 V.
  • Inverter: 260 × 1.25 = 325 W at least.
⚠️ Check units and order of magnitude
The array makes about Parray × PSH × η_sys = 474 × 5 × 0.75 = 1778 Wh a day, the same as E_dc: a quick back-substitution catches a slip. Battery energy for 2 days at 50 % DoD must be at least 4 times E_dc; here 8366/1778 = 4.7 including the efficiency. A result below 4 signals that DoD was multiplied instead of divided.

Key takeaways

  • E = Σ P·h gives daily energy; load factor LF = E/(Ppeak × 24 h). Energy sizes array and battery, the simultaneous peak sizes the inverter.
  • Peak sun hours equal insolation in kWh/m²/day. Array Wp = E_dc/(PSH × η_sys); use the tilted plane and, for a year-round stand-alone system, the worst month.
  • Battery Wh = E_dc × N/(DoD × ηb); Ah = Wh/Vbus. DoD and efficiency divide; the array does not depend on days of autonomy.
  • Inverter ≥ simultaneous AC load × safety factor (check motor surge); AC loads need E_dc = E_ac/ηinv. Charge controller current ≥ Np × Isc × 1.25.
  • Series sets bus voltage, parallel sets ampere-hours or current: a 48 V, 400 Ah bank from 12 V, 100 Ah units uses 4 × 4 = 16.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A load has an average power of 0.4 kW over a day and a peak demand of 2.0 kW. Its daily load factor is

    1. 5.0, because the peak is five times the average
    2. 0.20, the ratio of average to peak power
    3. 0.80, the peak minus the average divided by the peak
    4. 0.40, because the load factor equals the average power in kW
    Show answer

    Answer: B — 0.20, the ratio of average to peak power

    Load factor = average/peak = 0.4/2.0 = 0.20; it cannot exceed 1, which rules out the inverted ratio 5.0.

  2. At a site the daily insolation on the tilted array is 5.5 kWh/m²/day. In sizing a PV array this is used as

    1. A module efficiency of 5.5 %
    2. 5.5 hours between sunrise and sunset
    3. 5.5 hours for which the array delivers its rated power exactly
    4. 5.5 peak sun hours, the hours of 1 kW/m² that would give the same energy
    Show answer

    Answer: D — 5.5 peak sun hours, the hours of 1 kW/m² that would give the same energy

    Because STC irradiance is 1 kW/m², an insolation of 5.5 kWh/m²/day is numerically 5.5 PSH. It is neither the length of the day nor the hours of rated output, since the real irradiance varies and output carries losses.

  3. In a stand-alone system the permitted depth of discharge of the battery is lowered from 0.8 to 0.4 with every other quantity unchanged. The required nominal battery capacity

    1. doubles
    2. is unchanged, since the load is the same
    3. is halved
    4. rises by 0.4 times
    Show answer

    Answer: A — doubles

    Capacity ∝ 1/DoD, so 0.8/0.4 = 2: the bank doubles because only 40 % of each ampere-hour may be used. Halving would follow if DoD were mistakenly placed in the numerator.

  4. A designer raises the days of autonomy from 2 to 3 with load, PSH and all efficiencies unchanged. Which is correct?

    1. Both the battery and the array grow by 50 %, since both must cover the extra day
    2. Only the inverter must grow by 50 %, and the battery and array stay as they are
    3. The battery grows by 50 % and the array is unchanged, since the array covers a single day
    4. The array grows by 50 % to recharge the longer reserve, and the battery is unchanged
    Show answer

    Answer: C — The battery grows by 50 % and the array is unchanged, since the array covers a single day

    Battery energy ∝ N, giving 3/2 = 1.5. The array is sized for one day's energy divided by PSH, with no N in it, and the inverter depends on the simultaneous peak, which has not changed.

  5. A stand-alone system is sized with the annual average of 5.0 PSH. In the worst month the tilted-plane insolation is 3.5 kWh/m²/day. If the load is the same every month, the array needed for that month is larger than the average-based array by a factor of nearest

    1. 0.70
    2. 1.17
    3. 2.00
    4. 1.43
    Show answer

    Answer: D — 1.43

    Array ∝ 1/PSH, so the ratio is 5.0/3.5 = 1.43. The factor 0.70 is the inverse, which would be a smaller array.

  6. Which statement about sizing the inverter of a stand-alone system is correct?

    1. It is sized on the simultaneous AC load with a safety margin, and must handle motor starting surges
    2. It is sized from the number of days of autonomy, because they fix how long it must run without sun
    3. Its watt rating follows the daily energy in Wh, because it must supply all of that energy over the day
    4. It equals the array rating in every design
    Show answer

    Answer: A — It is sized on the simultaneous AC load with a safety margin, and must handle motor starting surges

    An inverter is a power device: its watt rating must cover the largest simultaneous load (plus surge, plus margin). Daily energy and autonomy belong to the array and battery, and the array rating is set by the energy chain, not by the inverter.

  7. A 48 V, 400 Ah bank is built from 12 V, 100 Ah batteries. The number of batteries needed is

    1. 4
    2. 8
    3. 16
    4. 32
    Show answer

    Answer: C — 16

    Series: 48/12 = 4 per string; parallel: 400/100 = 4 strings; total 4 × 4 = 16. Counting only the series set gives 4, and only the energy ratio would not need two separate checks.

  8. An array has 4 parallel strings of identical modules, each with Isc = 9 A and Imp = 8.3 A. The charge controller is rated 1.25 times the array short-circuit current. The minimum current rating, in amperes, is

    1. 33.2
    2. 45.0
    3. 41.5
    4. 36.0
    Show answer

    Answer: B — 45.0

    Isc,array = 4 × 9 = 36 A; × 1.25 = 45 A. The value 41.5 uses Imp (4 × 8.3 × 1.25), but the controller must stand the short-circuit current, and 36 A omits the margin.

  9. Two designs deliver the same daily energy to an AC load. Design X has a flat load (load factor 0.9); design Y has a peaky load (load factor 0.2). With the same efficiencies and PSH, which is correct?

    1. Y needs a larger inverter rating than X, while the array and battery energy are unchanged
    2. Y needs a larger battery than X, because its peak demand is higher and the bank must supply that peak
    3. Y needs a larger array than X, because its peak demand is higher and the array must follow it
    4. X needs a larger inverter rating than Y, because its load runs steadily for the whole day
    Show answer

    Answer: A — Y needs a larger inverter rating than X, while the array and battery energy are unchanged

    With equal E, the peak = E/(24 × LF), so Y's peak is 0.9/0.2 = 4.5 times X's and so is its inverter. The array and the battery depend on E alone, which is equal.

  10. For a stand-alone system with the other factors fixed, which of the following are correct?

    1. Battery capacity is directly proportional to the days of autonomy
    2. Battery capacity is inversely proportional to the permitted depth of discharge
    3. Array peak watts are inversely proportional to the peak sun hours
    4. Array peak watts are directly proportional to the days of autonomy
    Show answer

    Answer: A — Battery capacity is directly proportional to the days of autonomy; B — Battery capacity is inversely proportional to the permitted depth of discharge; C — Array peak watts are inversely proportional to the peak sun hours

    Battery Wh = E_dc N/(DoD ηb) gives the first two, and Parray = E_dc/(PSH η_sys) gives the third. N does not enter the array formula, so the statement tying array watts to days of autonomy is wrong.

  11. Which of the following statements about load factor are correct?

    1. Over a day it equals the daily energy divided by the peak power times 24 hours
    2. A perfectly constant load has a load factor of 1
    3. It is the ratio of peak power to average power
    4. For the same daily energy, a lower load factor means a higher peak demand
    Show answer

    Answer: A — Over a day it equals the daily energy divided by the peak power times 24 hours; B — A perfectly constant load has a load factor of 1; D — For the same daily energy, a lower load factor means a higher peak demand

    LF = E/(Ppeak × 24) = average/peak, so a flat load gives 1 and, for fixed E, a lower LF means a larger peak. The ratio of peak to average is the reciprocal and exceeds 1.

  12. Which of the following statements about the parts of PV systems are correct?

    1. A charge controller protects the battery from overcharge and from deep discharge
    2. A grid-connected system must contain a battery bank for the inverter to operate
    3. A stand-alone system that must work all year is sized on the month of poorest insolation or heaviest load
    4. Days of autonomy is the number of days the battery alone can supply the load
    Show answer

    Answer: A — A charge controller protects the battery from overcharge and from deep discharge; C — A stand-alone system that must work all year is sized on the month of poorest insolation or heaviest load; D — Days of autonomy is the number of days the battery alone can supply the load

    The statements on the charge controller, on sizing for the poorest month and on days of autonomy are standard definitions and design practice. A grid-tied inverter uses the grid as the store, so no battery is required.

  13. A household uses 7.2 kWh of energy in a day and its peak demand is 1.5 kW. Taking 24 h in a day, its daily load factor, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.20

    Average power = 7.2/24 = 0.30 kW; LF = 0.30/1.5 = 0.20.

  14. A load needs 1600 Wh/day of AC energy. The inverter efficiency is 0.90, the site has 5 peak sun hours and the combined array-side derating factor is 0.75. The array peak rating in watts needed, rounded to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 474

    E_dc = 1600/0.90 = 1777.8 Wh/day; Parray = 1777.8/(5 × 0.75) = 1777.8/3.75 = 474.07, i.e. 474 W. Verified in Python.

  15. The daily DC energy requirement is 1777.8 Wh. Design for 2 days of autonomy, depth of discharge 0.5 and battery efficiency 0.85, on a 24 V bus. The nominal battery capacity in ampere-hours, rounded to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 349

    Wh = 1777.8 × 2/(0.5 × 0.85) = 3555.6/0.425 = 8366.0 Wh; Ah = 8366.0/24 = 348.6, i.e. 349 Ah. Verified in Python.

  16. The connected AC loads total 1600 W, but at most 75 % of that runs at any one time. The inverter is to be rated at 1.25 times the largest simultaneous load. The minimum inverter rating in watts is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1500

    Simultaneous load = 0.75 × 1600 = 1200 W; × 1.25 = 1500 W. Using the full 1600 W would give 2000 W.

  17. A 3 kWp array at a site with 5.5 peak sun hours has an overall system derating factor of 0.78. The energy delivered per day, in kWh and to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 12.87

    Yield = rated kW × PSH × derating = 3 × 5.5 × 0.78 = 12.87 kWh/day. Ignoring the derating gives 16.5 kWh, the trap value.