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6.2 Maximum Demand Calculation and Diversity Factors for ≤45 kVA

Key Takeaways

  • Connected Load is the total sum of continuous power ratings of all installed electrical appliances, whereas Maximum Demand is the actual peak load expected during normal operation.

  • Applying diversity factors per SS 638 Appendix 4 accounts for the non-simultaneous usage of connected appliances, preventing unnecessary over-sizing of supply cables and main switchgear.

  • In Singapore, single-phase supplies are generally restricted to 30A or 60A (up to ~14 kVA); loads exceeding 14 kVA require upgrade to a 3-phase 400V supply rated up to 45 kVA (~63A per phase).

  • The three-phase apparent power formula S = √3 × VL × IL (where VL = 400V) converts line current to kVA: S(kVA) = 0.6928 × IL.

  • Phase balancing across L1, L2, and L3 is critical in 3-phase installations to minimize neutral current and optimize transformer efficiency.

Last updated: August 2026

6.2 Maximum Demand Calculation and Diversity Factors for ≤45 kVA

Quick Summary: Maximum Demand (MD) is the maximum load current or power drawn by an electrical installation from the supply grid during any specified operational interval. Because not all electrical appliances, lights, or socket outlets operate simultaneously at full rated capacity, calculating maximum demand using Diversity Factors under SS 638 Appendix 4 allows engineers to determine the realistic peak load. Proper maximum demand calculation prevents costly over-design while ensuring incoming supply cables, main cut-out fuses, and circuit breakers do not trip under normal peak operating conditions.


1. Connected Load vs. Maximum Demand

Before sizing intake cables and protective switchgear, a clear distinction must be made between Connected Load and Maximum Demand:

  • Total Connected Load (PconnP_{conn} or SconnS_{conn}): The mathematical sum of the full-load power ratings of all installed electrical equipment, luminaires, socket outlets, motors, and appliances connected to the installation.
  • Maximum Demand (ImdI_{md} or SmdS_{md}): The maximum current or apparent power actually drawn by the installation at any one time under normal operating conditions. It is calculated by applying standardized diversity factors to individual sub-circuits.
  • Diversity Factor: The ratio of the sum of individual maximum demands of sub-circuits to the maximum demand of the complete installation. In practical design, diversity factors are expressed as derating percentages or fixed allowances applied to connected loads.
Maximum Demand=∑(Connected Sub-Circuit Load×Diversity Factor)\text{Maximum Demand} = \sum \left( \text{Connected Sub-Circuit Load} \times \text{Diversity Factor} \right)

2. Supply Capacity Boundaries in Singapore (≤45 kVA Regime)

In Singapore, SP PowerGrid classifies low-voltage electricity supplies based on load capacity boundaries:

Single-Phase 230V Supply Limits

  • 30A Single-Phase: Equivalent to 30 A×230 V=6.9 kVA30\text{ A} \times 230\text{ V} = 6.9\text{ kVA}. Standard intake for small domestic apartments.
  • 60A Single-Phase: Equivalent to 60 A×230 V=13.8 kVA≈14 kVA60\text{ A} \times 230\text{ V} = 13.8\text{ kVA} \approx 14\text{ kVA}. Maximum permissible single-phase supply allocation for residential or commercial units.
  • Threshold Rule: If the calculated maximum demand exceeds 60 A60\text{ A} (or 14 kVA14\text{ kVA}), SP PowerGrid mandates an upgrade to a Three-Phase 400V supply to prevent severe phase imbalance on the distribution network.

Three-Phase 400V Supply up to 45 kVA

  • 45 kVA 3-Phase Rating: The maximum capacity boundary for standard small-commercial, food shop, office, or residential house installations operating on direct CT-less or standard 63A meter cut-outs.
  • Current per Phase Calculation: For a balanced 45 kVA45\text{ kVA} 3-phase supply at 400 V400\text{ V} line voltage:
S=3×VL×ILS = \sqrt{3} \times V_L \times I_L 45,000 VA=3×400 V×IL=692.82×IL45,000\text{ VA} = \sqrt{3} \times 400\text{ V} \times I_L = 692.82 \times I_L IL=45,000692.82=64.95 A≈63 A per phaseI_L = \frac{45,000}{692.82} = 64.95\text{ A} \approx 63\text{ A per phase}

Thus, a 45 kVA45\text{ kVA} 3-phase supply corresponds to a standard 63 A63\text{ A} 3-pole main isolator / MCCB per phase.


3. SS 638 Appendix 4 Diversity Factor Rules

Singapore Standard SS 638 Appendix 4 provides prescriptive rules for estimating maximum demand across domestic, commercial, and small industrial premises. The table below summarizes the mandatory diversity allowances:

Type of Load CircuitDomestic Premises Diversity AllowanceCommercial / Office Premises Diversity AllowanceSmall Industrial Premises Diversity Allowance
Lighting Circuits66% of total connected lighting load90% of total connected lighting load90% of total connected lighting load
13A Socket-Outlet Circuits100% of largest circuit + 40% of remaining socket circuits100% of largest circuit + 50% of remaining socket circuits100% of largest circuit + 50% of remaining socket circuits
Cooking Appliances10A + 30% of remaining full load + 5A for socket on switch unit100% of largest appliance + 80% of 2nd + 60% of remaining100% of largest appliance + 80% of 2nd + 60% of remaining
Instantaneous Water Heaters100% of largest heater + 100% of 2nd heater + 25% of remaining100% of largest heater + 100% of 2nd heater + 50% of remaining100% of largest heater + 100% of 2nd heater + 50% of remaining
Thermal Storage Water Heaters100% full load (No diversity allowed)100% full load (No diversity allowed)100% full load (No diversity allowed)
Air-Conditioning & Motors100% of largest AC unit + 80% of 2nd unit + 60% of remaining100% of largest motor/AC + 80% of 2nd motor + 60% of remaining100% of largest motor + 80% of 2nd motor + 60% of remaining

4. Single-Phase and Three-Phase Power Formulas

To compute phase currents accurately from connected active power (PP in kW) or apparent power (SS in kVA), use the standard electrical power equations:

Single-Phase Calculations (230V AC)

Active Power: P=V×I×cos⁡ϕ\text{Active Power: } P = V \times I \times \cos\phi Design Current: Ib=PV×cos⁡ϕ=P230×cos⁡ϕ\text{Design Current: } I_b = \frac{P}{V \times \cos\phi} = \frac{P}{230 \times \cos\phi} Apparent Power: S=V×I=230×Ib\text{Apparent Power: } S = V \times I = 230 \times I_b

Three-Phase Calculations (400V Line-to-Line / 230V Line-to-Neutral)

Active Power: P=3×VL×IL×cos⁡ϕ\text{Active Power: } P = \sqrt{3} \times V_L \times I_L \times \cos\phi Line Current: IL=P3×VL×cos⁡ϕ=P692.82×cos⁡ϕ\text{Line Current: } I_L = \frac{P}{\sqrt{3} \times V_L \times \cos\phi} = \frac{P}{692.82 \times \cos\phi} Apparent Power: S=3×VL×IL=0.6928×IL (kVA)\text{Apparent Power: } S = \sqrt{3} \times V_L \times I_L = 0.6928 \times I_L \text{ (kVA)}

Phase Balancing and Neutral Current

In a 3-phase 4-wire installation, single-phase loads must be distributed evenly across Phase L1, Phase L2, and Phase L3. If the phase currents I1,I2,I3I_1, I_2, I_3 are balanced (I1=I2=I3I_1 = I_2 = I_3), the neutral current IN=0 AI_N = 0\text{ A}. For unbalanced loads, the resultant vector neutral current is calculated using:

IN=I12+I22+I32−(I1I2+I2I3+I3I1)I_N = \sqrt{I_1^2 + I_2^2 + I_3^2 - (I_1 I_2 + I_2 I_3 + I_3 I_1)}

5. Comprehensive Step-by-Step Worked Maximum Demand Calculation

Example Scenario:

A small commercial office unit with a 3-phase 400V supply has the following connected loads:

  1. Lighting: 40×2×18W40 \times 2 \times 18\text{W} LED panel fittings (0.950.95 power factor, total 1.52 kW1.52\text{ kW}). Distributed equally: 0.507 kW0.507\text{ kW} per phase.
  2. Socket Outlets: 6 separate 13A13\text{A} twin ring final circuits (32A32\text{A} protection each). 2 ring circuits connected to L1, 2 to L2, 2 to L3.
  3. Water Heating: 2×3.5 kW2 \times 3.5\text{ kW} single-phase instantaneous water heaters (15.2 A15.2\text{ A} each at 230V230\text{V}). Heater 1 on L1, Heater 2 on L2.
  4. Air-Conditioning: 3×3HP3 \times 3\text{HP} 3-phase cassette air-conditioners (3 kW3\text{ kW} each, cos⁡ϕ=0.85\cos\phi = 0.85, total 9 kW9\text{ kW}). Line current per AC unit =30003×400×0.85=5.1 A= \frac{3000}{\sqrt{3} \times 400 \times 0.85} = 5.1\text{ A}.

Step-by-Step Calculation Solution:

Step 1: Calculate Diversity for Lighting (Commercial: 90% allowance)

  • Total Connected Lighting Power =1,520 W= 1,520\text{ W}
  • Maximum Demand Power =1,520 W×0.90=1,368 W= 1,520\text{ W} \times 0.90 = 1,368\text{ W}
  • Total 3-Phase Current =13683×400×0.95=2.08 A per phase= \frac{1368}{\sqrt{3} \times 400 \times 0.95} = 2.08\text{ A per phase}

Step 2: Calculate Diversity for 13A Socket Outlets (Commercial: 100% largest + 50% remaining)

  • Each phase has 2 ring circuits rated at 32A32\text{A} capacity.
  • Per Phase Socket Maximum Demand =100% of Circuit 1 (32A)+50% of Circuit 2 (32A×0.50=16A)= 100\% \text{ of Circuit 1 } (32\text{A}) + 50\% \text{ of Circuit 2 } (32\text{A} \times 0.50 = 16\text{A})
  • Per Phase Socket Maximum Demand =32A+16A=48A per phase= 32\text{A} + 16\text{A} = 48\text{A per phase}

Step 3: Calculate Diversity for Instantaneous Water Heaters (Commercial: 100% 1st + 100% 2nd)

  • Heater 1 on L1 =15.2 A= 15.2\text{ A}
  • Heater 2 on L2 =15.2 A= 15.2\text{ A}
  • L3 Water Heater Load =0 A= 0\text{ A}

Step 4: Calculate Diversity for Air-Conditioning Units (3-Phase Motors: 100% 1st + 80% 2nd + 60% 3rd)

  • Unit 1 (Largest) =100%×5.1 A=5.1 A= 100\% \times 5.1\text{ A} = 5.1\text{ A}
  • Unit 2 =80%×5.1 A=4.08 A= 80\% \times 5.1\text{ A} = 4.08\text{ A}
  • Unit 3 =60%×5.1 A=3.06 A= 60\% \times 5.1\text{ A} = 3.06\text{ A}
  • Total AC Maximum Demand per Phase =5.1+4.08+3.06=12.24 A per phase= 5.1 + 4.08 + 3.06 = 12.24\text{ A per phase}

Step 5: Tabulate Maximum Demand per Phase (L1, L2, L3)

Load CategoryPhase L1 Current (A)Phase L2 Current (A)Phase L3 Current (A)
Lighting2.08 A2.08\text{ A}2.08 A2.08\text{ A}2.08 A2.08\text{ A}
13A Sockets48.00 A48.00\text{ A}48.00 A48.00\text{ A}48.00 A48.00\text{ A}
Water Heaters15.20 A15.20\text{ A}15.20 A15.20\text{ A}0.00 A0.00\text{ A}
Air-Conditioning12.24 A12.24\text{ A}12.24 A12.24\text{ A}12.24 A12.24\text{ A}
Total Phase Demand77.52 A77.52\text{ A}77.52 A77.52\text{ A}62.32 A62.32\text{ A}

Step 6: Determine Overall Maximum Demand in kVA and Verify Supply Class

  • Peak Phase Line Current occurs on L1 / L2 at 77.52 A77.52\text{ A}.
  • Overall 3-Phase Apparent Power Demand:
Smd=3×VL×Imax=3×400 V×77.52 A=53,707 VA=53.71 kVAS_{md} = \sqrt{3} \times V_L \times I_{max} = \sqrt{3} \times 400\text{ V} \times 77.52\text{ A} = 53,707\text{ VA} = 53.71\text{ kVA}

Technical Conclusion & Design Verification:

Since the calculated Maximum Demand is 53.71 kVA53.71\text{ kVA} (77.52 A per phase77.52\text{ A per phase}), this exceeds the standard 45 kVA45\text{ kVA} / 63 A63\text{ A} supply limit. To bring this installation within the 45 kVA45\text{ kVA} limit, the LEW must either:

  1. Apply load management or re-group socket circuits (reducing socket demand to 32A32\text{A} per phase base), or
  2. Upgrade the intake submission to SP PowerGrid for a 100 A100\text{ A} 3-phase supply (69.3 kVA69.3\text{ kVA}).
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Maximum Demand & Diversity Derating Flow
Test Your Knowledge

What is the calculated balanced line current per phase for a 3-phase installation drawing its maximum rated capacity of 45 kVA at 400V line voltage?

A

64.95 A (standardized to 63A main protective device)

B

30.0 A

C

112.5 A

D

195.6 A

Test Your Knowledge

Under SS 638 Appendix 4, what is the diversity factor allowance for a commercial office installation with 10 installed 13A socket-outlet sub-circuits?

A

100% of total connected load across all 10 socket circuits

B

66% of total connected load across all socket circuits

C

100% of the largest socket circuit rating + 50% of the remaining socket circuits

D

100% of the first 2 socket circuits + 25% of the remaining socket circuits

Test Your Knowledge

According to SP PowerGrid supply guidelines, what is the maximum single-phase 230V supply allocation permitted before a mandatory 3-phase 400V upgrade is required?

A

30A (6.9 kVA)

B

60A (approx. 14 kVA)

C

100A (23 kVA)

D

150A (34.5 kVA)

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