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5th year peak load with population growth

Short answer

With 7 years of 1.8% population and 5% consumption growth, 70% coverage, individual peak 0.2345 kW, and contribution factors 1.0, 0.5, 0.1, the 5th-year peak load is 21.4 kW.

From the initial survey of particular load center, the following consumer data has been obtained.

Consumer classPotential consumerAvg. monthly consumptionNo. of effective days/month5th year contribution factor to peak5th year load factor
Domestic10030 kWh301.00.25
Commercial2030 kWh250.50.30
Non-commercial5020 kWh200.10.25

Assuming population growth factor of 1.8%, consumption growth factors of 5% in each year for each class and electrification coverage factor by 5th year is 70% for all consumers classes, determine 5th year peak load of the load center in kW. Also, project implementation period be 2 years.

The load centre peak is found by projecting the number of consumers and their consumption to the 5th year, converting energy to peak using the load factor, and combining the class peaks with their contribution factors.

Assumptions: The 5th year is counted after project completion, so with a 2-year implementation period the growth period from the survey is n=5+2=7n = 5 + 2 = 7 years for both population and consumption.

Formulae

N5=N0(1+p)n×coverageE5=E0(1+g)nPind=E5/effective days24×LFPLC=∑CF×N5×Pind\begin{aligned} N_5 &= N_0 (1 + p)^n \times \text{coverage} \\ E_5 &= E_0 (1 + g)^n \\ P_{ind} &= \frac{E_5 / \text{effective days}}{24 \times LF} \\ P_{LC} &= \sum \text{CF} \times N_5 \times P_{ind} \end{aligned}

Growth multipliers

(1.018)7=1.13301(1.05)7=1.40710\begin{aligned} (1.018)^7 &= 1.13301 \\ (1.05)^7 &= 1.40710 \end{aligned}

Number of consumers in 5th year

ClassN0N_0N0×1.13301×0.7N_0 \times 1.13301 \times 0.7
Domestic10079.31
Commercial2015.86
Non-commercial5039.66

Consumption and individual peak

ClassE5E_5 (kWh/month)Daily (kWh)LFPindP_{ind} (kW)
Domestic30 × 1.4071 = 42.21342.213/30 = 1.40710.250.23452
Commercial42.21342.213/25 = 1.68850.300.23452
Non-commercial20 × 1.4071 = 28.14228.142/20 = 1.40710.250.23452

Example (domestic): Pind=1.407124×0.25=0.23452P_{ind} = \dfrac{1.4071}{24 \times 0.25} = 0.23452 kW.

Class peak and contribution to load centre peak

ClassClass peak (kW)CFContribution (kW)
Domestic79.31 × 0.23452 = 18.6001.018.600
Commercial15.86 × 0.23452 = 3.7200.51.860
Non-commercial39.66 × 0.23452 = 9.3000.10.930
Total21.39

Answer: 5th year peak load of the load centre ≈ 21.4 kW.

(If the implementation period were ignored, i.e. n=5n = 5, the result would be lower; the 2-year period is included because consumers keep growing while the project is being built.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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