Growth is 5% in year 1 and increases by 1% each year. How is the 5th year consumption multiplier found?

1.05 x 1.06 x 1.07 x 1.08 x 1.09

How are 5th year consumer numbers obtained?

N0 x population growth (1.02)^5 x coverage factor

N0 x consumption growth x coverage

How is the individual consumer peak calculated?

E5 / (24 x LF) using daily energy E5/effective days

What do the contribution factors do?

Weight each class peak by how much it contributes at the system peak, before summing

Asked 7/20
2072 Asoj · 9 marks

5th year peak load, incremental growth rate

Short answer

Apply consumption growth multiplier (1.05×1.06×1.07×1.08×1.09=1.4019) and population multiplier (1.02^5=1.1041) with 70% coverage to get 5th-year consumers and monthly energy. Individual peak = (daily kWh)/(24×LF). Sum CF×N×P_ind yields peak demand ≈20.5 kW.

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

Consumer class	Potential consumer	Avg. monthly consumption	No. of effective days/month	5th year contribution factor to peak	5th year load factor
Domestic	100	30 kWh	30	1.0	0.25
Commercial	20	30 kWh	30	0.5	0.30
Non-commercial	50	20 kWh	20	0.1	0.25

Determine the 5th year peak load of load center in kW. Data for load forecast (for all consumer classes): Load growth factor: 5% for first year and an increment of 1% from 2nd year. Population growth factor: 2% each year. Percentage coverage factor: 70% by 5th year.

The 5th year peak is found by projecting consumer numbers and consumption, converting daily energy to individual peak with the load factor, and combining class peaks using contribution factors.

Growth multipliers

Consumption growth: 5% in year 1, increasing by 1% each year (6%, 7%, 8%, 9%):

\begin{aligned} G_c &= 1.05 \times 1.06 \times 1.07 \times 1.08 \times 1.09 = 1.40194 \\ G_p &= (1.02)^5 = 1.10408 \end{aligned}

Formulae

\begin{aligned} N_5 &= N_0 \times G_p \times \text{coverage} \\ E_5 &= E_0 \times G_c \\ P_{ind} &= \frac{E_5 / \text{effective days}}{24 \times LF} \\ P_{LC} &= \sum CF \times N_5 \times P_{ind} \end{aligned}

Number of consumers (5th year)

Class	$N_0$	$N_5 = N_0 \times 1.10408 \times 0.7$
Domestic	100	77.29
Commercial	20	15.46
Non-commercial	50	38.64

Consumption and individual peak

Class	$E_5$ (kWh/month)	Daily kWh	LF	$P_{ind}$ (kW)
Domestic	30 × 1.40194 = 42.058	42.058/30 = 1.4019	0.25	0.23366
Commercial	42.058	42.058/30 = 1.4019	0.30	0.19471
Non-commercial	20 × 1.40194 = 28.039	28.039/20 = 1.4019	0.25	0.23366

Example (commercial): $P_{ind} = \dfrac{1.4019}{24 \times 0.30} = 0.19471$ kW.

Contribution to load centre peak

Class	Class peak (kW)	CF	Contribution (kW)
Domestic	77.29 × 0.23366 = 18.058	1.0	18.058
Commercial	15.46 × 0.19471 = 3.010	0.5	1.505
Non-commercial	38.64 × 0.23366 = 9.029	0.1	0.903
Total			20.47

Answer: 5th year peak load of the load centre ≈ 20.5 kW (20.47 kW).

✓ Answer checked by Claude Opus 5.5 (high effort)

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