Connected Load, Average Load, and Maximum Demand Load

ElectronElectronics & ElectricalDigital Electronics

Connected Load

The connected load is defined as the sum of continuous ratings of all the equipment connected to the electrical power station.

An electric power station supplies the power to thousands of consumers. Each consumer has certain equipment in his premises. The sum of the continuous ratings of all the equipment in the consumer’s premises is the connected load of the consumer. The sum of the connected loads of all the consumers is called the connected load to the power station.

Average Load

The average load on a power station is defined as the average of loads occurring on the power station in a given period (either day or month or year).

The daily average load on the power station is given by,

$$\mathrm{Daily\: average\: load\:=\:\frac{Number\: of\: units \:generated \:in\: a\: day(in \:kWh)}{24\: Hours}}$$

The monthly average load on the power station is given by,

$$\mathrm{Monthly\: average\ load\:=\:\frac{Number\: of\: units\: generated\: in\: a \:month(in\: kWh)}{Number\: of\: hours\: in \:a\: month}}$$

The yearly average load on the power station is given by,

$$\mathrm{Yearly\: average\: load\:=\:\frac{Number\: of\: units \:generated \:in\: a \:year(in \:kWh)}{Number\: of\: hours \:in\: a \:year}}$$

Maximum Demand Load

The maximum demand on a power station is defined as the greatest demand of load on the power station during a given period.

The load on the power station varies from time to time. The maximum of all the demands that have occurred during a given period (let a day) is the maximum demand on the power station. Generally, the maximum demand on the power station is less than the connected load. It is expected because all the consumers do not switch on their connected load to the station at a time. The knowledge of maximum demand of a power station is very important because it is used to determine the installed capacity of the power station.

Numerical Example (1)

At the end of a power distribution system, a feeder supplies three distribution transformers, each one supplying a group of consumers, whose connected loads are given as −

TransformerLoadDemand FactorDiversity factor of group
Transformer I15 kW0.71.3
Transformer II10 kW0.653
Transformer III13 kW0.51.4

If the diversity factor among the transformers is 1.5, then determine the maximum load on the feeder.

Solution

The sum of maximum demands of consumers is given by,

$$\mathrm{\mathrm{Sum\: of\: max.demands\:=\:connected\:load\:\times \:demand\:factor}}$$

Therefore,

$$\mathrm{\mathrm{Sum\: of\: max.demands\: of \:consumers\: on\: Transformer\: I}\:=\:15\:\times 0.7\:=\:10.5\:\mathrm{kW}}$$

Similarly

$$\mathrm{\mathrm{Sum\: of\: max.demands\: of \:consumers\: on\: Transformer\:II }\:=\:10\:\times \:0.65\:=\:6.5\:\mathrm{kW}}$$

$$\mathrm{\mathrm{Sum\: of\: max.demands\: of \:consumers\: on\: Transformer\:III}\:=\:13\:\times \:0.5\:=\:6.5\:\mathrm{kW}}$$

Now, the maximum demand on the transform is given by,

$$\mathrm{Maximum \:demand\:=\:\frac{Sum\: of\: Max.demands}{Diversity\: factor}}$$

Therefore,

$$\mathrm{Maximum\: demand \:on\: Transform\: I \:=\:\frac{10.5}{1.3}\:=\:8.08\:kW}$$

$$\mathrm{Maximum\: demand \:on\: Transform\: II \:=\:\frac{6.5}{3}\:=\:2.167\:kW}$$

$$\mathrm{Maximum\: demand \:on\: Transform\: III \:=\:\frac{6.5}{1.4}\:=\:4.64\:kW}$$

As the diversity factor amount the transformers is 1.5. Therefore, the maximum demand on the feeder is given by,

$$\mathrm{Maximum\: demand\: on\: feeder\:=\:\frac{Sum\: of\: maximum \:demands}{Diversity\: factor}}$$

$$\mathrm{\therefore Maximum \:demand\: on \:feeder\:=\:\frac{8.08\:+\:2.167\:+4.64}{1.5}\:9.925\:kW}$$

Numerical Example (2)

If the number of units (kWh) generated per year by a power station is 45 ☓ 105, then determine the annual average load on the power station.

Solution

The annual average load on a power station is given by,

$$\mathrm{Annual \:average\: load\:=\:\frac{Number\: of\: units \:generated \:in \:a \:year\: (in kWh)}{Number\: of\: hours\: in\: a \:year}}$$

$$\mathrm{\therefore Annual\: average\: load\:=\:\frac{45\:\times 10^{5}}{8760}\:=\:513.7\:kW}$$

raja
Updated on 14-Feb-2022 06:48:46

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