Plant Capacity Factor, Plant Use Factor, and Units Generated per Annum

Units Generated per Annum

The kWh generated per annum or the energy generated per annum by a power station is determined as follows −

$$\mathrm{\because Load\: factor\:=\:\frac{Average\: load}{Maximum\: demand}}$$

$$\mathrm{\therefore Average\: load\:=\:Maximum\: demand\:\times \:Load\: factor}$$

Hence, the units generated per annum is given by,

$$\mathrm{Units\: generated \:per \:annum\:=\:Average\: load \:in\: kW\:\times \:Hours \:in\: a\: year}$$

$$\mathrm{=\:Maximum\: demand\: in\: kW\:\times \:Load\: factor\:\times\: 8760}$$

$$\mathrm{\Rightarrow Units\: generated\: per\: annum\:=\:Maximum\: demand \:in\: kW\:\times \:Load\: factor\:\times \:8760}$$

Plant Capacity Factor

The plant capacity factor of a power station is defined as the ratio of actual energy produced to the maximum possible energy that could have been produced during a given period, i.e.,

$$\mathrm{Plant \:Capacity \:Factor\:=\:\frac{Actual \:Energy\: Produced\:}{Maximum\: energy \:that\: could \:have\: been\: produced}}$$

$$\mathrm{\Rightarrow Plant\: Capacity \:Factor\:=\:\frac{Average\: demand\:\times \:T \:Hours}{Plant \:capacity\:\times T \:Hours}}$$

$$\mathrm{\therefore Plant \:Capacity \:Factor\:=\:\frac{Average\: demand}{Plant \:capacity}}$$

Therefore, if we consider the period (T) is one year, then

$$\mathrm{Plant\: Capacity \:Factor\:=\:\frac{Units\: generated/annum}{Plant \:capacity\:\times \:hours\: in\: a\: year}}$$

The plant capacity factor indicates the reserve capacity of the power station. An electric power station is so designed that it has some reserve capacity for meeting the increased load demand in future. Thus, the installed capacity of a power station is always somewhat greater than the maximum demand on the power station.

The reserve capacity of a power station is given by,

$$\mathrm{Reserve\: capacity\:=\:Plant \:capacity\:-\:Maximum\: demand}$$

Plant Use Factor or Utilisation Factor

The plant use factor of a power station is defined as the ratio of kWh generated to the product of plant capacity and the number of hours for which the plant was in operation, i.e.,

$$\mathrm{Plant\: use \:factor\:=\:\frac{kWh \:generated}{Plant \:capacity\:\times \:Hours\: of\: operation}}$$

$$\mathrm{\Rightarrow Plant\: use\: factor\:=\:\frac{Maximum\: demand}{Plant\: capacity}}$$

Numerical Example

A power station has a maximum demand of 14000 kW. The annual load factor is 60% and the plant capacity is 18750 kW. Determine the following −

• Plant capacity factor,

• Reserve capacity, and

• Plant use factor.

Solution

The energy generated per annum is given by,

$$\mathrm{kWh \:generated \:per\: annum\:=\:Maximum\: demand\:\times \:Load\: factor\:\times\: 8760}$$

$$\mathrm{\therefore kWh\:generated \:per\: annum\:=\:14000\:\times \:0.6\:\times \:8760\:73.584\:\times \:10^{6}\:kWh}$$

• Plant Capacity Factor −

$$\mathrm{Plant\: Capacity \:Factor\:=\:\frac{Units\: generated/annum}{Plant \:capacity\:\times \:hours\: in \:a \:year}}$$

$$\mathrm{\therefore Plant\: Capacity \:Factor\:=\:\frac{73.584\:\times 10^{6}}{18750\:\times 8760}\:=\:0.448\:=\:44.8\%}$$

• Reserve Capacity −

$$\mathrm{Reserve\: capacity\:=\:Plant\: capacity\:-\:Maximum\: demand}$$

$$\mathrm{\therefore Reserve\: capacity\:=\:18750\:-\:14000\:=\:4750\:kW}$$

• Plant Use Factor −

$$\mathrm{Plant\: use \:factor\:=\:\frac{Maximum\: demand}{Plant\:capacity}}$$

$$\mathrm{\therefore Plant\: use\: factor\:=\:\frac{14000}{18750}\:=\:0.7467\:=\:74.67\%}$$