Due to a shortage in energy where the reason behind is that the demand for electricity exceeds the available power is known as curtailment. Therefore, the term curtailment is one among many tools to maintain system energy balance. It should be known that there are hierarchies of customers. This type of hierarchy is clear where some might require a partial or total cut back of electricity before others. The industrial sectors (industrial customers) are usually curtailed before reducing the service to the residential customers. When electrical energy service has to be interrupted, it means that energy reserves have dropped or are expected to drop below a certain level (Billinton and Allan, 1996; Denholm and Mai, 2019; IEEE Standard 493–2007; Schermeyer et al., 2018).

$\begin{array}{c}\begin{array}{l}\text{Energy Curtailed = Total Energy − Total Time }\times \text{ Capacity IN}\\ \text{Expectation of Energy Curtailed = Probability }\times \text{ Energy Curtailed}\end{array}\end{array}$

Example (6.2):

The term LOLE is one of the reliability indices. During the observation of the power system load cycle, the expected number of hours (HLOLE: Hourly Loss of Load Expectation) or time period are calculated when insufficient generating capacity is available to serve the required load at a time. In another statement, it can be defined as the expected number of days per annual period in which the available generation capacity is insufficient to satisfy and serve the daily maximum load. It can be calculated from the daily peak load or annual peak load. In practical applications, the LOLE expectation index is more often used than the LOLP probability index. The relationship between both the LOLP and the LOLE is formulated as (IEEE Standard 493–2007):

$\begin{array}{c}\text{LOLE}=\text{LOLP}\text{.T}\end{array}$

In case the load model is an annual continuous load curve with day maximum load, then the LOLE unit is in days per year and T = 365 days.

In case the load model is an hourly load curve, then the LOLE unit is in hours per year and T = 8760 hours.

Example (6.4):

A power station is supplying a city in the Kingdom of Bahrain for a year. The weekly average loads have been recorded as such (in MW):

The power station supplying the city has three generating units, the first unit is working with a capacity of 25MW and 0.02 fail per week. The second unit has a capacity of 25 MW as well, with a failure rate equal to 0.05 f/week. The last unit (number 3) has a repair rate of 0.96 r/week, and its capacity is 50 MW. Calculate the LOLE in percentage.

Solution:

$\begin{array}{c}\text{LOLE}={\displaystyle \sum _{\text{i}=1}^{\text{N}}\left(\text{No}\text{. of Occurences}\right)\times {\text{P}}_{\text{i}}\left(\text{Maximum}\text{Available Power}\right)-\text{Peak}\text{Load}}\end{array}$

where: the maximum available power is the maximum capacity.

$\begin{array}{c}\begin{array}{c}\text{LOLE}=5\text{P}\left(100-77\right)+9\text{P}\left(100-57\right)+8\text{P}\left(100-52\right)+6\text{P}\left(100-46\right)+13\text{P}\left(100-41\right)+11\text{P}\left(100-34\right)\\ =5\text{P}\left(23\right)+9\text{P}\left(43\right)+8\text{P}\left(48\right)+6\text{P}\left(54\right)+13\text{P}\left(59\right)+11\text{P}\left(66\right)\\ =5\left(0.10624\right)+9\left(0.04096\right)+8\left(0.04096\right)+6\left(0.00276\right)+13\left(0.00276\right)+11\left(0.00276\right)\\ =0.5312+0.36864+0.32768+0.01656+0.03588+0.3036\\ =1.31022\text{week/year}\end{array}\end{array}$

The expected loss of load through the year is approximately equal to 1.3 week. This is equivalent to 9.17 days a year.

$\begin{array}{c}\text{LOLE in Percentage}=\left(1.31032/52\right)\times 100\%=2.519846\%\end{array}$

Example (6.5):

A power station is supplying a city in the Kingdom of Bahrain for a year. The weekly average loads have been recorded as such (in MW):

The power station supplying the city has three generating units, the first unit is working with a capacity of 50MW and 0.02 fail per week. The second unit has a capacity of 50MW as well, with a failure rate equal to 0.05 f/week. The third unit has a repair rate of 0.96 r/week, and its capacity is 100MW. Calculate the LOLE in percentage.

Solution:

$\begin{array}{c}LOLE={\displaystyle \sum _{i=1}^N\left(No.ofOccurences\right)\text{x}{P}_i\left(MaximumAvailablePower\right)-PeakLoad}\end{array}$

where: the maximum available power is the maximum capacity.

$\begin{array}{c}\begin{array}{c}\text{LOLE}=\text{13 P}\left(200-82\right)+\text{11 P}\left(200-68\right)+\text{9 P}\left(200-114\right)+\text{8 P}\left(200-104\right)+\text{6 P}\left(200-92\right)+5\text{P}\left(200-154\right)\\ =\text{13 P}\left(118\right)+\text{11 P}\left(132\right)+\text{9 P}\left(86\right)+\text{8 P}\left(96\right)+\text{6 P}\left(108\right)+\text{5 P}\left(46\right)\\ =13\left(0.00276\right)+11\left(0.00276\right)+9\left(0.04096\right)+8\left(0.04096\right)+6\left(0.00276\right)+5\left(0.10624\right)\\ =1.310322\text{week/year}\end{array}\end{array}$

The expected loss of load through the year is approximately equal to 1.3 week. This is equivalent to 9.17 days a year.

$\begin{array}{c}\text{LOLE in Percentage = (1}\text{.31032/52) }\times \text{ 100\% = 2}\text{.519846\%}\end{array}$

Example (6.6):

Three generators have been running for 52 weeks. These generators are represented by the following data summarized in the following table:

The table below shows the data of the peak loads it occurred and how frequent it is happening:

Calculate the LOLE in percentage.

Solution:

$\begin{array}{c}LOLE={\displaystyle \sum _{i=1}^N\left(No.ofOccurrences\right)\text{x}{P}_i\left(MaximumAvailablePower\right)-Peak\text{}Load}\end{array}$

where: the maximum available power is the maximum capacity.

$\begin{array}{c}\begin{array}{c}\mathrm{LOLE}=7\text{P}\left(210-200\right)+11\text{}\text{P}\left(210-160\right)+10\text{}\text{P}\left(210-125\right)+13\text{}\text{P}\left(210-95\right)+5\text{}\text{P}\left(210-80\right)+6\text{}\text{P}\left(210-75\right)\\ =7\text{}\text{P}\left(10\right)+11\text{}\text{P}\left(50\right)+10\text{}\text{P}\left(85\right)+13\text{}\text{P}\left(115\right)+5\text{P}\left(130\right)+6\text{}\text{P}\left(135\right)\\ =7\left(0.1493408\right)+11\left(0.1493408\right)+10\left(0.00800992\right)+13\left(0.00005592\right)+5\left(0.00005592\right)+6\left(0.00005592\right)\\ =2.76957568\text{week/year}\end{array}\end{array}$

The expected loss of load through the year is approximately equal to 2.77 weeks. This is equivalent to 19.4 days a year.

$\begin{array}{c}\text{LOLE in Percentage = (2}\text{.76957568/52) }\times \text{ 100\% = 5}\text{.326107077\%}\end{array}$

Example (6.7):

Three generators are running with the following data:

The three generators have been running for one year (365 days). The data for a period of one year with a maximum of 100MW are given in the following table:

Find the LOLE.

Solution:

To combine the three units together as one system:

$\begin{array}{c}LOLE\text{ = }{\displaystyle \sum _{i=1}^N\left(No\text{. }of\hspace{0.17em}Occurrences\right)}\text{x}{P}_i\left(Maximum\hspace{0.17em}Available\hspace{0.17em}Power\right)-PeakLoad\end{array}$

where: the maximum available power is the maximum capacity.

$\begin{array}{c}\begin{array}{l}\text{LOLE}=12\times \text{P}\left(100-57\right)+83\times \text{P}\left(100-52\right)+107\times \text{P}\left(100-46\right)+116\times \text{P}\left(100-41\right)+47\times \text{P}\left(100-34\right)\\ \text{LOLE}=\text{12}\times \text{P}\left(43\right)+\text{83}\times \text{P}\left(48\right)+\text{107}\times \text{P}\left(54\right)+\text{116}\times \text{P}\left(56\right)+\text{47}\times \text{P}\left(66\right)\\ \text{LOLE}=\text{12}\times \left(0.040576\right)+\text{83}\times \left(0.040576\right)+\text{107}\times \left(0.001976\right)+\text{116}\times \left(0.001976\right)+\text{47}\times \left(0.001976\right)\\ \text{LOLE}=\text{4}\text{.38824}\text{days/year}\end{array}\end{array}$

This figure shows that the system might loss power for 4.38824 days in a year. This means that the expected number of hours per year that the system’s electricity production cannot meet its demand.

The loss of load percentage = (4.38824 / 365) × 100 = 1.2327%

This index varies between the resource types and areas. The FOR can be also defined in a formula shape as (Billinton and Allan, 1992; IEEE Standard 493–2007):

$\begin{array}{c}FOR=\frac{Forced\hspace{0.17em}Outage\hspace{0.17em}Hours}{ForcedOutageHours+IN\text{}ServiceHours}\end{array}$

The availability (Davidov and Pantos, 2019) of the system has a number of formulas:

$\begin{array}{c}\mathrm{Availability}=\frac{\text{MTBF}}{\text{MTBF}+\text{MTTR}}\end{array}$

At the same time, when the system has two-states the availability of the system is given by

$\begin{array}{c}\begin{array}{c}\text{Availability of the system = Probability at any time when the system in the Up State}\\ \text{Unavailability = P (Down)}\end{array}\end{array}$

In case, the system has three states: up-state, derated-state, and downstate. Therefore,

$\begin{array}{c}\begin{array}{l}\text{Availability = P (Up) + P (Derated)}\\ \text{Unavailability = P (Down)}\end{array}\end{array}$

This means the availability is considered that the system is under operation. The unavailability means that the system is in the fail state.

Any power system is evaluated based on a number of points, these points are summarized as (Akhavein and Porkar, 2017; Al-Shaalan, 2012; Chaiamarit and Nuchprayoon, 2013; Zhong et al., 2017):