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Q.3 Assume a system of 5 x 40 MW units, each with a FOR of 0.01 supplying a linear characteristic load, with a maximum load of 160 MW and minimum load of 64 MW. If the load increases by 25%, what would be the loss of load expectation (LOLE), and determine the added generating unit that results in a LOLE of 0.12 days/year. Check if there is a different way to achieve the same LOLE, but with avoiding the capital cost of adding the generating unit.

Answer 1

& Bystem units = 5x40mW alinear characteristic load = 0.01st Day roolimum load = 160mw it also minimum load = Gumw The loud increase by = 25% The load loss of expectation (LOLE) Loss of Load Expectation(LOLE) = Loss of love probability (LOLP)*T where T= 365 days when the mastimum load is increased by | 25= 160+ 40 = 200mW! de Losset dead expectations The load model used depends on the required Yeliability index.s u se Lile. Another one yo ese sente the load using the videed hoursy load Velues. e daily Peak load Variation curve, the load had duration curve is created by arranging hourly lead values in de scending orded. 1 The 1055 of load me thud, The daily peak load aze combined with The capacity outage probabil method dur The CScanned with CamScanner

table to obtain the expected number of days in the given period in which the daily peak load cor hourly load ) exceed: The available capacity. if there is a difference way to achieve The Same LOL E. Enk-IPKTKE A Senk=1(tk-tk-1) PK To where nais The total number of capacity outage States. PR= is the individual Probability of the capacity outage State PK = is the cumulative probability of the - capacity outage state tksis The number of time units when There is a loss of load LOLE = 8.64 360000=0.12 Days/YEART The system can achieve the same LOLE by adeling The units of system. CScanned with CamScanner