Exercise 1. A feeder has three loads : street lighting, residential and commercial. Le load profile of each load is presented in the table below residential Commercial Time street lighting kw kw kw 12:00 AM 100 250 300 1:00 AM 100 250 300 2:00 AM 100 250 300 3:00 AM 100 250 300 4:00 AM 100 250 300 5:00 AM 100 250 300 6:00 AM 100 250 300 7:00 AM 0 350 300 8:00 AM 0 450 400 9:00 AM 0 550 600 10:00 AM 0 550 1100 11:00 AM 0 550 1100 12:00 PM 0 600 1100 1:00 PM 0 600 1100 2:00 PM 0 600 1300 3:00 PM 0 600 1300 4:00 PM 0 600 1300 5:00 PM 0 650 1300 6:00 PM 0 750 900 7:00 PM 0 900 500 8:00 PM 100 1100 500 9:00 PM 100 1100 500 10:00 PM 100 900 300 11:00 PM 100 700 300 12:00 AM 100 350 300 1.) Draw the load curve of the three loads and the system one. 2.) Calculate the daily average load of the system 3.) Give the maximum demand of each load (Di) as well as the feeder one (Dg) 4.) Calculate the Load factor (Fld) 1 5.) Assume that the total connected load are 150kW, 2500kW and 2500kW for the street lighting, residential and commercial respectively. Calculate the Demand factor (DF) of each load and for all the system. 6.) The peak power is a 5pm. calculate the contributions (Ci) factors for each load 7.) Calculate the Diversity Factor (FD) of the system 8.) Deduce the Coincidence Factor (Fc) 9.) Calculate the load diversity of the system Exercise 2 Repeat the same questions for the load below Time street lighting residential commercial 12:00 AM 120 200 350 1:00 AM 120 200 350 2:00 AM 120 200 350 3:00 AM 120 200 350 4:00 AM 120 200 350 5:00 AM 120 200 350 6:00 AM 120 200 350 7:00 AM 0 300 350 8:00 AM 0 500 450 9:00 AM 0 550 1200 10:00 AM 0 550 1200 11:00 AM 0 550 1200 12:00 PM 0 600 1200 1:00 PM 0 600 1200 2:00 PM 0 600 1350 3:00 PM 0 600 1350 4:00 PM 0 600 1350 5:00 PM 0 650 1350 6:00 PM 0 750 1000 7:00 PM 0 900 600 8:00 PM 120 1000 600 9:00 PM 120 1000 600 10:00 PM 120 900 350 11:00 PM 120 700 350 12:00 AM 120 350 350 2 Exercise 3: The feeder in the exercise 1 has a peak loss of 72kW at peak load and annual loss factor of 0.14 1.) Calculate the average power loss of the feeder 2.) The total annual energy loss of the feeder 3 Solutions : Exercise 1 The total can be calculated by adding all the loads Time 12:00 AM 1:00 AM 2:00 AM 3:00 AM 4:00 AM 5:00 AM 6:00 AM 7:00 AM 8:00 AM 9:00 AM 10:00 AM 11:00 AM 12:00 PM 1:00 PM 2:00 PM 3:00 PM 4:00 PM 5:00 PM 6:00 PM 7:00 PM 8:00 PM 9:00 PM 10:00 PM 11:00 PM 12:00 AM street lighting residencial commercial Total 100 250 300 650 100 250 300 650 100 250 300 650 100 250 300 650 100 250 300 650 100 250 300 650 100 250 300 650 0 350 300 650 0 450 400 850 0 550 600 1150 0 550 1100 1650 0 550 1100 1650 0 600 1100 1700 0 600 1100 1700 0 600 1300 1900 0 600 1300 1900 0 600 1300 1900 0 650 1300 1950 0 750 900 1650 0 900 500 1400 100 1100 500 1700 100 1100 500 1700 100 900 300 1300 100 700 300 1100 100 350 300 750 4 1.) Draw the load curve of the three loads and the system one. 5 As it can be seen the pick is at 5pm 2.) Calculate the daily average load of the system a) firs method : From the curve the average load can be given as ave. Load= (650*7+850+1150+1650*2+1700*2+1900*3+1950+1400+1700*2+1300+1100+750)/24 ave. Load=30500/24 ave. Load=1270.83kw b) second method from the table we do not count the first value ( because if we do the total hours will be 25) than we average the rest on 24 that gives the same value (650+650+650+650+650+650+650+850+1150+1650+1650+1700+1700+1900+1900+1900+1950+ 1650+1400+1700+1700+1300+1100+750)/24 ave. Load=1270.83kw 3.) Give the maximum demand of each load (Di) as well as the feeder one (Dg) From the curve or the table the maximum demand of each load (Di) as well as the feeder one (Dg) are given as: π·π π‘ππππ‘ = 100ππ€ π·πππ = 1100ππ€ π·πππ = 1300ππ€ π·π π¦π = 1950ππ€ 4.) Calculate the Load factor (Fld) From equation 6 6 πΉπΏπ· = 1270 = π. ππ 1950 5.) Assume that the total connected load are 150kW, 2500kW and 2500kW for the street lighting, residential and commercial respectively. Calculate the Demand factor (DF) of each load and for all the system. The demand factor is given from equation 1 π·πΉπ π‘ππππ‘ = 100 = π. ππ 150 π·πΉπππ = 1100 = π. ππ 2500 π·πΉπππ = 1300 = π. ππ 2500 π·πΉπ π¦π π‘ππ = 1950 = π. ππ 2500 + 2500 + 150 6.) The peak power is a 5pm. calculate the contributions (Ci) factors for each load The contributions (Ci) factors for each load is given by equation 18.a πΆπ π‘ππππ‘ = πΆπππ = 0 =π 100 650 = π. ππ 1100 πΆπππ = 1300 =π 1300 7.) Calculate the Diversity Factor (FD) of the system 7 a) first way : The Diversity Factor (FD) of the system is given by equation 11 πΉπ· = 100 + 1100 + 1300 = π. ππ 1950 b) second way: The Diversity Factor (FD) of the system is given by equation 11 and equation 18.b πΉπ· = 100 + 1100 + 1300 = π. ππ 0 × 100 + 0.59 × 1100 + 1 × 1300 8.) Deduce the Coincidence Factor (Fc) The Coincidence Factor (Fc) is given by equation 16 πΉπ = 1 = π. ππ 1.28 9.) Calculate the load diversity of the system The load density of the system is given by equation 17 8 πΏπ· = (100 + 1100 + 1300) − 1950 = πππππΎ All the results are given in this table Street Residential commercial Average load KW Total connected load: Given (kW) Individual maximum demand (Di) (kW) Load factor (Fld) Demand factor (DF) Contribution factors Ci at (the pick 5pm) Diversity Factor (FD) Diversity Factor (FD) using Ci Coincidence Factor (Fc) Load diversity LD (kW) 150 100 2500 1100 2500 1300 0,67 0 0,44 0,59 0,52 1 Total 1270,83 5150 1950 0,65 0,38 1,28 1,28 0,78 550 Exercise 2 The results are given by the same equations of exercise 1 Street Residential commercial Average load KW Total connected load: Given (kW) Individual maximum demand (Di) (kW) Load factor (Fld) Demand factor (DF) Contribution factors Ci at (the pick 5pm) Diversity Factor (FD) Diversity Factor (FD) using Ci Coincidence Factor (Fc) Load diversity LD (kW) 150 120 2500 1000 2500 1350 0,80 0,00 0,40 0,65 0,54 1,00 Total 1348,75 5150 2000 0,67 0,39 1,24 1,24 0,81 470 Exercise 3: 9 The feeder in the exercise 1 has a peak loss of 72kW at peak load and annual loss factor of 0.14 1.) Calculate the average power loss of the feeder The average power loss of the feeder is given by equation 25 ππ£πππππ πππ€ππ πππ π = πΉπΏπ × πππ€ππ πππ π ππ‘ ππππ ππππ ππ£πππππ πππ€ππ πππ π = 0.14 × 72 = ππ. πππΎ 2.) The total annual energy loss of the feeder To pass from the power to the energy we multiply by the time annual energy loss = 10.8 × 8760βπ/π¦πππ = πππππππΎπ 10