1. The daily load curve of a power system is given by the...

1. The daily load curve of a power system is given by the following data. Time..................12 3 6 8 10 11 12 1 3 4 5 6 7 9 12 Load, kw x $10^{-3}$......95 60 60 150 175 180 175 150 200 250 310 170 140 160 105 Plot the load curve to scales of 1cm=2 hr, 1cm= 50,000 kw. Determine the load factor. What is the daily load factor of the standby equipment that takes all peak load over 200,000 kw?

Transcript

00:01 Hello students in this question.

00:02 We are given the data of time and the load of consumer 1 and consumer 2 to given information consumer 1 is connected to 65 kilowatt and consumer 2 is connected to 75 kilowatt.

00:20 The transformer capacity is equal to 160.

00:24 Kilo volt ampere power factor is equal to 0 .90.

00:30 So first we calculate the demand factor.

00:33 So demand factor is the ratio between the maximum demand of the system to the sum of individual maximum demands.

00:39 So for consumer 1 maximum demand will be equal to 30 kilowatt, right? that is the maximum time slot from 14 to 20 hour time slot for consumer 2 maximum demand is at 40 kilowatt from 18 to 20 hours load.

00:57 So demand factor will be equal to 30 plus 40 that is equal to 70 by 140.

01:03 So that is equal to 0 .5.

01:07 That is a demand factor.

01:08 Now b part.

01:10 This is daily energy daily energy is the total energy consumed by the combined consumers in a day.

01:17 So for consumer 1 energy consumed is equal to 10 times 8 plus 12 times 2 plus 20 times 4 plus 30 times 6 plus 10 times 4.

01:33 So energy will be equal to the sum of all these things.

01:37 So that will give you 404 kilowatt hour now for consumer number 2.

01:44 So consumer number 2 energy consumed is equal to 15 times 5 plus 30 times 3 plus 20 times 10 plus 40 times 2 plus 10 times 4.

02:00 So the total energy consumed for consumer 1 this is for consumer 2.

02:05 It will be equal to 8485 kilowatt hour.

02:11 The total daily energy is equal to 444 not 4 plus 485 that is 889 kilowatt hour.

02:24 Now the c part the average load average load is the average power consumed by the combined consumers over a day.

02:32 So average load is equal to total daily energy divided by 24 that will give you 889 divided by 24 that will be equal to 37 .04 kilowatt...

Question

Please give Ace some feedback