Question

Homework 1 1. Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document. a. Calculate the total energy supplied by SPP on this date. b. Calculate the peak load for SPP on this date.

Answer 1

SOLUTION:

• The total Energy Supplied by a POWER STATION on a particular day /date is the Energy  supplied in 24 hour of that particular date/day.

And we know that

  

Energy = avg. (Power) Time Duration

SO Energy Supplied in one hour = avg. Hourly POWER   X 1 kWH or MWH

Because the Time duration = 1 hour ( as it is clear from the word' hourly '.)

if POWER is in kW then Energy in kWH ( kilo-watt-hour)

if Power is in MW then energy in MWH (Mega-Watt-hour)
To convert this in joule ,convert time into seconds and MW , kW in Watt.

Energy supplied in 24 hour of a day = Sum of Energy supplied in all the 24 hours

(a) From the given data

Total Energy Supplied by SPP on 8th April   $\huge =$    26739 X 1 + 27247 X 1 + 28610 X 1+  28654 X 1 + 28039X1 + 27333 X 1 + 26849 X1

+26637 X 1 + 26505 X 1+ 26512 X1

  + 26652 X1 + 27157 X 1 + 27950 X 1 + 28499 X 1

+ 29186 X 1 + 29395 X 1 + 29096 X 1 + 28768 X1 + 28376 X 1 + 28005 X 1 + 27617 X 1

+ 27528 X 1 + 27642 X1 + 27831 X 1

  
Energy Suplied = 666827 x 106......Watt – hour

To convert this in joule multiply it with 1 hour =3600 seconds

Energy Suplied = 666827 x 106 x 3600...... Joule

Total Energy Suplied = 2.4 x 10°.....MJ
ANSWER (a)

(b) PEAK Load is the maximum load that exists on the system in a particular hour.

From the Given data the Maximum load that exists = 29395 MW

but from the graph we can see that PEAK load =29400 MW

ANSWER (b)

Peak Load = 29400MW.............. ... Seethegraph)