Problem Set 1
1. A 100 MW powers stations delivers 100 MW for 2 hours, 50 MW for 8 hours and is
shut down for the rest of each day. It is also shut down for maintenance for 60 days
each year. Calculate its annual load factor.
Given:
π
ππ‘ππ πΆππππππ‘π¦ = 100 ππ
πΏ1 = 100 ππ
β1 = 2 βππ’ππ
πΏ2 = 50 ππ
β2 = 8 βππ’ππ
β′′ = 60
πππ¦π ⁄
π¦πππ
ππππππ = ππππ’ππ
Required:
πΏπΉππππ’ππ
Solution:
πΈπππππ¦ = πΏ1 β1 + πΏ2 β2
πΈπππππ¦ = (100 ππ)(2 βππ’ππ ) + (50 ππ)(8 βππ’ππ )
πΈπππππ¦ = 600 ππβ
πΈππππ’ππ = (πΏππ£π. πππππ¦ )(β′)
πΏππ£π. πππππ¦ =
600 ππβ
= 25 ππ
24 βππ’ππ
24 βππ’ππ
β′ = 365 πππ¦π − 60 πππ¦π = 305 πππ¦π (
) = 7320 βππ’ππ
1 πππ¦
πΈππππ’ππ = (25 ππ)(7320 βππ’ππ ) = 183,000 ππβ
πΏπΉππππ’ππ =
πΏππ£π. ππππ’ππ
πΏπππ₯
πΏππ£π. ππππ’ππ =
πΈππππ’ππ
β
24 βππ’ππ
β = 365 πππ¦π (
) = 8760 βππ’ππ
1 πππ¦
183,000 ππβ
πΏπΉππππ’ππ = 8760 βππ’ππ
100 ππ
πΏπΉππππ’ππ = 0.2089
2. A power station is to supply four regions of loads whose peak values are 10,000
kW, 5000 kW, 8000 kW and 7000 kW. The diversity factor of the load at the station
is 1.5 and the average annual load factor is 60%. Calculate the maximum demand
on the station and annual energy supplied from the station.
Given:
πΏ1 = 10000 ππ
πΏ2 = 5000 ππ
πΏ3 = 8000 ππ
πΏ4 = 7000 ππ
π·ππ£πππ ππ‘π¦ πΉπππ‘ππ = 1.5
πΏπΉππππ’ππ = 0.6
Required:
πΏπππ₯ πππ πΈππππ’ππ
Solution:
π·ππ£πππ ππ‘π¦ πΉπππ‘ππ =
1.5 =
πππππ£πππ’ππ ππππππ
π πππ’ππ‘πππππ’π ππππππ
(10000 + 5000 + 8000 + 7000) ππ
πΏπππ₯
πΏπππ₯ = 20,000 ππ
πΏπΉππππ’ππ =
0.6 =
πΏππ£π. ππππ’ππ
πΏπππ₯
πΏππ£π. ππππ’ππ
20,000 ππ
πΏππ£π. ππππ’ππ = 12,000 ππ
πΈππππ’ππ = (πΏππ£π. ππππ’ππ )(β)
πΈππππ’ππ = (12,000 ππ)(8760 βππ’ππ )
πΈππππ’ππ = 105,120,000 ππβ
3. A generating station supplies the following loads: 15000 kW, 12000 kW, 8500 kW,
6000 kW and 450 kW. The station has a maximum demand of 22000 kW. The
annual load factor of the station is 48%. Calculate (i) the number of units supplied
annually (ii) the diversity factor and (iii) the demand factor.
Given:
πΏ1 = 15,000 ππ
πΏ2 = 12,000 ππ
πΏ3 = 8,500 ππ
πΏ4 = 6,000 ππ
πΏ5 = 450 ππ
πΏ πππ₯ = 22,000 ππ
πΏπΉππππ’ππ = 0.48
Required:
πΈππππ’ππ
π·ππ£πππ ππ‘π¦ πΉπππ‘ππ
π·πππππ πΉπππ‘ππ
Solution:
0.48 =
πΏππ£πππππ
22,000 ππ
πΏππ£πππππ = 10,560 ππ
πΈππππ’ππ = (πΏππ£π. ππππ’ππ )(β)
πΈππππ’ππ = (10,560 ππ)(8,760 βππ’ππ )
πΈππππ’ππ = 92,505,600 ππβ
π·ππ£πππ ππ‘π¦ πΉπππ‘ππ =
π·ππ£πππ ππ‘π¦ πΉπππ‘ππ =
π π’π ππ πππππ£πππ’ππ ππππππ
π πππ’ππ‘πππππ’π ππππππ
(15,000 + 12,000 + 8,500 + 6,000 + 450) ππ
22,000 ππ
π·ππ£πππ ππ‘π¦ πΉπππ‘ππ = 1.906
π·πππππ πΉπππ‘ππ =
1
1
=
π·ππ£πππ ππ‘π¦ πΉπππ‘ππ 1.906
π·πππππ πΉπππ‘ππ = 0.52443
4. A substation supplies power by four feeders to its consumers. Feeder no. 1
supplies six consumers whose individual daily maximum demands are 70 kW, 90
kW, 20 kW, 50 kW, 10 kW and 20 kW while the maximum demand on the feeder is
200 kW. Feeder no. 2 supplies four consumers whose daily maximum demands
are 60 kW, 40 kW, 70 kW and 30 kW, while the maximum demand on the feeder is
160 kW. Feeder nos. 3 and 4 have a daily maximum demand of 150 kW and 200
kW respectively while the maximum demand on the station is 600 kW. Determine
the diversity factors for feeder no. 1. feeder no. 2 and for the four feeders.
Given:
πΉπππππ 1
πΏπΉ1.1 = 70 ππ
πΏπΉ1.2 = 90 ππ
πΏπΉ1.3 = 20 ππ
πΏπΉ1.4 = 50 ππ
πΏπΉ1.5 = 10 ππ
πΏπΉ1.6 = 20 ππ
πΏmax 1 = 200 ππ
πΉπππππ 2
πΏπΉ2.1 = 60 ππ
πΏπΉ2.2 = 40 ππ
πΏπΉ2.3 = 70 ππ
πΏπΉ2.4 = 30 ππ
πΏmax 2 = 160 ππ
πΏ3 = 150 ππ
πΏ4 = 200 ππ
πΏπππ₯ = 600 ππ
Required:
π·ππ£πππ ππ‘π¦ πΉπππ‘πππ
Solution:
π·ππ£πππ ππ‘π¦ πΉπππ‘ππ =
π π’π ππ πππππ£πππ’ππ ππππππ
π πππ’ππ‘πππππ’π ππππππ
πΉπππππ 1 π·ππ£πππ ππ‘π¦ πΉπππ‘ππ =
(70 + 90 + 20 + 50 + 10 + 20) ππ
200 ππ
πΉπππππ 1 π·ππ£πππ ππ‘π¦ πΉπππ‘ππ = 1.3
πΉπππππ 2 π·ππ£πππ ππ‘π¦ πΉπππ‘ππ =
(60 + 40 + 70 + 30) ππ
160 ππ
πΉπππππ 2 π·ππ£πππ ππ‘π¦ πΉπππ‘ππ = 1.25
πππ‘ππ π·ππ£πππ ππ‘π¦ πΉπππ‘ππ =
(160 + 200 + 150 + 200) ππ
600 ππ
πππ‘ππ π·ππ£πππ ππ‘π¦ πΉπππ‘ππ = 1.1833
5. The yearly load duration curve of a certain power station can be approximated as
a straight line; the maximum and minimum loads being 80 MW and 40 MW
respectively. To meet this load, three turbine generator units, two rated at 20 MW
each and one at 10 MW are installed. Determine (i) installed capacity (ii) plant
factor (iii) kWh output per year (iv) load factor.
Given:
πΏπππ₯ = 80 ππ
πΏπππ = 40 ππ
π
πΆ1+2 = 40 ππ
π
πΆ3 = 10 ππ
Required:
π
ππ‘ππ πΆππππππ‘π¦
πΆππππππ‘π¦ πΉπππ‘ππ
πΈππππ’ππ ππ ππβ
πΏπππ πΉπππ‘ππ
Solution:
π
ππ‘ππ πΆππππππ‘π¦ = π
πΆ1+2 + π
πΆ3
π
ππ‘ππ πΆππππππ‘π¦ = 40 ππ + 10 ππ
π
ππ‘ππ πΆππππππ‘π¦ = 50 ππ
πΈππππ’ππ = (πΏππ£π. ππππ’ππ )(β)
πΈππππ’ππ = (
πΈππππ’ππ = (
πΏπππ₯ + πΏπππ
) (β)
2
80,000 ππ + 40,000 ππ
) (8760 βππ’ππ )
2
πΈππππ’ππ = 525,600,000 ππβ
πΆππππππ‘π¦ πΉπππ‘ππ =
πΏππ£πππππ
πππ‘ππ πππππππ‘π¦
80 ππ + 40 ππ
)
2
πΆππππππ‘π¦ πΉπππ‘ππ =
50 ππ
(
πΆππππππ‘π¦ πΉπππ‘ππ = 1.2
πΏπππ πΉπππ‘ππ =
πΏππ£πππππ
πΏπππ₯
80 ππ + 40 ππ
)
2
πΏπππ πΉπππ‘ππ =
80 ππ
(
πΏπππ πΉπππ‘ππ = 0.75
Drawings:
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