138 POWER PLANT ENGINEERING

(b) Cost of energy according to flat rate

=

6

100



 × 6 × 10

(^6) = Rs. 360,000. Ans.
Example 8. Two lamps are to be compared:
(a) Cost of first lamp is Re. 1 and it takes 100 watts.
(b) Cost of second lamp is Rs. 4 and it takes 60 watts.
Both lamps are of equal candlepower and each has a useful life of 100 hours. Which lamp will
prove economical if the energy is charged at Rs. 70 per kW of maximum demand per year plus 5 paise
per kWh? At what load factor both the lamps will be equally advantageous?
Solution. (a) First Lamp
Cost of lamp per hour =
(1 100)
1000
×
= 0.1 paise
Maximum demand per hour =
100
1000
= 0.1 kW
Maximum demand charge per hour

0.1 (70 100)
7860
××
= 0.08 paise
Energy consumed per hour = 0.1 × 1 = 0.1 kWh
Energy charge per hour = 0.1 × 5 = 0.5 paise
Total cost per hour = 0.1 + 0.08 + 0.5 = 0.68 paise.
(b) Second Lamp
Cost of lamp per hour =
(4 100)
1000
×
= 0.4 paise
Maximum demand per hour =
60
1000
= 0.06 kW
Maximum demand charge per hour =
0.06 (70 100)
8760
××
= 0.048 paise
Energy consumed per hour = 0.06 × 1 = 0.06 kWh
Energy charge per hour = 0.06 × 5 = 0.3 paise
Total cost per hour = 0.4 + 0.048 + 0.3 = 0.748 paise
Therefore the first lamp is economical
Let x be the load factor at which both lamps become equally advantageous. Only maximum
demand charge changes with load factor.
0.1 +
0.08
x

• 0.5 = 0.4 +
  0.048
  x


• 0.3
  x = 0.32
  or 32%. Ans.