Power Plant Engineering

Ans.
Example 7. Any undertaking consumes 6 × 10
6
 kWh per year and its maximum demand is 2000
kW. It is offered two tariffs.
(a) Rs. 80 per kW of maximum demand plus 3 paise per kWh.
(b) A flat rate of 6 paise per kWh.
Calculate the annual cost of energy.
Solution.
(a) (According to first tariff the cost of energy)
= 2000 × 80 + 
3
100






× 6 × 10
6
= 160,000 + 180,000 = Rs. 340,000.
Ans.

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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