1.
Compare the annual cost of supplying a factory load having a maximum
demand of 1 MW and a load factor of 50% by energy obtained from.
a. A private oil engine generating plant and
b. Public supply.
(a). Private oil engine generating unit:
Capital Cost = Php 1,200,000
Cost of repair and maintenance = Php 0.005 per kWHr generated
Cost of Fuel = Php 1600 per 1000 kg
Interest and Depreciation = 10% per annum
Fuel Consumption = 0.3 kg/kWHr generated
Wages = Php 50,000 per annum
(b): Public Supply Company:
Php 150 per kW of maximum demand plus Php 0.15 per kWh
Solution:
Total Energy Produced ( TEP )=1000 kW∗0.5∗8760 Hrs=4.38 x 106 kWHr
For Private Oil Engine Generating Unit:
kg 6 6
Annual Fuel Consumption=0.3 ∗4.38 x 10 kWHr=1.314 x 10 kg
kWHr
1600 6
Cost of Fuel=Php ∗1.314 x 10 kg=Php 2,102,400.00
1000 kg
0.005 6
Annual Cost of Repair∧Maintenance=Php ∗4.38 x 10 kWHr=P hp 21,900.00
kWHr
50,000
Anual Wages=Php ∗1 year =Php50,000 .00
annum
1,200,000∗10 %
Anual Interest ∧Depreciation=Ph p =Php120,000 .00
annum
Total Anual Charge=Php 2,102,400+ Php 21,900+ Php 50,000+ Php 120,000
Total Anual Charge =Php 2,294,300.00
For Public Supply
150
Anual ¿ Charge=Php ∗1000 kW =Php 150,000.00
kW
0.15 6
Anual Running Charge=P hp ∗4.38 x 10 kWHr=Php 657,000 .00
kWHr
Total Anual C harge=Php 150,000+ Php 657,000=Php 807,000 .00
Conclusion:
The comparison between the Private Oil Engine generating unit and the
Public Supply reveals a significant cost disparity. The total annual charge for the
Private Oil Engine, at Php 2,294,300, is approximately 2.84 times higher than the
cost of Public Supply, which totals Php 807,000. This indicates that relying on
Public Supply is considerably more cost-effective for annual energy needs.
However, factors such as reliability, availability, and operational autonomy
should also be considered when evaluating the overall suitability of each option.
2. A hydro-electric plant costs Php 3000 per kW of installed capacity. The
total annual charges consist of 5% as interest; depreciation at 2%,
operation and maintenance at 2% and insurance, rent etc. 1.5%.
Determine a suitable two-part tariff if the losses in transmission and
distribution are 12.5% and diversity of load is 1.25. Assume that maximum
demand on the station is 80% of the capacity and annual load factor is
40%. What is the overall cost of generation per kWHr?
Solution:
Assume that the Installed capacity of the Station to be 100 kW
Maximum Demand=100 kW ∗0.8=80 kW
Average Demand=80 kW ∗0. 4=32 kW
3000
Capital cost of plant=Php ∗100 kW =Php 300,000.00
kW
Annual Fixed Charges
300,000∗(5+ 2 ) %
Annual ¿ Charges=P hp =Php 21,000.00
100 %
Aggregate of max demand=8 0 kW ∗1.25=100 kW
21,000
Annual ¿ Charges=Php =Php 210 per kW of maximum demand
100 kW
Annual Running Charges
300,000∗( 2+1.5 ) %
Annual Running Charges=Php =Php10,500.00
100 %
Units generated per annum=32 kW ∗8760 Hrs=280,320 kWHr
Units reaching the consumer=280 ,320 kWHr∗( 100−12.5 ) %=245,280 kWHr
10,500
Annual Running Charges=Php =Php0.0428 per kWHr
245,820 kWHr
Generation cost in two-part form is = Php (210 x kW + 0.0428 x kWHr)
Total Annual Charges=Php21,000+ Php 10,500=Php3 1,500.00
3 1 ,5 0 0
Cost per kWh=Php =Php 0.1281 per kWHr
245,820 kWHr
3. The annual working cost of a power station is represented by the formula (
a +b kW +c kWHr) where the various terms have their usual meaning.
Determine the values of a, b and c for a 60 MW station operating at annual
load factor of 50% from the following data:
a. Capital cost of building and equipment is Php 5,000,000
b. The annual cost of fuel, oil, taxation and wages of operating staff is
Php 900,000
c. The interest and depreciation on building and equipment are 10%
per annum
d. Annual cost of organization and interest on cost of site etc. is Php
500,000.
Solution:
Unit s generated per annum=Max . demand∗Load Factor∗Hours∈a year
3 6
Units generated per annum=60 x 10 kW ∗0.5∗8760 Hr=262.8 x 10 kWHr
Annual operating cost=Php(a+b∗kW + c∗kWHr )
a=annual ¿ cost
b∗kW =annual semi−¿ cost
c∗kWHr=annualrunning cost
Annual fixed cost (this is due to annual cost of organization and interest on
cost of site etc.)
a=Php500,000
Annual semi-fixed cost (this is due to the interest and depreciation on
building and equipment)
5,000,000∗10 %
Annual semi−¿ cost=Php =Php 500,000.00
100 %
3
b∗60 x 10 kW =Php 500,000
500,000
b=Php 3
=Php 8.3333 per kW
60 x 10 kW
Annual running cost (this is due to the annual cost of fuel, oil, taxation and
wages of operating staff)
C∗kW Hr generated=Php 900,000
6
C∗262.8 x 10 kWHr=Php 900,000
900,000
C=Php 6
=Php 0.0034 per kWHr
262.8 x 10 kWHr
