Electrical Energy Analysis and Project Economics Calculations

This assignment solution provides detailed answers to a series of energy-related problems. It begins with unit conversions, specifically from gigajoules to kilowatt-hours, and proceeds to calculate energy consumption by a motor, considering its operating hours and efficiency. The solution then analyzes a power consumption graph to determine daily energy consumption and peak demand. Further calculations involve the energy supplied by a gas plant, the total cost of a plant's operation over 20 years, and the average cost per kWh. The assignment also explores the energy generated by solar panels, calculating the number of panels needed, the total installation cost, and the payback period. Finally, the solution compares the cost-effectiveness of gas and solar energy generation, highlighting the economic and environmental benefits of solar power.

QUESTION ONE
Conversion from gigajoules (GJ) to kilowatt hour kWh
It’s done from the relation
1 GJ= 277.78 kWh, so for 500kWh its calculated below
? =500 kWh
(500kWh*1 GJ)/277.78 kWh= 1.8 GJ
QUESTION TWO
The amount of time in hours that the motor runs per day from 8a.m to 8p.m is,
2000hrs-800hrs=12
The amount of time in hours that the motor runs per year is evaluated by
multiplying the number of hours the motor runs per day by the days of the year
that is, 12*365= 4380
Energy consumed by the motor annually is given by
Energy=power consumed by motor* time the motor runs in a year
=12kW*4380h
=52560kWh
QUESTION THREE
Motor consumes electrical and produce a corresponding mechanical power.
However not all power consumed is found at the output due to losses that is why
the motors are not 100% efficient
Mechanical output power in kW= % efficiency*electrical output power in kW
= 0.75*12 kW

=9 kW
QUESTION FOUR
a) Energy consumed daily
Energy=power*time
The energy of the entire day is calculated by multiplying power (represented by
vertical axis) by time in hourly basis as shown on x-axis.
Energy consumed daily=
(1*3)+(2*1)+(3*1)+(5*1)+(3*1)+(2*2)+(4*1)+(2*1)+(3*4)+(2*2) Kw
Where first digit in every term is power and the second time
Energy consumed daily is thus 42 Kw
b) Peak demand
Peak demand in electrical power systems is the highest power demand in a
specified period of time. From the curve 5kW from the 6th hour of the day is the
peak demand
QUESTION FIVE
Energy supplied gas plant to the grid per year=maximum power*loading*hours in
a year
Where maximum power the plant can supply= 50*103 Kw, loading = 0.75 and
hours in a year is 24*365 h
= 50*103*0.75*24*365 kWh
=3285*105 kWh

QUESTION SIX
Plant’s cost of building is 175M dollars and cost of operation annually is 20M
dollars.
Cost of operation in 20 years= 20M dollars*20 years=400M dollars
Total cost of operation in 20 years= cost of building + operation costs for 20 years
= 175 + 400 M dollars
= 575 M dollars
Average cost for the 20 years per kWh= total cost of operation divided by energy
produced in the 20-year period
Where total cost is 575*106 and energy produced is 328.5*106 kWh
Average cost of production is thus =575*106 dollars/328.5*106kWh
=1.75 dollars/kWh
QUESTION SEVEN
Energy consumed by the company annually =2000 MWh
Energy to be supplied by solar annually is 50 % of the energy consumed by the
company and is computed as = 0.5*2000MWh
=1000 MWh
The 1000 MWh is the energy that is to generated by many solar panels, an
engineer has however given us the value of energy to be generated by one solar
panel to be 12000kWh
Minimum number of panels that can produce 1000MWh energy is calculated by
dividing the total energy to be produced by all solar panels divided by energy
produced by one solar panel

No of panels=1000*103/12000= 84
Therefore, the minimum number of solar panels are 84 solar panel
QUESTION EIGHT
Cost of installing one solar panel is 25,000 dollars
Total cost of installing 84 panels is found by multiplying the cost of one panel by
the number of panels installed
Total cost of installing all panels is 25,000*84= 2.1M dollars
Had the solar panels not have been installed, energy of (1 MWh) is to purchased
at a cost of 0.14 dollar per kWh
Cost per year that could have been incurred if energy supplied by solar was
purchased is found by multiplying the energy by the cost per kWh
=0.14 dollar/kWh*1*106 kWh
=0.14 M dollars.
Investment on installing panels will be paid back after a period of 2.1M/0.14M=15
years. After which cost of purchasing would be saved
QUESTION NINE
Energy consumed annually in the company is 2000MWh
Energy produced by the solar alone annually 1000MWh
Total cost of producing energy by solar in 20 years is 2.1 M dollars (from question
8)
Average cost of solar energy production for 20 years per kWh= total cost of solar
production divided by the total energy generated by the solar

Total cost of solar production is 2.1 M dollars and the total energy generated by
the solar is 20*1*106
Total cost is thus, 2.1 * 106 dollars/20*1*106 kWh
=0.105 dollars/kWh
QUESTION TEN
The cost per kWh for gas generation system is 1.75 dollars/kWh while that of
solar system is 0.105 dollars/kWh. Solar system is proved cheap. By using solar
system, the company cuts the cost of production increasing returns. Solar system
utilizes natural resources and is safe to the environment

REFERENCES
Capehart, B. L., Turner, W. C., & Kennedy, W. J. (2016). Guide to energy
management (8th ed.). International version. Lulburn, GA: The Fairmont Press,
Inc.
Smil, V. (2016). Energy: Beginner’s guide (2nd ed.). London, United Kingdom:
Oneworld Publications.
Georgescu-Roegen, N. (2018). Energy analysis and economic valuation. In Green
Accounting (pp. 75-110). Routledge.
Gomez-Exposito, A., Conejo, A. J., & Canizares, C. (2018). Electric energy systems:
analysis and operation. CRC press.