Electrical Energy and Power Plants Notes

We heavily rely on electricity for various essential functions.
Electricity is often taken for granted, yet it's invisible and difficult to define.
In physics, "electricity" refers to both electrical energy and the movement of charge.
Colloquial use of "electricity" often refers to the transfer or generation of electrical energy.
The law of conservation of energy applies to power plants, where different energy types convert to electrical energy.
- Hydroelectric plants: Kinetic energy of falling water → electrical energy.
- Natural gas/nuclear plants: Thermal energy → electrical energy.
Electricity is transferred via conducting wires in transmission lines.
Electrical devices transform electrical energy into other forms (kinetic, sound, thermal) to perform tasks.

Power $(P)$ is the rate at which energy is transformed.
Electrical power is the rate at which electrical energy is produced or consumed.
The terms "power" and "electrical power" are used interchangeably.
Equation: $P = \frac{\Delta E}{\Delta t}$ , where:
- $P$ = Power (in watts, $W$ )
- $\Delta E$ = Change in energy (in joules, $J$ )
- $\Delta t$ = Change in time (in seconds, $s$ )
Electrical devices have varying power ratings.

Problem: Calculate the power required to charge a cellphone if 740 J of energy are transferred in 1.0 min.
Given: $\Delta E = 740 \text{ J}$ , $\Delta t = 1.0 \text{ min}$
Required: $P$
Analysis: $P = \frac{\Delta E}{\Delta t}$
Solution:
- Convert minutes to seconds: $\Delta t = 1.0 \text{ min} \times \frac{60 \text{ s}}{1 \text{ min}} = 60 \text{ s}$
- $P = \frac{740 \text{ J}}{60 \text{ s}} = 12.33 \text{ W} \approx 12 \text{ W}$
Statement: The power required to charge the cellphone is approximately 12 W.

Different devices have different power needs (Table 1).
Power plants monitor and generate power "on demand."
Increased power demand (e.g., hot summer days due to air conditioning) requires increased power generation.
If generation capacity is reached, power may be purchased from external sources.
As a last resort, "brownouts" (temporary power interruptions) may occur.
Brownouts can damage electrical devices due to rapid power changes.
Electrical energy is generated on demand because storing it in large quantities is impractical.
Batteries are not feasible for large-scale storage due to the massive quantities needed.

Electrical energy is measured in kilowatt hours (kWh) because the joule (J) is too small for convenient measurement.
1 kWh = 3.6 million joules.
A typical Ontario home uses approximately 1000 kWh of electrical energy per month.
Power plants use megawatt hours (MWh) to describe electrical energy generation.
In 2007, Ontario generated over 158 million MWh of electrical energy (approximately 12,100 kWh per person per year).
Canadians generally use more electrical energy compared to some other countries (e.g., Chad: 9 kWh per person per year).

Problem: Calculate the energy needed by a 35 W halogen light bulb operating for 240 h; express the answer in both joules and kilowatt hours.
Given: $P = 35 \text{ W}$ , $\Delta t = 240 \text{ h}$
Required: $\Delta E$
Analysis: $P = \frac{\Delta E}{\Delta t} \implies \Delta E = P \Delta t$
Solution:
- Convert hours to seconds: $\Delta t = 240 \text{ h} \times \frac{3600 \text{ s}}{1 \text{ h}} = 864,000 \text{ s}$
- $\Delta E = (35 \text{ W}) \times (864,000 \text{ s}) = 3.024 \times 10^7 \text{ J}$
- Convert joules to kilowatt hours: $3.024 \times 10^7 \text{ J} \times \frac{1 \text{ kWh}}{3.6 \times 10^6 \text{ J}} = 8.4 \text{ kWh}$
Alternative solution:
- Convert watts to kilowatts: $35 \text{ W} \times \frac{1 \text{ kW}}{1000 \text{ W}} = 0.035 \text{ kW}$
- $\Delta E = (0.035 \text{ kW}) \times (240 \text{ h}) = 8.4 \text{ kWh}$
Statement: The halogen light needs $3.0 \times 10^7 \text{ J}$ or 8.4 kWh of energy to operate for 240 h.

Non-renewable energy sources are depleting, and current renewable sources are insufficient.
Two approaches to meet electrical energy demand:
- Conservation: Using less electrical energy.
- Efficiency: Generating energy more efficiently.
Newer devices are designed to consume less energy, and there's growing awareness about energy use.
Power plants transform source energy into electrical energy, but the process isn't 100% efficient.
Efficiency is a measure of how well energy is transformed.
Engineers design ways to minimize energy losses during transformation.
Example: Improving coal-fired plant efficiency by increasing steam pressure and temperature.
Modern coal-fired plants can reach steam temperatures of 600°C and pressures 250 times atmospheric pressure.

Table 2 shows the efficiency of different power plant technologies.
Hydro: 85% (e.g., Sir Adam Beck II: 1499 MW) - Energy lost as water isn't fully stopped.
Fossil Fuel: 45% (e.g., Nanticoke: 3640 MW) - Thermal energy not fully captured.
Wind: 40% (e.g., Huron Wind Farm: 9 MW) - Energy lost as not all wind is captured.
Nuclear: 35% (e.g., Darlington: 3500 MW) - Thermal energy not fully captured.
Solar: 15% (e.g., Sarnia Solar Farm: 20 MW) - Sunlight converted mostly to thermal energy instead of electrical.

Evaluating energy sources and power plant technologies requires considering environmental and societal aspects.
Environmental aspects: Source accessibility, dependability, plant location, atmospheric emissions, and pollution.
Societal impacts: Financial costs, employment opportunities, safety, and effects on nearby communities.
Efficiency impacts resource use and operational costs of power plants.
Operational efficiency: Describes the amount of time something is working compared to downtime.
Thermal efficiency: Describes how much thermal energy is used to perform a task versus wasted thermal energy.

Mechanical, thermal, and radiant energy are transformed into electrical energy in power plants.
Electrical power is the rate at which electrical energy is generated or transformed.
Electrical energy is measured in kilowatt hours (kWh) for homes and megawatt hours (MWh) for power plants.
Power plant technologies vary in efficiency and environmental/societal impacts.
Improving power plant efficiency can decrease resource use.