Power Systems Test 2

In pumped storage plant, synchronous generator is used as synchronous motor to pump the water back to the reservoir during off-peak period as pumped storage plant is mainly used for peak loads.

Baseload power station: Baseload is the minimum level of electricity demand required over a period of 24 hours. It is needed to provide power to components that keep running at all times (also referred to as continuous load). Plants that are running continuously over extended periods of time are said to be baseload power plants.

Example:

• Nuclear power plant
• Coal or thermal power plant
• Hydroelectric plant
• Geothermal plant
• Biogas plant
• Biomass plant
• Solar thermal with storage
• Ocean thermal energy conversion

Peak load power station: Peak load is the time of high demand. These peaking demands are often for only shorter durations. To cater to the demand peaks, peak load power plants are used. They are started up whenever there is a spike in demand and stopped when the demand recedes.

Example:

• Gas plant
• Solar power plants
• Wind turbines
• Pumped storage plants
• Diesel generators

Superheaters: It is used to dry the wet steam and raise the temperature of the steam above its saturation temperature. It is generally placed in the path of the furnace gases so as to utilize the heat from the hot flue gases.

Economizer: It is also known as a feedwater heater. It is a device in which the waste heat of the flue gases is utilized for heating the feed water.

Air Preheater: It is used to increase the temperature of the air before it enters the furnace. It is generally placed after the economizer; so that the flue gases pass through the economizer and then to the air preheater.

Overall efficiency of the station = 0.33 × 0.9 = 0.297

Coad used per annum \(= \frac{{30\; \times \;{{10}^5}}}{{300}} = {10^4}\;tons = {10^7}kg\)

Heat of combustion = Coal used/annum × calorific value

= 10⁷ × 5000 = 5 × 10¹⁰ kcal

Heat output = overall efficiency × heat of combustion

= 0.297 × 5 × 10¹⁰ = 1485 × 10⁷ kcal

Units generated/annum \(\frac{{1485\; \times \;{{10}^7}}}{{860}} = 17.26 \times {10^6}kWh\)

Average load on station = (units generated per annum)/ hours in a year

\(= \frac{{17.26\; \times \;{{10}^6}}}{{8760}} = 1971.1\;kW\)

Weight of water available/sec W = Q × 9.81 × 1000

= 9810 Q N

Average power produced = W × H × η

= 9810 × 20 × 0.8 × Q W

= 156.96 Q kW

Power developed in a day = 156.96 × 20 = 3139.2 kW

Units generated in a day = 3139.2 × 24 = 75,341 kWh

Load factor \(= \frac{{75341\; \times\; {{10}^3}}}{{10\; \times \;{{10}^6} \times 24}} \times 100 = 31.4\% \)

Load factor = (Average demand)/maximum demand

⇒ Average demand = 25 × 0.6 = 15 MW

Daily energy produced = Average demand × 24

= 15 × 24 = 360 MWh

Maximum energy that could be produced

= (Actual energy produced in a day)/(Plant use factor)

Energy generated/annum = Maximum demand × load factor × hours in a year

= 15000 × 0.5 × 8760 = 65.7 × 10⁶ kWh

Plant capacity factor = (units generated/annum)/(Plant capacity × hours in a year)

Plant capacity \(= \frac{{65.7\; \times \;{{10}^6}}}{{0.4\; \times \;8760}} = 18,750\;kW\)

Reserve capacity = plant capacity – maximum demand

= 18750 – 15000 = 3750 kW

The load factor percentage is derived by dividing the total kilowatt-hours (kWh) consumed in a designated period by the product of the maximum demand in kilowatts (kW) and the number of hours in the period

Maximum demand = 80 MW

Connected load = 150 MW

Units generated in one year = 400 × 10³ MW hr

Total number of hours in a year = 8760

Average load = 400 × 10³/8760 = 45.662 MW

\(Load\;Factor = \frac{{Average\;load}}{{Maximum\;demand}} = \frac{{45.662}}{{80}} = 0.57\)