## Glossary of terms |

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**Watt**- Watt (W) is the SI (International system of units) derived unit of power measuring the rate of energy consumption or production.
- Usually derived units:
- kW - kilowatt (thousand Watts)
- MW - megawatt (million Watts)
- GW - gigawatt (billion Watts)
- TW - terawatt (trillion Watts)

- Non-SI units adopted by the power industry:
- Electrical power: kWe, MWe, etc.; the subscript
*e*, for*electrical*, indicates the electrical power generation or load consumption. - Thermal power: kWt, MWt, etc.; the subscript
*t*(or th), for*thermal*, designates the thermal power output of a power plant. The heat produces steam that is used to drive the turbines that generate electricity. As a rule of thumb, thermal power is three times the amount of electrical power produced.

- Electrical power: kWe, MWe, etc.; the subscript
- Watt and Watt-hours
- Watts are the yardstick to measure power. Power plants are rated in Watts. A 1,000 MW power station is a plant with the capability of producing 1,000 MW. However, sales of energy are measured in MHh ( megawatt-hour, or alternatively kWh, etc. )
- Watt-hours designate the energy consumed over the period of 1 hour. For instance a bulb may consume 50 Watts of power when turned on. If it stays turned on for 2 hours, it will use 2 × 50 = 100 Watts-hour of energy.

**Conversion of Watts into Watts-hour***Example 1*: How many homes will a power station with a rated capability of 1,000 MW be able to supply, assuming that the average home consumes 600 kWh monthly ?- Since the normal budgeting cycle is 1 year, the calculations will be made on a yearly basis.
- Compute the effective power generation by taking into account the power plant efficiency. Indeed, power plants have stoppages for maintenance, inspection, upgrading, refuelling, etc. during which no power is produced. Efficiency is given by the "capacity factor", or the ratio of actual energy produced over a given period, to the rated capability of the plant. Assume a high 80% capacity factor.
- Multiply the actual capacity by the number of hours in one year.
- Multiply by 1,000 to have the result in kWh (1 MW = 1,000 kW).
- Lastly, divide by the average yearly kWh consumption (monthly 600 kWh × 12).

MW 1,000 × 0.80 capacity factor × 365 days × 24 hours × 1,000 kWh / ( 600 kWh × 12 months ) = 973,333 homes

- The station is capable of meeting the average yearly electricity demand of 973 thousand homes or roughly a city of 2.3 million inhabitants.
*Example 2*: You consider installing in your home a heating system made of storage heaters whose specs indicate the power of 2.55 kW. How much will cost you each device per month, assuming that it will be in operation on average 8 hours/day, and knowing that your utility company sells the kWh at €0.09 ?kW 2.55 × 8 hours × 30 days = kWh 612 monthly

kWh 612 × € 0.09 Euro = € 55.08 per month.

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*Ooops! electric power is becoming unaffordable...*)