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kilowatt-hour per 100 kilometres: kilowatt-hours per 100 kilometres: kilowatt-hour per 100 kilometers: kilowatt-hours per 100 kilometers: MJ/km kWh/mi: Kilowatt-hour: kWh/100 mi: kW⋅h/100 mi: 22.3694: kilowatt-hour per 100 miles: kilowatt-hours per 100 miles: mpge: Miles per gallon gasoline equivalent: MJ/100 km: MJ/100 km: 10: megajoule per ...
Efficiency of power plants, world total, 2008. Energy conversion efficiency (η) is the ratio between the useful output of an energy conversion machine and the input, in energy terms. The input, as well as the useful output may be chemical, electric power, mechanical work, light (radiation), or heat. The resulting value, η (eta), ranges ...
This thermal energy input of 1 kWh = 3.6 MJ = 3,412 Btu; Therefore, the heat rate of a 100% efficient plant is simply 1, or 1 kWh/kWh, or 3.6 MJ/kWh, or 3,412 Btu/kWh; To express the efficiency of a generator or power plant as a percentage, invert the value if dimensionless notation or same unit are used. For example: A heat rate value of 5 ...
A kilowatt-hour (unit symbol: kW⋅h or kW h; commonly written as kWh) is a non-SI unit of energy equal to 3.6 megajoules (MJ) in SI units, which is the energy delivered by one kilowatt of power for one hour. Kilowatt-hours are a common billing unit for electrical energy supplied by electric utilities.
The following unit conversions may be helpful when considering the data in the tables: 3.6 MJ = 1 kW⋅h ≈ 1.34 hp⋅h. Since 1 J = 10 −6 MJ and 1 m 3 = 10 3 L, divide joule/m 3 by 10 9 to get MJ/L = GJ/m 3. Divide MJ/L by 3.6 to get kW⋅h/L.
The red curve shows the power in the load, normalized relative to its maximum possible. The dark blue curve shows the efficiency η. The efficiency η is the ratio of the power dissipated by the load resistance R L to the total power dissipated by the circuit (which includes the voltage source's resistance of R S as well as R L):
In electrical engineering the load factor is defined as the average load divided by the peak load in a specified time period. [1] It is a measure of the utilization rate, or efficiency of electrical energy usage; a high load factor indicates that load is using the electric system more efficiently, whereas consumers or generators that underutilize the electric distribution will have a low load ...
The energy could now be seen as stored in the electric field. This process uses energy from the wire with power equal to the electric potential times the total charge divided by time. Where ℰ is the voltage or EMF. By defining the power we can calculate the work that is needed to create such an electric field.