# Kilowatts kW vs Kilowatt Hours kWh

Being careful and consistent when dealing with kilowatts `kW`

and kilowatt-hours `kWh`

is a basic for all energy professionals. Someone who misuses these shows either a lack of care or a lack of understanding - both of which can be fatal.

Understanding the difference between kilowatts `kW`

and kilowatt-hours `kWh`

is therefore a basic for energy industry professionals. The key to understanding the difference is understanding the how they relate to time.

An analogy with distance and speed can is helpful to grasp the concepts of `kW`

and `kWh`

. We measure distance in fixed amounts – such as 50 kilometers. Energy is measured in the same way using the unit `kWh`

. `50 kWh`

is a fixed amount of energy.

Speed is the rate at which we are covering distance – such as 50 km/h. The rate of energy generation or consumption is measured using units of kW. This rate is known as power. The `kW`

is a measurement of `kJ/s`

(`kJ`

is a different unit to measure energy. For reference `3,600 kJ = 1 kWh`

).

Table 1 – Analogy with speed and distance | |||
---|---|---|---|

Amount | Rate | ||

Distance | km | Speed | km/hr |

Energy | kWh | Power | kW |

If you ever see `kW/hr`

– this is almost certainly an error. Technically this is equivalent to acceleration – the rate at which we are changing our power. People often mistakenly use `kW/hr`

when they should be using `kWh`

– don’t be one of these people!

How then to relate `kW`

to `kWh`

? We need one more piece of information – the length of time over which we are producing or consuming energy.

If we were driving a car at a certain speed, to know how far we had driven we need to know how long we had been driving. Likewise if we are consuming energy at a certain rate, to know how much energy we had consumed we need to know how long we had been operating.

For example, if we had been consuming at 50 kW for 1 hour, we would have consumed 50 kWh. Two hours would lead to a consumption of 100 kWh, half an hour 25 kWh.

**Table 2 – Relationship between rate, length of time and consumption**

*How varying the rate (speed or power) affects consumption*

Time | Speed | Distance | Rate | Amount |
---|---|---|---|---|

1 hour | 50 km/hr | 50 km | 0.5 kW | 0.5 kWh |

1 hour | 75 km/hr | 75 km | 1 kW | 1 kWh |

1 hour | 100 km/hr | 100 km | 2 kW | 2 kWh |

*How varying the length of time affects consumption*

Time | Speed | Distance | Rate | Amount |
---|---|---|---|---|

0.5 hour | 100 km/hr | 50 km | 1 kW | 0.5 kWh |

1 hour | 100 km/hr | 100 km | 1 kW | 1 kWh |

2 hour | 100 km/hr | 200 km | 1 kW | 2 kWh |

These concepts need to be grasped forwards, backwards and side to side. You must be comfortable with moving from:

```
kW * hr = kWh
kWh / hr = kW
kWh / kW = hr
```

We can now use these concepts to analyze energy data.

Suppose you have the following data for a CHP scheme. This is a CHP scheme with the facility to dump heat (not all heat generated is necessarily recovered).

Table 3 – Sample data | ||
---|---|---|

Engine electric size | kWe | 400 |

Engine thermal size | kW | 400 |

Annual heat recovered | kWh | 1,809,798 |

Annual power generated | MWh | 2,571 |

Annual operating hours | hr | 5,638 |

What insights can we gain from this? I would look at the following:

*Annual heat recovered*

Calculate the maximum amount of heat our engine could generate in a year (i.e. assuming full load operation for the entire year).

```
maximum heat generation = engine thermal size * annual operating hours
22,55,200 kWh of heat = 400 kW * 5,638 hours
```

This validates the annual heat recovery as reasonable at 80% of the maximum available heat.

Calculate the average heat recovery

```
average heat recovery = annual heat recovery / annual operating hours
321 kW = 1,809,798 kWh / 5,638 hr
```

We do not why this number is low. It could be due to part load operation of the CHP or due to heat dumping.

*Annual power generated*

Again we dimension the power generation using the operating hours. Note the division to convert from kWh to MWh.

```
maximum power generated = engine electric size * annual operating hours
2,255 MWh = 400 kW * 5,638 hours / 1000
```

This flags up an error with the annual power generated value of `2,571 MWh`

as it is greater than our maximum!

We could also spot this error by calculating

```
average electric output = annual power generated / annual operating hours
456 kWe = 2,571 MWh / 5638 hr * 1000
```

This gives us an average engine electric output of `456 kWe`

- which is greater than the size of our engine.

## Summary

The two concepts of an amount of energy (kWh) and the rate of energy consumption/generation (kW) are related to each other by a length of time.

Understanding this allows energy industry professionals to check the validity of data, as well as calculate new data to evaluate performance. Both skills are valuable to engineers and non-engineers alike.

Thanks for reading!