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Efficiency

Efficiency is the ratio of what comes out at the back to what is put in at the front. Efficiency in all technical processes is less than 100%.⁷ When, for example, an electric motor converts 80% electrical energy into rotational energy (see chapter Energy and Supply), it is then 80% efficient. The remaining 20% of this electrical energy represents the incurred “losses”. According to the law of conservation of energy, these are not really lost, but converted into another form of energy: heat loss. If this isn’t properly dissipated, the engine will break down.

When you connect several technical processes in series, you can easily calculate their overall efficiency by multiplying the efficiencies (not the losses!) of the individual processes. If, for example, the efficiency of a coal-fired power plant is 40%, i.e. 0.4 (correct on average), the efficiency of the electric motor intended to drive something is then 80%, and the efficiency of the electric grid is then 92% (8% losses). You calculate this as follows: 0.4 × 0.8 × 0.92 = 0.294. In concrete terms, this means that 29.4% of the combustion energy of coal is converted into the rotational energy of this electric motor.

Efficiency is close to 100% only in a few technical processes. These are pure combustion processes, electrical processes and processes with incompressible fluids (water).

In all other technical processes (mostly so-called “circular processes”), which have to do with gases or where combustion is associated with a mechanical drive (gas turbine, internal combustion engine), efficiency is significantly below 100%.

This is a physical and not a technical limit.

Improvement is not possible.

The Last Percentages – Cutting Losses

“Nach fest kommt ab!” (Tight is followed by off!) says the German fitter when tightening a bolt. This means nothing else than that anything overdone makes no sense and is even counterproductive.

This can easily be illustrated: When we assume that a large electric motor has a connected load of 100 kW (it needs 100 kW from the electric grid), but the mechanical load is only 50 kW, it will then have 50 kW in losses (which are usually lost as heat). If you want to improve this motor and make it produce 1 kW more mechanical power, then you have to reduce the losses by 1 kW. 1 kW better at 50 kW corresponds to a 2% improvement. In doing so, the losses must be reduced by 2%. This seems simple.

Technical devices are better nowadays, however. Efficiencies are much closer to the maximum achievable value. Assuming that a modern motor with a 100 kW connection load converts 80 kW into a mechanical load, this means 20 kW are lost. If you want to improve this motor by 1 kW, this means a 1.25% improvement. The difficulty here: Any losses have to be reduced by 1 kW from 20 kW to 19 kW. That’s 5%. That’s a lot.

It’s therefore much more difficult to achieve any improvement when trying to optimize systems that are highly optimized already. Most of the time this kind of thing is associated with a lot of effort. It’s easy to reach an explosion in costs. That’s why you have to find a financially acceptable compromise most of the time.