Thank you very much ِdear Kralj your interest in helping.

I have a project for the study of energy and exergy consumption in residential buildings. There are many researchers dealing with the subject of energy and exergy consumption in different areas, but the subject of lighting is not yet clear to me.

The efficiency always depend on what kind of energy you assume to be useful. Since energy is always consereved, it is correct to say, as Aleš Kralj , that energy efficiency is 100% for any device. All energy that enters the system filnally gets out.

However, when luminaries are looked at, the useful energy is the light, that is the total energy of the photons that our eyes see. This light efficiency is generally given in lumens radiated per consumed watt. To convert this in energy efficiency, we should know the spectrum of the light source, that is to know the energy distribution of the radiated photons.

The best luminous efficiency is 683 lm/W for a monochromatic source at 555 nm wavelenght ( https://en.wikipedia.org/wiki/Luminous_efficacy ). So if a source radiates x lm and consumes P Watt, its energy efficiency is x/(683 P). So a LED having a liminous efficacy of 100 lm/W has an energy efficiency of 100/683 = 0.146 that is a bit less than 15 %.

The exergy efficiency is calculated by the work that can theoretically be obtained from the device divided by the consumed energy. It is not 100% even for electrical light sources, since the exergy shall be calculated from the energy converted by the source (that is generally light and heat). Since the light is pure energy (photons), it can theoretically be converted into work with a 100 % efficiency (but PV cells actually convert it with about 20% efficiency only). The remaining heat can be converted into exergy using the Carnot efficiency, based on the temperature of the light source and the ambient temperature. For example, the LED mendioned above has an exergy efficiency of less than 15% , since it is cold. A Tungsten light bulb at 2800 K can theoretically present a Carnot efficiency of (2800-300)/2800 = 90% but its light efficiency is close to 0 (1%), and the heat lost by the bulb is not at 2800 K but at about 400 K at the best, i.e. a Carnot efficiency of about 30%

Thank you very much dear Prof. Claude-Alain Rouletfor the helpful clarification.