{"id":20392,"date":"2026-05-14T11:00:00","date_gmt":"2026-05-14T10:00:00","guid":{"rendered":"https:\/\/giconmes.es\/how-to-convert-your-steam-boiler-output-into-revenue-via-cae\/"},"modified":"2026-05-21T04:20:43","modified_gmt":"2026-05-21T03:20:43","slug":"how-to-convert-your-steam-boiler-output-into-revenue-via-cae","status":"publish","type":"post","link":"https:\/\/giconmes.es\/en\/how-to-convert-your-steam-boiler-output-into-revenue-via-cae\/","title":{"rendered":"How to convert your steam boiler output into revenue via CAE"},"content":{"rendered":"\n

The efficiency of a boiler<\/a> is not a decorative data on the data sheet. It defines what percentage of the energy you pay for ends up as useful steam<\/strong> for your process, and what percentage escapes through the chimney, blowdown or radiation losses. Most Spanish industrial facilities still have combustion boilers operating at 75-85%. <\/p>\n\n

Modern electric generators<\/a>, such as those of Giconmes<\/strong>, achieve 99%<\/strong>. The difference, translated into kWh and euros, is significant, and from 2023 can also be monetized<\/strong> through the Energy Saving Certificate System (CAE).<\/strong> <\/p>\n\n

How to calculate the efficiency of a boiler<\/h2>\n\n

Let’s start by understanding how the efficiency of a boiler<\/strong> is calculated. Efficiency (\u03b7)<\/strong> is the ratio of the heat absorbed by the water<\/strong> – that which is actually incorporated into the steam – to the total heat supplied to the system<\/strong>: <\/p>\n\n

\u03b7 = (Heat absorbed by the water \/ Heat supplied to the system) \u00d7 100<\/strong><\/p>\n

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\"Steam<\/figure><\/div>\n

The heat absorbed<\/strong> depends on how much water evaporates and the enthalpy difference between the feed water and the steam produced:<\/p>\n\n

Q_absorbed = m \u00d7 (h\u2082 – h\u2081)<\/strong><\/p>\n\n

where m<\/strong> is the mass of steam generated (kg), h\u2081<\/strong> the enthalpy of the feedwater (kcal\/kg) and h\u2082 <\/strong>that of the steam (kcal\/kg).<\/p>\n\n

The heat input<\/strong> depends on the fuel consumed and its calorific value, in the case of a combustion boiler:<\/p>\n\n

Q_carried = m_f \u00d7 PCS<\/strong><\/p>\n\n

where m_f <\/strong>is the mass of fuel and PCS<\/strong> its gross calorific value (kcal\/kg). In an electric boiler, the heat input is simply the electrical energy consumed. <\/p>\n\n

Numerical example: 1,000 kg\/h of steam at 10 bar<\/h2>\n\n

Let’s take a typical industrial process<\/strong> that demands 1,000 kg\/h of saturated steam<\/strong> at 10 bar<\/strong>, starting with feed water at 20 \u00b0C:<\/strong><\/p>\n\n

m = 1,000 kg\/hh\u2082 \u2248 663 kcal\/kg (saturated steam at 10 bar)h\u2081 \u2248 20 kcal\/kg (water at 20 \u00b0C)Q_absorbed = 1.000 \u00d7 (663 – 20) = 643.000 kcal\/h \u2248 748 kWh\/h<\/strong><\/p>\n\n

That is the net energy<\/strong> that has to be incorporated into the steam<\/strong>. The energy that the equipment needs to consume will always be higher; how much more depends solely on the efficiency of the equipment. <\/p>\n\n

Gas vs. electric: where is the difference<\/h2>\n\n

Combustion boilers (natural gas, diesel, LPG) operate, at best, with efficiencies of 80-85 %. Losses are divided between: <\/p>\n\n