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Dear Minister of Energy, Belgium,
• Publié le 1 septembre 2019
Dirk Van de Voorde
Ambassadeur bij Thorium Energy World en lid van 100TWh, etc.
Dear Minister of Energy,
Allow me to demonstrate to you through objective mathematics that we need to take a different approach to solving the energy issue.
There are about 11,350,000 Belgians today, all of whom consume an average of 4,700 kg of oil equivalent per year. That is to say, a part of electricity, a part of gas, a bit of petrol, a bit of wood, coal, etc. ..but, for simplicity, we convert everything into kilograms of oil.
4700 kg oil/year x 11,350,000 Belgians = 53,345,000,000 kg oil/year is consumed in Belgium.
Converted into litres of oil, this comes to approximately 70,000,000,000 litres of oil per year.
Now there is 10kWh of energy in one litre of oil, i.e. by burning one litre of oil you can produce as much heat as an electric fire of 2000W burns for 5 hours.
If there is 10kWh of energy in one litre, 70.000.000.000 litres contain 700.000.000.000 kWh of energy which is needed to provide all Belgians with energy for one year, everything, driving, washing, food, lighting, heating and cooling.
Energy = Power x time expressed in kWh, kilo-Watt-hour
Power = Energy = 700,000,000,000 kWh = 700,000,000,000 kWh = 80,000,000 kW ................ time .........hours in one year ....... 365d x 24h/d
So in Belgium there is a continuous power available of 80 GigaWatt or 80,000 MegaWatt or 80,000,000 kiloWatt or 80,000,000,000 Watt (Power plants, wind, solar, hydro, nuclear, gas plants, our oil storage, heating oil, gasoline, gas supplies etc.).
Belgium has an area of approximately 30,680 km2 or 30,680,000,000 m2. Let us divide the total available capacity of Belgium by the total surface area of Belgium, you will soon find out why I am proposing this.
80,000,000,000W/30,680,000,000 = 2.60 W/m2
Belgium as thus continuously needs a power of 2.60 watts per square meter.
Why all the mathematically complicated stuff? What we need to do is to calculate how to put the budget in order and how to adjust our energy policy, this can only be done through mathematics. Biomass, hydropower, wind and solar energy can never offer a total solution in Belgium, because there is too little space.
BIOMASSA : You can harvest and produce 1500 litres of bio-fuel per hectare per year, depending on the type of crop. Sugar beet and potatoes are 5 times better than straw and hay. This is more or less 0.5 Watt per square meter. This means that you can fill Belgium with energy crops, you can never get the necessary 2.60 W/m2. Energy via biomass is not a total solution, but it can make a useful contribution on a small scale. Bio-fuel can be stored, easily distributed and perfectly adjusted to the demand for energy.
WINDTURBINES: Wind farms also need a lot of space and generate about 1 to 2.5 Watt per square metre. This is already one of the reasons why wind energy cannot be a total solution in Belgium. In addition, wind turbines produce their energy arbitrarily, capriciously and at the wrong time. Sometimes they don't run for days at a time, as was the case from 20 August to 31 August 2019. Energy that is produced at totally random times is useless unless storage or backup installations are provided. In Belgium this is negligible, good for a few minutes, that's all. Expanding this storage via batteries and/or hydropower plants is technically not feasible or unaffordable. If Belgium continues to invest in this wind turbine energy, it will be obliged to build additional gas-fired power stations that are simply ready for when the wind is not blowing and to jump in, and to continuously adjust and flatten all this volatile energy.
KERNENERGIE (including coal-fired and gas-fired power stations, but this is not allowed because they emit CO2) : Nuclear power plants yield 1000W/m2. Energy that can be perfectly regulated and adapted to demand.
I hope that in the end this fairly simple calculation is sufficient and that you can deduce the decision yourself.
Yours sincerely,
Dirk Van de Voorde
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M.v.g. Ronald