OUR IDEA
Our teams’ idea is to install a series of mirrors on the moon and use those to send a huge amount of energy in the form of solar rays to the earth via satellites. In these satellites the light will be bundled by means of lenses causing it to hit the earth’s surface perpendicular. This bundled light will be used on earth to heat water causing it to turn into steam, which will then be used to power turbines. This system will also be used to compress air into so called “gas fields”. This way it can be used later to keep generating electricity when the station is no longer receiving light from the moon, using aforementioned stored air.
STAGE 1
In stage 1 a moon base will be built using funding from governments, universities and companies/investors. We believe there will be interest in doing so due to the huge amount of profit this project can harvest (refer to the calculations sheet for more information) as well as the research that can also be performed on this base and the “credit” for building the first self-sustained moon base. All of these investors will also be mandatory for the project to succeed as the costs will be relatively high, but with a huge pay-off possibility. In this stage we will also launch four satellites into geostationary orbit. These satellites will function as the lenses that will bundle the light. The whole moon base will be built using materials from the moon itself, of which there are plenty.
STAGE 2
Stage 2 is the stage where the mirrors will be built using the moon base as a sort of factory and mining facility and made operational. The mirrors will also be built completely using materials present on the moon, greatly reducing costs. This stage is actually continuing as long as we want to produce energy using these mirrors (probably ad infinitum). The longer this stage continues, the more we will harvest from the big investments put into the project, as we will not only be producing huge amounts of energy but also be able to do research into other aspects of space travel and science, such as travelling to other planets from the moon in the (far) future.
USAGE OF THE LIGHT ON EARTH
We can use the light and the warmth sent from the mirrors on the moon on earth. This light will be rayed onto a station on earth where it will heat water not unlike the way a solar panel does.
Water flows through black pipes (these absorb the biggest amount of wavelength (ergo most heat) from the sunlight) until it turns to steam, then is used to power turbines. The water will flow through these tubes with quite a high velocity, producing a big amount of energy per second.
Only part of the steam produced will actually go into the generators. The remainder will be used to compress gas into, quite logically, so called “gas fields”. Then this compressed gas can be used to power the generators when the moon is not sending light to that particular station, thus creating a steady energy outflux.
Only part of the steam produced will actually go into the generators. The remainder will be used to compress gas into, quite logically, so called “gas fields”. Then this compressed gas can be used to power the generators when the moon is not sending light to that particular station, thus creating a steady energy outflux.
MIRROR DISFUNCTION OR TERRORIST ATTACK SAFETY SWITCH
In case of a scenario a little too much like a certain scene in Star Wars, there will be a safety switch implemented on several locations. Once this switch is either flipped or pressed (depending on what type of switch is used) all of the mirrors will more or less instantly stop collecting and bundling light by immediately turning away from the sun in such a way that a non-reflecting part of the mirror will be facing the sun. Once this happens all of the ground station will get an error message and a notification so that they can stop the energy producing process. The system will only be turned on again when all of the station, as well as the moon base give the command to do so. This to minimize the chance of any sort of terroristic attack.
WHAT WILL BE THE AREA OF THE MIRRORS?
The total surface of the moon is 38.000.000 km2.
To us, only 60% of that surface is visible; 0.6(%/100)*38.000.000(km2)=22.800.000km2.
Out of this 60%, 30% has an advantageous position for receiving solar energy; 0.3(%/100)*22.800.000(km2)=6.840.000km2.
Out of this 30%, we only use the 16% that consists of lunar mares for practicality reasons; 0.16(%/100)*6.840.000(km2)=1.094.400km2.
To us, only 60% of that surface is visible; 0.6(%/100)*38.000.000(km2)=22.800.000km2.
Out of this 60%, 30% has an advantageous position for receiving solar energy; 0.3(%/100)*22.800.000(km2)=6.840.000km2.
Out of this 30%, we only use the 16% that consists of lunar mares for practicality reasons; 0.16(%/100)*6.840.000(km2)=1.094.400km2.
HOW MUCH ENERGY CAN BE GATHERED THIS WAY?
Going from km2 to m2 we’ll have to multiply the surface by 1.000.000; 1.094.400(km2)*1.000.000=1.094.400.000.000m2 (1094,4*109).
To then determine the amount of energy we can transfer to the earth we’ll multiply the surface of the mirrors (in m2) by 1367 (=Max. W/m2 solar energy; source: http://en.wikipedia.org/wiki/Solar_constant); 1.094.400.000.000(m2)*1367(W/m2)=1.496.044.800.000.000J/s (1496*1012).
In the year of 2008 the total amount of energy consumed worldwide was 15.210.282.150.733J/s (15,21*1012). Our own produce of 1496*1012 J/s makes for a measly 9835,5% of that amount.
To then determine the amount of energy we can transfer to the earth we’ll multiply the surface of the mirrors (in m2) by 1367 (=Max. W/m2 solar energy; source: http://en.wikipedia.org/wiki/Solar_constant); 1.094.400.000.000(m2)*1367(W/m2)=1.496.044.800.000.000J/s (1496*1012).
In the year of 2008 the total amount of energy consumed worldwide was 15.210.282.150.733J/s (15,21*1012). Our own produce of 1496*1012 J/s makes for a measly 9835,5% of that amount.
HOW MUCH STEAM CAN BE PRODUCED THIS WAY?
Since we’re going to need superheated steam in order to power our generators, we’ll have to heat it to about 400°C or 752°F. (Source: http://en.wikipedia.org/wiki/Steam)
After doing some research on several websites the website http://www.stoommachine.info/tabel.html (excuses for it being in Dutch) stated that in order to achieve this with 1 litre of water we will have to insert 3263KJ of energy in said litre of water.
So, first of all let’s convert our energy in J to KJ; 1.496*1012(J/s)/1000=1496*109kJ/s.
This means that, with the amount of energy we produce, we can calculate how much water we can turn into useful steam in one second. To do this we divide the energy we produce by the 3263kJ needed to turn 1 litre into useful steam; 1496*109(kJ/s)/3263(kJ)= 458.473.797L steam/s (458,47*106), this is equivalent to 183,39 Olympic swimming pools per second (volume Olympic swimming pool: 2500m3 or 2500000dm3).
After doing some research on several websites the website http://www.stoommachine.info/tabel.html (excuses for it being in Dutch) stated that in order to achieve this with 1 litre of water we will have to insert 3263KJ of energy in said litre of water.
So, first of all let’s convert our energy in J to KJ; 1.496*1012(J/s)/1000=1496*109kJ/s.
This means that, with the amount of energy we produce, we can calculate how much water we can turn into useful steam in one second. To do this we divide the energy we produce by the 3263kJ needed to turn 1 litre into useful steam; 1496*109(kJ/s)/3263(kJ)= 458.473.797L steam/s (458,47*106), this is equivalent to 183,39 Olympic swimming pools per second (volume Olympic swimming pool: 2500m3 or 2500000dm3).
REFRACTION BY OUR SATELLITES
The sun sends light beams in all directions, some of which hit the moon. Part of these rays are then redirected by means of a mirror to a satellite in geostationary orbit around the earth. In the picture above three of these situations are depicted. In situation 1 the satellite will have to fracture the light ray 45° to the left, by means of a lens. In situation 2 the lens/satellite doesn’t have to do anything at all, as the light rays hit it under 0°. Situation 3 is the exact same as situation 1, except for the fact that it is mirrored. Using 4 satellites (one every 90°) there will be no problem with the lenses having to refract the light too strongly. Every station on earth will receive a beam of light for a quarter of a day, after which the beam will switch to another station.
BUT!
This, of course, is practically neither doable nor a worthwhile investment, so we decided to scale down a (tiny) bit.
The actual surface of our mirror on the moon will be 5.000 km2, equivalent to a diameter of about 80 km. Of course this will mean we will be capturing a lot less power send out by the sun, but it will still be a pretty decent amount, something a little bit more realistic.
Another thing that wasn’t covered in our previous calculations is energy loss. Now this is going to be quite a big factor, seeing as only 166W/m2 (http://nl.wikipedia.org/wiki/Zonneconstante) usually reaches the earth’s surface, an energy loss of (1367-166)/1367*100%=88,86%. However, since the light we will be sending through the atmosphere is bundled, we will only have an energy loss of 74,3%, meaning 1367-(0,743*1367)=351W/m2 will reach the earth’s surface.
The actual surface of our mirror on the moon will be 5.000 km2, equivalent to a diameter of about 80 km. Of course this will mean we will be capturing a lot less power send out by the sun, but it will still be a pretty decent amount, something a little bit more realistic.
Another thing that wasn’t covered in our previous calculations is energy loss. Now this is going to be quite a big factor, seeing as only 166W/m2 (http://nl.wikipedia.org/wiki/Zonneconstante) usually reaches the earth’s surface, an energy loss of (1367-166)/1367*100%=88,86%. However, since the light we will be sending through the atmosphere is bundled, we will only have an energy loss of 74,3%, meaning 1367-(0,743*1367)=351W/m2 will reach the earth’s surface.
HOW MUCH ENERGY CAN BE GATHERED THIS WAY?
Going from km2 to m2 we’ll have to multiply the surface by 1.000.000; 5.000(km2)*1.000.000=5.000.000.000m2 (5*109).
To then determine the amount of energy we can transfer to the earth we’ll multiply the surface of the mirrors (in m2) by 351 (W/m2 solar energy; source: http://en.wikipedia.org/wiki/Solar_constant); 5.000.000.000(m2)*351(W/m2)= 1.755.000.000.000J/s (1755*109).
In the year of 2008 the total amount of energy consumed worldwide was 15.210.282.150.733J/s (15,21*1012). Our own produce of 1.755.000.000.000J/s (1755*109) makes for a measly 11,5% of that amount.
To then determine the amount of energy we can transfer to the earth we’ll multiply the surface of the mirrors (in m2) by 351 (W/m2 solar energy; source: http://en.wikipedia.org/wiki/Solar_constant); 5.000.000.000(m2)*351(W/m2)= 1.755.000.000.000J/s (1755*109).
In the year of 2008 the total amount of energy consumed worldwide was 15.210.282.150.733J/s (15,21*1012). Our own produce of 1.755.000.000.000J/s (1755*109) makes for a measly 11,5% of that amount.
HOW MUCH STEAM CAN BE PRODUCED THIS WAY?
Since we’re going to need superheated steam in order to power our generators, we’ll have to heat it to about 400°C or 752°F. (Source: http://en.wikipedia.org/wiki/Steam)
After doing some research on several websites the website http://www.stoommachine.info/tabel.html (excuses for it being in Dutch) stated that in order to achieve this with 1 litre of water we will have to insert 3263KJ of energy in said litre of water.
So, first of all let’s convert our energy in J to KJ; 1755*109(J/s)/1000=1755*106kJ/s.
This means that, with the amount of energy we produce, we can calculate how much water we can turn into useful steam in one second. To do this we divide the energy we produce by the 3263kJ needed to turn 1 litre into useful steam; 1755*106(kJ/s)/3263(kJ)= 537.849L steam/s (538*103), this is equivalent to 0,22 Olympic swimming pools per second (volume Olympic swimming pool: 2500m3 or 2500000dm3).
After doing some research on several websites the website http://www.stoommachine.info/tabel.html (excuses for it being in Dutch) stated that in order to achieve this with 1 litre of water we will have to insert 3263KJ of energy in said litre of water.
So, first of all let’s convert our energy in J to KJ; 1755*109(J/s)/1000=1755*106kJ/s.
This means that, with the amount of energy we produce, we can calculate how much water we can turn into useful steam in one second. To do this we divide the energy we produce by the 3263kJ needed to turn 1 litre into useful steam; 1755*106(kJ/s)/3263(kJ)= 537.849L steam/s (538*103), this is equivalent to 0,22 Olympic swimming pools per second (volume Olympic swimming pool: 2500m3 or 2500000dm3).
THE SEIZE OF ONE POWER PLANT
The area which the water will flow through will be 750 meters (=7500dm) wide by 750 meters (=7500dm) long by 1 meter (=10dm) high, making for a total volume of 562.500m3 or 562.500.000dm3, equal to 562.500.000L.
With our energy influx we can warm this amount of water in (562.500.000(L)*3263(kJ/L))/1.755.000.000(kJ/s)=17,43 minutes, meaning we can, in the 6 hours of light per station, produce (6(h)*60(m/h))/17,43(m)=20,65 times 562.500.000(L)= 11.617.900.172L steam per station. This makes for a total amount of 4*11.617.900.172(L)=46.471.600.688L steam per day.
With our energy influx we can warm this amount of water in (562.500.000(L)*3263(kJ/L))/1.755.000.000(kJ/s)=17,43 minutes, meaning we can, in the 6 hours of light per station, produce (6(h)*60(m/h))/17,43(m)=20,65 times 562.500.000(L)= 11.617.900.172L steam per station. This makes for a total amount of 4*11.617.900.172(L)=46.471.600.688L steam per day.