Germany export 4MWh E-Energy although 8 Nuclear-Power-Stations turned off

[Shielder]

Are you trying to look stupid? (Sorry, you are, but that's beside the point).

How are you getting 10GW from a nuclear reactor that only generates 4.5GW? This figure is in the design specification as the thermal power output of the reactor! Are you seriously proposing that there is somehow a missing 5.5GW of thermal energy that is not being declared in the technical specifications for the reactors?

I know you take LSD, but it has seriously affected you if you think that you can somehow 'magic' up 5.5GW of heat from a power station that only generates 4.5GW.

And don't even bother trying to say about spent fuel heating up the spent fuel pond making up the 'missing' 5.5GW, you've lost that argument before it even starts.

[Tweenk]

We Discuses this in the past i calculate based on the (Nuclear)FUEL not on the heat on the FUEL its 4% efficiency.
Also i calculate the Thermal efficiency on the average heat usage per 1KW electric power and not the peak at the best point. (,,,)
I also calculate the energy lost on the Peak Load Backup because nuclear power plants can not service Peak Load.

You were shown repeatedly that the 4% figure is pulled straight out of your ass. The correct figure is 30-36%. Transmission losses at 6% reduce this to about 28-34% - still nowhere near your 4% number. Electrical output of any thermal power plant (including nuclear) is proportional to its thermal output, so your "average heat usage" argument is bogus as well.

Nuclear power plants can and do service peak loads, especially in France or on nuclear submarines. US subs can go from 0% to 100% reactor power in seconds. French EPR reactor can change output at 5% (80MW) per minute. Older French reactors have similar load following capabilities.

An alternative viewpoint about nuclear power is only valid if you put the nuclear power plant into a submarine and let it operate 200km away in the sea then the distance and the water makes sure no human are harmed.

Russian idea:
http://en.wikipedia.org/wiki/Russian..._power_station

[QUOTE]nuclear power plants pollute CO2[QUOTE]
How exactly? Radioactive unicorn farts?

Your level of denial is astounding. It's like debating a creationist - completely pointless, yet some people feel ideologically compelled to do it. I have no idea why we value our time so little to keep replying to him after he demonstrated time and again that he is immune to reality.

[Qaridarium]

Are you trying to look stupid? (Sorry, you are, but that's beside the point).
How are you getting 10GW from a nuclear reactor that only generates 4.5GW?
This figure is in the design specification as the thermal power output of the reactor! Are you seriously proposing that there is somehow a missing 5.5GW of thermal energy that is not being declared in the technical specifications for the reactors?

again and again and again... and i told you as many times its 4,5GW(thermic)+2GW(electric) +radiation and from the 2GW(electric) you lost 0,12GW on the cables.
you also only calculate Peak usage point i calculate the heat in the after usage cooling pond also the heat at disposal also the heat at maintenance
your numbers are only dream numbers you are "blind" whats about the loses on the cables and loses on the after usage cooling ponds and also the maintenance heat and disposal heat and the energy in the Radiation?

if somebody read this facts the only one look stupid are you!

I know you take LSD, but it has seriously affected you if you think that you can somehow 'magic' up 5.5GW of heat from a power station that only generates 4.5GW.
And don't even bother trying to say about spent fuel heating up the spent fuel pond making up the 'missing' 5.5GW, you've lost that argument before it even starts.

If some one calculates more factors than you he must be stupid and a Junkie.

Give me a calculation with 100% all factors means minding the source of the energy and clean and make ready the source then burn the source also heat after usage cooling pond also the heat at maintenance the heat at disposal and last also the energy lost in nuclear radiation.

Your Dream-numbers never ever hold a critical check.

[Qaridarium]

You were shown repeatedly that the 4% figure is pulled straight out of your ass. The correct figure is 30-36%. Transmission losses at 6% reduce this to about 28-34% - still nowhere near your 4% number. Electrical output of any thermal power plant (including nuclear) is proportional to its thermal output,

your claim of the correct figure "28-34%" is just calculating like a child.
but you prefer to end the calculation after thermal and transmission calculation.
come one try it like a real man! only 1 time!
Right now you are on "28-34%" now calculate the rest off the REAL WORLD!

"28-34%"-(mining the source)-(make ready the source)-(after usage cooling pond heat)-(maintenance heat)-(disposal heat)-(nuclear radiation)=X ?

Try it KID!

so your "average heat usage" argument is bogus as well.

sure it is in fact i help you to be a real "man" to Lift up your life.
so you can ad a 10% of 4,5GW in an updraft tower.
X+(10% of 4,5GW)

Nuclear power plants can and do service peak loads, especially in France or on nuclear submarines. US subs can go from 0% to 100% reactor power in seconds. French EPR reactor can change output at 5% (80MW) per minute. Older French reactors have similar load following capabilities.
Russian idea:
http://en.wikipedia.org/wiki/Russian..._power_station

Maybe but this critical usage also increase the failure rate.

and it drops down the efficiency.

"nuclear power plants pollute CO2"
How exactly? Radioactive unicorn farts?

mining the source,make ready the source,disposal

The CO2 Pollute of a nuclear power plant is higher than a Wind mill.

Your level of denial is astounding. It's like debating a creationist - completely pointless, yet some people feel ideologically compelled to do it. I have no idea why we value our time so little to keep replying to him after he demonstrated time and again that he is immune to reality.

stop blaming me and start a calculating like a real man KID!

"28-34%"-(mining the source)-(make ready the source)-(after usage cooling pond heat)-(maintenance heat)-(disposal heat)-(nuclear radiation)=X ? X+(10% of 4,5GW)

[Naib]

negative power plant

No such thing. such a term does not exist anywhere in power distribution circles. The moment you started using that, and justifying its usage, highlighted yourself as an ignorant narrowminded troll. your opinion on anything todo with power is dumpsterclass small son

*plunk*

[Qaridarium]

[/SIZE]

No such thing. such a term does not exist anywhere in power distribution circles. The moment you started using that, and justifying its usage, highlighted yourself as an ignorant narrowminded troll. your opinion on anything todo with power is dumpsterclass small son

*plunk*

its just a translation problem. i also do not found the english word for it.

The German words are: Laststeuerungs Regelleistung betriebender Intelligenter Energie Konsument der Ebergie in Energie Reiche Produkte Speichert.

Google translate translate it into bullshit: "Load balancing power control operation final consumer of energy in the Intelligent Energy-Energy Products Stores."

I try to fix it: Load balancing power control operation Intelligent energy consumer and this consumer produce Products and it Store energy in the Products."

A example is a methane syntesis or a aluminum smelter or a cold store

[Shielder]

again and again and again... and i told you as many times its 4,5GW(thermic)+2GW(electric) +radiation and from the 2GW(electric) you lost 0,12GW on the cables.

Erm, no. You don't get the efficiency by adding thermal power to the electrical power! Efficiency is electrical power out divided by thermal power.

Oh, and here I thought you were talking about 10GW from the plant when it was operating:

its just basic physic law: http://en.wikipedia.org/wiki/Conservation_of_energy

you can only cool down by "efficiency" because "efficiency" save CO2 and heat.

for example you can build a updraft tornado power plant on-top on a nuclear power plant.
this means you use the 10Gigawatt heat to updraft the air in your "tower" Tornado
a updraft power plant do have 10% efficiency this means 1Gigawatt extra electric energy from the same nuclear power plant.

So which is it?

Also:

"28-34%"-(mining the source)-(make ready the source)-(after usage cooling pond heat)-(maintenance heat)-(disposal heat)-(nuclear radiation)=X ? X+(10% of 4,5GW)

How about looking at it this way? Construction of wind turbines involves; mining the rare earth metals and other components, refining them, shipping them from China, fabricating the generators, fabricating the blades (I think they are some sort of fibreglass composite) transporting them to site and erecting the turbines. All this has to be done hundreds of times for a wind farm with an equivalent RATED output to a nuclear station. Then there is the 'smart grid' electronics that are required to account for the variability of wind generation, the extra 'stand-by' fossil powered stations ready for when the wind doesn't blow, and, in Germany's case, hundreds of km of extra cables to get the power from the wind powered north to the energy starved south.

Q, when you talk about the efficiency of your pet renewable/low-emissions energy source, you are simply quoting the same as me, (energy out)/(energy in) = efficiency

This is the standard way of defining efficiency. A combustion engine efficiency is calculated by taking the rated mechanical power output of the engine and dividing it by the heat generated by the combusion of the fuel. So a diesel engine producing 100kW of heat and having a mechanical power output of 40kW is 40% efficient. Similarly, a solar cell producing 700W from 1m² of surface is (approximately) 50% efficient (assuming 1.4kWm^-2 inbound solar flux.)

In a nuclear power plant, the thermal power output of the reactor is (for the EPR) 4500MW. The electrical power output is 1700MW, so it is approximately 37% efficient. This is a scientific method of calculating efficiency, not something you made up to try and make out that nuclear power is 1% efficient.

I'll indulge you for a second or two though.

You mentioned in one of the quotes above

(after usage cooling pond heat)-(maintenance heat)-(disposal heat)-(nuclear radiation)

so I'll just quantify all of these values for you shall I?

Cooling pond heat: Look at the following plot of decay heat against time from trip.

You can see that 1 day after shut down, the fuel is generating 0.5% of the full power. After 10 days it is 0.2%. The decay of radioactive nuclides decreases exponentially, so 100 days after shutdown, the decay heat will be less than 0.05% of full power, i.e. 2.25MW. About the power from a moderately sized wind turbine.

Maintenance heat? I'm assuming that you are talking about a statutory outage when the plant isn't generating? We call that load factor, which in a nuclear station is about 90%+.

Disposal heat? That, I assume, is the heat rejected to the environment post generation? That is (for the EPR) 4500-1700=2800MW, this is taken account of in the efficiency calculation.

Nuclear radiation? In the reactor, when it is operating, the energy released by the nuclear reactions all goes into heating up the fuel and hence the coolant flowing past the fuel elements. So, for example, looking again at the plot above, we can see that about 7% of the power output of a nuclear reactor is decay heat from all nuclear reactions.

Assuming you meant once the fuel is out of the reactor, then the nuclear radiation heats up the cooling pond water, which is already explained above. So, all of those factors you are blathering on about are already accounted for in the efficiency calculation, or are so small as to make no measurable difference to teh efficiency calculation.

As for mining and refining activities, I expect you to apply the same factors to your beloved renewable energies and see how efficient they are then.

[Qaridarium]

Erm, no. You don't get the efficiency by adding thermal power to the electrical power! Efficiency is electrical power out divided by thermal power.

Sorry its just my bad English i mean you have to calculate the 6% cable lost from the 2GW electric to.
and again you lost even more than i write for example you lost "Regelleistung" on a nuclear power plant. A nuclear power plant produce "off-peak power" in the night.
if you calculate the "off-peak power" lost in your calculation you lost even more!

Oh, and here I thought you were talking about 10GW from the plant when it was operating:

and again its only my bad English i talk about the heat you need in the average to make 2GW Nuclear powered electric energy.
This mean you have to calculate:
(off-peak power lost "Regelleistung"), (mining the source)-(make ready the source)-(after usage cooling pond heat)-(maintenance heat)-(disposal heat)-(nuclear radiation)

and in the end you produce more than 10GW heat for every 2GW el output.

How about looking at it this way? Construction of wind turbines involves; mining the rare earth metals and other components, refining them, shipping them from China, fabricating the generators, fabricating the blades (I think they are some sort of fibreglass composite) transporting them to site and erecting the turbines. All this has to be done hundreds of times for a wind farm with an equivalent RATED output to a nuclear station. Then there is the 'smart grid' electronics that are required to account for the variability of wind generation, the extra 'stand-by' fossil powered stations ready for when the wind doesn't blow, and, in Germany's case, hundreds of km of extra cables to get the power from the wind powered north to the energy starved south.

Sure you are right here but this do not mean wind power lose the overall calculation against nuclear power. In my knowledge Nuclear power loses the Overall calculation.
but yes you can lift me up with a studies about the overall energy usage and the overall crop factor and the CO2 polluting factor.

but you can't because all Studies i know show that wind wins over nuclear.

Q, when you talk about the efficiency of your pet renewable/low-emissions energy source, you are simply quoting the same as me, (energy out)/(energy in) = efficiency

sure but i calculate overall 100% of all stuff i also calculate the CO2 pollution of the nuclear staff in the nuclear power plant.

your claim about an nuclear power station do not pollute CO2 is 100% ignorance

This is the standard way of defining efficiency. A combustion engine efficiency is calculated by taking the rated mechanical power output of the engine and dividing it by the heat generated by the combusion of the fuel. So a diesel engine producing 100kW of heat and having a mechanical power output of 40kW is 40% efficient. Similarly, a solar cell producing 700W from 1m² of surface is (approximately) 50% efficient (assuming 1.4kWm^-2 inbound solar flux.)

NO WE DO NOT CALCULATE A COMBUSTION ENGINE POWER PLANT!
YOU always calculate based on the FUEL!
And the energy efficiency of nuclear FUEL is very low.

In a nuclear power plant, the thermal power output of the reactor is (for the EPR) 4500MW. The electrical power output is 1700MW, so it is approximately 37% efficient. This is a scientific method of calculating efficiency, not something you made up to try and make out that nuclear power is 1% efficient.

your error is simple A NUCLEAR POWER PLANT IS MORE COMPLEX TO CALCULATE TGAB A DIESEL ENGIN! Your Calculation do not cover particle radiation and the after glow heat after the usage. also it does not cover the Approach and shutdown and a nuclear power plant drop in effiency in Approach and shutdown!

I'll indulge you for a second or two though.

?????????????????????
sorry i don't understand you also Google translate don't help.

You mentioned in one of the quotes above so I'll just quantify all of these values for you shall I?

yes please and please also quantify the new suggestions from me " effiency in Approach and shutdown! " ,"the CO2 pollution of the nuclear staff",(off-peak power lost "Regelleistung")

Cooling pond heat: Look at the following plot of decay heat against time from trip.
You can see that 1 day after shut down, the fuel is generating 0.5% of the full power. After 10 days it is 0.2%. The decay of radioactive nuclides decreases exponentially, so 100 days after shutdown, the decay heat will be less than 0.05% of full power, i.e. 2.25MW. About the power from a moderately sized wind turbine.

wow now you are serious and yes it lower your efficiency.
but you try to play it down no 0,05-0,5% is not ZERO

YOUR CALCULATION BASED ON THE CLAIM THAT THIS IS ZERO!

Maintenance heat? I'm assuming that you are talking about a statutory outage when the plant isn't generating? We call that load factor, which in a nuclear station is about 90%+.

wow nice i learn english with you load factor yes you have to calculate the load factor to

But your CLAIM BASED ON 100% load factor.

Disposal heat? That, I assume, is the heat rejected to the environment post generation? That is (for the EPR) 4500-1700=2800MW, this is taken account of in the efficiency calculation.

no i mean the nuclear waste heat after glow.

[QUOTE=Shielder;235093]
Nuclear radiation? In the reactor, when it is operating, the energy released by the nuclear reactions all goes into heating up the fuel and hence the coolant flowing past the fuel elements. So, for example, looking again at the plot above, we can see that about 7% of the power output of a nuclear reactor is decay heat from all nuclear reactions.[QUOTE=Shielder;235093]

no i mean the Pollution of the Radiation and the Radiation in the Nuclear waste after the Life time of the Nuclear reactor.

Assuming you meant once the fuel is out of the reactor, then the nuclear radiation heats up the cooling pond water, which is already explained above. So, all of those factors you are blathering on about are already accounted for in the efficiency calculation, or are so small as to make no measurable difference to teh efficiency calculation.

there are also part load efficiency if you do load-balancing regular performance
because your numbers are only peak numbers.

As for mining and refining activities, I expect you to apply the same factors to your beloved renewable energies and see how efficient they are then.

sure i do. and the studies i know renewable energies beat them all.

[Shielder]

Off-peak power lost???

It is still being used somewhere (the lights don't go out overnight, hospitals still need power, big industry still needs power, heating and lighting still happens), so you are wrong, it isn't lost. It's called baseload generation.

Every country requires a certain amount of baseload generation. In the UK it is about 27GW, depending upon the time of year. Yes, we can shutdown gas and coal plants, but the nuclear plants go on generating, even in your 'off-peak' times because the power is still needed. If the power wasn't being used, then this would lead to grid instability, which can cause blackouts and more serious problems with the grid, including damaging equipment.

And it wasn't your bad english, how can you say "just my bad English i mean you have to calculate the 6% cable lost from the 2GW electric to" when you clearly state: "4,5GW(thermic)+2GW(electric) +radiation"

And how do you get "the 10Gigawatt heat to updraft the air in your "tower" Tornado" if you're now not talking about heat generated at the power station?

Oh, and the 6% cable loss will affect all large scale generation, including the transfer cables from the north of Germany to the south, so it isn't just a nuclear problem, it will affect your precious windmills too.

I had to laugh over your CO₂ contribution of the staff at the nuclear plant! What about the CO₂ produced by the servicing teams at the large windfarms? Off-shore wind farms are even worse because they require big ships to service them. However, using this as an argument that nuclear power stations produce CO₂ is laughable, just as it is laughable for wind farms.

With respect to efficiency of nuclear fuel, have you heard of reprocessing?

your error is simple A NUCLEAR POWER PLANT IS MORE COMPLEX TO CALCULATE TGAB A DIESEL ENGIN! Your Calculation do not cover particle radiation and the after glow heat after the usage. also it does not cover the Approach and shutdown and a nuclear power plant drop in effiency in Approach and shutdown!

I'm not sure I understand what you are saying here. A nuclear plant efficiency is just as easy to calculate: power generated divided by heat generated. How complicated can it be?

Particle radiation is what generates the heat in the reactor and the heat generated be spent fuel is a fraction of the heat generated at power. That's why spent fuel pools don't require as much cooling as the primary circuit. The "lost generation" you are talking about is less than 1%, if it was much greater, they'd be using the spent fuel as a power source!

What I don't understand is Approach and Shutdown. If you are talking about power raising, then that is a day or so of a 24 month power cycle. Hardly significant. It's the same with Shutdown. When there is an outage, it generally lasts less than a month, which is 1/24th or just over 4% of the time spent generating. So, assuming a perfect 100% power output, we have a load factor of at least 95%.

[Qaridarium]

Off-peak power lost???
It is still being used somewhere

i mean the "überschüssige Regelleistung" sorry my english fails here.
you can only use it in a storage power plant but thats my point your efficiency drop hardly if you need a storage power plant.
and again "Off-peak power lost???" its just a translation mintage.
i do not mean the used off-peak power i mean the not used off peak power or the power used in a storage power plant.

(the lights don't go out overnight, hospitals still need power, big industry still needs power, heating and lighting still happens), so you are wrong, it isn't lost. It's called baseload generation.

no you are wrong because there is still a not used off-peak power.

you need storage power plants because of this.

Every country requires a certain amount of baseload generation. In the UK it is about 27GW, depending upon the time of year. Yes, we can shutdown gas and coal plants, but the nuclear plants go on generating, even in your 'off-peak' times because the power is still needed.

no you are wrong nuclear power stations don't run at night because the power is needed nuclear power stations only run at night because they can not shut down at night.
because of this all water storage power plants use this energy in the night.
but you lost energy because of this because you can not shut it down!

If the power wasn't being used, then this would lead to grid instability, which can cause blackouts and more serious problems with the grid, including damaging equipment.

thats not my argument my argument is you lost efficiency based on the fact that water storage power plants use this energy to recharge but you lost energy because of the efficiency of the recharge.

And it wasn't your bad english, how can you say "just my bad English i mean you have to calculate the 6% cable lost from the 2GW electric to" when you clearly state: "4,5GW(thermic)+2GW(electric) +radiation"

some stuff is in German the other way around.
and because of this i lost the line.
but in fact its more than 4,5GW Thermic per 2 GW electric and that is my point.

And how do you get "the 10Gigawatt heat to updraft the air in your "tower" Tornado" if you're now not talking about heat generated at the power station?

i just don't think about you can not use all heat in the updraft tower. for example you can not use the cable lost.
for an updraft tower your "number" is useful.

Oh, and the 6% cable loss will affect all large scale generation,

right because of this all large scale power plants should be against the LAW !
decentralized micro heat and power power plants are the best way to go 90%+ efficiency.

"Big" Power plants are just bullshit.

even with coal.. if you use a alphaKat to make diesel out of coal and burn it in decentralized water-diesel micro heat and power power plants your total efficiency will be much higher than a coal power plant.

a alphakat with coal do have 80% efficiency to make diesel and a water-diesel engine with solar leading fuel(water)heater do have ~60% 1*0,8*0,6= 0,48 electric and ~1*0,8*0,4=0,32 heat usage
48%+32%=~80% overall efficiency.
Micro power plants beat the "BIG" Coal plants in electric efficiency 48% vs 40% and in overall efficiency 80% .

right now we drive coal power plants with only 40% efficiency and i vote for it to make this against the LAW!

including the transfer cables from the north of Germany to the south, so it isn't just a nuclear problem, it will affect your precious windmills too.

Wrong this don't effect windmills because you lose the energy of the wind anyway if you don't use it with windmills.

without windmills you do have 100% energy lost

I had to laugh over your CO₂ contribution of the staff at the nuclear plant!

sure but in fact a nuclear power plant pollute CO2.. directly and indirectly.

What about the CO₂ produced by the servicing teams at the large windfarms? Off-shore wind farms are even worse because they require big ships to service them. However, using this as an argument that nuclear power stations produce CO₂ is laughable, just as it is laughable for wind farms.

sure thats right. but in fact Windmills save more CO2 that nuclear power plans. all statistic show this.

With respect to efficiency of nuclear fuel, have you heard of reprocessing?

You are kidding only dangerous old Uran and Plutonium based nuclear power reactors need this modern thorium high-temperature nuclear power plants burn it in 1 round without reprocessing!
Also reprocessed fuel MOX Mixed oxide fuel are very high Dangerous. This means reprocessing is just a stupid idea.

I'm not sure I understand what you are saying here. A nuclear plant efficiency is just as easy to calculate: power generated divided by heat generated. How complicated can it be?

You can only calculate in that way if you calculate the size of a upstream power plant tower.
If you calculate the food print overall its much more complex than that.

Particle radiation is what generates the heat in the reactor and the heat generated be spent fuel is a fraction of the heat generated at power. That's why spent fuel pools don't require as much cooling as the primary circuit. The "lost generation" you are talking about is less than 1%, if it was much greater, they'd be using the spent fuel as a power source!

wow in your words: another 1% lost.
and the hole nuclear waste after the usage pollute Particle radiation
WOW another efficiency lost based on the Fuel...

What I don't understand is Approach and Shutdown. If you are talking about power raising, then that is a day or so of a 24 month power cycle. Hardly significant. It's the same with Shutdown. When there is an outage, it generally lasts less than a month, which is 1/24th or just over 4% of the time spent generating. So, assuming a perfect 100% power output, we have a load factor of at least 95%.

wow another efficiency lost....

even in your numbers the efficiency is less than your CLAIM!