Integral Fast Reactor: Why Did Funding Stop?

[mheslep]

Perhaps we should hold other forms of power generation to the same standard with the added requirement that one has to look at the whole fuel cycle as well as the construction, operation and decommissioning of the plants. By that standard, wind, and biofuels might come out on top, nuclear would be a close second, followed by hydro and natural gas. At the very bottom and by far the most dangerous and unsafe would be coal (whether conventional or "clean" coal).

Mining coal kills thousands of people per year. According to http://www.minesandcommunities.org/article.php?a=1155".

Coal is the most unsafe form of power generation there is but we don't hear any politician saying they will only approve safe coal plants. There is no such thing.

AM

Point taken about coal. Yes Chernobyl had only 56 direct deaths, less than the single Ukraine mining accident. But to be complete: the IAEA report predicts 4000 additional cancers from those highly exposed, and >300,000 people had to be long term relocated.
http://www.iaea.org/Publications/Booklets/Chernobyl/chernobyl.pdf

 

[Morbius]

Point taken about coal. Yes Chernobyl had only 56 direct deaths, less than the single Ukraine mining accident. But to be complete: the IAEA report predicts 4000 additional cancers from those highly exposed, and >300,000 people had to be long term relocated.
http://www.iaea.org/Publications/Booklets/Chernobyl/chernobyl.pdf

mheslep,

I don't understand why anyone thinks that the accident at Cernobyl is pertinent at ALL
to the current discussion of present day nuclear power.

The Chernobyl accident is like the crash of the Hindenberg for the airline industry. If the
subject under discussion was the safety of travel by airliner and the Boeing 777 in particular;
what does the crash of the Hindenberg have to do with the safety of a Boeing 777.

The Boeing 777 looks nothing like the Hindenberg. The design features that caused the crash
and fire of the Hindenberg are not present in a Boeing 777. If someone said that the Hindenberg
experience was pertinent to the question of the safety of a Boeing 777; people would consider
them an absolute IDIOT, a DOLT, a FOOL, a complete MORON; and they'd be correct.

Likewise, the Chernobyl RBMK reactor looks NOTHING like a US or western European light water
power reactor or the Integral Fast Reactor. The design features that caused the Chernobyl accident
are not present in LWR reactors, nor the IFR. So why would one even bring Chernobyl into the discussion
of LWR or IFR safety.

Dr. Gregory Greenman
Physicist

 

[Morbius]

Perhaps we should hold other forms of power generation to the same standard with the added requirement that one has to look at the whole fuel cycle as well as the construction, operation and decommissioning of the plants.

Andrew,

Even wind turbines have their dangers. Anyone that lives near a wind farm like I do knows that.

However, recently the dangers of wind turbines have been caught on video. Catastrophic wind turbine
failure in Denmark was featured on a recent edition of Discovery Channel's "Destroyed in Seconds".
You can view the video here:

http://www.boingboing.net/2008/02/25/wind-turbine-self-de.html

On the Discovery Channel the video also included a news interview with a nearby resident who
recounted the horror she experienced as pieces of wind turbine flew over her house.

This occurance is not unique; just Google "wind turbine failure" for additional videos.

Dr. Gregory Greenman
Physicist

 

[mheslep]

...So why would one even bring Chernobyl into the discussion
of LWR or IFR safety...

I did not 'bring' it in.

 

[Morbius]

I did not 'bring' it in.

mheslep,

Good. I'm glad Chernobyl isn't part of this discussion.

It's not pertinent to the safety question of either LWRs nor the IFR.

Dr. Gregory Greenman
Physicist

 

[QuantumPion]

I read an interesting statistic somewhere that stated we have enough depleted uranium; byproduct of the enrichment process for LEU LWR's; to supply all of the country's electricity in fast reactors for something like 100 years. Without even having to dig up any new uranium reserves!

 

[RobertW]

QuantumPion:

There is enough light water reactor (LWR) waste and enough processed uranium in inventory to furnish all of the U.S. power needs for 1,000 years if the LWR waste is reprocessed for use in the integral fast reactors (IFRs) using the pyroprocess, and if all of the reprocessed waste and uranium are "burned" in IFRs. The idea that the immense amount of power needed 24/7 by the U.S. can be provided by solar cells and wind turbines is a hallucination. The IFR also offers the potential to end our foreign oil dependence problem by helping retrieve shale oil for about $25 to $30 per barrel.

RobertW

 

[mheslep]

QuantumPion:

There is enough light water reactor (LWR) waste and enough processed uranium in inventory to furnish all of the U.S. power needs for 1,000 years if the LWR waste is reprocessed for use in the integral fast reactors (IFRs) using the pyroprocess, and if all of the reprocessed waste and uranium are "burned" in IFRs. The idea that the immense amount of power needed 24/7 by the U.S. can be provided by solar cells and wind turbines is a hallucination. The IFR also offers the potential to end our foreign oil dependence problem by helping retrieve shale oil for about $25 to $30 per barrel.

RobertW

No argument except on solar and wind. Solar/wind could provide the required 1000GW in the U.S. with current technology, just not economically and not with enough regularity.

 

[aquitaine]

No argument except on solar and wind. Solar/wind could provide the required 1000GW in the U.S. with current technology, just not economically and not with enough regularity.

Several western European countries tried this, but the most they could get out of them was 15~20% of their power needs, and now several of them are having looming power shortages because they tried to replace nuclear with wind/solar.

 

[RobertW]

Several western European countries tried this, but the most they could get out of them was 15~20% of their power needs, and now several of them are having looming power shortages because they tried to replace nuclear with wind/solar.

aquitaine:

Can you please name some of the countries you are referring to and provide a reference to information about the countries "having looming power shortages"?

RobertW

 

[mheslep]

aquitaine:

Can you please name some of the countries you are referring to and provide a reference to information about the countries "having looming power shortages"?

RobertW

Two different issues really. Denmark has ~20% wind with backup pulled from Scandanavian hydro and nuclear. They seem to be ok, though they're straining their transmission grid at 20%. Other countries, Belgium I believe qualifies, perhaps Germany too, have foregone more nuclear and not stepped up with sufficient alternatives so far.

 

[Andrew Mason]

No argument except on solar and wind. Solar/wind could provide the required 1000GW in the U.S. with current technology, just not economically and not with enough regularity.

To supply the equivalent of 1000 1000 MWe power plants using solar cells (ie. which would replace one nuclear plant) how much area would you need to have covered in solar cells?

Here is a ball-park calculation:

solar irradiance: 1367 w/m^2 (this is the solar energy falling on the Earth's upper atmosphere divided by $\pi R_e^2$. Of course, you have to take into account that the Earth is rotating and you cannot capture solar energy at this rate for 24 hours. And even if there are no clouds, some of this energy does not make it to the earth. The average on a sunny day would be no more than a quarter of this or about 340 w/m^2. On average, 30% would be reflected by clouds, so this reduces it to about 240 w/m^2.

Solar cells convert sunlight to electricity at a rate of around 20% with present technology. So a 1m^2 high efficiency solar panel could supply about 50 watts of electricity, on average. To produce 1000 GWe, you would need twenty billion of these panels. One square kilometre is 1 million square metres, so you would need 20,000 square kilometres of panels. This is about 140 km x 140 km to supply the entire country's electricity needs. If one used roofs of buildings you would not need to use valuable land.

Of course you would want to distribute these panels around the country. If you had 1000 sites of 20 square kilometers each, you could do it. You would also need some way of storing the electricity. But it does seem workable. At a cost of $2 per watt, the cost of the solar panels would be $2 trillion dollars. The cost of installation and infrastructure might be another trillion. Given the way governments seem to be throwing money around these days, that almost seems cheap.

AM

 

[vanesch]

To supply the equivalent of 1000 1000 MWe power plants using solar cells (ie. which would replace one nuclear plant) how much area would you need to have covered in solar cells?

Here is a ball-park calculation:

solar irradiance: 1367 w/m^2 (this is the solar energy falling on the Earth's upper atmosphere divided by $\pi R_e^2$. Of course, you have to take into account that the Earth is rotating and you cannot capture solar energy at this rate for 24 hours. And even if there are no clouds, some of this energy does not make it to the earth. The average on a sunny day would be no more than a quarter of this or about 340 w/m^2. On average, 30% would be reflected by clouds, so this reduces it to about 240 w/m^2.

Solar cells convert sunlight to electricity at a rate of around 20% with present technology. So a 1m^2 high efficiency solar panel could supply about 50 watts of electricity, on average. To produce 1000 GWe, you would need twenty billion of these panels. One square kilometre is 1 million square metres, so you would need 20,000 square kilometres of panels. This is about 140 km x 140 km to supply the entire country's electricity needs. If one used roofs of buildings you would not need to use valuable land.

Of course you would want to distribute these panels around the country. If you had 1000 sites of 20 square kilometers each, you could do it. You would also need some way of storing the electricity. But it does seem workable. At a cost of $2 per watt, the cost of the solar panels would be $2 trillion dollars. The cost of installation and infrastructure might be another trillion. Given the way governments seem to be throwing money around these days, that almost seems cheap.

AM

Although this is in the right ballpark, there are some extra's. The first problem is that your price: $2 per watt, is per watt PEAK POWER. However, depending on your location, there is a ratio of about 1:3 to 1:6 between peak power and average power (due to variability in solar illumination, cloudiness, etc...). So although you are about right that you could have 50W per square meter on average (I think it is closer to 30 W with affordable technologies and in temperate lattitudes, but ok), you would need to install about 300W peak power per square meter to obtain 50W average.

The second problem is the storage of electricity. It plays on 3 levels:
- day/night
- cloudy week/sunny week
- summer/winter

Now, there are techniques such as storage pumping stations and so on, but these are not always geographically possible, have a certain price, and are limited in capacity. They could eventually take care of day/night (at serious extra cost of about $2 per watt BTW).
However, it would be more problematic to have the cloudy week/sunny week variability.
And the only solution to the summer/winter variation would be to *increase* the solar capacity such that you reach average consumption not averaged over a year, but averaged over the darkest winter period. Now, depending on location, that can be a factor of up to 4 or 5 (meaning, averaged over, say, a month, the solar intensity during the darkest winter month is 4 or 5 times smaller than averaged over the whole year, summer included).
This means that you have to over-design your solar power system by the same factor in order to provide still the average power needed during winter (and have an equivalent surplus production in summer, with which you cannot do much for the moment, but which you could eventually use to do things like hydrogen production or anything else).

You could of course say that you will find a "long term storage" technique, like production of hydrogen or whatever in the summer, so that you can use it in winter. But then you will run into efficiency problems, which will give you easily a similar factor. So in any case, if solar is to provide for the bulk of the electricity, you will have to over-design your system (and its surface, and its cost) with a factor of 3-5 as compared to "yearly average production" AND provide for serious extra buffer capacity for the short-time variations (day/night ; cloudy/sunny).

EDIT: btw, your $2 per watt peak are optimistic: http://www.solarbuzz.com/ModulePrices.htm

 

[mheslep]

To supply the equivalent of 1000 1000 MWe power plants using solar cells (ie. which would replace one nuclear plant) how much area would you need to have covered in solar cells?

Here is a ball-park calculation:

solar irradiance: 1367 w/m^2 (this is the solar energy falling on the Earth's upper atmosphere divided by $\pi R_e^2$. Of course, you have to take into account that the Earth is rotating and you cannot capture solar energy at this rate for 24 hours. And even if there are no clouds, some of this energy does not make it to the earth. The average on a sunny day would be no more than a quarter of this or about 340 w/m^2. On average, 30% would be reflected by clouds, so this reduces it to about 240 w/m^2.

Solar cells convert sunlight to electricity at a rate of around 20% with present technology. So a 1m^2 high efficiency solar panel could supply about 50 watts of electricity, on average. To produce 1000 GWe, you would need twenty billion of these panels. One square kilometre is 1 million square metres, so you would need 20,000 square kilometres of panels. This is about 140 km x 140 km to supply the entire country's electricity needs.

Hah, very good. The annual w/m^2 guess agrees with actual measurements for the SW US. Several other threads have gone down this road. You could have saved your self the trouble here:
http://en.wikipedia.org/wiki/Image:Solar_land_area.png
The blue dots are 18x the land area size for the required US power.

If one used roofs of buildings you would not need to use valuable land.

Of course you would want to distribute these panels around the country. If you had 1000 sites of 20 square kilometers each, you could do it. You would also need some way of storing the electricity.

Its not just an 'also', storage is the currently insurmountable problem for attempting 100% variable renewables. Vanesch and others have been good about articulating the problem. See some of the energy threads:
https://www.physicsforums.com/showpost.php?p=1787254&postcount=83

But it does seem workable. At a cost of $2 per watt, the cost of the solar panels would be $2 trillion dollars. The cost of installation and infrastructure might be another trillion. Given the way governments seem to be throwing money around these days, that almost seems cheap.
AM

Its more than that for the 20% panels, and that's installed panel rating, not average. A better figure to use is energy cost for PV which is $0.22/kWh for large scale PV installations for sunny climates, double that cost in cloudier climates. This is mostly an amortization of the capital costs. This still doesn't cover the required transmission infrastructure which must be built. If you go with smaller rooftop residential installations to avoid transmission the cost jumps to $0.40/kWh. Then we still don't have a cost for the cloudy day/overnight storage infrastructure, whatever that might turnout to be.
http://www.solarbuzz.com/solarprices.htm

 

[Andrew Mason]

Although this is in the right ballpark, there are some extra's. The first problem is that your price: $2 per watt, is per watt PEAK POWER. However, depending on your location, there is a ratio of about 1:3 to 1:6 between peak power and average power (due to variability in solar illumination, cloudiness, etc...). So although you are about right that you could have 50W per square meter on average (I think it is closer to 30 W with affordable technologies and in temperate lattitudes, but ok), you would need to install about 300W peak power per square meter to obtain 50W average.

50W average represents around 270W peak. In daytime with no clouds and overhead sun, the solar irradiation would be the full amount (1367 w/m^2 less a small amount that does not make it to the surface). So the output at 20% efficiency would provide 270 watts/m^2.

The second problem is the storage of electricity. It plays on 3 levels:
- day/night
- cloudy week/sunny week
- summer/winter

Now, there are techniques such as storage pumping stations and so on, but these are not always geographically possible, have a certain price, and are limited in capacity. They could eventually take care of day/night (at serious extra cost of about $2 per watt BTW).
However, it would be more problematic to have the cloudy week/sunny week variability.
And the only solution to the summer/winter variation would be to *increase* the solar capacity such that you reach average consumption not averaged over a year, but averaged over the darkest winter period. Now, depending on location, that can be a factor of up to 4 or 5 (meaning, averaged over, say, a month, the solar intensity during the darkest winter month is 4 or 5 times smaller than averaged over the whole year, summer included).
This means that you have to over-design your solar power system by the same factor in order to provide still the average power needed during winter (and have an equivalent surplus production in summer, with which you cannot do much for the moment, but which you could eventually use to do things like hydrogen production or anything else).

You could of course say that you will find a "long term storage" technique, like production of hydrogen or whatever in the summer, so that you can use it in winter. But then you will run into efficiency problems, which will give you easily a similar factor. So in any case, if solar is to provide for the bulk of the electricity, you will have to over-design your system (and its surface, and its cost) with a factor of 3-5 as compared to "yearly average production" AND provide for serious extra buffer capacity for the short-time variations (day/night ; cloudy/sunny).

These are all good points. It was a ball-park figure. The $1 trillion for infrastructure may be a little low. If you distribute the solar panels over a large geographic area and keep them in the lower latitudes in areas that have more sunny days, you can avoid many of these problems. We are talking about providing all of the power needs for the largest consuming nation on earth. No one is going to do that with one single technology, of course But the cost appears to be competitive with nuclear, and the fuel is free.

EDIT: btw, your $2 per watt peak are optimistic: http://www.solarbuzz.com/ModulePrices.htm

That is low. But with large demand, you may be amazed what kind of efficiencies and cost reductions might become available. I have used the $2 figure based on today's figures. Where I may be wrong here is in assuming that this represents the price for average wattage and not peak wattage.

AM

 

[vanesch]

50W average represents around 270W peak. In daytime with no clouds and overhead sun, the solar irradiation would be the full amount (1367 w/m^2 less a small amount that does not make it to the surface). So the output at 20% efficiency would provide 270 watts/m^2.

ok, my 300 W wasn't too far off, no ? But that means that per square meter, you have to spend, at $2,- per peak watt, $ 530,- and not $100,- to get your 50W average.

These are all good points. It was a ball-park figure. The $1 trillion for infrastructure may be a little low. If you distribute the solar panels over a large geographic area and keep them in the lower latitudes in areas that have more sunny days, you can avoid many of these problems.

Well, you can attenuate them: the peak/year average can be as low as 3 (while in moderate regions, this is rather 5-6), and the summer/winter variation is probably smaller too. But you will still have at least a factor 2 or 3 over simply "yearly average", if solar is to provide a *large fraction* of the provided power (say, 70% or so of consumption) in a reliable way.

As long as solar (or wind or other erratic renewables) is a minority contributor (say, 15% or 20%), then this doesn't play a role, and the price per KWhr delivered will be much lower, as we can just use "yearly average". What renders this expensive is when we need reliability (which is not needed when it plays in the 15% ballpark, because reliability is then provided by the other technologies).

We are talking about providing all of the power needs for the largest consuming nation on earth. No one is going to do that with one single technology, of course But the cost appears to be competitive with nuclear, and the fuel is free.

I really don't think that, even as a minority contribution, at actual prices, solar PV is competitive with nuclear (you have 25c/KWhr for solar, while this is ~8c/KWhr for nuclear/coal). But even then this comes about because solar is not providing for an essential function in power delivery: reliability and load following.

That is low. But with large demand, you may be amazed what kind of efficiencies and cost reductions might become available. I have used the $2 figure based on today's figures. Where I may be wrong here is in assuming that this represents the price for average wattage and not peak wattage.

Well, current average retail price per peak W is ~$4.5 or so. So assuming this to be $2,- is already assuming a serious drop in price (for instance upscaling). In moderate regions, you have to multiply this with 6, and in sunny southern regions, with 3 to go from peak to yearly average.

 

[mheslep]

No need to estimate the solar irradiation. It has been measured, year round, cloudy days and not. In the US it varies from a high annual average of 292 W/m^2 to 167 W/m^2 (flat plate collection), depending on location.
http://www.nrel.gov/gis/images/us_pv_annual_may2004.jpg
The peak power at high surface elevations and low latitudes is about 1100 W/m^2

 

[Andrew Mason]

:
As long as solar (or wind or other erratic renewables) is a minority contributor (say, 15% or 20%), then this doesn't play a role, and the price per KWhr delivered will be much lower, as we can just use "yearly average". What renders this expensive is when we need reliability (which is not needed when it plays in the 15% ballpark, because reliability is then provided by the other technologies).

Good point. If you were to use solar electricity to produce hydrogen by electrolysis (for use in cars, say) an average supply works just fine.

I really don't think that, even as a minority contribution, at actual prices, solar PV is competitive with nuclear (you have 25c/KWhr for solar, while this is ~8c/KWhr for nuclear/coal). But even then this comes about because solar is not providing for an essential function in power delivery: reliability and load following.

The capital cost of the IFR must be huge on a per watt basis. A 1 GW conventional nuclear plant will run about $4 billion. which is $4 per Watt. An IFR would be at least double, maybe triple that so $8 - $12 billion not including development costs. While an IFR if very efficient, it does use fuel and has a significant operating cost. Accepting that my figures for cost may be out by a factor of 6, that puts solar at today's prices about $12/watt. So it I appears to me that solar would be competitive. A big advantage of solar would be the speed at which it could be implemented.


AM

 

[mgb_phys]

The obvious use for Solar power in most of the US would be to directly power AC units.
No need for supply infrastructure, power storage or baseline/peak load handling.
When it's sunny it generates more power - which drives more AC.

 

[mheslep]

. A 1 GW conventional nuclear plant will run about $4 billion. which is $4 per Watt. An IFR would be at least double, maybe triple that so $8 - $12 billion not including development...

Andrew Mason what is your source for this cost (at least the conventional plant)? I am just interested.

 

[Azael]

Andrew Mason what is your source for this cost (at least the conventional plant)? I am just interested.

Curious about that aswell, especialy the IFR estimate. The russians claim they can build BN fast reactors for only 25% higher capital costs than LWR's. Saying a IFR would cost tripple a LWR makes no sense. The acctual reactor is a small part of the total power plant costs.

From "Economic potential of modular reactor nuclear power plants based on the Chinese HTR-PM project", Nuclear Engineering and Design 237 (2007) 2265–2274

The total costs of all the PWR plant
are normalized to 100. Among them, reactor plant equipments
account for about 23–28%, depending on ways of delivery. Turbine
plant equipments take up about 12% and BOP is about 3%.
These are so called direct costs. Other costs include the costs for
design, engineering service, project management and financial
costs, etc.

Considering the total costs of the above-classified reactor
plant equipments, the costs of the RPV and the reactor internals
account for about 9%, the reactor auxiliary systems for about
23% and the I&C and electrical systems for about 26%. Thus,
the costs of RPV and reactor internals, compared to the total
plant cost will be about 9%×23% = 2%. This shows clearly
that the RPV and the reactor internals of PWR-plants exhibit
only a very limited influence on the total plant cost.

Since turbines, electrical system etc would be more or less the same in a IFR I don't se how the reactor itself could tripple or even come close to dubbling the power plant costs.

 

[vanesch]

Good point. If you were to use solar electricity to produce hydrogen by electrolysis (for use in cars, say) an average supply works just fine.

That is correct. I never said that solar (or wind for that matter) have no use, on the contrary. But when the main goal is to have fossil-free electricity production, then my claim is that as of now, technologically it is difficult to conceive how these techniques are going to be the major player if we want to keep things economical to a reasonable extend. This is where nuclear has a unique "ecological niche" for the moment. All this can change the day that we have cheap, reliable batteries of some kind.

If you would use lead batteries to make a totally reliable system, let's make a small estimate.
A 100 Ahr 12V deep cycle battery costs about $100,- and lives for about 4-5 years.
Now, that's about 1KWhr of storage. Imagine you have a 1KW average system, then you'd need about 12 batteries (12 hours light, 12 hours darkness) to average this out. On 30 years lifetime, you'd need to replace them at least 6 times, so that's 72 of these batteries, or $7200,-. That adds $7.2 per average watt, just to smoothen out the day/night cycle.
We didn't yet include the cloudy week/sunny week problem or the summer/winter problem.

The capital cost of the IFR must be huge on a per watt basis. A 1 GW conventional nuclear plant will run about $4 billion. which is $4 per Watt. An IFR would be at least double, maybe triple that so $8 - $12 billion not including development costs.

I don't see why an IFR, once it is produced on large scale, should be significantly more expensive than a classical PWR. The main material effort in a classical PWR goes into the confinement building - I suppose, although I don't have sources, that it is also a serious part of the cost. The other part, the pressure vessel of a PWR, is actually heavier and more difficult to make than the reactor vessel of an IFR which doesn't work under 150 bars of pressure. Of course, an IFR will have sodium-related piping and pumping and so on, this will then make it somewhat more expensive, but I would guess that this would be a small effect on the overall effect of the reactor. Concerning the pyroprocessing part, this will cost something, but that is then offset by the fact that you don't need much transportation, nor that you need enrichment. Also, the waste is less long-lived and smaller in amount (volume), so this will reduce the costs at the backend.

So I would think that the bulk will cost about the same, that some things are a bit more expensive, and that other things are less expensive. I don't see why this should be so much more expensive in the end.

While an IFR if very efficient, it does use fuel and has a significant operating cost. Accepting that my figures for cost may be out by a factor of 6, that puts solar at today's prices about $12/watt. So it I appears to me that solar would be competitive. A big advantage of solar would be the speed at which it could be implemented.

Your factor of 6 is the peak/yearly average (and including already a serious cost cut). If we want to have winter average, that adds a factor of 3, and if you include batteries, that adds $7,-. So we are around $40,- per installed watt for a reliable solar system which has *about equivalent functionality* as a reactor or a coal fired plant.

Solar is nice, but it will remain for some time in niche applications.

 

[mheslep]

...If you would use lead batteries to make a totally reliable system, let's make a small estimate.
A 100 Ahr 12V deep cycle battery costs about $100,- and lives for about 4-5 years.
Now, that's about 1KWhr of storage. Imagine you have a 1KW average system, then you'd need about 12 batteries (12 hours light, 12 hours darkness) to average this out. On 30 years lifetime, you'd need to replace them at least 6 times, so that's 72 of these batteries, or $7200,-. That adds $7.2 per average watt, just to smoothen out the day/night cycle.
.

Better to calculate in energy terms as the power cost would indicate up front costs, and for up front one would only pay for one set of batteries at a time. In energy terms this example is about 6c/kWhr: 21900 kWhrs over 5 years for $1358 ($1200,5%, 5years). Actual grid based battery systems (flow batteries,etc) appear to cost about http://www.leonardo-energy.org/drupal/node/959" r but I expect they would last 10 years with temperature controls. Such a system would then cost ~16c/kWhr

 

[Andrew Mason]

Andrew Mason what is your source for this cost (at least the conventional plant)? I am just interested.

The Bruce Power study for Saskatchewan was just released last week. Bruce Power is a private company that owns and operates nuclear plants in Ontario and New Brunswick. It was asked by the Government of Saskatchewan to do a preliminary assessment of the feasibility of adding 2.2 GW of nuclear power in Saskatchewan. Their http://www.brucepower.com/uc/GetDocument.aspx?docid=2771" .

In their report, they concluded (p. 15) that a two unit facility comprising two 1.085 GW Candu ACR-1000s (PTR), 2 1 GW Westinghouse AP1000s (PWR) or two 1.6 GW Areva EPR (PWR) reactors would cost $8 - $10 billion. This does not include the infrastructure needed to distribute the power to markets.

AM

 

[mheslep]

The Bruce Power study for Saskatchewan was just released last week. Bruce Power is a private company that owns and operates nuclear plants in Ontario and New Brunswick. It was asked by the Government of Saskatchewan to do a preliminary assessment of the feasibility of adding 2.2 GW of nuclear power in Saskatchewan. Their http://www.brucepower.com/uc/GetDocument.aspx?docid=2771" .

In their report, they concluded (p. 15) that a two unit facility comprising two 1.085 GW Candu ACR-1000s (PTR), 2 1 GW Westinghouse AP1000s (PWR) or two 1.6 GW Areva EPR (PWR) reactors would cost $8 - $10 billion. This does not include the infrastructure needed to distribute the power to markets.

AM

Interesting breakdown page 15: 200,000 cubic meters concrete, 10,000 tons steel per reactor. Some time ago Vanesch and I estimated 70,000 cubic meters concrete, 29,000 tons steel for the EPR, in a comparison of materials costs between wind and nuclear for equivalent power (wind requires a lot more steel than nuclear). We only estimated the reactor plant, so I can see how we were light on concrete, but I don't see how we could have been heavy on the steel for the pressure vessel and containment building.
https://www.physicsforums.com/showpost.php?p=1729108&postcount=242

 

[vanesch]

Interesting breakdown page 15: 200,000 cubic meters concrete, 10,000 tons steel per reactor. Some time ago Vanesch and I estimated 70,000 cubic meters concrete, 29,000 tons steel for the EPR, in a comparison of materials costs between wind and nuclear for equivalent power (wind requires a lot more steel than nuclear). We only estimated the reactor plant, so I can see how we were light on concrete, but I don't see how we could have been heavy on the steel for the pressure vessel and containment building.
https://www.physicsforums.com/showpost.php?p=1729108&postcount=242

I think that in fact not all concrete has rebar steel in it. I have to say that I'm more puzzled by the large amount of concrete here: where do they put it ?? I wonder if they didn't confuse cubic meters and tonnes...

BTW, there is a funny typo in that document elsewhere where they say that the AP-1000 (a PWR) can run on enriched or natural uranium

 

[Astronuc]

I think that in fact not all concrete has rebar steel in it. I have to say that I'm more puzzled by the large amount of concrete here: where do they put it ?? I wonder if they didn't confuse cubic meters and tonnes...

The concrete supports and fills the volume between the rebar. The EPR has a double containment. The external structure will have a higher density of rebar, and internal structures less so. The external structures much survive impact loads, while internal structures provide support primarily, and both must be designed for appropriate seismic loads.

BTW, there is a funny typo in that document elsewhere where they say that the AP-1000 (a PWR) can run on enriched or natural uranium

Presumably, the AP-1000 fueled with natural U would be a PHWR. That's feasible.


The actual cost of the plant will fluctuate with the costs of materials and labor.

The economics of Russia and China are different from the US, Canada and Europe.


One thing to consider in an IFR plant is the reprocessing facility, which is not found at a conventional NPP. Back in the early days of the commercial nuclear industry, it was invisioned that typical LWRs would operate on annual cycles with reloads of about 1/3 of the core (some used reloads of 1/4 core) - so an entire core's worth of fuel would be used every 3 to 4 years - and the fuel would be reprocessed so that MOX fuel would be used. Alternatively, the fuel could be reprocessed and the MOX fuel used in a fast reactor, and possibly the Pu produced in the breeder would be used in MOX fuel in LWRs. Well all that changed - reprocessing was abandoned and all that spent fuel began accumulating.

To reduce usage, plants have gone to longer cycles 18-24 mo, and many now reload about 45-50% of the core. Burnups are on the order of 4-5% FIMA (40-50 GWd/tU) on a batch average basis, with local burnups pushing 6-6.5% FIMA (~60-65 GWd/tU).

 

[mheslep]

The South African reactors projects in play with either Areva or Westinghouse are dead for now.

JOHANNESBURG -- South Africa's state power company, which has been forced to ration electricity to mines and smelters, Friday shelved plans to build the country's second nuclear power station, saying it can't afford to make the investment...
The utility had been expected to make a decision by the end of the year between two 1,650-megawatt reactors proposed by a consortium led by French nuclear-engineering giant Areva SA and three 1,140-megawatt reactors to be built by a group led by Toshiba Corp.'s Westinghouse.

http://online.wsj.com/articl/SB122868998183686411.html

and the Flamanville EPR project announced its price going up 20%, delayed a year.
http://www.reuters.com/article/rbssIndustryMaterialsUtilitiesNews/idUSB29884120081202

 

[Astronuc]

South Africa's state power company, which has been forced to ration electricity to mines and smelters,

Interesting. This evening, I was talking with a former classmate, who is involved with independent power supply systems. We discussed the potential in Africa, and he mentioned that various mines and plants in S. Africa are royal pissed at ESKOM. Apparently the companies are looking a building their own independent power supplies.

Both EPRs, TVO-3 and now Flamanville, are delayed. NPPs are big capital projects, and the new ones are supposed to take about 5 yrs (60 months) to build.

 

[RobertW]

Good point. If you were to use solar electricity to produce hydrogen by electrolysis (for use in cars, say) an average supply works just fine.


The capital cost of the IFR must be huge on a per watt basis. A 1 GW conventional nuclear plant will run about $4 billion. which is $4 per Watt. An IFR would be at least double, maybe triple that so $8 - $12 billion not including development costs. While an IFR if very efficient, it does use fuel and has a significant operating cost. Accepting that my figures for cost may be out by a factor of 6, that puts solar at today's prices about $12/watt. So it I appears to me that solar would be competitive. A big advantage of solar would be the speed at which it could be implemented.


AM

The estimated cost for a one-GW(e) IFR, if they are mass produced, is $1.5 to $2.0 billion. IFRs do not require huge cooling water reservoirs or cooling towers. A modest water supply would be required if steam is to be the operating fluid. If supercritical gas is used as the operating fluid, then water would only be required for sanitation and pyroprocessing purposes. An IFR is fueled only once - when it is built. An IFR creates only 1,700 POUNDS of waste per year that will be "safe in 300 to 400 years as opposed to 17,000 TONS of waste per year for a PWR or LWR that will be safe in 10,000 to 200,000 years. The LWR and PWR must be refuled every three to five years - a very costly maintenance item; no refueling required for the IFR. Fuel for the LWR and PWR must be trucked to the reactor and spent fuel trucked from the reactor. IFR fuel is reprocessed on site - a major safety consideration.
The IFR is 99.5% efficient; the PWR and LWR are typically 3% to 5% efficient.
You can calculate, manipulate, cogitate, and rationalize all you want, there currently is no cheaper, safer, more efficient way to reliably meet all energy needs of the U.S. on a 24/7 basis than the IFR. Suggest reading the whole thread.

 

[vanesch]

The concrete supports and fills the volume between the rebar. The EPR has a double containment. The external structure will have a higher density of rebar, and internal structures less so. The external structures much survive impact loads, while internal structures provide support primarily, and both must be designed for appropriate seismic loads.

Yes, but the calculation mheslep and I did was a geometry calculation of the primary grade kind: the volume of the double containment building walls, simplified as cylinders, with wall thickness of 1.3 meters (times two, for the two buildings).

I will retake it here (I even think we made a mistake back then):

cylinder of 55 meters high, 48 meters diameter:

Surface of wall: pi x 48 m x 55 m = 8289 m^2

Surface of bottom = surface of roof: pi * (48m/2)^2 = 1808 m^2

Total surface = 8289 m^2 + 2 x 1808 m^2 = 11907 m^2

That, times the double thickness of 2.6 m gives us a total wall volume of:
30957 m^3

So 31 000 m^3.

How do they come at 200 000 m^3 of concrete, which is almost 7 times more ??

 

[vanesch]

Presumably, the AP-1000 fueled with natural U would be a PHWR. That's feasible.

I think it was more a kind of typo, as they said that the CANDU needs to work on enriched U...

 

[gmax137]

Yes, but the calculation mheslep and I did was a geometry calculation of the primary grade kind: the volume of the double containment building walls, simplified as cylinders, with wall thickness of 1.3 meters (times two, for the two buildings).

I will retake it here (I even think we made a mistake back then):

cylinder of 55 meters high, 48 meters diameter:

Surface of wall: pi x 48 m x 55 m = 8289 m^2

Surface of bottom = surface of roof: pi * (48m/2)^2 = 1808 m^2

Total surface = 8289 m^2 + 2 x 1808 m^2 = 11907 m^2

That, times the double thickness of 2.6 m gives us a total wall volume of:
30957 m^3

So 31 000 m^3.

How do they come at 200 000 m^3 of concrete, which is almost 7 times more ??

I don't know the details, but the source says 200,000 m3 concrete for a two unit plant. Your calc of the containment concrete neglects any internal walls & floors within the containment; perhaps more importantly it neglects the auxiliary building, the tubine building, the intake structures and or cooling towers, the switchyard, the maintenence and admin buildings, parking lots, security structures etc etc...

 

[Andrew Mason]

I think it was more a kind of typo, as they said that the CANDU needs to work on enriched U...

Not a typo. This Candu is designed to use lightly enriched uranium.

AM

 

[mheslep]

The estimated cost for a one-GW(e) IFR, if they are mass produced, is $1.5 to $2.0 billion. ...

Source?