TO ROBERT A COOKgreenspun.com : LUSENET : TimeBomb 2000 (Y2000) : One Thread
On the issue of how long it takes to cool a reactor: You were right on the money. If anyone out there is spewing out my 4 month ranttings, I was WRONG. MR. COOK was RIGHT. Sorry to all!!!!!!!
-- Scotty (BLehman202@aol.com), March 10, 1999
and the answer is ...?
-- argh (email@example.com), March 10, 1999.
Well Scotty, that was by far "THE" dumbest message for every one to read!!!!! Many new readers don't know how long it takes to cool a nuclear core. There will be hundreds reading your post and will be asking themselves the question..... Well, how long does it really take??????? Try in the future to post more meaningfull posts! OK????
oR, I'Ll siCK DieTeR On YoU!!!!! YoU HyEnA!!! GoT ThAT?????
-- Whytokay (firstname.lastname@example.org), March 10, 1999.
Robert A. Cook has my respect! I never glance over his input. His stuff is really well worth reading!!! He probably is the smartest and most intelligent contributor to this Y2K forum! My hat is off to you Robert!! Keep up the good work!!!
-- Freddie the Freeloader (email@example.com), March 10, 1999.
When a fission reactor is shut down rapidly, as in a SCRAM, control rods are dropped into the core (failsafe via gravity) and absorb sufficient neutrons to bring the core reactivity below criticality. The reaction rate drops exponentially in less than 1 second, thereby reducing the power level to less than 1% of the normal full power output (e.g. from 3000 MWthermal to 30 MW thermal). The remaining heat generated by the core after the fission reaction is shut off is only due to radioactive decay (alpha, beta, and gamma) of unstable isotopes. Since there are many different radio-isotopes from the reaction products of fissioning U235 and PU239, there are correspondingly many different half-lives. The aggregate decay curves for the whole core depends, of course, on the actual burnup of the fuel, and, hence, on the exact mix of radio-isotopes. The total heat generated by the shut-down core decays over time; more quickly at first, and then more slowly later on. Standard textbooks by Duderstadt, or Lamarsh, have graphs showing these power decay curves. It is quite possible that the residual decay heating of the core could decline from 1% to 0.5% after 4 months or so. Obviously, the cooling required to cool only 15-30 MW of heat generation is vastly lower than for full power operation (3000 MW). Some reactor designs can cool the residual decay heat by natural convection only, not requiring any external cooling.
Robert D. Watson, Ph.
-- Bob Watson (firstname.lastname@example.org), March 10, 1999.
Well, Robert, in laymans terms, how long does it take??? Still puzzled!
-- Alisa (Alisa@y2k.com), March 10, 1999.
The actual cooldown (from operating conditions to room temperature) is slightly over 2 days. ( I've done it in just acouple of shifts on smaller plants, so allof these are nominal times.)
This usually requires outside power (from the grid) or local power (from the diesel gen. sets on the site.) This shutdown-cooldown process is repeated very often over the plant's lifecycle - and is the first step towards removal of the core pressure vessel cover for refueling. Very routine, very well covered in plant procedures. The vessel sits in an open "pool" up to 90+ feet deep, with room on both sides to remove and place core internals and steam separators, guide tubes, and other "stuff" in the way. (Actual arrangement depends on design, each plant will be a bit different - most especially between a boiling water reactor and a pressurized water reactor. Realistically, you don't probably care about the design details.)
To refuel the reactor core, the vessel "cover" must be out of the way: its studs are "un-tensioned", the nuts removed, and then the head pressure vessel head itself removed - the fuel bundles are then exposed and can be removed, replaced, re-arranged, or swapped with bundles from the storage pool. Water temperature will vary during this time - you can assume anything between 80 and 120 degrees and not be wrong. The operators want to keep the water as cool as possible - after all, they are on tight schedule and it is easier to work if the area is cooler. Small auxiliary heat exchangers and pumps - each with different backups and power supplies - are used for this.
Once the "plant" is at room temperature - you have to consider decay heat removal from two places: the fuel in core itself, and from the spent fuel, usually stored in a separate building in an open "swimming" type pool. Sizes and configurations vary - assume 120 feet by 50 feet deep by 60 feet wide. (Fuel bundles - long square assemblies of small tubes containing the fuel pellets themselves - about 30+ feet long, more or less 12" x 12" across, are moved from the reactor to the spent fuel pool underwater via deep channels and transfer "locks" by trolleys and specially certified cranes.)
So now you have spent fuel in two places: a relatively small amount that is generating heat in the very large water pool around the reactor itself that has been recently radiated. And much more (maybe up 3 to 4 times as much) that has been shut down (removed from the neutron flux) a much longer period ago over in the spent fuel pool. Both have to be kept cool - which can be easily done by ensuring they are simply kept covered with water. If they are covered, you can be guaranteed that they cannot overheat no matter what else happens: maximum water temperature possible is 212 F, normal operating is several hundred degrees higher than that. (Now, operating rules require that various backups and redundant cooling systems are always monitored, tested and kept available - the intent is never to let the cooling systems stop, much less let the water temperature rise, but the interest is in extreme measures - so I'll talk about the extremes - knowing the operators aren't going to let them get that hot.
-- Robert A. Cook, P.E. (Kennesaw, GA) (email@example.com), March 10, 1999.
Robert! That was fantastic! My hat is off to you too! All of us at the Forum thank you!
-- Freddie the Freeloader (firstname.lastname@example.org), March 10, 1999.
Well, I'll try again.
The radioactive core will stay "hot" for thousands of years, some elements such as plutonium, have half-lives longer than 10,000 years.
But, that's not what is important. What is important is the rate of heat generation vs time. As some point the core has decayed enough so that active cooling is not required. This would be the "passively safe" time, e.g. by natural convection.
I don't know what the passively safe time is. That depends on the reactor design, core size, etc.
Right after the core is shut down, the decay heating rate is at the highest and the need to actively cool the core is greatest.
-- Bob Watson (email@example.com), March 10, 1999.
Don't meddle in the affairs of dragons, for thou art crunchy and taste good.
Please pass the salt and pepper to Robert and Robert.
-- Dilbert (firstname.lastname@example.org), March 11, 1999.
Okay, I'll bite. If the above is the case, then who originally spread the nasty rumour that the nuclear plants had to be shut down in July if they weren't compliant by then? And why?
-- puzzled (email@example.com), March 11, 1999.
To: WHYTOKAY and anyone else that is to lazy to research this forum. It should be clear to anyone with half a brain that this was a topic on this forum. But again to all sorry. 2 to 3 days. Scotty
-- Scotty (BLehman202@aol.com), March 11, 1999.
We first read about reactors being shut down early on Gary North's site. I think it was a posting of an email he received. The writer's name was not posted. We have real hope, thanks to Robert Cook's explanations. (Thank you SIR!)
-- Helen (firstname.lastname@example.org), March 11, 1999.
Puzzled - The NRC is requiring that nukes show by July 1st that they are (or will be) compliant by the rollover. Gary's post says he was confused about the July 1st date until he "learned" that it would take 4-6 months to cool a nuclear plant, and assumed from that the NRC would have to decide by July 1st whether the plants must be shut down before the rollover so that they would have the benefit of external power. In the hope that I am not misrepresenting, I believe what we have learned on the Yourdon forum is that it can take 4-6 months to *decommission* a nuke, but the time required to safely cool the plant is far, far less than that. Assuming that cooling is something that can be handled by emergency generator (and I believe it is from what Roberts, et al have told us), then Gary's reason to shut the plants down before rollover goes away.
-- Brooks (email@example.com), March 11, 1999.
Continuing Dr. Watson's comments from above - you have to mentally segregate several things that are all going on at the same time, all at different rates, but all related and all having different effects on different things at different times - and the terminology (unfortunately) gets in the way. Particularly if the apparently clear "English" terms like "hot" get mixed up between the nuclear physics, the radiation health, and the heat transfer reactions.
Oh well, be patient with me - I'll try to make it more confusing by getting everybody look at the basics. Otherwise, when I claim someone can be both completely correct and dead wrong at the same time, it won't make sense.
First, think of heat only in terms of energy - if something moves, it has more energy, it is hotter. Mentally, think of radiation as an atom emitting smaller particles. The emitter may be at any temperature - it doesn't matter. The emiitted particle will be moving fast - obviously, so it will be physically "hot" - and so it can heat up (collide) with something else.
The physical effect of getting "hot" is completely different than the effect on your body - you can get radiation damage from particles hitting your skin that don't icrease the temperature of your skin. The radiation damage comes when a particle hits a cell and "breaks" it. Since that happens at a cell by cell level - the increase in energy caused by radiation can't be measured by a thermometer. There simply isn't enough energy to measure that way!
The following is a bit simplified, but it will work. As long as the emitted particle can't touch you - you are safe. The emitted particle can be one of three kinds: Alpha (an He nucleus) can't go through a sheet of paper. Beta (a high-speed electron) can't through a telephone book. Gamma (Xray) are usually a bit stronger (in general) - you may need up to 1-2 feet of concrete or 4-5 inches of steel (mass counts here - stand behind a fat guy!) to stop them depending on it energy. (Or several dozen feet of air.) But if the source particle remains in the power plant - it's emitted particles can't touch you - they can't harm you.
You may have heard about cosmic radiation - those are gamma rays emitted from other stars coming here. They (like any other particle) can get stopped by hitting the H2O, N2 and O2 in the atmosphere. Less atmosphere above you - less shielding between you and the source (the universe) and so more exposure. That's why pilots and flight crews tend to get more radiation than many nuclear-qualified workers. The nuclear workers, under the containment dome, have more shielding from a constant but minor source of radiation exposure. Some one living in CO will get more average exposure than someone at sea level on a concrete slab house - particularly if their basement in CO is dug into rock containing natural uranium or thorium.
So, we want to keep the source particles contained so they can't leave the power plant. (Same thing happens with a nuclear weapon - best way to contain it is not to explode the thing in the first place, or to explode them underground.)
Start with fission - because that begins the problem. A wandering neutron comes by from some previous fission, bouncing off of things like water molecules and the fuel. It loses energy slowly with each collision- imagine a bowling ball trying to roll through a tennis court covered with tennis balls. The tennis balls are going to move (speed up = heat up), the bowling ball is going to slow down (cool off). Eventually, the bowling ball will be slow enough to get captured by the U235 atom - if it does get captured, it has a certain chance of causing the U235 to split in half - this is the fission reaction. The split pieces (daughter products) are literally junk - very high speed atoms (= very hot atoms). We capture these newly created atoms inside the pelets inside the fuel rods so the new atoms themselves are contained by the pellets and rods- but this means the fuel rods get hot. (During regular critical operation, the water passing by removed this heat - created steam - turned turbine - makes electricity.) Almost all (97-98%) of the energy from fission is released immediately as physical energy (motion) of these new atoms. (Remember, this all relates to typical US designs; the Soviets used entirely different nuclear physics methods and cooling (and no containment) so they could produce Pu239 at Chernobyl and its sister reacotrs.)
Fine - we aren't critical now -as the good Dr. pointed out - after shutdown, the neutrons are absorbed by the control rods, so there are no wandering neutrons (after a few milliseconds) to be captured by the U235, there can be no more fissions. Once fission stops, only the remaining 2.5-3% continues to be generated - by some complicated internal processes happening automatically in the daughter products.
But - those d**n daughter products are still decaying after shutdown. And they are still producing a little bit of heat with each decay.
See, every one of the daughter products are themselves unstable (radioactive) and will themselves decay at different rates emitting more flavors of alpha, beta, and gamma particles at various energies - but they will not "split" any more. After each radiactive decay, a new atom results, which is itself radiactive, and will itself decay at different rates. While there is tremedous amount of energy (heat) in a moving daughter product immediately after fission, there is very, very little physical energy (heat) after radioactive decay because the emitted particles weigh less. (Unfortunately, adding lots of "little" heat energy does add up eventually to non-neglible amounts.)
It's the behavior and amounts of this non-neglible decay heat that causes future problems - if you don't treat it with the proper care. (The biggest advantage of fusion = no fission daughter products. The biggest disadvantage of fusion = we can't make a controlled fusion reaction yet.)
-- Robert A. Cook, P.E. (Kennesaw, GA) (firstname.lastname@example.org), March 11, 1999.
Here is a question for you. Just how many nuclear plants are located in Kennesaw GA? Do you drive to TN or southeastern GA every day?
-- (Q4U@gotta.know), March 11, 1999.
Robert E. Cook said: "It's the behavior and amounts of this non- neglible decay heat that causes future problems - if you don't treat it with the proper care." Robert, could you expand on those future problems and what the proper care would be for both spent fuel rods in a cooling pool as well as those rods in the containment chamber, especially vis a vis a loss of offsite power? Thanks.
-- argh (email@example.com), March 11, 1999.
Good question... How much power does it take to cool the core of a fission reactor after its been shutdown?
Let's assume that the operating power level was 3000 MW. Then, after shutdown, the decay heat generation rate might be 1% of that, or 30 MW.
It is well known that using forced convection cooling with water that the power required to run the pumps to circulate the water is about 1% of the heat that the water is cooling (because flowing water is a very effective coolant). So, 1% of 30 MW would be 0.3 MW, or 300 KW.
Assuming some inefficiencies, I could estimate that the backup diesel generators would be required to provide about 500 kW of electrial power to run the pumps to cool the core.
Over time, e.g. many months, this requirement decreases as the radioactivity of the core decays away.
For spent fuel rods which are stored in a cooling pool (e.g. swimming pool), the cooling requirements are less than for the core because the fuel rods/bundles are spread out in a much larger cooling pool than the fuel rods/bundles that are very closely packed inside the pressure vessel of the core. I'd estimate that the spent fuel cooling pond would require 1/5 to 1/2 of the pumping power for cooling that the shutdown core requires.
In summary, on the order of 500-1000 KW of backup generator capacity might be needed to cool the reactor site, for a 3000 MW size reactor.
I have no doubt that the existing power plants have more than enough backup generator capacity. The important question is how much fuel do they have to run these generators. If they only have 1 week of fuel, and the Grid is down for 2 weeks, then the core could easily overheat and melt over the time period of many hours
-- Bob Watson (firstname.lastname@example.org), March 11, 1999.
thanks, Robert. Also have to worry about mechanical failure of back- up generators aw well. argh
-- argh (email@example.com), March 11, 1999.
Okay - give me a few minutes to run some numbers - more later.
-- Robert A. Cook, P.E. (Kennesaw, GA) (firstname.lastname@example.org), March 11, 1999.
Now wait a minute, you are not going to have core cladding failure if you keep it under water right? So if the Super Criticality Reactor Ax Man rains on your parade and the poles go in the holes the plant is ok if you keep the core watered. But I seem to remember you need to keep the cooling water flowing for years!!! You must forgive me, I was a NUC many years ago and I've slept since then. As I remember it, it'll take at least 2 years to cool a plant which was at power for several years down to the point where you can remove the core without fear of failure.
But I'm just an old navy nuc, one of rickovers white nazi's out killing commies for christ.(Or SO WE SAID).
-- nine (email@example.com), March 12, 1999.
I posted this on one of the thousand other recent nuke threads, but here goes anyway...
My pet peve with nuke plants, besides the waste, is that the building of them seemed to assume that the society in which they're built will last forever...that there will never be any disasters or upheavals that would render them very dangerous. But that is precisely what we face: if not in USA then certainly in many of the other countries that have nukes. The safety of Russian plants, for example, has plummetted since their economic collapse, it is only by the Grace of.....(insert favourite deity here) that they haven't had a chernobyl repeat. The same will be the case as y2k screws the economies and social fabric of all them other countries. Are you feeling confident about Indian nuclear plants? Iranian ones? Lithuanian ones? Bulgarian ones? If we have a serious TEOTWAWKI rather than a mild one then these plants promise us disease and death, even though they probably seemed like a good idea at the time.
-- humptydumpty (firstname.lastname@example.org), March 12, 1999.
See new thread: The Nation magazine: The "Nuclear Nightmare"
-- (email@example.com), March 12, 1999.
Oops! That should be: The Nation magazine: The "Nightmare Scenario"
-- (firstname.lastname@example.org), March 12, 1999.
You're probably right, it would take a few years for the entire core to cool sufficiently to be able to be safely removed from the pressure vessel. But, that's really a maintenance issue, not a critical safety issue, such as a core meltdown.
-- Bob Watson (email@example.com), March 13, 1999.
In Doug Carmichael's current newsletter (Week 43), Mark Frautch=schi writes (in part):
"I spoke with Jared Wermiel of the NRC in the corridor about some of the basic physics of nuclear power plants and on site waste storage. Nuclear power requires off site power to cool the core and the stored spent fuel should the plant go black. If the grid is lost, the plant is required to shut down.
"Jared gave me a few basic facts about the operation of nuclear power plants.
"The core of a typical megawatt capacity power plant releases in the neighborhood of 6 megawatts of heat during normal operation. In the event of a failure, the plant SCRAMs, which means that the control rods fall (by gravity) into the core, among a number of other actions. The heat production drops exponentially, such that the thermal output within a few hours. Within a day or two, the residual heat production is on par with that required to keep the spent fuel cool, tenths of a megawatt, that is, one hundred, or so, thousand watts of heat to dissipate.
"To remove this waste heat, recirculation pumps are required. Since it is an open system (with the spent fuel) evaporation losses must be replenished. The power to run these pumps is a fraction of the residual losses. These pumps draw on the order of 2000 gallons per minute; let me call that 8000 liters in 60 seconds or 133 l/s. If I raise that 133 kg of water 50 meters (against gravity, g = 9.8 m/s^2) we have about 6,500 watts to operate a perfectly efficient pump. So call that 10,000 watts. That is in the realm of ordinary generators. You could buy three and use one for backup and one for maintenance. Say you lose all three, the water temperature slowly will rise and boil. (It will boil off in the case of the fuel - taking heat by vaporization and transport.) It will take time. Wermiel indicated it was the order of days. (If we knew the size of the storage vessel, we could estimate it."
A sufficient supply of water is needed, to replace evaporation losses.
-- Squire (Trelawney@AdmiralBenbow.com), March 13, 1999.
The following should resolve some the comments made above:
Let us continue with a topic of apparent great interest to many: given a heat source and water, what happens to water when it is heated, how long does it take to boil water, and what can you do to keep water from boiling away?
Disclaimer - the following numbers and values are simplified estimates and approximations for informal explanations and "internet-level" training purposes only to non-technical readers. Contact your local power plant or NRC office for their exact values and emergency procedures relative to your power plant. These values are intended to serve only as example of what could happen to an "unattended" nominal fuel pool under the stated conditions to serve as an example to laymen of certain emergency responses. Any comments or recommendations are welcomed, provided a text reference and specific formula or numerical value is included, and will be incorporated - if the change enhances the learning situation without further confusing a complicated situation. The user is invited to make his own calculations and assumptions, and is responsible for the results of those assumptions.
To resume: Much concern and misinformation has been expressed about the stored fuel in a shutdown power plant. Consider the following example to see what is involved.
Previously, it was proved based on personal observation that after reactor shutdown, the nuclear fuel rapidly can be cooled down in 2-3 days and the core safely exposed for work in the reactor vessel. This process is regularly done for refueling and reactor servicing. We may discuss separately the exact numbers for decay heat generation and fuel storage policies, suffice it to say here that at almost all US power plants there is more water and less decay heat in the reactor core in the pressure vessel than in the fuel pool. A fuel storage pool is typically smaller than the reactor service pool, and also has more fuel assemblies in it - you can reasonably assume two to four core "loads" in the spent fuel pool, and that each core load in the pool will have different power history and "time since shutdown". (Obviously, there is room for only one core load in the core itself - that's the definition of a "core load".)
Some fuel bundles may have been in storage for several years, others only a few weeks. Each fuel bundle also will have different thermal neutron exposure based on its previous core positions and history, and thus a slightly different decay heat generation pattern. Every fuel manager tracks the location and power history of his or her fuel bundles, and is responsible for their detailed history. We will assume averages here - as if the whole core was identically "exposed" to the thermal neutron flux. Therefore, if the "hottest" part of the "hottest" fuel bundle from highest heat load can be shown to be protected, then the colder parts of the colder bundles will also be protected.
Routine cooling (and the preferred emergency cooling process) is very simple: it involves a power supply and two small pumps. One to circulate water from the service water system to a small heat exchanger, the second pump to circulate water through that heat exchanger to the heat exchangers in the pool back to the service water system heat exchanger. All these pumps need to do is circulate water through an existing network of pipes: the throttle valves can be set manually and then essentially left alone. All the operator need do is record the various temperatures once an hour in her logbook, reading from analog thermometers already placed in the piping during construction. Simple and routine. The power required to size these pumps is easy to calculate and is of no particular interest here since it is unique to the size of pipes, number of bends, length of pipe, temperature of the water, flow rate needed, height of the heat exchangers and all pipes in the network, size and number and style of valves and fittings, type of steel and surface roughness of the pipes, type of joint and cleanliness in the pipes and fittings, etc. (This is a routine design calculation for a junior or senior fluid dynamics or thermodynamics class.) Once the pumps are running, the pool will not heat up. If the pool has already heated up, the heat exchangers are sized large enough to cool it down again, even under worst case conditions of highest possible fuel loading and poorest thermal conditions.
Power plant design requires both pumps have installed backup pumps, and that both sets of pumps be provided independent and redundant backup power supplies from independent sources. The power required by these pumps under shutdown conditions a few days after scram and plant cooldown is very small (perhaps as little as 15-25 HP) compared to what is needed for all the pumps in the rest of the plant under regular operating conditions. Under emergency conditions, any 440 V power supply could be used. If power to these two pumps is available, even if it is available only intermittently, the fuel pool will not, under any conditions, heat up. With respect to Year 2000 concerns, there are no automatic controls or moving parts in the system other than hot and cold water - given any source of power, the pumps will run and the heat exchangers will do their job unattended and without automatic controllers or processors.
Pure water boils at 212 F at sea level. The water in a fuel pool carries a lot of boron in solution that actually raises the boiling point slightly - we will neglect this in the following. The middle of a fuel pellet in a nominal 1/2" diameter fuel rod can be assumed to be about 15 - 25 degrees hotter than its surface - so in a boiling fuel pool at 212 degrees, the middle of the hottest fuel rod can be no hotter than 240 degrees. Regular "center of rod" operating conditions while at 100% power are up to or slightly higher than 500 degrees - so, at 240 degrees, the center of the hottest possible rod in the fuel pool is safe (far cooler than normal operating temperatures, much less design thermal conditions) as long as it is covered with water at atmospheric pressure. (Actually, depending on heat rate, you could even expose some of fuel bundles to air and rely on air circulation to cool the rods. But that won't work in all cases, and many of the bundles do require water cooling for a long period after removal from the core, so we can will make the conservative assumption and require the bundles be kept covered.) The standing water over the top of the bundles also serves as a radiation shield to the nuclear workers nearby, and they will appreciate the reduced exposure from the shielding.
Thus, if we can show what is needed to keep the fuel in the fuel pool safe (that is, kept covered with water), we can establish that any fuel remaining in the reactor also can be kept cool (and thus safe) under the postulated conditions of "non-attended" no-power circumstances. Further, since the heat generated in the fuel rods from decay of fission daughter products is constantly decreasing, if we can show that at some point in time, there is too little heat being generated to make up for ambient losses to the surrounding air and water, then we can show a degree of safety for unattended long-term storage - such as at Yucca Flats or other permanent "dry" contained storage not requiring water at all.
To check the following numbers and assumptions, you will need any good collegiate-level advanced thermodynamics text, a equivalent heat transfer textbook, and any one of several good standard engineering reference texts that covers nuclear power plant design and construction - see your local college bookstore to order a copy. A certified four-year engineering or technical college library should have one or more these in its technical documentation section - regular library, liberal arts or teacher-training college that I have checked do not carry them in their shelves or reference sections, but these places might be able to order one or more copies from a more-qualified library.
Assume the fuel pool is rectangular: 40 ft x 100 ft x 42 ft deep. If so, the surface area of the pool is 4000 sq. ft, the mass of water = 10.80 10^6 lbm, the heat transfer area of water (through the sides and bottom to the concrete (conc.)) is 15,760 sq. ft. In one ft of water in the pool, there is 256,000 lbm of water at STP. Cp of water assumed = 1.0 BTU/lbm-deg, Cp of conc. assumed 0.156 BTU/lbm-deg. (Volume of water displaced by the fuel bundles is neglected in this simplified analysis, but should be included in more refined calculations. Thermal expansion of water as temperature rises is also neglected, but could be included in the calculation if desired.)
Temp initial of water = 82 deg, Temp final of concrete = 65 deg ( this represents the heat sink of the "infinite" mass of concrete and dirt under the pool), thickness of conc. = 5 ft, H conc. ( for thermal conduction) = 1.05 BTU/(hr-ft^2-deg). At 212 F, enthalpy of water = 180.15 BTU/lbm, enthalpy (evaporate) = 970.4 BTU/lbm, enthalpy of steam = 1150.5 BTU/lbm. (Simplified somewhat, these numbers mean it only takes 1 BTU of energy to heat a pound of water from 82 degrees to 83 degrees, and approximately 1 BTU to heat one lbm of water from 211 degrees to 212 degrees. However, essentially translated, this means it also takes an additional 970.4 BTU's of energy to boil 1 pound of water at 212 degrees into steam at 212 degrees.)
Heat from the fuel rods will heat the water in the pool - which will lose heat several ways: by conduction through the concrete to the infinite mass of "concrete and dirt" under the pool, by conduction to the air above the pool (neglected), and by increasing temperature of the water in the pool.
The heat loss by conduction at any given time will be proportional to the current temperature of the water and the current temperature of the concrete, rock, and dirt that it touches, area of the heat exchange surface, thickness of concrete, thermal conduction and thermal retention factors in the concrete and fuel pool liner, etc. Under steady state conditions at 212 degrees, heat loss to the concrete is about (area pool walls x H conc. x (T water - T conc.) or 2.43 x 10^6 BTU per hr. (This assumes steady state conditions, and neglects the water-concrete and concrete-dirt thermal film coefficients, and a few other "minor" things of interest in an exact analysis, but this number is adequate as a first level approximation.) Thus, if the heat load from the stored fuel is less than 2.43 x 10^6 BTU, the water in the pool will stabilize at some temperature less than 212 degrees, and you need only replace the water lost by evaporation to prevent exposure of the fuel rods. Since the heat load created by the decay of daughter products in the spent fuel is constantly reducing over time, the fuel pool can be considered "safe" in all future conditions as long as the evaporated water is replaced.
So, one problem is solved, at low heat rates, the pool won't heat up high enough to boil water. We only need to replace the water lost to evaporation. And, if any source of power is available, we can drive two pumps and ensue there is no temperature increase at all, regardless of heat rates.
Now, let us see what happens with no power and higher heat rates.
Let us assume a heat load 6 times higher: 15 x 10^6 BTU per hour. As shown above, the water will eventually heat up to 212 degrees, and thermal losses will account for the same constant 2.43 x 10^6 BTU/hour of losses to the surrounding concrete once temperature = 212 F. Since the remaining heat energy (net heat load = 12.6 x 10^6 BTU/hr) will all go towards heating water, it is useful to know how long will it take to heat the water - first to boiling, then past boiling as water level goes down. (Any different heat load and water volume will change these predicted values, the user should feel free to make his own assumptions, and is responsible for justifying his assumptions, and for interpreting the results of his assumptions.)
Time to reach 212 = (Mass wtr x Cp wtr x(T final wtr - T initial wtr) /(net heat load) or 111.22 hours for these assumed conditions; a little over 4 days at 1.16 degrees per hour. Doubling the heat rate will cut this time in half to about 60 hours, reducing this assumed heat load by half will approximately double the time to reach 212 degrees to about 10 days. (This equation assumes steady state conditions, and of course, things aren't steady state if temperature is increasing. Regardless it is obvious that the pool will reach 212 degrees at some point in time after power is secured to the cooling pumps.) Note also that the thermal mass of the concrete needs to be heated up (the steady state heat loss through conduction of the pool walls assumes that the concrete is already at 212 degrees), and the thermal mass of the "cooler fuel bundles" needs to be heated up as well to the water temperature of 212 degrees. Both of these factors will slightly increase the time required, and can be calculated with some very intricate logarithmic equations if the student is interested.
Regardless of simplifying factors, now that we can reasonably expect to reach a point where the water can be shown to be boiling, what happens next? In other words, is this a catastrophe in the making or simply a problem that needs to be solved?
By the original definition of a BTU back in the 1800's, it takes 1 BTU of energy to raise 1 lbm of water one degree F. However, most people don't understand it takes an additional 970.4 BTU's of energy to boil 1 lbm of water at 212 F at atmospheric pressure (14.7 psia). In our case, we are worried about boiling the water in a "swimming pool" of surface area 4000 sq. ft. Thus it takes 248 x 10^6 BTU's of additional energy to boil 1 foot of water from the top of the fuel pool, or 19.77 hours at a net heat load of 12.6 x 10^6 BTU/hour. If the pool is 42 feet deep, the top of the fuel bundles themselves will not be exposed for several more weeks at this rate. (Again, assuming unattended operation and no power at all during any part of this time.)
But we still want to prevent that possibility of uncovering the bundles and letting them even partially "air-cool", so let us continue. That is, the upper few inches of a fuel bundle could be exposed to the air - and still be cooled by conduction down to the lower sections still cooled by contact with water. But we don't want that to have to happen because it is harder to analyze things and still assume a degree of safety, so we need to find a way to keep the bundles covered with water.
So, still assuming no power available several weeks later from either the grid or from either of the backup emergency diesel generators, or from alternate site supplies, or from emergency 440V sources from outside portable generators, what can you do to cool the fuel pool?
Simple. Add water and refill the pool. The alternative water source can be any of 5 or 6 "pure water" tanks or cooling water systems on-site already, or from any source brought to the site - the source of water should be as pure as possible from a chemical standpoint of minimizing reuse and cleaning the fuel pool later, but from a heat transfer standpoint, any source could be used. The fuel pools are typically physically lower than most on-site tanks, and so gravity drain is usually possible. If not, the extreme example of using a portable pump could be applied - the ultimate example would be a fire truck hooked up the site cooling pond to pump water into the pool periodically - about once a week would do the job.
Since the fuel storage building is sealed from the environment, one user recommended that the operators simply allow the boiled water to cool and condense as it hits the building walls. Once cooled, the water could be redirected through channels back down into the fuel pool. While this is theoretically possible, it is not considered a solution to the general problem and so is presented for observation only.
Student assignment: (extra credit) - Fill a saucepan with water, and place a pair of bolts horizontally in the bottom. Turn on the stove burner, and keep track of the temperature of the water with a high temperature (candy or meat cooking) thermometer: the water will slowly increase to 212 degrees (100 if metric) - then stabilize at 212. The bolts will, at that time, also be at 212 degrees. Your job is to keep the bolts covered with at least 1/4" water to keep them at 212 F. Once boiling occurs, notice the formation and behavior of the bubbles and naturally circulating water between hot (the bottom touching the burner) and cold regions (the walls of the saucepan.) Once water level goes down a little bit, add an ice cube to simulate powered cooling at a constant heat load. Look at what happens to the rest of the water, even if the ice cube only replaces a small percent of the total volume of the water in the whole pan. Let the water resume boiling: add tap water to refill the saucepan to the top, this will stop boiling and cool the overall combination of pot and bolt and water again. Repeat periodically - as you watch, you'll learn the lesson that most of nuclear power issues are very slow to develop and very slow to watch happen. You'll also see why the phrase "A watch pot never boils" was invented.
-- Robert A. Cook, P.E. (Kennesaw, GA) (firstname.lastname@example.org), March 15, 1999.
This question was asked in the middle of some other things, but it really "belongs" here as part of this group of comments.
I understand that there will be no "atomic explosion" as in a weapon, but living 50 miles from TMI I remember talk of the phrase 'China Syndrome' which I understood to mean that the reactor would melt it's way through the floor until it encountered the river water and the resulting steam explosion would cause a radiation risk to all downwind. Is my understanding correct? If so, are there any other facilities sited over water? And why would they site a plant in such a location. Thank you for this opportunity to ask questions that have been bugging me for decades.
-- kahley (email@example.com), March 13, 1999.
Your question is relevent and important - let me talk about it on the basis of several facts. Check these with Department of Nuclear Engineering at Penn. State, if you would rather not go to the TMI site itself and use their material and refernces.
The "China Syndrome" is a technical fiction started years ago by people who have a fear of nuclear so deep it can be considered a matter of "faith and religion" with them - you cannot discuss the technology or physics with those who cite this fear, because they fervently "want" to believe its falsehoods.
By the way, there is a varient of this theme that actually gave it the name: that the molten mass would resume criticality when it hit a "water table" below the power plant, then heat up again and melt its way through to China.
Let us assume the reactor does have an accident and reach a high enough temperature to receive core damage. First stage is swelling of the pellets, then stress on and then "stretching" or distortion of the fuel rod cadding. Further, let us continue to fail to cool the core - at some point in time, the part of fuel will melt and "slump" to the bottom of the reactor vessel. In US reacotrs, this slag has been found to cool rapidly and solidify at the bottom of the vessel - it simply doesn't have enough heat energy - even under worst case conditions - to melt the reactor vessel steel.
Now, even when the core damage, as you found at TMI, is extensive, the molten sections of core "parts and pieces" are contained in the steel vessel - which is 6-8" thick. After solidification, the resulting radioactive particles could get distributed from there later (as they most certainly did at TMI) but even there were kept completely within the containment dome and the room where the failed relief valve was dumping water.
This makes sense, the water can carry particles from one area to the next, but the fact that water must be present to carry the particles, and the small size of the particles themselves (so they can be carried at all) means that the "distributed" particles, though radioactive, are cool to the touch. (Touching is not recommended though.) They cannot melt anything - not even a sheet of aluminium foil - much less the 10 -20 feet of concrete under the reactor and the plant floors.
So, after core meltdown, you are left with a partially melted glob of sitting in the bottom of pressure vessel core doing nothing but continuning to cool down. It can't can't go critical again - the melted "junk" includes control rods, spacer plates, dividers, steam and waterflow guides, mounts, clips, locks, springs, etc. and small parts of the core (fuel) itself. This mixes together to destroy the carefully arranged geometry required by nuclear physics interactions between water, neutrons, U235, Zr, Hf, and the cladding and the steel and water "reflectors" in the vessel to resume criticality.
The preceeding statements you can check via outside channels: things are either known or can be found in the engineering references: thickness of vessel wall, size and shapes of the vessels, heat transfer coefficients of steel, concrete, mass of fuel, heat rates after core shutdown, etc. You will either have to trust me on these next statements, or take the advanced calculus, nuclear physics, nuclear engineering courses, and reactor operating courses to find it out yourself.
A core must have a very specific shape and metalurogical makeup to allow the neutrons to be captured and cause fission. You cannot vary this shape and remain critical. You cannot add material to the shape and remain critical. You cannot remove water and remain critical (thus the supposed "hit water table" criteria.) Water - once touching hot metal - tends to flash into steam and so is no longer able to support criticality.
Bottom line, even when accidents like you mention have occurred, and when they have been simulated in labs and by computers, you cannot melt through the steel and concrete below the plant to release contamination in US/Canada style reactors. You can create a tremendous mess inside the containment building - as they found out at TMI - but you cannot melt through.
The Soviets didn't (don't) care about their people, and didn't (don't) care about the environment around the plutonium production plants - thus they made no containment structures, and so they deliberately designed their plant so they could remain critical without water, and so they were surrounding a hot core with flamamble graphite (to enhance the fast neutron capture reactions in Pu), and so then they decided to operate their plant after deliberately removing or disconnecting several existing safeguards circuits to "test" the reactivity of the core.
The result was a steam explosion that blew the top off the building and exposed the melting core to the graphite (carbon) and the open air around Chernobyl. Guess what - hot carbon burns! The fire (releasing smoke particles) was what really spread the radioactivity into the atmosphere. In the meantime, because the core was designed differently, as it melted parts did spread below the vessel and were eventually solified into the concrete below.
Thank your friendly local Communistic Party for that one.
-- Robert A. Cook, P.E. (Kennesaw, GA) (firstname.lastname@example.org), March 17, 1999.
Anybody worried estimated radiation or heat generation levels after shutdown?
-- Robert A. Cook, P.E. (Kennesaw, GA) (email@example.com), March 23, 1999.
Yes, Robert. Please don't stop now. You're really cookin.
Best stuff on this board so far.
-- Greybear (firstname.lastname@example.org), March 23, 1999.
defintely second freddie the freeloader
always great stuff from Robert
-- dick of the dale (email@example.com), March 24, 1999.
Couple of good references you may want to look up to see both sides of the debate about nuclear power - these should be available from your local library -
Before It Is Too Late, Bernard Cohen, Plenum Press, 1983.
The Great Nuclear Power Debate, Gail Haines, Dodd, Meade & Co. 1985.
The first is more slighty more technical, but more thorough and defines very clearly the positives (including the tremedous radiation penalties from increasing coal-burning). The penalties of solar power, and other alternatives are also covered. The author often uses relative risk - so you've got to have at least algebra to follow the specifics of his arguments. If not, just skip the math, the message is still useable if you accept his logic.
Haines' book does a more complete job of addressing both pro & con arguments. It is less complicated, slightly easier to read, but thereby has less technical information.
The amounts of background radiation received (for example) just from the natural potassium in our bodies (25 mrem, where 1 mrem (millirem) = 1/1000 rem) is more than that allowed at the boundary of a nuclear plant. Cosmic radiation is 30 mrem, natural radioactive material in the ground and our buildings adds 30 mrem. Total from all background sources averages 85 mrem, a single dental X-ray is 10 mrem, a spinal X-ray averages 400 mrem. In general, assume you are getting hit by 15,000 radiactive particles of one or another every second. Moving to FL (at a lower elevation) will reduce this by about 15%, moving to CO (higher up = less atmospheric shielding; much more natural radiation in the rocks - which are also much closer to the surface) about doubles the national average amount of radiation you are exposed to daily.
Paradoxically, leukemia rates in CO are 86% the national average - you'd figure that if low levels of radiation inreased the risk of cancer, their rates would be higher than normal. Cancer rates in general up there are 35% below the national average.
Nuclear workers are allowed 5000 mrem per year (above background); but very, very few ever get this much. There is no study that clearly indicates that radiation workers get higher cancer rates than normal workers - one study about 4100 miners in poorly ventilated coal mines indicated they averaged 6000 mrem exposure to their lungs from radon gas, but only had 134 lung cancers higher than what could have been predicted from national averages. (Some think these cancers were influenced by smoking and black lung disease rather than radiation exposure. Realistically - no one can tell after the fact what caused any particular cancer in any particular body.)
775 women got 1,700,000 mrem from painting radium dial watch faces in the 1910's and 1920's - they 48 excess bone cancers above what was expected for their ages. 14,000 patients were given up to 300,000 mrem of Xrays in Britain in the 1935-1955 period to treat spinal problems - these 14,000 people got 80 excess cancers above what was expected. German patients in the 1950's were given 900,000 mrem doses to treat spinal tuberculosis (and other diseases) - of 900 patients, only 54 excess bone cancers were found.
Considering the above, it appears to be more dangerous to go to a hospital than to live near a power plant. 8<)
How much radiation is dangerous? It's hard to specifically say, because it (radiation exposure) becomes an emotional issue, not a scientific one. If you keep your extra exposure below background - less than 100 mrem/year - and of course, you can't stop the background radiation at all - there is nothing to indicate that your cancer rate will be higher than those people living in CO, UT, and ID. And evidently, they get less cancers than anybody less.
Can radiation cause cancer? Yes - if enough is received in a very short time. Dogs were exposed to 100,000,000 mrem doses - and most developed bone cancer within 3 years. Dogs exposed to lower amounts delevoped fewere cancers and took much longer: if the doses were 5,000,000 mrem, bone cancers were generally not found until 9 years after exposure. [It is estimated that 50% of average sized humans exposed to doses greater than 250,000,000 mrem at one time may die unusually early.]
Not a pleasant subject, but it's important to know the relative risks involved: 1 mrem of exposure is about the same risk as driving 3 miles more a year. Average exposure to the public outside TMI was 1.2 mrem, about the same from natural radiation in 5 days.
-- Robert A Cook, PE (Kennesaw, GA) (Cook.R@csaatl.com), April 01, 1999.
About nukes and the July 1999 NRC deadline for Y2K compliance.
Please, first, let me get some specific (detailed) references from the NRC offices here, and from the EPRI and INPO offices about Y2K, before I speak about the specific "regulations" you're talking about, and about the embedded chips in control circuits/safety circuits in the power plants. I'd rather confirm some estimates and look up a few things before I shoot myself in the toe. INPO is a "certification and training" organization sponsored by the industry, they are like the NRC, in that they serve in a QA and improvement role, but compliance is "voluntary" rather than "statutory" as in the CFR (Code of Federal Regulations). However, violations are taken equally seriously. EPRI (Electric Power Research Institute = www.epriweb.com) is a private company that provides general research and data exchanges services between its member (client or cooperative) companies. They have teamed with the several natural gas suppliers and several (most) natural gas distribution companies to serve as a data gathering and mutual exchange network (database) for Y2K issues in the power and natural gas industries. Good people, good information.
The deadline. Note though, unlike what has been implied in the other threads, that the deadline mentioned is a NRC "bureaucratic" one, not a physically or scientific one based on cooldown times. Reading between the lines in the published parts of the regulation, it appears that the NRC understands that 100% compliance cannot practically be demonstrated prior to actual implementation and testing. That is, they want a wide operating margin (chosen to be 6 months) before the actual crisis date. If problems are found, if implementation is delayed during testing, if operations can be done consistently and safely (while a certified backup (the current operating systems) are available), then the plant is in a more safe condition. That is their mandate - plant safety. So all their actions (at the NRC level) are geared that way.
So the certification for Y2K is set to provide a deadline, and a safe margin of operation after that deadline to continue testing and field use prior to exchanging the current process software with the revised software.
The reference also allows waivers (partial cert's, or restricted certification) based on conditions. Makes sense, now, from the headquarter's viewpoint, nobody can tell exactly what schedules will be met, and what waivers will be needed, by the middle of next summer. For example, one plant was granted a waiver to allow replacement of a feedwater reg valve during a scheduled outage in November. This plant will be technically "non-compliant" from July until Nov, but this same valve has been replaced before in a different plant, and the new valve regulator has been certified and tested in operation. So there is no technical risk in delaying installation until November. But without knowing the background, you cannot tell.
But, any plant operators not able to demonstrate compliance by the deadline (within this margin "allowed" by some sort of waiver), can be reasonably expected not to complete the remaining programming, "debugging", testing, re-programming, re-training, implementation and installation, and still have enough "operating" time before the deadline. These plants grossly missing the deadline, and not able to give realistic plans to complete remediation will be shut down.
Thus, the deadline was set early to give the biggest practical safety margin for operations, not for cooling the plants.
Nuclear plants in general with respect to Y2K?
I've lived with them, in them, and around them. I respect their power and capability, and know what parts can kill (depending on the time involved, different things are dangerous to different levels at different times), what parts are "dumb steel and concrete" and what parts are literally beautiful in their nuclear physics, design, simplicity and safety. A person scared of nuclear power may possibly scared to death of the very idea of the thing, and doesn't know enough to know where the "safe" limits are. Worse, he (and many others, don't want to understand where the relative risks are, and doesn't want to consider the relative risk involved..
For example, nuclear plants are among the first things every story about Y2K used to talk about, and they are perhaps the only part of the national grid I consider likely to be operational.
My opinion: you are more likely get high voltage power to the "end" of the distribution yard at a nuclear plant than any other place in the country.
The problem will be getting these volts downstream to the public.
Cooling the plant after shutdown? Cooling the things down is a little like radiation, there are degrees of difficulty involved. What I consider reasonable and practical in a power plant (when I walk around in them, or build them, or live next to one, or test them, or inspect them) probably isn't what a reporter for the local paper would consider rational. A reporter wants simplistic, first grade answers to questions that are based in physics.
So let me give you a "quickie" physics lesson to explain it; I figure you'll understand better that way. The plant generates heat while critical, this heat makes steam that drives the turbine that makes the electricity. Groovy. Same things happens in any steam power plant.
Difference is: in a steam power plant, when the burner turns off (like when there is no oil, or no coal, or no natural gas in the distribution network), the residual heat in a conventional plant quickly goes away. Like your car engine, the steel and water cool down at a measurable rate to room temperature. Say, in 18 to 24 hours, the pipes are cool enough in many areas to work on or repair. If cool water is flushed through them, they are also immediately ready to be worked on (safe to touch without getting burned.)
In a nuclear plant, about 5% of the "running" heat comes from the decay of radioactive daughter products in the core from earlier fissions. (This value was repeated above from the NRC rep, and actually refers to "immediate" decay heat always present under steady-state conditions.) So, a 1000 megawatt (electric) has a 3000 MWatt thermal load. The plant, immediately after a "scram" (shut down) still is CREATING heat: this is thermal energy that must be disappated. The fissions have stopped, but the decay heat is still being created, but at an ever decreasing rate. See numbers below.
As "nine fingers" pointed out above, if this decay heat is not removed from the core by passing "cooler" water through the core and auxiliary systems, then, after a certain period of time (several hours in most scenarios) if cooling is not provided, and if the water is not replaced (if there were a break in the pipes), the water originally covering the core may boil off. The core continues to heat up, but it is exposed to air not water, and damage may occur. [This is a VERY short description of part of what happened at Three Mile Island.]
So the key to safely shutting down a nuclear plant is to follow procedures, keep the core covered, and to keep emergency power and emergency water supplies always available (via backup generators, alternate pumps, alternate tanks, alternate supplies, alternate controllers, etc.) All these backup systems and emergency procedures are repeatedly covered, (and re-re-covered) in emergency drills and routine training. This training is also a reason to expect better reliability from a nuclear plant than a conventional one.
Nobody likes to work around decay heat, it is a problem I hope will go away when fusion plants come on line. But the nice thing about decay heat is that the heat load goes down exponentially over time after shutdown.
(The same thing happens with the radioactivity in the plant: but at a slightly different rate depending on which isotope, what power levels, what power history, etc. The radiation effect is also strongly dependent on the source and type of radiation measured.)
time = 0, decay heat equals 5% initial
time = 30 seconds, decay heat equals 4.2% initial
time = 1 minute, decay heat equals 3.6% initial
time = 2 minutes, decay heat equals 3.12% initial (This is the value most often used = 3%)
time = 5 minutes, decay heat equals 2.55% initial
time = 15 minutes, decay heat equals 1.90% initial
time = 60 minutes, decay heat equals 1.30% initial
time = 4 hours, decay heat equals 0.82% initial
time = 12 hours, decay heat equals 0.62% initial
time = 24 hours, decay heat equals 0.50% initial
time = 48 hours, decay heat equals 0.40% initial
time = 7 days, decay heat equals 0.28% initial
time = 30 days, decay heat equals 0.20% initial
time = 120 days, decay heat equals 0.10% initial
Using these values (the actual equation is complicated declining exponential decay function), after 7 days the decay heat is still being created, but it is being created at only 0.0028 the original amount (slightly over 8 MWatts). After one month, the decay heat is 0.0020 the original, and the energy from running a coolant pump contributes more heat to the primary water than the decay heat does: running the big pumps to circulate water tends to heat up the water from friction losses. In such cases, it is possible, as they did at TMI, to secure the pumps and relay on natural convection and ambient losses to continue cooling the reactor.
In fact, after enough time goes by, you can see that the outside of the containment building is receiving more thermal energy from the sun than it is being created from the reactor core inside.
Bottom line, the thing is still releasing heat "forever"- but not enough to worry about in a Y2K scenario.
For example, during every refueling shutdown, the cover of every reactor can be removed and the core exposed for refueling after a very short amount of time: 2-3 days is typical. [All this exposure happens inside the containment building, above a "swimming pool" type of open tank.]
So what a politician or reporter talks about as a "Y2K problem" should NOT (and I will shout here: [in this PARTICULAR case, and ONLY in this SINGLE case of power plant cooling]) be worried about for Y2K.
That leaves 9,999,999 things left that will cause problems that the reporters are not talking about .....
-- Robert A Cook, PE (Kennesaw, GA) (Cook.R@csaatl.com), April 02, 1999.