Rod Motion -> power level calc +loss of all station air

HELLO PEOPLE! Sorry for the enthusiasm but, this is my first post. I am in the process of getting my masters in material science and engineering and I liked my nuclear materials course so I decided to jump right into the integration of mechanical systems in nuclear engineering this summer. I have understood everything from a theoretical standpoint but, my B.S. is in biochemistry and I am still catching up on some of this engineering math and there is one question that I have not been able to find a lot on.

first things first: MATH! I am still coming to terms with the fact that metric is not being used in this course and I have combed through our notes and I feel like my professor just assumed that I knew how to solve this.....
I feel like it's an easier problem than I am making it out to be but I just can't seem to find a set of equations putting all this together.

1.) How many steps of Bank D rod motion will be required to change power level from 5% to 18%? Assume that the full power Delta T from primary to secondary is 50 degrees F. Assume the moderator temperature coefficient is -5 pcm/degree F. Assume the Bank D rod worth is 5 pcm/step. Neglect fuel temperature changes.

Also, I have been asked this question:

The Beaver Valley Station is at 100% power. Discuss what would happen to the plant on a total loss of station air.

I have found information regarding what valves would close at different pressure levels in the event of a problem with the air system but, I feel as though there is a bigger picture answer mixed in with those supporting details that I have not hit the button on yet.

ANY HELP IS GREATLY APPRECIATED!!

Astronuc
Staff Emeritus
1.) How many steps of Bank D rod motion will be required to change power level from 5% to 18%? Assume that the full power Delta T from primary to secondary is 50 degrees F. Assume the moderator temperature coefficient is -5 pcm/degree F. Assume the Bank D rod worth is 5 pcm/step.
Does one understand the question? Neglect fuel temperature changes means Doppler effect is not significant. Is there other information regarding core inlet temperature, or feedwater temperature?

Strife_Cloud
jim hardy
Gold Member
Dearly Missed
The Beaver Valley Station is at 100% power. Discuss what would happen to the plant on a total loss of station air.

I have found information regarding what valves would close at different pressure levels in the event of a problem with the air system but, I feel as though there is a bigger picture answer mixed in with those supporting details that I have not hit the button on yet.
To what degree of detail ?
At what air pressure can you no longer keep feedwater valves far enough open to maintain steam generator level ?

Strife_Cloud
Good questions,

For the first question, I know that I copied and pasted it exactly because I also thought I was missing a detail needed to answer it.
-I am going to look through the questions and make sure a side note was not put anywhere saying (Assume ( )... for the next # of questions.) I don't think there was but, I was pretty sleep deprived by the time I got through the questions and saw I had skipped these 2.

As far as the second question, I believe that they are looking for somewhat of a larger picture view. I started my answer by indicating the valves they listed in the notes responding to the air pressure drop. Then, I stated the different systems that would try and save the station's air pressure, such as, the diesel powered air compressor. I tried to conclude that the reactors (Unit 1: BWR, Unit 2: PWR) would trip. I was trying to layout the path of signals for the reactor trip when I came across an accident occurring at a plant that had both of its units on the same air and an accident caused complete loss of the station air system. One of the reactors was forced to initiate shutdown while the other reactor was held steady at (I want to say) 40% power. This confused me because my line of thinking had been both would have to trip. I put the question on this post before diving deeper into the BWR vs. PWR possible difference in loss of station air.

I'm sorry, I believe I have explained too much before checking any of my collected resources but, I hadn't been able to get back on this topic since I posted.

I will figure out where the accident I am trying to reference happened and double-check the fine print to see if I did miss a detail for the first question.

Thank you both very much for responding!

OK the first question that I posted correlates with this one,

The Unit is at 5% power using the Steam Dump system (steam bypass to condenser). The steam dump system is in the steam pressure mode of operation controlling steam header pressure at 1005 psig. The station wants to raise reactor power to 15-18% to allow synchronization to the grid. Briefly explain how this is done including changes in the reactor plant and secondary plant.

and, after re-reading my answer to this question I feel as though I did not put enough information.

Here are my thoughts:
The steam dump system in steam pressure mode allows for the reactor to be started up prior to synchronization while also providing a means of keeping system stability during the synchronization process.
In order to increase reactor power, would the operator begin by reducing control rod length to increase the amount of reactivity in the core?

I think this may be counter-intuitive because I know that if the exhaust steam system becomes isolated, the steam gets dumped to the condenser and away from the RCS re-heaters, causing the condensate to lose heating, which results in an increased temperature differential (positive reactivity addition). Ultimately increasing reactor power. Although, since all of the steam is currently being dumped, I suppose those factors are already in play.

Instead of first making an adjustment on the primary side, would an operator adjust something on the secondary side first? I know that manual feedwater control should come into play here but, I am not sure if I am getting ahead of myself. The notes on synchronization are a bit brief.

Thank you very much, Astronuc, for making me take a more thorough look back at that. I did not answer the full question. I simply explained a lot about the steam dump system in that mode and did not finish with the steps to raise power and then synchronization. I will do more research.

As for the second question I posted. Here are the notes I wrote down about the accident I was trying to reference:

October 26, 1992 – Sequoyah Unit 1 scrammed and Unit 2
ran back 40%. Corrosion plugged a drain line and allowed an
air receiver tank to fill with water. When both units lined up to
share the tank, 1,000 gallons of water were blown into the

I have checked and both of these units had ice condenser containments so, I believe they were both PWRs. Maybe the BWR vs. PWR loss of station air assessment is too in depth. Here are my notes I have collected so far:

The plant would experience a station air compressor trip. Assuming that all capabilities of returning system air pressure to normal were unavailable, the plant would have to determine if they were able to restore the compressed air system to operable in around twelve hours or experience a “hot shutdown”. In the case that an extreme accident occurred compromising the compressed air system such as at the Sequoyah site, one of the units may have to be scrammed.

NOP cycles between 100-110 psig. Standby compressor starts at 95 psig. 2SAS-AOV105 auto closes at 86 psig. CP air compressor starts at 90 psig. Diesel-driven air compressor starts at 82 psig. AOVs begin to drift shut at around 40 psig.

2SAS-AOV105 → station main air header to service air header isolation. → isolates station air from instrument air.

LOSS OF AIR SAFETY RELATED VALVES → fail to the safe position depending on function. Major pump revirculation valves fail open for pump protection → main FW and condensate PP’s.

STATION AIR SUPPLIED TO CONTAINMENT: containment pressure will need monitored carefully, any small leaks into containment will increase containment pressure. → CONTAINMENT ISOLATION VALVES

The condensate feedwater heater bypass valve → fails “as is” on a loss of air or power.

When the station instrument air system is lost at % power, the condensate system experiences the gland steam condenser bypass fails open, condensator polish bypass fails open, and the recirculation flow control valve fails open.

OK! I found the answer to that one:

During a plant startup or cooldown, the steam dump system is operated in the steam pressure mode. As shown in Figure 11.2-1, when the steam dump mode selector switch is placed in the steam pressure position, the relay associated with the mode selector switch energizes and closes two contacts. Closing the contact shown to the right of the steam pressure controller places the steam pressure controller in service, while closing the middle contact in the set of three parallel contacts completes the electrical circuit which "arms" the steam dumps. This arming contact, once closed, allows dc power to be applied to the two solenoid valves in series. The inputs to the steam pressure controller are steam header pressure (measured at the main steam bypass header) and a variable setpoint which is selected by the control room operator. The steam pressure controller is a proportional-plus-integral controller. The steam pressure mode is selected during hot standby operation, reactor startup, and initial loading of the turbine. The steam dump valves act as a load on the primary system by removing heat from the reactor coolant. As the steam pressure is maintained by the steam dumps at the selected setpoint, the temperature of the reactor coolant is maintained at a value corresponding to the saturation temperature for the setpoint pressure. For example, if the steam pressure setpoint is selected to 1092 psig, the temperature of the reactor coolant will be maintained at approximately 557°F. At the end of a plant heat-up, the heat added to the reactor coolant by the reactor coolant pumps is transferred to the steam generators and removed by the steam dump system to maintain the reactor coolant at 557°F (no-load Tavg). During the subsequent reactor startup and low power operation, withdrawing the control rods causes reactor power to increase, which in turn causes the reactor coolant temperature and the pressure in the steam system to increase. The increasing steam pressure inputs an error signal (the actual steam pressure is greater than the setpoint) into the steam pressure controller. The controller converts this error signal into a valve positioning demand that, once converted to control air pressures by the individual valve positioners, modulates open the steam dump valves. The integral USNRC HRTD 11.2-4 Rev 0403 action of the controller will increase the valve opening demand if the steam pressure error persists for an extended period. The number of open valves and the degree of modulation are determined by the magnitude of the controller output. A steam pressure error, after integration, equivalent to 100 psid results in the full opening of all four valve groups. When the reactor power is approximately 10 - 15%, the turbine is placed in service. The steam dump system is still selected to the steam pressure mode, and the dump valves are passing approximately 10 - 15% of full-power steam flow. During the initial turbine startup, the steam dump system maintains a relatively constant total steam flow in the main steam system. The steam dump valves are open and dumping steam to the main condenser to maintain the desired pressure in the main steam system. As the turbine governor valves open, steam is admitted to the high pressure turbine, which causes the steam pressure in the main steam header to decrease. This reduces the magnitude of the pressure error generated by the steam pressure controller, and the controller=s decreasing output causes the steam dump valves to modulate closed. The net result of this evolution is that the total steam flow remains relatively constant. This process of transferring the main steam flow from the steam dumps to the turbine governor valves continues as the turbine generator is loaded until the steam dump valves are fully closed and the turbine governor valves are passing the entire 10 - 15% main steam flow. Since the total steam flow is not affected (i.e., the heat removal from the reactor coolant does not change), the temperature of the reactor coolant does not change significantly, which reduces the need to move control rods. Also, the heat transfer in the steam generators remains unchanged, which allows for a constant feed flow rate to the steam generators, making feedwater control easier for the feed station operator.

Many thanks to the Westinghouse Technology systems manual

I am still confused how all of that will get me to number of steps of rod motion........ hmmmm

Astronuc
Staff Emeritus
The 557°F (no-load Tavg) is hot zero power (HZP) coolant temperature. If one knew the ΔT across the core, which is probably about 60 - 70°F at full power, then one could estimate an average moderator temperature. I'm trying to understand "Assume that the full power Delta T from primary to secondary is 50 degrees F." Is that the ΔT between primary and secondary systems, or is that the ΔT across the steam generator - hot let to cold leg.

If one can determine the ΔT of the moderator, then to offset the change in moderator temperature coefficient, one withdraws Control Bank D as many steps to offset the MTC. For example, if the moderator temperature increased 10°F, then the negative reactivity is simply 10°F * -5 pcm/°F or -50 pcm. Then to offset -50 pcm in the moderator, one needs to add 50 pcm by withdrawing Control Bank D. If the worth of Bank D is 5 pcm/step, one would move Bank D by 10 steps, or 5 pcm/step * 10 steps = 50 pcm.

So to solve the problem, one needs to determine by how much the moderator temperature changes from 5% to 18% power.

Also, to change (increase) power, one needs to have positive reactivity in the core, i.e., k > 1.

Last edited:
Strife_Cloud
jim hardy
Gold Member
Dearly Missed
Sequoyah is a Westinghouse PWR somewhat newer vintage than mine.
Is that the ΔT between primary and secondary systems, or is that the ΔT across the steam generator - hot let to cold leg.

My Westinghouse plant used ~constant cold leg temperature
it's a compromise ,
Reactor design guys would like constant T-average but that makes T-cold fall and steam pressure droops so far that Turbine guys complain
Turbine design guys would like constant throttle pressure but that requires T-cold to rise, and T-hot must rise even more which pushes it up uncomfortably close to critical where there's no phase anymore.... which makes reactor guys complain
so they settled for Tcold constant , as power increases Thot will rise some and steam pressure will droop some.

Since it's not given in the problem statement how much steam pressure droops or Taverage rises,

i'd advise OP to apprise teacher that he used 50 degrees as Δhotleg to cold leg, meaning T-average rises by 25 degrees and secondary temperature drops by 25 degrees
giving full power steam temperature of 535 degF, which is 885psi
and Thot of 607, (my plant was 602) well below critical of 705, leaving enough headroom for pressurizer

..............................................................

@Strife_Cloud

The plant is basically a mover of heat. The medium is water. Primary side water carries heat from reactor to boilers(aka steam generators), and secondary side water carries heat from boilers to turbine(as steam).
Flow rate on primary side is not actively regulated.
Flow rate on secondary side is actively regulated. The level in the boilers is measured and used to control rate of feedwater inflow.
An upset to feedwater flow very quickly pushes boiler level out of bounds and the reactor trips so as to stop production of heat,
in anticipation of losing ability to move it.
Feedwater flow is controlled by air operated valves .
Feedwater valves are pushed opened by air , but a stout spring pushes them in 'close' direction. That's so they'll fail safe, in direction to not overcool the reactor.
So when you lose air you lose the ability to hold the feedwater valves open. That causes feedwater flow to drop, lowering boiler level, and will soon cause a plant trip.
That imho will be among the very early events in your 'loss of air' scenario.

I'd wager there's an ORNL study for loss of air, try a search .

Strife_Cloud
I was definitely trying to overdue the reactor power calculation. Thank you both very much Astronuc and Jim Hardy! I believe I was able to answer both of those questions correctly!
I really appreciate your time and your thoughts in the effort to help me out!
I have one last quick personal question if you guys don't mind (or until next week's homework, haha). I love science and find myself able to get wrapped up in most of its many categories. For instance, I was a biochemistry major that got a job in a specialty steel mill.... and then loved being able to jump into something I knew little about and try to conquer the steelmaking process. I am currently 25 and have a lot of questions I feel like I'm going to have to answer about where I want to go with my career, and life after I am done with my master's degree. How did you both make decisions about moving forward in your careers?
I mean, I feel like I work as hard as possible all of the time (or, as much time as my wife will allow me too, haha, she is my "ground" wire). Is there a better way than FORWARD! with the best opportunity presenting itself? Or is that just life's (or entropy's) way of moving me in the right direction, at the right time?
I didn't mean to get spiritual or make anyone uncomfortable but, I feel like my own perspective could use a little growth.

Astronuc
Staff Emeritus
How did you both make decisions about moving forward in your careers?
My first job came by serendipity. A colleague (in grad school) was approached by a manager at a consulting firm, but the colleague was interested in opportunities elsewhere. He tossed a business card on the table at a meeting. I was curious, so I took the card and contacted the firm. I sent my resume to the company, but didn't hear anything for a couple of months (they were relocating their office). I was contacted about an interview and arranged a trip. I had a good meeting with various managers, and the VP called me the next morning with an offer, which I accepted.

During my first job, I built a good reputation in industry, and after 10 years, I got two nice offers, one from a competitor, and one from a mutual client. I'd had received other offers from customers/clients, but wasn't interested at the time. I recently changed jobs after I was invited to apply for my current job.

More or less, the jobs found me.

Specialty steel should make for an interesting and rewarding career. Some of my current work involves specialty steels in nuclear systems.

Strife_Cloud
jim hardy
Gold Member
Dearly Missed
How did you both make decisions about moving forward in your careers?

I don't know that i "moved forward" , i started as a maintenance man and thirty years later was still a maintenance man.
Life is what happens to you while you're planning your life, i think .
I wanted to work for an airline fixing radios like my next door neighbor's dad.

In high school i had an extraordinary electronics teacher who got us boys competent in vacuum tubes, circuit analysis and radio. So it seemed natural to take EE in college, which i did. I found i preferred power to communications and did better in machinery courses than fields.
Every day i walked past the school's windowless reactor building and was curious, so one day knocked on the door and asked if i could see it. Mr Alva Elliot, the fellow who ran it welcomed me and gave me a tour. It looked like a fascinating machine.. So i went to see my guidance counselor and asked if i could apply some 300 level electives in nuclear engineering toward my EE degree , explaining that my hometown utility had announced plans to build a nuke plant. Counselor said it sounded like a good idea.
So i took reactor physics which i just loved. And reactor operation which consisted of startups and shutdowns under the watchful eye of Mr Elliott. Since i was EE he made me learn every resistor in the electronic reactor instruments, which was easy for i'd learned tubes in high school.

When i got home after graduation i went to my local utility and asked if they were hiring engineers for that new plant and showed the personnel guy my diploma. He called his supervisor Mr H, and i heard him say "I have a young man here just graduated in EE and has some nuclear background, his name is ..."
Then he said "Mr H wants to speak to you" and handed me the phone
Well - small world that it is, Mr H remembered me. His daughter Nancy and i were in same second grade class and we'd won a lot of spelling bees. He told me to go to the plant and introduce myself they'd be expecting me.

They apprenticed me to an old timer with vast experience to whom i am forever indebted. He taught me volumes .
But the times they were a'changin. Together he and i learned operational amplifiers , digital IC's, computer assembly language, in between working some major bugs out of the new plant.
Over the years i tried a couple stints at office work, found paperwork depressing guess i'm just not cut out for a desk job
nothing against office workers though, they are bright and excel at communication and organization .
I on the other hand like you get wrapped up in things, in my case obsessively .
So i had the great fortune to become a knowledge worker and spend a career around machinery and the good people who tend to it .
That suits my quirks of personality , i'd rather be a good technician than a bad manager.

I didn't mean to get spiritual or make anyone uncomfortable but, I feel like my own perspective could use a little growth.
Read the first couple chapters of Moby Dick, where Ishmael explains why he always goes to sea as an ordinary seaman.
Finally, I always go to sea as a sailor, because of the wholesome exercise and pure air of the fore-castle deck. For as in this world, head winds are far more prevalent than winds from astern (that is, if you never violate the Pythagorean maxim), so for the most part the Commodore on the quarter-deck gets his atmosphere at second hand from the sailors on the forecastle. He thinks he breathes it first; but not so. In much the same way do the commonalty lead their leaders in many other things, at the same time that the leaders little suspect it. But wherefore it was that after having repeatedly smelt the sea as a merchant sailor, I should now take it into my head to go on a whaling voyage; this the invisible police officer of the Fates, who has the constant surveillance of me, and secretly dogs me, and influences me in some unaccountable way--he can better answer than any one else.

Now - specialty steel sounds fascinating to me . As does a steel refinery. I was intrigued by the exotic steels around the reactor, in particular the magnetic alloy used in control rod drive shafts. It taught me a lot about magnetics. And the reactor head bolts - i think something like 4140 tempered to 170k tensile - is that possible ?
Before you abandon specialty steel see what markets they sell to. You may be sitting amidst acres of diamonds.

old jim

Strife_Cloud
1.) How many steps of Bank D rod motion will be required to change power level from 5% to 18%? Assume that the full power Delta T from primary to secondary is 50 degrees F. Assume the moderator temperature coefficient is -5 pcm/degree F. Assume the Bank D rod worth is 5 pcm/step. Neglect fuel temperature changes.

First, fuel temperature changes cannot be neglected. Reactivity feedback, Δρ/Δ% power, comes primarily from changes in fuel temperature. All you have to know is the total power coefficient, which includes both, the change in moderator temperature and fuel temperature:

Total power coefficient of reactivity is calculated: ∂ρ/∂P = (∂ρ/∂ Tf) (dTf /dP) + (∂ρ/∂ TAVG)(dTAVG /dP)
See: http://www.nrc.gov/docs/ML1214/ML12142A130.pdf

Example from nuclear-power.net:

Let assume that the reactor is critical at 75% of rated power and that the plant operator wants to increase power to 100% of rated power. The reactor operator must first bring the reactor supercritical by insertion of a positive reactivity (e.g. by control rod withdrawal or boron dilution). As the thermal power increases, moderator temperature and fuel temperature increase, causing a negative reactivity effect (from the power coefficient) and the reactor returns to the critical condition. In order to keep the power to be increasing, positive reactivity must be continuously inserted (via control rods or chemical shim). After each reactivity insertion, the reactor power stabilize itself proportionately to the reactivity inserted. The total amount of feedback reactivity that must be offset by control rod withdrawal or boron dilution during the power increase (from ~1% – 100%) is known as the power defect.

Let assume:

• the power coefficient: Δρ/Δ% = -20pcm/% of rated power
• differential worth of control rods: Δρ/Δstep = 10pcm/step
• worth of boric acid: -11pcm/ppm
• desired trend of power decrease: 1% per minute
75% → ↑ 20 steps or ↓ 18 ppm of boric acid within 10 minutes → 85% → next ↑ 20 steps or ↓ 18 ppm within 10 minutes → 95% → final ↑ 10 steps or ↓ 9 ppm within 5 minutes → 100%

Strife_Cloud
You mention a BWR. On loss of air, the main steam isolation valves will fail closed causing a loss of condenser and turbine driven feed pumps and operators will use RCIC and HPCI for level and pressure control, augmenting with SRV/EMRV actuations as necessary.

Decay heat is to the suppression pool with the residual heat removal heat exchangers in service cooling the pool.

Strife_Cloud and jim hardy
@Astronuc and @jim hardy for your sharing your experiences. I will keep them in mind as I move forward. I would say that my current job and my entrance into the realm of materials engineering came by serendipity. I mean, my dad has sold steel since he graduated college but, I never considered that path. I joined a Temporary job placement company for chemists after graduating to try and get a better feel for what kind of jobs I qualified for and what I might want to focus on in grad school. Interviewing at the mill was suppose to be my "practice interview" and only last two months if they picked me. Funny to think that was over three years ago haha.

and Jim- next time I get some free time to read for purposes not concerning class I will have to pick that book up. I like the selection you posted. I can honestly say that I have never read it.
and about those acres of diamonds.... one of the reasons I took this class in particular is that my one of my nuclear materials professors (from the Bettis Laboratory) said that he thought there was still a lot of material optimization work left to do with nuclear power systems. Specifically on some of the less focused on systems. His example: "if you can take a pump or even a bolt that they normally have to replace every year or every other and make one that you would replace every five, even though you are using a material 2x the price of the original, that's a part you can sell because it comes out cheaper in the long run." We went over the development of Zircaloy and other major material developments but, I wanted specifics to better relate to what systems he may have been talking about haha. If anyone has any ideas on that subject, feel free to comment. I have asked if any of the steel we currently produce ends up in nuclear applications. A few of the higher-ups have told me that it wouldn't be until a few customers down the line, so they were unsure and would look into it. (Example: steelmaking produces a steel ingot --> a forge further shapes the ingot --> multiple fabrication steps--> surface engineering (EX:nitriding) --> final customer)

@Michal Kovac unfortunately, I understand your point and thank you for the information but, I copied that question word for word from the ones I have been given by my professors. Disregarding fuel temperature changes happened on their orders. I am guessing that will be covered in one of the next five weeks of the course. It seems like the logical evolution of the problem I originally posted. I have no doubt I will need that in the future.

@Hiddencamper Thank you very much for responding. BWR systems are mentioned as a side note at the end of lectures in this class, so I have not been able to develop my understanding of them on the same level as the PWR. In your opinion, would the loss of station air incident lead to a reactor trip, and what indicator would trip first?
Also, what effect would that have on the turbine end. I am thinking, loss of condenser, increased steam pressure, reduction in boiling, decreased reactivity in core? does the turbine have the same type of governor valves found with the PWR? and I am also going with an eventual decrease in power but, how drastic of a reduction would have to take place with a loss of air, if the reactor is not tripped?

I greatly appreciate everyone's posts!

Depends on the particular BWR. On a loss of instrument air, your scram pilot air header pressure will drop, causing the scram pilot valves to drift open, and the scram discharge volume vents and drains to close. The moment the operator sees air pressure low on the scram pilot air header, or the scram discharge volume filling, or any unexpected rod motion, they are required to manually trip the reactor. If no manual operator action is taken, the partial rod scrams that start to occur will cause the scram discharge volume to fill up, and the reactor will trip on high level in the scram discharge volume. This is an anticipatory trip to ensure the reactor scrams before the volume is too full and a hydraulic lock ATWS occurs. (Actually happened at Browns ferry).

The main steam isolation valves will drift shut (they are air powered to open and spring loaded closed). If for some reason the MSIVs start to drift close before the scram pilot air header goes low, then when any two steam lines have their MSIVs off the full open limit switches, the reactor will scram on MSIV closure signals. If the MSIV closure signal fail the BWR will scram on high pressure or high neutron flux.

One way or the other, a loss of all station air will result in a pretty fast trip. When my plant (BWR/6) lost air to containment a few years ago, it was about 10 minutes before we had the low scram pilot air header alarm and the operator manually scrammed. If you have an air leak it would be much faster. Our MSIVs went closed around an hour later. We had just replaced hoses and fittings on the MSIVs so they actually were fairly leak tight, compare this to 20 years ago when a loss of air caused the MSIVs to shut in 5-10 minutes. Huge difference. We dropped pressure to 500-600 psig after the scram to remove decay heat from the containment, so when the MSIVs went shut, we were fortunately at a point where decay heat and reactor pressure were low enough to allow us to not lift any safety relief valves.

On the secondary side of the plant, the turbine uses an electro hydraulic control system. The turbine operates in steam pressure control mode, that is, the governor valves automatically throttle to maintain a given steam header pressure plus error. The steam header pressure is within 30 psig of your no load steam pressure set point, and as power goes up, pressure goes up. It's a little different from a PWR. The loss of air wouldn't affect turbine load until the reactor tripped, as turbine load follows reactor power automatically by monitoring steam header pressure. The turbine's automatic load follow and pressure control functions provide reactivity feedback to prevent the core from voiding out.

It's different for each plant, how the secondary responds. Bwrs aren't required to have an automatic or fail safe water isolation, and I don't think any have them activated anymore. It's preferred in a BWR to maintain feed water as long as possible. On loss of air, my feed water min flow valves to open, so level control becomes sluggish after the scram. We would have to manually isolate the min flows. The other issue is turbine driven feed pumps will only have steam supply until the MSIVs go shut, a few minutes later they will stall out. For plants like mine with high pressure booster pumps (725 psig shutoff head) and a motor driven startup feed pump, we can provide at least 600 gpm to the core at rated pressure, and lower reactor pressure as necessary to maintain feed. Otherwise it's the RCIC/HPCI/HPCS/SRV game, using alternate pressure control strategies as available. Another possible outcome for a scram, is if your air pressure rapidly drops in the turbine building, the loss of air will cause feed water min flows to drift open at power, a low suction pressure trip of the feed pumps, and as reactor level powers you'll initially get a reactor runback followed by a low level scram.

You'll have to break vacuum due to a loss of gland sealing steam. So this will complicate recovery efforts.

Strife_Cloud and jim hardy
Astronuc
Staff Emeritus
there was still a lot of material optimization work left to do with nuclear power systems.
I agree. That's one of the areas of my research, in addition to radiation effects on materials. Zr-alloys is another area.

316L and 304L are common in nuclear systems. They are found in control rods (PWRs) and control blades (BWRs). The industry uses high purity alloys with low S and P. Stainless steels are also found in upper and lower nozzles (PWR fuel) and tie plates (BWR fuel). The nozzle and tie plates form the top and bottom interfaces with the core support structures.

Some core internals and surrounding structures are made of stainless steels.

jim hardy
Gold Member
Dearly Missed
neutron embrittlement of reactor vessels would be an interesting avocation
small things of the earth confound the mighty ......
like: welding wire comes with a thin coating of copper to keep it from rusting before use
that tiny amount of copper in vessel welds was a "what if du jour" worry forty years ago , i think we actually hardness tested some vessel welds in the 90's cant quite recall...

They do place metal samples inside the vessel to be removed and tested as it ages.

Metallurgy - another of those worlds unto itself that i only glimpsed into over the shoulders of specialists.

old jim