Designing a Core Lattice: Thermal-Hydraulics & Neutronics Considerations

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In designing a core lattice, the interplay between thermal-hydraulics and neutronics is crucial, particularly concerning power distribution and thermal boundary conditions. The moderation-to-fuel ratio and critical heat flux are key parameters to balance, impacting neutron leakage and heat flux. Regulatory guidelines, such as 10 CFR 50 and General Design Criteria, must be adhered to, influencing fuel system design. Fast reactors typically utilize hexagonal lattices for better thermal-hydraulic performance, while accurate burnup calculations are essential for fuel performance and regulatory compliance. Understanding the dynamics of coolant flow, including pressure drop and buoyancy, is vital for optimizing reactor design and operation.
  • #31
I think pressure at the horizontal line is the same.
I cannot picture the second question, therefore I don't know where the pressure is high in the tube.
 
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  • #32
candice_84 said:
I think pressure at the horizontal line is the same.
I cannot picture the second question, therefore I don't know where the pressure is high in the tube.
The primary cooling circuit or loop of a PWR is more or less a closed system, with the exeception of lines use to infuse or extract water for cooling pumps, boric acid, coolant sampling lines. The primary cooling system consists of the reactor core and pressure vessel (PV) internals, hot and cold legs, steam generators (headers and tubing), and cross over legs from the steam generator to reactor coolant pump on the cold leg. The pressurizer is attached to one of the hot legs and provides pressure to the system.

The highest pressure in the primary circuit is not the pressurizer, but at the outlet of the reactor coolant pump (RCP). The further one moves down stream from the RCP, the lower the pressure. The lowest pressure is at the inlet of the RCP. It's a bit like a voltage source in an electric circuit, the greatest electrical potential is at the + terminal, and the potential falls as one moves downstream around the circuit until returns to the - terminal (Kirchhoff's voltage law). The pressure drop across the RCP from inlet to outlet equals the pressure drop around the entire loop (continuity).

The pressure drop across the reactor core is about 25-26 psid. After the water enters the RPV through the nozzle with the cold leg, it then flows through the downcomer, through the lower plenum, up through the core support plate, through the core (flow channels between fuel rods), up through the upper core guide structure (where control rods are located) where it turns out to the hot nozzles.

Across the core, there is a pressure drop (pressure drops because of flow resistance). The temperature increases from the inlet (~280-292°C) to exit (~320-330°C). The reactor components in contact with the coolant are at those local temperatures. There may also be nucleate boiling in the hottest (combination of local coolant enthalpy/temperature and heat flux). The density of the coolant also changes.

The hot and cold nozzles are at about the same elevation, or the cold nozzles may be slightly lower than the hot nozzles. The coolant density is a function of temperature. The static head in the cold leg is of course greater than the static head in the reactor where there is a temperature gradient - and the hottest temperature is at the core exit.

So what are the advantages and disadvantages of downflow in the core vs the conventional upflow.

Think about component operating temperature, thermal hydraulics in the core, and safety issues.
 
  • #33
candice_84 said:
... But the actual flow in the pwr goes down in the downcomer and then flows upward which i don't think its a good design but there has to be a trade off that I am not aware of it.

Aside from any of the hydraulic effects Astronuc is discussing, think about what would happen if one of the main coolant pipes were to break. The vessel design with the downcomer allows the safety injection system to refill the vessel and flood the core. If you eliminate the downcomer, you would need to have the cold leg piping penetrate the vessel bottom - and a break in that cold leg piping would just drain away any safety injection flow. (And if you had flow 'up', it would be the hot leg piping that penetrates the lower head).

I think this is the main reason for the downcomer design (though there may be other reasons, like having the water provide shielding to the vessel material). Notice that the ECCS rules specify breaks in the main coolant loop piping - a break in the vessel itself is not a postulated design basis event.
 
  • #34
I didn't specifically call out the ECCS, but that comes under safety considerations.

One might be interested in 3.9.5 Reactor Pressure Vessel Internals in AP1000 DCD Rev. 17
http://adamswebsearch2.nrc.gov/idmws/doccontent.dll?library=PU_ADAMS^PBNTAD01&ID=083251077

The entire DCD can be found
http://adamswebsearch2.nrc.gov/idmws/ViewDocByAccession.asp?AccessionNumber=ML083230868

Chapter 4 contains the information on the reactor.
Chapter 3 Design of Structures, Components, Equipment and Systems
Chapter 5 Reactor Coolant System and Connected Systems

Earlier version of DCD is found at - http://www.nrc.gov/reactors/new-reactors/design-cert/ap1000.html#dcd
 
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  • #35
Can you guide me through calculating actual nuclear plants efficiency?
 
  • #36
candice_84 said:
Can you guide me through calculating actual nuclear plants efficiency?

OK. You start, by telling us what you mean by 'efficiency.'
 
  • #37
candice_84 said:
Can you guide me through calculating actual nuclear plants efficiency?
One could start with Carnot efficiency.

Somewhere you'd want to compare the MWt (thermal energy produced) vs the useful electrical (or turbine mechanical energy to generator) available to sell.
 
  • #38
Ok, for example ap1000 produces 3030 MWth and 1000 MWe. I'd like to know how do we start from 3030 and produce 1000 MWe. I am aware of the fact that is Rankine Cycle and we reuse the steam from High pressure turbine to low pressure turbine. I am interested to know which formula to be used step by step.
 
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  • #39
candice_84 said:
Ok, for example ap1000 produces 3030 MWth and 1000 MWe. I'd like to know how do we start from 3030 and produce 1000 MWe. I am aware of the fact that is Rankine Cycle and we reuse the steam from High pressure turbine to low pressure turbine. I am interested to know which formula to be used step by step.
Well one could take 1000/3030 = ~0.33 or 33% efficiency.

One has to look at the thermal power generated, which input into the primary coolant. So the thermal power must equal the mass flow rate times the change in specific enthalpy.

The energy is transferred through the steam generator, so one has the know the primary circuit mass flow rate and the decrease in the specific enthalpy of the primary coolant in the steam generator, and the mass flow rate and change in specific enthalpy in the secondary circuit.

Then one has to look at the turbines HP, IP (if applicable) and LP turbines. Some fraction (x) of the specific enthalpy will be lost through the HP turbine, and some fraction (1-x) through the IP and LP turbines. One has to know the inlet and exit conditions of the steam, and then the thermal to mechanical conversion efficiency in the turbines.

\dot{W} (power) out of a turbine = η\dot{Q}, where η is the efficiency and \dot{Q} is the rate of change of enthalpy, or \dot{m}Δh.

One could also use the conditions of the condenser, since much of the thermal energy lost directly to the environment is passed through the condenser.

So recent modern turbines enable PWR plants to approach 37% efficiency.

Also one has to differentiate between gross and net efficiency. The NPP must use some of the electrical energy it produces in order to drive pumps and operate various electrical systems.
 
  • #40
If you just want to calculate a value for the efficiency, I think Astronuc has given you a good start. The power companies that operate the plants generally think in terms of 'heat rate' which has units of Btu per kw-hr. This is the ratio of the heat input to the electric output. It is really the same concept as efficiency, only it is the inverse - higher efficiency gives a lower heat rate. You can see how the two are related by noting that one kw-hour is equivalent to 3412 Btu; 33% thermo efficiency would be a heat rate of 10,340 Btu/kw-hr.

If you want to know how the particular value comes to be (as a result of the plant design), well that is a much more detailed subject that can't really be explained fully in this format, it is more like the subject of a mechanical engineering course. But essentially, the plant architect-engineer is usually responsible for designing the so-called 'secondary side' or 'balance of plant.' The design of the entire Rankine cycle is typically represented on a drawing called a 'heat balance diagram.' This shows the path of the steam from the reactor (BWR) or steam generator (PWR) through the turbines and condenser and back through the feedwater system. It includes all of the extraction steam lines and flow paths through the feedwater heaters, as well as the reheat steam between the high and low pressure turbines. The flow rate and thermodynamic state (pressure and enthalpy) at the exit of each component is determined based on the design characteristics of the components. The details of the calculations (for a FW heater, for example) would be based on parameters such as number of tubes, tube area, tube thickness & conductivity, etc., along with the flow rates on the inside & outside of the tubes. The designer's job is to juggle all of this in a way that gives the lowest heat rate, keeping in mind the capital cost to build the unit and the equipment operating and maintenance costs.

I suspect there are people on physicsforums that know a lot more about it than I do; but you might have better luck finding them in the General Engineering section than here in Nuclear land.
 
  • #41
I am actually interested in the balance of plant section :) I am studying thermodynamics and nothing is more interesting than learing how to do this calculation...
 
  • #42
I think you will have better luck getting posts from the people who really know about designing the rankine cycle if you ask this (as a new thread) in the General Engineering forum - those guys probably haven't tuned into this thread ("Reactor Design").
 
  • #43
Does small lattice pitch make -reactivity?
 
  • #44
candice_84 said:
Does small lattice pitch make -reactivity?
Reactivity within a fuel lattice is a function of enrichment (i.e., concentration of fissile nuclides), the geometry and composition of structural material and coolant, concentration of burnable (depletable) neutron absorbers, and temperature. Also, in an LWR, pressure and temperature of the coolant determine its density which has an effect on moderation and neutron spectra.

There is an optimal pitch in which reactivity is maximized. In an LWR, a smaller than optimal pitch would produce undermoderation, while a larger pitch would produce overmoderation.
 
  • #45
I have heard that in some reactors by reducing the size of lattice pitch, one could reduce the positive reactivity.
And it actually make sense because the neutrons that are born in the fuel, will have less moderation (since the lattice is small) and I assume that a little bit of moderation from the small lattice, will keep them in the range of resonance absorption energy range. Therefore all the neutrons that are lowered only to the resonance energy range, after passing the lattice region they will reach the neighboring fuel which has U238, and since the neutrons are in resonance region and U238 have high resonance at that range it will absorb those neutrons, therefore we get negative reactivity as a result. Am I right about this?
 
  • #46
LWRs are optimally designed to moderate fast fission neutrons from energies of about 1 to several MeV to thermal energies, ~ 0.024 MeV. If one reduces the pitch while keeping the rod diameter the same, then there is less moderation, and the spectrum becomes a little harder, i.e., the energy spectrum would shift up a little. Looking at the cross-section data,

U-235
total sigma - http://www.nndc.bnl.gov/sigma/getPlot.jsp?evalid=4551&mf=3&mt=1&nsub=10
total fission sigma - http://www.nndc.bnl.gov/sigma/getPlot.jsp?evalid=4551&mf=3&mt=18&nsub=10

U-238
total sigma - http://www.nndc.bnl.gov/sigma/getPlot.jsp?evalid=4554&mf=3&mt=1&nsub=10
total fission sigma - http://www.nndc.bnl.gov/sigma/getPlot.jsp?evalid=4554&mf=3&mt=18&nsub=10

One can see where the fission cross-section dominates and where absorption (non-fission) dominates. The primary effect for an undermoderated system is the fact that the fission cross-section has 1/v dependency, i.e., as the neutron energy increases the fission cross-section decreases, and so the probability of non-fission absorption increases.

The fission process is controlled by virtue of delayed neutrons, which are neutrons that are emitted from various fission products. Since some neutrons are delayed by several seconds, these enable an LWR system to change power in a controlled manner. Prompt criticality is to be avoided, and introduction of more than $1 of reactivity is also to be avoided.
 
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  • #47
Delayed neutrons are very few in compare to prompt neutrons, but how could they be so effective in controlling the reactor. When neutrons fission, 2.47neutrons are released. most of those neutron are 95% prompt. By the time the delayed neutrons emission (few millisecond), the prompt neutrons would have many fission chain cycle. so how could that delayed neutron be so effective in terms of control?
 
  • #48
candice_84 said:
Delayed neutrons are very few in compare to prompt neutrons, but how could they be so effective in controlling the reactor. When neutrons fission, 2.47neutrons are released. most of those neutron are 95% prompt. By the time the delayed neutrons emission (few millisecond), the prompt neutrons would have many fission chain cycle. so how could that delayed neutron be so effective in terms of control?
Delayed neutrons account for about 0.0065 or 0.65% of the neutrons in the given population. Basically the reactor goes critical with delayed neutrons. They last long enough to effectively increase the mean lifetime of neutrons which allows for controlled power changes.

That fraction (0.0065) of neutrons is also equivalent to $1 of reactivity. If one adds $1 of reactivity to the system, i.e., k ~ 1.0065+ or kprompt > 1, the reactor becomes prompt critical and the power ascension can be dramatic - the power multiplies rapidly in fractions of a second - faster than the reactor can scram.

A reactivity insertion accident is a big deal (potentially significant exposure of plant personnel and increased risk to public, and damage or disruption of the core) and that's why there are strict controls on reactivity increases.

Hopefully when one takes a course in reactor kinetics, it will become apparent and very well appreciated!
 
  • #49
Do fast reactors lower the amount of waste? since U238 fissions. but we only convert a little portion of the U238 to energy.
 
  • #50
candice_84 said:
Do fast reactors lower the amount of waste? since U238 fissions. but we only convert a little portion of the U238 to energy.
Fast reactors produce fission products as well. They can push fuel to higher burnups - maybe 50-100 GWd/tHM (HM = heavy metal) or 5-10% FIMA (fission(s) per initial metal atom), and even higher. If one considers spent fuel, then fast could produce more energy for the same amount of waste. On the other hand, fast reactor fuel is usually of higher enrichment, so there is generally more up front DU produced for a given mass of fast reactor fuel. The selling point of fast reactors is the recycling of Pu and transuranics - but that complicates the fuel cycle because the fuel then has to be manufactured and handled remotely - which makes it much more expensive than conventional LWR fuel.
 

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