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REACTOR KINETICS: • The study of time-dependent phenomena in nuclear reactor systems is referred to as reactor kinetics or reactor dynamics. • Until now, we have considered critical systems. We now examine systems which are super-critical or sub-critical. • Some neutrons emitted as a result of fission appear in the system seconds after fission (as a result of fission product decay) resulting in a dominant influence on kinetic behavior of the system. PROMPT NEUTRON LIFETIME: In general, a neutron produced by fission will experience scattering interactions which will decrease its energy. Eventually it will either be absorbed or it will leak out of the system. The average time a neutron exists in an infinite system can be estimated by noting that the average distance of travel without being 1 absorbed is the mean free path, a a If the neutron has an average speed, it will take a time, l, to interact. a 1 l V V a Here l is the prompt neutron lifetime. The lifetime is often broken into slowing-down and thermal lifetime with the average results: lSD ~ 10-7 sec lTh ~ 10-4 sec Where lSD -fast reactor lTh - thermal reactor l = lSD+ lTh ~ lTh = 10-4 sec 1 In a finite system, if a fraction of the neutrons leak from the system, this will result in a decrease of neutron lifetime. l l finite D 1 B2 a D B2 -leakage fraction a (for large reactors, leakage fraction is only a few percent) The average neutron lifetime is the characteristic multiplication time or generation time for a prompt neutron system. If there are N(t) Neutrons in the system at a given time, t, at a time, l, later there will be N(t+l) neutrons. The magnitude of N(t+l) will depend on keff . N(t+l) = N(t)keff or dN t N t l ) N t l N t keff dt Rearranging the above (Taylor expanded) expression for N(t+l) gives: dN t keff 1 keff t t dt l l Where keff is the excess multiplication excess over the critical multiplication factor keff = 1. This differential equation form has a general solution: keff t N t N 0 e For neutron population, N(t), at time t, based on an initial population, 1 N(0) at arbitrary time t=0. When t , the population will change by a keff factor of e; this quantity is called the period: 2 1 Prompt Neutron Period = TP keff 3 For a thermal reactor where l ~ 10-4 an excess multiplication of 0.1% (keff = 1.00; keff = 1.001-1.00 = 0.001) will give 1 x104 sec TP 0.1sec 1 x103 or t N t N 0 e 0.1sec so that one second 1.0sec N t N 0 e 0.1sec N 0 e10 22,026 N 0 An increase in power of a factor of over 20,000 in one second. For fast reactors where l ~ 10-7, the power increase would, of course, be substantially more rapid. On the basis of such typical "time constants", reactor systems would be very difficult to control if only prompt neutrons were available. 4 DELAYED NEUTRONS: Recall that of the v neutrons associated with each fission event, a small fraction result from fission product decay, denoting this fraction by the symbol ; we then have: Delayed Neutrons/fission V Prompt Neutrons/fission (1-V Then, considering the average time between the fissions and the absorption (or leakage) of the neutrons, we have an effective lifetime based on both prompt and delayed neutrons of: I i leff 1 l j 1 i where i is the fraction of the v neutrons associated with each of the I precursor decay chains. KINETIC EQUATIONS: Recall that the general neutron balance is [net increase] = [production] - [absorption] - [leakage] Using the diffusion theory approximation theory for the leakage term and assuming that the spatial and time dependents are separable, we can write: dN t S t DB 2 t t dt Here the parameters are arranged over space, energy and the source term will be composed of prompt and delayed neutron parts: 6 S t 1 V t t i C i t i 1 5 where concentration of the i precursor CI(t) must be determined from the equation. 6 For convenience, the twenty-or-so known precursor nuclides are often represented by six "effective" delayed neutron groups: 6 i leff 1 l i 1 i Where the number of delayed neutrons per fission, and the precursor half-lives are shown below for the important fissile nuclides. To calculate leff it is convenient to convert as per the table for U235 with the result that for a thermal system with l ~ 10-4 leff = (1-0.0065)x10-4 + 0.084 ~ 0.084 sec Using leff rather than l as a characteristic multiplication time will greatly enhance the stability of the system to small changes in multiplication. dC i t iV t C i t ai t dt Fission Product Decay Rate Absorption Rate Due to the fact that these nuclides normally have small absorption cross sections, the last term can usually be neglected. By using the definitions of k00 and keff from diffusion theory along with the following: t vN t keff 1 keff l 1 Neutron Generation l k k v a 7 We can rewrite the equations for neutron and precursor population in a simplified form. dN t 6 N t iCi t dt l i 1 dC i t i N t i Ci t dt l This form is known as the point kinetic. Seven linear, coupled differential equations which can be solved in a straight forward manner. One delayed neutron group: Some insight into the kinetic behavior of a reactor can be gained by considering the effect of a single "average" group of delayed neutrons, where: i i 6 i i 1 i dN t N t C t dt l dC t N t C t dt l Using a general solution to form e wt for both N(t) and C(t) and an assumption of an initially critical system with "step-change" of reactivity. We have: t TP N t N 0 e e l l TP delayed prompt 8 Note that the equation can be very different depending on the relationship between and . 1. The second term will be positive as will the exponential; l x is small, the increase will be rapid, out-stripping the delayed neutron effect in the first term. : This system is said to be prompt critical (critical on prompt neutrons only.) : This system is prompt super critical since the relationship between and is significant, a reactivity unit, the dollar was defined: $1 so $1 is prompt critical 1 yes, $0.01 = 1 cent 100 2. The first term will dominate falling a prompt-jump due to the second term. The system in not critical on prompt neutrons alone (the additional prompt neutrons produce less than one net prompt neutron in the next generation and must wait for the delayed neutrons to continue the supercritical increase. This system will establish a stable period characteristic of the delayed neutron period. 3. The second term causes a prompt-drop due to decreased fission rate; the first term then dominates since the delayed precursor nuclides must decay out for the neutron population to die out completely. In this case the period of decay is that of the one delayed group. 9 SIX DELAYED GROUPS: With all six groups effective groups, the procedures and results are similar. 1. prompt critical 2. established a stable period, TP, after the transient terms (6 ??) dies out. l 6 i TP i 1 1 iTP 3. stable period of decrease established by the longest precursor half-life: t1/2 = 56 seconds TP = 80 10 PROBLEM 1: It has been determined that the reactor is in a condition such that keff is equal to 0.95. Do you have to add or subtract reactivity to make the reactor critical? SOLUTION: Add. k 1 k How much reactivity do you have to add to make the reactor critical? k = 0.95 k 1 0.95 1 0.053 k 0.95 k=1 k 0 0.053 0.053 k 11 PROBLEM 2: Reactor power is 100 watts and increasing on a 30 second period. How long will it take for the reactor power to reach 500 kilowatts? t t N t N 0 e P T N t N 0 e TP t P P0 e T t P eT P0 t P ln TP P0 P t TP ln P0 = 30 ln(500,000 / 100) = 255 sec = 4.25 min 12 PROBLEM 3: Calculate the reactor period for the given conditions (assume that eff = 0.007.) keff = 1.0001 for a prompt critical reactor k eff 1 0.0001 k keff k l T l TP l*-provided @ (0.0001 sec) 0.0001 0.0001 1sec What is the reactor period for the same reactivity insertion if the reactor is not Prompt Critical? T = 0.01 0.0075 0.0001 0.1 0.0001 690sec 13 PROBLEM 4: At t=0 the reactor power at 100 kw. Thirty seconds later the power level is 150 kw. What is the reactivity? t P t P 0 e T t P t ln T P 0 P t T t / ln P 0 30 150 ln 100 74sec 1 from P 1 T TP 0.0075 1 0.1 74 k 0.00089 k 14 PROBLEM 5: Calculate the final stable reactor period if a rod is inserted with a k reactivity of 0.005 . k 0.0065 0.005 TP 0.1 0.005 23sec Not So! = -80 seconds! 15 PROBLEM 6: A reactor is made subcritical by a large negative reactivity insertion. (a) Determine the stable negative period that is reached during shut- down. Reactor neutron level is maintained by the decay of delayed neutron sources during shutdown. After the shorter-lived fission products decay away, neutron population is maintained by the decay of Br-87 with a 55.72 second half-life. This decay causes neutron level to decline on an apparent period that is determined by: t 0.693 t P P0 e P0 e T T2 t 0.693 t eT e T2 t 0.693T2 T T2 T 0.693 55.72sec 0.693 80sec (b) How long does it take to establish the stable T=-80 sec after the rods are inserted? The next longest-lived precursor is I-137 with a 24.4 second half-life. Using the seven half-life rule, I-137 will decay away to less than 1% of its steady state operating value 7x24.4 = 171 sec ~`3 min. 16 REACTIVITY CONTROL: Objective: to control neutron population; introduce material that absorbs neutrons. The system may use one or more of the following to provide absorption: •control rods •soluble poison (boric acid) •burnable poison Short Term Changes: 1. Fuel temperature 2. Moderate and/or coolant temperature 3. Fuel motion Each of these is often quantified in terms of a reactivity coefficient. 17 TEMPERATURE EFFECTS: •Dominant absorption effect is known as the Doppler: 1-1000 ev •Coolant and Moderator Effects: •Changes due to: •temperature •pressure •density * •void contact *Decrease Density •increase leakage •decrease moderation •decrease moderator absorption •Fuel movement: •for example, Bowing 18 Limiting Conditions Reactivity and Period Reactivity Limit Reactor Period Reactor Condition k/k (sec) > eff * T l / Prompt critcal eff > > T eff / Super critical T Critical 0 > eff/1-80 T Sub critical eff/1-80 T Sub critical 19 REACTOR SHUTDOWN: • reactor period limited by the radioactive decay of the longest lived neutron precursor Br-80 Show that the shortest period for shutdown is -80 sec. .693t t N t N0 e t 1/2 N 0 e T for this to be true then . 693t t t 1/2 T and T t 1/2 T . 693 by substituting Br-80 half-life 55.72 sec results in 55. 72 sec T 80 sec .693 20 XENON AND SAMARIUM POISONING: • All fission products absorb neutrons. • The accumulation of fission products reduce the multiplication factor k • This reduction can be described with reactivity aP f p In general the mass balance for any fission fragment being produced in a reactor is: accumulation = production - destruction production = fission product + decay product fi-1 destruction = decay + absorption a Therefore f N i1 a N dN dt 21 For Xenon (11sec ) (6.7hour) (9.2hour) (10 6 years ) Te135 I135 Xe135 135 Ba 135 (stable) Cs fission fission fission While the reactor is in operation and combining the fission yields for I and Te the following system of equations results dI I f I dt x f I x aX X dX dt At equilibrium: I I f I I X f X X aX After reactor shutdown X t X 0 e t X II0 I X e t X e t I 22 For Samarium By Simplifying the Sm chain, During reactor operation the concentration of Sm and Pm are: dP P f PP dt dS P P aS S dt at equilibrium: Pf P P S P f aS After Shutdown St S 0 P 0 1 e tP 23 TEMPERATURE EFFECTS ON REACTIVITY: d T dT Nuclear Doppler Effect • absorption cross sections vary with temperature • nuclei are in atoms which are continual motion due to their thermal energy As a result of these thermal motions even of beam of monoenergetic neutrons appear to have a continuous spread of energies. Therefore the resonance peak widens with temperature. (shape changes but the area under the curve remains essentially constant i 1 prompt ln 2 T p(300) Moderator coefficent • thermal cross sections change • physical density of the moderator changes mod T f T (p) T ( P) Void coefficient In a liquid moderator and/or cooled reactor when the liquid boils the volume occupied by the vapor is a void. These voids effect reacivity. 24 d v dx EXAMPLE: Determine the limiting temperature of a reactor that has a excess reactivity of 0.50 % at 80 F and a temperature coefficient of reactivity of -0.000040/F. Assume the coefficient is constant throughout the temperature range involved d T dT Rearrange T T Substitute . 0050 T 125F . 000040 Therefore the reactor will become subcritical if the temperature rises above 205F (80+125) 25 EXAMPLE: A triga operating at 100 kW is inadvertantly exposed to a positive reactivity of $.90 for 5 seconds when the operator recognizes the unplanned reactivity increase and scrams the reactor decreasing the power level to 10% of its original power. What reacivity was inserted to scram the reactor? t1 t2 P1 1 T1 P2 1 T2 e and e P 0 1 1 $ P1 1 2 $ by substitution t2 t1 P2 1 1 T 2 T1 e e P 0 1 1 $ 1 2 $ Solve for p2$ t1 t2 P 1 T T 2 $ 1 0 e 1 e 2 P2 1 1 $ Identify Knowns 1 $ 0. 95 t 1 5sec t 2 60 sec Scram = fastest period a reactor can shut down is -80 sec therefore T2=-80 sec Still need T1 26 1 1 $ T1 1 $ By substitution 1. 90 T1 1.39 sec . 08 . 90 Now solve for p2$ 5 60 100 1 1.39 80 2 $ 1 e e 1723 10 1. 90 What is P1 t1 5 1 T1 1 P1 P0 e P1 100e 1.39 1 1 $ 1. 95 P1=72986 kW 27 CONTROL RODS: Black rod - strongly absorbing rod Grey rod - less strongly absorbing rod For a Black Rod the reactivity can be expressed by 7. 43M2 1 T R d w 0.116 ln 0 T 1 B 2 M 2 R 2 2. 405a a Incremental reactivity for control rods x 1 2 x w (x ) w (H) sin H 2 H Typical Geometry’s Cluster Control Rods Cruciform Control Rods Chemical Shim Affects the thermal utilization factor f = absorption in the fuel divided by the absorption in the fuel + everything else. aB reactivity is w 1 f 0 aM where f0 is the thermal utilization factor without the chemical 28

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posted: | 2/24/2010 |

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