Exercises 3
- The figure below shows the neutron cycle for a finite, critical reactor. The notes in the figure refer to an arbitrary unit of time. Calculate the number of epithermal and fast neutrons capturedper unit time.
Solution:
Thermal fissions =1000/2.5=400
Fast fissions =0.025*400=10 Neutrons from fast fission = 10*2.7=27
Total number of fission neutrons =1027
Number of fission neutrons which escape leakage = 1027-7=1020
Thermal captures =1.25*400=500
Thermal absorptions in fuel = 500+400=900
Total number of thermal neutrons captured =900+62=962
Thermal leakage = 3*7=21
Total number of thermal neutrons = 962+21=983
Epithermal + Fast neutrons captured = 1020-983=37 (final answer)
2. Consider a reactor with a mean neutron-generation time of 0.0014 s and 1 delayed-neutron-precursor group with a delayed-neutron fraction of 0.0055 and a decay constant of 0.012 s-1.
The reactor is running in steady state with a neutron density n =4*108 cm-3 and at a neutron power of 1 (arbitrary units), when a perturbation suddenly inserts positive reactivity, and the neutron power jumps to a value of 1.15.
(a) What was the reactivity inserted by the perturbation?
Solution:
(b) If thisreactivity is not countered or changed, when will the neutron power reach 1.35?
Solution:
(c) In the initial steady state, what was the value of the delayed source, in cm-3.ms-1?
Solution:
3. A research reactor fuelled with 235U is operated in critical steady state at a power of 40 MW. It is in the shape of a sphere with a diameter of 2.2 m. The average one-group properties are = 2.4, f= 0.0025 cm-1, and D=1.1 cm.
(a) Calculate the material buckling.
Solution:
(b) What is the value of a?
Solution:
(c) If one fission releases 200 MeV (and 1 MeV = 1.6*10-13 J), what is the average value of the thermal flux?
Solution:
(d) At what rate is 235U being consumed by fission?
Solution:
(e) What fraction of neutrons leaks out of the reactor?
Solution:
4.A subcritical reactor is on its approach to critical. At some point in time, a detector in the core has a reading of 5 (arbitrary units). The approach to critical is halted, and to determine the reactivity value, a control rod whose reactivity is known to be -0.2 mk is inserted into the core; the detector then reads 4.5 units.
(a) What was the reactivity of the core without the control rod?
Solution:
(b) If the control rod is removed and the approach to critical is resumed, what will be the reading of the detector when the core reactivity reaches -0.6 mk?
Solution:
(c) What is the general criterion which must be satisfied for the calculation method which you used in parts (a) and (b) to be reasonably accurate?
Solution: The reactivity inserted with the control rod must not be so large as to perturb the flux shape significantly.
5. Consider a reactor with an average neutron-generation time = 1 ms, and 4 delayed-neutron-precursor groups, with the following properties:
1 = 0.0002, 1= 0.0133 s-1
2 = 0.0014, 2= 0.09 s-1
3 = 0.0035, 3= 0.5 s-1
4 = 0.0017, 4= 2.0 s-1
(a) How many terms will there be in the general solution for the time evolution of the flux in a transient in this reactor?
Solution: Number of (exponential) terms = 4+1=5
(b) How many of these terms will have a negative exponent, how many a positive exponent, how many a 0 exponent?
Solution: 4 will definitely have a negative exponent. The 5th exponent will be positive, 0, or negative if the reactivity insertion is > 0, =0, or < 0.
(c) If a very large negative reactivity is inserted in this reactor, what will be the stable period?
Solution:
After a very large negative reactivity insertion, the stable period will be the inverse of the lambda with the smallest absolute value, i.e., = -1/1= 1/(-0.0133s-1) = -75.2 s
(d) What reactivity insertion in this reactor would result in prompt criticality?
Solution: Prompt criticality would occur at a reactivity insertion of
= 0.0002+0.0014+0.0035+0.0017 = 0.0068 =6.8 mk
6. The decay constants and direct yields in fission of I-135 and Xe-135 are:
- I = 2.92*10-5 s-1
- X = 2.10*10-5 s-1
- I = 0.0638
- X = 0.0025
The Xe-135 neutron-absorption cross section and the fission cross section are:
- X = 3.20*10-18 cm2
- f = 0.002 cm-1
At a certain point in a transient, the Xe-135 concentration is 2*1013 nuclides.cm-3, and the rates of production of Xe-135 from I-135 decay and directly from fission are 6*109 cm-3.s-1 and 3*108 cm-3.s-1respectively.
(a)What is the instantaneous I-135 concentration?
Solution:
(b)What is the instantaneous flux?
Solution:
(c)What are the instantaneous net rates of change of the I-135 and Xe-135 concentrations?
Solution:
7. One day, Hollywood puts out a “catastrophe” movie based on the premise of a giant uncontrolled xenon oscillation in the zero-power research reactor ZED-2 at Chalk River (Hollywood is very smart and knows that ZED-2 does not have the devices in referred to in part a). But why is this premise flawed anyway?
Solution: There is no xenon concentration at 0 power, therefore there cannot be xenon oscillations in a zero-power research reactor.
8. (a) What are the units and the meaning of the coolant-temperature reactivity coefficient?
Solution:
Units: mk/(g.cm-3 of coolant)
Meaning: The derivative of the reactivity with respect to the coolant temperature.
(b) A certain “standard” CANDU reactor has a fuel-temperature reactivity coefficient of -0.009 mk/oC. When the reactor is at full power (FP), the average fuel temperature is 690 oC. At hot shutdown (0% neutron power), the average fuel temperature is 270 oC. If the reactor power is increased from FP to 105% FP, estimate the change in reactivity (including its sign), assuming that there is no other change besides a fuel-temperature change.
Solution:
(c) If coolant-density changes are also considered, do you think that the change in reactivity as the power is increased to 105% FP is likely to be algebraically greater or smaller than the value you calculated in (b)? Explain why.
Solution: The change in reactivity would be algebraically larger, because at the higher power there may be some coolant boiling (or density decrease), which would bring in a part of the positive “coolant-void reactivity”.
9. Consider nuclear material with the following properties in 2 energy groups:
a1 = 0.0012 cm-1, 12 = 0.007 cm-1, a2 = 0.0045 cm-1, f2 = 0.0056 cm-1
D1= 1.08 cm, D2 = 0.90 cm
(a) Calculate thereactivityof an infinitelattice made of this material.
Solution:
(b)Now let’s make a homogeneous cylindrical reactor with this material. Its dimensions are diameter 5.2 m and height 5 m [neglect the extrapolation distance]. Calculate the fast and thermal non-leakage probabilities for this reactor, and its reactivity.
Solution:
10. Does a shutdown system reduce reactor power more quickly or less quickly, in the presence of delayed neutrons (compared to the case without delayed neutrons)? What is the physical reason?
Solution: Less quickly, because delayed neutrons “hold back” any power change.