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Original public paper: Physical PDF pages 826–828. All three questions and all alternatives in question 2 are answered. Optical amplitudes and quadratures use the normalization printed in the paper; factors of two depend on that convention.

Original paper: 22 April 2015, physical PDF pages 751–754.

Original paper: 23 April 2014, physical PDF pages 678–680.

At the stated threshold the idealized system admits a self-sustaining neutron mode. Production and losses balance without the external source. A nonzero continuing source then makes the assumed linear, time-independent response singular rather than selecting a finite equilibrium amplitude.

An apparent infinite steady flux is therefore a failure of that steady linear approximation at the threshold, not a physical prediction that an actual neutron population instantly becomes infinite. Time dependence and feedback must be considered to discuss subsequent behaviour.

Write $R_e=R^{+}$ and apply the stated constant-coefficient boundary value problem. A constant particular solution is $-S/(D{B}^2)$. Regularity at the origin removes $\cos (Br)/r$, soThe outer boundary condition gives $A=S{R}_e/[D{B}^2\sin (B{R}_e)]$, henceAt $r=0$, use the limit $\sin (Br)/r\to B$ rather than substitute into the apparent $0/0$. The result satisfies both the differential equation and $\varphi (R_e)=0$ provided $\sin (B{R}_e)\ne 0$. This is the ideal extrapolated-boundary model specified in the question; a separately resolved outer transport layer would need its own matching conditions.

This part is a radial differential-equation identity. Expanding the spherical derivative givesPut $\psi =r\varphi$. Since ${\psi }^{\prime \prime }=r{\varphi }^{\prime \prime }+2{\varphi }^{\prime }$, multiplication of the equation by $r$ yields ${\psi }^{\prime \prime }+B^2\psi =0$. Its real solution isFor a regular field at the centre, the cosine-over-radius contribution must vanish. The sine-over-radius term is regular because $\sin (Br)/r\to B$ as $r\to 0$.

With no delayed neutrons, set $\beta =0$ and omit the precursor equation. The source-free model reduces to $\dot{n}=(\rho /\Lambda )n$, givingThe delayed-neutron model's dominant e-folding time is about $121$ times larger. This comparison illustrates why a small delayed neutron fraction strongly alters the kinetics. It is a comparison of the supplied idealized models at the same imposed positive reactivity, not a control prescription or a claim that every transient has the same timescale.

The precursor exponential from part (e) has inverse period $\rho \lambda /(\beta -\rho )$. HenceThis is approximately $0.96\%$ shorter than the exact dominant period, so it slightly overestimates growth for these stipulated constants. Neglecting the small $\Lambda$ contribution is more accurate here than replacing the complete frequency dependence by its tangent at the origin.

Differentiating the supplied relation givesThus the tangent approximation is $\rho \simeq \omega (\Lambda +\beta /\lambda )$ andIt is about $16.4\%$ longer than $60.58\mathrm{s}$ and therefore predicts slower growth. For this positive step it is not conservative as a growth-rate estimate. The local expansion needs $|\omega |\ll \lambda$; here the actual ratio is approximately $0.165$, so its error is appreciable.

The idealized isotope mass is $1.63\times 10^{-2}\mathrm{kg}$, about $16.3\mathrm{g}$. Both the alpha particle and daughter recoil energy are recovered in the stipulated heat balance. Thus the recovered energy per transformation isand the nuclear decay constant isWith $N_0=(m/M_{\mathrm{mol}})N_A$, thermal power is $P_0=\lambda N_0{E}_h$. ConsequentlyThe conversion of $210\mathrm{kg}\mathrm{kmo}{\mathrm{l}}^{-1}$ to $0.210\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$ is essential. This is an isotope-only energy-accounting result under the paper's prescribed branch and perfect heat-recovery assumptions; it is not the mass or design of a complete generator.

After three half-lives, $P(T)=P_0/8$. Therefore the minimum initial thermal power isEquality gives the stated terminal power exactly in the idealized exponential decay model. The associated duration is $415.2$ days. No thermal-to-electrical efficiency is needed because both specified powers are thermal.

The approximate population peaks about $11.3\mathrm{h}$ after flux removal. Differentiate the preceding difference of exponentials. The stationary condition isso for distinct nuclear decay constants,Using hours consistently,and thereforeThe approximate population is positive for $t>0$, starts at zero, tends back to zero and has only this one stationary point, so it is a maximum.

The delayed maximum corresponds to a delayed maximum in the magnitude of the absorber's negative contribution to reactor reactivity, within the fixed model. It does not determine a permissible restart time, which would require information absent here. The estimate also inherits the high-flux initial-population approximation of part (ii); the exact solution can peak at a different time when that approximation is unsuitable.

The general solution contains both decay of the initial xenon and continued feeding from the initial iodine. Use $t=0$ for removal of the flux, and retain $\varphi$ as the pre-shutdown value in the initial conditions. Neglect post-shutdown nuclear fission and neutron capture. ThenMultiply this first-order linear differential equation by the integrating factor $e^{{\lambda }_{X}t}$ and integrate:With the preceding steady initial conditions, ${\lambda }_I{I}_0=G={\gamma }_I{\Sigma }_f\varphi$ and $X_0=G/({\lambda }_X+\varphi {\sigma }_{aX})$, this isIf ${\lambda }_X+\varphi {\sigma }_{aX}\gg |{\lambda }_X-{\lambda }_I|$, the coefficient of the first exponential term is small compared with the coefficients of the feeding term, yieldingThis approximation neglects the finite initial xenon population and is consequently not a small relative error at $t=0$, where its value is zero while the exact $X_0$ is positive. The exact expression supplies the initial value and satisfies $\dot{X}(0)=G-{\lambda }_X{X}_0$. For equal nuclear decay constants, the continuous limit of the exact solution is $X(t)=(X_0+{\lambda }_I{I}_{0}t)e^{-{\lambda }_{I}t}$.

Let $G={\gamma }_I{\Sigma }_f\varphi$ be the constant precursor production rate per volume. Here ${\gamma }_I$ is the precursor yield per nuclear fission, and $I$ and $X$ are number densities. At a steady state,For fixed homogeneous coefficients and spatial shape, the reactivity contribution of an added absorption coefficient ${\Sigma }_{aX}={\sigma }_{aX}{X}_0$ is $-{\Sigma }_{aX}/(\nu {\Sigma }_f)$. Thus the model givesWhen capture removal dominates xenon radioactive decay, $\varphi {\sigma }_{aX}\gg {\lambda }_X$, and henceThe limit is independent of the xenon neutron absorption cross section: at high neutron flux, nearly every xenon atom supplied by the precursor ultimately consumes a neutron. The assumptions include constant pre-shutdown neutron flux, sufficient time for precursor and xenon equilibrium, the supplied single-precursor model with no direct xenon fission source, and fixed parameters for translating absorption into reactor reactivity.

The average moderator temperature is $725{}^{\circ }\mathrm{C}$. First the mean coolant temperature in the lumped thermal model isAt a steady state, the same thermal power crosses both thermal resistances. If $R_1=R_2$, thenThusIn particular $R_1=R_2=(800-650)/(2P)=3.17797\times 10^{-7}\mathrm{K}{\mathrm{W}}^{-1}$ at the specified state.

The coolant mass flow rate is approximately $983\mathrm{kg}{\mathrm{s}}^{-1}$. All thermal power leaves in the steady coolant energy rise:With the stated constant specific heat capacity,The $100{}^{\circ }\mathrm{C}$ difference equals a $100\mathrm{K}$ difference. Thermal power is used directly; no electrical conversion efficiency enters.

Substitute the separated solution into the one-group neutron diffusion equation. The radial and axial Laplacian contributions are $-{\alpha }^2\varphi$ and $-{\beta }^2\varphi$. A nonzero solution therefore requiresThe left side is the geometric buckling; the right side is fixed by the stipulated homogeneous material. Their equality is the model's nuclear criticality condition, expressing the balance of production, absorption and leakage. Both sides have inverse-length-squared units.

The fundamental nonnegative mode is ${\varphi }_0{J}_0(2.405r/R)\cos (\pi z/H)$. Put the origin at the centre of the cylinder, so $-H/2\le z\le H/2$. Neglecting extrapolation imposes zero neutron flux at $r=R$ and $z=\pm H/2$. WriteFor an axisymmetric field, separation of variables in cylindrical coordinates givesChoose the axial separation constant so $Z^{\prime \prime }+{\beta }^{2}Z=0$. Midplane symmetry and the lowest positive axial mode give $Z(z)=\cos (\beta z)$, withThe radial equation isThe solution regular on the axis is the zero-order Bessel function of the first kind, $F(r)\propto J_0(\alpha r)$; the independent second-kind solution is singular there. The first radial zero gives $\alpha R=j_{0,1}\simeq 2.405$. Higher zeros would introduce sign changes and do not describe the fundamental nonnegative neutron flux. Normalize by the central value, using $J_0(0)=1$:The printed $2.405$ is a rounded Bessel function zero, rather than an exact root.

Define ${\gamma }_{SV}$, ${\gamma }_{SL}$ and ${\gamma }_{LV}$ as the solid-vapor, solid-liquid and liquid-vapor interfacial free energies per actual area. For the corresponding planar material, Young's contact-angle equation isConsider advancing the macroscopic liquid footprint over projected area $A$, while retaining the specified wetting only to each sphere's equator. The newly wetted upper hemispheres have actual area $2\varphi A$. The gaps occupy projected area $(1-\varphi )A$, across which a liquid-vapor interface remains instead of solid-liquid contact.

The microscopic free-energy change, apart from the outer macroscopic droplet surface, isEquating this with $-{\gamma }_{LV}\cos {\theta }^{\ast }$, the macroscopic wetting contribution, yieldsThe factor $2$ comes from the curved hemisphere area; the negative gap term is the trapped liquid-vapor interface. This combines Wenzel roughness on the wetted solid fraction with Cassie-type air gaps. It differs from both homogeneous Wenzel wetting and a Cassie formula that treats the solid contacts as flat.

For $\varphi =0.9069$ and $\theta =68^{\circ }$, this gives $\cos {\theta }^{\ast }\simeq 0.5864$ and ${\theta }^{\ast }\simeq 54.1^{\circ }$. The hybrid state is less wettable than the fully penetrated hemispherical model, but still not hydrophobic. The result is conditional on the assumed equatorial meniscus and retained air: pinning or metastability may be required, and allowing further penetration can change the equilibrium geometry.

Let $A$ be projected area. Close-packed circular footprints cover $\varphi A$ and the exposed flat substrate covers $(1-\varphi )A$. A hemisphere's curved area is twice its circular footprint, so the actual wetted solid area isThe Wenzel roughness factor is therefore $r=A_{\mathrm{wet}}/A=1+\varphi =1.9069$, not $\varphi$ or $2$.

Using the Wenzel wetting relation,The nanostructure makes this surface more wettable, because planar PMMA already has $\theta <90^{\circ }$. Under complete Wenzel wetting, roughening amplifies the underlying hydrophilic character; it does not turn this material hydrophobic.

The calculation assumes identical chemistry on spheres and substrate, complete liquid penetration, ideal hemispheres, a macroscopic contact angle and negligible pinning/line-tension effects. Actual dried particles may have different contact geometry, deformation, contamination or trapped air. A hydrophobic coating with planar angle above $90^{\circ }$ would make roughness increase the angle instead. Designing a texture that retains air in a Cassie–Baxter state is another route, but its stability against liquid penetration must be checked.