Question 8.7: A uniform cylindrical reactor core has a height-to-diameter ......

Part a: For a reflector savings of M, in Eq. (7.21) we add M to the radius of the reactor and 2M to the height. Thus Eq. (7.21) is replaced by

\phi(r,z)=C J_{0}(2.405r/(R+M))\cos(\pi z/(2R+2M))

where H=2R. Correspondingly, Eqs. (8.16) and (8.17) become

C_{r}\,\frac{2}{R^{2}}\int_{0}^{R}\,J_{0}(2.405r\,/(R+M))r d r=1

and

C_{z}\,\frac{1}{2R}\int_{-R}^{R}\,\mathrm{cos}(\pi z\,/(2R+2M))d z=1

where integrals are taken over the core dimensions, and not the extrapolated dimensions. next solve for C_r and C_z. Making use of the Bessel integral B-7 in Appendix B and that fact that J_1(0) = 0

1=C_{r}\,\frac{2}{R^{2}}\int_{0}^{R}\,J_{0}(2.405r/( R+M))r d r=C_{r}\,\frac{2}{R^{2}}\bigg(\frac{R+M}{2.405}\bigg)R J_{1}\bigg(\,2.405\frac{R}{R+M}\bigg)

or

C_{r}=\frac{2.405}{2}\frac{(R/M)(1+R/M)^{-1}}{J_{1}\bigl(2.405(R/M)(1+R/M)^{-1})\big)}\,.

Using the trigonometric integral in Appendix A, along with sin(0) = 0:

1=C_{z}\,{\frac{1}{2R}}{\int_0^R\cos(\pi z/(2R+2M))}dz = C_{z}{\frac{1}{R}}{\frac{(2R+2M)}{\pi}}\mathrm{sin}\biggl({\frac{\pi R}{2(R+M)}}\biggr)

or

C_{z}=\frac{(\pi\,/\,2)(R\,/\,M)(1+R\,/\,M)^{-1}}{\sin\left(\,(\pi\,/\,2\,)(R\,/\,M)(1+R\,/\,M)^{-1}\right)}\,.

Then since the maximum values of both J_0 and cos are equal to one, F_{q}=C_{r}C_{z} and the expression for F_{q} given above results.

Part B:

Without the reflector the result is 3.63 as given in equation (7.30), regardless of the value of R/M.