Question 8.8: Suppose the reflected reactor in problem [8.7] is a sodium-c......
Suppose the reflected reactor in problem [8.7] is a sodium-cooled fast reactor with M = 18.0 cm and a power of 2000 MW(t). If the thermal design limits the maximum allowable power density to 450 W/cm³,
a. What is (1) the minimum value that the core radius can have, (2) the corresponding value of the core volume, and (3) the required value of k_∞ to maintain criticality?
b. Suppose that to increase the thermal safety margins, it is decided to reduce the maximum permissible power density by 10%. What are the percentage changes for the reflected reactor in radius, volume, and k_∞? (Assume that M remains the same.)
Part a. From Eq. (8.3) we may write
V=F_{q}P/ P_{\mathrm{max}}^{\prime\prime \prime}
and with a height to diameter ratio of one V = 2πR³. Since F_q is given in terms of the ratio x ≡ R / M, we rewrite the volume equation as
x^{3}=\frac{{P}}{2\pi M^{3}P_{\mathrm{max}}^{\prime\prime \prime}}\,F_{q}(x)=\frac{2000\cdot10^{6}}{2\pi18^{3}\cdot450}\,F_{q}(x)=121.3F_{q}(x)
Plugging in the peaking factor and simplifying
x={\frac{229.1(1+x)^{-2}}{{J}_{1}\left[2.405\left(1+x\right)^{-1}x\right]{\mathrm{sin}}\left[(\pi / \,2)\left(1+x\right)^{-1}x\right]}}
We must solve this transcendental equation for x. The value of F_q must be in the range of 1 < F_q < 3.63 , since 3.63 is the value derived in Chapter 7 for a bare reactor. Since x^{3}=12\displaystyle1.\ \!3F_{q}(x)\mathrm{~or~}x=\left(12\displaystyle1.\ \!3F_{q}(x)\right)^{1/3} , we look for a value in the range 4.95 < x < 7.60
The transcendental equation is solved using MathCad by plotting the left and right hand sides, as shown in the figure. The solution is x = 6.78
Therefore, R = xM = (6.78)(18.0 cm) = 122 cm.
Part b. The earlier equation becomes
x^{3}=\frac{P}{2\pi M^{3}P_{\mathrm{max}}^{\prime\prime \prime}\cdot(0.90)}F_{q}(x){=}\frac{121.3}{0.90}F_{q}(x)
which reduces to
0.90x={\frac{229.1{\bigl(}1+x\bigr)^{-2}}{J_{1}\biggl[2.405\bigl(1+x\bigr)^{-1}\,x\biggr] \text{sin}\bigg[ (\pi /2)(1 + x)^{-1}x\bigg]}}
We can obtain the solution off of the same graph by replacing the left hand side with 0.90x. The solution is x = 7.05.
Therefore, R = xM = (7.05)(18.0 cm) = 127 cm.
\mathbf{y}(\mathbf{x}):={\frac{229.1\cdot\left(1+\mathbf{x}\right)^{-2}}{\mathrm{J}1\left[2.405\left(1+\mathbf{x}\right)^{-1}\mathbf{x}\right]\cdot\sin\left[{\frac{\pi}{2}}\cdot\left(1+\mathbf{x}\right)^{-1}\cdot\mathbf{x}\right]}}
\mathbf{y}_{1}(\mathbf{x}):=\mathbf{x}
\mathbf{y}_{2}(\mathbf{x}):=0.90.\mathbf{x}
