Question 8.6: You are to design a 3000 MW(t) pressurized water reactor. Th......
You are to design a 3000 MW(t) pressurized water reactor. The reactor is a uniform bare cylinder with a height-to-diameter ratio of one. The coolant to fuel volume ratio is two to one in a square lattice. The volumes occupied by control and structural materials, as well as the extrapolation distances, can be neglected. The core inlet temperature is 290 °C. The reactor must operate under three thermal constraints:
1. maximum power density = 250 W/cm³
2. maximum cladding surface heat flux = 125 W/cm²
3. maximum core outlet temperature = 330 °C
Determine:
a. the reactor dimensions and volume
b. the fuel element diameter and lattice pitch
c. the approximate number of fuel elements
d. the mass flow rate and average coolant velocity
Part a. The peak to average power density in a uniform cylindrical reactor is 3.63 Thus
P_{\mathrm{max}}^{\prime \prime \prime}=3.63{\overline{{P}}}^{\prime \prime \prime}=3.63 \ \ P /V
or
V=3.63P/P_{\mathrm{max}}^{\prime \prime \prime}=3.63\cdot3000~\mathrm{MW}/\,250(\mathrm{MW/m^{3}})
V=43.6\operatorname{m}^{3}
V=\pi{\frac{D^{2}}{4}}\,H=\pi{\frac{D^{3}}{4}}
D=H=(4V/\pi)^{1/3}=(4\cdot43.6/\pi)^{1/3}=3.81\,\mathrm{m}
Part b. We also Have for the cladding heat flux q_{\mathrm{max}}^{\prime \prime}=3.63{\overline{{q}}}^{\prime \prime} or
{\overline{{q}}}^{\prime\prime}=q_{\mathrm{max}}^{\prime \prime}\ /3.63=125/3.63=34.4~\mathrm{W/cm}
but if N is the number of rods and a is their radius, the total heat transfer surface is 2πaHN, and
2\pi a H N\overline{{{q}}}^{\prime \prime}=P or
a N={\frac{P}{2{\pi}{ H}\overline{{{q}}}^{\prime \prime}}}={\frac{3\cdot10^{9}\,\mathrm{W}}{2{\pi}\cdot381~\mathrm{cm}\cdot34.4\,\mathrm{W/cm}}}=3.64\cdot10^{4}\,\,\mathrm{cm}
However, if p is the lattice pitch, then
N p^{2}=\pi D^{2}\;/\,4=\pi D^{2}\;/\,4=11.4\;\mathrm{m^{2}}=11.4\cdot10^{4}\;c\mathrm{m^{2}}
Divide this equation by aN:
{\frac{p^{2}}{a}}={\frac{N p^{2}}{a N}}={\frac{11.4\cdot10^{4}}{3.64\cdot10^{4}}}=3.1\,\mathrm{cm}
Also
{\frac{p^{2}-\pi a^{2}}{\pi a^{2}}}={\frac{1}{\pi}}\left({p\ /a}\right)^{2}-1=2.0 (moderator to fuel volume ratio)
p/a={\sqrt{3\pi}}=3.07
Thus
{\frac{p^{2}}{a}}=p\left({\frac{p}{a}}\right) or {3}\cdot{1}=p\cdot{3}.07~\mathrm{or}~p={1.01~c m} for the lattice pitch
The fuel diameter d=2a=2(a/p)p={\frac{2\cdot1.01}{3.07}}=0.658\,\mathrm{cm}
Part c.
N=11.4\cdot10^{4}\ /\ p^{2}=11.4\cdot10^{4}\ /1.01^{2}=112\cdot10^{3} fuel elements
Part d: Since for a uniform cylindrical core F_{r}=2.32 and from Eq.(8.42)
\left.W=\frac{F_{r}P}{c_{p}\left(T_{0}|_{\mathrm{max}}-T_{i}\right)}\right.
Take c_{_P}=6.4\cdot10^{3}\ \mathrm{J/K}g°C (see part e of the PWR example in section 8.3)
W=\frac{2.32\cdot3\cdot10^{9}\mathrm{~J~}/s}{6.4\cdot10^{3}\left(\mathrm{J/Kg^{o}C}\right)(330^{o}C-290^{\circ}C)}=27.2\cdot10^{3}~\mathrm{Kg/S}
Taking the density as 0.676 10^{-3} kg/cm³ (see part e of the PWR example)
{\overline{\text{v}}}={\frac{\mathit{W}}{\rho N(p^{2}-\pi a^{2})}}={\frac{27.2\cdot10^{3}}{0.676\cdot10^{-3}\cdot112\cdot10^{3}(1.01^{2}-\pi\cdot0.329^{2})}}
\overline{\text{v}}=528\mathrm{~}\mathrm{cm/s}=5.28\mathrm{~m/s}