Question 9.8: In the prismatic block form of the graphite moderated gas-co......

Part a. See solution to problem [8.14] for transient analysis. By setting the time derivatives to zero, we obtain the steady state heat transfer:

$P={\frac{1}{R_{1}}}\left[{\overline{{T}}}_{f}-{\overline{{T}}}_{m}\right]$

$\frac{1}{R_{1}}[\overline{{{T}}}_{f}-\overline{{{T}}}_{m}]=\frac{1}{R_{2}}[\overline{{{T}}}_{m}-\overline{{{T}}}_{c}]$

${\frac{1}{R_2}}[{\overline{{T}}}_{m}-{\overline{{T}}_{c}}]=2W c_{c}\ [{\overline{{T}}}_{c}-T_{i}]$

from which we may write the steady state model as:

$\overline{{{T}}}_{c}=\frac{1}{2W{\it c}_{c}}P+T_{i}$

$\overline{{{T}}}_{m}=\left(\,R_{2}+\frac{1}{2W c_{p}}\,\right)P+T_{i}$

$\overline{{{T}}}_{f}=\left(R_{1}+R_{2}+\frac{1}{2W c_{p}}\right)P+T_{i}$

Part b: Let $α_f , α_m$ and  $α_c$ be the fuel, moderator and coolant temperature coefficients. The isothermal temperature coefficient is them simply:

$\alpha_{T}=\alpha_{_{f}}+{{{\alpha}}}_{m}+\alpha_{c}$

Part c: The power coefficient is just a generalization of Eqs. (9.30) through (9.33) to include the third region:

$\alpha_{P}=\alpha_{f}\,\frac{d}{dP}\,\overline{{{T}}}_{f}+\alpha_{m}\,\frac{d}{d P}\,\overline{{{T}}}_{m}+\alpha_{c}\,\frac{d}{d P}\,\overline{{{T}}}_{c}$

assuming a constant inlet temperature. Find the power derivatives from the three final equations in part a. We then have

$\alpha_{P}=\left(R_{1}+R_{2}+{\frac{1}{2W c_{p}}}\right)\alpha_{f}+\left(R_{2}+{\frac{1}{2W c_{p}}}\right)\alpha_{m}+{\frac{1}{2W c_{p}}}\alpha_{c}$