A few more remarks

chapters/chapter3/monoblocks.tex

@@ -249,7 +249,7 @@ \subsection{Influence of cycling}\labsec{influence of cycling}
 Simulating these transient cycles would require stepsizes of $\approx \SI{10}{s}$ in order to capture the ramp-up and ramp-down phases.
 Simulating one cycle would therefore require more than 60 steps (excluding the resting phase).
 
-On the other hand, FESTIM has an adaptive stepsize feature allowing the stepsize to increase (resp. decrease) when steps are solved in less (resp. more) than 5 Newton iterations.
+On the other hand, FESTIM has an adaptive stepsize feature allowing the stepsize to increase (resp. decrease) when steps are solved in less (resp. more) than five Newton iterations.
 Therefore, if a continuous plasma exposure was simulated, the adaptive stepsize would allow the stepsize to increase up to thousands of seconds, reducing a lot the simulation time.
 
 To verify the validity of the continuous exposure approximation, 1D simulations were run with plasma cycles or continuous exposure.

chapters/chapter4/divertor.tex

@@ -155,7 +155,7 @@ \section{ITER results}
 \begin{figure*}[h!]
     \captionsetup[subfigure]{format=plain,singlelinecheck=true}  % needed to center the subcaptions
     \centering
-    \begin{subfigure}{0.42\linewidth}
+    \begin{subfigure}{0.40\linewidth}
         \includegraphics[width=\linewidth]{Figures/Chapter4/ITER/inventory_along_inner_divertor.pdf}
         \caption{Inner Vertical Target.}
     \end{subfigure}%

chapters/chapter5/He_transport_in_PFCs/model.tex

@@ -177,7 +177,7 @@ \subsection{Grouped approach}
     \langle r_b \rangle = r_{\mathrm{He}_0 \mathrm{V}_1} + \left(\frac{3}{4 \pi} \frac{a_0^3}{2} \right)^{1/3} \frac{1}{c_b}\sum\limits_{i=N+1}^\infty c_i(\frac{i}{4})^{1/3} - \left(\frac{3}{4 \pi} \frac{a_0^3}{2} \right)^{1/3}
 \end{equation}
 
-Assuming $c_i$ follows a narrow gaussian distribution, $\frac{1}{c_b}\sum\limits_{i=N+1}^\infty c_i(\frac{i}{4})^{1/3} \approx \left( \frac{1}{c_b}\sum\limits_{i=N+1}^\infty c_i\frac{i}{4} \right)^{1/3} $ (see \reffig{sum of powers approximation}).
+Assuming $c_i$ follows a narrow gaussian distribution \cite{faney_spatially_2015}, $\frac{1}{c_b}\sum\limits_{i=N+1}^\infty c_i(\frac{i}{4})^{1/3} \approx \left( \frac{1}{c_b}\sum\limits_{i=N+1}^\infty c_i\frac{i}{4} \right)^{1/3} $ (see \reffig{sum of powers approximation}).
 % is assumed to be only dependent on $\langle i_b \rangle$.
 
 \begin{figure}