I am solving one problem, where the sub term contains a constant $L_v$ that converts the units of $P$ from units of millimeter per hour to watts per square meter. The following is the equation given in the paper.
\begin{equation*} \nabla . v_1 = \frac{L_v(P-\bar{P})+ L_v(Q_{vc}-\bar{Q_{vc}})+(F_s-\bar{F_s})-\xi}{M_s} \end{equation*}
I need this value while computing the $v_1$ term. Can anyone please help me with this. The following is the text given in the paper.
I tried to find the answer of the problem. \begin{eqnarray*} 1 \frac{Watt}{m^2}= \frac{1}{2454000}&=&4.075\times10^{-7} kg. m^{-2}.s^{-1}\\ &=&4.075\times10^{-7} \times 60 \times 60\\ &=& 1.4670\times10^{-3} \frac{mm}{h} \end{eqnarray*} So, $1 \frac{mm}{h} = \frac{1}{1.4670\times10^{-3}}\frac{Watt}{m^2}$.
Is my attempt correct?