14. Given a cube $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, the centers of faces $A B C D$ and $A D D^{\prime} A^{\prime}$ are $M$ and $N$, respectively. Then the sine of the angle formed by the skew lines $M N$ and $B D^{\prime}$ is $\qquad$
14. $\frac{\sqrt{3}}{3}$.

Let the edge length of the cube be 1, and establish a spatial rectangular coordinate system with $\overrightarrow{D A} 、 \overrightarrow{D C} 、 \overrightarrow{D D^{\prime}}$ as the basis. Then $B(1,1,0)$. Therefore,
$$
D^{\prime}(0,0,1) 、 M\left(\frac{1}{2}, \frac{1}{2}, 0\right) 、 N\left(\frac{1}{2}, 0, \frac{1}{2}\right) \text {. }
$$

Thus, $\overrightarrow{B D^{\prime}}=(-1,-1,1)$,
$$
\begin{array}{l}
\overrightarrow{M N}=\left(0,-\frac{1}{2}, \frac{1}{2}\right) . \\
\text { Since } \overrightarrow{B D^{\prime}} \cdot \overrightarrow{M N}=1,\left|\overrightarrow{B D^{\prime}}\right|=\sqrt{3},|\overrightarrow{M N}|=\frac{\sqrt{2}}{2},
\end{array}
$$

Therefore,
$$
\begin{array}{l}
\cos \left\langle\overrightarrow{B D^{\prime}}, \overrightarrow{M N}\right\rangle=\frac{\overrightarrow{B D^{\prime}} \cdot \overrightarrow{M N}}{\left|\overrightarrow{B D^{\prime}}\right| \cdot|\overrightarrow{M N}|}=\frac{\sqrt{6}}{3} \\
\Rightarrow \sin \left\langle\overrightarrow{B D^{\prime}}, \overrightarrow{M N}\right\rangle=\frac{\sqrt{3}}{3} .
\end{array}
$$