${}_2(1+cos)=2\sqrt[\phantom{2}]{{}_1^d{}_2^d}+2\sqrt[\phantom{2}]{{}_1^p{}_2^p}$ Owens$-$Wendt$-$geometric mean

${}_2={}_2^d+{}_2^p,{}_3={}_3^d+{}_3^p$

${}_2^d=22\mathrm{.}0m\frac{J}{m^2}$ DI water Surface energy $-$ dispersion term

${}_2^p=50\mathrm{.}2m\frac{J}{m^2}$ DI water Surface energy $-$ polar term

${}_3^d=48\mathrm{.}5m\frac{J}{m^2}$ DIIO Surface energy $-$ dispersion term

${}_3^p=2\mathrm{.}3m\frac{J}{m^2}$ DIIO Surface energy $-$ polar term

Under the Condition of Phase; $1=$ Solid $2=$ Water $3=$ Diiodomethane

${}_2(1+cos{}_2)=2\sqrt[\phantom{2}]{{}_1^d{}_2^d}+2\sqrt[\phantom{2}]{{}_1^p{}_2^p}$

${}_3(1+cos{}_3)=2\sqrt[\phantom{2}]{{}_1^d{}_3^d}+2\sqrt[\phantom{2}]{{}_1^p{}_3^p}$

With the value of $,{}_2^d,{}_2^d,{}_3^d,{}_3^d,$ we can know ${}_s={}_1^d+{}_1^d$

$2-1)$ Calculate the surface energy of each substrate $($SiO$2,$ HMDS$)$ with Owens$-$Wendt geometric mean.

$($a$)$ Surface Energy polar term of Bare SiO$2$ substrate $({}_1^p)$

$($b$)$ Surface Energy dispersion term Bare SiO$2$ substrate $({}_1^d)$

$($c$)$ Surface Energy of Bare SiO$2$ substrate $({}_1)$

$($d$)$ Surface Energy polar term of HMDS$-$treated SiO$2$ substrate $({}_1^p)$

$($e$)$ Surface Energy dispersion term HMDS$-$treated SiO$2$ substrate $({}_1^d)$

$($f$)$ Surface Energy of HMDS$-$treated SiO$2$ substrate $({}_1)$