Fluid Mechanics

To determine the energy released when 216 small liquid drops coalesce into a larger drop, follow these steps: Number of small drops $=216$Surface area of each small drop $=A=4 \pi r^2$Surface tension of the liquid $=T$Calculations:Volumes:Volume of one small drop: $\;\frac{4}{3} \pi r^3$Total volume for 216 small drops: $216 \times \;\frac{4}{3} \pi r^3$Volume of the large drop:Since volume is conserved, the volume of the large drop is:$\;\frac{4}{3} \pi R^3 = 216 \times \;\frac{4}{3} \pi r^3$$R^3 = 6 r^3 \;\; \Rightarrow \;\; R = 6 r$Surface Areas:Total surface area of 216 small drops:$216 \times 4 \pi r^2 = 864 \pi r^2$Surface area of the large drop:$4 \pi (6 r)^2 = 144 \pi r^2$Decrease in Surface Area:$\Delta A = 864 \pi r^2 - 144 \pi r^2 = 720 \pi r^2$Energy Released:Using the formula $E = T \cdot \Delta A$ :$\;E = T \cdot 720 \pi r^2$$\;E = 180 \cdot T \cdot 4 \pi r^2$$\;E = 180 A T$Thus, the energy released during the coalescence process is $180 A T$.