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=4πr2
Surface tension of the liquid =T
Calculations:
Volumes:
Volume of one small drop: ‌

4

3

πr3
Total volume for 216 small drops: 216×‌

4

3

πr3
Volume of the large drop:
Since volume is conserved, the volume of the large drop is:
‌‌

4

3

πR3=216×‌

4

3

πr3
‌R3=6r3‌‌⇒‌‌R=6r
Surface Areas:
Total surface area of 216 small drops:
216×4πr2=864πr2
Surface area of the large drop:
4π(6r)2=144πr2
Decrease in Surface Area:
∆A=864πr2−144πr2=720πr2
Energy Released:
Using the formula E=T⋅∆A :
‌E=T⋅720πr2
‌E=180⋅T⋅4πr2
‌E=180AT
Thus, the energy released during the coalescence process is 180AT.