Workdone in spraying a liquid droplet of radius r into n droplet
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If the droplet is sprayed into multiple droplets, the total volume is unchanged. Thus, we can write
n(4/3)πr³ = (4/3)πR³
where R is the radius of the initial drop and r is the radius of the droplets.
We can divide the equation by (4/3)π and we get
nr³ = R³
or
r = R(1/n)¹/³
The total surface area of the droplets is n4πr².
We rewrite it by using the last equation
n4πr² = n4πR²(1/n)²/³ = 4πR²n¹/³
4πR² is the surface area of the original drop, so the total area has increased by a factor of n¹/³.
The surface energy is the work required to increase the surface area of a substance by unit area. Then, the surface energy is proportional to surface area.
Hence, the surface energy has increased by a factor n¹/³ and is n¹/³u
The original surface energy was u, so the increase in surface energy is
n¹/³u – u = [n¹/³ – 1]u
If n = 1000 then n¹/³ = 10
and [n¹/³ – 1]u = [10 – 1]u = 9u
n(4/3)πr³ = (4/3)πR³
where R is the radius of the initial drop and r is the radius of the droplets.
We can divide the equation by (4/3)π and we get
nr³ = R³
or
r = R(1/n)¹/³
The total surface area of the droplets is n4πr².
We rewrite it by using the last equation
n4πr² = n4πR²(1/n)²/³ = 4πR²n¹/³
4πR² is the surface area of the original drop, so the total area has increased by a factor of n¹/³.
The surface energy is the work required to increase the surface area of a substance by unit area. Then, the surface energy is proportional to surface area.
Hence, the surface energy has increased by a factor n¹/³ and is n¹/³u
The original surface energy was u, so the increase in surface energy is
n¹/³u – u = [n¹/³ – 1]u
If n = 1000 then n¹/³ = 10
and [n¹/³ – 1]u = [10 – 1]u = 9u
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