I'm practicing for a transport phenomena exam and I came across this question:

A mothball with a diameter of 1.0 cm is hung (by a thread) in stationary air. Mothballs consist of pure naphthalene. How long does it take for the diameter of the mothball to decrease to one half of its original size? Assume Sh= 2.

In the solutions manual the question is solved using a dynamic mass balance:

  1. $\frac{dM}{dt} = \rho\frac{d}{dt}V = \rho\cdot\frac16\pi\frac{d}{dt}D^3 = -F_M$
  2. $\rho\cdot\frac16\pi\cdot3D^2\frac{d}{dt}D = -Sh \cdot \pi \cdot \Bbb{D} \cdot D \cdot \frac{p^*M_{naph}}{RT}$
  3. (rest of the solution omitted for the sake of brevity)

Legend: $M$ = mass, $\rho$ = density, $V$ = volume (of sphere), $D$ = diameter (of sphere), $F_M$ = mass flow rate, $Sh$ = Sherwood number, $\Bbb{D}$ = diffusion coefficient, $p^*$ = vapour pressure of naphthalene, $R$ = gas constant, $T$ = temperature, $M_{naph}$ = molar mass of naphtalene

My question is about step 2 in the solution, specifically the left hand site of the equation. Why did they take the derivative of the diameter? I've tried to reason about it in the context of the question, but I couldn't think of a logical explanation. To me, it just seems like some arcane math trickery to make the solution correct.

  • $\begingroup$ Although one can guess it: Please add what all these variables are supposed to be. $\endgroup$ – ACuriousMind Jun 18 '15 at 10:16
  • $\begingroup$ Done. Also is this the right stackexchange? There are no tags for "mass balance", "transport phenomena", "sherwood" or "evaporation"... $\endgroup$ – user1870238 Jun 18 '15 at 11:03
  • $\begingroup$ This is the right SE, but given that the question relates to purely pedagogical study it should carry the tag 'homework-and-exercises'. Note that questions should also relate to a specific concept in physics, rather than asking for the solution to one particular example problem. Whether your question satisfies this criterion is a moot point, but given that it's received an answer already I think it should be left as it stands. $\endgroup$ – tok3rat0r Jun 18 '15 at 11:35

They're just applying the chain rule.

$$\frac{d}{dx}f^{n}(x)=nf^{n-1}(x)\frac{df}{dx} \Rightarrow \frac{d}{dt}D^{3}(t)=3D^{2}(t)\frac{dD}{dt}$$

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  • $\begingroup$ Hmmm, makes sense I guess $\endgroup$ – user1870238 Jun 18 '15 at 11:33

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