# Mass Transfer from an Oxygen Bubble

I think there is an error in this questions solution and I want someone to check:

Example 8.1-4:Mass transfer from an oxygen bubble A bubble of oxygen originally 0.1 cm in diameter is injected into excess stirred water, as shown schematically in Fig. 8.1-1(d). After 7 min, the bubble is 0.054 cm in diameter. What is the mass transfer coefficient? The solution goes on to state:

$$\frac{d}{dt}(c_1\frac{4}{3}\pi r^3)=AN_1=-4\pi r^2k[c_1(\text{sat})-0]$$ $$\frac{dr}{dt}=-k\frac{c_1(\text{sat})}{c_1}=-0.034k$$

Shouldn't it be: $$\frac{dr}{dt}=-3k\frac{c_1(\text{sat})}{c_1}=-0.102k$$

• How did you get your answer? Mar 24 '16 at 15:44
• @Chuck In the original problem set numerical values are given for $c_1(sat) = 1.5 * 10^{-3}$ and $c_1$ is 1/22.4 mol/L. Check google books if you would like to read the original text w/ solution. Diffusion: Mass Transfer in Fluid Systems by E. L. Cussler, p.242 Mar 24 '16 at 15:49

Your approach is wrong because the left hand side represents the derivative of the composition of two time dependent functions. You completely ignored that fact and started cancelling terms. I try to give a simpler example:

$\frac{d 4x^3}{dx} = kx^2 \Leftrightarrow 12x^2=kx^2 \Leftrightarrow 12=k$

This is not the same as

$\frac{d 4x}{dx} = k \Leftrightarrow 4=k$

Therefore use the chain rule

$\frac{d(c \cdot V)}{dt} = V\frac{dc}{dt}+\frac{dV}{dt}c=-Ak\Delta c$

I would now assume a stationary diffusion, hence

$V\frac{dc}{dt} = 0$

$\frac{dV}{dt}c=-Ak\Delta c$

$A = \frac{dV}{dr}$

$\frac{dV}{dt}c=-\frac{dV}{dr}k\Delta c$

For this step I hope there are no mathematicians around, I cancel the differentials

$\frac{dr}{dt}c=-k\Delta c$

$\frac{dr}{dt}=-k\frac{\Delta c}{c}$

• Thanks, your method makes a lot of sense. But what did I do wrong previously? Mar 24 '16 at 16:52
• @MathsIsHard I updated my answer. Mar 24 '16 at 17:07