Question

A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is

Answer 1

The question is incomplete. The complete question is :

A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s) and the dissolution rate (kg/s).

Solution :

From flow over sphere, the mass transfer equation can be written as :

$Sh = 2 + 0.6 Re^{1/2} Sc^{1/3}$

where, Sherood number, $Sh = \frac{K_L d}{D_{eff}}$

            Reynolds number, $Re=\frac{Vd\rho}{\mu}$

            Schmid number, $Sc= \frac{\mu}{\rho D_{eff}}$

So,

$\frac{K_L d}{D_{eff}}=2+0.6 \left( \frac{V d \rho}{\mu} \right)^{1/2} \ \left( \frac{\mu}{\rho D_{eff}} \right)^{1/3}$

Diameter, d = 1 cm = $1 \times 10^{-2}$ m

                 V = 1 m/s

                 $\rho = 1000 \ kg/m^3$

                 $\mu = 10^{-3} \ kg/m/s$

                 $D_{eff} = 2 \times 10^{-9} \ m^2/s$

$\frac{K_L \times 10^{-2}}{2 \times 10^{-9}}=2+0.6 \left( \frac{1 \times 10^{-2} \times 10^3}{10^{-3}} \right)^{1/2} \ \left( \frac{10^{-3}}{10^3 \times 2 \times 10^{-9}} \right)^{1/3}$

$K_L \times 5 \times 10^6=478.22$

$K_L=9.5644 \times 10^{-5}$ m/s

So the mass transfer coefficient is 9.5644 $\times 10^{-5}$ m/s. It is given solubility,

$\Delta C = 2 \ kg/m^3$

$N = Md^2 \times \Delta C \times K_L$

$N= M \times (10^{-2})^2 \times 2 \times 9.5644 \times 10^{-5}$

$N= 6 \times 10^{-8}$ kg/s (dissolution rate)