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,
`$Sh = (K_L d)/(D_(eff))$`

Reynolds number,
`$Re=(Vd\rho)/(\mu)$`

Schmid number,
`$Sc= (\mu)/(\rho D_(eff))$`

So,

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

Diameter, d = 1 cm =
`$1 * 10^(-2)$` m

V = 1 m/s

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

`$\mu = 10^(-3) \ kg/m/s$`

`$D_(eff) = 2 * 10^(-9) \ m^2/s$`

`$(K_L * 10^(-2))/(2 * 10^(-9))=2+0.6 \left( (1 * 10^(-2) * 10^3)/(10^(-3)) \right)^(1/2) \ \left( (10^(-3))/(10^3 * 2 * 10^(-9)) \right)^(1/3)$`

`$K_L * 5 * 10^6=478.22$`

`$K_L=9.5644 * 10^(-5)$` m/s

So the mass transfer coefficient is 9.5644
`$* 10^(-5)$` m/s. It is given solubility,

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

`$N = Md^2 * \Delta C * K_L$`

`$N= M * (10^(-2))^2 * 2 * 9.5644 * 10^(-5)$`

`$N= 6 * 10^(-8)$` kg/s (dissolution rate)