Question

Calculate the value of the mass-transfer coefficient for mass transfer from a sphere of naphthalene to air at 45 oC and 1 atm abs flowing at a velocity of 0.305 m/s. The diameter of the sphere is 25.4 mm. The diffusivity of naphthalene in air at 45 oC is 6.92 x 10-6 m2/s and the vapor pressure of solid naphthalene is 0.555 mm Hg. Viscosity of air is 1.93 x 10-5 Pa.s, density of air is 1.113 kg/m3.

Answer 1

Let A denote naphthalene and B denote air Since the mole fraction of naphthalene is very

the physical properties Of air at and I Will be used the gas mixture:

$\mu=\mu_B=1.93*10^{-5} Pa.s\\\rho=\rho_B=1.113kg/m^3$

We now evaluate the dimensionless numbers:

$Sc={\mu\over \rho D_{AB}}={1.93 \times10^{-5}\over(1.113)(6.92 \times10^{-6}})=2.506$

$R_e={\rho VD\over\mu}={(1.113)(0.0254)(0.305)\over 1.93 \times10{-5}}=446.8$

$K_c={(21)(6.92\times 10^{-6}\over0.0254}=5.72\times10^{-3}$

$N_A={k_c \over (1-y_A)}(c_Ai-c_A)\\={k_c\over(1-y_A)_lmRT}(p_Ai-p_A)$

$N_A={5.72X10^{-3}\over(8314)(318)}({0.555X1.013X10^5\over760}-0)\\$

$=1.60\times10^{-7}$ kmol/m^2/s

The mass transfer rate from the sphere is then

$W_A=\pi D^2N_A=3.24\times10^{-10}$ kmol/s