Question

The sherwood number for vapor transport from the surface of high moisture food product to the surrounding air is 2.78. compute the convective mass transfer coefficient, when the dimension of the product in the direction of air movement is 15cm, and the mass diffusivity of water vapor in air is 1.8×10^-5m²/s.

Answer 1

The convective mass transfer coefficient is calculated as 0.00333 m/s using the provided Sherwood number of 2.78, the characteristic length of 0.15 m, and the mass diffusivity of 1.8×10−5 m²/s.

Step-by-step explanation:

To compute the convective mass transfer coefficient using the given Sherwood number, the following formula from mass transfer theory can be used:

Sherwood number (Sh) = hL/D where h is the convective mass transfer coefficient, L is the characteristic length dimension, and D is the diffusivity.

We are given that the Sherwood number (Sh) for vapor transport is 2.78, the characteristic length dimension (L) is 15 cm (which is 0.15 m when converted to meters), and the mass diffusivity of water vapor in air (D) is 1.8×10−5m²/s.

By rearranging the formula to solve for the convective mass transfer coefficient (h), we get:

h = Sh × D / L

Substituting the known values, we have:

h = 2.78 × 1.8×10−5 m²/s / 0.15 m

h = 0.00333 m/s

Thus, the convective mass transfer coefficient is 0.00333 m/s.