CHEMICAL ENGINEERING

312 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

may be written as:

hD[1 ̨ 1 Sc]Dh/Cp[1 ̨ 1 Pr] (equation 12.121) (i)

The Schmidt group,ScD/DD 1 ð 10 ^3 / 1000 ð 5 ð 10 ^9 D 200

The Prandtl group,PrDCp/kD 4 ð 103 ð 1 ð 10 ^3 / 0. 6 D 6. 67

and: h/CpD 4000 / 4 ð 103 ð 1000 D 0. 001

Thus, in equation (i):

hD[1 ̨ 1  200 ]D 0 .001[1 ̨ 1  6. 67 ]

and: hDD 0. 001  1 C 5. 67 ̨/ 1 C 199 ̨m/s (ii)

When ̨D 0 .2, then from equation (ii),

hDD 0. 001  1 C 1. 134 / 1 C 39. 8 D 5. 2 ð 10 ^5 m/s

When ̨D 0 .6, then from equation (ii),

hDD 0. 001  1 C 3. 402 / 1 C 119. 4 D 3. 6 ð 10 ^5 m/s

It is worth noting that even with a very large variation in ̨(threefold in fact) the
change in the mass transfer coefficient is less than 50%.

PROBLEM 12.20

By using the simple Reynolds analogy, obtain the relation between the heat transfer
coefficient and the mass transfer coefficient for the gas phase for the absorption of a
soluble component from a mixture of gases. If the heat transfer coefficient is 100 W/m^2 K,
what will the mass transfer coefficient be for a gas of specific heat capacity 1.5 kJ/kg K
and density 1.5 kg/m^3? The concentration of the gas is sufficiently low for bulk flow
effects to be negligible.

Solution

From Section 12.8.1, the heat transfer coefficient is given by:

R/u^2 Dh/Cpus (equation 12.102)

and the mass transfer coefficient by:R/u^2 DhD/us (equation 12.103)

Hence: hDDh/Cp (equation 12.105)

In this case: hDD 100 / 1. 5 ð 103 ð 1. 5 D 0 .044 m/s

PROBLEM 12.21

The velocity profile in the neighbourhood of a surface for a Newtonian fluid may be
expressed in terms of a dimensionless velocityuCand a dimensionless distanceyCfrom