CHEMICAL ENGINEERING

MOMENTUM, HEAT AND MASS TRANSFER 301

PROBLEM 12.6

Obtain an expression for the simple Reynolds analogy between heat transfer and friction.
Indicate the assumptions which are made in the derivation and the conditions under which
you would expect the relation to be applicable.
The Reynolds number of a gas flowing at 2.5 kg/m^2 s through a smooth pipe is 20,000.
If the specific heat of the gas at constant pressure is 1.67 kJ/kg K, what will the heat
transfer coefficient be?

Solution

The derivation of the simple Reynolds analogy and its application is presented in detail
in Section 12.8.
For a Reynolds number of 2. 0 ð 104 , from Fig. 3.7,R/u^2 D 0 .0032 for a smooth
pipe.

From equation 12.102: h/CpuD 0. 0032

uD 2 .5 kg/m^2 s

and hence: hD 0. 0032 ð 1670 ð 2. 5 D 13 .4W/m^2 K

PROBLEM 12.7

Explain Prandtl’s concept of a ‘mixing length’. What parallels may be drawn between
the mixing length and the mean free path of the molecules in a gas?
The ratio of the mixing length to the distance from the pipe wall has a constant value
of 0.4 for the turbulent flow of a fluid in a pipe. What is the value of the pipe friction
factor if the ratio of the mean velocity to the axial velocity is 0.8?

Solution

Transfer by molecular diffusion is discussed in Section 12.2 and the concept of the mixing
length in Section 12.3.2. By analogy with kinetic theory, the eddy kinematic viscosity,
E, is given by:
E/EuE (equation 12.18)

whereEis the mixing length anduEis some measure of the linear velocity of the fluid
in the eddies.

As shown in equation 12.21: uE/Ejdux/dyj

Combining this with the previous equation:E/EEjdux/dyj

Putting the proportionality constant equal to unity,ED^2 Ejdux/dyj (equation 12.23)

In the absence of momentum transfer by molecular movement, the shear stress is
given by:

RyDEdux/dyD^2 Ejdux/dyjdux/dy. (equation 12.20)