Page 63 - Modelling in Transport Phenomena A Conceptual Approach

44 CHAPTER 3. INTERPHASE TRANSPORT

The drag force can be calculated from Eq. (3.1-7) as

1
(1.2)(27.78)2 (0.24) = 333.4N

The use of Eq. (3.1-11) gives the power consumed as

w = FDV&
= (333.4)(27.78) = 9262 W

b) In this cae the characteristic velocity is
(i:::)
=
vCh = (100 + 30) - 36.11 m/s
Therefore, the drag force and the power consumed are
I
[‘
FD = (2 x 1.5) 5 (1.2)(36.11)2 (0.24) = 563.3N

rir = (563.3)(36.11) = 20,341 W

c) In this case the characteristic velocity is
(El)
v,h = (100 - 30) - 19.44m/s
=
Therefore, the drag force and the power consumed are
1
[ f
FD = (2 x 1.5) - (1.2)(19.44)2 (0.24) = 163.3N

rir = (163.3)(19.44) = 3175W

3.1.1 Physical Interpretation of Friction Factor

According to Newton’s law of viscosity, Eq. (2.1-2), the shear stress at the wall is
expressed as
(3.1-12)

The minus sign is omitted in Eq. (3.1-12) because the value of v, increases as the
distance y increases. Substitution of Eq. (3.1-12) into h. (3.1-4) gives

(3.1-13)