Page 121 - Bird R.B. Transport phenomena

106 Chapter 3 The Equations of Change for Isothermal Systems

Fig. 3B.4. Creeping flow in the re-
gion between two stationary con-
centric spheres.

УХ/К/Х/К/К/Х/Х7Ш
X A A A
(я) У у- Н
,..л / Fluid out

3B.4 Creeping flow between two concentric spheres
/ ' * (Fig. 3B.4). A very viscous Newtonian fluid flows in the
space between two concentric spheres, as shown in the fig-
(Ъ) ure. It is desired to find the rate of flow in the system as a
function of the imposed pressure difference. Neglect end
Fig. 3B.2. (a) Compact heat-exchanger element, showing effects and postulate that v depends only on r and в with
e
channels of a triangular cross section; (fr) coordinate sys- the other velocity components zero.
tem for an equilateral-triangular duct.
(a) Using the equation of continuity, show that v sin в =
e
u(r), where u(r) is a function of r to be determined.
(b) From Eq. 3B.2-1 find the average velocity, maximum (b) Write the ^-component of the equation of motion for
velocity, and mass flow rate.
this system, assuming the flow to be slow enough that the
Answers: (b) (v ) = 7 ^ = — v z>max ; [v • Vv] term is negligible. Show that this gives
z
oU/xL ZU
Г 1 1 d ( idu\\
r
2
^l sin в r dr\ drj\ (3B.4-1)
w = 180/nL
(c) Separate this into two equations
3B.3 Laminar flow in a square duct.
(a) A straight duct extends in the z direction for a length L OB.4-2,3)
and has a square cross section, bordered by the lines x = dr
±B and у = ±В. A colleague has told you that the velocity where В is the separation constant, and solve the two
distribution is given by equations to get
(2P 0 - 2P )B 2
L
1 - £ - Ш (ЗВ.З-1)
4/JLL B = (3B.4-4)
2 In
Since this colleague has occasionally given you wrong ad- 0
<3> 2 )R
vice in the past, you feel obliged to check the result. Does it u(r) = - cotte/2)[ИМ-*)] "*
satisfy the relevant boundary conditions and the relevant 4/xln
differential equation? where 2^ and ^ a r e the values of the modified pressure at
2
(b) According to the review article by Berker, 3 the mass в = s and в = IT - e, respectively.
rate of flow in a square duct is given by (d) Use the results above to get the mass rate of flow
3
3
(3B.3-2) - 9> )R (1 - к) р
2
w = (3B.4-6)
12/1 In cotte/2)
Compare the coefficient in this expression with the coeffi-
cient that one obtains from Eq. 3B.3-1. 3B.5 Parallel-disk viscometer (Fig. 3B.5). A fluid, whose
viscosity is to be measured, is placed in the gap of thick-
ness В between the two disks of radius R. One measures
3 the torque T required to turn the upper disk at an angular
R. Berker, Handbuch der Physik, Vol. VIII/2, Springer, Berlin z
(1963); see pp. 67-77 for laminar flow in conduits of noncircular cross velocity П. Develop the formula for deducing the viscosity
sections. See also W. E. Stewart, AIChE Journal 8,425-428 (1962). from these measurements. Assume creeping flow.