Page 242 - Bird R.B. Transport phenomena

226 Chapter 7 Macroscopic Balances for Isothermal Flow Systems

7В.1 Velocity averages from the \ power law. Evaluate the velocity ratios in Problem 7A.7 ac-
cording to the velocity distribution in Eq. 5.1-4.
7B.2 Relation between force and viscous loss for flow in conduits of variable cross section.
Equation 7.5-6 gives the relation F^ s = pSE v between the drag force and viscous loss for
straight conduits of arbitrary, but constant, cross section. Here we consider a straight channel
whose cross section varies gradually with the downstream distance. We restrict ourselves to
axisymmetrical channels, so that the force F^ s is axially directed.
If the cross section and pressure at the entrance are S^ and p v and those at the exit are S 2
and p , then prove that the relation analogous to Eq. 7.5-7 is
2
= P S m E v - S 2 ) (7B.2-1)
where

(7B.2-2)
S 2
P2S2
Vm = (7B.2-3)
Si + S 2
Interpret the results.

7B.3 Flow through a sudden enlargement (Fig. 7.6-1). A fluid is flowing through a sudden en-
largement, in which the initial and final diameters are D^ and D respectively. At what ratio
2
D /D will the pressure rise p - p^ be a maximum for a given value of v ?
2 } 2 }
Answer: D /D } = V2
2
7B.4 Flow between two tanks (Fig. 7B.4). Case I: A fluid flows between two tanks A and В because
p A > p . The tanks are at the same elevation and there is no pump in the line. The connecting
B
line has a cross-sectional area S and the mass rate of flow is w for a pressure drop of (p A - p \.
{
B
Case II: It is desired to replace the connecting line by two lines, each with cross section S =
u
lSi. What pressure difference (p A - p ) is needed to give the same total mass flow rate as in
B u
Case I? Assume turbulent flow and use the Blasius formula (Eq. 6.2-12) for the friction factor.
Neglect entrance and exit losses.
Answer: (p A - p ) /(p A - p \ = 2 5/ 8
B n
B
Circular tube of Circular tubes of
cross section Sj cross section S n
A В
Mass flow rate w
Sum of mass
flow rates is w
Fig. 7B.4. Flow between two tanks.

7B.5 Revised design of an air duct (Fig. 7B.5). A straight, horizontal air duct was to be installed in
a factory. The duct was supposed to be 4 ft X 4 ft in cross section. Because of an obstruction,
the duct may be only 2 ft high, but it may have any width. How wide should the duct be to
have the same terminal pressures and same volume rate of flow? Assume that the flow is tur-
bulent and that the Blasius formula (Eq. 6.2-12) is satisfactory for this calculation. Air can be
regarded as incompressible in this situation.
(a) Write the simplified versions of the mechanical energy balance for ducts I and II.
(b) Equate the pressure drops for the two ducts and obtain an equation relating the widths
and heights of the two ducts.
(c) Solve the equation in (b) numerically to find the width that should be used for duct II.
Answer: (c) 9.2 ft