Page 228 - Bird R.B. Transport phenomena

212 Chapter 7 Macroscopic Balances for Isothermal Flow Systems

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EXAMPLE 7.6-3 Water at 95°C is flowing at a rate of 2.0 ft /s through a 60° bend, in which there is a contrac-
tion from 4 to 3 in. internal diameter (see Fig. 7.6-3). Compute the force exerted on the bend if
Thrust on a Pipe Bend ^ pressure at the downstream end is 1.1 atm. The density and viscosity of water at the con-
е
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ditions of the system are 0.962 g/cm and 0.299 cp, respectively.
SOLUTION The Reynolds number for the flow in the 3-in. pipe is
D(v)p 4
Re = W
M 77-D/x
3
4(2.0 X (12 X 2.54) )(0.962) 6
= 3 X 10 (7.6-18)
77(3 X 2.54X0.00299)
At this Reynolds number the flow is highly turbulent, and the assumption of flat velocity pro-
files is reasonable.
(a) Mass balance. For steady-state flow, w = w . If the density is constant throughout,
x 2
^ = J - P (7.6-19)

in which p is the ratio of the smaller to the larger cross section.
(b) Mechanical energy balance. For steady, incompressible flow, Eq. (d) of Table 7.6-1 be-
comes, for this problem,

M ~ v\) + g(h 2 A (p 2 - E = 0 (7.6-20)
o
According to Table 7.5-1 and Eq. 7.5-4, we can take the friction loss as approximately Щр?) =
\v\. Inserting this into Eq. 7.6-20 and using the mass balance we get

P! - p 2 = pv\{\ - {0 1 + g) + pg(h - hj (7.6-21)
2
This is the pressure drop through the bend in terms of the known velocity v 2 and the known
geometrical factor /3.
(c) Momentum balance. We now have to consider both the x- and y-components of the mo-
mentum balance. The inlet and outlet unit vectors will have x- and y-components given by
u lx = 1, щ у = 0, u 2x = cos в, and u 2y = sin в.

Fluid out
Plane 2
3" internal
diameter

Fluid in ——t* 1
Force exerted by
fluid on elbow
4" internal Fig. 7.6-3. Reaction force at a reducing
diameter bend in a pipe.