4AA Comparison of exact and aproximate proles for ow along a at
plate. ' '
Let IL, = 121/1) and Y = ydvoo / 112:; then Eq. 4.418 gives the approximation
3 13 1 13 32 3
v-avzsay (2750) Y
= 0.323209Y 00
4A.2 Velocity near a moving sphere.
From Eq. 42-14 at 9 = 1r / 2, the uid velocity relative to the approach velocity
falls to- 1% of 000. at '09 = ~03va relative to the sphere, giving
1 [1- =2- (51) $
6A.2 Pressure difference required for ow in pipe with elevation change.
In this problem the pipe diameter is (3.068 in.)/(39.37 in./m) = 0.07793 m, the
mass ow rate is
w = (18/60 gal/s)(3.78531it/gal)
6A.3 Flow rate for a given pressure drop.
The quantities needed for this calculation are as follows, in units of lbm, ft, and s:
p0 ~ 131, = (0.25 lbf/in2)(144 inz/ft2)(32.174 lbmft/sz-lbf)
= 1.16 x 1
. 4A.6 Use of boundary-layer formulas.
The data for this problem are:
W = 10 ft
L=3
v00 = 20 ft/s
From Table 1.12 and Appendix F:
u = (0.1505 cmz/s)/(12 x 2.54 cm/ft)2
= 1.62 x 10"4 ftz/s
,u = (0.0181
5B.1 Average ow velocity in turbulent tubeow.
a. The ratio of average to maximum velocity is obtained as
follows for the power-law expression:
<vz> WAIRQJYZW
mm fjjrdrde R2
=2L:(1-)1/"<d =2jgcl/"(1
6A.1 Pressure drop required for a pipe with ttings.
The average velocity at the given conditions is
_ 4(w/p) _ (4)(197 m3/8) _ S
(v) "D2 (7r)(0.25 m)2 _ 40.1 m/
and the Reynolds number is
D (11)
V
=
4B.6'Potential ow near a stagnation point
a. At the origin of coordinates (z _= O) the complex Velocity
(170/ dz = Zvoz is zero, which is a stagnation point. _
b. By taking the real and imaginary part
_ 4B.5 Steady potential flow around a stationary sphere
a. The boundary conditions are:
(i) as r > oo, v > 0,082, or by using Eqs. A.6-28 and 29
as r> oo, 0, = v, c056 and 1249 = v(, sine
(ii)atr=R,v,