106
CHAPTER 4. HYDRODYNAMIC MODELS
For similitude we must have
Z/AX = (HA
\ Ht / p
\ Ht / m
therefore
Ht) = 58.14
m
However, it is important to first check the model Reynolds number to determine which
transmission equations to use. Applying Froude scaling to the prototype parameters
and noting a different model value for kinematic viscosity gives the following model
parameters:
P - 0.38
um
-
1.3563(10)“6 m2 / s
(Hi)m - 0.045 m
hm
- 0.10 m
Tm
-
1.0 s
Lm
- 0.924 m
Dm ~ 0.003 m (estimate for use in Reynolds number)
(△Z)m
- 0.14 m
The model Reynolds number is calculated from Eqn. 4.40 as
0.38(0.045 m)(0.924 m)(0.003 m)
n “ 2(1.3563(10)-6 m2/s)(0.1 m)(1.0 s) “
Because the model Reynolds number is less than 2000, solve for D using Eqn. 4.38
and Eqn. 4.39. Begin by solving Eqn. 4.38 for ym, i.e.,
(58.14)^ = l +
y2(0.1 m)
2/3
0.14 m X
0.924 m )
7m — 250
Now substitute
into Eqn. 4.39 and solve for Dm
750 = (°-38)~4 f l-3563(10)~6 m2 / s (1.0 s) V/3 Z0.924 mX
1-52
\
D (0.924 m)
)
\
D
) '
f 9.806 m/s2(0.1 m)(1.0 3)2V/3
V
(0.924 m)2
)
which eventually yields
= 8 mm.
The factor A is found by substituting the length scale and the prototype and
model quarryrun diameter into Eqn. 4.34, i.e.,