# H

```493
Chapter 6 x Viscous Flow in Ducts
For sheet steel, take H|0.00015 ft, hence H/Dh | 0.000346. Now relate everything to the
input power:
Power 1 hp
550
ftlbf
s
UgQh f
(0.00234)(32.2)Q[54.4fQ 2 ],
or: fQ3 | 134 with Q in ft 3 /s
Guess f | 0.02, Q
1/3
(134/0.02)
ft 3
, Re
| 18.9
s
U(Q/A)D h
| 209000
P
Iterate: fbetter | 0.0179, Qbetter | 19.6 ft3/s, Rebetter | 216500. The process converges to
f | 0.01784, V | 80.4 ft/s, Q | 19.6 ft3/s.
6.91 Heat exchangers often consist of
many triangular passages. Typical is Fig.
P6.91, with L 60 cm and an isoscelestriangle cross section of side length a
2 cm and included angle E 80q. If the
average velocity is V 2 m/s and the fluid
is SAE 10 oil at 20qC, estimate the
pressure drop.
Ans.
Fig. P6.91
Solution: For SAE 10 oil, take U 870 kg/m3 and P 0.104 kg/ms. The Reynolds
number based on side length a is Re UVa/P | 335, so the flow is laminar. The bottom
side of the triangle is 2(2 cm)sin40q | 2.57 cm. Calculate hydraulic diameter:
A
Re Dh
1
(2.57)(2 cos 40q) | 1.97 cm2 ; P
2
UVD h
P
Then f
6.57 cm; Dh
870(2.0)(0.0120)
| 201; from Table 6.4, T
0.104
52.9
| 0.263, 'p
201
f
L U 2
V
Dh 2
| 23000 Pa
4A
| 1.20 cm
P
40q, fRe | 52.9
&sect; 0.6 &middot; &sect; 870 &middot; 2
(0.263) &uml;
(2)
&copy; 0.012 &cedil;&sup1; &uml;&copy; 2 &cedil;&sup1;
Ans.
```