```483
Chapter 6 x Viscous Flow in Ducts
6.75 You wish to water your garden with
5
100 ft of 8 -in-diameter hose whose roughness is 0.011 in. What will be the delivery,
in ft3/s, if the gage pressure at the faucet is
60 lbf/in2? If there is no nozzle (just an
open hose exit), what is the maximum
horizontal distance the exit jet will carry?
Fig. P6.75
Solution: For water, take U 1.94 slug/ft3 and P 2.09E5 slug/fts. We are given H/d
0.011/(5/8) | 0.0176. For constant area hose, V1 V2 and energy yields
pfaucet
Ug
60 u 144 psf
h f , or:
1.94(32.2)
138 ft
or fV 2 | 4.64. Guess f | ffully rough
L V2
f
d 2g
100
V2
f
,
(5/8)/12 2(32.2)
0.0463, V | 10.0
ft
, Re | 48400
s
then fbetter | 0.0472, Vfinal | 9.91 ft/s (converged)
The hose delivery then is Q
(S/4)(5/8/12)2(9.91)
0.0211 ft3/s. Ans. (a)
From elementary particle-trajectory theory, the maximum horizontal distance X travelled
by the jet occurs at T 45q (see figure) and is X V2/g (9.91)2/(32.2) | 3.05 ft Ans. (b),
which is pitiful. You need a nozzle on the hose to increase the exit velocity.
6.76 The small turbine in Fig. P6.76
extracts 400 W of power from the water
flow. Both pipes are wrought iron. Compute
the flow rate Q m3/h. Sketch the EGL and
HGL accurately.
Solution: For water, take U 998 kg/m3
and P 0.001 kg/ms. For wrought iron,
take H | 0.046 mm, hence H/d1 0.046/60
| 0.000767 and H/d2 0.046/40 | 0.00115.
The energy equation, with V1 | 0 and p1
p2, gives
V22
z1 z 2 20 m
h f2 h f1 h turbine , h f1
2g
Also, h turbine
P
UgQ
400 W
998(9.81)Q
Fig. P6.76
L1 V12
f1
d1 2g
and Q
S
4
and h f2
d12 V1
S
4
L 2 V22
f2
d 2 2g
d 22 V2
```