Chapter 3 x Integral Relations for a Control Volume
289
3.173 The horizontal wye fitting in Fig.
P3.173 splits the 20°C water flow rate
equally, if Q1 5 ft3/s and p1 25 lbf/in2
(gage) and losses are neglected, estimate
(a) p2, (b) p3, and (c) the vector force
required to keep the wye in place.
Solution: First calculate the velocities:
Fig. P3.173
V1
Q
A1
5.0
(S /4)(6/12)2
25.46
ft
; V2
s
2.5
(S /4)(3/12)2
50.93
ft
, V3
s
28.65
ft
s
Then apply Bernoulli from 1 to 2 and then again from 1 to 3, assuming 'z | 0:
p2
p1 p3
p1 U
V
2
U
2
25(144) 1.94
[(25.46)2 (50.93)2 ] | 1713 psfg
2
Ans. (a)
(V12 V 23 ) 25(144) 1.94
[(25.46)2 (28.65)2 ] | 3433 psfg
2
Ans. (b)
2
1
V 22
(c) to compute the support force R (see figure above), put a CV around the entire wye:
¦ Fx
R x p1A1 p2 A 2 sin30q p3 A 3 sin50q
U Q 2 V2 sin30q U Q 3 V3 sin50q U Q1V1
R x 707 42 229 124 106 247, or: R x
¦ Fy
R y p2 A 2 cos30q p3 A3 cos 50q
lbf (to left) Ans. (c)
U Q 2 V2 cos30q U Q3 ( V3 ) cos 50q
R y 73 193 214 89, or: R y | 5 lbf (up) Ans. (c)
3.174 In Fig. P3.174 the piston drives
water at 20°C. Neglecting losses, estimate
the exit velocity V2 ft/s. If D2 is further
constricted, what is the maximum possible
value of V2?
Solution: Find p1 from a freebody of the
piston:
¦ Fx
F pa A1 p1A1 , or: p1 pa
Fig. P3.174
10.0 lbf
lbf
| 28.65 2
2
(S /4)(8/12)
ft
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P3.173 splits the 20°C water flow rate 5 ft 25 lbf/in