Download 5.46

Problem 5.46
Figure P5.46 shows a lateral pipe fitting. This particular fitting has a mainline diameter of 4.0 in.
The diameter of the lateral is 3.0 in., and the lateral angle is 45°; 60 °F water is flowing in the
lateral. Measurements show that the pressure at point 1 is 34.0 psig, the pressure at point 2 is
35.0 psig, the pressure at point 3 is 33.5 psig, and the flow rate at point 2 is 1.0 ft3/s. Determine
the horizontal and vertical force components (Fx and Fy) required to hold the lateral fitting
stationary. Neglect gravity. Q 1 = 1.63 ft3/s.
Solution
First apply the continuity equation
Q1 = Q2 + Q3 ⇒ Q3 = Q1 − Q2 =1.63
ft 3
ft 3
ft 3
− 1.0
= 0.63
s
s
s
For the “x” direction
∑ Fx =Fx + p1 A1 − p2 A2 − p3 A3 cos 45o =ρQ2V2 + ρQ3V3 cos 45o − ρQ1V1
so
Fx = ρ Q2V2 + ρ Q3V3 cos 45o − ρ Q1V1 + p2 A2 + p3 A3 cos 45o − p1 A1
π 2
Q
4Q
but
A
D=
and V =
so V
=
π D2
4
A
4 ρ Q22 4 ρ Q32
4 ρ Q12
ppp
o
substituting
+
+ p2 D22 + p3 D32 cos 45o − p1 D12
Fx =
cos 45 −
2
2
2
ppp
D2
D3
D2
4
4
4
gathering terms and substituting numerical values and conversion factors gives
Q12  p
4 ρ  Q22 Q32
o
F=
+
cos
45
−
+ ( p2 D22 + p3 D32 cos 45o − p1 D12 )
 2
x
2
2 
p  D2 D3
D2  4
6
6
6
slugs 

2 ft
2 ft
2 ft
1.0
0.63
1.63
3 

2
2
2
ft
s +
s cos 45o −
s  + π  35.0 lb ×16in 2 + 33.5 lb × 9in 2 cos 45o − 34.0 lb ×16in 2 



2
2
2
(0.333 ft )  4 
π
in 2
in 2
in 2

 (0.333 ft ) (0.25 ft )




4 ×1.94
Fx
Fx = +154.2lb (acts to the right)
For the “y” direction
∑F
= Fy − p3 A3 = ρ Q3V3 sin 45o − 0 ⇒ Fy = p3 A3 + ρ Q3V3 sin 45o
The weight of the fitting and the water in it have been neglected. Substituting as before
y
p
4 ρ Q32
=
Fy p3 D +
sin 45o
2
4
p D3
6
slugs

4 ×1.94 3  0.632 ft

2
ft
s sin 45o  + π  33.5 lb × 9in 2 sin 45o 
=
Fy



2
π
in 2

 (0.25 ft )
 4




Fy = +28.0lb (acts up)
2
3