EXAMPLE 4.5. Water with a density of 998 kg/m³ (62.3 lb/ft³) enters a 50 mm (1.969-in.) pipe fitting horizontally, as shown in Fig. 4.10, at a steady velocity of 1.0 m/s (3.28 ft/s) and a gauge pressure of 100 kN/m² (14.48 lb,/in.2). It leaves the fitting horizontally, at the same elevation, at an angle of 45° with the entrance direc- tion. The diameter at the outlet is 20 mm (0.787 in.). Assuming the fluid density is con- stant, the kinetic energy and momentum correction factors at both entrance and exit are unity, and the friction loss in the fitting is negligible, calculate (a) the gauge pressure at the exit of the fitting and (b) the forces in the x and y directions exerted by the fitting on the fluid.
EXAMPLE 4.5. Water with a density of 998 kg/m³ (62.3 lb/ft³) enters a 50 mm (1.969-in.) pipe fitting horizontally, as shown in Fig. 4.10, at a steady velocity of 1.0 m/s (3.28 ft/s) and a gauge pressure of 100 kN/m² (14.48 lb,/in.2). It leaves the fitting horizontally, at the same elevation, at an angle of 45° with the entrance direc- tion. The diameter at the outlet is 20 mm (0.787 in.). Assuming the fluid density is con- stant, the kinetic energy and momentum correction factors at both entrance and exit are unity, and the friction loss in the fitting is negligible, calculate (a) the gauge pressure at the exit of the fitting and (b) the forces in the x and y directions exerted by the fitting on the fluid.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
I want to know the exeac process of answering Sb,x.
Is it Sb,x=Sb cosθ right?
I'm confusing It's Sb,x=Sb sinθ or cosθ
but I think it's wrong.... Please solve this problem .
Could you show me the exact solution with calculatuion?

Transcribed Image Text:EXAMPLE 4.5. Water with a density of 998 kg/m³ (62.3 lb/ft³) enters a 50 mm
(1.969-in.) pipe fitting horizontally, as shown in Fig. 4.10, at a steady velocity of
1.0 m/s (3.28 ft/s) and a gauge pressure of 100 kN/m² (14.48 lb,/in.²). It leaves the
fitting horizontally, at the same elevation, at an angle of 45° with the entrance direc-
tion. The diameter at the outlet is 20 mm (0.787 in.). Assuming the fluid density is con-
stant, the kinetic energy and momentum correction factors at both entrance and exit are
unity, and the friction loss in the fitting is negligible, calculate (a) the gauge pressure
at the exit of the fitting and (b) the forces in the x and y directions exerted by the fitting
on the fluid.
Solution
(a) V₁ = 1.0 m/s. From Eq. (4.14),
from which
50
20
Pa = 100 kN/m²
The outlet pressure p, is found from Eq. (4.71). Since Z = Z, and h, may be neglected,
Eq. (4.71) becomes
V² - V²
V₂
Da
V₁ = V. (D.) ²³ =
Va
Po
= 1.0
Pa - Pb
P
P(V² - V²)
2
Vb.Y
Pb = Pa
998(6.25² - 1.0²)
1,000 x 2
= 100-18.99 = 81.01 kN/m² (11.75 lb//in.²)
= 100
(b) The forces acting on the fluid are found by combining Eqs. (4.51) and (4.52).
For the x direction, since F, = 0 for horizontal flow, this gives
m(BbVb.xBaVa.x) = Pa Sax - PbSb.x + Fw, x
2
V₂
= 6.25 m/s
Pb
(4.72)
FIGURE 4.10
Flow through reducing fitting.
viewed from the top,
Example 4.5.

Transcribed Image Text:Sb Cose
4
0₁
Sb-Sing
Sb.
x
교재
: Sb₁x = Sb. cose
교수님
: Sbik = Sb.sine.
6.
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