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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.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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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?

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