University:
California State University, Northridge
Course:
ME 692 | Computational Fluid Dynamics
Academic year:

2023
Views:

488

Pages:

15
Author:

ilikemyicons0y5

Problem 8.19 A fluid flows through a horizontal 01-in.- diameter pipe. When the Reynolds number is 1500, the head loss over a 20-ft length of the pipe is 6.4 ft. Determine the fluid velocity. Solution 8.19 h_{L} = f * l/D * (V ^ 2)/(2g) where since Rc = 1500 < 2100 the flow is laminar. Thus, so that f = 64/(Rc) = 64/1500 = 0.0427 6.4ft = 0.0427 * (20ft)/(0.1/12 * fr) * (v ^ 2)/(2(32.2 * A/(s ^ 2))) V = 2.01 * (ft)/c Problem 8.34 During a heavy rainstorm, water from a parking lot completely fills an 18-in.- diameter, smooth, concrete storm sewer. If the flowrate is determine the pressure drop in a 100-ft horizontal section of the pipe. Repeat the problem if there is a 2-ft change in elevation of the pipe per 100 ft of its length. 10 * (f * t ^ 3)/s Solution 8.34 l = 10011 222 V_{1} V_{2} (1) Dain In general p 1 y + V 1 ^ 2 2g +z 1 = p 2 y + V 2 ^ 2 2g + z_{2} + f * i/D * (V ^ 2)/(2g) where p_{1} - p_{2} = Delta*p gamma = pg and V = V_{1} = V_{2} = Q/A = (10 * (R ^ 3)/s)/(pi/4 * (18/12 * R) ^ 2) = 5.66 * R/8 (1) (2) Delta*p = gamma(z_{2} - z_{1}) + f * i/D * 1/2 * rho * V ^ 2 For a smooth concrete pipe E/D = (0.001ft)/(18/12 * ft) = 6.67 * 10 ^ - 4 (see the table below) Equivalent Roughness, Pipe Feet Mülimeters Rivered steel 0.003-0.03 0.9.9.0 Concrete 0.001-0.01 0.3-3.0 Wood stave 0.0006-0.003 0.18-0.9 Cast iron 0.00085 0.26 Galvanized iron 0.0005 0.15 Commercial steel 0.00015 0.045 or wrought iron 0.000005 0.0015 Drawn tubing 0.0 (smooth) 0.0 (smooth) Plastic, glass and Re = (VD)/v = (5.66 hbar 8 )( 18 12 l hbar) 1.21*10^ -S hbar! 8 =7.02*10^ 5 Thus, from the figure below f = 0.0185

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