Fluid Mechanics Free Essay Example, 2000 words

1.60 cfs c. 1.28 cfs b. 1.43 cfs d. 1.15 cfs Computations d = 8† = 0.67 ft, r = 0.333 ft L = 5,400.00 ft H = 40.5 ft n = 0.013; n value for new cast iron pipe Solve for C using Mannings formula C =  Area = Ï€(r)2 =Ï€(0.333)2 = 0.3483 ft2 R =  R =  R =  =  = 0.1665 Q = Av Q = A  Q = 0.3483 Q = 1.12 cfs 10. a symmetrical trapezoid plate has the following dimensions: the width of the parallel sides are, respectively, 2.50 ft and 4.50 ft. The perpendicular distance between those is 1.50 ft. The plate is submerged in a liquid in a vertical position with the parallel sides horizontal and the shorter parallel side at the top and exactly in the surface of the liquid will be a. 5.43 cu. We will write a custom essay sample on Fluid Mechanics or any topic specifically for you Only $17.96 $11.86/pageorder now Ft c. 6.93 cu. Ft b. 4.31 cu. Ft d. 7.68 cu. Ft Computations: Static Moment of the plate w/ respect to the surface of the liquid Static moment of the rectangle = 2.5 (1.5)  = 2.8125 Static moment of the 2 triangles = (1)(1.5) = 1.50 Total Static moments = 2.8125 + 1.50 = 4.31 cu. ft 11. The diameter of a new cast iron pipe in which water is flowing is 6.0 in. ,and the estimated velocity is 5.83 fps. The friction factor, determined from table 4 and expressed to four decimal places, is a. 0.0239 c. 0.0233 b. 0.0236 d. 0.0129 The answer of this problem can be taken from table 4 in the textbook. 12. the pipe described in question 11 contains four 90 ° elbows and two y’s. The sum of the minor losses in head caused by these fittings is a. 7.20 ft c. 4.33 ft b. 6.00 ft d. 3.80 ft Computations: Diameter = 6 inches = 0.5 ft. , r =. 25 ft A = Ï€ r2 A = 3.1416(0.25) A = o. 20 ft2 Q = Av Q = 0.20 ( 5.83) Q = 1.20 cfs Qtotal = sum of Q1 - Q6 Q total = 7.20 cfs 13. In the Hazen-William formula, the value of the factor C for a certain pipe may be taken as 145. The diameter of the pipe is 16 in. Its length 6400 ft, and the head tending the cause flow is 52.0 ft. The rate of discharge for the pipe in gallons per minute is a. 4100.00 gpm c. 2650.00 gpm b. 3380.00 gpm d. 1900.00 gpm Computation: Diameter = 16 inches, radius = 8 inches = 0.666 ft. Area = Ï€ r2 = 1.3932 ft2 R =  R =  R = 0.3326 V = 1.318 C (R)0.63 (S)0.54 V =  V = 7.0904 fps Q = Av Q = 0.4435 (7.0904) Q = 9.88 cfs = 9.88 (7.477) (60) Q = 4432 gpm ≈ 4100.00 gpm 14.