CIVE 530 - OPEN-CHANNEL HYDRAULICS HOMEWORK 6, PROBLEM 4

PERMISSIBLE TRACTIVE FORCE METHOD: EXAMPLE A (sides and bottom are the same) Given: Q = 750 cfs; b = ?; z = 3; S= 0.0028; n = 0.024; sides and bottom: noncohesive material, slightly angular, d25 = 0.7 in. Assume b/y = 6. Assume tractive force on sides is critical (as opposed to tractive force on level ground). With b/y and z, enter Fig. 7-7 left to determine Cs = 0.78 (Assume value for z= 2). With d25 and grain shape, find angle of repose θ from Fig. 7-9: θ = 36o. Calculate φ from: tanφ = 1/z φ = tan-1 (1/z) = 18.435o. Calculate K from: K = [1 - (sin2φ/sin2θ)]1/2 = 0.843 Determine permissible unit tractive force on level ground τL from Fig. 7-10. τL = 0.4 × d25 (in) = 0.4 × 1.0 = 0.4 psf. Calculate the permissible unit tractive force on the sides: τs = K τL = 0.843 × 0.4 = 0.337 psf. Set permissible and acting unit tractive forces equal: τs = Ts τs = 0.337 = Csγ y S = 0.78 × 62.4 × y × 0.0028 Solve for flow depth y: y = τs/(CsγS) = 0.337 / (0.78 × 62.4 × 0.0028) = 2.47 ft. With y and b/y, calculate b = y (b/y) = 2.47 × 6 = 14.8 ft. Assume b = 15 ft. With Q = 750, b = 15, z = 3, S = 0.0028, and n = 0.024 known, use CHANNEL to find yn = 4.19 ft. Test to confirm that: yn = 4.19 > y = 2.47. Normal depth too high! If not satisfied, assumed b/y is too small. Assume a greater value and return. Assume b/y = 19. Then b = 46.9 ≅ 48 ft. With Q = 750, b = 48, z = 3, S = 0.0028, and n = 0.024 known, use CHANNEL to find yn = 2.47 ft. Test to confirm that: yn = 2.47 = y Normal depth now OK! With b/y = 19 enter Fig. 7-7 to determine Cb = 1.0 Calculate TL = Cbγ yn S = 1.0 × 62.4 × 2.47 × 0.0028 = 0.43 psf. Compare acting unit tractive force TL with permissible unit tractive force on level ground τL. TL = 0.43 > τL = 0.40. Therefore, the bottom control the design. The design is not OK. Force TL = 0.4. Then: 0.4 = Cbγ yn S = 1.0 × 62.4 × yn × 0.0028 Solve for new yn: yn = 0.4/(1.0 × 62.4 × 0.0028) = 2.29 ft. Solve for new b by trial and error: With Q = 750, b = 55, z = 3, S = 0.0028, and n = 0.024 known, use CHANNEL to find yn = 2.29 ft. b = 55 ft, with bottom control. ANSWER. PERMISSIBLE TRACTIVE FORCE METHOD: EXAMPLE B (sides and bottom are different) Given: Q = 750 cfs; b = ?; z = 2; S= 0.001; n = 0.022; sides: noncohesive material, slightly angular, d25 = 1.0 in; bottom: noncohesive material, with d50 = 0.8 mm, with high content of fine sediment in the water. Assume b/y = 6. Assume tractive force on sides is critical (as opposed to tractive force on level ground). With b/y and z, enter Fig. 7-7 left to determine Cs = 0.78 (Assume value for z= 2). With d25 and grain shape, find angle of repose θ from Fig. 7-9: θ = 36o. Calculate φ from: tanφ = 1/z φ = tan-1 (1/z) = 21.8o Calculate K from: K = [1 - (sin2φ/sin2θ)]1/2 = 0.775 Determine permissible unit tractive force on level ground τL from Fig. 7-10. τLs = 0.4 × d25 (in) = 0.4 × 1.0 = 0.40 psf. τLb = 0.09 psf. Calculate the permissible unit tractive force on the sides: τs = K τLs = 0.775 × 0.4 = 0.31 psf. Set permissible and acting unit tractive forces equal: τs = Ts τs = 0.31 = Csγ y S = 0.78 × 62.4 × y × 0.0028 Solve for flow depth y: y = τs/(CsγS) = 0.31 / (0.78 × 62.4 × 0.0028) = 2.27 ft. With y and b/y, calculate b = y (b/y) = 2.27 × 6 = 13.6 ft. Assume b = 14 ft. With Q = 750, b = 14, z = 2.5, S = 0.0028, and n = 0.024 known, use CHANNEL to find yn = 4.44 ft. Test to confirm that: yn = 4.44 > y = 2.27. Normal depth too high! If not satisfied, assumed b/y is too small. Assume a greater value and return. Assume b/y = 20. Then b = 45.4 ≅ 45 ft. With Q = 750, b = 45, z = 2.5, S = 0.0028, and n = 0.024 known, use CHANNEL to find yn = 2.58 ft. Test to confirm that: yn = 2.58 > y = 2.27. Normal depth too high! Assume b/y = 25. Then b = 56.7 ≅ 57 ft. With Q = 750, b = 57, z = 2.5, S = 0.0028, and n = 0.024 known, use CHANNEL to find yn = 2.26 ft. Test to confirm that: yn = 2.26 < y = 2.27. Normal depth now OK! With b/y = 25, enter Fig. 7-7 to determine Cb = 1.0 Calculate TL = Cbγ yn S = 1.0 × 62.4 × 2.26 × 0.0028 = 0.395 psf. Compare acting unit tractive force TL with permissible unit tractive force on level ground τL calculated in step 7. TL = 0.395 > τLb = 0.09. Therefore, the bottom controls the design. The design is not OK. Force TL = 0.09. Then: 0.09 = Cbγ yn S = 1.0 × 62.4 × yn × 0.0028 Solve for new yn: yn = 0.09/(1.0 × 62.4 × 0.0028) = 0.51 ft. Solve for new b by trial and error: Assume b/y = 60; b = 136.2; say 140 ft. With Q = 750, b = 140, z = 2.5, S = 0.0028, and n = 0.024 known, use CHANNEL to find yn = 1.33 ft. Too high. Assume b/y = 120; b = 272.4; say 280 ft. With Q = 750, b = 144, z = 2.5, S = 0.0028, and n = 0.024 known, use CHANNEL to find yn = 0.88 ft. Still too high. Assume b/y = 300; b = 681; say 700 ft. With Q = 750, b = 680, z = 2.5, S = 0.0028, and n = 0.024 known, use CHANNEL to find yn = 0.51 ft. OK now. b = 700 ft, with bottom control. ANSWER.

051004