CIV E 444 - APPLIED HYDRAULICS
FALL 2009
HOMEWORK No. 5 SOLUTION

Use ONLINE CHANNEL 15 to calculate flow conditions u/s:
Q = 65.92 cfs; A_{n, u/s} = 20 ft²; V_{n, u/s} = 3.296 fps; T_{n, u/s} = 12 ft.
With Q = 65.92 cfs, use ONLINE CHANNEL 01 to calculate flow conditions d/s:
y_{n, d/s} = 2.533 ft; A_{n, d/s} = 10.13 ft²; V_{n, d/s} = 6.506 fps; T_{n, d/s} = 4 ft.
L_transition = [(12/2) - (4/2)] / tan 27.5^o = 7.68 ft.
For design, assume L_transition = 8 ft. Calculate flow conditions at A = 2, B = 4, and C = 6 ft.
z₁ + y₁ + α₁V₁²/(2g) = z₂ + y₂ + α₂V₂²/(2g)+ h_L
1000 + 2 + [1.2 × (3.296)  ²/(2 × 32.17)] =
z₂ + 2.533 + [1.1 × (6.506)  ²/(2 × 32.17)] + 0.2 × [(6.506)  ²/(2 × 32.17)]
1002 + 0.203 = z₂ + 2.533 + 0.724 + 0.132
z₂ = 998.814 ft.
y₂ = 2.533 ft.
z₂ + y₂ = 1001.347 ft.
Three sections A @ 2 ft, B @ 4 ft, and C at 6 ft.
d_A = 2 - 0.25 (1002 - 1001.347) + 0.25 (1000 - 998.814) =
d_A = 2 - 0.25 (1002 - 1001.347 - 1000 + 998.814)
d_A = 2 + 0.25 (0.533) = 2.133 ft.
d_B = 2 + 0.5 (0.533) = 2.267 ft.
d_C = 2 + 0.75 (0.533) = 2.400 ft.
d_d/s = 2.533 ft.
Vertical depth at section A: 0.25 × 2.533 = 0.633 ft.
Vertical depth at section B: 0.50 × 2.533 = 1.267 ft.
Vertical depth at section C: 0.75 × 2.533 = 1.900 ft.
Triangular (nonvertical) depth at section A: 2.133 - 0.633 = 1.50 ft.
Triangular (nonvertical) depth at section B: 2.267 - 1.267 = 1.00 ft.
Triangular (nonvertical) depth at section C: 2.400 - 1.900 = 0.50 ft.
Flow area in section A: [7 + (1.5 × 2)] 2.133 - 1.5  ² = 19.08 ft²
Flow area in section B: [6 + (1.0 × 2)] 2.267 - 1.0  ² = 17.14 ft²
Flow area in section C: [5 + (0.5 × 2)] 2.400 - 0.5  ² = 14.15 ft²
V_A = 65.92/19.08 = 3.454 fps.
V_B = 65.92/17.14 = 3.846 fps.
V_C = 65.92/14.15 = 4.659 fps.
z₁ + y₁ + &alpha₁V₁²/(2g) = z_A + y_A + α_AV_A²/(2g) + h_{L (1 → A)}
1000 + 2 + [1.2 × (3.296)  ²/(2 × 32.17)] =
z_A + y_A + [1.15 × (3.454)²/(2 × 32.17)] + {0.25 × 0.2 × [(6.506)²/(2 × 32.17)]}
z_A + y_A = 1002 + 0.203 - 0.213 - (0.25 × 0.132) = 1001.957 ft.
z_A = 1001.957 - 2.133 = 999.824 ft.
z₁ + y₁ + &alpha₁V₁²/(2g) = z_B + y_B + α_BV_B²/(2g) + h_{L (1 → B)}
1000 + 2 + [1.2 × (3.296)²/(2 × 32.17)] =
z_B + y_B + [1.15 × (3.846)  ²/(2 × 32.17)] + {0.50 × 0.2 × [(6.506)²/(2 × 32.17)]}
z_B + y_B = 1002 + 0.203 - 0.264 - (0.50 × 0.132) = 1001.873 ft.
z_B = 1001.873 - 2.267 = 999.606 ft.
z₁ + y₁ + &alpha₁V₁²/(2g) = z_C + y_C + α_CV_C²/(2g) + h_{L (1 → C)}
1000 + 2 + [1.2 × (3.296)  ²/(2 × 32.17)] =
z_C + y_C + [1.15 × (4.659)  ²/(2 × 32.17)] + {0.75 × 0.2 × [(6.506)²/(2 × 32.17)]}
z_C + y_C = 1002 + 0.203 - 0.388- (0.75 × 0.132) = 1001.716 ft.
z_C = 1001.716 - 2.400 = 999.316 ft.
Summary:
z₁ = 1000 ft.
z_A = 999.824 ft.
z_B = 999.606 ft.
z_C = 999.316 ft.
z₂ = 998.82 ft.
(a) The water-surface profile is M₁.
(b) Use ONLINE CHANNEL 01 to calculate the normal depth.
y_n = 1.668 m
The u/s depth to calculate length is 1.668 × 1.01 = 1.684 m.
Use ONLINE WSP21 to calculate M₁ water surface profile, with given data. Leave [number of computational intervals] n blank [default n = 100] and set number of tabular output intervals m = 100.
(c) From the output table, the total length is L = 7279.4 m.
(d) From the output table, for y_{1.01y_n} = 1.684 m, L = 4779.5 m.
(e) Percentage: 100 (4779.5/7279.4) = 65.66 %
The ratio H/(H + P) = (0.5)/(0.5 + 2) = 0.2.
The uncorrected empirical coefficient (Fig. 4-37) is: C' = 0.86.
The coefficient, corrected for sloping upstream face: C' = 0.86 × 1.1 = 0.946
The discharge: Q = 0.385 (2g)^1/2 C' L H^3/2 = 3.088 × 0.946 × 5 × 0.5  ^3/2
Q = 5.164 cfs.
Q = (8/15) K (2g)^1/2 tan (θ/2) H^5/2
Assume a low (conservative) value of K = 0.57 for head within 0.2-2.0 ft.
Q = (8/15) × 0.57 × (2 × 32.17)^1/2 tan (45^o) × 1.2^5/2 = 3.846 cfs.