Holooly Plus Logo

Question 6.2: A trapezoidal channel having a bottom slope of 0.001 is carr......

A trapezoidal channel having a bottom slope of 0.001 is carrying a flow of 30 m³ /s. The bottom width is 10.0 m and the side slopes are 2H to 1V. At the downstream end, a control structure raises the water depth to 5.0 m. Determine the water-surface levels at 1, 2, and 4 km upstream of control structure. The Manning n for the flow surfaces is 0.013, α = 1.0, and the elevation of the channel bottom at the downstream end is 0.0.

Given:

Bottom slope, S_{o} = 0.001;
Discharge, Q = 30 m³ /s;
Channel width, B_{o} = 10.0 m;
Manning n = 0.013;
Depth at the downstream end (i.e., at x = 0) = 5.0 m;
α = 1.0.

Determine:

Water-surface levels at 1, 2, and 4 km upstream of the control structure.

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

Let us call the distance at the control structure as x = 0. Since the distance in the downstream flow direction is considered positive, the upstream distances where we want to determine the flow depths are negative.
\quad\quadTable 6-2 lists the calculations using a trial-and-error procedure.
\quad\quadThe following explanatory comments should help to understand the computations of Table 6-2. We estimate the flow depth at 1 km upstream of the control structure and we then check whether this estimated flow depth satisfies Eq. 6-22 or not. If it does not satisfy this equation, then we discard the calculations corresponding to this estimated depth and estimate another value. This process is continued until a solution is obtained. Then, we consider each distance where we want to determine the water level (i.e., 2 and 4 km) one by one and repeat the above procedure.

\quad\quad\quad\quad F(y_{2})=y_{2}+\frac{\alpha_{2}Q^{2}}{2gA^{2}_{2}}+\frac{1}{2}S_{f2}(x_{2}-x_{1})\\\quad\quad\quad\quad\quad\quad\quad\quad\quad+z_{2}-H_{1}+\frac{1}{2}S_{f1}(x_{2}-x_{1})=0 (6-22)

Column 1, x. This is the specified location at which flow depth, y, is to be computed.
Column 2, y. This is the estimated flow depth.
Column 3, A. This is the flow area, A, for the flow depth of column 2.
Column 4, P. This is the wetted perimeter, P, for the flow depth of column 2.
Column 5, R. This is the hydraulic radius, R, corresponding to the flow depth of column 2 obtained by dividing column 3 by column 4.
Column 6, R^{4/3}. This column lists the value of R raised to the power 4/3.
Column 7, V . Flow velocity, V = 30/A, where A is listed in column 3. Velocity head corresponding to this depth is shown in column 8.
Column 8, V²/(2g). This column lists the value of velocity head, V²/(2g).
Column 9, z. This is the elevation of the channel bottom. It is computed from the known bottom elevation (z_{d} = 0) at the downstream end (x_{d} = 0) and the known channel bottom slope, S_{o}, of 0.001; i.e., z = z_{d}-S_{o}(x-x_{d}).
Column 10, y. Total head of column 10 corresponds to the flow depth of column 2, velocity head of column 8, and the channel bottom level of column 9, i.e., H = z + y + αV²/(2g).
Column 11, S_{f}. This is the slope of the energy grade line. It is computed by using the velocity of column 7, R^{4/3} of column 6 and the known value of Manning n from the equation, S_{f} = n²V²/(C_{o}^{2}R^{4/3}).
Column 12, \overline{S}_{f}. This is the average of the S_{f} value for the flow depth at the current distance and that for the flow depth at the previous distance.
Column 13, Δx. This is the distance between the current location where we want to determine the flow depth and the location where we determined the flow depth during the previous step.
Column 14, h_{f}. The head losses in distance Δx are computed from the equation h_{f} = \overline{S}_{f}Δx, where \overline{S}_{f} is given in column 12 and Δx is given in column 13.
Column 15, H. This is the elevation of the energy grade line computed by adding the head losses (h_{f} of column 14) to the elevation of the energy grade line (i.e., H of column 10 for the previous step) at the location where the flow depth was computed during the previous step.
Column 16. If the values of H in columns 10 and 15 are within an acceptable tolerance, then the estimated depth of column 2 is the flow depth at the location under consideration. We then proceed to compute the flow depth at the next selected location. However, if these values are not within the specified tolerance, then we discard the values corresponding to this flow depth and repeat the process with another value for the estimated depth.

Table 6-2 Standard step method
 S_{o}= 0.001; Q = 30 m³ /s; B_{o}= 10.0 m; n = 0.013; y_{o}= 5.0 m; α = 1; s = 2

x

(1)

y

(2)

A

(3)

P

(4)

R

(5)

R^{1.33}

(6)

V

(7)

V²/(2g)

(8)

z

(9)

H

(10)

S_{f}

(11)

\overline{S}_{f}

(12)

Δx

(13)

h_{f}

(14)

 H

(15)

Difference

(16)

 Remarks

(17)

0 5.000 100.0 32.36 30.090 4.484 0.300 0.0046 0.0 5.0046 0.00000379  

 

-1000 4.001 72.0 27.89 2.582 3.542 0.417 0.0088 1.0 5.0095 0.00000828 0.00000583 -1000 0.00583 5.0104 0.0009
 

 

4.002 72.04 27.9 2.583 3.543 0.416 0.0088 1.0 5.0104 0.00000827 0.00000583 -1000 0.00583 5.0104 0.0000 ok
-2000 3.003 48.10 23.43 0.052 2.607 0.624 0.0198 2.0 5.0233 0.00002525 0.00001676 -1000 0.01676 5.0272 0.0038
 

 

3.007 48.16 23.45 2.054 2.611 0.623 0.0198 2.0 5.0271 0.00002512 0.00001670 -1000 0.01670 5.0271 0.0000 ok
-3000 2.014 28.30 19.01 1.487 1.697 1.062 0.0574 3.0 5.0719 0.00011224 0.00006868 -1000 0.68680 5.0958 0.2390
 

 

2.038 28.68 19.11 1.501 1.718 1.046 0.0558 3.0 5.0935 0.00010763 0.00006638 -1000 0.00664 5.0935 0.0000  ok
-4000 1.078 13.10 14.82 0.884 0.848 2.290 0.2674 4.0 5.3450 0.00104508 0.00057635 -1000 0.57635 5.6698 0.3249
1.232 15.40 15.51 0.990 0.986 1.954 0.1947 4.0 5.4263 0.00065429 0.00038096 -1000 0.38096 5.4744 0.0481
1.263 15.83 15.65 1.011 1.015 1.896 0.1832 4.0 5.4465 0.00059837 0.00035300 -1000 0.35300 5.4465 0.0000 ok

Related Answered Questions