1. Determine the maximum depth in a 5-m wide rectangular channel if the flow is to be supercritical...

Question

1. Determine the maximum depth in a 5-m wide rectangular channel if the flow is to be supercritical with a flow rate of Q = 40 m3 /s. What is the critical slope corresponding to the above flow rate? (n = XXXXXXXXXXA flow of 28 m3 s-1 occurs in an earth-lined canal having a base width of 3 m, side slopes of 1(vertical) to 2 (horizontal) and the Manning’s n value of XXXXXXXXXXCalculate the critical depth and critical slope. 3. a. A hydraulic jump occurs in a rectangular open channel. The water depths before and after the jump are 0.6 m and 1.5 m, respectively. Calculate the critical depth. b. A hydraulic jump occurs in a rectangular channel 3 m wide having a discharge of 9 m3 s-1 . The approach depth is 0.75 m. What is the downstream depth, and what fraction of the upstream energy is dissipated in the jump? 4. Water enters a very wide rectangular channel (Manning’s n = XXXXXXXXXXat a velocity of 12.2 ms-1 and a depth of 0.61 m. An H3 curve forms downstream and the water jumps to a depth of 2.4 m some distance downstream. Find the distance downstream to the location of the hydraulic jump. 5. A long wide rectangular channel carries a steady flow rate of 35 m2 s-1 (flow per unit width) and passes through an abrupt change in slope from XXXXXXXXXXupstream of the change) to XXXXXXXXXXdownstream). Manning’s n for the channel is XXXXXXXXXXIs there a hydraulic jump in the channel and if so where? Describe the water surface profiles at the various parts of the channel. 6. Question from Hydrology section Note: (i) The numerical values for the variables in the questions are based on your ID number and you must use the appropriate values for answering the questions. Failure to do so will result in no marks being awarded to those questions. The following variable is used in the questions: R If the last two digits of your ID is greater than or equal to 45 then R = the last two digits If the last two digits of your ID is less than 45, then R = 45 + the last two digits Example: If the ID = XXXXXXXXXX, then R = 67 If the ID = XXXXXXXXXX, then R = XXXXXXXXXX = 79 Note that your R value can only be between 45 and 99 (both inclusive) _____________________________________________________________________ 6 (a) Assuming that the following flood data follows a Gumbel distribution, plot the data on a Gumbel probability scale and fit a linear trend to your plot. Determine the Flood magnitude of ARI (return period) equal to R years from your fitted linear trend and compare with that from the equation; (x = ß + a.y) Year XXXXXXXXXX XXXXXXXXXX XXXXXXXXXXPeak flow in m3/s XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX XXXXXXXXXXln[ ln(1 )] ( 0.4) /( 0.2) y AEP AEP m N = - - - = XXXXXXXXXXb) Flood peaks of a river for 5-yr and 100-yr return periods are given as XXXXXXXXXXR) and XXXXXXXXXXR) m3 /s respectively: i. Calculate mean ( X ) and standard deviation ( S ) of the river floods assuming that Gumbel distribution fits the data ii. Using Normal distribution calculate 50 and 100 year return period flood peaks, iii. Calculate 50 and 100 year return period flood peaks using 2- parameter Log-Normal distribution. The mean (Y ), standard deviation (Sy ) and skewness coefficient (gy) of the logarithms (base 10) of the annual flood peaks are given as follows: Y = 2 + R/1000, Sy = 0.4 + R/1000 , gy= -(R /100) iv. For statistics of the log transformed data in (iii), calculate flood peak for 50 and 100 year return periods using the Log-Pearson distribution

Robert · Accepted Answer

Problem-1:- 
Here, the flow is supercritical. Hence Froude No. Fr >1. 
Hence , ∗ 1 
Q
A g ∗ y 1 
Q b ∗ y ∗ g ∗ y 
Hence, y 	 ∗√  
Hence, 0 y Qb∗ g
2 3 
Since, we want to maximize y we have, 
y Qb∗ g
2 3 	 405∗ 9.81
2 3 1.868	   
Hence, taking ymax = 1.86 m. 
Check:- Fr is greater than 1. 
Now, at critical slope, the depth is critical depth i.e. 1.87 m as seen from above. 
Now, from manning’s eq. v = (1/n) * R2/3 * S00.5. 
Also at critical flow Fr = 1. Thus v= (g*y)0.5 
Now, for the given channel parameters, Area = b*y = 9.35 m2. 
Wetted Perimeter P = b + 2y = 8.74 m. 
Hence, hydraulic radius R = (A/P) = 1.07 m. 
Now, equating manning’s eq. and Fr. No. eq.,we get 
The value of critical slope as 0.002. 
Thus, the max. depth is 1.86 m and the critical slope is 0.002.
Problem-2:- 
At critical depth , Fr = 1. 
Hence, v = (g*y)0.5. 
Hence, Q A ∗ g ∗ y 
For a trapezoidal channel of base width b of 3 m, side slope of m=2 horizontal to 1 
vertical and manning’s roughness coefficient n of 0.022, we have the following 
geometric properties:- 
Area A 	 ∗ 2b my  
Wetted Perimeter P 2 ∗ b y ∗ √1 m my  
Also given is that Q = 28 m3/s. Putting these values in the above eq. of Froude’s No. 
gives that  
Q y2 ∗ 2b my ∗ g ∗ y giving on simplification the following eq.:- 
56 	√9.81 ∗ y . ∗ 6 2y  
Solving the above eq. by trial and error or any other suitable method gives the 
following value of critical depth ycr =y=1.57 m. 
ycr = 1.57 m. 
Hence, A=7.15 m2. P = 16.13 m. This gives R = A/P = 0.443 m. 
Hence, as per manning’s eq. we have 
v = (1/n) * R2/3 * S00.5 
Puttings, velocity v = Discharge Q/Area A gives , 
S
∗
 = 0.022. 
Thus, the answers are critical depth is 1.57 m and critical slope is 0.022.
Problem 3.(A):- 
Here, y1 = 0.6 m. 
y2 = 1.5 m. 
ycr = ?. 
Hence, y ∗ ∗ 0.98	  
Critical depth corresponding to the above alternate depths is 0.98 m.
Problem-3(B) 
We know that, for two alternate depths y1 and y2 and discharge per unit width q the 
following formula. 
2 ∗ 	 ∗ ∗  
Here, q = 9/3 = 3 m2/s. 
y1 = 0.75 m. 
y2 = ? 
Putting the above values in the above e q. gives the following quadratic eq. 
0.75 ∗ 0.5625 ∗ 1.83 0 
Solving gives, y2 = 1.23 m. 
Now, initial energy E1 	 ∗ 	 	 ∗ ∗  = 1.

1. Determine the maximum depth in a 5-m wide rectangular channel if the flow is to be supercritical with a flow rate of Q = 40 m3 /s. What is the critical slope corresponding to the above flow rate?...

Solution

Answer To This Question Is Available To Download

Related Questions & Answers