Question of the Day 8 Aug 06

[jeb6294]

This is from the WR module of a sample exam that I have....

 

[RIP - VTEnviro]

Haha, I ain't touching that one with a stick without my hydrology notes on me.

I understand the procedure for doing it, but I suck at geometry, and that's a serious geometry problem.

 

[RIP - VTEnviro]

I immediately ruled out "B" because it has 3.7 depth and it can't exceed 3 feet.

Even I caught that one! "oldtimer"

 

[DVINNY]

Okay, using mannings equation and conveyance factor K.
K = ((1+m(d/ b ))^5/3)/((1+2(d/ b )(SQRT(1+m^2))^2/3

m = COT of 3:1 slope or m = 3

d = depth; b = width

I immediately ruled out "B" because it has 3.7 depth and it can't exceed 3 feet.

So, I plugged in the d and b answers in options "A", "C", and "D" and got to solve for K.

Then I used mannings equation:

Q = K (b^(8/3))(SQRT(s) / n

and determined the Q for each of the three remaining dimensions to get

"A" :  Q=1500 cfs

"C" :  Q=4254 cfs

"D" :  Q=285 cfs

So, my answer is "D"

Sounds good to me. :claps:

 

[Timber]

This is a good example. I like how you can immediately eliminate choice "C" from the possible answers. To deep at 3.7'

Ok, now you are looking at a 33% chance of simply guessing between A,B, D.

Quickly determine the cross sectional areas of each seperate channel;

A: 10.2 (x) 3 + [1/2 (9 (x) 3) + 1/2 (9 (x) 3)] = 30.6 + 27 = 57.6 ft^2

B: N/A (See Above)

C: 16 (x) 2 + [1/2 (6 (x) 2) + 1/2 (6 (x) 2)] = 32.0 + 12.0 = 44.0 ft^2

D: 4.6 (x) 3 + [1/2 (9 (x) 3) + 1/2 (9 (x) 3)] = 13.8 + 27 = 40.8 ft^2

You need to be identify the channel is a trapezoid, and then determine the hydraulic radius accordingly.

This is demonstrated by the following equation;

Hr = (Base (x) Depth (x) SIN (T) + D^2 (x) COS(T)) / (Base (x) SIN(T) +2 (x) Depth) (Recheck this equation for accuracy in the CERM)

Notice T = 71.56 degrees, this is derived from the given 3:1 slope.

From the equation referenced above we can determine the HR.

A: 2.033

B: N/A Triangle

C: 1.648 ft

D: 1.537 ft

Now cross compare the resulting Q from Mannings Eqn using the calculated HR from above.

A: Q= 392.5 cfs

B: N/A

C: Q= 259.79 cfs

D: Q= 231.00 cfs

The Answer is D, since A & C are > than 250 cfs, and B is > than 3.0' in height.

 

[jeb6294]

By George I think you've got it. I think that's how I ended up doing this one...once you can rule out "B" as an answer, I think it's just faster to plug in the information from the three remaining choices into mannings.

 

[jeb6294]

But, I figure they all need to carry at least 250cfs, and according to your solution, "D" would not carry it. So, you would need to go to the next higher, which would be "C"

You were right...the answer is "D"...and you even solved it the fancy way they give in the solutions. When I was working this one, I think I did it the same way Timber did. Probably just a math error somewhere.

 

[Timber]

I tried solving the problem without refernce. I may have busted the equation. What result of cfs did the solution provide for channel "D". Good example, the more problems you do, you are certain to build confidence in selecting the correct answer.

Best of luck to all who are preparing for the exam this October.

Later,

Timber

 

[Timber]

I found my error obove, the angle is 18.43 degrees, not 71.56 degrees.

I originally had the channel modeled too steep. Notice 90 - 71.56 = 18.41, 3:1 not 1:3

OK, so check this out.

The corresponding hydraulic radius' now equal;

A: 1.97 ft

B: N/A

C: 1.53 ft

D: 1.73 ft

Then the corresponding Q values equal;

A: 384 cfs

B: Who cares

C: 248 cfs

D: 250 cfs

The Answer is "D" - 250 cfs as per the design flow.

 

[cement]

I loaned my CERM out, but there were some really nice tables in the back for solving open channel for depth given the Q, the width and slope and pitch of the sides or vice versa. I found them two days before the exam.

 

[cantaloup]

Yes the answer is D ; I solved this in less than 2 minutes by using HP-33s equation solver. It's wrong for some guys putting two decimal places in the answers like Timber. If the numbers in the question have 2 decimal places , I would select A for answer. Why ? Because I used the HP-33s to solve for B, I got B=4.61 ft , rounded off to have 4.6 ft which is the correct answer. Had they give: D/ 4.60 then I have to pick A for answer (with 4.60 ft for depth, the Q would be only 249.7 cfs)

 

[cantaloup]

This is a good example. I like how you can immediately eliminate choice "C" from the possible answers. To deep at 3.7'
Ok, now you are looking at a 33% chance of simply guessing between A,B, D.

Quickly determine the cross sectional areas of each seperate channel;

A: 10.2 (x) 3 + [1/2 (9 (x) 3) + 1/2 (9 (x) 3)] = 30.6 + 27 = 57.6 ft^2

B: N/A (See Above)

C: 16 (x) 2 + [1/2 (6 (x) 2) + 1/2 (6 (x) 2)] = 32.0 + 12.0 = 44.0 ft^2

D: 4.6 (x) 3 + [1/2 (9 (x) 3) + 1/2 (9 (x) 3)] = 13.8 + 27 = 40.8 ft^2

You need to be identify the channel is a trapezoid, and then determine the hydraulic radius accordingly.

This is demonstrated by the following equation;

Hr = (Base (x) Depth (x) SIN (T) + D^2 (x) COS(T)) / (Base (x) SIN(T) +2 (x) Depth) (Recheck this equation for accuracy in the CERM)

Notice T = 71.56 degrees, this is derived from the given 3:1 slope.

From the equation referenced above we can determine the HR.

A: 2.033

B: N/A Triangle

C: 1.648 ft

D: 1.537 ft

Now cross compare the resulting Q from Mannings Eqn using the calculated HR from above.

A: Q= 392.5 cfs

B: N/A

C: Q= 259.79 cfs

D: Q= 231.00 cfs

The Answer is D, since A & C are > than 250 cfs, and B is > than 3.0' in height.

Timber: It's great to solve this problem by analyze it, solving step by step. But going to the exam, we have no time, plus it's easy to make mistake in step calculation. Use the calculator's feature as much as you can. Whatchagonna do if they give you 3-section pipe in series or 3-section pipe in parallel ? You have no time to plug numbers in an equation 3 times. Or calculate depth of flow (or side slope) of traperzoid shape open channel ? You have no time for "solve by trial" type of thing. Remember that engineers need to utilize all the tools available to solve the prolems quick and correctly.

 

[3gorgesdam]

I found my error obove, the angle is 18.43 degrees, not 71.56 degrees.
I originally had the channel modeled too steep. Notice 90 - 71.56 = 18.41, 3:1 not 1:3

OK, so check this out.

The corresponding hydraulic radius' now equal;

A: 1.97 ft

B: N/A

C: 1.53 ft

D: 1.73 ft

Then the corresponding Q values equal;

A: 384 cfs

B: Who cares

C: 248 cfs

D: 250 cfs

The Answer is "D" - 250 cfs as per the design flow.

I got about the same answer as yours. I think to solve this problem it is better to just plug in the answer and see which one is close to the given condition. I am surprised that answer C and D are that close. I don't have a programmable calculator.