Determine the normal depth of flow and reynold number???

[mhanna632]

Hello I was practicing old exam questions for my fluids and hydraulics coming exam, when i got the following question:

A trapezodial channel has a bed slope of 0.001. the base width is 10m, the side slopes are 1:3 and the discharge is 30m3/sec. Mannings n can be taken as 0.015

Find the normal depth of flow and reynolds number.

I have done similar ones to this with finding other unknowns, but without the depth value i have no idea where to begin-

My first general step is:

Q=

A 5/2 X S 1/2

-------------------

N x P 2/3

Any help please???

 

[Happy]

Q = (1/n)(A)(R^.666666)(S^.5) - CERM equation 19.13

You have Q, n, S and b. With a slope of 1H:3V (you don't specify which is which but I'll assume), the angle, @, is 71.56deg.

From CERN table 19.2, I see that

A=(b+d/tan@)d

R=(bdsin@ + d^2cos@)/(bsin@ + 2d)

So, you need to plug all your number in equation 19.13 and solve for d, the depth of flow with Q=30m^3/s. A rathern onerous equation to muddle through. If I saw that on the exam I'd start plugging in the numbers they give for answers until I found the correct one.

I get about 1.3 meters.

 

[Happy]

As for the Re,

From the above equation for Area, I got 13.56 m^2.

With flow 30, V=2.21 m/s

Re=DV/v

I haven't seen a Reynolds problem for open channel trapezoidal flow. Do you use the depth from above for the flow diameter, d=1.3m. That is what I would assume given no other info. You also didn't give the water temp for viscosity, so I'll assume its 50 deg and therefore v=.0000141. Plugging in those numbers I get and Re of 2X10^5.

 

[sac_engineer]

You're trying to use Chezy Manning with an unknown depth which can result in many instances of "d" in your equation. You can eliminate calculation interpolations by using Kings tables, which can provide d/b ratios given everything else required in a flow scenario (Q, n, base or depth, etc.).

Use the equation: K = (Q x n) / (k x b^8/3 x S^0.5)

I got K = 0.0307. In the Kings tables, use m = 3 (for the side slopes), and find the closest d/b value.

Just by crude interpolation, I found that the corresponding d/b value is 0.11, which means that d = 1.1 meters.

Are you sure the question is asking for the Reynolds number (Re) and not the wetted perimeter ®? Please check. Reynolds number is applied to closed conduit flows to determine laminar or turbulent conditions.

 

[Happy]

Good to know about the slope. Hopefully on an exam they make it clear enough. The OP wrote 1:3 though, so I assume I am correct in saying 1H:3V.

And I'm pretty sure if that is the case, the answer is d=1.3m. Plugging 1.1 meters into the equation gives a flow of 27.3 (not the 30 given)

 

[sac_engineer]

Good to know about the slope. Hopefully on an exam they make it clear enough. The OP wrote 1:3 though, so I assume I am correct in saying 1H:3V.
And I'm pretty sure if that is the case, the answer is d=1.3m. Plugging 1.1 meters into the equation gives a flow of 27.3 (not the 30 given)

1:3 implies 1V:3H, not the other way around.

 

[mhanna632]

Task 2

a) A trapezoidal channel has a bed slope of 0.001. The base width is 10 m, the side slopes are 1:3 and the discharge is 30 m3 / sec. Manning’s n can be taken as 0.015. Determine the normal depth of flow and Reynold’s Number.

B) A symmetric trapezoidal section, 1900 mm deep and with top and bottom widths 3 m and 0.7 m respectively carries water at a rate of 3 m3/sec. Manning’s n may be taken as 0.012. Find:

i. The normal depth at a slope of 1 in 1500

ii. The Froude number at the normal depth

That is the exact wording of the question.

Im sorry but can you simplify as to how you obtain y? I felt there was a big jump in between? (Apologies if I sound idiotic!?!)

Again thanks everyone for taking time out of your schedules. Much appreciated

 

[Happy]

Well, by my calculations, as per my post #2 above, if I was to use 1V:3H then the angle @ would be 18.43 deg, and running the numbers through again I get

d=1.144 m

My calculation method still works groovy and more accurately than the Kings tables, but I'm a little biased as since I'm taking trans depth I'm only bringing in the CERM and my calculator to help answer water questions.

 

[djgint]

Normal Depth = 1.1438 m

Froude No. = 0.6532

 

[structuraleit]

ic

 

[civilized_naah]

This is what I am getting -

Problem 1

S = 0.001; b = 10 m; m = 3; Q = 30 m3/s; n = 0.015

K = Qn/b^(8/3)S^(1/2) = 30 x 0.015/(10^2.67 x 0.001^0.5) = 0.0307

d/b = 0.116 (linear interpolation from table - approximate)

d = 1.16

Check: b = 10; Top width T = 16.96; Area A = 15.64; Wetted Perimeter P = 17.34; Hydraulic Radius R = 0.902; Velocity V = 1.968; Flow rate Q = 30.8

Reynolds number calculation is not meaningful for open channel flows. 

Problem 2

The only thing to take away from the “1900 mm deep and with top and bottom widths 3 m and 0.7 m” is the side slope, which must be a rise of 1.9 m in a horizontal flare of 1.15 m. This is 0.605 H: 1 V. This and the bottom width = 0.7 m are meaningful to the rest of the question, the depth of 1.9 m and top width of 3 m are not.

b = 0.7, side slope parameter m = 0.605; Q = 3 m3/s; n = 0.012; S = 1/1500 = 0.00066667 will lead to

K = Qn/ b^(8/3)S^(1/2) = 3 x 0.012/(0.7^2.67 x 0.0006667^0.5) = 3.6093

Approximating between columns for m = 0.5 and m = 0.75; d/b = 2 (approx)

Therefore normal depth = 2x0.7 = 1.4 m

Top width = 0.7+2x1.4x0.605 = 2.394

Area = 2.166; Average velocity = 3/2.166 = 1.385

Hydraulic depth = 2.166/2.394 = 0.905 m

Froude number = V/sqrt(gdh) = 1.385/√(9.81x0.905) = 0.465