WR module topic: Energy/Continuity equation

[cantaloup]

The next topic on the list of WR module is Energy/Continuity equation. There was a post some days ago about Bernouilli equation and it's a kind of plug and chug solution. Well, they are not that stupid to give out many freebies. Here is a problem which is exam type (this problem is very simplified, real question in the exam may have descriptive/narative wording blah blah blah and some numbers listed which are not related to the question just to distract/confuse you):

For the 50-ft wide rectangular channel shown in the attachment, determine the depth of flow and the velocity at section 2 using the energy principle. Neglect the energy losses. Assume that alpha1 = alpha2 = 1

Gupta_prob_10_4_sketch.pdf

 

[cantaloup]

The next topic on the list of WR module is Energy/Continuity equation. There was a post some days ago about Bernouilli equation and it's a kind of plug and chug solution. Well, they are not that stupid to give out many freebies. Here is a problem which is exam type (this problem is very simplified, real question in the exam may have descriptive/narative wording blah blah blah and some numbers listed which are not related to the question just to distract/confuse you):
For the 50-ft wide rectangular channel shown in the attachment, determine the depth of flow and the velocity at section 2 using the energy principle. Neglect the energy losses. Assume that alpha1 = alpha2 = 1

Sorry , please add: The depth of water at section 1 is 4.55 ft

 

[owillis28]

d2=2.81 ft

v2=13.8 ft/sec

Cantaloup - How did I do?

 

[cantaloup]

d2=2.81 ft
v2=13.8 ft/sec

Cantaloup - How did I do?

The answers are correct. You're on the right track!

 

[jartgo]

for some reason I'm coming up with v2=8.20 and d2=4.75.

I knew that this is wrong because the depth should be less and the velocity greater due to the slope....but that's what the math is giving me....where is my mistake?

z1=100*0.001=0.1

d1=4.55

V1=8.55

E1=5.78

z2=0

d2=?

V2=?

V1*50*D1 = V2*50*D2

1945 = V2*50*D2

I then solved everything in terms of V2...but I keep getting 8.2.

 

[IlPadrino]

for some reason I'm coming up with v2=8.20 and d2=4.75.
I knew that this is wrong because the depth should be less and the velocity greater due to the slope....but that's what the math is giving me....where is my mistake?

z1=100*0.001=0.1

d1=4.55

V1=8.55

E1=5.78

z2=0

d2=?

V2=?

V1*50*D1 = V2*50*D2

1945 = V2*50*D2

I then solved everything in terms of V2...but I keep getting 8.2.

To make sure we're on the same page, you're using the Energy Equation, right?



This is Bernouli's Equation in the case of fluids where the temperature doesn't change (I1 = I2), no heat is added (eh=0), no pump is involved (em=0, and losses are small (hL=0).

The pressure head at the surface is equal at both points, so I'd cancel the first terms leaving a very simple equation for solving v2. Unfortunately, this answer (8.92 ft/sec) isn't the answer owillis gave and cantaloup confirmed (13.8 ft/sec). So I'm curious to learn what they did.

If the Energy Equation wasn't mentioned I'd have thought about using Manning's Formula to solve for flow (assuming an n value) - this might have been a mistake.

 

[cantaloup]

SOLUTION

Let datum be the bottom of location 2

Z1 + P1/gamma +(V1)^2/(2*g) = Z2 + P2/gamma +(V2)^2/(2*g) + hloss

hloss = 0; P1 = P2

==> Z1 + 8.55^2/64.4 = Z2 + (V2)^2/64.4 (1)

but Z1 = 4.55 ft + 100*1/1000 = 4.65 ft

Also V1*A1 = V2*A2

Same bottom width ==> V1*Z1 = V2*Z2

Thus V2*Z2 = 8.55*4.65 = 39.8 (2)

(1) and (2) yield: (V2)^3 -372.6*(V2) +2505 = 0

Use your Casio calculator to solve the cubic equation:

V2 = 13.84 fps; V2 = -22.05 fps; V2 = 8.2 fps

V2 = 13.84 is correct

 

[jartgo]

I used energy and continuity....my solution (albeit incorrect) is attached if anyone can point me to my mistake, I'd appreciate it.

disregard...I see now that I only found one solution with my hp calculator....too bad it wasn't the correct one. I am happy to see that I was on the right track though and I knew my answer wasn't correct.

Thanks cantaloup

In addition I just checked the cubic equation capabilities of my HP 33s, it only requires six and a half pages of code in order to write the program to do it. I'm assuming the casio has a built in function?

 

[IlPadrino]

SOLUTIONLet datum be the bottom of location 2

Z1 + P1/gamma +(V1)^2/(2*g) = Z2 + P2/gamma +(V2)^2/(2*g) + hloss

hloss = 0; P1 = P2

==> Z1 + 8.55^2/64.4 = Z2 + (V2)^2/64.4 (1)

but Z1 = 4.55 ft + 100*1/1000 = 4.65 ft

Also V1*A1 = V2*A2

Same bottom width ==> V1*Z1 = V2*Z2

Thus V2*Z2 = 8.55*4.65 = 39.8 (2)

(1) and (2) yield: (V2)^3 -372.6*(V2) +2505 = 0

Use your Casio calculator to solve the cubic equation:

V2 = 13.84 fps; V2 = -22.05 fps; V2 = 8.2 fps

V2 = 13.84 is correct

Hmmm... I think my oversight was comparing two points at the surface of the water, not the bottom of the channel. So the pressure head doesn't cancel (P1=d1 and P2=d2). Z1=0.1 ft and Z2=0.

[Edit:] If the two points at the surface are compared, then P1=P2=0 but z2=d2 and z1=4.65 ft)

Thanks!

 

[IlPadrino]

SOLUTIONLet datum be the bottom of location 2

Z1 + P1/gamma +(V1)^2/(2*g) = Z2 + P2/gamma +(V2)^2/(2*g) + hloss

hloss = 0; P1 = P2

==> Z1 + 8.55^2/64.4 = Z2 + (V2)^2/64.4 (1)

but Z1 = 4.55 ft + 100*1/1000 = 4.65 ft

Also V1*A1 = V2*A2

Same bottom width ==> V1*Z1 = V2*Z2

Thus V2*Z2 = 8.55*4.65 = 39.8 (2)

(1) and (2) yield: (V2)^3 -372.6*(V2) +2505 = 0

Use your Casio calculator to solve the cubic equation:

V2 = 13.84 fps; V2 = -22.05 fps; V2 = 8.2 fps

V2 = 13.84 is correct

Cantaloup,

Thanks again for the solution, and I want to clarify a point...

You chose (as did I) the datum of 0 ft to be at the bottom of location 2. If you want the pressures to be equal, you have to choose two points at the same depth. Did you choose the surface?

I don't recall seeing any problems on the exam that would require solving a cube root, but if you did see one, I'd recommend trying each of the four answers given until one fits.

 

[zxu]

anyone can tell me where the 4.55ft comes from?

SOLUTIONLet datum be the bottom of location 2

Z1 + P1/gamma +(V1)^2/(2*g) = Z2 + P2/gamma +(V2)^2/(2*g) + hloss

hloss = 0; P1 = P2

==> Z1 + 8.55^2/64.4 = Z2 + (V2)^2/64.4 (1)

but Z1 = 4.55 ft + 100*1/1000 = 4.65 ft

Also V1*A1 = V2*A2

Same bottom width ==> V1*Z1 = V2*Z2

Thus V2*Z2 = 8.55*4.65 = 39.8 (2)

(1) and (2) yield: (V2)^3 -372.6*(V2) +2505 = 0

Use your Casio calculator to solve the cubic equation:

V2 = 13.84 fps; V2 = -22.05 fps; V2 = 8.2 fps

V2 = 13.84 is correct

 

[cantaloup]

anyone can tell me where the 4.55ft comes from?

The sketch was missing the depth at location 2, which is 4.55 ft (the second post after the first with sketch has correction).

Interestingly, when I try to solve this problem with the back water curve equation, with depth at 4.55 ft and 2.81 ft, with n=0.013 (assuming there is some friction loss) I got the distance = 96.7 ft (originally it's 100 ft).

The points of reference at location 1 and 2 for the energy equation are chosen at surface level so that P1 is nearly equal to P2.

 

[owillis28]

It is the depth of water at section 1. It was not given in the original problem (see pdf in first post). I believe the second post from this topic gives this information.

owillis

 

[IlPadrino]

The points of reference at location 1 and 2 for the energy equation are chosen at surface level so that P1 is nearly equal to P2.

Do you agree that at the surface, P1=P2=0? I don't think there's any nearly about it.

 

[cantaloup]

Do you agree that at the surface, P1=P2=0? I don't think there's any nearly about it.

That's right. At surface the pressure is at atmospheric.