CERM Practice Problem (11-th edition) solution #17-11 (question #7)..Help

[Calixico]

This question has got me stump. Its an iterative 3-reservoirs solution. Ok the solution leads into four equations & four unknows. there are listed below:

0.05134V(a-d) + 0.5476V(b-d) - 0.08441V(d-c)= 0 .........(eq #1)

V(a-d)= sqrt[25.73 - 0.0165Pd).........(eq #2)

V(b-d) = sqrt[ 80.66 - 0.0862Pd].........(eq #3)

V(d-c) = sqrt[3.24 + 0.0173Pd].........(eq #4)

where V(a-d) is the velocity of flow from a to d, etc.. and Pd is the pressure at point d in the system, etc...

The solution states that the four equations should be solved simultaneously. Then it said, to do this assume a value for pd (pressure at d). Then these value then determine all three velocities in eqn # 2, 3 & 4. After that the velocities from eqn # 2, 3 & 4 is then substituted in eqn #1. By trial and error, the answer should be found to be:

V(a-d)= 3.21 ft/sec

V(b-d) = 0.408 ft/sec

V(d-c) = 4.40 ft/sec

and the pressure was discoved to be 933.8 psf

The confusion for me is, if I assume a Pd of say 1 for simplicity, I come up with different answers. What are the parameters, or what I am trying to do here to figure out what value to chose for(Pd); and if it's too large or small to get a correct answer for the velocities.

Second question, are there questions on the exam that mimick this iterative process?

Third:

btw what is also the short cut to solvine this question on the P.E. exam?

Any help would be appreciated on this matter...

Thank you!

 

[IlPadrino]

The solution states that the four equations should be solved simultaneously. Then it said, to do this assume a value for pd (pressure at d). Then these value then determine all three velocities in eqn # 2, 3 & 4. After that the velocities from eqn # 2, 3 & 4 is then substituted in eqn #1. By trial and error, the answer should be found to be:
.

.

.

The confusion for me is, if I assume a Pd of say 1 for simplicity, I come up with different answers. What are the parameters, or what I am trying to do here to figure out what value to chose for(Pd); and if it's too large or small to get a correct answer for the velocities.

If you choose a Pd of 1, then solve Eqs 2-4, and finally plug those solutions into Eq 1, you do NOT get a net zero... you got something like 5.02. So that tells you the solution isn't Pd=1. Here's my approach:

Pd=10 --> Eq1=4.99 (not much difference than Pd=1, so the next guess should be much bigger!)

Pd=100 --> Eq1=4.71 (holy *****, this sure ain't linear! but you knew that by the sqrt)

Pd=500 --> Eq1=3.27 (now we're getting closer)

Pd=1000 --> Opps! Vbd becomes undefined (except for EEs!), or rather becomes negative.

OK... let's take a pause... where does Vbd get to zero? That's easy enough... at Pd=935.73. So Pd must be less than 936

Pd=900 --> Eq1=0.76 (almost there)

Pd=930 --> Eq1=0.18 (really close... let's see if we can overshoot)

Pd=935 --> Eq1=-0.07 (finally! we know the answer is between 930 and 935 - probably good enough)

Pd=933 --> Eq1=0.06 (one more try)

Pd=934 --> Eq1=0.005

Call it done! Pd=934 (just a little bit over).

Second question, are there questions on the exam that mimick this iterative process? Third:

btw what is also the short cut to solvine this question on the P.E. exam?

Yes, you'll likely find a few questions that you can't easily solve algebraically. The key to a trial-and-error approach is to get the bracket smaller and smaller. In this problem, it took four tries to get the high end then another five tries to get an acceptable answer. That's nine run through the equations - which might take a bit more than six minutes!

BUT... here's why this is almost always easier on the exam: you're given four possible answers, one of which is the right one! Try each in the equations. If you're a good guesser, it takes just ONCE. If you're a horrible guesser, it still only takes FOUR (or three, if you're real confident... because if the first three don't work, and you didn't make a mistake, it's got to be the last). But certainly less than NINE tries like I had to do on this problem.

Make sense?

 

[Calixico]

OK... let's take a pause... where does Vbd get to zero? That's easy enough... at Pd=935.73. So Pd must be less than 936

Pd=900 --> Eq1=0.76 (almost there)

Pd=930 --> Eq1=0.18 (really close... let's see if we can overshoot)

Pd=935 --> Eq1=-0.07 (finally! we know the answer is between 930 and 935 - probably good enough)

Pd=933 --> Eq1=0.06 (one more try)

Pd=934 --> Eq1=0.005

Call it done! Pd=934 (just a little bit over).

Sraymond:

Thank for your guidance to this problem. Appreciated. I followed you logics up until Pd must be less than 936. Now you say plug Pd= 900 into equation #1. But equation #1 variables are all velocities. Could you show me how you got to Eq1=0.76, so I can complete these claculations. I am still in the dark.Could you expand a bit more on this slow learner...

Thanks again...

 

[IlPadrino]

Thank for your guidance to this problem. Appreciated. I followed you logics up until Pd must be less than 936. Now you say plug Pd= 900 into equation #1. But equation #1 variables are all velocities. Could you show me how you got to Eq1=0.76, so I can complete these claculations. I am still in the dark.Could you expand a bit more on this slow learner...
Thanks again...

No problem... if this explanation doesn't work, let me know and we'll try a different tack.

Do you see where Vad and Vbd become imaginary (square root of a negative number) for some values of Pd? That's obviously a bad thing and think of it as the point where the velocities go down to zero and then pass into the realm of negative velocity. This represents the upper bound of a choice for Pd.

To answer your specific question (Could you show me how you got to Eq1=0.76, so I can complete these calculations.): You're picking (guessing!) a value of Pd and then plugging into Eqs 2-4. Now you're got all the unknowns for Eq 1... but is it true given these values? If you pick a Pd of 900, you get the left side of Eq 1 to be 0.76. That does not equal 0, so the guess of Pd is WRONG. But you're getting closer to zero... so you pick another value of Pd and see if you can get still closer.

So, maybe this is the part I'm not explaining good enough: Eqs 2-4 depend on Pd only... you pick a value of Pd and calculate the velocities. Eq 1 depends on the velocities calculated in Eqs 2-4, so you can see if Eq 1 holds true for choice of Pd that gave you the results of Eqs 2-4.

In general, iterative solutions are appropriate when the equations are too difficult to solve algebraically - it's what calculators and computers do. For this problem, you should set up a table like this along the top:

Pd    Vab    Vbd   Vbc     Eq 1 (left side only)

Now pick a value of Pd, and then calculate Vab, Vbd, Vbc, and Eq1. You're trying to drive the left side of Eq 1 to zero. Keep picking values of Pd that get you closer.

But... and this is a big BUT... for the exam, so long as the iterative solution is the final step, you should NOT waste time solving. You should just try each of the answer choices to see which one works.

 

[Calixico]

No problem... if this explanation doesn't work, let me know and we'll try a different tack.
Do you see where Vad and Vbd become imaginary (square root of a negative number) for some values of Pd? That's obviously a bad thing and think of it as the point where the velocities go down to zero and then pass into the realm of negative velocity. This represents the upper bound of a choice for Pd.

To answer your specific question (Could you show me how you got to Eq1=0.76, so I can complete these calculations.): You're picking (guessing!) a value of Pd and then plugging into Eqs 2-4. Now you're got all the unknowns for Eq 1... but is it true given these values? If you pick a Pd of 900, you get the left side of Eq 1 to be 0.76. That does not equal 0, so the guess of Pd is WRONG. But you're getting closer to zero... so you pick another value of Pd and see if you can get still closer.

So, maybe this is the part I'm not explaining good enough: Eqs 2-4 depend on Pd only... you pick a value of Pd and calculate the velocities. Eq 1 depends on the velocities calculated in Eqs 2-4, so you can see if Eq 1 holds true for choice of Pd that gave you the results of Eqs 2-4.

In general, iterative solutions are appropriate when the equations are too difficult to solve algebraically - it's what calculators and computers do. For this problem, you should set up a table like this along the top:

Pd    Vab    Vbd   Vbc     Eq 1 (left side only)

Now pick a value of Pd, and then calculate Vab, Vbd, Vbc, and Eq1. You're trying to drive the left side of Eq 1 to zero. Keep picking values of Pd that get you closer.

But... and this is a big BUT... for the exam, so long as the iterative solution is the final step, you should NOT waste time solving. You should just try each of the answer choices to see which one works.

Bingo!

I got it now..Thanks a bunch IlPadrino.

It's really hard to believe that these kinds of questions could be on the exam. This is a question that I started working in the practice problem booklet. This is a three reservoirs problems where it asked for the direction of flow and pressure at a point in the system. I cannot figure out how this smae question could be asked, as it took me the third timne around, 12 minutes to solve it. These types, and the parallel pipes problems has me scared as h*ll of the exam.

 

[IlPadrino]

I cannot figure out how this smae question could be asked, as it took me the third timne around, 12 minutes to solve it. These types, and the parallel pipes problems has me scared as h*ll of the exam.

You shouldn't expect to answer every problem on the exam. I'd skip pipe networks and culverts altogether - or at the very least approach them last. Remember, this test is not about being 100% right, it is just about being 100% right at least about 70% of the time.