Need help with a Hydraulics Water Question

[kdiehl]

Hello. I am new to the forum. I need help with a question I cannot understand the answer to. Just a quick background. I am a Class3 Water Distribution Operator(or will be if I pass my test). My test is this Friday and I have prepared for it by reading every available AWWA book, Taking online classes and using as many online resources as I can find. I have done my homework. I ran into a question on the practice exams that has me concerned as I believe I answered it correctly but the Answer key gave a different answer due to rounding. I ran it by my Asst Water super, OUr Chief Chemist and the Asst Town engineer and they all agreed with me but that doesn't help me with my exam Saturday. I asked the Agency giving the actual test and they told me the book was correct due to Significant Digits rule but I fail to see how. Can anyone look at the question and both the answer I chose vs the answer the book gave and tell me why the book is right ? If the book is right then I must be missing some sort of Water Hydraulic Engineering Discipline. I have read the AWWA Basic Science concept 4th ed from cover to cover and found nothing in it regarding this.
I will copy the questions and the answer the book gave. It is the AWWA Operator Training guide 6th ed.

A small cylinder on a hydraulic jack is 10 inches in diameter.A force of 130lb is applied to the small cylinder.If the diameter of the large cylinder is 2.5 ft, what is the total lifting force ?
a.)1170 lb
b.)1200 lb
c.)1250 lb
d.)1300 lb.
By using Force=PxA I came up with 238.49 for P in the 10" cyl
Then using that to figure out the 2.5ft cyl I had 238.49x4.91=1170.1
The book did it slightly different and had 1170.926 but then said now round to 1200.
Why round that to 1200 ?? I don't understand and I fear I am missing something that will affect my test Friday.
Thanks for any help anyone can offer.

 

[rconner]

Whoever wrote the "book" answer would be right, only if the accuracy of the force to be applied ("130" lb) or the diameter of the "10" inch or the "2.5" feet cylinder were only accurate to two significant figures. If on the other hand the size of the 10" and 2.5 feet cylinders are both accurate e.g at least to the nearest tenth inch, and the force is accurate also to the nearest pound, he/she would not be right.
While good lessons in significant figures are important, I don't really like this particular trick question however, as I believe many cylinders are machined to more accurate dimensions, as are many modern scales.

 

[kdiehl]

So based on the information given in the question should I have arrived at the book answer ? I'm not seeing it. I just don't want to mistake this on the test.

 

[kdiehl]

OK... Every number given in the question has 2 or less significant digits:
10...1 significant digit
130...2 sig digs
2.5...2 sig digs

So, we are limited to 2 significant digits for this question. If we were to write 1170 with 2 significant digits, it would be 1200.That part I get now But it's kind of BS to require you to guess that you need to use significant digits for this question. Is there some way of knowing that I should use this ? Some standard I am missing ? In other words how would I know from the information given in the question that it needed to be answered with Significant digits vs just assuming 1170 as the closest answer ?

 

[rconner]

The author of the "right" answer will defend his/her answer, not due to "rounding" but instead that (they say) the force could be anywhere from 125-134 and still could be called by some inexact soul/weigher 130, or the diameter of the pistons could similarly be off some from the two digits given e.g. from 9.5-10.4 and still called "10" in the problem description. If he/she is "given" that etc., they are "right", and the answer can be no more accurate than two figures. An answer can indeed be no more accurate than that of any component part, and the most egregious violations came with the electronic calculator and computers.
If I were writing this question, I would on the other hand also tell the young minds full of mush that the numbers given upfront are only accurate to the two significant figures shown in comparable units (then we probably wouldn't be arguing).
While again I don't like this particular somewhat impractical example, I hope you get the lesson ( Learn this - the world is not fair, and sometimes it is just best to understand where the book/teacher is coming from or trying to say!)

 

[cvg]

the answers are all rounded to the nearest 10 lbs. you would never round up to the nearest 100 lbs. I believe the book is incorrect.

 

[itsmoked]

Lame question indeed. Idiots. However they will defend their answer(s) and you should expect them to be counting sig figs and expecting you to answer appropriately. It's exactly as you last describe.

Nothing to panic about just realize they're choosing to be anal and so you will need to be too.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[fel3]

I agree that this is a stupid way to write an exam question. As discussed, the real problem with the question is that the data has unrealistically limited precision, which leads to an answer with unrealistically limited precision. As rconner indicated, the type of data used in the question should be known to more sigfigs than were presented.

If we use the tolerance limits for each number to come up with the extreme values for the answer, it gets even harder to justify rounding off to 1200 lbf. In fact, since the diameter of the small cylinder is known to only one sgifig, I think 1000 lbf (one sigfigs) is a better way to round of 1170 lbf than 1200 lbf (two sigfigs).

Here's my take, based on adding one sigfig for each number to establish the tolerance limits:
-- Force on small cylinder, Fs = 130 lbf => range = 125 lbf to 135- lbf
-- Diameter of small cylinder, Ds = 10 in => range = 5 in to 15- in. However, if the diameter is reported as 10. in, then we have two sigfigs and the range is either 9.5 in to 10.5- in (two and three sigfigs) or 9.95 in to 10.05- in (three and four sigfigs), depending on how you were taught to handle data centered on a power of 10. For the calculation below, I will assume that the decimal point does not exist.
-- Diameter of large cylinder, Dl = 2.5 ft => range = 2.45 ft to 2.55- ft
-- Maximum lifting force, F(max) = Fa(max)*(Dl(max)/Ds(min))^2 = (135- lbf)*((2.55- ft)(12 in/ft)/(5 in))^2 = 5,056- lbf => 5,000 lbf (to one sigfig)
-- Minimum lifting force, F(min) = Fa(min)*(Dl(min)/Ds(max))^2 = (125 lbf)*((2.45 ft)(12 in/ft)/(15- in))^2 = 480 lbf => 0 lbf (to one sigfig). Thus, this hydraulic jack is defective.


==========
"Is it the only lesson of history that mankind is unteachable?"
--Winston S. Churchill

 

[itsmoked]

LOL You're actually right!

Keith Cress
kcress -

http://www.flaminsystems.com

 

[kdiehl]

Ok. Thank for the help. I get how the author can justify the answer but when I sit down Friday for my test and run into a similar question how can I tell if that is how I should determine whether to apply significant digits rule or not ?
Or will it be basically a lucky Guess ?
I suppose I could look at then answer choices and if there is no answer based on significant digits but if there is ??

 

[rconner]

I agree with your supposition (if same guy/gal wrote exam, might better watch out -- they are apparently at least trying to teach a little more than hydraulics)

 

[coloeng]

kdiehl,

I've taken some of the actual tests and have yet to find one of the questions that had two answers that could be correct, one based solely on significant digits and the other based on given data. If you've studied hard you will do well and don't worry about missing one question because of this scenario.

Good luck on your test.

 

[bimr]

The 2 significant digits is correct. The problem with the example is the choice you have for answers. Are you taking a operator test or a math test?

In carrying out calculations, the general rule is that the accuracy of a calculated result is limited by the least accurate measurement involved in the calculation.

In addition and subtraction, the result is rounded off to the last common digit occurring furthest to the right in all components. Another way to state this rule is as follows: in addition and subtraction, the result is rounded off so that it has the same number of digits as the measurement having the fewest decimal places (counting from left to right). For example,

200 (assume 3 significant figures) + 14.753 (5 significant figures) = 214.753,

which should be rounded to 214 (3 significant figures). Note, however, that it is possible two numbers have no common digits (significant figures in the same digit column).

In multiplication and division, the result should be rounded off so as to have the same number of significant figures as in the component with the least number of significant figures. For example,

6.0 (2 significant figures ) × 23.80 (4 significant figures) = 142.8000

which should be rounded to 140 (2 significant figures).

Some numbers are exact because they are known with complete certainty.

Most exact numbers are integers: exactly 12 inches are in a foot, there might be exactly 23 students in a class. Exact numbers are often found as conversion factors or as counts of objects.

Exact numbers can be considered to have an infinite number of significant figures. Thus, the number of apparent significant figures in any exact number can be ignored as a limiting factor in determining the number of significant figures in the result of a calculation.

Weights are not exact numbers.

I agree with Coloeng, if you have done the preparation, you have nothing to worry about.

 

[kdiehl]

I am taking the operator cert level 3.
I hope the questions are more specific than in the Practice books. They are loaded with typos besides and what is scary is they are co written by ABC who also writes the test.