Transmitter has a range

Discussion in 'Homework Help' started by bwd111, Aug 5, 2013.

  1. bwd111

    Thread Starter Member

    Jul 24, 2013
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    1
    A given transmitter has a range of 0 to 200 degrees Fahrenheit. Maintains an output signal of 3 psig at 0 Fahrenheit and 15 psig at 200Fahrenheit what is the output change per unit of controlled variable.
    My answer was 0.075 psig can some check and let me know
    0.06psig/
    Fahrenheit


    0.075psig/
    Fahrenheit


    12.00psig/
    Fahrenheit

    16.67psig/
    Fahrenheit
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,757
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    An answer of 0.075psig is definitely wrong. Look at the units.

    You need to show your work.

    Would you get the same answer if the problem had stated that the output was 14psig at 0°F?

    If so, does it make sense that both transmitters could have the same sensitivity?
     
  3. paulktreg

    Distinguished Member

    Jun 2, 2008
    612
    120
    Both transmitters?

    I ain't really sure what's going on here but my maths is pretty good.

    3psig at 0F

    15psig at 200F

    therefore 0F to 200F is 12psig which is 12/200=0.06psig/F :eek:
     
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  4. WBahn

    Moderator

    Mar 31, 2012
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    And thank you, Paul, for contributing to the Homework Done For You forum, where you don't have to think or show your work because someone will come along and serve up the answer on a silver platter for you.
     
  5. paulktreg

    Distinguished Member

    Jun 2, 2008
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    You're welcome!

    If you'd have started the OP out on the right foot I wouldn't have bothered.
     
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  6. WBahn

    Moderator

    Mar 31, 2012
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    So asking the OP to show some work is the wrong foot?

    Asking the OP to consider the very thing that he failed to take into account is the wrong foot?

    Just spoon feeding someone the solution is the right foot?
     
  7. paulktreg

    Distinguished Member

    Jun 2, 2008
    612
    120
    An answer of 0.075psig is definitely wrong. Look at the units.

    OK he forgot the /F. Would 0.06psig have been definitely wrong? Well perhaps you'd have said the same?

    You need to show your work.

    I'm all for that!

    Would you get the same answer if the problem had stated that the output was 14psig at 0°F?

    Why should he get the same answer? He'd still get a different wrong answer based on his first?

    If so, does it make sense that both transmitters could have the same sensitivity?

    Who mentioned two transmitters?
     
  8. bwd111

    Thread Starter Member

    Jul 24, 2013
    117
    1
    I tried to show my work but had a different calc and had a hard time posting my work. But I do learn and work the calc out for myself as I have 7 other questions that this calc will be used. This was the easy question
     
  9. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    I would have expanded a bit more. The point here is that the units were wrong. The fact that the answer was also numerically wrong meant that there wasn't any real need to distinguish it further.

    The evidence is very much to the contrary. He is asked to show his work and before he even makes any attempt to do so you throw the solution at him. A solution, I might add, that neglects proper tracking of units and reinforces the notion that it is okay to just magically tack units onto the answer out of nowhere.

    No, he would almost certainly get the SAME wrong answer based on his first. It's impossible to be sure, of course, since he didn't show any work and you didn't give him the opportunity to.

    Did you give any thought as to how he came up with his original answer of 0.075psig/°F? He took the one endpoint, 15psig at 200°F, and simply divided them. 15psig/200°F=0.075psig/°F.

    Hence, with a slightly changed problem in which the lower point has a very different output, his original method would have yielded the same answer, despite the fact that it, hopefully, should be pretty obvious that the two transmitters should yield very different values. So, what piece of information changed between the two? That's a big hint that perhaps that piece of information plays a role in the computation.

    I did!

    The first transmitter is the original one with an output of 3psig at 0°F and the second one is the transmitter that I introduced with an output of 14psig at 0°F. Both transmitters produce 15psig at 200°F.
     
  10. bwd111

    Thread Starter Member

    Jul 24, 2013
    117
    1
    Actually this really helped out a lot as this was going to be the other calculation I was going to try. This is just one of 85 question I need to answer. But thank you
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Hope you don't want all 85 answers verified!
     
  12. LDC3

    Active Member

    Apr 27, 2013
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    Why not, you don't have to read them if you don't want to. ;)
     
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  13. bwd111

    Thread Starter Member

    Jul 24, 2013
    117
    1
    No! If I needed that I would just quit! Need help with about 6 questions
     
  14. bwd111

    Thread Starter Member

    Jul 24, 2013
    117
    1
    That's A good one
     
  15. WBahn

    Moderator

    Mar 31, 2012
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    4,800
    While I can only recommend, I do recommend that for those questions you post your work. You may not believe and it may not make sense, but you really will learn more if you have to struggle to find the solution and if others restrict themselves to providing hints and asking leading questions to steer you toward figuring things out mostly on your own.

    Figure that, in many cases, someone that is struggling with something has read the explanations and the examples in the text and has heard the explanations and seen the examples worked by the instructor in class. Despite this, there remains some block, be it big or tiny, that is preventing the person from having it click and all come together. It is unlikely, in that situation, that seeing yet one more example worked out by someone else is going to address that. Yes, it can and does happen, but more often than not it only creates the illusion of understanding and as soon as you encounter a similar problem again you are back right where you were. But when you struggle with a problem you have to come face-to-face with whatever is preventing it from clicking and the smaller and more incremental the hints and suggestions are the greater the chance that you hit that roadblock square on and have an "ah-ah" moment.
     
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  16. bwd111

    Thread Starter Member

    Jul 24, 2013
    117
    1
    Thank you and you are correct and I want to learn the correct ways as I love the lessons I have enrolled in and learning is what I like. I have been learning and has helped a lot
     
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