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Math Help - vacuum pump poser

  1. #1
    Newbie
    Joined
    Nov 2009
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    vacuum pump poser

    hello again.

    i have a question i wonder whether some one will give a hint as to how best solve it ?

    A vacuum pump removes 15% of the air that is in a container at the


    beginning of a stroke. If
    mr is the mass of air in the container after the r th
    stroke:
    (i) express


    m2 and m3 in terms of the original mass m0
    (ii) how many strokes are required to obtain 1% of the original mass?


    so assuming mr is the mass of air after rth stroke i am assuming m2 must be the 2nd stroke =m1(15%) +m2,m3 must be m2 +m1 so we have a number series going on.

    if mo is total air mass ie mo =100% can i express m2 and m3 as a percentage of mo ie m2 must equal 30% and m3 must equal 45% of mo respectively?not sure thats accurate enough.

    qii. i will need to think more about,if every stroke assuming its in equilibrium removes 15% of air every time 100/15 =approx 7 strokes required to get to this point maybe thats also too simplistic.


    apologies if i have got this in the wrong forum!

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  2. #2
    Senior Member
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    2
    If the pump removes 15% then it leaves behind 85%.
    So, if the original mass of air is m0, the mass left after 1 stroke m1 = m0 x 0.85.
    The mass left after 2 strokes: m2 = m1 x 0.85 = m0 x 0.85^2, etc...
    and so m3 = .... I'll leave that up to you!
    This forms a geometrice progression.
    So the mass left after r strokes mr = m0 x 0.85^r.

    For the second part, your approach is incorrect. You need to find r such that mr = 0.01 x m0, ie you need to find r such that 0.85^r = 0.01. Hint: That will involve logs!!
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  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    49
    A vacuum pump removes 15% of the air that is in a container at the beginning of a stroke. If mr is the mass of air in the container after the r th stroke:
    (i) express m2 and m3 in terms of the original mass m0
    (ii) how many strokes are required to obtain 1% of the original mass?

    so assuming mr is the mass of air after rth stroke i am assuming m2 must be the 2nd stroke =m1(15%) +m2,m3 must be m2 +m1 so we have a number series going on.


    if mo is total air mass ie mo =100% can i express m2 and m3 as a percentage of mo ie m2 must equal 30% and m3 must equal 45% of mo respectively?not sure thats accurate enough.

    qii. i will need to think more about,if every stroke assuming its in equilibrium removes 15% of air every time 100/15 =approx 7 strokes required to get to this point maybe thats also too simplistic.
    ----------------------------------------------------------

    your assuming the vacuum takes out 15 percent of the full tank. it actually removes 15% of whats left.
    e.g
    M=(100% full tank),
    M-Mx15% will give you the air left after first stroke,
    which can be written as Mx0.85, now on the second stroke you remove another 15% hence (Mx0.85)x0.85.... etc
    now for r strokes the mass (should really be volume) of air in the tank is given by Mx0.85^r

    (i) express m2 and m3 in terms of the original mass m0

    m2 is the mass of air after the second stroke hence m0x0.85^2, since m0 = M
    m3 = m0x0.85^3

    (ii) how many strokes are required to obtain 1% of the original mass?

    now this part requires logs so i doubt you can do this.
    but heres how it goes:

    Mx0.85^r = 1%, where M is a full tank (100%)

    1x0.85^r=1/100

    Log(base(0.85)) 1/100 = r

    r=28.3.....

    so it would take about 29 strokes to get 1% fuel.


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  4. #4
    Newbie
    Joined
    Nov 2009
    Posts
    17
    many thanks for reply and "words of encouragement", its obviously a bit more complicated than i anticipated,then you prob guessed that.

    i shall have to try and redo this lesson again in the hope i can understand it better, hate logs and powers why are they so complicated?.
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