1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Material Balance

  1. Mar 15, 2014 #1
    1. The problem statement, all variables and given/known data
    Question 1.
    An industry uses a long drainage ditch to break down their wastes and in particular to remove the odour. The waste travels along the ditch at a velocity of 0.5 m/h. The odour is reduced as a first order reaction with a reaction constant of k=0.30 day-1 The company must reduce the original odour emissions of the waste by 90% to reach the environmental guideline for acceptable odour.
    What is the minimum ditch length (in m) to ensure the wastes reach the environmental guideline for odour?

    2. Relevant equations



    3. The attempt at a solution
    I was thinking because the velocity is 0.5m/h = 12m/day and the 90% would mean the odour would flow for 3 days, 12*3 = 36m. However I'm not sure what equation should be used or if I'm approaching this completely wrong..
     
  2. jcsd
  3. Mar 15, 2014 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi dimens! :smile:
    "reaction constant" has a definition, and that definition is an equation

    what is that equation? :wink:
     
  4. Mar 15, 2014 #3
    Reaction constant is the speed of which a reaction creates or reducts? I always thought it was r = k[a]??
     
  5. Mar 15, 2014 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    eek! :eek:

    what is r ? what is a ? :confused:

    (is r a derivative? if so, of what?)
     
  6. Mar 15, 2014 #5
    [a] concentration of substance in a first order reaction
    r = reaction?

    ???
    I'm feeling so dumb and confused right now lol
     
  7. Mar 15, 2014 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    (have you done calculus?)

    r is the rate at which a (the amount) is getting smaller

    so r = -da/dt,

    and so da/dt = -ka :wink:
     
  8. Mar 15, 2014 #7
    Haven't done calculus for a few years and even still I was pretty pedestrian with it. Referring back to the initial question, does that mean we substitute our own values in and assume? There's no amount given in the original question only a percentage they want reduced.
     
  9. Mar 15, 2014 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    sorry, but you're going to need to dust off those calculus books for this course! :wink:

    to get you started, the solution to da/dt = -ka is a(t) = a(0)e-kt :smile:
     
  10. Mar 15, 2014 #9
    In tiny-tim's equation, t is the cumulative residence time in the ditch. So t = x/v, where x is the distance along the ditch.

    Chet
     
  11. Mar 16, 2014 #10
    Making a little more sense, but do we substitute values in for potential concentrations? As there's none stated only percentage we want to lose? So we've got 100% initially going into the lake, then 10% in the output.

    100pm= 1000ppm * e ^ -(0.30/day*(x*0.5m/hr)
     
    Last edited: Mar 16, 2014
  12. Mar 16, 2014 #11

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    you need to think more in terms of equations

    your basic equation is a(t) = a(0)e-kt

    so the equation the question needs you to solve is a(t) = a(0)*0.1,

    which becomes e-kt = 0.1 :wink:

    (or kt = -ln(0.1) = ln(10))
     
  13. Mar 16, 2014 #12
    Really appreciate the help guys, I think I'm almost there.

    But I'm having trouble with the units, it should only be in meters these give me a weird set of units.
     
    Last edited: Mar 16, 2014
  14. Mar 16, 2014 #13

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    let's see :smile:

    k = 0.3/day

    kt = ln(10), so t = ln(10)/k = ln(10)/0.3 days

    and now multiply by metres/day to get metres …

    what weird units were you getting? ​
     
  15. Mar 16, 2014 #14
    Thanks for the reply, just realized my original workings are wrong...

    t = ln(10)/0.3days
    t = 7.67/days
    t = x/v
    x/v=7.67/days
    x = 7.67/day * (0.5 m/hour * 24 hours/day)
    x = 7.67/day * (12m/day)
    x = 92m/day^2??
     
  16. Mar 16, 2014 #15

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    careful!! :rolleyes:

    t = 7.67 days

    (what else could the units of t be? :wink:)​
     
  17. Mar 16, 2014 #16
    Should they all be in SI units? So seconds?
     
  18. Mar 16, 2014 #17

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    no, the units of time can be anything, days seconds, hours, it doesn't matter

    but you wrote /days instead of days, which is why you finished with /days2
     
  19. Mar 16, 2014 #18
    Definitely feeling silly and like I need to polish up my maths skills right now. With my fix up of my units:


    t = ln(10)/0.3days
    t = 7.67 days
    t = x/v
    x/v=7.67 days
    x = 7.67 day * (0.5 m/hour * 24 hours/day)
    x = 7.67 day * (12m/day)
    x = 92m

    Cheers, much appreciated and thankyou.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted