Material Balance: Determine Minimum Ditch Length for 90% Odour Reduction

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dimens
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Homework Statement


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?

Homework Equations





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..
 
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Reaction constant is the speed of which a reaction creates or reducts? I always thought it was r = k[a]??
 
[a] concentration of substance in a first order reaction
r = reaction?

?
I'm feeling so dumb and confused right now lol
 
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.
 
dimens said:
Haven't done calculus for a few years …

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:
 
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:
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))
 
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.
 
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dimens said:
But I'm having trouble with the units, it should only be in meters these give me a weird set of units.

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?​
 
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??
 
Should they all be in SI units? So seconds?
 
dimens said:
Should they all be in SI units? So seconds?

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
 
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.