Material Balance

1. Mar 15, 2014

dimens

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. Mar 15, 2014

tiny-tim

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

what is that equation?

3. Mar 15, 2014

dimens

Reaction constant is the speed of which a reaction creates or reducts? I always thought it was r = k[a]??

4. Mar 15, 2014

tiny-tim

eek!

what is r ? what is a ?

(is r a derivative? if so, of what?)

5. Mar 15, 2014

dimens

[a] concentration of substance in a first order reaction
r = reaction?

???
I'm feeling so dumb and confused right now lol

6. Mar 15, 2014

tiny-tim

(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

7. Mar 15, 2014

dimens

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.

8. Mar 15, 2014

tiny-tim

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

to get you started, the solution to da/dt = -ka is a(t) = a(0)e-kt

9. Mar 15, 2014

Staff: Mentor

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

10. Mar 16, 2014

dimens

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
11. Mar 16, 2014

tiny-tim

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

(or kt = -ln(0.1) = ln(10))

12. Mar 16, 2014

dimens

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
13. Mar 16, 2014

tiny-tim

let's see

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? ​

14. Mar 16, 2014

dimens

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??

15. Mar 16, 2014

tiny-tim

careful!!

t = 7.67 days

(what else could the units of t be? )​

16. Mar 16, 2014

dimens

Should they all be in SI units? So seconds?

17. Mar 16, 2014

tiny-tim

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

18. Mar 16, 2014

dimens

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.