Material Balance Lead in a lake

AI Thread Summary
Tetraethyl lead was used in petrol in Australia from 1950 to 2002, impacting a lake used for drinking water. The lake has a steady state volume of 8.2 x 10^7 m3 and an inflow and outflow of 6.0 x 10^7 m3/year, with an initial lead concentration of 40 ppb in inflow and a deposition rate of 120 kg/year. Participants in the discussion are attempting to calculate the steady state concentration of lead in the lake water, with some confusion about the correct equations and the need for a transient material balance. One user calculated a concentration of 4.22 x 10^-5 μg/L but is uncertain about the accuracy of their approach. The conversation emphasizes the importance of including the lake's volume and the rate of lead accumulation in the calculations.
dimens
Messages
43
Reaction score
0

Homework Statement



Question 2.
Tetraethyl lead was used in early model cars to help reduce engine knocking, boost octane ratings, and help with wear and tear on valveseats within the motor. In Australia, our petrol
contained lead between approximately 1950 and 2002. Consider the impact of lead in petrol on a lake used for drinking water supply. The lake has a steady state water volume of 8.2 x 107m3 with an inflow and outflow of 6.0 x 107m3/year. From 1950, the lead concentration in the lake inflow water was 40 ppb and the deposition rate of lead onto the lake surface was 120 kg/y. Assume the lead concentration in the lake water was zero when lead first started being added to petrol and that the lake is completely mixed.

a.Calculate the steady state concentration of lead in
the lake water (in μg/L) for the time period when lead was included in petrol.


Homework Equations


I was thinking since output rate = reaction rate... QC = G-kCV


The Attempt at a Solution



We've got pretty much all of the variables.. Except for C and k...

Rearranging for C...

C = G / (Q + kV)

To find k... k = G/VC

(120kg/year) / 8.2x10^7 * 40 ppb

= 120 kg/year / 8.2*10^10L * 4x10^-8 kg/L
= 120/year / 3936
= 0.0366/year

...

c = 120 kg/year / 6.0*10^7 m3/year + 0.0366/year * 8.2*10^7m3

However I get some ridiculous number.
 
Physics news on Phys.org
It's hard to follow your calculations because you omit so many units.
 
There is no reaction taking place, so forget that. You need to do a transient (time dependent) material balance on the lead in the lake.

Let Q = volumetric water flow rate into and out of the lake.
Let C0 = concentration of lead in the water flowing into the lake.
Since the lake is well-mixed, let C = concentration of lead in the water exiting the lake, and also the concentration of lead within the lake (at any time t)
Let R be the rate of lead being deposited into the lake from above.
Write a time dependent differential mass balance for the lead in the lake.

Chet
 
  • Like
Likes 1 person
input rate = output rate + deposition
Qin*C0 = Qout * C + Rt
C(t) = (Qin*C0 - Rt) / Qout

Something like this?
 
Chestermiller said:
There is no reaction taking place, so forget that. You need to do a transient (time dependent) material balance on the lead in the lake.

Let Q = volumetric water flow rate into and out of the lake.
Let C0 = concentration of lead in the water flowing into the lake.
Since the lake is well-mixed, let C = concentration of lead in the water exiting the lake, and also the concentration of lead within the lake (at any time t)
Let R be the rate of lead being deposited into the lake from above.
Write a time dependent differential mass balance for the lead in the lake.

Chet

Hi Chester, i made an account just for this as i have the same question.

i have converted the kg of lead being deposited onto the lake into litres, 12.346 per year and have included that into the inputs. I have calculated the concentration of the water exiting the lake at year 1 as 4.22x10^-5 μg/L, however I am still struggling to find the correct equation as i do not think i have think that is correct answer. The equation i used was (Qin x Cin + Rin)/Qout = Cout. Any help would be appreciated.
 
Connor Hadley said:
Hi Chester, i made an account just for this as i have the same question.

i have converted the kg of lead being deposited onto the lake into litres, 12.346 per year and have included that into the inputs. I have calculated the concentration of the water exiting the lake at year 1 as 4.22x10^-5 μg/L, however I am still struggling to find the correct equation as i do not think i have think that is correct answer. The equation i used was (Qin x Cin + Rin)/Qout = Cout. Any help would be appreciated.
Why did you convert the lead deposited into liters?

Your equation omits the rate of accumulation of lead in the lake. Why did you think they told you the volume of the lake?
 

Similar threads

Back
Top