Contaminated water is passed through a filtration system that removes contaminan

Question

Contaminated water is passed through a filtration system that removes contaminants. Assume that the concentration of the contaminants decays exponentially with respect to the amount of time the water spends in the filtration system. This particular filtration system is able to reduce the concentration of the contaminants by 70% over a 24 hour period.

(a) By what percent does the concentration of the contaminants decrease by each hour that the water spends in the filtration system?

(b) Find the continuous hourly rate at which contaminants are removed by the filtration system.

(c) How long does it take for the filtration system to reduce the concentration of contaminants by 50%?

(d) Suppose a contaminant begins leaking into a reservoir. Before the leak, the contaminant was present at a level of 2 ppm (parts per million). After the leak occured, the concentration of the contaminant increases at rate of 1 ppm every 3 hours.

The leak continues for 3 days. At this point, it is stopped and the filtration system described above is used to remove the contaminant from the water.

Find a formula for C(t), the concentration of the contaminant in the reservoir t hours after the leak began. (Note: C(t) will be a piecewise function.)

Suppose the water is unsafe when the concentration of the contaminant is greater than 17 ppm. How long was the contaminant present in the reservoir at unsafe levels?

Explanation / Answer

Since it is exponential decay we know that ,

Y= X e-vt

where Y = concentration of contaminant at time t

X = concentration of contaminant at time t = 0

t = time spend

v= exponential decay

As per question it reduces concentration by 70% with in 24 hrs.This means that concentration is reduced to 30% of original concentration

Let original concentration be X

=> Y= 30% of X = 0.3X

t= 24 hrs

=> Y= X e-vt

=> 0.3X = X e-24v

=> 0.3= e-24v

=> -24v = ln(0.3)

=> v= -ln(0.3)/24

a) t=1 hour

Y = X e-v.1

Y= Xe-v

using value of v

Y= Xe-(-ln(0.3)/24)

Y= X eln(0.3)/24

Y= X e-0.05016

Y=0.951X

Percentage Reduction = (X- 0.951X)*100/X= 4.89%

b)

For Exponential Decay continuous rate= dX/dt= -vX= -0.05016X

c) For concentration to be reduced by 50%,half lifetime is needed. For Exponential decay, half lifetime

t= ln(2)/v

using value of v which is exponential constant

t= ln(2)/0.05016=13.81 hours

d) C(t) = 2 + (1/3*t) ppm

where t is hours after the leakage.

Leakage continued for 3 days which means 72 hours.

C(72) = 2 + (1/3 * 72) = 2 + 24 = 26 ppm

using formula Y = X e-0.05016t

water will become safe when Y= 17 ppm

=> 17 =26 e-0.05016t

=> e-0.05016t= 17/26

=> t=8.470 hours

this is the time taken by filter to reduce contamination level to safe one.

Time for which contamination level was unsafe before starting filtering = t = 72- time taken by contamination to reach a 17ppm level

t= 72 - ((C(t)-2)*3)

t = 72 - ((17-2)*3)

t= 72 - (15*3)

t= 72 - 45

t = 27 hours

total time for which water remained unsafe = 27 + 8.470 = 35.470 hours

while time for which water remained unsafe after starting filtering = 8.470 hours

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