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Using of differential equation

I've to solve the following problem:

Volume of a swimming pool is 500m3500 m^3. Correct level of pollutions should not be higher than 1%1\%. Due to failure of the purification system, the level of pollutions reached level 10%10\%. How much time is needed to lower level of pollutions to correct value, assuming the purification system is pumping in and out 5m3h5 \frac{m^3}{h}?

I don't know what equation should I create. I can solve differential equations so it will be enough, when you would just create the equation, without solving it.

Of course I know that 1%=5m31\%=5 m^3 and 10%=50m310\%=50 m^3 :biggrin:

Thanks in advance :wink:
Reply 1
Avatar for aznkid66
aznkid66
1
Let y denote the level of pollution, in percentage of total volume (500 cubic metres).
Let p denote the volume of pollutants, in cubic metres.

Equation 1: y = p/500
(Note: setting the total volume to a constant assumes the purification system pumps in clean water with equal volume to any pollutants removed.)

Now, let t denote the time since the purification system has started, in hours.
Let dp/dt denote the change in volume of pollutants over time, in cubic metres per hour.

Equation 2: dp/dt = -5.

Last but not least, note that when t=0, p=50.

That should be everything you need :P



And note how much easier it is to do without calculus XD
(edited 11 years ago)
Reply 2
Avatar for the bear
the bear
20
not sure if method in post 2 is correct... calculus is needed...

let the amount of clean water in the pool = C ( at the start it is 450 )

the proportion of clean water which is pumped in is 100% so 5 m3per hour

the proportion of clean water which is pumped out is C/500 so 5*C/500 m3 clean water is removed per hour

thus the rate of change of clean water is

dC/dt = 5 - 5*(C/500) m3hr-1

now solve this equation using the initial values t = 0 and C = 450

then substitute C = 495 to find the value of t when just 1% of filth remains
(edited 11 years ago)
Reply 3
Avatar for pawellogrd
pawellogrd
OP
Thank you both for replies.

In first case got t=9t=9 hours.
In second case got t=100ln⁡(10)≈230,26t=100 \ln(10) \approx 230,26 hours.

Not sure, which result is correct..

My calculations:

First case:

dpdt=−5\frac{dp}{dt}=-5

dp=−5dtdp=-5dt

∫dp=−5∫dt\int dp = -5 \int dt

p=−5t+Cp=-5t+C, here CC is constant of integration of course

From initial values, when p=50, then t=0:

50=−5⋅0+C50=-5 \cdot 0 + C

Thus C=50C=50

p=−5t+C=−5t+50p=-5t+C=-5t+50

−p5=t−10-\frac{p}{5} = t - 10

t=10−p5t = 10 - \frac{p}{5}

So my reply is: t=10−55=9t=10-\frac{5}{5}=9

---

Second case:

dCdt=5−5⋅C500\frac{dC}{dt} = 5-5 \cdot \frac{C}{500}

dCdt=500−C100\frac{dC}{dt} = \frac{500-C}{100}

100500−CdC=dt\frac{100}{500-C} dC = dt

100∫1500−CdC=∫dt100 \int \frac{1}{500-C} dC = \int dt

−100ln⁡(500−C)+Cons=t-100 \ln(500-C) + Cons = t, here ConsCons is constant of integration

From the initial values, when t=0, then C=450:

−100ln⁡(500−450)+Cons=0-100\ln(500-450)+Cons=0

Thus Cons=100ln⁡(50)Cons=100\ln(50)

t=−100ln⁡(500−C)+Cons=−100ln⁡(500−C)+100ln⁡(50)t=-100\ln(500-C)+Cons=-100\ln(500-C)+100\ln(50)

So my reply is: t=−100ln⁡(500−495)+100ln⁡(50)=100(ln⁡(50)−ln⁡(5))=100ln⁡(10)t=-100\ln(500-495)+100\ln(50)=100(\ln(50)-\ln(5))=100\ln(10)
Reply 4
Avatar for aznkid66
aznkid66
1
Right, I made the mistaken assumption that the purification system could take out pure pollutants rather than only a solution of polluted water.

I checked your working for the second method, and everything seems correct.
Reply 5
Avatar for pawellogrd
pawellogrd
OP
Ok, thank you :smile:

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