Question 1: Consider a lake of constant volume V containing...

Question 1: Consider a lake of constant volume V containing at time t an amount Q(t) of pollutant, evenly distributed throughout the lake with a concentration c(t), where c(t) = Q(t)/ V. Assume that water containing a concentration k of pollutant enters the lake at a rate r, and that water leaves the lake at the same rate. Suppose that pollutants are also added directly to the lake at a constant rate P. Note that the given assumptions neglect a number of factors that may, in some cases, be important; for example, the water added or lost by precipitation; absorption; and evaporation; the stratifying effect of temperature differences in a deep lake; the tendency of irregularities in the coastline to produce sheltered bays; and the fact that pollutants are not deposited evenly throughout the lake, but (usually) at isolated point around its periphery. The result below must be interpreted in the light of the neglect of such factors as these. (a) If at time t = 0 the concentration of pollutant is c0, find an expression for the concentration c(t) at any time. What is the limiting concentration as t ? ?. (b) If the addition of pollutant to the lake is terminated (k = 0 and P = 0 for t > 0), determine the time interval T that must elapse before the concentration of pollutants is reduced to 50% of its original value; to 10% of its original value. (c) Table below contain data for several of the Great Lakes. Using these data determine from part b the time T necessary to reduce the contamination of each of these lakes to 10% of the original value. Lake | V (km3x103) | r (km3/year) A | 12.2 | 65.2 B | 4.9 | 158 C | 0.46 | 175 D | 1.6 | 209

Transcript

00:01 Hi there, consider a lake with constant volume v containing at the time t an amount of pollutants given by qt which is evenly distributed throughout the lake with a concentration c t which is equal to qt divided by v.

00:19 Assume that water containing concentration k of pollutants enter the lake at the rate r and that the water leaves the lake at the same rate.

00:29 Suppose that the pollutants are also added directly to the lake with a concentrate p.

00:37 If at time t is equal to 0, the concentration of pollutants was c0, then we want to find the expression of concentration c t.

00:46 By the law of conservation, we can say that the rate of mass accumulation is equal to rate of mass in minus rate of mass out.

01:15 From here we know that the rate of mass in is the sum of rk plus p and the rate of mass out is r multiplied by c t so we have d q divided by d t is equal to rk is equal to rk plus p minus r multiplied by c of t which can be written as r k plus p minus r d divided by v q of t this gives us the differential equation d q divided by d t plus r divided by v q is equal to r k plus p this is a first order linear in homogeneous differential equation and we will solve it by multiplying both sides by the integrating factor i which is given by e raised to the power integral r divided by v d t now since d divided by d x of x is equal to 1 this and by that integral 1 d x will be equal to x so we have i is equal to e raised to the power r t divided by v now we multiply the equation by the integrating factor i and we get e raised to the power r t divided by v d q divided by d t plus e e raise to the power r t divided by d t plus e raise to the power r t the power r t divided by v multiplied by r divided by v multiplied by r divided by q is equal to e to the power r d divided by v multiplied by r k plus p now notice that since d divided by d d of e x to the power r t divided by v of q is equal to by the product rule of differentiation we have d divided by d t e raised to the power r t divided by v multiplied by q plus e r t divided r t divided by v d d t of q which is equal to r d divided by v e raised to the power r t divided by v of q plus e to the power r t divided by v d q divided by d t so we can write the differential equation as d divided by d t of e raised to the power r t divided by v multiplied by q is equal to e raised to the power r t divided by v multiplied by r k plus p so this can be written as d of e raised to the power r t divided by v multiplied by q is equal to e raised to the power r t divided by v multiplied by r k plus e raise to the power r t divided by v multiplied by p t t integrating both sides we get e -raged to the power r t divided by v multiplied by cube is equal to integration is linear we can write it as integral e -raise to the bar r t divided by v multiplied by r k d t plus integral e -raise to the power r t divided by v multiplied by d t since d divided by d x of e raise to the power a x where a sum constant is equal to a, e to the power a x, this implies that integral e raised to the bar a x, dx will be equal to e raised to the bar a x divided by a.

05:58 So we have this equal to e raised to the bar rt divided by v divided by r divided by v multiplied by r k plus e -rae -rae -to -the -par -t divided by v multiplied by p divided by r divided by v plus some constant c so we get e -ra -t to the part r t divided by v multiplied by q will be equal to r divided by r is one kb e -rae -raise to the part r t divided by v plus p v divided by r e raised to the power rt divided by v plus some constant c.

06:46 Now we divide both sides by e raise to the power r2 divided by v and we get q is equal to kv plus pv divided by r plus c multiplied by e raised to the power minus rt divided by v.

07:06 Dividing both sides with the volume b we obtained the concentration c t to be equal to k plus p divided by r plus c divided by v e raised to the power minus r t divided by v now since c of zero is equal to c not we get c not is equal to k plus b divided by r plus c divided by v e raised to the path 0.

07:45 This implies that c divided by v will be equal to c0 minus k minus p divided by r.

07:55 Therefore, concentration at any time t will be given by c of t is equal to k plus p divided by r plus c0 minus k minus p divided by r it divided by e raised to the power minus r t divided by v now we want to know the limiting concentration as t tends towards infinity that will be limit t t tending towards infinity c of t will be equal to limit k tense towards infinity k plus p divided by r plus c0 minus k minus p divided by r multiplied by e raised to the power minus r t divided by p.

08:48 Limit of sum is equal to the sum of limits.

08:51 This will be equal to limit t t t tending towards infinity of k plus limit t t t t t t t t t t turning towards infinity p divided by r plus limit t t t tending towards infinity c not minus k minus p divided by r multiplied by one divided by e x to the power r t divided by v which is equal to k plus p divided by r plus c not minus k minus p divided by r multiplied by 1 divided by infinity which is equal to k plus p divided by r and since 1 divided by infinity is 0 this is 2 plus 0 as t tends towards infinity concentration c of t 10ths towards k plus p divided by r now if the addition of the polluted to the lake is terminated, that is if k is equal to 0 and p is equal to 0, we wish to determine the time interval t that must elapse before the concentration of the pollutants is reduced to 50 % of the original value and when it is reduced to 10 % of its original value.

10:25 Now if the pollution stops then the rate of mass in must be 0.

10:30 So we have rate of mass accumulation is equal to rate of mass in minus rate of mass out.

10:58 And since this is zero, we get dq divided by dt is equal to minus r multiplied by c of d.

11:12 R divided by v or deprived by q of t this implies that dq divided by q is equal to minus r divided by v t t now integrating both sides since d divided by d x of ln x is equal to one divided by x this implies that integral one divided by x v x this implies that integral one divided by x v x is equal to lnx.

11:47 So integral dq divided by q will be lnq which is equal to integral minus r divided by v dt that is minus r divided by vt plus some constant c.

12:05 We can write it as q is equal to e raise to the power minus r divided by v multiplied by t plus c plus c.

12:17 This can be written as e raised to the par c multiplied by e raised to the power minus rt divided by v...

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