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Chapter 3. Analysis of Environmental System 3.1 Analysis of a Simple Environmental System A simple environmental system can be analyzed using simple differential equations. Let us look over a basic mechanism arising in environmental system analysis. 3.1.1 Analysis of a Lake We may assume that a lake is a Completely Mixed Flow Reactor (CMFR). Then the phenomena may be described in a differential equation as follows. Fig. 3.1 Schematic diagram of a lake If we write a mass balance for the lake regarding a pollutant which is being discharged into the lake, it may be expressed as the following. V dC (3.1.1) QC0 W1 W2 W3 ...... QC VkC dt where V : volume of a lake (m 3 ) C0 : concentration of a pollutant ( Example : P, N ( g / m 3 )) Q : flow rate(m 3 / day) C : concentration of a pollu tan t in lake( g / m 3 ) W1 , W2 , ... : disch arg ed amount of pollu tan t ( g / day) k : reaction coefficien t (day1 ) Rearranging Eq.(3.1.1), V dC (Q Vk )C QC 0 W1 W2 ...... dt (3.1.2) Dividing both sides of Eq.(3.1.2), dC Q Q 1 k C (W1 W2 ......) dt V V V (3.1.3) Eq.(3.1.3) may be solved without any difficulty if we follow the process as we have done so far. Especially, when discharging rate of a pollutant into the lake is constant, it is much easier because the condition is steady state, dC/dt = 0. 3.1.2 Analysis of Chemical Reaction Using Differential Equations Chemical reactions arising in an environmental system may be solved using differential equations. Let us take a look various chemical reactions. ka A B (3.1.4) kb As shown in reaction (3.1.4), a chemical reaction may be expressed using a differential reaction. This chemical reaction is a reversible reaction where A is converted to B, and at the same time B is converted to A. dC A A k a C A kb C B dt (3.1.5) dC B B k a C A kbCB dt (3.1.6) If we solve the equations, (3.1.5), and (3.1.6), simultaneously, we may understand the way which the chemical is converting from A to B. In equilibrium state, reaction coefficients, Ka, and Kb are the same, and where dC A dC D 0.0 . dt dt C A : Concentration of A (mg/l) C B : Concentration of B (mg/l) t : time (min) A : reaction reate of A (mg/l min) B : reaction rate of B (mg/l min) k a : reactio rate coefficient of A(min -1 ) k b : reaction rate coefficient of B(min -1 ) 3.2 Chemical Reactions A general type of chemical reaction taking place in environmental system may be expressed like Eq.(3.2.1). k1 aA bB cC dD k (3.2.1) 2 Reaction equation (3.2.1) means that a moles of material A and b moles of material B react chemically to produce c moles of material C, and d moles of material D. At the same time, c moles of material C and d moles of material D react to produce a mole of material A, and b moles of material B. To understand the way of changing of the materials with time, their chemical reaction may be expressed in differential equations as the followings, which are mass balances of each species. 1 d [ A] k1[ A]a [ B ]b k 2 [C ]c [ D ]d a dt (3.2.2) 1 d [ B] k1[ A]a [ B ]b k 2 [C ]c [ D ]d b dt (3.2.3) 1 d [C ] k1[ A]a [ B ]b k 2 [C ]c [ D ]d c dt (3.2.4) 1 d [ D] k1[ A]a [ B ]b k 2 [C ]c [ D ]d d dt (3.2.5) From Eq. (3.2.2) ~ Eq(3.2.5), we may deduce the following relationships in Eq.(3.2.6). 1 d [ A] a dt 1 d [ B] b dt 1 d [C ] c dt 1 d [ D] d dt (3.2.6) In reaction formula (3.2.1), according to a, b, c, and d, coefficients of Eq. (3.2.2.) ~ Eq. [3.2.5] and order of reactions are determined. The followings are the some examples. 1. 0th Order Reaction k A product (3.2.7) A mass Balance equation for the reaction (3.2.7) is as follows. d [ A] k[ A]0 dt d [ A] k dt (3.2.8) When the type of changing of material A is like Eq. (3.2.8), the reaction is called 0th order reaction. 2. 1st order reaction k A product (3.2.9) Mass Balance for the reaction (3.2.9) may be written as follows. d [ A] k[ A]1 dt (3.2.10) When material A is changing by Eq.(3.2.10), the reaction is called 1st order reaction. 3. 2nd order reaction k 2A B product (3.2.11) Mass Balance equation of the chemical reaction for Eq.(3.2.11) can be expressed as the followings. 1 d [ A] k[ A]2 2 dt (3.2.13) d [ B] k[ A]2 dt (3.2.13) When chemical reaction for (3.2.11) is expressed like Eq. (3.2.12), and Eq. (3.2.13), the reactions are 2nd order reaction. As shown so far, order of chemical reaction may be determined by type of chemical reaction. In water chemistry, chemical reaction rates generally follows to Eq.(3.2.2)~Eq.(3.2.5). However, the reaction rate may not follow the rule which we have investigated in combustion environment related with air pollution. 3.3 Application of Differential Equations in Solving Environmental System Problem 1> A process reaction in a completely mixed lake follows first order kinetics with a rate constant K=0.1day -1. The influent concentration in stream is Cio and the volume of the Lake in which the process occurs in 4,000m3. Determine the detention time required in the lake to promote a 80% conversion of reaction. Sol) V Q V (det ention time) dC W QC kVC dt QCio QC kVC At steady state, V dC 0 dt QCio QC kVC 0 Q Q Cio C kC 0 V V Q (Cio C ) kC V 1 (Cio C ) kC Cio C kC 80% Conversion C 0.2Cio Problem 2> Cio C C 0.2Cio io kC 0.1 0.2 Cio 0. 8 40 (day) 0.02 A first order reversion reaction k1 A B k2 1) It is known that when CA=1mole/liter , rA CB=0.5mole/liter and d A 0.2(mole /(liter sec)). dt 2) rA 0.9(mole /(liter sec)) when CA=3.5mole/liter , Find k1 and k2. Sol) CB=1.5mole/liter. If we write mass balance for the reaction, d A 1 1 rA k1 A k 2 B 0.2, dt when C A 1, C B 0.5 k111 k 2 0.51 0.2 k1 0.5k 2 0.2 d A 1 1 rA k1 A k 2 B 0.9, dt when C A 3.5, C B 1.5 k1 3.51 k 2 1.51 0.9 3.5k1 1.5k 2 0.9 k1 0.5k 2 0.2 (1) 3.5k1 1.5k 2 0.9 (2) (1) 3 (2) 0.3 0.5k1 k1 0.6 (sec1 ) Plugging k1 to (1) 0.4 0.5k 2 Problem 3> k 2 0.8 (sec1 ) The reaction K A 2B Is observed as a 3/2reaction. When the reaction starts, CA= 2 moles/l CB=0.0 moles/l. Plot CB and CB vs. time. Where Sol) d A 3/ 2 k A dt rA Separaing var iables d A kdt A3 / 2 d A A k dt 3/ 2 2 A 1 / 2 A 2 kt C at t 0 C 2 A 1 / 2 2 2 2 kt 2 kt 2 2 2 2 A kt 2 A 1 / 2 k=0.05sec-1(mole/l)-1/2. and In aA bB reaction, relationship between reaction rates may be written as, 1 1 rA rB a b Therefore in the given reaction A 2B, reaction rate can be written, 1 d A 1 d B 1 dt 2 dt d A 2 d B 1 2 A [ B ] C When t 0, A 2, and B 0 C4 B 4 2A 2 B 4 2 kt 2 2