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Unformatted text preview: MESMG Essay 13 THE WELL—STIRRED REACTOR In the previous reactor examples. the reactor housed a fixed mass. Here. focus is extended to a reactor into which a stream flows steadily to supply feedstock. In the present instance. the feedstock consists of componndsA and B whose mass flow rates at the inlet to the reactor are at and mm. The {1'th temperattu'e of these compounds is Tm and the pressure pl.Pr =1at:1n. Within the reactor proper. a chemical reaction occurs which transforms A and B into C and D. The stoichiometric equation for this reaction is aA+bB—>CC+dD (1} The rates of depletion of A and B are perfectly balanced by the rates at which feedstock is supplied to the reactor. Therefore. the reactor operates at steady state. The operating temperature is It;i . “11 mg I: Tm
: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E
é) Z
3123 "faﬁa ﬂ‘ "EC. “ED :
elf aft {if eff
Magnetic T99
Stiner
an .n'ig
é
______________________________________________ __ E} I
mCmD Rotating lk'laguetic Field Generator The contents of the reactor are maintained in a wellstirred condition by means of a magnetic stirring
unit that is shown schematically in the diagram. The practical meaning of the wellstirred condition is that the contents of the reactor are homogeneous. The efﬂuent ﬁ‘om the reactor is identical in composition to that of the mixture that occupies the reactor proper. The products of the reactor are rite and ran. and it is the goal of this analysis to predict these quantities. MASS BALANCE To begin the analysis. the conservation of mass principle is applied for all of the compoundsA. B. C and D. Fora representative compoundr' +3}? = ram, (2) m Lilli
In this equation. 3,. is a volmnetric source of mi. If mass is being created. 3,. is a positive number. while if mass is being depleted. 5i: is a negative number. The units of 5": are kgfmgs . To obtain homogeneous units for all terms in Eq. [2). it is necessary to multiply SI. by the volume V. Suppose that i = A. then the rate of depletion in terms of the reaction rate is off).
’M. :8. 3
dt ' ' ( } in which M, is the molecular weight of the compound I" in units of kgfmofe. Similar expressions are appropriate for if: A. B. C and D. From reaction rate knowledge. Eq. (3} can be rewritten for compoundA. %=i (4)
1‘ MA Similar equations can be written for all of the other participating compounds.
FIRST LAW OF THER‘tIODYNAMICS In view of the very large energy release rate associated with the chemical reaction. it is permissible
to neglect the changes in both the kinetic energy and the potential energy between the inlet and the exit. Consequently. in the absence of work. the first law can be written as ma." + Q = mam (5) u‘here tit denotes the mass flow rate of the mixture. and hm and how are the mixture enthalpies at the
inlet and outlet. respectively. The quantity is the rate of heat transfer in watts. Since
Might." = mdjuh + thmhh". Eq. [5) becomes .41"
mas"fin" + “Tswana + = who (5)
A similar rewriting of the righthand side leads to
mas"ham + marshals + = mglpm‘h .Lam' + mB.crurhE.m + mC:th.put + mD:awhD:m The next step is to expand the enthalpies h]. to take account of the heat of formation FT For example.
rhdﬁlhijn = ﬁtmyjn = Flam + Elia)— 'l—IA (If (3)
mdpurhdput = ﬁdpu't'l—fdpur = ﬁdpm‘ + IIIl [Tia )_ Eel u‘here Tf is the temperature of formation. equal to 25:6“ and Tm is the actual temperature of the entering components. The pressure at formation is 1 atm. With = —Tmb R. and u‘ith the prototype equations. Eqs. (3) and [9). the first laux Eq. can be fully written as Malia + 31 [Ta Elli} + Flam [fair + falling]— E1.1(T.r]l+ (Ts: ‘Tmbl/R
= flaw + E: (T5; 1' Ha r” + Flam + E13 l — Ha (10) + Fprutlﬂjiﬂ + Eclﬂ)‘ hC‘le]l+ flaw: fifth + £13th fall} Since the given flour quantities are mass flow rates tit and not mole flow rates 1’1. it is appropriate to transform the latter into the former by using the respective molecular weights Ml.
mi = HAMJ. m3 = HBMB. mt = nCMC. mp = HDMD (ll) Substitution of Eq. (1 1} into (10) results in mid" { +h{(T,")_ Eliif] + mam [Ens + £3 (Tm )_ Ea )l'l' (Ts: — Tomb )er Add =  tag
filial+a<2:5)—a(23)l+%tasany—sail] (12>
+%lfi~c+s(2;>—a(rfll+%tsnwas} stall J. C J. D Equatlon (12) contains ﬁve unknowns: T” _ min”. m3”. map”. and mam . In order to match the number of equations and the nmnber of unknowns. it is necessary to ﬁnd four more equations. These a 1'8 +5Eiiagr (13) m. =mi=m+SiV=mm d! Lou: which results from the combination of Eqs. (2) and It is next necessary to find dﬁfdt. For componentA Eniﬂizlmd
as Add dr V343 as (14) The quantity dmj fat is the rate of depletion of componentA by the chemical reaction. In order that the process proceeds at steady state. it is necessary that ' =tﬁ¢m (15)  =__a, (15) so that the reaction rate r' is In”:1 (1?) r_ thfA By following the same procedure for component B as was demonstrated for componentA. there follows dpﬁ _ mBjJ = 18
as V313 ( )
For components C and D
E _ c}, [C a] min}! and
d5” =a'r= (dialah. (20) The ﬁnal form of the governing equation is obtained by ﬁrst substituting Eq. (13) into (12) and then replacing HEIde from Eqs. [16) and (1820) in the resulting equation. The end result of these substitutions is ta. + n22.)— atal+ [ta + stat—stat is: sat/R
i .1 A 3 1 ta...+21.(2:.)—ata)t+ 1.33 to... +2i(2"..)— stat (22>
+[§f;;jn.]2t2;.+an.>—2.t2.)t+{%2..]2tai van—sat The outcome of the successive substitutions has yielded one equation for the one unknown T55 . The solution method is. necessarily. iterative. Once a value of I}: is guessed. for example. the enthalpies that depend on T55 can be obtained by table lookups. The end result of this operation is 2:.” = 2n. + do.) (22) where f represents the functional form of the known terms in Eq. [21). A possible next step is to use +1 . .. . . . . .
T ,1. as the next input to Eq. (22}. Since there maybe nonlinearities 111 the system. it may prudent to use damped iteration. which may defined a s .55 [Li‘1l=2::+r2(2::'l—T"l 22211 (22) where is the next input to Eq. (21} instead of 3:1 from Eq. (22 . ...
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This note was uploaded on 01/29/2012 for the course ME 5446 taught by Professor Sparrow during the Spring '11 term at Minnesota State University, Mankato.
 Spring '11
 Sparrow
 Combustion

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