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Unformatted text preview: ME 5446
Es say 10 COMBUSTION IN A CONSIiLVTPRE SSL'RE REACIOR A constantpressure reactor is pictured at the right. The reaction occurs in a space bounded above by a freely mos'ing piston. The ambient air and the weight of the piston, taken together. exert a Free” * a. moving constant pressure on the reaction zone. There is a spark plug situated % Wm”
. . . . . N at the top of the combustion chamber that is used to initiate the E at." Reaction . . N. Zone
chemical reaction. *2 s §
I I I I . \\\\“Q
The reaction to be investigated is
nA+£iB—>cC+o'D [1) ts‘hereA, B, C, and D are chemical compounds or elements, and a, b, c. and r3 are stoichiometric coefficients. Prior to the initiation of the reaction, the reaction space contains (rig), and (rig), moles of :f. aiidB. respectively. [n addition, the initial. precomloustion s‘olume is If , and the initial temperature is I, . The mass in of the contents of the reaction zone is constant. Heat :2 is exchanged between the reaction zone and the space outside of the reactor. The energy transfers are gos‘enied the first 1mm of thennodynamics. If the piston motion is very slow, the rate at which work is done by the gaseous mixture on the piston is ,mﬁci‘l’fdt) . With this, the first law can be written as d—E=QTf’=Qntdl’i’a'o [2)
(1 From the definition of the enthalpy, it = h— pt’, and ii = L'lilm . With these, Eq. (3] becomes an [33.
arr—d! pair—m ' ri'r so that, n'i'r _ E _ [4} E lit: Ihe enthalpy I? of the mixture is the sum of the enthalpies of the component compounds, so that . 1 l — _
r'?=;Zm_.h_ +5214}; [3) and after differentiation there results _ _ i'
ﬁ=iTnﬂ+iTam is) 0’: mf' ' art in?l ' (it The next step is to eliminate the MI in favor of El _ which gives, after noting that HI = PE, dial. _ V nip—JI _ JV — _ _ "— ?
n'r r1": ' air [ 2'
When Eqs. (6] and are substituted into the first law, Eq. [4). there follows
all—r —. (ﬁr—1; _ n'V O = T I". —' T I’ ' — — 3
i f p. ff: .K d! a d? E) Furthermore, since the enthalpy is a unique function of temperature, it is advantageous to write (a? GE. a'r arr —=—— =? _— 9
n'r d1" dr P" (ﬁr i 2'
With this. Eq. (8] becomes
I [WET __ .— ﬁrllli—j (ﬂ? _—_ T — —V r' . ' —"_— ' .— I'l’
‘9‘? r +pl‘?.'+V;”£ ﬁr! Igr? Enrpr— a." R (In) [t is appropriate to take stock of the unknowns that appear in E4]. (JO). In this regard, it may be
noted that and ER.. are functions of temperature. Therefore, the unlmowns in Eq. [ID] are: Ht], Wt),
and EU] . Since r' = A. B, C, and B: there are six tutlmowris To find additional equations to match the number of mimosaus, use can be made of the reaction rate equations and of the ideal gas law. For the latter. if the molar form is used: [\J PER? {11) Since the pressure is constant. it follows that. ___ _ _ __"£"L _ :0 — (9.4 + .03 + pa + pa )— E" I {12)
and. after differentiation. ME. : 0'53 : art—ac + dﬁs‘tzt’ p: 1 df {m L d: dr (1’! (it ,t' LEIx T1 at, ' It only remains to write the reaction rate equations for A. B. C. and D. They are (fat—9.4 , — a — w? {a}?! — a'— .ﬁ' (if =m = ak(f}(p_tl [pal =rIAe (to) {pg} {14) (f— _.. _ . ,_  ﬁ = am. “Matt‘th t a 5] dr (1'— _ . ﬁwe “Vomit—73f (161 (it L2” = dAe‘Ecmwoit {m There are six governing equations. Eqs. [10). (13). and [Isl1?). to match the six unknowns. These equations are to be solved numerically subjected to six initial conditions. which are V=V[O). I=T{0]. ﬁat—alto} (LS) The nutnerical solution proceeds as follows: Step 1: ‘With the known initial conditions. the slopes art—)3 [alt . ($153 / df. dﬁc fdt . and (ﬁt—JD [alt can be determined from Eqs. {141 T"). Step '2: The output of step 1 is then used in conjunction with Eq. (13) to find dT/a't .
Step 3: Then. (fl/fa?! is obtained from Eq. (10).
Step 4: A time step At is chosen such that t1 = O + At . Step 5: Calculate the values off. V. ,5}. . at II from T61] = + Mir/(£050 ~ V0?) = _ (dV/df)..<. 5.103] = )t +(dEB/derﬁ‘ {19) Step 6: Repeat steps 1 through 3 by using the values of T._ V, and . Step 7': Continue this marching procedure. ...
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