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Unformatted text preview: 267 7.2. ADIABATIC, ISOCHORIC KINETICS consider A and B to be isomers of an identical molecular species. So we have N = 2 species
reacting in J = 1 reactions. The number of elements L here is irrelevant.
7.2.1.1 Onestep reversible kinetics Let us insist our reaction process be isochoric and adiabatic, and commence with only A
present. The reaction kinetics with β = 0 are
−E
RT dρA
= − a exp
dt ρA − 1
ρ,
Kc B (7.259) =k
=r dρB
dt −E
RT = a exp ρA − 1
ρ,
Kc B (7.260) =k
=r ρA (0) = ρA ,
ρB (0) = 0. (7.261)
(7.262) For our alternate compact linear algebra based form, we note that
r = a exp −E
RT ρA − 1
,
ρ
Kc B (7.263) and that
d
dt ρA
ρB = −1
(r ).
1 (7.264) Performing the decomposition yields
d
dt ρA
ρA + ρB −1
(r ).
0 = (7.265) Expanded, this is
10
11 d
dt ρA
ρB = −1
(r ).
0 (7.266) Combining Eqs. (7.2597.260) and integrating yields
d
(ρ + ρ B ) = 0 ,
dt A
ρA + ρB = ρA ,
ρB = ρA − ρA . (7.267)
(7.268)
(7.269) CC BYNCND. 18 November 2011, J. M. Powers. 268 CHAPTER 7. KINETICS IN SOME MORE DETAIL Thus, Eq. (7.259) reduces to
−E
RT dρA
= −a exp
dt 1
ρ − ρA
Kc A ρA − . (7.270) Scaling, Eq. (7.270) can be rewritten as
d
d(at)
7.2.1.2 ρA
ρA = − exp − E
1
RTo T /To ρA
1
−
ρA Kc 1− ρA
ρA . (7.271) First law of thermodynamics Recall the ﬁrst law of thermodynamics and neglecting potential and kinetic energy changes:
dU
˙
˙
= Q − W.
(7.272)
dt
˙
Here E is the total internal energy. Because we insist the problem is adiabatic Q = 0.
˙ = 0. Thus we have
Because we insist the problem is isochoric, there is no work done, so W
dU
= 0.
(7.273)
dt
Thus, we ﬁnd
U = Uo . (7.274) Recall the total internal energy for a mixture of two calorically perfect ideal gases is
U = nA uA + nB uB ,
nA
nB
=V
uA +
uB ,
V
V
= V (ρA uA + ρB uB ) ,
PA
PB
= V ρA hA −
+ ρB (hB −
ρA
ρB
= V ρA hA − RT + ρB (hB − RT , (7.275)
(7.276)
(7.277)
, (7.278)
(7.279) o o = V ρA cP (T − To ) + hTo ,A − RT + ρB (cP (T − To ) + hTo ,B − RT
o o = V (ρA + ρB )(cP (T − To ) − RT ) + ρA hTo ,A + ρB hTo ,B ,
o (7.281) o = V (ρA + ρB )((cP − R)T − cP To ) + ρA hTo ,A + ρB hTo ,B ,
o (7.282)
o = V (ρA + ρB )((cP − R)T − (cP − R + R)To ) + ρA hTo ,A + ρB hTo ,B ,
o o = V (ρA + ρB )(cv T − (cv + R)To ) + ρA hTo ,A + ρB hTo ,B ,
o o = V (ρA + ρB )cv (T − To ) + ρA (hTo ,A − RTo ) + ρB (hTo ,B − RTo ) ,
= V (ρA + ρB )cv (T − To ) + ρA uo o ,A + ρB uo o ,B .
T
T CC BYNCND. 18 November 2011, J. M. Powers. ,(7.280) (7.283)
(7.284)
(7.285)
(7.286) ...
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This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Parksou during the Fall '11 term at FSU.
 Fall '11
 ParkSou
 Dynamics

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