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HEMISKI Zbirka re{eni zada~i REAKTORI 3

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238<br />

Fio Fi() Fi(X) pi = (Fi/FAo)P<br />

A FAo FAo–2 FAo(1–X) pA = (1–X)<br />

B 0 FAoX/2 pB = X/2<br />

C 0 FAoX/2 pC = X/2<br />

FTo = FAo FT = FAo P vkupen = 1<br />

Sega ja p<strong>re</strong>smetuvame numeri~kata v<strong>re</strong>dnost na brzinskata<br />

konstanta na 760 o C:<br />

8 1<br />

k ( 1033K)<br />

2,<br />

4 10<br />

exp( 15200<br />

/ 1033)<br />

97,<br />

683h<br />

Numeri~kata v<strong>re</strong>dnost na ramnote`nata konstanta e poznata,<br />

pa zaedno so stehiometriskata tablica se formira sledniov<br />

brzinski izraz kako funkcija od konverzijata:<br />

<br />

2<br />

2 X <br />

2<br />

( rA ) 97,<br />

683(<br />

1<br />

X ) 97,<br />

68312X0,<br />

2X<br />

(mol/l)/h.<br />

2 <br />

<br />

2 0,<br />

312<br />

(1)<br />

a) So brzinskiot izraz (1) mo`eme da ja p<strong>re</strong>smetame ramnote`nata<br />

konverzija na 760 o C i, isto taka, da go kombinirame<br />

so ravenkata za dizajn.<br />

Ramnote`nata konverzija }e ima v<strong>re</strong>dnost:<br />

* 2 *<br />

*<br />

X 10X<br />

5 0 0,<br />

528.<br />

( rA ) 0 <br />

X<br />

(2)<br />

Od <strong>re</strong>zultatot za ramnote`nata konverzija proizleguva<br />

deka zavisnosta V / FAo<br />

f ( X ) }e ima smisla samo do X 0,<br />

528 .<br />

Eksplicitnata forma na ovaa zavisnost se dobiva od ravenkata<br />

za dizajn (58):<br />

V<br />

F<br />

Ao<br />

<br />

*<br />

X X<br />

<br />

0<br />

V<br />

F<br />

Ao<br />

dX<br />

<br />

( r<br />

)<br />

<br />

A<br />

0,<br />

01024<br />

*<br />

X X<br />

<br />

0<br />

*<br />

X X<br />

<br />

0<br />

dX<br />

97,<br />

683(<br />

1<br />

2X<br />

0,<br />

2X<br />

( 0,<br />

2<br />

2<br />

2X<br />

1)<br />

i mo`e da se <strong>re</strong>{i so primena na tabli~en integral:<br />

X<br />

dX<br />

2<br />

,<br />

)<br />

.<br />

(3)

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