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HEMISKI Zbirka re{eni zada~i REAKTORI 3

HEMISKI Zbirka re{eni zada~i REAKTORI 3

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Zada~a 3<br />

326<br />

Reakcija vo gasna faza vo adijabatski PBR<br />

Reakcijata A + 0,5B C se izveduva vo gasna faza vo ceven<br />

<strong>re</strong>aktor so fiksen sloj katalizator pod adijabatski uslovi.<br />

Ako operacionata linija za adijabatskata rabota na <strong>re</strong>aktorot<br />

e T 688 211<br />

X , da se p<strong>re</strong>smeta pot<strong>re</strong>bnata koli~ina<br />

katalizator za postignuvawe 60% konverzija na A.<br />

Drugite pot<strong>re</strong>bni podatoci se:<br />

K<br />

F<br />

y<br />

k <br />

To<br />

Ao<br />

612 kmol/h;<br />

Po<br />

1atm;<br />

To<br />

688K<br />

;<br />

0,<br />

07;<br />

y 0,<br />

11;<br />

y 0,<br />

82;<br />

Bo<br />

I<br />

pC<br />

kp A pB<br />

( 1<br />

)<br />

0,<br />

5<br />

p A pB<br />

K<br />

( rA ) <br />

( kmol /kg )/h;<br />

2 A kat p<br />

22,<br />

414(<br />

1<br />

K p K p )<br />

1<br />

exp( 12,<br />

16<br />

<br />

exp( 9,<br />

953 <br />

1<br />

A<br />

2<br />

( 8619 / T )); K<br />

C<br />

( 5473 / T )); K exp( 10,<br />

86 ( 11300 / T ));<br />

2<br />

exp( 71,<br />

745 <br />

i<br />

( atm);<br />

( 52596 / T )); T ( K)<br />

.<br />

Re{<strong>eni</strong>e:<br />

Sè {to t<strong>re</strong>ba da se napravi vo ovaa zada~a e da se izvede<br />

brzinski izraz p<strong>re</strong>ku konverzija i da se kombinira so ravenkata<br />

za dizajn i operacionata linija na <strong>re</strong>aktorot.<br />

Stehiometriskata tablica i primenata na podatocite od<br />

tablicata vo brzinskiot izraz se kako {to sledi:<br />

pC<br />

kpA<br />

pB<br />

( 1<br />

)<br />

0.<br />

5<br />

p A pB<br />

K<br />

( rA<br />

) <br />

22.<br />

414(<br />

1<br />

K p K p )<br />

1<br />

A<br />

2<br />

C<br />

2<br />

0,<br />

5<br />

( 1<br />

X )( 1,<br />

57 0,<br />

5X<br />

) <br />

<br />

X 14,<br />

28(<br />

1<br />

0,<br />

035X<br />

)( 1<br />

0,<br />

35X<br />

)<br />

k<br />

1<br />

2<br />

2<br />

0<br />

14,<br />

28 ( 1<br />

0,<br />

035X<br />

)<br />

<br />

K 14,<br />

28(<br />

1<br />

0,<br />

035X<br />

)( 1<br />

X )( 1,<br />

57 0,<br />

5X<br />

)<br />

<br />

2<br />

( 1<br />

X )<br />

X <br />

22,<br />

414<br />

1<br />

K1<br />

K 2<br />

14,<br />

28(<br />

1 0,<br />

035 ) 14,<br />

28(<br />

1 0,<br />

035 )<br />

<br />

<br />

X<br />

X <br />

<br />

, 5<br />

<br />

<br />

<br />

<br />

.

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