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HEMISKI Zbirka re{eni zada~i REAKTORI 3

HEMISKI Zbirka re{eni zada~i REAKTORI 3

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Vtor del. [ar`en <strong>re</strong>aktor<br />

b) Ako se poznati site konstanti i po~etniot volumen,<br />

V 2,<br />

832 P ( litri);<br />

P ( atm)<br />

, Vo<br />

4,<br />

2477 litri;<br />

o<br />

2<br />

T 60 C 333 K ; k 3908 (l/kmol) /s,<br />

da se p<strong>re</strong>smeta v<strong>re</strong>me na <strong>re</strong>akcija za koe volumenot na smesata }e<br />

ja dostigne v<strong>re</strong>dnosta V = 5,66 litri.<br />

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

a) Formata na brzinskiot izraz e poznata, <strong>re</strong>akcijata se<br />

slu~uva izotermno, pa sè {to e pot<strong>re</strong>bno za da se op<strong>re</strong>deli izrazot<br />

za brzinata na <strong>re</strong>akcijata p<strong>re</strong>ku konverzijata na <strong>re</strong>aktantot<br />

A e da se sostavi stehiometriska tablica:<br />

nio ni() ni(X) Ci(X) = ni(X)/V(X)<br />

( 1<br />

X )<br />

A nAo nAo – nAo(1 – X) C A C Ao 0,<br />

5<br />

( 1<br />

3X<br />

)<br />

B nBo = nAo nAo – nAo(1 – X) CB = CA<br />

X<br />

C 0 8 8nAoX C A 8C<br />

Ao 0,<br />

5<br />

( 1<br />

3X<br />

)<br />

nTo = 2nAo nT = 2nAo+6 nT = 2nAo(1+3X)<br />

Kon stehiometriskata tablica:<br />

n Ao n A<br />

X <br />

n Ao<br />

n Ao X <br />

nT<br />

V Vo<br />

nTo<br />

Po<br />

P<br />

Po<br />

Vo<br />

/ a<br />

Vo<br />

( 1 X<br />

) Vo<br />

( 1 X<br />

)<br />

P<br />

V / a<br />

2<br />

V<br />

2<br />

V ( 1 X<br />

) V V<br />

0,<br />

5<br />

( 1 X<br />

)<br />

o<br />

<br />

y Ao<br />

o<br />

<br />

i 1 8 1<br />

1<br />

<br />

3<br />

( <br />

) 1 1 ( (<br />

1))<br />

A<br />

V Vo<br />

( 1 3X<br />

)<br />

(1)<br />

Brzinskiot izraz p<strong>re</strong>ku konverzijata }e izgleda vaka:<br />

3<br />

2<br />

3 3 ( 1 X )<br />

( rA ) k C AC<br />

B k C A kC Ao 1,<br />

5<br />

( 1 3X<br />

)<br />

( kmol/l)/s.<br />

(2)<br />

0,<br />

5<br />

89

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