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HEMISKI Zbirka re{eni zada~i REAKTORI 3

HEMISKI Zbirka re{eni zada~i REAKTORI 3

HEMISKI Zbirka re{eni zada~i REAKTORI 3

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simumot na krivata XMB(T) za a = 3, kade {to X = 0,4, upatuva na<br />

barawe uslovi kako da se postigne taa v<strong>re</strong>dnost ako ne e problem<br />

da se raboti so golem <strong>re</strong>aktor (a = 3, V = 150 dm 3 ). O~igledno<br />

e deka toa bi se postignalo so zgolemuvawe na naklonot<br />

~<br />

na XEB(T), i toa ili so pogolema v<strong>re</strong>dnost za iC<br />

P,<br />

i ili so namaluvawe<br />

na temperaturnata razlika vlez‡izlez. Prvoto se postignuva<br />

so pove}e inerti vo vleznata struja, a vtoroto so odveduvawe<br />

toplina od sistemot.<br />

180<br />

1<br />

0.8 X<br />

0.6<br />

*<br />

XMB<br />

XEB<br />

0.4<br />

0.2<br />

Zada~a 13<br />

a = 0,5<br />

0<br />

300 350 400 450 500 550<br />

T T(K)<br />

Neelementarna <strong>re</strong>akcija vo gasna faza vo izotermen CSTR,<br />

vo staciona<strong>re</strong>n i nestaciona<strong>re</strong>n <strong>re</strong>`im na rabota<br />

Neelementarna <strong>re</strong>akcija vo gasna faza se odviva vo CSTR<br />

so volumen V = 60,64 m 3 . Stehiometrijata i kinetikata se:<br />

1<br />

XMB(T)<br />

k1C<br />

A<br />

( rA<br />

) <br />

( 1<br />

k C<br />

A B <br />

2<br />

k 0,<br />

253 min<br />

1<br />

A<br />

;<br />

)<br />

k<br />

2<br />

2<br />

C<br />

X * (T)<br />

a = 3<br />

a = 2<br />

a = 1<br />

XEB(T)<br />

4<br />

(mol/m<br />

3<br />

) / min<br />

3<br />

0,<br />

429 m /mol.<br />

1<br />

2 3

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