Page 193 - Modelling in Transport Phenomena A Conceptual Approach

6.3. CONSERVATION OF ENERGY 173

c) For this problem, Eq. (6.3-42) simplifies to

(9)

The standard heat of reaction at 298K is
AH,&(298) = Cai(AB;)i
i
= (-I)(-60)+(-I)(-75)+(2)(-90) = -45kJ/mol

The standard heat of reaction at 338K is given by Eq. (6.3-29)

AH&,(338) = AH&, (298 K) + 1;: Aeg dT

where
AC$ = CLY~@,
i
= (- 1)(175) + (- 1)(130) + (2)(110) = -85 J/mol. K
Hence

AH,0,,(338) = - 45,000 + (- 85)(338 - 298) = - 48,400 J/ mol

On the other hand, the use of Eq. (6.3-43) gives
(CP>in = C(~i)inepi

i
= (2000)(175) + (2400)(130) = 662,000 J/ m3. K

Therefore, substitution of the numerical values into Eq. (9) yields

Qint = - (0.256)(662,000)(25 - 65)
- (14) [(9.15 x 10-5)(400)(800)] (48,400) = - 13 x lo6 J/min
The minus sign indicates that the system, i.e., reactor, loses energy to the sur-
roundings.
d) The application of Newton's law of cooling gives

13 x lo6
AH = = 4.1 m2
(1050)(65 - 15)(60)