Question

(40 marks) Question 1 An ethanol dehydrogenation reaction is carried out with the feed entering at 300 °c. The feed containing 90.0 mole % ethanol (C2H5OH) with the balance acetaldehyde(CH3CHO) enters a reactor at a rate of 150 mol/s. Heat is transferred to the reactor to keep the temperature from dropping too much and thereby decreasing the reaction rate to an unacceptably low level. When the heat addition is 2440 kl/s, the outlet temperature is 253 °C. Develop the block diagram of the process (5 marks) Calculate the degree of freedom of the process (5 marks) Calculate the fractional conversion of ethanol achieved in the reactor. (30 marks) CzH,он ) CH,CнO + Н2) (1 (v) The enthalpies of formation of each component are given below Component С-H, Он CH Cно H2 AH298K (kJ/mol) -235.31 -166.20 0.00 The heat capacity of each component is a function of temperature as shown below: (20.1 x 10-5)T-(12.78 x 10-) 7 (15.72 x 10-)T -(8.749 x 10-)72+(19.83 x 10-12)79 (0.00765 x 10-5)7T- (34.76 x 10-12)7 CP(CHCHO 0.07196 CPICHO) 0.06134 - +(0.3288 x 10-)72-(0.8698 x 10-12)73 CP(H 0.02884+ X where Cp is the heat capacity in J/(mol K) and T is the temperature in °C.

Please only solve this problem if you really can solve it and if it would be self-explanatory.

Answer 1

degree of freedom (f) is calculated as

f = c - p + 2

where c = components = 3

p is phases = 1

Therefore f = 3-1+2 = 4 ; degrees of freedom.

Block diagram is as follows:

ful Prod R 30o'c 253'e H S 15 lk CHseHe 136 na/s CH3HO +4ear א 2ס4י

Lets assume x mol/s of Ethanol, y mol/s of Acetone and z mol/s of H2 is present in the outlet stream.

Molecular weights of Ethanol = 46, acetone = 44 and hydrogen = 2

Therefore by Mass balance at inlet and outlet, we have:

(135 mol/s x 46 + 15 mol/s x 44) = (46 x + 44 y + 2 z) mol/s

=> 6870 mol/s = (46 x + 44 y + 2 z) mol/s - Eq A

By molar balance we have 150 mol/s = (x + y + z) mol/s - Eq B

Solving the two equations A & B above we get: 44 x + 42 y = 6570. - Eq C

Now calculating the Heat of the reactants at 300 deg C using the Cp values, putting the temperature in the Cp equations, we get

Cp in (Tot) = 16.13275 J/s ( mind it that this is in J and not kJ)

Similarly calculating the Heat of the products at 253 deg C using the Cp values, putting the temperature in the Cp equations, we get

Cp out (Tot) = (0.06 x + 0.1152 y + 0.029 z) J/s

Now Heat of reaction , deltaH r = deltaH (prod) - deltaH (react)

= [ 1000 x {( -235.31 x + (-166.2 y) + 0z) - (135 x -235.31 + (-166 x 15))}] J/s

(Note 1000 is multiplied to convert kJ to J)

Now from energy balance we get

Heat of reactants + Heat of reaction + Heat provided (2440 kJ) = Heat of reactants

=> 16.13275 + (-34259550 + 235310 x + 166200 y) + 2440000 = 0.06 x + 0.1152 y + 0.029 z

=> 235310 x + 166200 y = 31819533.9

=> 1.416 x + y = 191.45 - Eq D

From Eq C we get (dividing by 42 throughout ) => 1.047 x + y = 156.43 - Eq E

Solving Eq D and Eq E, we get

=> 0.3684 x = 35.02

ie => x = 95.06 mol/s

Therefore conversion is 95/135 ~ 70 % approx