1. Dr. F. IskanderaniChE 201 Spring 2003/20042 REACTOR

2. Dr. F. IskanderaniChE 201 Spring 2003/20043 Q : Is total moles in = total moles out ? Q : Is total mass in = total mass out ? Mass in mass out 2x28 + 32 = 2x44 88 = 88

3. Dr. F. IskanderaniChE 201 Spring 2003/20044 Let us now look at the elements (atoms) C and O: IN OUTGene ration Consump tion Accumu lation C 2 2000 O In stream1 2 In stream2 1 x 22 x 2 000

4. Dr. F. IskanderaniChE 201 Spring 2003/20045 Example 2 (Note : All calculations are in moles) 10 moles of CH 4 are burned completely with 80 moles of O 2. Find all components out. CH 4 + 2O 2 CO 2 + 2H 2 O accu = in - out + generation - consumption CH 4 0 = 10 - 0 + 0 - consumption Consumption of CH 4 = 10

5. Dr. F. IskanderaniChE 201 Spring 2003/20046 CH 4 + 2O 2 CO 2 + 2H 2 O accu = in - out + generation - consumption O 2 0 = 80 - ? + 0 - O 2 OUT = 80 - 20 = 60 10 x 2/1

6. Dr. F. IskanderaniChE 201 Spring 2003/20047 CH 4 + 2O 2 CO 2 + 2H 2 O accu = in - out + generation - consumption CO 2 0 = 0 - ? + - 0 CO 2 OUT = 10 10 x 1/1

7. Dr. F. IskanderaniChE 201 Spring 2003/20048 CH 4 + 2O 2 CO 2 + 2H 2 O accu = in - out + generation - consumption H 2 O 0 = 0 - ? + - 0 H 2 O OUT = 20 10 x 2/1

8. Dr. F. IskanderaniChE 201 Spring 2003/20049 SUMMARY CH 4 + 2O 2 CO 2 + 2H 2 O In 10 80 Consumed 10 20 Generated 0 0 10 20. Out 0 60 10 20

9. Dr. F. IskanderaniChE 201 Spring 2003/200410 (Now Solve using ATOMIC BALANCES) IN OUT CH 4 O 2 CO 2 H 2 O O 2 IN = OUT C bal 10 + 0 = + 0 + 0 Moles C =1x moles of CO 2 1 x n CO2 THEREFORE n CO2 OUT = 10 or x 1

10. Dr. F. IskanderaniChE 201 Spring 2003/200411 ATOMIC BALANCES IN OUT CH 4 O 2 CO 2 H 2 O O 2 IN = OUT H bal 4x10 + 0 = 0 +2n H2O + 0 Thus n H2O = 40/2 = 20 moles

11. Dr. F. IskanderaniChE 201 Spring 2003/200412 ATOMIC BALANCES IN OUT CH 4 O 2 CO 2 H 2 O O 2 IN = OUT O bal 0 + 2x80= Moles O =2x moles of CO 2 2 x n CO2 +1n H2O + 2n O2,out Moles O =1x moles of H 2 O 80 moles O 2

12. Dr. F. IskanderaniChE 201 Spring 2003/200413 IN OUT CH 4 O 2 CO 2 H 2 O O 2 IN = OUT O bal 0 + 2x80= Moles O =2x moles of CO 2 2 x n CO2 +1n H2O + 2n O2,out Moles O =1x moles of H 2 O 80 moles O 2 Thus, n O2,out = (160 - 20 - 20)/2 =60 moles

13. Dr. F. IskanderaniChE 201 Spring 2003/200414 FOR REACTION PROBLEMS, with no accuulation, the following is true: 1. total mass in = total mass out 2. for every element (atom) present in the problem, Mass of element in = Mass of element out Moles of element in = Moles of element out 3. Any inert compounds that are present and not involved in the reaction, component balances can be carried out for them