Presentation is loading. Please wait.

Presentation is loading. Please wait.

Lecture18 Thursday 3/13/08 Solution to Tuesdays In-class Problem. User Friendly Energy Balance Derivations Adiabatic (Tuesday’s lecture). Heat Exchange.

Similar presentations


Presentation on theme: "Lecture18 Thursday 3/13/08 Solution to Tuesdays In-class Problem. User Friendly Energy Balance Derivations Adiabatic (Tuesday’s lecture). Heat Exchange."— Presentation transcript:

1 Lecture18 Thursday 3/13/08 Solution to Tuesdays In-class Problem. User Friendly Energy Balance Derivations Adiabatic (Tuesday’s lecture). Heat Exchange Constant Ta. Heat Exchange Variable Ta Co-current. Heat Exchange Variable Ta Counter Current

2 Adiabatic Operation Elementary liquid phase reaction carried out in a CSTR The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1. (a)Assuming the reaction is irreversible, A  B, (K C = 0) what reactor volume is necessary to achieve 80% conversion? (b)If the exiting temperature to the reactor is 360K, what is the corresponding reactor volume? (c)Make a Levenspiel Plot and then determine the PFR reactor volume for 60% conversion and 95% conversion. Compare with the CSTR volumes at these conversions (d)Now assume the reaction is reversible, make a plot of the equilibrium conversion as a function of temperature between 290K and 400K.

3 CSTR Adiabatic Mole Balance Rate Law Stoichiometry

4 Energy Balance – Adiabatic,  C P = 0

5 Irreversible for Parts (a) through (c) (a) Given X = 0.8, find T and V

6 (b)

7 (c) Levenspiel Plot

8 CSTR X = 0.6 T = 360 CSTR X = 0.95 T = 395

9 PFR X = 0.6 PFR X = 0.95

10 Summary CSTRX = 0.6T = 360V = 2.05 dm 3 PFRX = 0.6T exit = 360V = 5.28 dm 3 CSTRX = 0.95T = 395V = 7.59 dm 3 PFRX = 0.95T exit = 395V = 6.62 dm 3

11 (d) At Equilibrium

12 (e) T e = 358 X e = 0.59

13

14 User Friendly Equations Relate T and X or F i

15

16 Heat Exchange Elementary liquid phase reaction carried out in a PFR The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1.

17 Rate Law(2) (3) (4) Stoichiometry(5) (6) Parameters(7) – (15) Mole Balance(1)

18 Energy Balance Adiabatic and (16A) Additional Parameters (17A) & (17B) Heat Exchange (16B)

19 A. Constant T a (17B) T a = 300K Additional Parameters (18B – (20B): B. Variable T a Co-Current (17C) C. Variable T a Counter Current (18C) Guess T a at V = 0 to match T a = T ao at exit, i.e., V = V f

20

21

22

23

24

25

26 More heat effects T TaTa V+ΔVV mcHCmcHC F A, F i V+ΔV V FiFi T TaTa FiFi In - Out + Heat Added = 0

27 Heat removed Heat generated

28 Example: Constant T a Find conversion, X eq and T as a function of reactor volume 1)Mole balance 2)Rates 3)Stoich 4)Heat effects V X V T V rate X X eq Parameters

29 Example: Variable T a CoCurrent Coolant balance: In - Out + Heat Added = 0 All equations can be used from before except T a parameter, use differential T a instead, adding m C and C PC Example: Variable T a Counter - Current In - Out + Heat Added = 0 All equations can be used from before except dT a /dV which must be changed to a negative to arrive at the correct integration we must guess the T a value at V=0, integrate and see if T a0 matches; if not, reguess the value for T a at V=0


Download ppt "Lecture18 Thursday 3/13/08 Solution to Tuesdays In-class Problem. User Friendly Energy Balance Derivations Adiabatic (Tuesday’s lecture). Heat Exchange."

Similar presentations


Ads by Google