Homework 8 Solutions

ChBE 424 - HW #8 Solution Set P8-5. Given: The endothermic liquid-phase elementary reaction: A+B 2C A B C Feed Fi0 (lbmol/hr) 10.0 10.0 0.0 Feed Ti0, (°F) 80 80 --Specific heat CPi, (Btu/lbmol⋅°F) 51.0 44.0 47.5 MW (lb/lbmol) 128 94 222 Density ρi, (lb/ft3) 63.0 67.2 65.0 o ΔH Rx = 20,000 Btu/lbmol A, independent of temperature Calculate the steady-state reactor temperature assuming that the reaction goes to completion in a single steam-jacketed continuously-stirred tank reactor. Solution: The steady-state energy balance for a well-mixed system is: o Q − Ws − FA0 ∑ Θi CPi (T − Ti 0 ) − ΔH Rx FA0 X A,final = 0

Eq. (8-27)

where

Θ A = 1, Θ B =

FB0 10 = = 1, FA0 10

X A,final = 1

Q = UA(TS − T )

Btu ⎛ 2545 Btu ⎞ −Ws = ( 25 hp ) ⎜ = 63,525 ⎟ hr ⎝ hp-hr ⎠ Substituting into the energy balance gives o UA(TS − T ) − Ws − FA0 (CPA + CPB )(T − T0 ) − ΔH Rx FA0 = 0 or

o UA(TS − T0 ) − Ws − ( FA0 (CPA + CPB ) + UA ) (T − T0 ) − ΔH Rx FA0 = 0

Solving for the temperature T gives UA(TS − T0 ) − Ws − ΔH Rox FA0 T = T0 + FA0 (CPA + CPB ) + UA T = 80 o F +

T = 199 o F

(150 Btu/hr ⋅ ft 2 ⋅ o F)(10 ft 2 )(365.9 o F − 80 o F) + (63,525 Btu/hr) − (20, 000 Btu/lbmol)(10 lbmol/hr) (10 lbmol/hr)((51 + 44) Btu/lbmol ⋅ o F) + (150 Btu/hr ⋅ ft 2 ⋅ o F)(10 ft 2 )

P8-8(a) Using POLYMATH and following the example in Fogler on pages 506-507 for adiabatic operation gives: Calculated values of the DEQ variables Variable W X T Po R E k To R2 P Cao Ca ra vo Fao HCr HAr HBr DHr Tr Cpa

initial value 0 0 450 10 0.0083144 31.4 0.133 450 0.082 10 0.2710027 0.2710027 -0.0360434 20 5.4200542 -40 -70 -50 -20 273 0.04

minimal value 0 0 450 10 0.0083144 31.4 0.133 450 0.082 10 0.2710027 0.0066567 -0.1566938 20 5.4200542 -40 -70 -50 -20 273 0.04

maximal value 50 0.9060474 903.02371 10 0.0083144 31.4 8.9606698 450 0.082 10 0.2710027 0.2710027 -0.0360434 20 5.4200542 -40 -70 -50 -20 273 0.04

final value 50 0.9060474 903.02371 10 0.0083144 31.4 8.9606698 450 0.082 10 0.2710027 0.0066567 -0.0596489 20 5.4200542 -40 -70 -50 -20 273 0.04

(kg) (K) (atm) (kJ/mol-K) (kJ/mol) (L/kg-s) (K) (L-atm/mol-K) (atm) (mol/L) (mol/L) (mol/kg-s) (L/s) (mol/s) (kJ/mol) (kJ/mol) (kJ/mol) (kJ/mol) (K) (kJ/mol-K)

ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(W) = -ra/Fao [2] d(T)/d(W) = (-ra)*(-DHr)/(Fao*Cpa)

(DCp = 25 + 15 - 40 = 0)

Explicit equations as entered by the user [1] Po = 10 [2] R = 0.0083144 [3] E = 31.4 [4] k = 0.133*exp(E/R*((1/450)-(1/T))) [5] To = 450 [6] R2 = 0.082 [7] P = Po [8] Cao = Po/(R2*To) [9] Ca = Cao*(1-X)/(1+X)*(P/Po)*(To/T) [10] ra = -k*Ca [11] vo = 20 [12] Fao = Cao*vo [13] HCr = -40 [14] HAr = -70 [15] HBr = -50 [16] DHr = HBr+HCr-HAr [17] Tr = 273 [18] Cpa = 40/1000

(atm) (kJ/mol-K) (kJ/mol) (L/kg-s) (K) (L-atm/mol-K) (atm) (mol/L) (mol/L) (mol/kg-s) (L/s) (mol/s) (kJ/mol) (kJ/mol) (kJ/mol) (kJ/mol) (K) (kJ/mol-K)

Comments [3] ra = -k*Ca mol/(kg cat*s) [4] k = 0.133*exp(E/R*((1/450)-(1/T))) dm3/(kg cat*s)

[5] E = 31.4 kJ/mol [6] R = 0.0083144 kJ/(mol*K) [7] Ca = Cao*(1-X)/(1+X)*(P/Po)*(To/T) mol/L [8] Cao = Po/(R2*To) mol/L [9] Po = 10 atm [10] To = 450 K [11] P = Po atm [12] R2 = 0.082 L*atm/(mol*K) [13] Fao = Cao*vo mol/s [14] vo = 20 L/s [15] DHr = HBr+HCr-HAr kJ/mol [16] HAr = -70 kJ/mol [17] HBr = -50 kJ/mol [18] HCr = -40 kJ/mol [19] Tr = 273 K [20] Cpa = 40/1000 kJ/(mol*K)

P8-8(b) Following the example on page 494; F X FA 0 X WCSTR = A0 = − rA ⎛ 1 − X ⎞⎛ P ⎞⎛ T0 ⎞ kC A0 ⎜ ⎟⎜⎜ ⎟⎟⎜ ⎟ ⎝ 1 + X ⎠⎝ P0 ⎠⎝ T ⎠ From the total energy balance, we have: o Q − W S − F A 0 ∑ Θ i C Pi ( T − Ti 0 ) − F A 0 X [ Δ H Rx ( T R ) + Δ C P ( T − T R )] = 0 From the problem statement, Adiabatic: Q = 0 Negligible stirring work: W = 0 S

ΔC P = C PB + C PC − C PA = 25 + 15 − 40 = 0 o ΔH Rx (TR ) = H Bo + H Co − H Ao = −50 − 40 + 70 = −20 kJ/mol P 10 atm mol C A0 = A0 = = 0.271 L RT0 ⎛ L ⋅ atm ⎞ ⎜ 0.082 mol ⋅ K ⎟ ( 450 K ) ⎝ ⎠ mol ⎞⎛ L ⎞ mol ⎛ FA0 = C A0 v0 = ⎜ 0.271 20 ⎟ = 5.42 ⎟⎜ L ⎠⎝ s ⎠ s ⎝ Insert into the total energy balance, rearranging, & inserting into the CSTR design equation to give: o o ( − ΔH Rx )X ( − ΔH Rx )X − ( −20 kJ/mol)(0.8) T = T0 + = T0 + = 450 K + = 850 K (0.04 kJ/mol ⋅ K ) ΣΘi C Pi C PA

kJ ⎡ ⎤ 31.4 1 ⎞⎤ 1 ⎞⎥ L ⎡E ⎛ 1 ⎢ ⎛ 1 mol k = 0.133exp ⎢ ⎜ − ⎟ ⎥ = 0.133exp ⎢ ⎜ 450 K − 850 K ⎟ ⎥ = 6.903 kg cat ⋅ s 450 kJ R T ⎠⎦ ⎝ ⎠⎥ ⎣ ⎝ ⎢ 0.0083144 mol ⋅ K ⎣ ⎦ mol ⎞ ⎛ 5.42 ( 0.8) ⎜ s ⎟⎠ FA0 X ⎝ WCSTR = = = 39.4 kg cat L mol ⎞⎛ 1 − 0.8 ⎞ ⎛ 10 atm ⎞ ⎛ 450 K ⎞ ⎞⎛ ⎛ 1 − X ⎞ ⎛ P ⎞ ⎛ T0 ⎞ ⎛ kC A0 ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ 6.903 kg cat ⋅ s ⎟ ⎜ 0.271 L ⎟⎜ 1 + 0.8 ⎟ ⎜ 10 atm ⎟⎜ 850 K ⎟ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ 1 + X ⎠⎝ P0 ⎠ ⎝ T ⎠ ⎝

P8-8(c) An example question is “How would your answer in part (a) change if pressure drop was taken into account?” This question requires critical thinking since it is the fifth type of Socratic question listed in Table P-1 on page xx. P8-8d) Repeat P8-8a using POLYMATH, this time including the pressure differential equation and appropriate alpha value. For alpha = 0.0075/kg cat for particle diameter D2: Calculated values of the DEQ variables Variable W X T P Po R E k To R2 Cao Ca ra vo Fao HCr HAr HBr DHr Cpa alpha

initial value 0 0 450 10 10 0.0083144 31.4 0.133 450 0.082 0.2710027 0.2710027 -0.0360434 20 5.4200542 -40 -70 -50 -20 0.04 0.0075

minimal value 0 0 450 5.3620913 10 0.0083144 31.4 0.133 450 0.082 0.2710027 0.0098888 -0.1212363 20 5.4200542 -40 -70 -50 -20 0.04 0.0075

ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(W) = -ra/Fao [2] d(T)/d(W) = (-ra)*(-DHr)/(Fao*Cpa) [3] d(P)/d(W) = -(alpha/2)*(T/To)*(Po/(P/Po))*(1+X) Explicit equations as entered by the user [1] Po = 10 [2] R = 0.0083144 [3] E = 31.4 [4] k = 0.133*exp(E/R*((1/450)-(1/T))) [5] To = 450

maximal value 50 0.7751552 837.57761 10 10 0.0083144 31.4 6.462801 450 0.082 0.2710027 0.2710027 -0.0360434 20 5.4200542 -40 -70 -50 -20 0.04 0.0075

final value 50 0.7751552 837.57761 5.3620913 10 0.0083144 31.4 6.462801 450 0.082 0.2710027 0.0098888 -0.0639091 20 5.4200542 -40 -70 -50 -20 0.04 0.0075

[6] R2 = 0.082 [7] Cao = Po/(R2*To) [8] Ca = Cao*(1-X)/(1+X)*(P/Po)*(To/T) [9] ra = -k*Ca [10] vo = 20 [11] Fao = Cao*vo [12] HCr = -40 [13] HAr = -70 [14] HBr = -50 [15] DHr = HBr+HCr-HAr [16] Cpa = 40/1000 [17] alpha = 0.0075 Comments [1] d(X)/d(W) = -ra/Fao 1/kg cat [2] d(T)/d(W) = (-ra)*(-DHr)/(Fao*Cpa) K/kg cat [3] d(P)/d(W) = -alpha/2*(T/To)*(Po/(P/Po))*(1+X) atm/kg cat [4] ra = -k*Ca mol/(kg cat*s) [5] k = 0.133*exp(E/R*((1/450)-(1/T))) dm3/(kg cat*s) [6] E = 31.4 kJ/mol [7] R = 0.0083144 kJ/(mol*K) [8] Ca = Cao*(1-X)/(1+X)*(P/Po)*(To/T) mol/L [9] Cao = Po/(R2*To) mol/L [10] Po = 10 atm [11] To = 450 K [12] R2 = 0.082 L*atm/(mol*K) [13] Fao = Cao*vo mol/s [14] vo = 20 L/s [15] DHr = HBr+HCr-HAr kJ/mol [16] HAr = -70 kJ/mol [17] HBr = -50 kJ/mol [18] HCr = -40 kJ/mol [19] Cpa = 40/1000 kJ/(mol*K) [20] alpha = 0.0075 1/kg cat

The conversion is lower for a reactor with 50 kg of catalyst when pressure drop is taken into account, which makes sense, since the lower pressure results in lower concentration of reactant A, and lower reaction rate. For the range of alpha between 0.0075 and 0.0123, decreasing the alpha value increases the conversion X, whereas increasing the alpha value lowers the pressure P which lowers the concentration of reactant A, which lowers the reaction rate which lowers the conversion. A value of alpha > 0.0123 results in pressure P reaching zero before the end of the reactor, in which case an inlet pressure of 10 atm is not large enough to force flow through the reactor. Such a reactor would need a higher inlet pressure, or a more loosely packed catalyst bed (which would result in a lower value for alpha). In POLYMATH, entering a value of alpha larger than 0.0123 results in numerical problems when the pressure P reaches zero in the reactor with 50 kg of catalyst, which is not unexpected since the model equations make no physical sense when the pressure becomes negative. Hence the listed value of alpha of 0.019 for a particle diameter of D1 leads to POLYMATH convergence problems, unless the catalyst weight W is reduced so that P > 0 throughout the reactor.