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Published byBeatriz Escobar Salazar Modified over 6 years ago
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P8-8 The elementary gas phase reaction A B + C is carried out adiabatically in PFR packed with catalyst. Pure A enters the reactor at a volumetric flow rate of 20dm3/s at pressure of 10atm and temperature of 450K. Plot the conversion and temperature down the PFR until 80% conversion is reached. The max. catalyst can be packed in the reactor is 50kg and ΔP=0.
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P8-8
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P8-9 If the heat is removed using heat exchanger jacking the reactor with high flow rate of coolant that the ambient exchanger temperature remain constant at 50C.
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Steady-state Nonisothermal reactor Design Part II
Reactor Staging Complex Reactions with Heat Exchange
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Equilibrium Conversion
The highest conversion is the Xe for reversible reactions which increase with temperature for endothermic reactions. For exothermic reactions it decrease with increase temperature. Endothermic Xe T exothermic
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To determine the max conversion achieved in exothermic reaction find the intersection of XEB with Xe
If entering temp increased the Xe decreased T T01 T T adiabatic
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Example 8-6 Elementary solid catalyzed liquid phase reaction A B
Make a plot of equilibrium conversion as function of temp. Determine the adiabatic temp and conversion when pure A fed to the reactor at 300K HAo(298) = cal/mol HBo(298) = cal/mol CPA=50cal/mol.K CPB=50cal/mol.K Ke=100,000 at 298
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Solution
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T Ke Xe 298 100,000 1 350 400 425 450 475 500
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T(K) XEB X Xe 0.42 XEB T 465K
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Reactor Staging with Interstate Cooling
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Endothermic Reaction Reactor heater Xe T
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Consider adiabatic reactor with reversible exothermic reaction.
Xe Consider adiabatic reactor with reversible exothermic reaction. Vary the feed temperature. At high T, the reaction rate is high but will reach the Xe which is small At low feed T, reaction rate is small and it need longer time to reach the Xe What is the suitable operation temp? Optimum Temp? T T T01 T T03 T02 T01 X W
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Optimization T0 X
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Example 8-7: Interstage cooling for high exothermic reactions
What conversion could be achieved in example 8-6 if 2 interstage coolers that have the capacity to cool the exit stream to 350K were available? Also, determine the heat duty of each exchanger for a molar feed rate of A of 40mol/s. Assume that: 95% of equilibrium conversion is achieved in each reactor. The feed temperature to the 1st reactor is 300K Coolant can not be heated over 400K CPC=18cal/mol.K
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Solution Calculate the exit temperature for the first reactor
For entering T=300K, Xe was Consider 95% of equilibrium conversion , therefore, Texit for 1st reactor at 0.4 T= X=460K X Xe 0.42 XEB 0.40 T 300 K
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Solution 2- Now cool the exit stream to 350K in heat exchanger
Calculate the heat load W=0 At the exchanger Fi(in) = Fi(out) and CPA=CPB Energy Balance Q=(FA+FB)(CPA)(T2 - T1) FA0=(FA+FB) X Xe 0.40 XEB T 460K Q = (40)(50)(350 – 460)= – 220kcal/s – 220kcal/s must be removed in the heat exchanger
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Calculate the coolant flow rate
Q = m CPC (Tout – Tin) T coolant (in) = 270K but can not be heated over 400K CPC=18cal/mol.K Calculate the heat Exchanger Area
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95% of the Xe in the 2nd reactor = (0.63)(0.95)=0.6
Second reactor X 0.76 95% of the Xe in the 2nd reactor = (0.63)(0.95)=0.6 T= (0.6 – 0.4) T=430K 0.60 0.40 T 430K 460K The heat duty to cool the exit of 2nd reactor from 430 to 350K is
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P8-13 The reaction A+BC+D is carried out adiabatically in a series of packed bed reactors with interstage cooling. The lowest temp to which the reactant stream may be cooled to 27C. The feed is equal molar in A and B. The feed enters at 27C. If 4 reactors and 3 coolers are available what conversion may be achieved . ΔHRx = - 30,000cal/molA CPA=CPB=CPC=CPD=25cal/gmol.K Ke(50C)=500,000 FA0=10molA/min
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T Ke Xe 323 500,000 1 350 0.98 400 0.887 425 450 475 500 0.153
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Multiple Reactions with Heat Effect
To account for heat effects in the multiple reactions, we simply replace the term (ΔHRx)(-rA)
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Multiple Reactions with Heat Effect
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Example 8-10 Parallel Reactions in PFR with heat Effect
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Solution
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Solution
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