Microsoft Word - _98_動Chap 4.doc

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殿炫任
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4-1. 前言 hapter 4 Ideal Reactors for a Single Reaction ( 本章為課本的第四章與第五章 1.

Ideal reactor types: (1 Batch reactor: 批式反應器 BR (2 連續式反應器 (continuous reactor: 反應過程是

目前最常用的名稱為混合流動反應器 MFR (Mixed flow reactor 其他名稱, 依常用情況依序為 STR or FSTR or * constant

1 4-1. 前言 hapter 4 Ideal Reactors for a Single Reaction ( 本章為課本的第四章與第五章 1. 在 hapter 3 所討論的是 batch reactor, 但本 chapter 開始討論 PFR 與 MFR 2. Ideal reactor types: (1 Batch reactor: 批式反應器 BR (2 連續式反應器 (continuous reactor: 反應過程是 stedy state, 有 PFR 和 MFR 兩種 Plug flow reactor: 塞流式反應器 PFR Mixed flow reactor: 目前最常用的名稱為混合流動反應器 MFR (Mixed flow reactor 其他名稱, 依常用情況依序為 STR or FSTR or * constant flow stirred tank reactor 的簡稱 Southern mixed reactor Taiwan backmix reactor University the ideal stirred tank reactor 均勻的攪拌 product 4-1

2 (3 半分批式反應器 ((semi-batch reactor 4-2. Material balance Element of reactor volume Reactant enters Reactant leaves Reactant 可能有一部分因反應而消失 Reactant 可能有一部分累積在系統內 Material balance or Energy balance 的基本概念 Element of reactor volume Reactant enters Reactant leaves (1 Input 1 mol/hr 的 (2 Output 2 mol/hr 的 (3 累積在反應器 (system 中 1 mol/hr 的 (4 因反應而消失的 mol/hr 1 = accumulation = Input Output disappearance disappearance 前面已經取負號消失的量是取正值 4-2

3 accumulation = Input Output + disappearance disappearance 前面已經取正號消失的量是取負值 1 = Input = output + disappearance + accumulation 明顯的, 此表示法的消失的量是取正值 4-3. Ideal batch reactor 1. 在第三章 ( 3-1, 我們探討的 onstant-volume Batch Reactor, product, 進行 n-order 的反應 : (1 則 - r = - d = k n d = - t k n d n = 1 - r = - n 1 a d n = k -ln = - t k = kt 1n 1-n = (n-1kt d (2 但是若以 - r = - t = - n = 1 - r 的觀點來看, 則 d or t = - d = k -ln = kt n 1 - r = k n 1n 1-n = (n-1kt 所以,onstant-Volume Batch Reactor t = - d --- 另一種通式 Southern 2. 以 mass balance 的角度來看 Taiwan batch reactor: University (1 在反應過程 no reactant enters or leaves the batch reactor. Input = output + disappearance + accumulation or Rate of reactant loss due to chemical reaction within the element of reactor volume = - Rate of accumulation of reactant in the element of reactor volume 4-3

4 (2 Disappearance of by reaction (mole/time = (- V = moles reacting (time(volume of fluid x volume of fluid accumulation of dn = d[ N = (1 ] d = -N (mole/time d (- V = -(- N t = t x = N d ( V 通式 Eq. 3 onstant-volume Batch Reactor x t = N d ( V = V N x d = x d = - d ---Eq. 4 Varying-Volume Batch Reactor 根據 3-2 Varying-Volume Batch Reactor V=V (1+ε x t = N d ( V = N x d x = V (1 d (1 ( ( -----Eq. 5 Eq. 3~5 are applicable to both isothermal and nonisothermal operations. 4-4.Space-time and space-velocity Southern 在 batch reactor 中計算反應所進行的時間是理所當然的 Taiwan University 在 continuous reactor 中反應時間不易計算便用 Space-time and space-velocity 作為衡量的方式 1. Space-time: 1 Time required to process one reactor volume τ= = = [time] s of feed measured at specified conditions τ= 2 min. 在某條件下, 處理一個反應器體積的進料需 2 分鐘 4-4

2. Space-velocity: 1 Number of reactor volumes of feed at specified s = = = [time -1 ] conditions which can be treated in unit time s = 5 hour -1 在某條件下, 每小時可處理 5 個反應器體積的進料 τ= s 1 =.

5 2. Space-velocity: 1 Number of reactor volumes of feed at specified s = = = [time -1 ] conditions which can be treated in unit time s = 5 hour -1 在某條件下, 每小時可處理 5 個反應器體積的進料 τ= s 1 =.2 hour 處理一個反應器體積的進料需.2 小時 3. V = volume of reactor ( 例如 1 liters N = 的進料 mole 數 = mole = 的進料濃度 = (moles entering/(volume of feed [ 如 : = 5 M = 5 mol/liter] F = 單位時間, 的進料 mole 數 (molar feed rate of, 如 :1 mol/min. v o = 單位時間, 的進料體積 (volumetric feed rate of, 其單為是 liter/min, F F = v o v o = τ= s 1 = vo V = V F ft 3 /sec 等 v o = mol 5 x 1 τ= mol = 5 min. = 1 min mol 1 min mol 5 V = v o 1 2 min = 2 min 4-5. Southern Steady-state mixed flow Taiwan reactor University 1.Steady-state mixed flow reactor (1 Input = output + disappearance + accumulation input of = F (1- = F (moles/time output of = F (1- = F (moles/time disappearance of by reaction = (- V (moles/time --- 可參看 page

6 (2 F = F (1- + (- V F = (- V V = = F 1 V τ= = = V s v F and - = any are measured at exit stream conditions, which are the same as the conditions within the reactor. 若進料的轉化率不是則以 i 表示而出口的轉化率不以表示而以 f 表示則 V = F 應寫成更廣泛的式子 : V f i = ( f F V τ= F = ( f ( f i Special case of constant-density systems ( 即 = = 1 / V = F = V τ= = = F ( d for first-order reaction - = - τ= kτ = k = k = 1 當然, 若是要選廣泛的 first-order reaction (1 則 V=V (1+ε & = (1 ( 1 kτ = for any 1 for second-order reaction - = - d = 2 k kτ = 2 14 k or = 2k 1 4-6

7 Example 4-1: 以 2 liters/hr 的速率輸送初濃度為 1 mol/l 的氣體反應物到一個 mixed flow reactor 中進行 2 R (- r =.2 2 的反應, 若測得出口尚有氣體 1 mol/l 且反應系統是 constant-pressure system, 則此 MFR 的體積應是多少? Example 4-2: Pure gases reactant ( = 1 millimol/liter is fed at a steady rate into a mixed flow reactor (V =.1 liter where it dimerizes ( 2 R. For different gas feed rates the following data are obtained: Run number v o, liter/hr af, millimol/liter Find a rate equation for this reaction. Sol: - r = - d = k n log (- r = log k + n log 或 ln (- r = ln k + n ln 以 log (- r 對 log 作圖 [ 或 ln (- r 對 ln 作圖 ] 得一直線, 其 slope 為 n; 截距 = log k ( 或 ln k 先求 - r V 由 τ= = V = - r = v F 再由 = ( 1 & =

8 得 = ( = 1 2 No τ r 8 4 log(- r log log (- r 2.5 y = x log 4-8

9 Example 4-3: The elementary liquid-phase reaction + 2 B with rate equation k1 k2 R - = - 1 = (12.5 liter 2 /mol 2.min 2 B 2 B - (1.5 min -l R, [ mol/(liter min] is to take place in a 6-liter steady-state mixed flow reactor. Two feed streams, one containing 2.8 mol /liter and the other containing 1.6 mol B/liter, are to be introduced at equal volumetric flow rates into the reactor, and 75% conversion of limiting component is desired. What should be the flow rate of each stream? ssume a constant density throughout. 4-6.Steady-state plug flow reactor 1.(1In a plug flow reactor, the composition of the fluid varies from point to point along a flow path. (2Material balance 選一個 differential element of volume dv 作分析 2. (1 Input = output + disappearance + accumulation input of = F (moles/time 4-9

10 output of = F + df (moles/time disappearance of by reaction 注意左三式與 MFR(page 4-5 = (- dv (moles/time 的不同 (2 Material balance F = (F + df + (- dv df = d[f (1- ] = -F d F d = (- dv 這只是對一個微量體積 dv 作分析, 若考 V dv F 慮整個反應器, 當然要做積分 f d = V f d = F = V τ= = V = v F f d any 與 MFR(page 4-4 的差異在哪裡? MFR 的在 reactor 中的任何位置是相同的定值 PFR 的在 reactor 中, 隨著離進料處的距離遠近而變, 所以, 需用積分的方式若進料的轉化率不是則以 i 表示而出口的轉化率不以表示而以 f 表示 V f d 則 = = F i or τ= V = V = v F f d i Special case of constant-density systems ( 即 = d = 1 / d = - V f d = = F = - 1 f d V V τ= = = f d F = - f d v for zero-order homogeneous reaction - = - d = k f d d τ= = f kτ= k = k First-order irreversible reaction, anyε 1 dn - r = - (1 = k = k V (1 4-1

11 kτ= -(1+ε ln (1- - ε 若 ε = τ= - ln (1- /k 在 PFR 中, 若 ε = zero-order reaction τ= /k first-order reaction τ= - ln (1- /k 在 chapter 3 的 batch reactor 中, 若 ε = zero-order reaction t = /k (pgae 3-7 first-order reaction t= - ln (1- /k (pgae 3-3 的 For system of constant density: τ for plug flow reactor is equivalent to t for the batch rector. For system of changing density: The correct eq. must be used for each particular situation. Example 4-4: homogeneous gas reaction 3R has a reported rate at r = 1-2 ½ [mol/liter sec] Find a space time needed for 8% conversion of a 5% & 5% inert feed to a plug flow reactor operating at 215 and 5 atm( =.625 mol/liter. 4-11

12 Example 4-5: The homogeneous gas decomposition of phosphine 4 PH 3(g P 4(g + 6 H 2(g prcceeds at 649 with the first-order rate - r = (1/hr PH3 What size of plug flow reactor operating at 649 and 46 kpa can produce 8% conversion of a feed consisting of 4 mol of pure phosphine per hour? 1 atm = pa = kpa 4-7. Holding time and space time for flow reactors 1.Define: (1space time: 1 V τ= = = = V reactor volume of feed measured s at specified conditions v F (2holding time: Time required to process one Mean residence time of t = = f d flowing material in the reactor ( 1 2.For constant density system (all liquids and constant density gases τ= t = V v 4-12

13 3.For changing density system τ t and t 4. 如圖所示 : V v 將玉米花原料以 1 liter/min 的進料送入一個體積為 1 liter 的加熱器 ( 反應器中, 由於玉米花原料遇熱會膨脹, 所以, 若以爆開的玉米花來看, 其前進的體積流率為 28 liters/min 假設玉米花爆開的位置如下圖所示, 則 : (1space-time (2holding time: t = 6 sec V = = Y Z = = 1 liter/(1 liter/min. = 1 min. = 6 sec. v t Y = 1 liter/(28 liter/min. =.357 min. = 2.14 sec. t Z = 2.14 ~ 6 sec 之間 Example 4-6: pure gas (reaction 2 R with volumetric flow rate.1 m 3 /min is introduced into a mixed flow reactor. The reactor volume needed for 5% conversion of is.5 m 3. Find the corresponding ( space-time (B mean residence time for this system. 4-13

14 Example 4-7: 均勻液相反應 : R,- r = k 2,in STR, =.5 ( 若改成一 6 倍大的 STR, 條件不變, =? (B 若為一相等大小的 PFR, 條件不變, =? Example 4-8: Ethylene is produced by the dehydrogenation of ethane 2 H 6(g 2 H 4(g + H 2(g Determine the plug flow reactor volume necessary to produce 1 million kilograms of ethylene a year from the above reaction. The reaction is irreversible and elementary. We want to achieve 8% conversion of ethane, operating the reactor isothermally at 11 K and a pressure of 6 atm. The rate constant k at 11 K is 3.7 sec -1. <Hint> 1 d = (1+ε ln [1/(1-]- ε

15 Example 4-9: onsider the following batch, elementary reaction of ethyl acetate in an aqueous solution of sodium hydroxide at 25 : H 3 OO 2 H 5 + NaOH k H 3 OONa + 2 H 5 OH where k equals 6.48 l/mol min. The reaction is commenced with equal concentration (.1 mol/l of ethyl acetate and sodium hydroxide. Please calculate concentration of ethyl acetate 25 minutes after the start of the reaction. Example 4-1: n irreversible reaction R was carried out in a plug flow reactor. equation is The rate 1 = 2 ( 1 The flow rate of feed is.2 m 3 /sec. The concentration of in the feed is 1 kmol/m 3. What is the reactor volume to get the 99% conversion of? xdx Hint: 2 ( a bx 1 = 2 b [ ln (a+bx + a a bx ] 4-15

16 Example 4-11: second-order liquid phase reaction takes place with 5% conversion in a STR. What will be the conversion if this reactor is replaced by one PFR twice as large as STR? ll else remain unchanged? 8% Example 4-12: first order irreversible gas-phase reaction B + 2 take place in a STR. The operation is isothermal and isobaric. The reaction rate constant is.2 min -1, and the reactor volume is 5L. The feed contains a 6 mol% of and a 4 mol% of an inert compound. The molar flow rate of the reactant is 4 mol/min. and the initial concentration of the reactant is 4 mol/l. What is the conversion at the exit of this STR? 4-16

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untitled 1-1-1 1-1-3 1-4 1-5 1-6 1-7 1-8 (differential equation) F(,, ') = (first order differential equation) ' + + 4= 1 + = e hapter 1 (separable equation) A( ) d = B( ) d (1.1) ò ò Ad ( ) = Bd ( ) 1 - cos =