CHE654Modeling Chemical Processes Using MATLAB Homework Set #5 Solutions Class-16 Prepared by Dr. Hong-ming Ku King Mongkut’s University of Technology Thonburi Chemical Engineering Department Chemical Engineering Practice School July 2004-2012 – Use with Permission of the Author Only 1 53. Boiling and Draining of Ethanol in a Spherical Vessel Consider the boiling of pure ethanol in a jacketed spherical vessel with a radius R = 1 meter, as shown in the diagram. The liquid is drained at the bottom of the vessel while some of it is boiled and escapes as vapor through the top of the vessel. Steam TS 1 1 atm v z CV P0 The following data are available: λ(TB) = 3.858x107 J/kmol MW = 46.07 ρ = 16.575 kmol/m3 CP = 1.4682x105 J/kmol-°C vap log10 P = 8.04494 − __1554.30__ T in °C and Pvap in mmHg T + 222.65 3 -1/2 CV = 4.5 m -atm /hr = valve constant TS = 100 °C U = 2.0x106 J/hr-m2-°C P0 = 1 atm z0 = z(t=0) = 1 m (a) Model this operation and use MATLAB to determine the time it takes for the vessel to completely empty, assuming that initially the liquid is at its boiling point. Note that the heat transfer area AT is not constant. (b)Repeat the calculations in Part (a), assuming that there is no draining of the liquid at the bottom (i.e. the liquid leaves the vessel only through boiling). Determine the solution analytically (an exact solution is possible in this case). Useful conversion factors and formulae: g = gravitational constant = 9.807 m/s2 1 atm = 1.01325x105 N/m2 1 N = 1 kg-m/s2 The volume of liquid V(z) in a spherical vessel as a function of its height z is given by V(z) = πRz2 − πz3 3 while the surface area S(z) is given by 2 S(z) = 4πRz − πz ∫ [a + bx − a ln(a + bx)] x dx = a + bx b2 Solution: (a) First, calculate TB: log10 (760) = 8.04494 – 1554.3/(TB + 222.65) TB = 78.33 °C Overall mass balance: ρ (dV/dt) = −v − ρCv√ P – P0 dV/dt = −0.06033 v − 4.5√ P – 1 But P = P0 + φgz = 1 atm + (763.61 kg/m3) ( 1 atm ) (9.807 m/s2)( 1 N ) 1.01325x105 N/m2 P = 1 + 0.073908z dV/dt = −0.06033 v − 1.22337√z Energy balance: (2.0x106)(100 – 78.33)AT v = q/λ = = 1.12338(4πRz – πz2) 3.858x107 (2πz – πz2) (dz/dt) = –0.27109πz + 0.06777πz2 – 1.22337√z z(2 – z)(dz/dt) = –0.27109z + 0.06777z2 – 0.38941√z dz dt = 0.27109 – 0.06777z + 0.38941z-1/2 (z – 2) 3 (1 kg-m/s2) drain_t.t. end [t.01.1) > 0 drain_t = drain_t + 0.drain_t.drain_t.initZ).06777*z(1)+0. z] = ode45('spherea'.01.1) > 0 drain_t = drain_t + 0. z(t = 0) = 1 m Determine t at z = 0 using MATLAB t = 1.38941*z^(-0. initZ = 1. 4 .z) dzdt = zeros(1. [t.drain_t. while z(length(z).5))/(z(1)-2).6966 hours or 101.initZ).8 minutes to empty The MATLAB Script File: % % CHE654: Problem #53 % % Boiling and Draining of Ethanol in a Spherical Vessel % clc clear format long drain_t = 1.initZ). while z(length(z). end [t.s.z) title('Boiling and Draining of Ethanol in a Spherical Tank') xlabel('Time in Hours') ylabel('Ethanol Height in Meters') % Part (b) drain_t = 1. [t. z] = ode45('sphereb'. dzdt(1) = (0.1). z] The MATLAB M-File for the ODE: function dzdt = ode(t.initZ). z] plot(t.27109+0. z] = ode45('sphereb'. z] = ode45('spherea'. % Part (a) [t. [t. 06777 t 0 4 – z – 2ln(4–z) = 0.(b) Repeat the calculations in Part (a) analytically if there is no drainage.06777z (z – 2)(dz/dt) = 0.06777(4 – z) 0 ∫ 1 0 z dz 4–z – ∫ 1 t 2 dz 4–z = ∫ 0.06777πz2 (2 – z)(dz/dt) = –0.06033 v (2πz – πz2) (dz/dt) = –0.06777 dt 0 [4 – z – 4ln(4–z)] + 2ln(4–z) = 0.27109πz + 0. dV/dt = −0.06777 t 1 5 .27109 + 0. 06777*(4-z(1))/(z(1)-2).1).06777 t t = 6.22741 – 0.1. 6 . The MATLAB M-File for the ODE: function dzdt = ode(t.266 hours or 375.27 hours.802775 = 0.9 minutes MATLAB gives an answer of about 6. dzdt(1) = 0.z) dzdt = zeros(1. 015 kg/kmol φ (mass density) = 1000 kg/m3 λ (heat of vaporization) = 2256.84 kJ/kg CP (liquid heat capacity) = 4. Boiling of Water in a Closed Cylindrical Vessel A closed cylindrical vessel (R = 0. there is also 5 kg of water vapor (or steam) in the gas phase of the vessel. Initially. (b) Derive an analytical expression of the water gas pressure PG as a function of time. The following data are also known about water: MW (molecular weight) = 18. Hint: You may check the correctness of your analytical solution by checking PG at t = 0 and at t when all water has been vaporized. Answer the following questions: How long does it take for water in Part 1 to boil off completely? ____________ minutes 7 . water is being boiled by a heating coil at TS = 200 °C with a constant heat transfer area A = 2. the water (500 kg) is already at 100 °C and starts to boil. Let’s assume that initially at t = 0.0 m2 and an overall heat transfer coefficient U = 100 kJ/min-m2-°C as shown in the figure for Part 1.5 m and H =5.15 K) as shown in the figures below. We assume the water vapor is an ideal gas and that the temperature in the gas and liquid phases are always in thermal equilibrium at 100 °C.08206 atm-m3/kmol-K Steam PG z VG Water PG v VL VG Water Figure for Part 1 v VL Figure for Part 2 This problem is divided into two parts. (a) Derive an analytical expression of the liquid height z as a function of time. namely Part 1 and Part 2. Plot your analytical solution in MATLAB and run the model until all the water has vaporized.0 m) contains 500 kg of water at its boiling temperature of 100 °C (373.1813 kJ/kg-K R (universal gas constant) = 0. Part 1: In this part.54. We assume U and TS to be the same as in Part 1. (a) Derive an analytical expression of liquid volume VL as a function of time. In this case. (b) Derive a single ODE for the gas pressure PG using the analytical expression in Part 2(a). The heating now comes from the surrounding wall and the bottom of the vessel. let’s study the same system in which the cylindrical vessel is being heated by steam in a jacketed chamber as shown in the figure for Part 2.What is the gas pressure in Part 1 at the following conditions? (i) PG = ____________ atm after 30 minute (ii) PG = ____________ atm when water has boiled off completely Part 2: Now. the heat transfer area decreases with the liquid height and is no longer constant. Note that the final value of PG when the vessel contains only gas in Part 2(b) should be equal to that in Part 1(b). Answer the following questions: How long does it take for water in Part 2 to boil off completely? ____________ minutes What is the gas pressure in Part 2 at the following conditions? (i) PG = ____________ atm after 30 minutes (ii) PG = ____________ atm when water has boiled off completely Solution: Part 1(a): Constant heat transfer area Mass balance for the liquid phase: φ dV L = −v dt Energy balance for the liquid phase: d (φVL C P TB ) = UA(TS − TB ) − v(λ + C P TB ) dt φC P TB v= dVL = UA(TS − TB ) − vλ − vC P TB dt UA(TS − TB ) λ 8 . Then use MATLAB (ode45) to solve the ODE for PG from t = 0 to t when all liquid is gone. and make a plot of the PG profile. t = 56.6366 − 0.6366 m dP dV L vRTB + (VT − V L ) G = dt dt ( MW ) 9 .The mass balance equation then becomes: πR 2 UA(TS − TB ) dz =− dt φλ z 0 = z (t = 0) = z = z0 − → UA(TS − TB ) dz =− dt φλπR 2 (500 kg ) (1000 kg / m 3 )(π )(0.5) 2 z = 0.44 minutes Part 1(b): Constant heat transfer area Mass balance for the vapor phase: MW UA(TS − TB ) dn =v= dt λ PGVG = nRTB → n= PGVG RTB d ( PGVG ) vRTB = dt ( MW ) PG dVG dP vRTB + VG G = dt dt ( MW ) PG dP d (VT − V L ) vRTB + (VT − V L ) G = dt dt ( MW ) − PG = 0.5 m) 2 UA(TS − TB ) (100)(2)(200 − 100) t = 0.84)(π )(0.01128t When z = 0.6366 − t 2 φλπR (1000)(2256. 5 − 0.427 + 0.15) = 8.733 − PG 3.08206)(373.015 1000 dPG = 0.493 = 2.927 − 0.733 − 5816.478 atm (3.493 3.427 1697.008862t Let’s check PG at t = 0 and when water completely vaporized: At t = 0.008862t ln = ln 3.427 10 .008862t ) dPG P (0.427 = → = 1699.427 + 0.427 + 0.5) 1697.008862t VT = πR 2 H = π (0.15) = 2.427 + 0.PG v φ + (VT − V L ) dPG vRTB = dt ( MW ) UA(TS − TB ) dV L =− φλ dt But and → VL = VL 0 − UA(TS − TB ) φλ = 0.8619 − G dt 18.427 + 0.5 + 0.427 + 0.733 − P 0 3.008862t PG = 1699.008862[1699.015)(0.733 − PG 1699.255 3.255 3.0 / 18.5) 2 (5) = 3. PG = 1699.255 3.008862t G ∫ But PG PG 0 = (5.478 atm 3.733 − PG 1697.733 − PG ] dt t dPG dt = 0.733 − 5816.008862t ) (3.427 1699.008862t 3.927 (VT − V L ) dPG RTB P UA(TS − TB ) RTB P = v − G= − G dt λ ( MW ) φ ( MW ) φ (3.927 − 0.08206)(373.008862∫ PG 0 1699. PG(t+1) = 1699.67 atm MATLAB script to plot the derived equation: clc clear time(1) = 0.008862*t). PG = 124.733-5816.08206)(373.44) PG = (500 + 5)(0. end time(58) = 56.At t = 56.PG) title('Gas Pressure vs. for t = 1:56 time(t+1) = t. plot(time.58 atm (18.008862(56.733-5816.493/(3.15) = 218.0 min.493 = 218.427+0.478. PG(58) = 1699.927) At t = 30.008862*56. Time') xlabel('Time in Minutes') ylabel('Gas Pressure in atm') 11 .427 + 0.44 min.44.64 atm 3.44).733 − Check against the ideal gas law: 5816.015)(3. PG(1) = 2.493/(3.427+0. PG = 1699. 5) 2 + L (200 − 100) dV L 0 .7854 + 4V L ) dt (1000)(2256.19635 Time for water to completely vaporize = 71.01772t ) − 0.7854 + 4V L = −0.01772t (0.44 min when VL = 0 Part 2(b): Varying heat transfer area 12 .7854 VL = 0.5 =− = −0.004431dt → ln = −0.69635 exp(−0.Part 2(a): Varying heat transfer area UA(TS − TB ) dV L =− dt φλ → 2V U πR 2 + L (TS − TB ) dV L R =− dt φλ 2V (100) π (0.7854 + 4V L ) 2.84) dV L 0.004431(0. 84 (18.t.478 .7854 + 4VL ) 1.927 − VL ) dPG P = 4.5 (3.015) 1000 (3.4310(0.927 − VL ) G = − G dt 2256.5) 2 + L (200 − 100) dP (0.01772 t PG 1. PG (0) = 2.69635e −0.12335 − 0.6997 − 1000 13 s.08206)(373.6997 − G dt 1000 dPG 12.3421e −0.01772t = dt 4.(VT − V L ) dPG UA(TS − TB ) RTB P = − G λ dt ( MW ) φ 2V U πR 2 + L (TS − TB ) dP P RTB R (VT − VL ) G = − G dt λ ( MW ) φ 2V (100) π (0.15) P 0 . [t. simtim.6997-PG(1)/1000)/(4. initPG = [2.62 atm 14 . PG] plot (t.71 atm At t = 71. PG) fPG = zeros(1.69635* .MATLAB Script File: clc clear format short simtim = [0:0.44]. [t. PG = 133.. initPG).PG) title ('Gas Pressure vs Time When Heat Tranfer Area Is Not Constant') xlabel ('Time in Minutes') ylabel ('Gas Pressure in Atm') function fPG = ode(t. PG = 218.478]. exp(-0.1).3421*exp(-0. PG] = ode45('pg'. At t = 30 min.12335-0.. fPG(1) = 12.01772*t)).44 min.01:71.01772*t)*(1. e. (c) Solve the ODE in Part (b) analytically to find x2 as a function of τ.58. and CC represent the concentrations (mol/volume) of components A. (d)Using the following data: k2f = 1. CB. CB.0 min-1). Repeat the above calculations for the case of k1 = 3 min-1.0 min-1. α. and CC. Use both sets of rate constants (i.0 min-1 Solve for CB analytically as a function of time. prove it mathematically.5 min-1. and C. α.0 min-1. (a) Assuming that each of the reactions is first-order and constant volume. Dimensionless concentration of B. k2r = 1. Your ODE must contain only dimensionless quantities (x2. find the maximum concentration of B and the reaction time at this concentration.5 min-1. Conversion of A. respectively. τ.5 min-1. Ratio of forward and reverse rate constants. τ x1 x2 α β = = = = = k1 t (CA0 – CA) / CA0 CB / CA0 k2f / k1 k2r / k1 Derive a second-order ODE for the dimensionless concentration of B. while k1 represents the rate constant for the conversion of A to B.5 min-1 k1 = 1. and β). If no such maximum exists. k2r = 2. k1 = 1. k2f = 1.0 mol/liter CB0 = CB (t=0) = 0 mol/liter CC0 = CC (t=0) = 0 mol/liter k2r = 2.0 min-1 CA0 = CA (t = 0) = 3. Batch Reactor with a Series Reaction Consider a batch reactor with a series reaction where component A reacts to form component B. 15 . k2f = 1. Ratio of rate constants. B. write down the 3 modeling equations for CA. and C. k2r = 1 min-1 while the initial conditions remain the same.0 min-1 and k1 = 3. (f) Validate your analytical solutions by solving the differential equations in Part (a) with MATLAB and plot the time profiles of components A. B. and β. The reaction scheme can be characterized by: k2f k1 A B C k2r Here k2f and k2r represent the kinetic rate constants for the forward and reverse reactions for the conversion of B to C. where CA. (e) Given the data in Part (d). (b)Using the following definitions: Dimensionless time. k2f = 1. Component B can also react reversibly to form component C. Solution: k1 A k2f B C k2r (a) There are 3 modeling equations which are : dCA/dt = –k1CA dCB/dt = k1CA – k2fCB + k2rCC dCC/dt = k2fCB – k2rCC (b) Define τ = k1 t x1 = (CA0 – CA) / CA0 x2 = CB / CA0 α = k2f / k1 β = k2r / k1 Need a 2nd-order ODE for dimensionless concentration of B starting with: CA = CA0 e-k1t or ⇒ 1 – x1 = e-τ x1 = 1 – e-τ d2CB/dt2 = k1dCA/dt – k2fdCB/dt + k2rdCC/dt d2CB/dt2 = –k12CA0 e-k1t – k2f dCB/dt + k2r(k2f CB – k2rCC) = –k12CA0 e-k1t – k2f dCB/dt + k2r k2f CB – k2r[dCB/dt + k2f CB – k1 CA0 e-k1t] = ( k1k2r – k12) CA0 e-k1t – (k2r + k2f)dCB/dt Introducing dimensionless variables: d[dCB/dt]/dt = [1/(1/k1)dτ] [CA0dx2/(1/k1)dτ] = k12CA0d2x2/dτ2 –(k2r + k2f)dCB/dt = –k1(k2r + k2f)CA0dx2/dτ = –(α+β)k12CA0dx2/dt 16 . β. CB = 0 ⇒ τ = 0 . λ2 = –(α+β) ∴ x2(τ) = C1 + C2e-(α+β)τ + C3e-τ At t = 0 .(k1k2r – k12)CA0 e-k1t = (β-1)k12CA0e-τ ⇒ k12CA0d2x2/dτ2 = (β–1) k12CA0e-τ – (α+β)k12CA0dx2/dτ ⇒ d2x2/dτ2 + (α+β)dx2/dτ + (1–β)e-τ = 0 d2x2/dτ2 + (α+β)dx2/dτ = (β–1)e-ττ (c) Solve for x2 analytically as a function of α. and τ x2(τ) = C1eλ1τ + C2eλ2τ + C3e-τ λ2 + (α+β)λ = 0 ⇒ λ1 = 0 . dCB/dt = k1CA(t=0) – k2fCB(t=0) + k2rCC(t=0) = k1CA0 17 . x2 = 0 C1 + C2 + C3 = 0 dx2/dτ = –(α+β)C2e-(α+β)τ – C3e-τ d2x2/dτ2 = –(α+β)2C2e–(α+β)τ + C3e-τ d2x2/dτ2 + (α+β)dx2/dτ = C3(1–α–β)e-τ = (β–1)e-τ ⇒ C3 = (1–β)/(α+β–1) C1 + C2 + (1–β)/(α+β–1) = 0 At t = 0 or τ = 0 . k2r = 2 min-1 CA0 = 3 mol/liter .5 min-1 .or dx2/dτ = 1 1 = –(α+β)C2 – C3 C2 = [–1/(α+β)][1+(1–β)/(α+β–1)] = –(1/α+β)[(α+β–1+1–β)/(α+β–1)] = α / [(α+β)(1–α–β)] C1 = –C2–C3 = α / [(α+β)(1–α–β)] + (β–1)/( α+β–1) = [α+(α+β)(β–1)]/[(α+β)(α+β –1)] ∴ x2(τ) = β/(α+β ) + α/[(α+β )(1–α–β )] e-(αα+ββ)ττ + (1–β )/(α+β –1) e-ττ (c) If k1 = 1 min-1 .51429e-3. CB = CC = 0 when t = 0 ⇒ α = 1. k2f = 1.5t – 1.5e-3.5 and β = 2 x2(τ) = 0. CB does not go through any maximum.5) – 3.17143e-3.2e-t = 0 ⇒ –1.57143 – 0.5t = –t ⇒ no solution / root to the above equation! Therefore.5t = e-t ⇒ In (–1.71429 – 0.5t + 1.80e-3.2e-t (d) Find CB.max and tmax: dCB/dt = 1.5τ – 0. 18 .4e-τ CB(t) = 1. 'B'.5t = In (1.6e-0. [t.c] = ode45('batchr2'.initC). [t.0.c) title('Batch Reactor with a Series Reaction') xlabel('Time in Minutes') ylabel('Concentration of A.0].initC).c) title('Batch Reactor with a Series Reaction') xlabel('Time in Minutes') ylabel('Concentration of A.max = 1.8333τ – 4e-τ CB(t) = 1.5t –12e-3t dCB/dt = –27e-2.c] plot(t.Now k1 = 3 min-1 k2f = 1. and C in Mole/Liter') legend('A'. B. initC=[3.1:5].5 β = 1/3 x2(τ) = 0. and C in Mole/Liter') legend('A'.c] = ode45('batchr1'. [t.c] plot(t.8e-2.'B'.40 – 3.simtime.simtime.5 min-1 k2r = 1 min-1 α = 0.5t + 36e-3t = 0 ⇒ –2.'C') 19 .2 + 10. B.3333) – 3t ⇒ tmax = 0.627 mol/liter (f) The MATLAB Script File: % % CHE654: Problem #58 % % Batch Reactor with a Series Reaction % clc clear simtime=[0:0.575 min ⇒ CB.'C') % [t. fv(2) = 3*c(1)-1.6000 0.5341 1.5000 2.0674 0.5*c(2)-2*c(3).5*c(2)+2*c(3).2000 0.c) fv = zeros(3.0194 0.2044 1.3674 0.1824 0 0.1).2833 0.8000 1.The MATLAB M-Files for ODE: function fv=batchr1(t.9000 2.4060 0.5811 1.4487 0.2000 2. fv(3) = 1.7000 2.6391 0.3608 0.5*c(2)-c(3).2573 1.5*c(2)+c(3).9036 0.9242 0.6157 1.3000 1.7398 0.0865 1.5759 0.5000 1.0172 1.4438 1.3039 1.7512 0.6694 0.6000 1.1036 0.1763 20 .3324 0.2016 0.2225 2.9891 1.3480 1. function fv=batchr2(t.8000 0.3818 1.6000 2.6251 1.1).4000 1.4562 2.6975 0.7000 1.8196 1.5938 1.4702 1.2197 1.7831 0.0426 1.3000 2.5082 0.0000 2.3008 0.2661 0.1054 1.3000 0.4000 0.5670 1.0110 1.3452 1.1000 1.4898 1.8457 0.7145 2.5480 0.2722 0.8868 0.0000 2.8006 0.1381 1.4938 1.2059 0.9986 0.6413 0 0.1000 2.2463 0. fv(3) = 1. fv(1) = -c(1).0656 1.1648 1.4959 0.5150 1. fv(2) = c(1)-1.4000 2.7000 0.1226 1.9000 1. fv(1) = -3*c(1).2000 1.2228 0.1521 1.c) fv = zeros(3.0740 1.6464 1.5000 0.1322 0.4764 0.8176 0.6453 0.1000 0.6053 1.8971 0.4363 0.9928 1.8000 3.9582 0.6336 1.0000 1.1438 1.6057 0.4145 1. The MATLAB Output: >> ans = 0 0.5514 1. 4000 4.6482 1.5730 1.5000 0.8778 1.3000 1.0450 0.8000 0.9035 0.2193 1.2365 1.3758 1.6846 1.0000 0.0055 2.5319 1.6694 0.9000 0.0000 4.4098 1.2693 1.7747 21 .2613 1.2000 1.2225 0.6923 1.7495 1.7000 4.0607 0.7062 1.0407 0.7000 0.3674 0.7022 1.6763 1.4000 0.5376 1.0333 0.3616 1.0273 0.1107 0.2559 1.0254 1.2412 1.1000 2.3000 4.1651 0.7000 3.1961 1.6874 1.0247 1.6000 0.0074 2.6545 1.0549 0.0030 0 0 0.4478 1.2000 3.1867 1.6264 0.4227 0.3000 3.4000 3.7053 1.0100 2.1106 1.2636 1.7139 1.1494 0.5000 3.6000 0.7034 1.1648 0.6742 1.0223 0.6460 1.7000 0.2.5000 4.7681 1.0450 1.0742 0.9000 3.6088 1.8000 4.0368 0.6780 1.1223 0.2223 1.1001 0.3000 0.2665 1.2314 1.6000 4.2533 1.2278 1.2809 1.7010 1.1000 0.2527 1.6996 1.4890 1.6464 0.0183 1.6944 1.7367 1.5082 1.9000 0.0000 3.3206 1.6815 1.0000 0.6000 3.0136 1.2493 1.2723 1.6073 1.0333 1.4958 0.3576 1.7207 1.2256 1.0671 0.1548 1.1000 4.2658 1.2990 1.0000 0.0000 0.7008 1.7599 1.4415 1.1494 1.6700 1.0563 1.0820 0.3000 0.1351 0.0302 0.2346 1.2197 0.7212 0.2000 0.2124 1.0247 0.7044 1.9000 5.2736 ans = 0 3.2000 4.6963 1.2016 1.9000 4.2587 1.6602 1.0041 2.2721 0.4000 0.0820 1.2709 1.3462 1.0608 1.6900 1.6654 1.0497 0.2657 1.0906 0.8000 0.5634 1.2046 1.8000 3.2430 1.6980 1.2000 0.5000 0.1888 1.1000 3.2676 1.6167 0.0202 1.1000 0.5589 0.2454 1. 7901 1.2090 1.2000 1.0000 3.7000 2.7981 1.0000 0.8000 3.2001 1.2000 4.0002 0.0000 0.5000 2.5000 4.2.0017 0.2003 1.7985 1.2000 3.2011 1.0009 0.2113 1.4000 4.2022 1.2007 1.0000 0.2036 1.2045 1.5000 3.7994 1.0005 0.7841 1.2001 1.7999 1.2005 1.7951 1.0001 0.9000 4.7976 1.2001 1.2000 1.2009 1.7799 1.0002 0.7000 3.7962 1.2014 1.0000 0.7998 1.2178 1.2071 1.6000 2.9000 3.8000 4.7998 1.2018 1.0000 0.7999 1.2004 1.3000 3.0012 0.8000 22 .7996 1.0000 0.8000 2.0007 0.8000 1.0003 0.7938 1.1000 4.2028 1.0000 4.6000 4.0000 0.7000 4.0000 0.2003 1.0004 0.7997 1.7997 1.2057 1.7989 1.7991 1.2002 1.0000 1.0022 0.0001 0.0000 0.3000 4.0000 0.2142 1.4000 3.7970 1.4000 2.7999 1.7993 1.2002 1.7922 1.2001 1.0000 0.7874 1.7999 1.0000 0.0000 0.6000 3.0000 0.1000 3.9000 5.0001 0. 23 . 0 0.28 1. the reactor now operates in a batch mode.52 1.40 1.500 moles of a new component called D (same density as components A. respectively. the 2 feeds to the reactor are suddenly shut off. The value of k2 has been measured to be 1.0 7. CB. Derive analytically the concentration of A as a function of time.5 1. 2 moles of component A react with one k1 mole of component B to form one mole of component C: 2A + B -----> C.5 12. Isothermal Semi-Batch Reactor (a) Consider an isothermal semi-batch reactor where a single reaction takes place in a solvent S.0 CA(mol/liter) CB(mol/liter) 2.06 (b) At the end of 20 minutes.0 40. The following experimental data have been obtained for this reaction when carried out in a batch reactor: Time(minute) 0 5. the concentrations (moles/liter) of A. which is inert. and C. Use ode45 in MATLAB to solve for and plot (in a single graph) the concentrations of the 3 components. The reaction rate does not conform to the stoichiometry but is 1st-order with respect to each reactant as follows: rA = −k1CACB FA = 15 liter/min FB = 10 liter/min Initially (t = 0 min).60.17 1. B. and CC. Assuming that all components have the same density of 60 mol/liter. That is.65 1.5 minute. and compute CA at steady state based 24 .0 32. Component D reacts with component C to form A and B.10 1. and 4. In this reaction.5 25.0 min-1. the reactor contains 100 liters of solution and 300 moles of A. derive 3 ODE equations needed to compute CA. and C) are charged to the reactor. Run the model for 20 minutes with an increment of 0. and the reaction now looks as follows: k1 2A + B C + D k2 with a reaction rate of rA = −k1CACB + k2CD (2nd -order forward and 1st-order reverse). B.0 15. 5CA0)+0.on your derived equation. Integrate by partial fractions or use a table of integrals: CA – [1/(CB0–0.5CA0)+0.5CA] Integrate from CA0 to CA on the LHS and from 0 to t on the RHS. dCA = 2dCB CA – CA0 = 2(CB – CB0) (dCA/dt) = –kCACB CB = CB0 + ½ (CA – CA0) dCA = – kdt CA[(CB0–0.5CA0– CB0)kt 25 .5CA0)] ln[(CB0–0. ∫ Useful Integrals: ω tan −1 γ γ 2 + ∫ dx = B dx 1 a + bx = − ln x( a + bx) a x − 2 if γ = 0 ω − where if γ > 0 ω 2 tanh −1 −γ −γ if γ < 0 B = a + bx + cx2 γ = 4ac – b2 ω = b + 2cx Solution: (a) Must first determine the rate constant k using the graphical method.5CA)/CA] = –kt CA0 ln(CA/CB) – ln(CA0/CB0) = (0. Your final expression should be simplified as much as possible and should not contain any parameters except t (time) and CA. 5 – 0.t.1084 liter/mole-min V = 100 + 25t d(VCA)/dt = –kVCACB + ρAFA because 1 dNA = dCA = –kCACB V dt dt d(VCA)/dt = V(dCA/dt) + CA(dV/dt) = V(dCA/dt) + 25CA = –0.5CA0– CB0)kt to obtain the value of k.So plot ln(CA/CB) vs.0)] + (60)(15) = –0.1084CA[0.5CA) + (900 – 25CA)/(100 + 25t) s. CA(t = 0) = 3.5CA – 1.5]V + 900 dCA/dt = 0. get slope = 0. From MATLAB.5(CA – 3.1084CA(1.0 26 .1084VCA[0 + 0. (0.054203 dV = FA + FB = 15 + 10 = 25 dt k = 0. 06]. CB = 0. logy = log(CA.5CA – 1. CB(t = 0) = 0 d(VCC)/dt = 0.t.0 15.5 25.5CA) + (600 – 25CB)/(100 + 25t) s.0542CA(0.5*(CA-2.5 12.d(VCB)/dt = –0. CA = [2. 0.0 7.0 40.28 1.10 1.0).5kVCACB dCC/dt = 0.logy.1). % plot(t. initC).S] = polyfit(t. model the semi-batch reactor for 20 minutes % format short simtime = [0:0. z] 27 .0 32.y) % % Now. CC(t = 0) = 0 The MATLAB Script File: % % CHE654: Problem #60 % % Isothermal Semi-Batch Reactor % clc clear format short e % % First.'o'.5+0. determine the value of k t = [0 5.0.65 1.5 – 0.17 1./CB).5:20].52 1. [t.logy. 0].0].5kVCACB + ρBFB dCA/dt = 0. P(1) k = P(1)/0. y = P(1)*x+P(2). initC = [3. simtime.t. [t.0 1.5 x = [0:2:40].5) – 25CC/(100 + 25t) s. [P.0542CA(1.x. z] = ode45('fexam'.40 1. 4203e-002 k= 1.1620 10.9453 9.0000 7.0000 10.0000 8.7977 8.4107 1.6730 7.8299 8.8452 7.z) xlabel('Time in Minutes') ylabel('Concentration in Moles/Liter') legend('CA'.4809 6.1).5000 8. fC(3) = 0.7984 7.5000 6.0000 8.0000 9.2585 7.7616 8.4138 7.1280 4.7749 6.6719 7.0000 8.5-0.5*C(1)) + (900-25*C(1))/(100+25*t).2869 5.1084*C(1)*(1.0053 1.0000 10.5000 7.2384 7.0000 9.3955 9.9270 9.9069 9.7855 7.5000 10.5*C(1)) + (600-25*C(2))/(100+25*t).2776 6.2396 7.2278 6.5635 5.5548 6.8844 9.5000 8.4190 7.0106 2.2751 9.7650 3.1014 3.2122 7.9756 4.'CB'.5000 8.5387 0.9947 6.5443 2.1425 3.7198 7.5000 9.7616 0 0 2.0545 10.5) .0841e-001 ans = 0 3.5223 7.9517 11. >> >> ans = 5.5207 6.8112 4.6175 7.9878 4.0542*C(1)*(0.5*C(1)-1.4559 2.1095 6.5000 8.3932 5.1556 4.0301 7.6570 8.plot(t.9324 7. fC(2) = 0.9402 7.5000 7.0000 9.25*C(3)/(100+25*t).4731 1.5000 9.0000 8.8611 28 .0000 8.4530 5.2365 0.8588 9. C) fC = zeros(3.6008 7.3176 1.5232 8.6326 5.5-0.0542*C(1)*(1.9493 7.8541 11.4724 7.5000 9.0000 0. fC(1) = 0.8444 7.'CC') The MATLAB M-File for the ODE: function fC = ode(t.5000 9. 0768)(600)/675 = 7.6535 7.5885 7.9232 >> (b) After 20 minutes.0000 19.6654 11.6535)(600)/675 = 5.0009 8.3735 11.0551 8.4750 11.5000 17.12.8025 6.5000 16.0000 14.5000 18.5134 10.9142 CB = (8.0262 6.5000 19. the initial conditions for the batch reactor are: The total liquid volume = 100+(25)(20) +4500/60 = 675 liters CA = (6.1564 11.0000 17.5000 13.0713 8.0000 13.2848 7.0000 7.2163 7.6552 10.0208 8.7016 6.5722 11.0114 8.8557 6.4303 7.0642 8.2058 10.5000 15.3562 7.9105 6.0436 8.0873 7.0000 18.0679 8.7900 10.5000 14.2674 11.0401 11.0000 15.0368 8.0000 16.1794 29 .9616 7.9761 7.1506 7.0292 8.5000 20.3639 10.0497 8.6733 7.9672 6.0599 8.5075 7.9891 8.8407 11.0384 10.7512 6.0768 10.9182 11.0742 8.7548 11.0000 12. 6238 – CA[0.9577] dCA = dt 7.9541 mol/liter from the above equation.5700 –(2/√–γ) tanh-1 (ω/√–γ) = –(2/√–γ) tanh-1 [(b+2cCA)/√–γ] –1.6667 + ½ (5.24756 tanh-1 [–0.6238–CA[0.6238.0542CA + 0.9142 – CA)] = CA[−0.1084CA+0.9577CA–0.6238 = 7.9142 mol/liter.9577)] – 4.62378(0.6667 mol/liter dCA = 2dCB = –2dCD CA – CA0 = 2(CB – CB0) and CA – CA0 = 2(CD0 – CD) CB = CB0 + ½ (CA – CA0) CD = CD0 + ½ (CA0 – CA) (dCA/dt) = −k1CACB + k2CD = −k1CA[CB0 + ½ (CA – CA0)] + k2[CD0 + ½ (CA0 – CA)] = −0.0542CA+0.9142)] + 1.24756 tanh-1 [–0.1084CA[7.62378(0. From the ODE: (dCA/dt) = 7.32055 – 0. CA does equal to 5.9577] = 0 ------> CA = 5.CD = 4500/675 = 4.77825 – 0.9577] dCA = dt 7.9541 mol/liter same answer! 30 .6238–0. CA = 5.5] + 7.0542CA + 0. c = –0.0542CA2 a = 7. When ∝.0542CA + 0.1084CA+0.6238 – CA[0.0542 γ = 4ac – b2 = –2.0[4.9577)] = dt –1.1206 = t Check: when t = 0.1794 + ½ (CA – 5. b = –0.9577. dCB/dt + p(t)CB = q(t) but do not solve this ODE.14 mol/liter at t = 0.0 0.62. dCA/dt = −k1CA1/2 dCB/dt = k1 CA1/2 − k2CB 2 Derive an analytical expression for CB as a function of time. Determine the values of k1 and k2.30 CA(mol/liter) 1. derive analytically a 1st-order ODE for CB.35 0.1 hr.0 0. 31 .05 0. and C.85 1.30 0.03 0. and CC represent the concentrations (mol/liter) of components A.76 0.20 0. (b) The following experimental data were obtained for component CA: Time(hr) 0 0.06 CB was also measured to be 0.06 0. CB = 0.e.54 0. i. Determine the values of k1 and k2 and the time tmax at which CB is at its maximum. the following experimental data were obtained for CA: Time(hr) 0 0. Isothermal Batch Reactor with a Series Reaction Consider an isothermal batch reactor with a series reaction where 2 moles of component A react to form one mole of component B. i.51 0.79 0. (c) Now. suppose the order of the above series reaction conforms to the stoichiometry.25 It was also observed that CB reached a maximum of 0.60 0. where CA. CA = CA0. and CC = 0.93 0.e. The reaction scheme can be characterized as follows: k2 k1 2A B C Initially (t = 0 hr).33 0. respectively. Component B also reacts to form component C. CB. (d) When the reaction order conforms to the stoichiometry.15 0. (a) Assume constant volume and that the first reaction 2A → B is one-half order and the second reaction B → C is first-order.0 CA(mol/liter) 1.15 0.13 0. Hint: Use Polyfit function in MATLAB to help determine k1 and k2.10 0.39 0.63 0. B.11 mol/liter at t = 1 hr. t.t.Solution: k1 k2 2A B C (a) If rA = dCA/dt = -k1CA½ s. CA(0) = CA0 rB = dCB/dt = (k1/2)CA½ − k2CB s. CC(0) = 0 Derive analytically CB(t): CA ∫ t dCA/CA½ = −k1 CA0 ∫ dt ⇒ 2(CA½ − CA0½) = −k1t 0 CA(t) = (√CA0 − ½k1t)2 dCB/dt + k2CB CB(t) = (k1/2)(√CA0 − ½k1t) = C1e-k2t + C2t + C3 At t = 0 0 = C1 + C3 = −C1e-k2t + C2 dCB/dt dCB/dt + k2CB = −k2C1e-k2t + C2 + k2C1e-k2t + k2C2t + k2C3 k2C2t + (C2 + k2C3) = (k1/2)√CA0 − (k12/4)t ⇒ k2C2 = −(k12/4) C2 + k2C3 or C2 = −(k12/4k2) = k12√CA0 32 . CB(0) = 0 rC = dCC/dt = k2CB s.t. 9818 hr-1 CB(t) = 0.06 Has been observed: CB (t = 1 hr) = 0.5732/k2)t CB(t = 1 hr) = 0.3 0.C3 = (1/k2)[(k1/2)√CA0 + (k12/4k2)] = (k1/2k2)√CA0 + (k1/2k2)2 = (k1/2k2)(√CA0 + k1/2k2) C1 = −(k1/2k2)(√CA0 + k1/2k2) ∴ CB(t) = (k1/2k2)(√CA0 + (k1/2k2) [1 − e-k2t] − k12/4k2 t (b) Time (hr) 0 0.3184(1 – e-2.5732/k2 Use MATLAB to solve the above nonlinear equation for k2 k2 = 2.05 0.5142 CB(t) = (0.5729 ⇒ k1 = 1.5146t + 1.15 0.5138 So take the average value: k1 = 1.7571/k2)(1 – e-k2t) – (0.0 0.9818t – 0.93 0.7571/k2)(1+0.5729t2 – 1.0023 Can obtain 2 values of k1 as : k1 = 1.13 1.11 = (0.60 CA (mole/liter) 1.9818t) – 0.7571/k2)(1 – e-k2) – 0.11 mole/liter CA(t) = (1 − ½k1t)2 = 1 − k1t + (k12/4)t2 Use Polyfit in MATLAB to find the CA vs. t) From MATLAB : CA = 0.35 0.5146 and k12/4 = 0. t data to a 2nd-order degree polynomial (or plot CA1/2 vs.0 0.9494e-2.79 0.54 0.7571/k2)(1+0.85 0.1922t dCB/dt = 0.1922 = 0 33 . 33 0.2 0.0002 k1 = 10.39 CB.0676t + 1.63 0.max = 0.76 0.max occurs at dCB/dt = 0 ⇒ k2CB.0676/2)(10.e.151 mole/liter (c) The reaction order conforms to stoichiometry dCA/dt = –k1CA2 s.25 .⇒ e-2.3 0.15 0.0676tmax + 1)–2 k2 = 8. t by fitting y = ax + b Slope of the straight line is k1 ⇒ y = 10.51 0. same initial conditions dCB/dt = (k1/2) CA2 – k2CB Derive 1st-order ODE for CB.06 0.1 0.max has been observed at 0.03 0.2024 tmax = 0.0676 CB.9818t = 0. i.0 0.max = (10. dCB/dt + p (t)CB = q(t) ∫ CA t dCA/CA2 = –k1 CA0 ∫ dt ⇒ 1/CA – 1/CA0 = k1t 0 ⇒ CA(t) = CA0/(CA0k1t + 1) ∴ dCB/dt + k2CB = (k1CA02/2)[CA0k1t + 1]–2 (d) Time (hr) CA (mole/liter) 0 1.9285 hr-1 34 0.t.1 hr 1/CA = k1t + 1 Use Polyfit to regress 1/CA vs.536 hr CB.14 mol/lit at t = 0. 0 mole% n-pentane and 40.00 log10 PVAP = 6.85221 − __1064. Solution: (a) Start with Raoult’s Law for ideal liquid: yc5P = xc5PC5VAP (1.63/(T+232.366 PVAP in mmHG T in °C (b) Derive an analytical expression relating the amount of liquid left in the batch still L as a function of x (mole fraction of n-pentane in the still) and compute the value of x after 90% of liquid has been vaporized.8804 xc52 )P = xc5 10[6. Assume ideal gas and ideal liquid. compute the total system pressure P at the system temperature.67. log10 PVAP = 6. determine T in °C at which the above equation is valid).8804 xc5 – 0.8804 x2 Note that this equation is only valid for n-pentane.e.53_ T + 224. V (gmol/min) yi xi L (gmoles) (a) Calculate the system temperature at which the above equilibrium relationship was established for the given liquid mixture (i. and the vapor pressures of the two components are: For n-pentane.8804 x – 0. II A liquid mixture containing 60. For n-hexane.630__ T + 232.87776 − __1171.0 mole% n-hexane is to be distilled in a batch still. and is initially charged with 100 gmoles of the mixture. and not necessarily for n-hexane.85221 35 – 1064.0)] . Batch Distillation. The equilibrium relationship between the mole fraction x of n-pentane in the liquid and that in the vapor y has been correlated to the following equation: y = 1. Also. 0)] ) = 0 Use MATLAB for solve for the root or T Nonlinear equation in T Get T = 55.8804 ln(L) 0.6 to x and L = 100 to L L 1 + 1 x = dL 1–x L ln (x) – ln(1 – x) = 0.8804x – 0.69 = 1042.11322 (10[6.13585 dx = dL x( 1 – x) 1.0)] 0.86 + 196.0)] + 0. P = 845.8113[0.6 100 36 .32452 (10[6.63/(T+232.6 and P = xc5PC5VAP + xc6PC6VAP 0.55 °C Also get.85221 – 1064.366)]] – 1064.8804x – 0.63/(T+232.53/(T+224.55 mmHg (b) Start with mass balances: dL/dt = –V d(xL)/dt = –yV or x(dL/dt) + L(dx/dt) = (1.6*10[6.13585 Integrate from x = 0.But xc5 = 0.366)]) – 0.87776 – 1171.85221 – 1064.8804x2)dL 1.8804x2)(dL/dt) xdL + Ldx = (1.53/(T+224.85221 = 0.87776 – 1171.63/(T+232.6*10[6.8804x2)dL Ldx = (0.8804 ln(L) x ln x 1 – x L = 0.8804x – 0.4*10[6. 8804 L 100 = 1 – x When 90% of the liquid L is vaporized 0.8804 L 100 L = 10 37 x = 0.ln 0.165 .6667 x = ln 1 – x 0.6667 x 0.