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ContinuousStirred Tank Reactors - Research Paper Example

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This work called "Continuous Stirred Tank Reactors" describes the experiments to simulate the conversion of reactant in a cascade of three continuous stirred tanks. The author outlines the effect of flowrate (i.e. residence time) on the conversion of reactant in one tank in a cascade of continuous stirred tanks…
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ContinuousStirred Tank Reactors
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CONTINUOUS STIRRED TANK REACTORS Effect of Residence Time on Conversion Division of Chemical and Petroleum Engineering Objectives: 1. To simulate the conversion of reactant in a cascade of three continuous stirred tanks. 2. To determine the effect of flowrate (i.e. residence time) on the conversion of reactant in one tank in a cascade of continuous stirred tanks. Introduction: Many processes in the chemical industry involve the addition of reactants to a cascade of several continuous reactors in series, i.e. CSTR. The liquid feed enters the first tank and overflows to each of the subsequent tanks in the cascade. The rate of reaction in each tank is affected by the concentration of the reactants at any specific time. Liquid phase reactions are carried out on small to medium scale, for the manufacture of organic chemicals and pharmaceutical products. For example, ethanol is produced by a reaction between sodium hydroxide and ethyl acetate. In this experiment, no chemical reaction takes place. The residence time behaviour of a dissolved salt through a cascade of three continuous stirred tanks is examined at different flowrates for a constant starting concentration of salt. The experimentally determined variation of slat concentration with time in each tank is compared with results predicted from theoretical equations, allowing deviations from ideal behaviour to be observed. Conversion of reactant is then estimated by graphical integration for both the experimental results and theoretical data. The general effects of the transient concentration of reactants are similar to those induced in a chemical reactor, subject to variation in throughput. Equipment: 1. Three equal sized Perspex tanks with provision for overflow from tank 1 to tank 2 then to tank 3, and for the overflow from drain. Each tank measures 152mm by 254mm deep with the overflow weirs being placed 228mm from the base. The tanks are stirred by propellers driven by the same drive motor. A rotameter controls the feed to tank 1. 2. A portable conductivity meter is available to measure the conductivity of the solution in each tank. It has four scales, which can be adjusted to give a reading in the desired range. For dilute solutions, conductivity is approximately proportional to salt concentration. The units on the meter must be checked. Symbols: Q = volumetric flowrate m3∙s-1 V = volume of tank m3 C = solution conductivity Siemens m-1 C0 = initial salt concentration in tank 1 Siemens m-1 C1 = conductivity in tank 1 at time t Siemens m-1 C2 = conductivity in tank 2 at time t Siemens m-1 C3 = conductivity in tank 3 at time t Siemens m-1 t = the time after start up s τ = mean residence time in each tank s Theory: In this experiment it is assumed that: a) salt concentration is proportional to solution conductivity (units: Siemens∙m-1) for dilute solution and does not vary very much with temperature b) complete mixing takes place in each tank instantaneously c) zero concentration corresponds to the concentration of the raw water supply, about 800 microSiemens∙cm-1 Tank 1: Water enters a tank flowrate, Q and a salt concentration of zero. The salt concentration in the tank is C0 at zero time and decreases with time as it is diluted by influent water. Taking an incremental mass balance for the slat when the concentration changes by  in time : i.e. Rearranging we get: Or: Where  Rearranging  gives: Integrating for time between 0 and t; and concentration between C0 and C1 gives: Rearrange to give: i.e.  decreases with time. Tank 2: By similar differential mass balance and integration, we get: Tank 3: Similarly, the time dependent concentration in tank 3 is given by: i.e. C3 increases initially with time then decreases. Procedure: Experiment 1 1. Fill the tanks with cold water directly from the tap initially, and then through the rotameter unit all tanks are full and flowing freely down the cascade. Stop the water flow and remove half of the water from the first tank. Place 50g of sodium chloride in a 500ml beaker, add some cold water and stir with a glass rod to dissolve the salt. Pour the solution into the first tank and start the stirrers to mix completely. 2. Start the water flow at 1.0 litre per min into the first tank using the rotameter (range 0.2 to 3.4 L/min). The water levels in each tank will automatically adjust until the flow in and out of each tank is the same. 3. The test starts when the water starts to overflow from tank 1 to tank 2. Start the clock at this point and take conductivity readings in each tank at various time intervals, e.g. every 1 or 2 minutes, so that graphs of conductivity in each tank can be drawn as a function of time. The solution conductivity is measured by dipping the probe into the tank and allowing a few seconds before taking a reading. It is not necessary to wash the probe between readings. Also it is not necessary to sample all tanks at exactly the same time provided that the correct time is recorded for each reading. 4. Take a sample of tap water in a clean beaker and measure the conductivity. (Typically the conductivity of the tap water about 700 to 800 microSiemens cm-1). Obtain standardised values of the conductivity in each tank by deducting the conductivity of tap water from the values measured in step 3. For example, when the salt solution is added to tank 1 in step 1 and the tank is filled with water to the weir from the rotameter, the conductivity should be about 2000 microSiemens cm-1. 1. 2. For tank 1 only, we have measured the standardised conductivity vs time. The profile for the standardised conductivity is as expected. The profile is a function of flowrate. 2.- For the experimental curves only, calculate the area under the graph for tank 1 up to t=5τ. Comment on the changes in values of area under curves with flowrate in your discussion. Area under the curve in figure 5 (solid-red line) with flowrate = 0.5 L/min is given as:  Experimentally Area under the curve in figure 5 (solid-green line) with flowrate = 1.0 L/min is given as:  Experimentally Area under the curve in figure 5 (solid-blue line) with flowrate = 2.0 L/min is given as:  Experimentally Discussion: In order to study the effect of residence time on conversion we have used a chemical process that involves the addition of reactants to a cascade of several continuous reactors in series i.e. CSTR. To simulate the conversion of reactant in a cascade of three continuous stirred tanks, we have measured/calculated the conductivity for three reactors in cascade. As we expected, we got three shapes in agreement with theory, i.e. we get the standardised conductivity vs time that obeys the equations for tank 1, tank 2 and tank 3 respectively. In order to describe the three conductivity profile. Initially, tank 1 expected an exponential decay of the standardised conductivity characterized by mean residence time that depends in volumetric flowrate and volume of tank. Reference: F.A.Holland and F.S.Chapman "Liquid Mixing and Processing in Stirred Tanks", p.109 (Chapman & Hall, 1996). J.JCarberry. "Chemical & Catalytic Reaction Engineering", p.92 (McGraw-Hill, 1976). 5. Deducting the conductivity of tap water (e.g. 800 mS∙cm-1) from this value gives the standardised value of the initial concentration: C0=(2000-800)=1200 mS∙cm-1 Experiment 2 Repeat the procedure in experiment 1 at water flowrates of 0.5 L/min and 2 L/min respectively, but measure the time dependent conductivity in tank 1 only. Results: Experiment 1 1. Plot graphs of standardised conductivity vs. time for each tank on the same graph and compare the shapes of the graphs. The figure 1 shows standardised conductivity vs time for each tank. The solid-red line shows tank 1 standardised conductivity vs time. The solid-blue line shows tank 2 standardised conductivity vs time. Whereas, the solid-green line shows tank 3 standardised conductivity vs time. The three standardised conductivity curves show the expected conductivity profiles with the expected mean residence time, . 2. Calculate theoretical values of conductivity for each tank using the relevant theoretical equations and the measured standardised value of C0. For each tank, plot theoretical and experimental conductivities on the same graph and compare the shapes of the graphs. The theoretical values of conductivity for each tank calculated from the equations vi, vii,viii is shown in the table below. For each tank, we show the theoretical and experimental conductivities in the figure 2, figure 3 and figure 3. 3.- For both the experimental and theoretical curves calculate the area under the graph for each tank up to t=5τ (you may use any numerical integration method, such as Simpson’s rule). The area represents reactant conversion and should be approximately the same for all tanks. Comment on the reasons for any differences in your discussion. Reactant conversion factor. Area under the curve in figure 2 (tank 1) is given as: Theoretically  Experimentally Area under the curve in figure 3 (tank 2) is given as: Theoretically  Experimenetally Area under the curve in the figure 4 (tank 3) is given as: Theoretically Experimentally With  and  Experiment 2 1.- For each flowrate (including the flowrate used in experiment 1), plot graphs of standardised conductivity vs. time for tank 1 only, on the same graph and compare the shapes of the graphs. 3. 4. 5. 6. 7. The theoretical values of conductivity for each tank is in excellent agreement with theoretical predictions, with standardised conductivity  and  Furthermore, we have calculated the area under the graph for each tank, in order to do so, we have implemented the numerical integration method Simpson’s rule using MATLAB. Both the experimental and theoretical area under the graph for each tank up to t=5τ. As we know, the area represents reactant conversion and should be approximately the same for all tanks. Because, theoretically it is equal to  affected by a numerical factor when we integrated over a finite range, for example: Tank 1, area under the curve over  Tank 2, area under the curve over  Tank 3, area under the curve over  We concluded that when the upper limit tends to infinity, we have the area under the curve equal to . The experimental reactant conversion is within an acceptable error 5.85%, 2.00% and 1.50% for tank 1, tank2 and tank 3, respectively. Finally, in experiment 2, we studied both the reactant conversion and standardised profile as a function of flowrate. We confirm the expected conductivity profile as a function of flowrate, however when the flowrate is equal to 2 L/min it seems that the conductivity is overestimated. We list all the MATLAB script used in the numerical analysis. Results Experiment 1 Flowrate = 1.0 L/min Measured C0 = 18.24 Tap water conductivity = 0.69 Standardised C0 = 17.55 Time (s) Tank 1 Measured conductivity (micro-Siemens/cm)   Tank 2 Measured conductivity (micro-Siemens/cm)   Tank 3 Measured conductivity (micro-Siemens/cm) Tank 1 Standardised conductivity (micro-Siemens/cm)   Tank 2 Standardised conductivity (micro-Siemens/cm)   Tank 3 Standardised conductivity (micro-Siemens/cm) 0 18.24 0.69 0.69 17.55 0 0 60 15.72 3.38 1.04 15.03 2.69 0.35 120 13.19 5.13 1.69 12.5 4.44 1 180 11.22 6.21 2.35 10.53 5.52 1.66 240 9.46 6.9 3.01 8.77 6.21 2.32 300 7.99 7.1 3.66 7.3 6.41 2.97 360 6.71 7.1 4.26 6.02 6.41 3.57 420 5.73 6.92 4.72 5.04 6.23 4.03 480 4.85 6.62 5.05 4.16 5.93 4.36 540 4.15 6.27 5.29 3.46 5.58 4.6 600 3.57 5.81 5.39 2.88 5.12 4.7 660 3.09 5.41 5.42 2.4 4.72 4.73 720 2.69 4.92 5.37 2 4.23 4.68 780 2.36 4.55 5.25 1.67 3.86 4.56 840 2.06 4.19 5.11 1.37 3.5 4.42 900 1.82 3.8 4.9 1.13 3.11 4.21 960 1.64 3.45 4.68 0.95 2.76 3.99 1020 1.45 3.13 4.43 0.76 2.44 3.74 1080 1.3 2.84 4.17 0.61 2.15 3.48 1140 1.17 2.56 3.91 0.48 1.87 3.22 1200 1.08 2.27 3.62 0.39 1.58 2.93 1260 1 2.1 3.4 0.31 1.41 2.71 1320 0.93 1.88 3.1 0.24 1.19 2.41 1380 0.89 1.66 2.91 0.2 0.97 2.22 1440 0.81 1.56 2.75 0.12 0.87 2.06 1500 0.76 1.45 2.53 0.07 0.76 1.84 1560 0.73 1.32 2.34 0.04 0.63 1.65 1620 0.69 1.22 2.14 0 0.53 1.45 Experiment 2 Measured C0 = 15.15 Tap water conductivity = 0.69 Standardised C0 = 14.46, 17.55,14.98 Time (s) Flowrate = 0.5 L/min Measured conductivity (micro-Siemens/cm)   Flowrate = 1.0 L/min Measured conductivity (micro-Siemens/cm)   Flowrate = 2.0 L/min Measured conductivity (micro-Siemens/cm) Flowrate = 0.5 L/min Standardised conductivity (micro-Siemens/cm)   Flowrate = 1.0 L/min Standardised conductivity (micro-Siemens/cm)   Flowrate = 2.0 L/min Standardised conductivity (micro-Siemens/cm) 0 15.15 18.24 15.67 14.46 17.55 14.98  120 13.24 13.19  13.82 12.55 12.50 13.13  240 11.42  9.46 11.8 10.73 8.77 11.11  360 9.87  6.71 9.95 9.18 6.02 9.26  480 8.56  4.85 8.37 7.87 4.16 7.68  600 7.4  3.57 7.12 6.71 2.88 6.43  720 6.44  2.69 6.03 5.75 2.00 5.34  840 5.62  2.06 5.16 4.93 1.37 4.47  960 4.88  1.64 4.38 4.19 0.95 3.69  1080 4.26  1.3 3.73 3.57 0.61 3.04  1200 3.71  1.08 3.21 3.02 0.39 2.52  1320 3.22  0.93 2.78 2.53 0.24 2.09  1440 2.84  0.81 2.42 2.15 0.12 1.73  1560 2.5  0.73 2.11 1.81 0.04 1.42  1680 2.21   1.85 1.52   1.16  1800 1.97   1.61 1.28   0.92  1920 1.75   1.44 1.06   0.75  2040 1.6   1.31 0.91   0.62  2160 1.42   1.17 0.73   0.48  2280 1.3   1.05 0.61   0.36  2400 1.2   0.98 0.51   0.29  2520 1.1   0.91 0.41   0.22  2640 1.03   0.86 0.34   0.17  2760 0.97   0.76 0.28   0.07  2880 0.9   0.72 0.21   0.03  3000 0.86   0.69 0.17   0.00  3120 0.82     0.13      3240 0.79     0.10     Revised: Dr S Larkai/Dr F Jahanzad (January 2012) Appendix MATLAB Scripts plot1.m close all; tau=0:60:1620; C1 = [ 17.55 15.03 12.5 10.53 8.77 7.3 6.02 5.04 4.16 3.46 2.88 2.4 2 1.67 1.37 1.13 0.95 0.76 0.61 0.48 0.39 0.31 0.24 0.2 0.12 0.07 0.04 0 ]; C2 = [ 0 2.69 4.44 5.52 6.21 6.41 6.41 6.23 5.93 5.58 5.12 4.72 4.23 3.86 3.5 3.11 2.76 2.44 2.15 1.87 1.58 1.41 1.19 0.97 0.87 0.76 0.63 0.53 ]; C3 = [ 0 0.35 1 1.66 2.32 2.97 3.57 4.03 4.36 4.6 4.7 4.73 4.68 4.56 4.42 4.21 3.99 3.74 3.48 3.22 2.93 2.71 2.41 2.22 2.06 1.84 1.65 1.45 ]; figure(1); hold on; plot(tau,C1,r,LineWidth,2); plot(tau,C2,b,LineWidth,2); plot(tau,C3,g,LineWidth,2); xlabel(Time (secods)); ylabel(Standardised Conductivity, C (microSiemens/cm)); title(Standardised Conductivity); legend(Tank 1, C_1,Tank 2, C_2,Tank 3, C_3); plot2a.m function plot2a clc; close all; clear all; time=0:60:1620; C1 = [ 17.55 15.03 12.5 10.53 8.77 7.3 6.02 5.04 4.16 3.46 2.88 2.4 2 1.67 1.37 1.13 0.95 0.76 0.61 0.48 0.39 0.31 0.24 0.2 0.12 0.07 0.04 0 ]; Q = 1.0*1.66666*10^(-5.0); % 1.0 litre/min = m^3/s V = 152.0 * 152.0 * 228.0 * (1e-3)*(1e-3)*(1e-3); tau = V/Q t=0:0.1:1600.0; c1=C1(1)*exp(-t/tau); figure(1); hold on; plot(time,C1,r,LineWidth,2); plot(t,c1,g,LineWidth,2); xlabel(Time (secods)); ylabel(Standardised Conductivity, C_1 (microSiemens/cm)); title(Standardised Conductivity, Tank 1); legend(Measured, C_1,Theoretical, C_1); %Simpsons Rule Integration % t = 0, 5tau % t = 0, 5tau t = 0.0 integral = 0.5* C1(1)*exp(-t/tau); while t < 5.0*tau integral = integral + C1(1)*exp(-t/tau); t = t+0.1; end integral = integral + 0.5* C1(1)*exp(-5.0*tau/tau); integral = integral*0.1; integral %integral = 5511.3 %integral = 5509.548325 t = 0.0; m_time = [t]; data = [C1(1)]; integral = 0.5* data(1); t=t+0.1; while t < 5.0*tau m_data = linealC(t,C1); integral = integral + m_data; data = [data m_data]; t=t+0.1; m_time = [m_time t]; end m_data = linealC(5.0*tau,C1); integral = integral + 0.5*m_data; integral = integral*0.1; integral figure(2); hold on; plot(time,C1,r,LineWidth,2); plot(m_time,data); end function C = linealC(value,C1) index = floor(value/60) + 1; C = (C1(index+1)-C1(index))/60.0 * (value - (index-1)*60.0) + C1(index); end plot2b.m function plot2b clc; close all; clear all; time=0:60:1620; C1 = [ 17.55 15.03 12.5 10.53 8.77 7.3 6.02 5.04 4.16 3.46 2.88 2.4 2 1.67 1.37 1.13 0.95 0.76 0.61 0.48 0.39 0.31 0.24 0.2 0.12 0.07 0.04 0 ]; C2 = [ 0 2.69 4.44 5.52 6.21 6.41 6.41 6.23 5.93 5.58 5.12 4.72 4.23 3.86 3.5 3.11 2.76 2.44 2.15 1.87 1.58 1.41 1.19 0.97 0.87 0.76 0.63 0.53 ]; Q = 1.0*1.66666*10^(-5.0); % 1.0 litre/min = m^3/s V = 152.0 * 152.0 * 228.0 * (1e-3)*(1e-3)*(1e-3); tau = V/Q t=0:0.1:1600.0; c2=C1(1)*(t/tau).*exp(-t/tau); figure(1); hold on; plot(time,C2,r,LineWidth,2); plot(t,c2,g,LineWidth,2); xlabel(Time (secods)); ylabel(Standardised Conductivity, C_2 (microSiemens/cm)); title(Standardised Conductivity, Tank 2); legend(Measured, C_2,Theoretical, C_2); %Simpsons Rule Integration % t = 0, 5tau % t = 0, 5tau t = 0.0 integral = 0.5* C1(1)*(t/tau).*exp(-t/tau); while t < 5.0*tau integral = integral + C1(1)*(t/tau).*exp(-t/tau); t = t+0.1; end integral = integral + 0.5* C1(1)*(t/tau).*exp(-t/tau); integral = integral*0.1; integral %integral = 5322.7 %integral = 5322.673952 t = 0.0; m_time = [t]; data = [C2(1)]; integral = 0.5* data(1); t=t+0.1; while t < 5.0*tau m_data = linealC(t,C2); integral = integral + m_data; data = [data m_data]; t=t+0.1; m_time = [m_time t]; end m_data = linealC(5.0*tau,C2); integral = integral + 0.5*m_data; integral = integral*0.1; integral figure(2); hold on; plot(time,C2,r,LineWidth,2); plot(m_time,data); end function C = linealC(value,C2) index = floor(value/60) + 1; C = (C2(index+1)-C2(index))/60.0 * (value - (index-1)*60.0) + C2(index); end plot2c.m function plot2c clc; close all; clear all; time=0:60:1620; C1 = [ 17.55 15.03 12.5 10.53 8.77 7.3 6.02 5.04 4.16 3.46 2.88 2.4 2 1.67 1.37 1.13 0.95 0.76 0.61 0.48 0.39 0.31 0.24 0.2 0.12 0.07 0.04 0 ]; C3 = [ 0 0.35 1 1.66 2.32 2.97 3.57 4.03 4.36 4.6 4.7 4.73 4.68 4.56 4.42 4.21 3.99 3.74 3.48 3.22 2.93 2.71 2.41 2.22 2.06 1.84 1.65 1.45 ]; Q = 1.0*1.66666*10^(-5.0); % 1.0 litre/min = m^3/s V = 152.0 * 152.0 * 228.0 * (1e-3)*(1e-3)*(1e-3); tau = V/Q t=0:0.1:1600.0; c3=0.5*C1(1)*((t/tau).*(t/tau)).*exp(-t/tau); figure(1); hold on; plot(time,C3,r,LineWidth,2); plot(t,c3,g,LineWidth,2); xlabel(Time (secods)); ylabel(Standardised Conductivity, C_3 (microSiemens/cm)); title(Standardised Conductivity, Tank 3); legend(Measured, C_3,Theoretical, C_3); %Simpsons Rule Integration % t = 0, 5tau % t = 0, 5tau t = 0.0 integral = 0.5* 0.5*C1(1)*((t/tau).*(t/tau)).*exp(-t/tau); while t < 5.0*tau integral = integral + 0.5*C1(1)*((t/tau).*(t/tau)).*exp(-t/tau); t = t+0.1; end integral = integral + 0.5* 0.5*C1(1)*((t/tau).*(t/tau)).*exp(-t/tau); integral = integral*0.1; integral %integral = 4855.6 %integral = 4855.488021 t = 0.0; m_time = [t]; data = [C3(1)]; integral = 0.5* data(1); t=t+0.1; while t < 5.0*tau m_data = linealC(t,C3); integral = integral + m_data; data = [data m_data]; t=t+0.1; m_time = [m_time t]; end m_data = linealC(5.0*tau,C3); integral = integral + 0.5*m_data; integral = integral*0.1; integral figure(2); hold on; plot(time,C3,r,LineWidth,2); plot(m_time,data); end function C = linealC(value,C3) index = floor(value/60) + 1; C = (C3(index+1)-C3(index))/60.0 * (value - (index-1)*60.0) + C3(index); end plot3.m function plot3c clc; close all; clear all; Q = 2.0*1.66666*10^(-5.0); % 1.0 litre/min = m^3/s V = 152.0 * 152.0 * 228.0 * (1e-3)*(1e-3)*(1e-3); tau = V/Q time1=0:120:3240; C1 = [ 14.46 12.55 10.73 9.18 7.87 6.71 5.75 4.93 4.19 3.57 3.02 2.53 2.15 1.81 1.52 1.28 1.06 0.91 0.73 0.61 0.51 0.41 0.34 0.28 0.21 0.17 0.13 0.10 ]; time2=0:120:1560; C2 = [ 17.55 12.50 8.77 6.02 4.16 2.88 2.00 1.37 0.95 0.61 0.39 0.24 0.12 0.04 ]; time3=0:120:3000; C3 = [ 14.98 13.13 11.11 9.26 7.68 6.43 5.34 4.47 3.69 3.04 2.52 2.09 1.73 1.42 1.16 0.92 0.75 0.62 0.48 0.36 0.29 0.22 0.17 0.07 0.03 0.00 ]; figure(1); hold on; plot(time1,C1,r,LineWidth,2); plot(time2,C2,g,LineWidth,2); plot(time3,C3,b,LineWidth,2); xlabel(Time (secods)); ylabel(Standardised Conductivity, C (microSiemens/cm)); title(Standardised Conductivity); legend(Measured, Tank 1, Flowrate = 0.5 L/min,... Measured, Tank 1, Flowrate = 1 L/min,... Measured, Tank 1, Flowrate = 2 L/min); plot3a.m function plot3a clc; close all; clear all; time=0:120:3240; C1 = [ 14.46 12.55 10.73 9.18 7.87 6.71 5.75 4.93 4.19 3.57 3.02 2.53 2.15 1.81 1.52 1.28 1.06 0.91 0.73 0.61 0.51 0.41 0.34 0.28 0.21 0.17 0.13 0.10 ]; Q = 0.5 * 1.66666*10^(-5.0); % 0.5 litre/min = m^3/s V = 152.0 * 152.0 * 228.0 * (1e-3)*(1e-3)*(1e-3); tau = V/Q t=0:0.1:3240.0; c1=C1(1)*exp(-t/tau); figure(1); hold on; plot(time,C1,r,LineWidth,2); plot(t,c1,g,LineWidth,2); xlabel(Time (secods)); ylabel(Standardised Conductivity, C_1 (microSiemens/cm)); title(Standardised Conductivity, Tank 1, Flowrate = 0.5 L/min); legend(Measured, C_1,Theoretical, C_1); %Simpsons Rule Integration % t = 0, 5tau % t = 0, 5tau t = 0.0 integral = 0.5* C1(1)*exp(-t/tau); while t < 5.0*tau integral = integral + C1(1)*exp(-t/tau); t = t+0.1; end integral = integral + 0.5* C1(1)*exp(-5.0*tau/tau); integral = integral*0.1; integral %integral = 5511.3 %integral = 5509.548325 t = 0.0; m_time = [t]; data = [C1(1)]; integral = 0.5* data(1); t=t+0.1; while t < 5.0*tau m_data = linealC(t,C1); integral = integral + m_data; data = [data m_data]; t=t+0.1; m_time = [m_time t]; end m_data = linealC(5.0*tau,C1); integral = integral + 0.5*m_data; integral = integral*0.1; integral figure(2); hold on; plot(time,C1,r,LineWidth,2); plot(m_time,data); end function C = linealC(value,C1) index = floor(value/120) + 1; C = (C1(index+1)-C1(index))/120.0 * (value - (index-1)*120.0) + C1(index); end plot3b.c function plot3b clc; close all; clear all; time=0:120:1560; C1 = [ 17.55 12.50 8.77 6.02 4.16 2.88 2.00 1.37 0.95 0.61 0.39 0.24 0.12 0.04 ]; Q = 1.0*1.66666*10^(-5.0); % 1.0 litre/min = m^3/s V = 152.0 * 152.0 * 228.0 * (1e-3)*(1e-3)*(1e-3); tau = V/Q t=0:0.1:1560.0; c1=C1(1)*exp(-t/tau); figure(1); hold on; plot(time,C1,r,LineWidth,2); plot(t,c1,g,LineWidth,2); xlabel(Time (secods)); ylabel(Standardised Conductivity, C_1 (microSiemens/cm)); title(Standardised Conductivity, Tank 1, Flowrate = 1 L/min); legend(Measured, C_1,Theoretical, C_1); %Simpsons Rule Integration % t = 0, 5tau % t = 0, 5tau t = 0.0 integral = 0.5* C1(1)*exp(-t/tau); while t < 5.0*tau integral = integral + C1(1)*exp(-t/tau); t = t+0.1; end integral = integral + 0.5* C1(1)*exp(-t/tau); integral = integral*0.1; integral %integral = 5322.7 %integral = 5322.673952 t = 0.0; m_time = [t]; data = [C1(1)]; integral = 0.5* data(1); t=t+0.1; while t < 1560 %5.0*tau m_data = linealC(t,C1); integral = integral + m_data; data = [data m_data]; t=t+0.1; m_time = [m_time t]; end m_data = linealC(1560-0.1,C1); integral = integral + 0.5*m_data; integral = integral*0.1; integral figure(2); hold on; plot(time,C1,r,LineWidth,2); plot(m_time,data); end function C = linealC(value,C1) index = floor(value/120) + 1; C = (C1(index+1)-C1(index))/120.0 * (value - (ind - (index-1)*120.0) + C1(index); end Read More
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