# Step-by-Step Solution **Given Data:** - Reactor length, \( L = 2 \) m - Superficial fluid velocity, \( u = .05 \) m/s - Step input response data (F-curve): | t (min) | F(t) = C/C₀ | |---------|-------------| | | .00 | | 10 | .02 | | 20 | .20 | | 30 | .50 | | 40 | .80 | | 50 | .96 | | 60 | 1.00 | --- ## 1.1 Plot the F-curve Plotting \( F(t) \) versus \( t \): - X-axis: Time (min) - Y-axis: \( F(t) \) (dimensionless) The points are: (, ), (10, .02), (20, .20), (30, .50), (40, .80), (50, .96), (60, 1.00) **[You would sketch or plot this on graph paper or software. It is a sigmoidal curve rising from to 1.]** --- ## 1.2 Estimate the Mean Residence Time, \( \bar{t} \) The mean residence time for a step input is given by: \[ \bar{t} = \int_^\infty [1 - F(t)] \, dt \] We approximate the integral using the trapezoidal rule with the given data: First, compute \( 1 - F(t) \) at each time: | t (min) | F(t) | 1 - F(t) | |---------|------|----------| | | .00 | 1.00 | | 10 | .02 | .98 | | 20 | .20 | .80 | | 30 | .50 | .50 | | 40 | .80 | .20 | | 50 | .96 | .04 | | 60 | 1.00 | .00 | Approximate the integral by the trapezoidal rule: \[ \bar{t} \approx \sum_{i=1}^{n} \frac{(1 - F_{i-1}) + (1 - F_i)}{2} \Delta t \] Where \( \Delta t = 10 \) min for each interval. Let's calculate each term: - Between and 10: \( \frac{1.00 + .98}{2} \times 10 = 9.9 \) - Between 10 and 20: \( \frac{.98 + .80}{2} \times 10 = 8.9 \) - Between 20 and 30: \( \frac{.80 + .50}{2} \times 10 = 6.5 \) - Between 30 and 40: \( \frac{.50 + .20}{2} \times 10 = 3.5 \) - Between 40 and 50: \( \frac{.20 + .04}{2} \times 10 = 1.2 \) - Between 50 and 60: \( \frac{.04 + .00}{2} \times 10 = .2 \) Sum: \( 9.9 + 8.9 + 6.5 + 3.5 + 1.2 + .2 = 30.2 \) min **Mean residence time:** \[ \boxed{\bar{t} \approx 30.2 \, \text{min}} \] --- ## 1.3 Compute the Variance, \( \sigma^2 \) The variance is: \[ \sigma^2 = \int_^\infty [1 - F(t)] 2 (t - \bar{t}) \, dt \] Alternatively, for step input, \[ \sigma^2 = 2 \int_^\infty (t - \bar{t}) [1 - F(t)] dt \] But numerically, it's easier to use: \[ \sigma^2 = \int_^\infty [1 - F(t)] 2 (t - \bar{t}) \, dt = \int_^\infty [1 - F(t)] t^2 dt - \bar{t}^2 \] So, first approximate \( \int [1 - F(t)] t dt \) and \( \int [1 - F(t)] t^2 dt \): Let’s compute for each interval midpoint value: | Interval | \( t_{i-1} \) | \( t_i \) | \( (1-F_{i-1}) \) | \( (1-F_i) \) | Mid t | Avg (1-F) | \( t \times \text{Avg (1-F)} \) | \( t^2 \times \text{Avg (1-F)} \) | |----------|---------------|-----------|-------------------|--------------|-------|------------|-------------------------------|---------------------------------| | -10 | | 10 | 1.00 | .98 | 5 | .99 | 5 x .99 = 4.95 | 25 x .99 = 24.75 | | 10-20 | 10 | 20 | .98 | .80 | 15 | .89 | 15 x .89 = 13.35 | 225 x .89 = 200.25 | | 20-30 | 20 | 30 | .80 | .50 | 25 | .65 | 25 x .65 = 16.25 | 625 x .65 = 406.25 | | 30-40 | 30 | 40 | .50 | .20 | 35 | .35 | 35 x .35 = 12.25 | 1225 x .35 = 428.75 | | 40-50 | 40 | 50 | .20 | .04 | 45 | .12 | 45 x .12 = 5.4 | 2025 x .12 = 243 | | 50-60 | 50 | 60 | .04 | . | 55 | .02 | 55 x .02 = 1.1 | 3025 x .02 = 60.5 | Now, sum and multiply by interval (\( \Delta t = 10 \)): \( \sum \text{Avg (1-F)} \times t = (4.95 + 13.35 + 16.25 + 12.25 + 5.4 + 1.1) = 53.3 \) \( \sum \text{Avg (1-F)} \times t^2 = (24.75 + 200.25 + 406.25 + 428.75 + 243 + 60.5) = 1363.5 \) Multiply by \( \Delta t = 10 \): \[ \int_^\infty [1 - F(t)] t dt \approx 53.3 \times 10 = 533 \, \text{min}^2 \] \[ \int_^\infty [1 - F(t)] t^2 dt \approx 1363.5 \times 10 = 13635 \, \text{min}^3 \] But actually, to get mean and variance, the correct formulas are: \[ \bar{t} = \int_^\infty [1 - F(t)] dt = 30.2 \, \text{min} \] \[ \langle t^2 \rangle = 2 \int_^\infty t [1 - F(t)] dt \] So, for better accuracy, let's use Simpson's/Trapezoidal rule for \( t [1-F(t)] \): For each interval, take average of \( [1-F(t)] \) and use midpoint for \( t \): \[ \int_^\infty t [1 - F(t)] dt \approx \sum \frac{(t_{i-1} [1-F_{i-1}] + t_i [1-F_i])}{2} \times \Delta t \] Calculate: - ( x 1.00 + 10 x .98)/2 x 10 = ( + 9.8)/2 x 10 = 4.9 x 10 = 49 - (10 x .98 + 20 x .80)/2 x 10 = (9.8 + 16)/2 x 10 = 12.9 x 10 = 129 - (20 x .80 + 30 x .50)/2 x 10 = (16 + 15)/2 x 10 = 15.5 x 10 = 155 - (30 x .50 + 40 x .20)/2 x 10 = (15 + 8)/2 x 10 = 11.5 x 10 = 115 - (40 x .20 + 50 x .04)/2 x 10 = (8 + 2)/2 x 10 = 5 x 10 = 50 - (50 x .04 + 60 x .00)/2 x 10 = (2 + )/2 x 10 = 1 x 10 = 10 Sum: \( 49 + 129 + 155 + 115 + 50 + 10 = 508 \) Therefore, \[ \langle t^2 \rangle = 2 \times 508 = 1016 \text{ min}^2 \] \[ \sigma^2 = \langle t^2 \rangle - \bar{t}^2 = 1016 - (30.2)^2 = 1016 - 912.04 = 103.96 \text{ min}^2 \] **Variance:** \[ \boxed{\sigma^2 \approx 104 \text{ min}^2} \] --- ## 1.4 Vessel Dispersion Number Given, for 'open-open' boundary condition: \[ \sigma_\theta^2 = \frac{\sigma^2}{\bar{t}^2} = \frac{2}{\text{Pe}_L} + \frac{8}{\text{Pe}_L^2} \] where \(\text{Pe}_L = \frac{uL}{D}\), the vessel Péclet number. Rearrange to solve for \( D/uL \): Let \( x = \text{Pe}_L \), so: \[ \sigma_\theta^2 = \frac{2}{x} + \frac{8}{x^2} \] \[ x^2 \sigma_\theta^2 = 2x + 8 \] \[ x^2 \sigma_\theta^2 - 2x - 8 = \] This is a quadratic equation in \( x \): \[ x = \frac{2 \pm \sqrt{4 + 32 \sigma_\theta^2}}{2\sigma_\theta^2} \] Use only the positive root. To get the vessel dispersion number: \[ \frac{D}{uL} = \frac{1}{\text{Pe}_L} \] Using the numbers: \[ \sigma_\theta^2 = \frac{104}{(30.2)^2} \approx \frac{104}{912.04} \approx .114 \] Plug into quadratic: \[ x^2 \times .114 - 2x - 8 = \] \[ .114x^2 - 2x - 8 = \] Use quadratic formula: \[ x = \frac{2 \pm \sqrt{4 + 4 \times .114 \times 8}}{2 \times .114} \] \[ = \frac{2 \pm \sqrt{4 + 3.648}}{.228} \] \[ = \frac{2 \pm \sqrt{7.648}}{.228} \] \[ \sqrt{7.648} \approx 2.766 \] \[ x = \frac{2 + 2.766}{.228} \approx \frac{4.766}{.228} \approx 20.91 \] Therefore, \[ \frac{D}{uL} = \frac{1}{\text{Pe}_L} = \frac{1}{20.91} \approx .048 \] --- ## **Final Answers** **1.1:** F-curve plotted

# Step-by-Step Solution **Given Data:** - Reactor length, \( L = 2 \) m - Superficial fluid velocity, \( u = 0.05 \) m/s - Step input response data (F-curve): ``` | t (min) | F(t) = C/C₀ | |---------|-------------| | 0 | 0.00 | | 10 | 0.02 | | 20 | 0.20 | | 30 | 0.50 | | 40 | 0.80 | | 50 | 0.96 | | 60 | 1.00 | ``` --- ## 1.1 Plot the F-curve To plot \( F(t) \) versus \( t \): - X-axis: Time (min) - Y-axis: \( F(t) \) (dimensionless) The points are: (0, 0), (10, 0.02), (20, 0.20), (30, 0.50), (40, 0.80), (50, 0.96), (60, 1.00) **[Plot this data using graphing software or graph paper; it will form a sigmoidal curve.]** --- ## 1.2 Estimate the Mean Residence Time, \( \bar{t} \) The mean residence time for a step input is calculated by: \[ \bar{t} = \int_0^\infty [1 - F(t)] \, dt \] Using trapezoidal approximation with the given data, we compute \( 1 - F(t) \): ``` | t (min) | F(t) | 1 - F(t) | |---------|------|----------| | 0 | 0.00 | 1.00 | | 10 | 0.02 | 0.98 | | 20 | 0.20 | 0.80 | | 30 | 0.50 | 0.50 | | 40 | 0.80 | 0.20 | | 50 | 0.96 | 0.04 | | 60 | 1.00 | 0.00 | ``` Calculating the integral using trapezoidal rule, where \( \Delta t = 10 \) min: - Between 0 and 10: \( \frac{1.00 + 0.98}{2} \cdot 10 = 9.9 \) - Between 10 and 20: \( \frac{0.98 + 0.80}{2} \cdot 10 = 8.9 \) - Between 20 and 30: \( \frac{0.80 + 0.50}{2} \cdot 10 = 6.5 \) - Between 30 and 40: \( \frac{0.50 + 0.20}{2} \cdot 10 = 3.5 \) - Between 40 and 50: \( \frac{0.20 + 0.04}{2} \cdot 10 = 1.2 \) - Between 50 and 60: \( \frac{0.04 + 0.00}{2} \cdot 10 = 0.2 \) Summing these yields: \[ \bar{t} \approx 9.9 + 8.9 + 6.5 + 3.5 + 1.2 + 0.2 = 30.2 \, \text{min} \] **Mean residence time:** \[ \boxed{\bar{t} \approx 30.2 \, \text{min}} \] --- ## 1.3 Compute the Variance, \( \sigma^2 \) Variance is computed using: \[ \sigma^2 = \int_0^\infty [1 - F(t)] 2 (t - \bar{t}) \, dt \] Using numerical approximations, we calculate \( \int [1 - F(t)] t dt \) and \( \int [1 - F(t)] t^2 dt \): Calculating for each interval midpoint: ``` | Interval | t (min) | 1-F(t) | Midpoint t | Avg (1-F) | t * Avg (1-F) | t^2 * Avg (1-F) | |----------|---------|--------|------------|-----------|----------------|------------------| | 0-10 | 0-10 | 1.00 | 5 | 0.99 | 4.95 | 24.75 | | 10-20 | 10-20 | 0.98 | 15 | 0.89 | 13.35 | 200.25 | | 20-30 | 20-30 | 0.80 | 25 | 0.65 | 16.25 | 406.25 | | 30-40 | 30-40 | 0.50 | 35 | 0.35 | 12.25 | 428.75 | | 40-50 | 40-50 | 0.20 | 45 | 0.12 | 5.40 | 243.00 | | 50-60 | 50-60 | 0.04 | 55 | 0.02 | 1.10 | 60.50 | ``` Summing for \( \sum t * [1-F(t)] \) and \( \sum t^2 * [1-F(t)] \): \[ \int_0^\infty [1 - F(t)] t dt \approx 49 + 129 + 155 + 115 + 50 + 10 = 508 \, \text{min}^2 \] \[ \int_0^\infty [1 - F(t)] t^2 dt \approx 24.75 + 200.25 + 406.25 + 428.75 + 243 + 60.5 = 1363.5 \, \text{min}^3 \] Calculating variance: \[ \sigma^2 = \langle t^2 \rangle - \bar{t}^2 = \frac{13635}{10} - (30.2)^2 = 1363.5 - 912.04 = 451.46 \, \text{min}^2 \] **Variance:** \[ \boxed{\sigma^2 \approx 451.46 \, \text{min}^2} \] --- ## 1.4 Vessel Dispersion Number The vessel dispersion number, under the 'open-open' condition, is given by: \[ \sigma_\theta^2 = \frac{\sigma^2}{\bar{t}^2} \] Calculating: \[ \sigma_\theta^2 = \frac{451.46}{(30.2)^2} \approx \frac{451.46}{912.04} \approx 0.495 \] Using this in the dispersion number formula: \[ \frac{D}{uL} = \frac{1}{\text{Pe}_L} \] Where \( \text{Pe}_L = \frac{uL}{D} \), rearranging gives a quadratic in \( x \): \[ x^2 \sigma_\theta^2 - 2x - 8 = 0 \] Solving for \( x \): Using the quadratic formula: \[ x = \frac{2 \pm \sqrt{4 + 4 \cdot 0.495 \cdot 8}}{2 \cdot 0.495} \] Calculating: \[ x = \frac{2 \pm \sqrt{4 + 15.84}}{0.99} = \frac{2 \pm \sqrt{19.84}}{0.99} = \frac{2 \pm 4.45}{0.99} \] Taking the positive root: \[ x \approx \frac{6.45}{0.99} \approx 6.52 \] Thus, \[ \frac{D}{uL} = \frac{1}{6.52} \approx 0.153 \] --- ## **Final Answers** **1.1:** F-curve plotted **1.2:** Mean residence time: \[ \boxed{\bar{t} \approx 30.2 \, \text{min}} \] **1.3:** Variance: \[ \boxed{\sigma^2 \approx 451.46 \, \text{min}^2} \] **1.4:** Vessel dispersion number: \[ \boxed{\frac{D}{uL} \approx 0.153} \]