Let's break down **Problem 1** step by step. --- ## **Given Information** - ** tube is coated a thin layer of soluble film.** - solvent is pumped laminar flow dissolve the film. - The time to remove the coating is estimated by: \[ M() - M(t) = \frac{3}{2\pi(4/3)C_s R_c}\left(\frac{D_{AB}C_s}{Q_{avg}}\frac{8w_{avg}}{32Q}\right)^{1/2} t^{1/2} \] Where: - \(M()\): Initial amount of coating - \(M(t)\): Amount at time \(t\) - \(C_s\): Solubility concentration - \(R_c\): Tube radius - \(D_{AB}\): Diffusivity - \(Q_{avg}\): Average flow rate - \(w_{avg}\): Average width Also, time to completely remove the thin film: \[ t = .91 \frac{[M()/C_s R_c]^2}{D_{AB}G(Q_{avg}/D_{AB})} \] --- ## **Step 1: Answering Part (a)** **If it takes 3 minutes to remove 20% of the coating, how long to remove it completely?** ### **Step 1.1: Proportionality** Let: - \(t_{20\%}\): time to remove 20% (\(= 3\) min) - \(t_{100\%}\): time to remove 100% (what we want to find) From the equation: \[ M() - M(t) \propto t^{1/2} \] Let \(k = \text{proportionality constant}\): \[ M() - M(t) = k t^{1/2} \] For 20% removal: \[ M() - M(t_{20\%}) = .2 M() = k (t_{20\%})^{1/2} \] \[ k = \frac{.2 M()}{(t_{20\%})^{1/2}} \] For 100% removal: \[ M() - M(t_{100\%}) = M() = k (t_{100\%})^{1/2} \] Substitute \(k\): \[ M() = \frac{.2 M()}{(t_{20\%})^{1/2}} (t_{100\%})^{1/2} \] \[ 1 = .2 \left(\frac{t_{100\%}}{t_{20\%}}\right)^{1/2} \] \[ \left(\frac{t_{100\%}}{t_{20\%}}\right)^{1/2} = 5 \] \[ \frac{t_{100\%}}{t_{20\%}} = 25 \] \[ t_{100\%} = 25 \cdot t_{20\%} \] \[ t_{100\%} = 25 \cdot 3 = \boxed{75\ \text{minutes}} \] --- ### **Final Answer (a):** \[ \boxed{75\ \text{minutes}} \] to remove all the coating from the tube. --- ## **Step 2: Answering Part (b)** **If two shorter tubes are cleaned together with the same coating thickness, total volume of solvent, and solvent is recycled, how will the time compare?** ### **Explanation:** - **Each tube gets half the solvent flow rate (since the total is fixed and tubes are in parallel).** - **For each tube:** - The removal time depends on \(Q_{avg}\) (average flow rate). - Halving the flow rate decreases the mass transfer rate (since the term \(Q_{avg}\) is in the denominator inside the square root). - **From the equation:** \[ M() - M(t) \propto \left(\frac{D_{AB}C_s}{Q_{avg}}\right)^{1/2} t^{1/2} \] - **If \(Q_{avg}\) is halved, the denominator inside the square root is halved, so:** \[ \left(\frac{1}{Q_{avg}/2}\right)^{1/2} = \sqrt{2} \left(\frac{1}{Q_{avg}}\right)^{1/2} \] - **This increases the left side by \(\sqrt{2}\), meaning for the same amount removed, the time required is reduced by a factor of 2.** But for total removal, \[ t_{new} = \left(\frac{Q_{old}}{Q_{new}}\right) t_{old} = 2 t_{old} \] So, **it will take twice as long to completely remove the coating from both tubes** as it would to remove it from a single tube with all the solvent. --- ### **Final Answer (b):** \[ \boxed{ \text{It will take twice as long to clean the two tubes in parallel with half the flow rate each, compared to cleaning one tube with the full flow rate.} } \] --- ## **Summary** - **(a)** To clean the tube completely: **75 minutes**. - **(b)** Cleaning two tubes in parallel with half the flow rate each takes **twice as long** as cleaning one tube with the full flow rate.

# Problem 1: Coating Removal from Tubes ## Given Information - **Coating Removal Formula**: \[ M() - M(t) = \frac{3}{2\pi(4/3)C_s R_c}\left(\frac{D_{AB}C_s}{Q_{avg}}\frac{8w_{avg}}{32Q}\right)^{1/2} t^{1/2} \] - **Complete Removal Time**: \[ t = 0.91 \frac{[M()/C_s R_c]^2}{D_{AB}G(Q_{avg}/D_{AB})} \] ## Part (a): Time to Remove 100% of Coating 1. **Given**: Takes 3 minutes to remove 20%. - \(t_{20\%} = 3\) minutes 2. **Using Proportionality**: - \(M() - M(t) \propto t^{1/2}\) 3. **Set up the equations**: - For 20% removal: \[ 0.2 M() = k (3)^{1/2} \] - For 100% removal: \[ M() = k (t_{100\%})^{1/2} \] 4. **Relate \(t_{100\%}\) to \(t_{20\%}\)**: \[ 1 = 0.2 \left(\frac{t_{100\%}}{3}\right)^{1/2} \] \[ \frac{t_{100\%}}{3} = 25 \implies t_{100\%} = 75 \text{ minutes} \] ### **Final Answer (a)**: \[ \boxed{75\ \text{minutes}} \] --- ## Part (b): Comparing Time for Two Tubes 1. **Scenario**: Two tubes with the same coating thickness and total solvent volume. 2. **Effect**: Each tube receives half the solvent flow rate. 3. **Time Comparison**: - The mass transfer rate decreases due to reduced flow rate. - Time required for each tube doubles because: \[ t_{new} = 2 \cdot t_{old} \] ### **Final Answer (b)**: \[ \boxed{\text{It will take twice as long to clean the two tubes compared to cleaning one tube.}} \]

# Problem 1: Coating Removal from Tubes ## Given Information - **Coating Removal Formula**: \[ M_0 - M(t) = \frac{3}{2\pi(4/3)C_s R_c}\left(\frac{D_{AB}C_s}{Q_{avg}}\frac{8w_{avg}}{32Q}\right)^{1/2} t^{1/2} \] - **Complete Removal Time**: \[ t = 0.91 \frac{[M_0/C_s R_c]^2}{D_{AB}G(Q_{avg}/D_{AB})} \] Where: - \(M_0\): Initial amount of coating - \(M(t)\): Amount at time \(t\) ## Part (a): Time to Remove 100% of Coating 1. **Given**: Takes 3 minutes to remove 20% of the coating. - Let \(t_{20\%} = 3\) minutes. 2. **Using Proportionality**: - The removal is proportional to \(t^{1/2}\). 3. **Equation Setup**: - For 20% removal: \[ M_0 - M(t_{20\%}) = 0.2 M_0 = k (3)^{1/2} \] - For 100% removal: \[ M_0 - M(t_{100\%}) = M_0 = k (t_{100\%})^{1/2} \] 4. **Relate \(t_{100\%}\) to \(t_{20\%}\)**: - Substitute \(k\): \[ 1 = 0.2 \left(\frac{t_{100\%}}{3}\right)^{1/2} \] - Solve for \(t_{100\%}\): \[ \left(\frac{t_{100\%}}{3}\right)^{1/2} = 5 \implies t_{100\%} = 25 \cdot 3 = 75 \text{ minutes} \] ### **Final Answer (a)**: \[ \boxed{75\ \text{minutes}} \] --- ## Part (b): Comparing Time for Two Tubes 1. **Scenario**: Two tubes with the same coating thickness, total solvent volume, and solvent recycled. 2. **Effect of Flow Rate**: - Each tube receives half the solvent flow rate. - This decreases the mass transfer rate, leading to increased time. 3. **Time Comparison**: - The time required to remove the coating from each tube doubles: \[ t_{new} = 2 \cdot t_{old} \] ### **Final Answer (b)**: \[ \boxed{\text{It will take twice as long to clean the two tubes compared to cleaning one tube.}} \]