Let's solve the given questions step by step. --- # Step-by-Step Solution ## Given Data | Time (min) | Cumulative deformation (μm) | |------------|-----------------------------| | . | 240 | | .10 | 318 | | .25 | 340 | | .50 | 360 | | 1.00 | 385 | | 2.00 | 415 | | 4.00 | 464 | | 8.00 | 530 | | 15. | 622 | | 30. | 738 | | 60. | 842 | | 120 | 930 | ### Notation - Initial deformation, \( \delta_ = 240 \) μm - Final deformation, \( \delta_f = 930 \) μm --- ## (a) Estimate the coefficient of consolidation (\( C_v \)) in cm²/s ### 1. Calculate Degree of Consolidation (\( U \)) \[ U = \frac{\delta_t - \delta_}{\delta_f - \delta_} \] ### 2. Find time for 50% consolidation (\( t_{50} \)) First, calculate the deformation at 50% consolidation: \[ U_{50} = .5 \] \[ \delta_{50} = \delta_ + .5 (\delta_f - \delta_) = 240 + .5 \times (930 - 240) = 240 + 345 = 585 \text{ μm} \] Now, interpolate the time at which cumulative deformation is 585 μm. From the table: - At 8.00 min: 530 μm - At 15. min: 622 μm So, between 8 and 15 min, use linear interpolation: \[ t_{50} = 8 + \frac{585 - 530}{622 - 530} \times (15 - 8) \] \[ = 8 + \frac{55}{92} \times 7 \] \[ = 8 + .5978 \times 7 \] \[ = 8 + 4.185 \] \[ = 12.19 \text{ min} \] ### 3. Use the Formula for \( C_v \) For a standard consolidation test, the degree of consolidation at 50% (\( U_{50} \)) corresponds to: \[ T_{v,50} = .197 \] Where \( T_v \) is the time factor: \[ T_v = \frac{C_v t}{H^2} \] - \( t = t_{50} \) (in seconds) - \( H \) = maximum drainage path (half the sample height if double drainage, full height if single) **Assume sample height \( H = 1 \) cm (if not given, use symbolically):** \[ C_v = \frac{T_{v,50} H^2}{t_{50}} \] Convert \( t_{50} \) to seconds: \[ t_{50} = 12.19 \text{ min} = 731.4 \text{ s} \] \[ C_v = \frac{.197 \cdot H^2}{731.4} \] If \( H = 1 \) cm, \[ C_v = \frac{.197 \cdot 1^2}{731.4} = 2.69 \times 10^{-4} \text{ cm}^2/\text{s} \] --- ## (b) Time for 60% consolidation (\( U_{60} = .6 \)), in years ### 1. Find deformation at 60% consolidation \[ \delta_{60} = \delta_ + .6 (\delta_f - \delta_) = 240 + .6 \times (930 - 240) = 240 + .6 \times 690 = 240 + 414 = 654 \text{ μm} \] From the table: - At 15. min: 622 μm - At 30. min: 738 μm Linear interpolation: \[ t_{60} = 15 + \frac{654 - 622}{738 - 622} \times (30 - 15) \] \[ = 15 + \frac{32}{116} \times 15 \] \[ = 15 + .276 \times 15 \] \[ = 15 + 4.14 \] \[ = 19.14 \text{ min} \] Convert to seconds: \[ t_{60} = 19.14 \times 60 = 1148.4 \text{ s} \] ### 2. Use time factor for 60% consolidation For \( U = .6 \): - \( T_{v,60} \approx .283 \) \[ T_v = \frac{C_v t}{H^2} \] \[ t_{60} = \frac{T_{v,60} H^2}{C_v} \] Using \( C_v = 2.69 \times 10^{-4} \) cm²/s (from earlier): \[ t_{60} = \frac{.283 H^2}{2.69 \times 10^{-4}} \] If \( H = 1 \) cm, \[ t_{60} = \frac{.283}{2.69 \times 10^{-4}} = 1052 \text{ s} \] Convert seconds to years: \[ 1 \text{ year} = 365 \times 24 \times 360 = 31,536,000 \text{ s} \] So, \[ \text{Time in years} = \frac{1052}{31,536,000} = 3.34 \times 10^{-5} \text{ years} \] --- ## **Final Answers** ### (a) Coefficient of consolidation (\( C_v \)): \[ \boxed{2.69 \times 10^{-4} \text{ cm}^2/\text{s}} \] *(Assuming sample height, \(H = 1\) cm; scale for your actual sample height if given)* ### (b) Time for 60% consolidation: \[ \boxed{3.34 \times 10^{-5} \text{ years} \approx 18 \text{ minutes}} \] *(For \(H = 1\) cm; scale for your actual sample height if given)* --- **If you provide the actual sample height, update \(H\) in the formulas above!**