Question

For a clayey soil stratum, the time required for degree of consolidation from 25 % to 50 % is 30 days.

The total time (in days) required for 90 % degree of consolidation of the same soil stratum is (rounded off to the nearest integer).

Hint:

Use the parabolic $T_v = (\pi/4)U^2$ relation for the 25% and 50% points to find the fixed $d^2/C_v$ constant, then switch to the $1.781-0.933\log_{10}(100-U)$ formula for the 90% point (since U exceeds 60%).

Answer 1

Correct Answer: 173

Step 1: Use a ratio method instead of solving for $d^2/C_v$ as a separate number first.
Since $t = T_v \cdot (d^2/C_v)$ for every degree of consolidation, and $d^2/C_v$ is the same fixed constant throughout (it depends only on the soil and drainage path, not on how far consolidation has progressed), we can write a ratio that connects any two stages directly:
\[ \frac{t_{50}-t_{25}}{T_{v50}-T_{v25}} = \frac{t_{90}}{T_{v90}} = \frac{d^2}{C_v} \]

Step 2: Compute $T_{v25}$ and $T_{v50}$.
\[ T_{v25} = \frac{\pi}{4}(0.25)^2 = 0.7854 \times 0.0625 = 0.04909 \]
\[ T_{v50} = \frac{\pi}{4}(0.50)^2 = 0.7854 \times 0.25 = 0.19635 \]
\[ T_{v50}-T_{v25} = 0.19635 - 0.04909 = 0.14726 \]

Step 3: Compute $T_{v90}$ using the greater-than-60% formula.
\[ T_{v90} = 1.781 - 0.933\log_{10}(10) = 0.848 \]

Step 4: Set up the ratio equation and solve for $t_{90}$ directly, without ever isolating $d^2/C_v$ as a standalone number.
\[ \frac{30}{0.14726} = \frac{t_{90}}{0.848} \]
\[ t_{90} = 0.848 \times \frac{30}{0.14726} = 0.848 \times 203.72 \]
\[ t_{90} = 172.75 \text{ days} \]

Step 5: Round.
This ratio route reaches the same figure as computing the drainage constant explicitly, since it is really the same algebra rearranged, just skipping the naming of $d^2/C_v$ as an intermediate quantity.
\[ \boxed{t_{90} \approx 173 \text{ days}} \]