Question

What is the equivalent depth of water contained in a soil profile of 1 m deep, if the mass wetness of the upper 0.4 m is 15% and that of the lower 0.6 m is 25%? Assume a bulk density of 1200 kg/m$^3$ in the upper layer and 1400 kg/m$^3$ in the lower layer.

• A. 0.072
• B. 0.210
• C. 0.350
• D. 0.282

Hint:

Always convert mass wetness (\( w \)) to volumetric water content (\( \theta \)) first, using the relation \( \theta = w \cdot \text{Specific Gravity of bulk soil} \).
Specific gravity \( = \frac{\rho_b}{\rho_w} \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To find the equivalent depth of water in a soil profile, we must convert the gravimetric water content (mass wetness, $w$) to volumetric water content ($\theta$) for each layer, and then multiply by the thickness of the respective layer.
Key Formula or Approach:
1. Volumetric water content ($\theta$) = $w \times \frac{\rho_b}{\rho_w}$, where $\rho_b$ is the bulk density of soil and $\rho_w$ is the density of water ($1000 \text{ kg/m}^3$)
2. Equivalent depth of water ($D$) = $\theta \times d$, where $d$ is the thickness of the layer.
3. Total depth of water = $D_1 + D_2$

Step 2: Detailed Explanation:
Let us calculate the water depth for each layer:
1. For the Upper Layer (thickness $d_1 = 0.4 \text{ m}$):
- Mass wetness ($w_1$) = $15\% = 0.15$
- Soil bulk density ($\rho_{b1}$) = $1200 \text{ kg/m}^3$
- Volumetric water content ($\theta_1$):
\[ \theta_1 = 0.15 \times \frac{1200}{1000} = 0.15 \times 1.2 = 0.18 \] - Equivalent water depth ($D_1$):
\[ D_1 = \theta_1 \times d_1 = 0.18 \times 0.4 \text{ m} = 0.072 \text{ m} \] 2. For the Lower Layer (thickness $d_2 = 0.6 \text{ m}$):
- Mass wetness ($w_2$) = $25\% = 0.25$
- Soil bulk density ($\rho_{b2}$) = $1400 \text{ kg/m}^3$
- Volumetric water content ($\theta_2$):
\[ \theta_2 = 0.25 \times \frac{1400}{1000} = 0.25 \times 1.4 = 0.35 \] - Equivalent water depth ($D_2$):
\[ D_2 = \theta_2 \times d_2 = 0.35 \times 0.6 \text{ m} = 0.210 \text{ m} \] 3. Total Equivalent Depth of Water ($D$):
\[ D = D_1 + D_2 = 0.072 \text{ m} + 0.210 \text{ m} = 0.282 \text{ m} \]

Step 3: Final Answer:
The total equivalent depth of water in the 1 m profile is 0.282 m, corresponding to option (D).