A field crop is irrigated when the available soil water reduces to 60%. The moisture content at field capacity and wilting point are 32% and 12%, respectively. The bulk density of the soil is 1.5 g/cm³. The field water application efficiency is 75%, and the crop root zone depth is 50 cm. The gross depth of irrigation required to bring soil moisture content to field capacity in cm is

To calculate the gross depth of irrigation required to bring the soil moisture content back to field capacity, we can follow these steps:

### Step 1: Determine the available water content (AWC)
The available water content is the difference between the moisture content at field capacity and the moisture content at the wilting point.

\[
\text{AWC} = \text{Field Capacity} - \text{Wilting Point} 
\]
\[
\text{AWC} = 32\% - 12\% = 20\%
\]

### Step 2: Determine the moisture depletion at 60% available water
The crop is irrigated when the available soil water is reduced to 60%. So, the soil has only 60% of the available water and needs to be refilled to field capacity (100% available water).

\[
\text{Moisture depletion} = (100\% - 60\%) \times \text{AWC}
\]
\[
\text{Moisture depletion} = 40\% \times 20\% = 8\%
\]

Thus, the soil moisture deficit that needs to be refilled is 8%.

### Step 3: Calculate the depth of irrigation water required
To calculate the required irrigation depth, we use the following formula:

\[
\text{Net Irrigation Depth (d)} = \frac{\text{Moisture Deficit} \times \text{Root Zone Depth} \times \text{Bulk Density}}{1}
\]

Given:
- Moisture Deficit = 8% = 0.08
- Root Zone Depth = 50 cm
- Bulk Density = 1.5 g/cm³ = 1.5 g/cm³ × 1 cm³/g = 1.5 cm

\[
\text{d} = 0.08 \times 50 \times 1.5 = 6 \, \text{cm}
\]

So, the **net depth of irrigation** is 6 cm.

### Step 4: Calculate the gross irrigation depth
Since the water application efficiency is 75%, we need to account for water losses:

\[
\text{Gross Irrigation Depth} = \frac{\text{Net Irrigation Depth}}{\text{Application Efficiency}}
\]
\[
\text{Gross Irrigation Depth} = \frac{6}{0.75} = 8 \, \text{cm}
\]

Thus, the gross depth of irrigation required is **8 cm**.

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Would you like more details on any of the steps or a further breakdown?

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Here are 5 related questions to expand on this topic:
1. How does soil bulk density affect irrigation calculations?
2. What factors can influence field water application efficiency?
3. How would the calculation change if the field water application

Learn how to calculate the gross depth of irrigation required to bring soil moisture content to field capacity for a field crop. This problem involves key concepts like available water content, soil bulk density, and irrigation efficiency. Suitable for Grades 11-12 students.

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