Irrigation Water Volume Calculation
The question asks for the volume of water needed to irrigate a wheat crop field based on soil moisture characteristics and field size.
Calculating Required Water Volume
The calculation involves determining the amount of water needed to bring the soil moisture from the current depleted level back to field capacity.
Step 1: Calculate Available Moisture (AM) on a weight basisAvailable Moisture is the difference between Field Capacity (FC) and Wilting Point (WP).
Given: $FC = 32\%$ (weight basis), $WP = 20\%$ (weight basis)
$AM = FC - WP$
$AM = 32\% - 20\% = 12\% \text{ (weight basis)}$
Step 2: Determine the depth of water to apply ($d_a$)The field is irrigated when 40% of the Available Moisture (AM) is depleted. Therefore, we need to replenish 40% of the AM.
Moisture to replenish (weight basis) $= 40\% \times AM = 0.40 \times 12\% = 4.8\%$
To find the depth ($d_a$), we convert this percentage to a volume per unit area using the bulk density ($\rho_b$), water density ($\rho_w$), and root zone depth (RZD).
Given: Bulk Density ($\rho_b$) = $1400 \text{ kg/m}^3$, Root Zone Depth (RZD) = $0.8 \text{ m}$. Assume water density ($\rho_w$) = $1000 \text{ kg/m}^3$.
$d_a = \left( \frac{\text{Moisture to replenish}}{100} \right) \times \frac{\rho_b}{\rho_w} \times RZD$
$d_a = \left( \frac{4.8}{100} \right) \times \frac{1400 \text{ kg/m}^3}{1000 \text{ kg/m}^3} \times 0.8 \text{ m}$
$d_a = 0.048 \times 1.4 \times 0.8 \text{ m}$
$d_a = 0.05376 \text{ m}$
Step 3: Calculate the total volume of water requiredThe volume of water (V) is the product of the field area (A) and the depth of water to apply ($d_a$).
Given: Field Area $A = 5 \text{ ha} = 5 \times 10000 \text{ m}^2 = 50000 \text{ m}^2$.
$V = A \times d_a$
$V = 50000 \text{ m}^2 \times 0.05376 \text{ m}$
$V = 2688 \text{ m}^3$
The calculated volume is $2688 \text{ m}^3$. This value is closest to Option A (2700 $m^3$). The information regarding the pump's capacity and irrigation duration is not needed for this specific calculation.
Final Answer SelectionThe calculated volume ($2688 \text{ m}^3$) is approximately equal to Option A.
Correct Answer: Option A (2700)
Important Questions from Soil Moisture Measurement
- An undisturbed soil sample is collected from a field with core cylinder having an internal diameter of $60\text{ mm}$ and length of $100\text{ mm}$. The weights of moist soil and oven dried soil are $0.48\text{ kg}$ and $0.44\text{ kg}$, respectively. Assuming density of water as $1000\text{ kg.m}^{-3}$, the water depth (in metres per metre depth of soil) is ________
(Rounded off to two decimal places) - In-situ volumetric soil moisture content measuring method(s) is/are:
- The elevations of pressure gauge and porous cup of the tensiometer installed in unsaturated zone are 145.8 m and 144.2 m, respectively. Pressure measured at the gauge is $-19.62\times10^3$ $N$ $m^{-2}$. The specific weight of water is 9810 $N$ $m^{-3}$. The estimated pressure at the porous cup is ________ $N$ $m^{-2}$.
- Tensiometer installed in the soil measures
Undisturbed soil sample is collected from a field when the soil moisture is at field capacity. The inside diameter of the core sampler is 7.5 cm with a height of 15 cm. Weight of the core sampling cylinder with moist soil is 2.81 kg and that with oven dry soil is 2.61 kg. The weight of the core sampling cylinder is 1.56 kg. Assuming $\pi = 3.14$, the water depth in centimeter per meter depth of soil (rounded off to two decimal places) is__________.
