Tile Drainage Discharge Calculation

Problem Analysis

The problem asks for the average discharge from a tile drainage system given its drainage coefficient and the area it serves.

• Drainage Coefficient (DC): $25\text{ mm}$ (Assumed to be per day, a standard unit in drainage engineering).
• Area (A): $0.2\text{ km}^2$.
• Required Output: Average discharge ($Q$) in $\text{m}^3\text{.s}^{-1}$, rounded to three decimal places.

Calculation Steps

1. Convert Area to Square Meters: The area needs to be in metric units compatible with the discharge rate. $A = 0.2\text{ km}^2 = 0.2 \times (1000 \text{ m})^2 = 0.2 \times 1,000,000 \text{ m}^2 = 200,000 \text{ m}^2$.
2. Convert Drainage Coefficient to Meters per Second: The drainage coefficient represents the depth of water to be removed. We assume it's $25\text{ mm/day}$. Convert mm to meters: $25\text{ mm} = 0.025\text{ m}$. Convert days to seconds: $1 \text{ day} = 24 \text{ hours} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} = 86400 \text{ s}$. Drainage Rate ($R$) = $\frac{0.025 \text{ m}}{86400 \text{ s}}$.
3. Calculate Discharge: Discharge ($Q$) is the product of Area ($A$) and Drainage Rate ($R$). $Q = A \times R$ $Q = 200,000 \text{ m}^2 \times \frac{0.025 \text{ m}}{86400 \text{ s}}$ $Q = \frac{200,000 \times 0.025}{86400} \text{ m}^3\text{.s}^{-1}$ $Q = \frac{5000}{86400} \text{ m}^3\text{.s}^{-1}$ $Q \approx 0.05787037...\text{ m}^3\text{.s}^{-1}$
4. Round the Result: Rounding the discharge to three decimal places: $Q \approx 0.058 \text{ m}^3\text{.s}^{-1}$.

Result

The calculated average discharge is approximately $0.058 \text{ m}^3\text{.s}^{-1}$. This value falls within the range specified in the correct answer (between 0.055 and 0.06).