Question

A tube well is used to irrigate the field. It pumps 9000 m³ of water, out of which 8100 m³ reaches the field head. Calculate the farm irrigation system efficiency, if the irrigation water application efficiency is 60%.

• A. 90%
• B. 65.5%
• C. 54%
• D. 75%

Hint:

To find the overall efficiency of a system, convert the individual percentage efficiencies to decimals, multiply them together, and then multiply by 100:
$\text{Overall Efficiency} = 0.90 \times 0.60 = 0.54 \rightarrow 54\%$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The overall farm irrigation system efficiency ($\eta_s$) is the product of the individual efficiencies of the system, primarily the water conveyance efficiency ($\eta_c$) and the water application efficiency ($\eta_a$).
Key Formula or Approach:
The formulas are:
- Water conveyance efficiency: \[ \eta_c = \frac{\text{Water reaching the field head}}{\text{Water pumped from well}} \times 100 \] - Overall system efficiency: \[ \eta_s = \eta_c \times \eta_a \]

Step 2: Detailed Explanation:
Given values from the problem:
- Water pumped = $9000 \text{ m}^3$
- Water reaching field head = $8100 \text{ m}^3$
- Water application efficiency ($\eta_a$) = $60\% = 0.60$
First, calculate the water conveyance efficiency ($\eta_c$): \[ \eta_c = \frac{8100}{9000} \times 100 = 90\% = 0.90 \] Second, calculate the overall farm irrigation system efficiency ($\eta_s$): \[ \eta_s = \eta_c \times \eta_a \] \[ \eta_s = 0.90 \times 0.60 = 0.54 = 54\% \] Thus, the overall farm irrigation system efficiency is $54\%$.

Step 3: Final Answer:
The overall efficiency is $54\%$, which corresponds to Option (C).