Question Bank • QB_FM_ADV_038 NAT • 2 Marks

Section 2: Farm Machinery

Farm Machinery

Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses

Question Statement:

A tractor-mounted hydraulic boom sprayer is equipped with $16$ flat-fan nozzles spaced at $w = 0.5\text{ m}$ intervals along the boom. The sprayer travels at a uniform forward speed of $v = 5.4\text{ km/h}$. Each nozzle has a measured discharge rate of $q = 0.90\text{ L/min}$ at operating pressure. The chemical application rate of the sprayer in $\text{L/ha}$ is ________ (answer in integer).

Answer Key:

200

Step-by-Step Solution & Explanation

``` Boom Sprayer Swath Geometry: Nozzle spacing w = 0.5 m, Speed v = 5.4 km/h Total Boom Width = 16 * 0.5 = 8.0 m ``` **Method 1: Standard Ag-Sprayer Formula** $Q = \frac{600 \times q}{w \times v}$ Where: - $q = 0.90\text{ L/min}$ (discharge per nozzle) - $w = 0.5\text{ m}$ (nozzle spacing) - $v = 5.4\text{ km/h}$ (forward speed) $Q = \frac{600 \times 0.90}{0.5 \times 5.4} = \frac{540}{2.70} = 200\text{ L/ha}$ **Method 2: First Principles Area-Volume Balance** 1. Area covered per hour: $\text{Area} = \frac{\text{Boom Width (m)} \times \text{Speed (km/h)}}{10} = \frac{(16 \times 0.5) \times 5.4}{10} = \frac{8.0 \times 5.4}{10} = 4.32\text{ ha/h}$ 2. Total spray volume emitted per hour: $V = 16 \text{ nozzles} \times 0.90\text{ L/min} \times 60\text{ min/h} = 864\text{ L/h}$ 3. Application rate: $Q = \frac{864\text{ L/h}}{4.32\text{ ha/h}} = 200\text{ L/ha}$
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