This set of Advanced Farm Machinery Questions and Answers focuses on “Plant Protection Equipments – Sprayers – 2”.

1. A sprayer having 0.4 litre per minute nozzle discharge rate and 50 cm width of coverage, is required to apply 0.8 kg of active ingredient per hectare. If one kg of active ingredient is contained in 120 litres of spray solution, determine the speed of the sprayer.

a) 4 km/h

b) 3 km/h

c) 5 km/h

d) 6 km/h

View Answer

Explanation: 1 kg active ingredient is dissolved in 120 L water

0.8 kg active ingredient is dissolved in 120 * 0.8 = 96 L

Application rate = 96 L/ha

Area covered = \(\frac{0.5*s}{10} \) ha/h

Time required to cover one ha land = \(\frac{10}{0.5*s} \) h

Time required to spray one ha land = \(\frac{96}{0.4*60} \) h

Equating,

\(\frac{10}{0.5*s}\) h = \(\frac{96}{0.4*60} \)h

S = 5 km/h

2. An air blast of sprayer is to be operated at 3 km/h and application rate is 18 l/tree. The spacing is 9m*9m and each nozzle delivers 5 lit/min and operating process of 4*10^{5} N/m^{2}. If one half row is sprayed from each side of the machine. How many sprayers will be needed?

a) 30

b) 50

c) 20

d) 40

View Answer

Explanation: Time of spacing/tree = \(\frac{9*18*60}{3*5} \) h

Time of spray/tree = \(\frac{Q}{n*Qr} \)

\(\frac{9*18*60}{3*5} = \frac{18}{n*5} \)

N = 20

For both sides = 2*20 = 40

3. During a discharge measured through a hydraulic spray nozzle was 450*10^{-6}m^{3}/min at nozzle operating pressure of 284 kPa. If the pressure is increased by 10%. The percent increase in discharge will be _______

a) 4.9%

b) 5.1%

c) 3.6%

d) 6.4%

View Answer

Explanation: \(\frac{Q1}{Q2}=\Big\{\big(\frac{P1}{P2}\big)^{\frac{1}{2}}\Big\} \)

\(\frac{450*10^{-6}}{Q2} = \{\Big(\frac{284}{312.4}\Big)^{\frac{1}{2}}\} \)

Q2= 471.96*10

^{-6}m

^{3}/min

% increase = \(\frac{Q2-Q1}{Q1}\)*100 = \(\frac{(471.96-450)}{450} \)*100 = 4.9%

4. What is the Nozzle Flow Rate (NFR)?

a) NFR = application rate + area covered by single nozzle

b) NFR = application rate * area covered by single nozzle

c) NFR = application rate – area covered by single nozzle

d) NFR = \(\frac{application \, rate}{area \, covered \, by \, single \, nozzle} \)

View Answer

Explanation:

NFR = application rate * area covered by single nozzle

NFR = n (L/ha) * y (ha/h)

Area covered by single nozzle = S’ * S

= (km/h) * m

= \(\frac{(10^3*10)m^2}{10 h} \)

= \(\frac{1ha}{10h} \)

5. A hydraulic sprayer fitted with hollow nozzle is required to be calibrated for application rate 150 L/ha. The sprayer speed is 6.5 km/h. Nozzle spacing is 50 cm. The available 0.787 mm orifice diameter. Nozzle is rated at 0.473 L/min. What is the nozzle flow rate?

a) 49.51 \(\frac{L}{h} \)

b) 54.44 \(\frac{L}{h} \)

c) 48.75 \(\frac{L}{h} \)

d) 24.10 \(\frac{L}{h} \)

View Answer

Explanation: NFR = application flow rate * area covered by single nozzle

= 150*6.5/10 * 0.5

= 48.75 \(\frac{L}{h} \)

6. A sprayer having 0.4 L/min nozzle discharge rate and 50 cm width of coverage is required to apply 0.8 kg of active ingredient per hectare. If 1 kg of active ingredient is content in 120 L of spray solution. What will be the speed of travel of sprayer?

a) 2 km/h

b) 5 km/h

c) 3 km/h

d) 4 km/h

View Answer

Explanation: NFR = 0.4 \(\frac{L}{min} \)

= 24 \(\frac{L}{h} \)

1 kg of spray ingredient is containing 120 L solution

0.8 will contain 120*0.8 L

Application rate = 96 \(\frac{L}{ha} \)

NFR = Application rate * S’ * S

24 = \(\frac{96*S’*0.5}{10} \)

S’ = \(\frac{24*10}{96*0.5} \)

S’ = 5 km/h

7. What will be the diameter of base of a spray cone in a boom sprayer?

a) D = 2h tan \(\frac{θ}{2} \)

b) D = 2h/tan \(\frac{θ}{2} \)

c) D = 2h + tan \(\frac{θ}{2} \)

d) D = 2h – tan \(\frac{θ}{2} \)

View Answer

Explanation: A boom sprayer is a pipe with attached nozzles for distributing spray from a tank.

For overlapping of spray cone

Case 1: If height is given

S = (1-O) D

Case 2: If height is not given

D = (1+O) S

8. What is the formula used for calculating VMD, when pressure is less than 675 kPa?

a) \(\frac{(VMD1)}{VMD2}=\Big\{(\frac{P1}{P2})^{\frac{2}{3}} \Big\} \)

b) \(\frac{(VMD1)}{VMD2}=\Big\{(\frac{P2}{P1})^{\frac{2}{3}} \Big\} \)

c) \(\frac{(VMD2)}{VMD1}=\Big\{(\frac{P1}{P2})^{\frac{2}{3}} \Big\} \)

d) \(\frac{(VMD2)}{VMD1}=\Big\{(\frac{P1}{P2})^{\frac{1}{3}} \Big\} \)

View Answer

Explanation: The droplets size that device the spray into two equal parts by volume. One half containing smaller than this diameter and other half of the volume containing larger droplets. In other words, VMD divides the droplet in two portion such that the total volume of all droplets smaller the VMD is equal to the total volume of all droplets larger than VMD.

9. A hydraulic sprayer fitted with hollow cone nozzle is required to be calibrated for application rate 150 l/ha. The sprayer speed is 6.5/h, nozzle spacing is 50 cm, and the available 0.787 mm orifice diameter nozzle is rated at 0.473 l/min. What will be the nozzle flow rate?

a) 0.0135 l/s

b) 0.0136 l/s

c) 0.0137 l/s

d) 0.0138 l/s

View Answer

Explanation: Flow rate = application rate*area covered by single nozzle

Area covered by single nozzle = speed * nozzle spacing = 6.5*1000*\(\frac{50}{100} \) = 0.325 ha/hr

Flow rate = 150*0.325 = 48.75 l/hr

= 0.0136 l/sec.

10. A field sprayer having a horizontal boom with 20 nozzles spaced 40 cm apart is to be designed for maximum application rate of 650 l/ha at nozzle pressure of kPa and forward speed of km/h. What will be the pump capacity in l/min?

a) 44

b) 46

c) 48

d) 50

View Answer

Explanation: Flow rate = application rate*area covered by single nozzle

Area covered by single nozzle = speed * width = \(\frac{40}{100}*\frac{5*1000}{60} \) = 0.0333 ha/min

Total flow rate = 20*650*0.00333 = 43.29 l/min

Pump capacity = \(\frac{21.6452}{\frac{90}{100}} \) = 48 l/min.

11. During a test, discharge measured through a hydraulic spray nozzle was 45010^{-6} m^{3}/min at nozzle operating pressure of 284 kPa. If the pressure is increased by 10%, the percent increase in discharge will be ________

a) 2.82%

b) 3.87%

c) 4.87%

d) 5.87%

View Answer

Explanation: \(\frac{Q1}{Q2} = \sqrt{\frac{P1}{P2}} \)

\(\frac{45*10^{-6}}{Q2} = \sqrt{\frac{284}{312.4}} \)

Q2 = 4.719*10

^{-4}m

^{3}/min

% increase in discharge = \(\frac{Q2-Q1}{Q1}*100 = \frac{4.719*10^{-4}-450*10^{-6}}{450*10^{-6}} \)*100 = 4.87%.

12. An air blast of sprayers is to be operated at 3 km/hour and desired application rate is 18 l/tree. The tree spacing is 9*9 m and nozzle delivers 5 l/min at operating pressure of 4*10^{5} N/m^{2}. How many nozzles would be needed?

a) 10

b) 20

c) 30

d) 40

View Answer

Explanation:

Time of spraying/tree = \(\frac{spacing}{forward \, speed} = \frac{(application \, width)}{number \, of \, nozzles*discharge \, rate} \)

\(\frac{9*60}{3*1000} = \frac{18}{n*5} \)

n = \(\frac{18*3*1000}{9*60*5} \)

n = 20.

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