Question

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%

I had been asked this question during an online interview.

This question is from Plant Protection Equipments topic in division Plant Protection Equipments of Farm Machinery

Answer 1

The correct choice is (c) 4.87%

Best 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%.