The cylindrical tube of a spray pump has a cross-section of $8.0 ext{ cm}^2$, one end of which has

Question

The cylindrical tube of a spray pump has a cross-section of $8.0	ext{ cm}^2$, one end of which has $40$ fine holes each of diameter $1.0	ext{ mm}$. If the liquid flow inside the tube is $1.5	ext{ m/min}$, what is the speed of ejection of the liquid through the holes?

Professor PC Thomas Classes · Answer

By the equation of continuity, $A_1 v_1 = n A_2 v_2$. $A_1 = 8 	imes 10^{-4}	ext{ m}^2$, $v_1 = \frac{1.5}{60} = 0.025	ext{ m/s}$. $A_2 = \pi r^2 = \pi \left(0.5 	imes 10^{-3}ight)^2 \approx 0.785 	imes 10^{-6}	ext{ m}^2$. $v_2 = \frac{A_1 v_1}{n A_2} = \frac{8 	imes 10^{-4} 	imes 0.025}{40 	imes 0.785 	imes 10^{-6}} \approx 0.64	ext{ m/s}$.