The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min–1, what is the speed of ejection of the liquid through the holes?
Answers
Answered by
59
CSA of tube (A) = 8 cm²
= 8×10^-4 m²
no of holes ( N) = 40
Diametre of each hole = 1mm
radius of each hole (r) = 0.5 mm
= 5 × 10^-4 m
Velocity of liquid flow in liquid (v) = 1.5 m/min
= 1.5 m/60 s
= 2.5 × 10^-2 m/s
total area of hole (A) = no of holes × area of circle .
= 40 × πr² = 40πr²
= 40 × 3.14 × (5×10^-4)²
= 3.14 × 10^-5 m²
According to equation of continuity ,
A1v1 = A2v2
v2 = A1v1/A2
= 8 × 10^-4 × 2.5 × 10^-2/3.14 × 10^-5
= 2/31.4 = 0.64 m/s
= 8×10^-4 m²
no of holes ( N) = 40
Diametre of each hole = 1mm
radius of each hole (r) = 0.5 mm
= 5 × 10^-4 m
Velocity of liquid flow in liquid (v) = 1.5 m/min
= 1.5 m/60 s
= 2.5 × 10^-2 m/s
total area of hole (A) = no of holes × area of circle .
= 40 × πr² = 40πr²
= 40 × 3.14 × (5×10^-4)²
= 3.14 × 10^-5 m²
According to equation of continuity ,
A1v1 = A2v2
v2 = A1v1/A2
= 8 × 10^-4 × 2.5 × 10^-2/3.14 × 10^-5
= 2/31.4 = 0.64 m/s
Answered by
29
this is the answer to your question
Attachments:
Similar questions