Question

A 1.0 cm diameter pipe widens to 6.0 cm, then narrows to 0.4 cm. Liquid flows through the first segment at a speed of 5.0 m/s. (a) What is the speed in the second and third segments?

second segment m/s

third segment m/s

(b) What is the volume flow rate through the pipe?

m3/s

Answer 1

from the equation of continuit

AV= constant

a= area of crossection

V= velicity

a) i. from the equation of continuity

$A_{1}V_{1}=A_{2}V_{2}$

$V_{2}=\frac{A_{1}V_{1}}{A_{2}}$

$V_{2}=\frac{\frac{\pi d_{1}^{2}}{4}V_{1}}{\frac{\pi d_{2}^{2}}{4}}$

$V_{2}=\frac{d_{1}^{2}V_{1}}{d_{2}^{2}}$

$V_{2}=\frac{0.01^{2}\times 5}{0.06^{2}}m/sec$

$V_{2}=\0.139m/sec$

ii.

$A_{2}V_{2}=A_{3}V_{3}$

similarly

$V_{3}=\frac{d_{2}^{2}V_{2}}{d_{3}^{2}}$

$V_{3}=\frac{0.06^{2}\times 0.139}{0.04^{2}}$

$V_{3}=\0.3125m/sec$

in three significant figure

$V_{3}=\0.313m/sec$

b) volume flow

$A_{3}V_{3}=0.313m/sec\frac{\pi d_{3}^{2}}{4} L/sec$

$A_{3}V_{3}=0.313\times \frac{\pi \times 0.04^{2}}{4} L/sec$

$A_{3}V_{3}=3.93\times 10^{-4} L/sec$