Question

An ideal fluid is flowing at a rate of 1.8 m3/s through a pipe with a radius of 

2.0 cm. (a) What is the speed of the fluid through the pipe? Suppose that in one 

portion, the pipe is clogged, and the radius decreased to 1.7 cm.

Answer 1

By definition, flow rate is given as follows:

$$Q = vA$$

where $v$ is the speed of fluid, and $A = \pi r^2$ is the cross-sectional area of a pipe. Since in first case $r_1 = 2cm = 0.02m$, and $Q = 1.8m^3/s$ the speed will be:

$$v_1 = \dfrac{Q}{\pi r_1^2}\\ v_1 = \dfrac{1.8}{\pi \cdot 0.02^2}\approx 1432m/s$$

Since for ideal fluid its flow rate is the same at any point (see http://labman.phys.utk.edu/phys221core/modules/m8/ideal_fluid_dynamics.html), the speed at the location of the clog will be:

$$v_2 = \dfrac{Q}{\pi r_2^2}$$

where $r_2 = 1.7cm = 0.017m$. Thus, obtain:

$$v_2 = \dfrac{1.8}{\pi\cdot 0.017^2} \approx 1983m/s$$

Answer. (a) 1432 m/s, (b) 1983 m/s.