Question

The circular stream of water from a faucet is observed to taper from a diameter of 20 mm to 10 mm in a distance of 50 cm. Determine the flowrate.

Answer 1

Solution:

From Bernoulli equation

$\frac{1}{2}$ρ$V_{1}^{2}$ + ρgh = $\frac{1}{2}$ρ$V_{2}^{2}$ , where ρ is density of water, h – height difference and $V_{1}$ and $V_{2}$  are the velocities in upper and lower cross sections correspondingly. We take into account that the pressure in both cross-sections is the same and equal to the atmospheric one.

From the continuity and assuming water incompressible:

$\pi r_{1}^{2}$$V_{1}$ = $\pi r_{2}^{2}$$V_{2}$ , where $A_{1}$ and $A_{2}$ are the corresponding cross-sections. As the lower diameter is twice as low as the upper one, we can conclude that $V_{1} = \frac{V_{2}}{4}$

Inserting it back into Bernoulli equations produces:

$V_{1} = \sqrt{\frac{2}{15}gh} = 0.26m/s$ and the flow rate is

$\pi r_{1}^{2}V_{1} = \pi r_{2}^{2}V_{2} = 8.10^{-5} \frac{m^{3} }{s}$