Question

Your low-flow showerhead is delivering water at 1.2×10−4m3/s, about 2.0 gallons per minute.

Answer 1

To solve this problem it is necessary to apply the fluid mechanics equations related to continuity, for which the proportion of the input flow is equal to the output flow, in other words:

$Q_1 = Q_2$

We know that the flow rate is equivalent to the velocity of the fluid in its area, that is,

$Q = VA$

Where

V = Velocity

A = Cross-sectional Area

Our values are given as

$Q_2 = 1.2*10^{-4}m^3/s$

$d = 0.021m$

$r = \frac{0.021}{2} = 0.0105m$

Since there is continuity we have now that,

$V_1A_1 = Q_2$

$V_1A_1 = 1.2*10^{-4}$

$V_1 = \frac{1.2*10^{-4}}{A_2}$

$V_1 = \frac{1.2*10^{-4}}{\pi r^2}$

$V_1 = \frac{1.2*10^{-4}}{\pi (0.0105)^2}$

$V_1 =0.347m/s$

Therefore the speed of the water's house supply line is 0.347m/s