Question

1. Consider the steady flow in a water pipe joint shown in the diagram. The areas are:

Answer 1

Answer:

-4.5 m/s

Explanation:

Assuming steady and incompressible flow and uniform properties at each section

$V_1A_1+V_2A_2+A_3V_3+Q_4=0$

Here V is velocity of flow and A is area, Q is flow rate out of the leak, subscript 1-4 represent different sections

At the surface,  is negative hence the equation above will be

$-V_1A_1+V_2A_2+A_3V_3+Q_4=0$

Making  the subject of the formula then

$V_2=\frac {V_1A_1-A_3V_3-Q_4}{A_2}$

Substituting the given values then

$V_2=\frac {(5\times 0.2)-(12\times 0.15)-0.1}{0.2}=-4.5 m/s\\V_2=-4.5m/s$