Question

An incompressible liquid travels as shown in figure. Calculate the speed of the fluid in lower branch.

Answer 1

1 m/s

Answer 2

Using the continuity equation for an incompressible fluid, the flow rate must be constant.

1. At the inlet:

   $Q_{\text{in}} = A_1 \cdot v_1 = 0.12 \, \text{m}^2 \times 3 \, \text{m/s} = 0.36 \, \text{m}^3/\text{s}$

2. At the horizontal branch:

   $Q_{\text{horizontal}} = 0.12 \, \text{m}^2 \times 1.5 \, \text{m/s} = 0.18 \, \text{m}^3/\text{s}$

3. At the lower branch (unknown velocity $V$):

   Let

   $Q_{\text{lower}} = A_2 \cdot V = 0.18 \, \text{m}^2 \times V$

Since the incoming flow splits into the two branches:

$Q_{\text{in}} = Q_{\text{horizontal}} + Q_{\text{lower}}$

$0.36 = 0.18 + 0.18V$

$0.18V = 0.36 - 0.18 = 0.18$

$V = \frac{0.18}{0.18} = 1 \, \text{m/s}$