The flow rate through Branch 2 is 8·√2 ml/min
Reasons:
Radius of Branch 1 = 0.5 cm
Radius of Branch 2 = 1.0 cm
Flow rate, Q = Velocity, V × Area, A
The area of the Branch 1 = π × 0.5² = 0.25·π
We have;
[tex]\dfrac{v_1^2}{r_1} = \dfrac{v_2^2}{r_2}[/tex]
Which gives;
[tex]v_2 = \mathbf{\sqrt{\dfrac{v_1^2}{r_1} \times r_2}}[/tex]
[tex]The \ velocity \ of \ Branch 1 , \ v_1 = \dfrac{2}{\pi \times 0.5^2} = \dfrac{8}{\pi}[/tex]
Therefore;
[tex]v_2 = \sqrt{\dfrac{\left(\dfrac{8}{\pi} \right) ^2}{0.5} \times1} = 8 \cdot\dfrac{\sqrt{2} }{\pi}[/tex]
[tex]The \ flow \ rate, \ Q_2 = 8 \cdot\dfrac{\sqrt{2} }{\pi} \times \pi \times 1^2 = 8 \times \sqrt{2}[/tex]
The flow rate through Branch 2 = 8·√2 ml/min
Learn more here:
https://brainly.com/question/13868524
