# Velocity of fluid dividing to two different paths [closed] Fluid of density $$\rho$$ is flowing through a pipe with area $$A$$ with velocity of $$V$$. The pipe divides into two different pipes of area $$\frac{1}{2}A$$ and $$\frac{3}{2}A$$ with the right end opened. All the pipes are at the same height. How do I represent the velocity of fluid at each paths?

I tried using Bernoulli's equation and continuity equation and got the following results: $$p_0+\frac{1}{2}\rho V^2 = p_1+\frac{1}{2}\rho V_1^2 = p_2+\frac{1}{2}\rho V_2^2$$, $$V_1 + 3V_2 = 2V$$.

However, I can't find any more equations from here. Also, I'm stuck at the second question too, where the two pipes are separated by height $$H$$ and have the same area $$A$$. Thank you in advance.

## closed as off-topic by Kyle Kanos, Gert, GiorgioP, John Rennie, ZeroTheHeroApr 16 at 13:39

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• Add equations: in the first case $p_1=p_2$, in the second case $p_1=p_2+\rho gH$ – Alex Trounev Apr 15 at 14:08
• @AlexTrounev How do you know that $p_1 = p_2$? – Erno Apr 16 at 2:51
• Are the outlets of the pipe discharging into atmosphere (or some other common reservoir)? – Deep Apr 16 at 4:44