Solving Liquid Flow Problem: Speed in Lower Pipe

[thursdaytbs]

"A liquid is flowing through a horizontal pipe whose radius is 0.10m. The pipe bends straight up for 10.0m and joins another horizontal pipe whose radius is 0.20m. It was found that the pressure in the lower pipe and the upper pipe were the same. What is the speed in the lower pipe? [An image shows this that the higher pipe is 10m above it."

I'm just checking if this is right, any help - thanks.

I used Bernoulli's Equation of
$$P_{1} + \frac{1}{2} \rho v_{1}^2 + \rho gy_{1} = P_{2} + \frac{1}{2} \rho v_{2}^2 + \rho gy_{2}$$
[Initial conditions = Final conditions, where P=pressure, v=velocity, $$\rho$$=density, and y=height]

Since it said "a liquid" i knew $$\rho$$=$$\rho$$, and the pressure is the same so $$P_{1}$$=$$P_{1}$$. So I removed Pressure from both sides, and removed $$\rho$$.

The new equation:

$$\frac{1}{2} v_{1}^2 + gy_{1} = \frac{1}{2} v_{2}^2 + gy_{2}$$

since $$y_{1} = 0$$, it becomes: $$\frac{1}{2}v_{1}^2 = \frac{1}{2}v_{2}^2 + (9.8)(10)$$

then I plugged that equation into $$Av_{1} = Av_{2}$$ [area times velocity = area times velocity, 2equations-2unknowns.]

I solved for the first velocity getting 14.44m/s.


**Can anyone confirm if this is the right way? -Thanks in advance.

 

[Pyrrhus]

Yes, this is what i got

$$\frac{1}{2}v_{2}^2 + gh = \frac{1}{2}v_{1}^2$$

For solving this you need the Flow Continuity equation

$$A_{1}v_{1} = A_{2}v_{2}$$

Just like you did

Good work!

 

[thursdaytbs]

Awsome, thank you so much.