(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Water flows steadily from an open tank. The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is 0.0480 m[tex]^2[/tex]; at point 3 it is 0.0160 m[tex]^2[/tex]. The area of the tank is very large compared with the cross-sectional area of the pipe.

Part 1- Assuming that Bernoulli's equation applies, compute the discharge rate in cubic meters per second.

Part 2- What is the gauge pressure at point 2

2. Relevant equations

[tex]A_1v_1=A_2v_2[/tex]

[tex]p_1+\rho gh_1+\frac{1}{2}\rho v_1^2=p_2+\rho gh_2+\frac{1}{2}\rho v_2^2[/tex]

3. The attempt at a solution

I have already calculated the discharge rate (at point 3) to be 0.200 m[tex]^3[/tex]/s which I know to be correct but I am stuck with the second part. I used Bernoulli's equation with points 2 and 3 and came up with an answer of 17.8 Pa but this appears to be wrong. I took [tex]h_2=h_3=0[/tex] to simplify Bernoulli's equation to get

[tex]p_3-p_2=\frac{1}{2}\rho v_2^2-\frac{1}{2}\rho v_3^2[/tex]

and used the volume flow rate equation to get the velocity at point 2 to be 0.067m/s and then used that in Bernoulli's equation to get the answer I have, am I doing something wrong here or should I even be using Bernoulli's equation to solve this part of the question. [tex]p_3-p_2[/tex] should be equal to the gauge pressure shouldn't it? Any guidance whatsoever would be greatly appreciated.

**Physics Forums - The Fusion of Science and Community**

# Fluids problem - gauge pressure

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Fluids problem - gauge pressure

Loading...

**Physics Forums - The Fusion of Science and Community**