# Fluids problem - gauge pressure

by CurtisB
Tags: fluids, gauge, pressure
 P: 16 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$$^2$$; at point 3 it is 0.0160 m$$^2$$. 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 $$A_1v_1=A_2v_2$$ $$p_1+\rho gh_1+\frac{1}{2}\rho v_1^2=p_2+\rho gh_2+\frac{1}{2}\rho v_2^2$$ 3. The attempt at a solution I have already calculated the discharge rate (at point 3) to be 0.200 m$$^3$$/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 $$h_2=h_3=0$$ to simplify Bernoulli's equation to get $$p_3-p_2=\frac{1}{2}\rho v_2^2-\frac{1}{2}\rho v_3^2$$ 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. $$p_3-p_2$$ should be equal to the gauge pressure shouldn't it? Any guidance whatsoever would be greatly appreciated.
Mentor
P: 41,315
 Quote by CurtisB ...and used the volume flow rate equation to get the velocity at point 2 to be 0.067m/s ...
Revisit this calculation.
 P: 16 Oh, thank you, I accidentally put in the volume flow rate in as v instead of the velocity of the fluid, I have the correct answer now, Thanks.
P: 150
Fluids problem - gauge pressure

 Quote by Doc Al Revisit this calculation.
I still don't understand... If v_2 = 4.17 and v_3 = 12.5, then what is this p_3 that I need in order to solve for p_2?? (I also know that rho=density of water=1000 kg/(m^3))
 Mentor P: 41,315 Note that point 3 is in an open stream.
 P: 150 Then p_3 is atmospheric pressure .. is it 101.3 kPa? so then p_2 = -(.5(1000(4.7^2-12.5^2)))-101300) = 168380 Pa?
 P: 16 The question asked for gauge pressure, which is the pressure at 2 minus atmospheric pressure given by p_2-p_3 in the top equation, which is 6.97*10^4. So yeah, the absolute pressure at 2 would be the gauge pressure plus atmospheric which is pretty much what you got.
 P: 150 OHHHH i finally get it!

 Related Discussions Introductory Physics Homework 1 Introductory Physics Homework 5 Introductory Physics Homework 1 Introductory Physics Homework 5 Introductory Physics Homework 1