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. 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 [tex]p_1+\rho gh_1+\frac{1}{2}\rho v_1^2 = 0+(\rho g*10.0 m)+\frac{1}{2}\rho(0 m/s)^2[/tex] Is this correct for the left side of the equation? I figure p_1=0 because it's at the top of the tank and v_1=0 because it's not flowing. I'm not sure what to put for the right side because I don't know p_2. I know that [itex]p=\rho gh[/itex] but that doesn't make sense in the equation.
The tank is open to the atmosphere. Would it make sense that there is no pressure on it? You got the speed part correct, this is due to the area of the tank being much larger, a usual approximation. Also, what exactly is the question here?
Compute (A) the discharge rate in cubic meters per second and (B) the gauge pressure at point 2. "You got the speed part correct, this is due to the area of the tank being much larger, a usual approximation." I don't understand. What does the area of the tank have to do with the speed at point 1? Point 1 is at the very top of the water level. How could it be anything other that zero, regardless of the area of the tank? The left side of the equation is: [tex]= 1.013\cdot 10^5 Pa + (\rho g\cdot 10.0 m)+\frac{1}{2}\rho0^2[/tex] What would p_2 be? Atmospheric pressure as well?
Since volume flow rate has to be continuous, if you have a small tank that springs a leak, the water at the top probably will have a speed other than zero. P_{3} would be atmospheric pressure since it comes out into the air. Answer a) first.
When you have a large tank and a small leak, it can take forever for the tank to drain (this is because it has a large cross-sectional area on top so the height of the liquid only decreases a little when a lot of volume goes out). Alternatively, when you have a small tank and a large leak, the tank could take seconds to drain due to the small cross sectional area (by which I mean the part that is exposed to the air in your diagram) of the liquid. For the pressure at point two, it would have to be calculated from the known pressure at point one as well as with the assumption that the velocity at point one is zero (we do not have to use this assumption if the cross sectional area of the tank was given but you would find the answer changes very little even if the tank's area term was included.)
[tex]1.013\cdot 10^5 Pa+\rho g\cdot 10 m+\frac{1}{2}\rho \cdot 0^2=1.013\cdot 10^5 Pa+\rho g\cdot 2 m+\frac{1}{2}\rho v_2^2 \Rightarrow v_2=12.522 m/s[/tex] [tex]12.522 m/s \cdot 0.016 m^2 = 0.200 \frac{m^3}{s}[/tex] I think [itex]0.200 m^3/s[/itex] is correct but I don't know why it works? I know it say "assume the area of the tank is very large"...so it's ok to just completely ignore [itex]A_1v_2[/itex] in the formula [itex]A_1v_2=A_2v_2[/itex] and the [itex]0.0480 m^2[/itex] cross sectional area?
No. The only volume flow rate you ignore is the one of the large tank. You don't always have to use the continuity equation (A*v = A*v) to find the volume flow rate (as you just saw by solving problem A). To solve part B, however, you do need to use the continuity equation comparing points 2 and 3. Then use Bernoulli's equation to find the abs. pressure at point 2 and then getting the gauge pressure is a simple step away.
[itex]0.0480 m^2v_2=0.200 m^3/s \Rightarrow v_2=4.17 m/s[/itex] [itex]0.0160 m^2v_3=0.200 m^3/s \Rightarrow v_3=12.5 m/s[/itex] [tex]1.013\cdot 10^5 Pa + \frac{1}{2}\rho (12.5 m/s)^2=p_2+\frac{1}{2}\rho (4.17 m/s)^2 \Rightarrow p_2 = 179416.306 Pa[/tex] Is that correct? And then gauge pressure would be [itex]179416.306 Pa - 101300 Pa = 78116.306 Pa[/itex]?