Water flowing from an open tank

In summary, the volume flow rate of water from an open tank with an elevation of 10.0 m and cross-sectional areas of 0.0480 m^2 and 0.0160 m^2 is 0.200 m^3/s. The gauge pressure at point 2, where the cross-sectional area is 0.0480 m^2, is 78116.306 Pa.
  • #1
cdotter
305
0

Homework Statement


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.

[PLAIN]http://img132.imageshack.us/img132/4245/xcvbb.jpg [Broken]

Homework 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]

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.
 
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  • #2
cdotter said:
I figure p_1=0 because it's at the top of the tank and v_1=0 because it's not flowing.

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?
 
  • #3
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?
 
  • #4
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.

P3 would be atmospheric pressure since it comes out into the air. Answer a) first.
 
  • #5
cdotter said:
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?

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.)
 
  • #6
AtticusFinch said:
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.

P3 would be atmospheric pressure since it comes out into the air. Answer a) first.

[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?
 
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  • #7
cdotter said:
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.
 
  • #8
AtticusFinch said:
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]?
 
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  • #9
Looks good to me.
 
  • #10
Thank you for your help.
 
  • #11
I need help!
 
  • #12
please!can someone give me the result of the part A and B??
 

1. How does water flow from an open tank?

Water flows from an open tank due to the force of gravity. The water in the tank has potential energy, and when the tank is opened, this potential energy is converted into kinetic energy, causing the water to flow out of the tank.

2. What factors affect the rate of water flow from an open tank?

The rate of water flow from an open tank is affected by several factors, including the size and shape of the tank opening, the height of the tank, and the viscosity of the water. Other external factors such as air pressure and temperature can also impact the rate of flow.

3. Can the flow of water from an open tank be controlled?

Yes, the flow of water from an open tank can be controlled by adjusting the size of the tank opening or using a valve to regulate the flow. Additionally, the height of the tank and the viscosity of the water can also be manipulated to control the flow rate.

4. What happens to the water level in an open tank as it flows out?

As water flows out of an open tank, the water level decreases. This is because the volume of water in the tank is decreasing, while the surface area of the water remains the same. The water level will continue to decrease until the tank is empty or the flow rate is stopped.

5. Can the flow of water from an open tank be used to generate electricity?

Yes, the flow of water from an open tank can be used to generate electricity through a process called hydroelectric power. As the water flows out of the tank, it turns a turbine, which then converts the kinetic energy of the flowing water into electrical energy.

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