You are working as a summer intern for the Illinois Department of Natural Resources (DNR) and are assigned to some initial work on water resources project. The department will be overseeing the construction of a dam to create a large fresh water lake that will be approximately 18 meters deep. A horizontal pipe 1.2 meters long and 4 cm in diameter will pass through the dam at a depth of 7 meters to allow for release of the water in emergencies and for sampling. In normal situations, a plug will secure the pipe opening.
d) If the plug were to be removed, how much water would flow out in 3 hours?
ΔVOL = m3
* Physics - drwls, Friday, April 30, 2010 at 2:09am
a) A*[Po + (rho) g H]
Po is atmospheric pressure
rho is the mass density of water, 1000 kg/m^3
H is the water height above the hole, 7 m
A is the pipe inner area, pi*D^2/4, in square meters
b) A* Po
c) (rho) g H
d) A*(10,800 s)*V
where V is the speed of wather leaving the pipe.
Use Bernoulli's principle to get V.
Thanks for your help but i really cant understand what to plug in bernoulli's equation. i know the equation which is P1+.5(p)v1^2=P2+.5(p)v2^2
this problem was very hard, even the student tutors at my college were unable to figure out D. so help me with it if you can since you have got me almost to the end as i got a b and c.
(1/2) rho v^2 = rho g H
(1/2) v^2 = 9.8 * 7
alrite love you man.
To calculate the flow rate of water out of the pipe in this scenario, you can use Bernoulli's principle. Here's how you can find the answer:
1. Start by understanding Bernoulli's principle. It states that the sum of the pressure, kinetic energy, and potential energy per unit volume of a fluid remains constant along a streamline. In this case, we can neglect the potential energy term since the water is not moving up or down.
2. Choose two points along the streamline. Let's call the point where the water enters the pipe Point 1 and the point where it exits Point 2. The pressure at Point 1 is the atmospheric pressure (Po), and the pressure at Point 2 is also atmospheric pressure without the plug.
3. Apply Bernoulli's principle to the chosen points. The equation you mentioned, P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2, is correct. However, keep in mind that the velocity (v) in this equation refers to the speed of the fluid, not just the speed of water coming out of the pipe.
4. Convert the given values into appropriate units. In the case of the pipe length (L), diameter (D), and time (t), ensure that they are in meters, centimeters, and seconds, respectively.
5. To get the cross-sectional area of the pipe (A), use the formula A = π(D/2)^2.
6. Rearrange Bernoulli's equation to solve for the velocity (v2) at Point 2:
v2 = sqrt((P1 - P2 + 0.5ρv1^2)*2/ρ)
7. Use the calculated velocity (v2) to determine the flow rate of water, which is the volume of water flowing per unit time (ΔVOL/Δt). The formula is:
ΔVOL/Δt = A * v2
8. Substitute the given values into the formula:
ΔVOL/Δt = A * sqrt((P1 - P2 + 0.5ρv1^2)*2/ρ)
9. Finally, calculate the volume of water that would flow out in 3 hours by multiplying the flow rate (ΔVOL/Δt) by the duration (3 hours converted to seconds).
Remember to use the appropriate units along the way and ensure that all the necessary conversions are made.
I hope this explanation helps you solve part D of the problem. If you have any further questions, feel free to ask!