Filling a Tank with Water: Solving for Time

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Homework Help Overview

The problem involves calculating the time required to fill a tank with a volume of 180 m3 using water flowing from a pipe. The water exits the pipe at a velocity of 22 m/s, and the diameter of the pipe is 48 cm. Participants are discussing the relevant equations and concepts related to fluid dynamics and volumetric flow rates.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are exploring the use of the continuity equation and Bernoulli's principle, questioning which equations are appropriate for the scenario. There are discussions about calculating the volumetric flow rate based on the cross-sectional area of the pipe and the velocity of water. Some participants suggest focusing on the geometry of the problem to find the volume of water flowing through the pipe each second.

Discussion Status

The discussion is ongoing, with various participants providing insights and suggestions for approaching the problem. Some have offered guidance on calculating the volumetric flow rate and the volume of water flowing through the pipe, while others are questioning the assumptions and figures used in the calculations. There is no explicit consensus yet, but productive lines of reasoning are being explored.

Contextual Notes

Participants are grappling with the correct application of fluid dynamics equations and the implications of the given parameters, such as the diameter of the pipe and the velocity of water. There is a focus on ensuring that units are consistent and that the calculations are based on the correct interpretation of the problem's setup.

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Homework Statement


The water from a pipe flows into a tank of volume 180 m^3. If the tank is initially empty, how long will it take to fill the tank?

I know that water is coming out of the pipe with a velocity of 22 m/s, and the diameter of this end of the pipe is 48 cm.


Homework Equations


This is what I don't know.
Is it v1A1=v2A2? or p1+1/2ρv1^2+ρgy1=p2+1/2ρv2^2+ρgy2?


The Attempt at a Solution


v1A1=v2A2

22 m/s*.10898 (A1)=0?
 
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You're kind of using the formula wrong. I might suggest calculating it with all the units explicitly stated and then look at the question again.
 
If you know the cross sectional area of the pipe and the average velocity of the water coming out of the pipe, to you know the formula for calculating the volumetric throughput rate coming out of the pipe?
 
Or - don't worry about the equation: just look at the geometry.
What is the volume of water flowing through the pipe each second?

You know how to find the volume of a cylinder right?
 
paisiello2 said:
You're kind of using the formula wrong. I might suggest calculating it with all the units explicitly stated and then look at the question again.

Okay so my units on each side will be meters^3/sec. I don't understand how to just get m^3 without taking out the velocity (which I am going to need to use right?)
 
Chestermiller said:
If you know the cross sectional area of the pipe and the average velocity of the water coming out of the pipe, to you know the formula for calculating the volumetric throughput rate coming out of the pipe?

The only formula I can find that I feel applies is V1A1=V2A2.
Is this not the correct equation to use?
Even if it is, my confusion is what goes on the right side of the equation? Being that V2 and A2 don't really apply to the information I am given.
 
Simon Bridge said:
Or - don't worry about the equation: just look at the geometry.
What is the volume of water flowing through the pipe each second?

You know how to find the volume of a cylinder right?

V=∏r^2h
∏(.325m)^2(.65m)=.2157m^3
How does this tell me the flow rate per second?
 
I'm sure your text or course notes talked about the formula for volumetric flow (you actually already stated it in context)?

Suppose you were standing next to the 48cm dia. pipe and you measured an h=1m length of pipe. You could wait for t=1 second and measure the volume of water V flowing by in that time and this would give you Q=V/t (the volumetric flow rate).

But V = Ah (volume)
and v = h/t (speed)
therefore Q = vA

Hopefully you can take it from there.
 
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Since you are told that the flow rate is constant, the volume of water through the pipe in one second is the flow rate.

Don't worry about these terms right now. I am trying to get you off your dependence on memorizing equations.

V=∏r^2h
... that's the volume of a cylinder radius r and length h - good.
(It's better if you use "π" instead of "∏" there, but I can see what you mean.)

∏(.325m)^2(.65m)=.2157m^3
You used r=0.325m and h=0.65m ... why? Where do these figures come from?

How does this tell me the flow rate per second?
If the numbers you use have no relation to the problem then it won't of course.

The pipe is cylindrical with diameter d=0.48m
The water flows in the pipe at speed v=22m/s

The water leaving the pipe is basically a cylinder that grows in length... or it would be if it didn't fall etc. So imagine water pouring out the pipe in a long straight cylinder of water. You can work out the volume of the water leaving the pipe by working out the volume of the "cylinder" of water.

I'm asking you to use these figures, and the equation for the volume of a cylinder, to work out the volume of water that left the pipe in 1 sec.

If you know the volume of water that flows out the pipe in 1 second, then you can work out howm many seconds are needed to fill the tank.
 
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Chase11 said:

Homework Statement


The water from a pipe flows into a tank of volume 180 m^3. If the tank is initially empty, how long will it take to fill the tank?

I know that water is coming out of the pipe with a velocity of 22 m/s, and the diameter of this end of the pipe is 48 cm.

Homework Equations


This is what I don't know.
Is it v1A1=v2A2? or p1+1/2ρv1^2+ρgy1=p2+1/2ρv2^2+ρgy2?

The Attempt at a Solution


v1A1=v2A2

22 m/s*.10898 (A1)=0?
solution:
rate of flow can be determined by,
wkt v=22m/s area of hole:3.14*(0.24)*(0.24)
rate of flow =22*3.14*(0.24)*(0.24)m^3/s
time taken : 180m^3/(22*3.14*(0.24)*(0.24))m^3/s
 

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