pump problem.

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kpx001

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ok basiclly i have no physics knowlledge and the profeesor assumes that we all know how to do these types of problems so he skipped it. I dont have resource materials but this is all i know.

I know i need to take slab.
so area = pi(5)^2
so the area is 25pi
The density of liquid is 40
F = 25pi * 40 = 1000
Work = F* D

the tank is half full so i start at 4 initialy and im moving up. (x+4) is the distance i use then (x+2) because of the spout thing? i dont understand how to get distance and what changes. Could anyone explain this part as if i've never taken physics before. thanks.
 


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kpx001 said:


ok basiclly i have no physics knowlledge and the profeesor assumes that we all know how to do these types of problems so he skipped it. I dont have resource materials but this is all i know.

I know i need to take slab.
so area = pi(5)^2
so the area is 25pi
The density of liquid is 40
F = 25pi * 40 = 1000
Work = F* D

the tank is half full so i start at 4 initialy and im moving up. (x+4) is the distance i use then (x+2) because of the spout thing? i dont understand how to get distance and what changes. Could anyone explain this part as if i've never taken physics before. thanks.
Click to expand...

Work is mgd cos(phi). However, since you are given "weight" the acceleration due to gravity, g, is included in the definition of weight. You are told the liquid has a weight of 40 pounds / ft^3 times the number of ft^3 that you have (100 pi). And since you're moving it straight up (180 degrees with respect to gravity), cos(phi) is just -1 (lifting gives us a negative by definition). I realize the "negative" work done is still tiring, but that's physics for you.

But, to continue with the explanation, since you're pumping all the liquid out, you don't actually start from the top of the liquid to compute the distance. Computing it that way would just take the surface of the liquid up to the top. That's a lot of work, I know, but you've still got about twice that amount to do. Logically, you have to do enough work to move the entire mass of liquid through the entire displacement. That's where distance comes from.
 
Was physics a prereq. for this class??!!

If it wasn't - then you should complain to the professor - vehemently.

If it was - then you should drop this class, take physics - and come back to it.
 
\(\displaystyle work = \int WALT\)

W = weight density
A = cross-sectional area as a function of y
L = lift distance for a representative "slab" of liquid in terms of y
T = "slab" thickness ... dy

for your problem ...

\(\displaystyle W = 40\)
\(\displaystyle A = 25\pi\) ... a constant
\(\displaystyle L = 10 - y\)
\(\displaystyle T = dy\)

\(\displaystyle work = \int_0^4 40(25\pi)(10-y)dy\)
\(\displaystyle work = \int_0^4 1000\pi(10-y)dy\)
 
How did u get the function, and also the distance? its half full so i understand 0 to 4 to go to the top of the cylinder but dont i take in account the additional +2 feet i have to go up?
and for the function did u split it into a graph so u have the points 0,5 and and 5, 8 ? thanks
 
kpx001 said:
How did u get the function, and also the distance? its half full so i understand 0 to 4 to go to the top of the cylinder but dont i take in account the additional +2 feet i have to go up?
and for the function did u split it into a graph so u have the points 0,5 and and 5, 8 ? thanks
Click to expand...

If the function wasn't provided for you, you need to check the prerequisites for this class or complain to your professor.

But you are taking into account a height of 10 feet. The limits of integration are for the function that gives the force required to move this liquid through a distance. The force is variable here, and that is the function we're integrating.

What I wrote earlier was only partially correct, and I apologize for the confusion (I still think lifting is negative by definition, but this is not the forum for that discussion).

I wrote that you needed to move the entire mass of liquid through the entire distance. That's not exactly correct, because once the first drop comes out of the tube, the force decreases (less liquid doesn't weigh as much as your original volume). Weight starts coming off. The limits of integration are from the top of the cylinder of water, which is where the force is at its maximum value still, to the bottom (4 feet lower), where the force would be 0, because there is no liquid left.

The force function starts decreasing once the liquid weighs less. So, that's why we need the integration.

The distance over which that force changes is represented as 10-y. The 10 is sort of a maximum value that you will have to pump any drop of water with the entire force of the column bearing down on you (the diameter of the tube doesn't affect how much WORK you do, so just add that height). I'm sure someone will correct me if I'm wrong, but I think it should always be written as ((top-to-bottom-distance) - y) with this type of setup. As I said, if physics was not a prerequisite and you weren't given this function, complain.
 
How did u get the function, and also the distance? its half full so i understand 0 to 4 to go to the top of the cylinder but dont i take in account the additional +2 feet i have to go up?
Click to expand...

0 to 4 is not to get to the top of the cylinder.

Just focus on Skeeter’s explanation and your tank drawing.

“y” is the height of a slab from the bottom of the tank. The liquid is only in the bottom 4 feet of the tank, so the limits of integration will be from 0-4 along the y axis.

Now consider a random slab somewhere in the liquid. It’s height from the bottom of the tank is “y”, and 10 is the height the liquid must be raised to, so the distance each slab must be raised is “10-y”.

So, where do we see Work = Force x Distance in all this? The distance is the (10-y) and the force is the 1000pi(dy).

Hope that helps.
 
i somewhat get it. i thought that since its half full thats like similar to natural length. i get the 10-y because that changes. so i would be moving the bottom 0 to 4 feet of water up an additional 4 feet, which is 0 to 4. to help my understanding lets say i have 2 tanks, one empty and one full. the empty intergration would be from 0 to 8. and the full tank would be 0, because there would not need to be any work required to move it up because its already there?
 
“i somewhat get it. i thought that since its half full thats like similar to natural length. i get the 10-y because that changes. so i would be moving the bottom 0 to 4 feet of water up an additional 4 feet, which is 0 to 4. to help my understanding lets say i have 2 tanks, one empty and one full. the empty intergration would be from 0 to 8. and the full tank would be 0, because there would not need to be any work required to move it up because its already there?”
Click to expand...

No. The limits of integration are stating where the liquid is when you start – not where you are moving it to/where it ends up.

Consider a slab of liquid at the bottom of the tank. Its thickness is dy and its location is at y = 0. Every slab needs to end up at the 10 foot level, so this slab needs to be moved 10 – 0 = 10 feet straight up.

Now consider a different slab, this time a the top of the liquid. Its location is at y = 4, so it needs to be moved a distance of 10 – 4 = 6 feet straight up.

The integration calculates the work needed to move all the slabs (the entire volume of liquid) up to the 10 foot level.

***
Re your empty and full tank examples:

On an empty tank, the limits of integration would be from 0 to 0 because there is no liquid.

On a full tank, the limits of integration would be from 0 to 8.
 
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