Work to Pump Water Out of Tank: Find W w/ 62.5 lb/ft3

In summary: So that's why you have to subtract y from 5. The integral of 9(12/5)(5-y)dy from y=0 to 5 is 50625 ft-lb.So the key is what's the height in terms of y. Not the other way around. So you have it backwards. The height is 5-y. Not y.[/QUOTE]In summary, the problem involves finding the work W required to pump water out of a tank given that the tank is full of water and water weighs 62.5 lb/ft3. The solution involves setting up a formula for a rectangular box of water and using the relationship between the two rectangles to find the area
  • #1
RedBarchetta
50
1

Homework Statement


A tank is full of water. Find the work W required to pump the water out of the spout. Use the fact that water weighs 62.5 lb/ft3.
a = 5
b = 9
c = 12

6-4-024alt.gif


The Attempt at a Solution


So for this one, you have to set up a formula for a rectangular box of water.

The biggest rectangle is 5(or y) by 9, in relation to a smaller rectangle that is w by 9. I can also setup a relationship between the two to find that y/w=9/9 or y=w. So therefore we can produce an equation for the area of one rectangle.

dA=9y

Now the thickness of this rectangle is extremely small so the thickness is dy.

dV=9y dy

The density of water is 62.5 lb/ft^3, so this one rectangular-box of water is:

dM=562.5y dy

To find the force, we multiply by the force of gravity, I used 32.2 ft/sec^2.

dF=18112.5y dy

So our work is dF times the distance. This is where I become lost. Would the distance be vertical or horizontal? would it be (5-y) or (12-y)?


Sorry to bother you guys again! :yuck: Thanks for all of the help so far!
 
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  • #2
RedBarchetta said:

Homework Statement


A tank is full of water. Find the work W required to pump the water out of the spout. Use the fact that water weighs 62.5 lb/ft3.
a = 5
b = 9
c = 12

6-4-024alt.gif


The Attempt at a Solution


So for this one, you have to set up a formula for a rectangular box of water.

The biggest rectangle is 5(or y) by 9, in relation to a smaller rectangle that is w by 9. I can also setup a relationship between the two to find that y/w=9/9 or y=w. So therefore we can produce an equation for the area of one rectangle.
Please, please, please tell us exactly what you are doing! I can guess that "y" represents the height in the pool measured vertically and that "w" is the width parallel to the b= 9 ft long side.

If that's what you are doing, then your setup is wrong. Each "layer" of water, at a constant height is a rectangle with 2 edges parallel to the b= 9 foot side and 2 edges parallel to the c= 12 foot side. You do NOT want to work with vertical layers as you appear to be doing. You want to work with "layers" of water that are all at the same depth and so have to be lifted the same height. You need rectangles with two sides parallel to the 9 foot sides and two sides parallel to the long 12 foot sides. But since the bottom of the pool is sloping up, that long sides varies with depth. Looking at the triangular side of the pool (5 by 12) and using "similar triangles" you get y/5= w/12 so that w= (12/5) y. Now, each "layer" of water has area 9(12/5)y and volume 9(12/5)y dy. How much does each such layer weight? To what is the height each such layer must be lifted? How much work is that? Then integrate as y goes from 0 to 5.

dA=9y

Now the thickness of this rectangle is extremely small so the thickness is dy.

dV=9y dy

The density of water is 62.5 lb/ft^3, so this one rectangular-box of water is:

dM=562.5y dy

To find the force, we multiply by the force of gravity, I used 32.2 ft/sec^2.

dF=18112.5y dy

So our work is dF times the distance. This is where I become lost. Would the distance be vertical or horizontal? would it be (5-y) or (12-y)?


Sorry to bother you guys again! :yuck: Thanks for all of the help so far![/QUOTE]
 
  • #3
To what is the height each such layer must be lifted?

Is that 5-y?
 
  • #4
Well, after working the problem I came up with 50625 ft-lb. Wrong.

This problem is a nightmare. :mad:

The triangular relationship stuff didn't work, so I tried a linear relationship.y=12/5y+12

dV=9(12/5y+12) dy
dV=108/5*y + 108

I multiplied by the density of water(62.5 lb/ft^3). d=m/v then d*v=m.

dV=1350y+6750 dy

Now I integrate from zero to five.

[675x^2+6750x](zero to five)=50625 ft-lb...
 
Last edited:
  • #5
If ((12/5)*y+12) is supposed to be the length of the pool along the 'c' direction, then if y=0, that's 12. If y=5, that's 24. If y=(-5), that's 0. You consistently have been refusing to actually say where the origin of the y coordinates are (just like in your previous problem). Where is it? Your expression for some quantity in terms of y doesn't match up with your integration limits, because you don't seem to be connecting the two things.
 
  • #6
RedBarchetta said:
The triangular relationship stuff didn't work, so I tried a linear relationship.y=12/5y+12

dV=9(12/5y+12) dy
dV=108/5*y + 108

I multiplied by the density of water(62.5 lb/ft^3). d=m/v then d*v=m.

dV=1350y+6750 dy

Now I integrate from zero to five.

[675x^2+6750x](zero to five)=50625 ft-lb...
It's hard to follow your working here. y goes from 0 to 5 right? Then this means that at the top, y=0, bottom y=5. But you should note that at the bottom, dV is zero since the volume of each infinitesimal depth decreases the lower you go. This strongly suggests that dV = 9(12/5)(5-y)dy. The 5-y term is because when y=5 at the bottom, dV would then be zero. The height for the formula mgh=Work done, would then be y itself, except we are taking the top as 0.
 

1. How do you calculate the weight of water in a tank?

The weight of water in a tank can be calculated by multiplying the volume of water in the tank by its density. In this case, the density of water is given as 62.5 lb/ft3. So, if the volume of water in the tank is known, the weight can be easily calculated using this formula.

2. What is the formula for calculating the volume of water in a tank?

The volume of water in a tank can be calculated by multiplying the length, width, and height of the tank. This is known as the formula for finding the volume of a rectangular prism. If the tank is not a perfect rectangular shape, the volume can be approximated by dividing the tank into smaller rectangular prisms and adding their volumes together.

3. How does the density of water affect the weight of water in a tank?

The density of water directly affects the weight of water in a tank. The higher the density, the heavier the water will be. This is because density is a measure of how compact the molecules of a substance are. Water has a relatively high density compared to other liquids, which is why it is commonly used as a benchmark for density measurements.

4. Can the weight of water in a tank change over time?

Yes, the weight of water in a tank can change over time. This can happen due to factors such as evaporation, which causes the water to lose weight. Additionally, if the water is being pumped in or out of the tank, the weight will also change. Other factors such as temperature and pressure can also affect the weight of water in a tank.

5. How can the weight of water in a tank be measured accurately?

The weight of water in a tank can be measured accurately using a scale or balance. The tank can be placed on the scale, and the weight can be recorded. Alternatively, the volume and density of the water can be measured separately, and the weight can be calculated using the formula mentioned earlier. It is important to ensure that the measurements are precise and accurate to get an accurate weight of the water in the tank.

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