Physics problem involving work on pump

In summary, the conversation discusses the work needed to pump kerosene from a buried conical tank to the surface. The work is calculated by summing the work needed to bring a "slice" of kerosene from the bottom-most level to the top of the tank. The distance in part A is 4 + (8-h), and it is recommended to label the axes for better understanding.
  • #1
enh89
7
0

Homework Statement


A conical tank filled with kerosene is buried 4 feet underground. The density of kerosene is 51.2 lbs/ft3. The kerosene is pumped out until the level drops 5 feet. How much work is needed to pump the kerosene to the surface if the variable is given as:

A. The distance between the vertex of the cone and the “slice”?
B. The distance between the top of the cone and the “slice”?*
C. The distance between the surface and the “slice”?*
D. The distance between the final level of the kerosene and the “slice”?*
2qsnzip.jpg

Homework Equations


W = Force*distance
Force=vol*density

The Attempt at a Solution


My first question is- do I have the schematic set up right? Am I integrating the volume of the slice as indicated in A or B (I used A to set up my integral):
2mcwoja.jpg


*I am struggling with where to put the displacement (distance):
A. The distance the slice must travel is the full length minus h (since it is only emptied to a certain depth) plus the 4 ft above ground; so d = (8-h) + 4
B. I would say distance the slice must travel is up the top of the cone (so the origin is at the rim, and h goes down to the slice), plus the 4 ft above ground, so d = h+4
C. I would say the distance the slice must travel is the full length plus the additional 4 ft, subtracted from the depth of h, so d = 12 - h [but why would it be the same as in part A?]
D. The distance here is what made me think I initially set up my integral completely wrong, if so- would the distance then become (5-3)-h = 2-h?
 
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  • #2
enh89 said:

Homework Statement


A conical tank filled with kerosene is buried 4 feet underground. The density of kerosene is 51.2 lbs/ft3. The kerosene is pumped out until the level drops 5 feet. How much work is needed to pump the kerosene to the surface if the variable is given as:

A. The distance between the vertex of the cone and the “slice”?
B. The distance between the top of the cone and the “slice”?*
C. The distance between the surface and the “slice”?*
D. The distance between the final level of the kerosene and the “slice”?*
2qsnzip.jpg

Homework Equations


W = Force*distance
Force=vol*density

The Attempt at a Solution


My first question is- do I have the schematic set up right? Am I integrating the volume of the slice as indicated in A or B (I used A to set up my integral):
You're summing (by integrating) the work needed to bring a "slice" of kerosene to the surface. The limits of integration will be the bottom-most level that's pumped (lower limit) to the top of the tank (upper limit).

It would be helpful in your drawing to explicitly label the axes, especially the vertical axis. That will help you get a handle on the various distances for the four parts of this problem.
enh89 said:
2mcwoja.jpg


*I am struggling with where to put the displacement (distance):
A. The distance the slice must travel is the full length minus h (since it is only emptied to a certain depth) plus the 4 ft above ground; so d = (8-h) + 4
B. I would say distance the slice must travel is up the top of the cone (so the origin is at the rim, and h goes down to the slice), plus the 4 ft above ground, so d = h+4
C. I would say the distance the slice must travel is the full length plus the additional 4 ft, subtracted from the depth of h, so d = 12 - h [but why would it be the same as in part A?]
D. The distance here is what made me think I initially set up my integral completely wrong, if so- would the distance then become (5-3)-h = 2-h?
 
  • #3
Thanks for your tip, Mark!
So I corrected the limits to be from just the slab that is moving out 3 ft - 8 ft, and the distance as 4 + (8-h).
Does this seem correct to you, or would you have any other comments?

2m3n2nm.jpg



Mark44 said:
You're summing (by integrating) the work needed to bring a "slice" of kerosene to the surface. The limits of integration will be the bottom-most level that's pumped (lower limit) to the top of the tank (upper limit).

It would be helpful in your drawing to explicitly label the axes, especially the vertical axis. That will help you get a handle on the various distances for the four parts of this problem.
 
  • #4
Which part of the problem (a, b, c, or d) is this the work for?
 
  • #5
Mark44 said:
Which part of the problem (a, b, c, or d) is this the work for?
Sorry, just A
 

1. What is the definition of work in physics?

In physics, work is defined as the amount of force applied over a distance. It is measured in joules (J).

2. How is work calculated in a pump problem?

In a pump problem, work is calculated by multiplying the force applied to the pump by the distance the pump is moved. The formula for work is W = F x d, where W is work, F is force, and d is distance.

3. What is the unit of measurement for work?

The unit of measurement for work is joules (J). However, in certain situations, other units such as foot-pounds (ft-lb) or calories (cal) may be used.

4. How does the work done on a pump affect its efficiency?

The work done on a pump directly affects its efficiency. The more work that is done on the pump, the more energy it will use, resulting in a decrease in efficiency. This is because some of the work done is lost to friction and other factors, reducing the overall efficiency of the pump.

5. Can the work done on a pump ever be negative?

Yes, the work done on a pump can be negative in certain situations. This can occur when the pump is acting as a generator, converting the energy of the fluid into mechanical energy. In this case, the work is being done by the pump rather than on it, resulting in a negative value for work.

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