Physics problem involving work on pump

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Homework Help Overview

The problem involves calculating the work required to pump kerosene from a conical tank that is buried underground. The tank's dimensions and the density of kerosene are provided, along with various interpretations of the distances involved in the work calculation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem, particularly the integration limits and the distances that need to be considered for different parts of the problem. There are questions about whether the schematic is correct and how to accurately express the distance each slice of kerosene must travel.

Discussion Status

Some participants have provided feedback on the setup, suggesting corrections to the limits of integration and the distances involved. There is ongoing exploration of the different interpretations of the problem parts, but no consensus has been reached on a single approach.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information available for discussion. There is an emphasis on understanding the setup and assumptions rather than arriving at a final solution.

enh89
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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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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?
 
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.
 
Which part of the problem (a, b, c, or d) is this the work for?
 
Mark44 said:
Which part of the problem (a, b, c, or d) is this the work for?
Sorry, just A
 

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