Work required to pump water out of conical tank

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Homework Help Overview

The problem involves calculating the work required to pump water out of a conical reservoir with specific dimensions and initial water depth. The context is centered around concepts of integration and physical principles related to work and fluid density.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the integral needed to calculate work, including the limits of integration and the expression for the radius as a function of depth. Questions are raised about the use of specific constants and the formulation of the integral.

Discussion Status

There is an ongoing exploration of the integral setup, with some participants providing clarifications and adjustments to the original approach. A new integral formulation has been suggested, and there appears to be a positive reception to the adjustments made by the original poster.

Contextual Notes

Participants note the importance of using the correct density of water as specified by the instructor and the implications of keeping values positive for ease of integration. The discussion reflects on the assumptions made regarding the geometry of the cone and the positioning of the coordinate system.

akbar786
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Homework Statement


Find the work done in pumping all the water out of a conical reservoir of radius 10ft at the top and altitude 8ft if at the beginning the reservoir is filled to a depth of 5ft and the water is pumped just to the top of the reservoir.

Homework Equations


None

The Attempt at a Solution


This is my work so far. I am taking the integral from -8 to -3. The top center of the cone is on (0,0). I have the integral going from -8 to -3 of (100 pi) *(62.4) (0-y) dy. The 100 is from the radius squared. The (0-y) is the distance the water has to travel up any given y. Any help?
 
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You want to integrate pi*r^2*(-y)*dy from -8 to -3, right? So you want to express r as a function of y, don't you? Why are you using 100, the radius squared at the top and where did 62.4 come from?
 
Dick said:
You want to integrate pi*r^2*(-y)*dy from -8 to -3, right? So you want to express r as a function of y, don't you? Why are you using 100, the radius squared at the top and where did 62.4 come from?

62.4 is the density of water the teacher wanted us to use.I decided to keep all my numbers positive which will also make it much easier to integrate.Here is my new integral with expressing r as a function of y. 62.4*pi * integral from 0 to 5 of ((5/4y)^2) * (8-y)
8-y is the distance the water has to travel given the generic y and on each of those y's the radius will be 5/4 of the y term. Is this right? I am solving for work and 62.4 is my density for water
 
akbar786 said:
62.4 is the density of water the teacher wanted us to use.I decided to keep all my numbers positive which will also make it much easier to integrate.Here is my new integral with expressing r as a function of y. 62.4*pi * integral from 0 to 5 of ((5/4y)^2) * (8-y)
8-y is the distance the water has to travel given the generic y and on each of those y's the radius will be 5/4 of the y term. Is this right? I am solving for work and 62.4 is my density for water

Now that looks right to me. Putting the origin at the bottom does make it much less confusing.
 
Awesome, thanks a lot for your help.
 

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