How to Calculate the Work Needed to Pump Water Out of a Hemispherical Tank?

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Discussion Overview

The discussion revolves around calculating the work required to pump water out of a hemispherical tank. Participants explore the mathematical formulation of the problem, including volume integrals and the geometry involved in the setup.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant describes an integral they attempted to evaluate but expresses confusion about their approach.
  • Another suggests using a volume integral and emphasizes the importance of correctly accounting for the work done against gravity.
  • A participant acknowledges an incorrect assumption in their previous reasoning.
  • Another participant proposes a method involving circular disks and provides a specific integral formulation for calculating work.
  • There is a discussion about the relationship between the radius and height in the context of the hemispherical geometry, referencing the Pythagorean theorem.
  • One participant encourages another to think through a geometric relationship independently.
  • A later reply expresses gratitude for the clarification received, indicating some understanding of the concepts discussed.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the problem, as multiple methods and interpretations are presented. Some participants express uncertainty about their calculations and assumptions.

Contextual Notes

There are unresolved mathematical steps and assumptions regarding the setup of the integral and the geometry of the hemispherical tank. The discussion reflects varying levels of understanding and approaches to the problem.

Who May Find This Useful

Students and individuals interested in physics or mathematics, particularly those studying fluid mechanics or integral calculus, may find this discussion relevant.

drklrdbill
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A hemispherical tank of radius 5 feet is situated so that its flat face is on top. It is full of water. Water weighs 62.5 pounds per cubic foot. The work needed to pump the water out of the lip of the tank is ? foot-pounds.


I tried evaluating the integral of pi(125x/3)(5-x)^2 from 0 to 5.

What am I doign wrong? I cannot figure this out for the life of me, spent about the last hour on a seemingly simple problem that continues to stump me.
 
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It's a volume integral you'll need. Suppose the hemisphere is situated so that it is the bottom half of the sphere centred on the origin.

Let the deinsity by p. A tiny volume dV at point (x,y,z) will require you to do work -zpdV to get it to the top of the lip. Now integrate over the volume of the sphere. Note, I can never get my minus signs correct, so watch out for that.
 
woops, my work was based on an incorrect assumption.
sorry.
 
Last edited:
drklrdbill said:
What am I doign wrong? I cannot figure this out for the life of me, spent about the last hour on a seemingly simple problem that continues to stump me.
Here's how I would calculate the work done. First imagine the volume as a stack of circular disks of area [itex]\pi r^2[/itex] and thickness dx. The work needed to lift each disk is [itex]\rho \pi r^2 x dx[/itex], where [itex]\rho[/itex] is the weight per unit volume and [itex]x[/itex] is the height it needs to be lifted. Of course, [itex]r^2 = R^2 - x^2[/itex], so the integral you need is:
[tex]W = \pi \rho \int_{0}^{R} (R^2 - x^2)x dx[/tex]
 
Doc Al said:
[itex]r^2 = R^2 - x^2[/itex]
Where are you getting this relation from?

*edit
is that because of the circle formula x^2 + y^2 = r^2?
 
Last edited:
ShawnD said:
Where are you getting this relation from?
The Pythagorean theorem!

It's a right triangle: R is hypotenuse, x is vertical side, r is horizontal side.
 
Think about it a bit, Shawn. You can answer that on your own. =]

cookiemonster
 
thanks

Thanks for the help. That makes a ton of sense.

peace
 

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