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Ch. 6 - Applications of Definite Integrals
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 6 - Applications of Definite IntegralsProblem 6.PE.33
Chapter 6, Problem 6.PE.33

Work
Pumping a conical tank A right-circular conical tank, point down, with top radius 5 ft and height 10 ft, is filled with a liquid whose weight-density is 60lb/ft³. How much work does it take to pump the liquid to a point 2 ft above the tank? If the pump is driven by a motor rated at 275ft-lb/sec (1/2 hp), how long will it take to empty the tank? 

Verified step by step guidance
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First, set up a coordinate system to describe the height of the liquid in the tank. Let the vertical axis \( y \) measure the distance from the vertex (point) of the cone upward, with \( y=0 \) at the tip and \( y=10 \) ft at the top of the tank.
Express the radius \( r \) of the tank at height \( y \) using similar triangles. Since the radius is 5 ft at the top (\( y=10 \)), the radius varies linearly with height: \( r(y) = \frac{5}{10} y = \frac{y}{2} \).
Consider a thin horizontal slice of liquid at height \( y \) with thickness \( dy \). The volume of this slice is the area of the circular cross-section times \( dy \): \( dV = \pi r(y)^2 dy = \pi \left( \frac{y}{2} \right)^2 dy = \pi \frac{y^2}{4} dy \).
Calculate the weight of this slice using the weight-density \( w = 60 \) lb/ft³: \( dW_{weight} = w \cdot dV = 60 \pi \frac{y^2}{4} dy = 15 \pi y^2 dy \).
Determine the distance the slice must be lifted to reach 2 ft above the tank. Since the top is at 10 ft, the liquid must be pumped to height \( 12 \) ft. The distance lifted for the slice at height \( y \) is \( 12 - y \). The work done to lift this slice is \( dW = (\text{weight of slice}) \times (\text{distance lifted}) = 15 \pi y^2 (12 - y) dy \). Integrate this expression from \( y=0 \) to \( y=10 \) to find the total work:
\[ W = \int_0^{10} 15 \pi y^2 (12 - y) \, dy \]
To find the time to empty the tank, first compute the total work \( W \) from the integral above. Then, use the motor power rating of 275 ft-lb/sec. The time \( t \) in seconds is given by:
\[ t = \frac{W}{275} \]

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Work in Pumping Liquids

Work is the energy required to move a liquid against gravity. For pumping problems, it is calculated by integrating the force (weight of liquid) times the distance each slice of liquid is lifted. The weight depends on the volume and weight-density, and the distance varies with the height of the liquid slice.
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Volume of a Conical Slice

To find the volume of a thin horizontal slice of liquid in a conical tank, use the formula for the volume of a disk: area times thickness. The radius of the slice changes linearly with height due to the cone’s shape, so similar triangles help express radius as a function of height.
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Power and Time Relationship

Power is the rate of doing work, measured in ft-lb/sec here. To find the time to empty the tank, divide the total work by the power output of the motor. This gives the duration needed to pump all the liquid out at the given power.
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Related Practice
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