Question

Writing to Learn You are in charge of the evacuation and repair of the storage tank shown here. The tank is a hemisphere of radius 10 \(\mathrm{ft}\) and is full of benzene weighing 56 \(\mathrm{lb} / \mathrm{ft}^{3}\) . A firm you contacted says it can empty the tank for 1\(/ 2\) cent per foot-pound of work. Find the work required to empty the tank by pumping the benzene to an outlet 2 ft above the tank. If you have budgeted \(\$ 5000\) for the job, can you afford to hire the firm?

Short answer

Using the given data, we can calculate the work done to pump out the benzene and then multiply it by the cost per foot-pound to check if it is within the budget.

Step-by-step solution

Defining variables and constants

Let's define the necessary variables: height (\(h\)) from top of the tank is ranging from 0 to 10, volume of each slice of benzene (\(dV\)) which is \(2 \pi rh dr\) where \(r\) is \(\sqrt{10^2-h^2}\), weight of the benzene (\(W\)) which is 56 lb/ft^3, distance to pump each slice (\(D\)) which is \(h+2\) feet and gravity (\(g\)) which we'll use 9.8 m/s^2.

Formulate the equation for work

The work to pump out each slice of benzene is \(dW=WdVgD\). Substitute for weight of benzene (\(W = 56lb/ft^3\)), the slice volume (\(dV = 2 \pi rh dr\)), and the slice pumping distance (\(D = h+2\)), we get \(dW = 56 \cdot 2 \pi \cdot \sqrt{10^2 - h^2} \cdot h \cdot dr \cdot g \cdot (h+2)\).

Integrate to find the total work

Let's integrate from 0 to 10 (from top to bottom of the tank) to find the total work done: \(W = \int_0^{10} 56 \cdot 2 \pi \cdot \sqrt{10^2 - h^2} \cdot h \cdot dr \cdot g \cdot (h+2)\). Solving this equation will give us the total work in foot-pounds.

Calculate the cost

To find out the cost of pumping the benzene out of the tank, we need to multiply the total work with the cost per foot-pound which is $0.005. If this cost is less than or equal to the budget of $5000, the firm can be hired.

Key concepts

Work-energy principle

The work-energy principle is a fundamental concept in physics, particularly in the field of mechanics. It asserts that the work done by the forces on an object results in a change in its kinetic energy. However, when dealing with scenarios like pumping liquids from a tank, we concern ourselves with the potential energy as well. In the given exercise, the problem calculates the work needed to pump benzene out of a tank, which can be understood as transferring the potential energy of the benzene at a certain height to a higher elevation, overcoming gravitational forces.

The work done to move an object (in this case, a segment of benzene) a certain distance is given by the formula:
\[ W = F \times D \]
where \( W \) is the work, \( F \) is the force applied, and \( D \) is the distance through which the force is applied. When dealing with fluids, the force involved is due to the weight of the fluid being moved. Therefore, the work to pump out the fluid involves lifting the weight of the fluid through the gravitational field to a certain height.

Integration in physics

Integration is a mathematical tool that plays a vital role in physics, especially when dealing with continuous systems. In the case of pumping benzene out of a hemispherical tank, integration enables us to sum up the work done on an infinite number of infinitesimally small slices or segments of the liquid.

Each slice of liquid has its own small volume \( dV \) and requires a small amount of work to be lifted \( dW \). By integrating over the entire volume of the tank, we aggregate the work needed for each slice, calculating the total work. In calculus terms, integration allows us find the area under the curve of the work function with respect to distance, which in this problem’s case is from the bottom of the tank to the outlet.

Volume of a hemisphere

Understanding the volume of a hemisphere is crucial when solving physics problems involving hemispherical containers like the one in our exercise. A hemisphere is essentially half of a sphere, and therefore to find its volume, one must know the volume formula for a sphere and divide it by two.

The formula for the volume of a sphere with radius \( r \) is: \[ V_{\text{sphere}} = \frac{4}{3} \times \pi \times r^3 \]
Hence, the formula for the volume of a hemisphere is half that amount: \[ V_{\text{hemisphere}} = \frac{1}{2} \times \left( \frac{4}{3} \times \pi \times r^3 \right) = \frac{2}{3} \times \pi \times r^3 \]
In the problem at hand, we use calculus to find the volume of infinitesimally thin slices of benzene, which are effectively the differential volume elements of a hemisphere, given by the formula \( dV =2 \pi rh dr \).

Pumping liquids calculus

The calculus involved in pumping liquids relates to the work required to move the liquid from one place to another, accounting for gravity, the fluid's density, and geometric dimensions of the container. Here, the benzene must be pumped up and out of the hemispherical tank, which involves calculating the amount of work done for each infinitesimal layer of liquid and summing it up using integration.

The total work, \( W \), involved in pumping is found by integrating the work done on each individual slice, as determined by the formula: \[ dW = WdVgD \] where \( dW \) represents the differential work done to move a slice with volume \( dV \), and \( D \) is the distance the liquid must be moved vertically. The integration bounds reflect the physical dimensions of the tank, with the integration carried out from the top to the bottom. The result then gives us the total work which, when multiplied by the cost to convert to actual currency, will determine the feasibility of the project within the budget.