Problem 60 - University Physics with Modern Physics

Question

On the afternoon of January 15, 1919, an unusually warm day in Boston, a 17.7-m- high, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-m- deep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of 1600 \(\mathrm{kg} / \mathrm{m}^{3} .\) If the tank was full before the accident, what was the total outward force the molasses exerted on its sides? (Hint: Consider the outward force on a circular ring of the tank wall of width \(d y\) and at a depth \(y\) below the surface. Integrate to find the total outward force. Assume that before the tank ruptured, the pressure at the surface of the molasses was equal to the air pressure outside the tank.)

Step-by-Step Solution

Answer

The total outward force exerted by the molasses was approximately \(3.14 \times 10^7 \, \mathrm{N}\).

Step 1: Understand the Problem

The problem involves calculating the total outward force exerted by molasses on the side of a cylindrical tank before it ruptured. We need to use principles of fluid pressure and integration to find this force, considering the tank's dimensions and molasses density.

Step 2: Define the Key Variables

Identify the following variables:- Density of molasses, \( \rho = 1600 \, \mathrm{kg/m^3} \)- Height of the tank, \( h = 17.7 \, \mathrm{m} \)- Radius of the tank, \( r = \frac{27.4}{2} \, \mathrm{m} = 13.7 \, \mathrm{m} \)

Step 3: Formula for Pressure at Depth

The pressure at a depth \( y \) in a fluid column of height \( h \) is given by the formula: \( P = \rho g y \), where \( g = 9.81 \, \mathrm{m/s^2} \) is the acceleration due to gravity. This applies since the surface pressure equals atmospheric pressure, which cancels out as per the problem's setup.

Step 4: Element of Force on Tank Wall

Consider an elemental strip at depth \( y \) with thickness \( dy \) and height \( y \). The surface area of this strip is \( dA = 2 \pi r dy \). The force on this strip is \( dF = P \cdot dA = \rho g y \cdot 2 \pi r \cdot dy \).

Step 5: Integrate to Find Total Force

Integrate the expression for \( dF \) over the entire height of the tank to get the total force:\[F = \int_{0}^{h} 2 \pi r \rho g y \, dy\]Substitute \( r = 13.7 \, \mathrm{m}, \rho = 1600 \, \mathrm{kg/m^3}, g = 9.81 \, \mathrm{m/s^2}, h = 17.7 \, \mathrm{m} \) and calculate.Compute the definite integral:\[F = 2 \pi \rho g r \left[ \frac{y^2}{2} \right]_{0}^{17.7} = 2 \pi \rho g r \left( \frac{17.7^2}{2} \right)\]Calculate the numerical value.

Step 6: Calculate the Numerical Value

Plug in the numbers and compute:\[ F = 2 \times 3.1416 \times 1600 \times 9.81 \times 13.7 \times \frac{17.7^2}{2} \]\[ F \approx 3.14 \times 10^7 \, \mathrm{N} \]

Key Concepts

Cylindrical Tank Rupture

In fluid mechanics, understanding how forces operate on cylindrical tanks helps explain failure events like the 1919 molasses disaster. This happens when the structural limits of the tank are exceeded, often due to pressure differences within and outside the tank. The cylindrical shape is usually preferred for storing liquids as it offers uniform pressure distribution. However, when a rupture occurs, the entire stored fluid can be released catastrophically, as seen in the Boston incident. The molasses within created an outward force on the tank walls. Due to the tank's shape, stress was distributed around the circular walls. This incident highlights the importance of adequate design considerations against possible overpressure situations.
The molasses tank in Boston was vulnerable to rupture because of its limited capacity to withstand internal fluid pressure. When the strength of the tank wall isn't sufficient for the weight of the stored liquid, breakage can be sudden. This creates hazards not just from the material inside the tank but also from its kinetic energy once released.
Learning how such forces develop in cylindrical tanks helps engineering professionals to devise better containment systems that can withstand similar stresses in various scenarios.

Pressure Calculation

Calculating fluid pressure in the context of tanks involves understanding how depth affects the exerted force. The force at any given point inside a fluid contained within a tank is influenced by fluid density, gravitational force, and the depth of the point from the fluid's surface.
For cylindrical tanks, pressure is uniform at a given depth. However, as you move deeper, the pressure increases linearly. This principle is captured by the formula \( P = \rho g y \), wherein \( P \) is pressure, \( \rho \) represents fluid density, \( g \) is gravity, and \( y \) is depth. In our example, with molasses having a density of 1600 \( \text{kg/m}^3 \), it exerts substantial pressure at the bottom due to both its density and the tank’s full height.
Such calculations are vital not only for assessing the attractiveness of certain storage designs but also for predicting pressure-related issues that might necessitate structural support adjustments. It ensures that engineers can preemptively address potential failure risks by calculating maximum pressures during the design phase.

Integration in Physics

In physics, integration is a key mathematical tool for summing infinitesimally small contributions across an object’s dimensions—such as calculating the total force exerted by a fluid on tank walls. It allows for determining the summed effects from a continuous distribution of pressures, forces, or other characteristics.
The exercise in calculating the force from molasses illustrates the integration process in a real-world context. The small ring-like elements of the tank at various depths contribute differently to the overall force based on their respective pressures. By integrating the expression \( dF = \rho g y \cdot 2 \pi r \cdot dy \), one can find the total force exerted across the entire depth of the tank. This technique sums up all the individual small forces starting from the bottom to the top level of the molasses.
This approach is critical in engineering practice, as it helps accurately model physical systems where properties such as pressure or force change with position. Here, integration gives a comprehensive assessment of total stresses, crucial for designing safe and effective storage systems.