Problem 5 - Applied Physics

Question

A pump is rated to deliver \(50.0\) gal/min. Find the velocity of water in (a) a 6.00-in.-diameter pipe and (b) a 3.00-in.-diameter pipe.

Step-by-Step Solution

Answer

6.00-in. pipe: 0.567 ft/s; 3.00-in. pipe: 2.271 ft/s.

Step 1: Find the volumetric flow rate

The volumetric flow rate of the pump is given as \(50.0\) gallons per minute. To work in standard units, first convert this flow rate to cubic feet per second (cfs).\[\text{1 gallon} = 0.13368 \text{ ft}^3\text{, so } 50.0 \text{ gal/min} = 50.0 \times 0.13368 \text{ ft}^3/\text{min} = 6.684 \text{ ft}^3/\text{min}.\] Next, convert minutes to seconds: \[6.684 \div 60 \approx 0.1114 \text{ cfs}.\]

Step 2: Calculate the cross-sectional area of the 6.00-in.-diameter pipe

The diameter of the pipe is \(6.00\) inches, which can be converted to feet by dividing by \(12\):\[\frac{6.00}{12} = 0.500 \text{ ft}.\]Now, calculate the cross-sectional area using the formula for the area of a circle, \(A = \pi r^2\), with \(r = \frac{d}{2}\):\[A = \pi \left(\frac{0.500}{2}\right)^2 = \pi (0.25)^2 \approx 0.1963 \text{ ft}^2.\]

Step 3: Calculate the velocity in the 6.00-in.-diameter pipe

Use the formula \( V = \frac{Q}{A} \), where \( V \) is the velocity, \( Q \) is the flow rate, and \( A \) is the cross-sectional area.\[V = \frac{0.1114}{0.1963} \approx 0.567 \text{ ft/s}.\]

Step 4: Calculate the cross-sectional area of the 3.00-in.-diameter pipe

The diameter of this pipe is \(3.00\) inches, which converts to feet: \[\frac{3.00}{12} = 0.250 \text{ ft}.\] Calculate the cross-sectional area: \[A = \pi \left(\frac{0.250}{2}\right)^2 = \pi (0.125)^2 \approx 0.0491 \text{ ft}^2.\]

Step 5: Calculate the velocity in the 3.00-in.-diameter pipe

Use the formula for velocity again: \( V = \frac{Q}{A} \).\[V = \frac{0.1114}{0.0491} \approx 2.271 \text{ ft/s}.\]

Key Concepts

Volumetric Flow Rate

When we're discussing fluid dynamics, one important concept is the volumetric flow rate. This measures how much volume of fluid passes through a point in the system per unit of time, usually given in gallons per minute (gal/min) or cubic feet per second (cfs).

• It's essential to convert units depending on calculations. For example, converting from gallons per minute to cubic feet per second is a critical step in many engineering problems.
• The conversion factor from gallons to cubic feet is: \( \text{1 gallon} = 0.13368 \text{ ft}^3 \).
• Furthermore, converting minutes to seconds typically involves dividing by 60.

In our example, a pump delivers \(50.0\) gal/min, which needs conversion to cubic feet per second (cfs) for consistency with other calculations, resulting in approximately \(0.1114\) cfs.

Pipe Diameter

The diameter of a pipe is straightforward but critical in flow dynamics. It directly influences the cross-sectional area through which the fluid flows.

• Conversion to feet is common, as calculations often use feet for distance and area measurements.
• Remember, to convert from inches to feet, divide by 12 since there are 12 inches in a foot.

In our examples, we have pipes of \(6.00\) inches and \(3.00\) inches in diameter, which convert to \(0.500\) feet and \(0.250\) feet respectively.

Cross-Sectional Area

The cross-sectional area of a pipe significantly impacts the flow rate and velocity. It is the slice or section through the pipe perpendicular to the flow and is computed using the formula for the area of a circle, \( A = \pi r^2 \).

• The radius \( r \) is half the diameter \( d \) of the pipe, \( r = \frac{d}{2} \).
• For fluid dynamics problems, it's calculated in square feet (ft²) when working with fluid flow in feet per second (ft/s).

For a \(6.00\)-inch diameter pipe, the cross-sectional area turns out to be approximately \(0.1963\) ft². For a \(3.00\)-inch diameter pipe, it's \(0.0491\) ft².

Velocity Calculation

Calculating the velocity of a fluid inside a pipe involves determining how fast the fluid moves through the pipe's cross-section. The velocity \( V \) can be found using the quotient of volumetric flow rate \( Q \) and the cross-sectional area \( A \), given by the formula:

• \( V = \frac{Q}{A} \).
• A smaller cross-sectional area results in a higher velocity, given the same flow rate, akin to pinching a garden hose.

Applying this to our example, the velocity of water in a \(6.00\)-inch diameter pipe comes to approximately \(0.567\) ft/s, whereas in a \(3.00\)-inch pipe, it reaches about \(2.271\) ft/s, demonstrating how diameter changes impact velocity.