QuestionJune 22, 2025

What is the required pressure to push two inches of water in order to pass through the trap seal? three inches two inches one inch

What is the required pressure to push two inches of water in order to pass through the trap seal? three inches two inches one inch
What is the required pressure to push two inches of water in order to pass through the trap seal? three inches two inches one inch

Solution
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Answer

55.808 lb/ft² Explanation 1. Understand the Problem Determine the pressure needed to push water through a trap seal of given height. 2. Use Hydrostatic Pressure Formula **Pressure (P) = \rho \cdot g \cdot h**, where \rho is the density of water, g is acceleration due to gravity, and h is the height of water column. 3. Calculate for Two Inches Convert inches to feet: 2 inches = \frac{2}{12} feet. Use \rho = 62.4 \, \text{lb/ft}^3, g = 32.2 \, \text{ft/s}^2. P = 62.4 \cdot 32.2 \cdot \frac{2}{12} 4. Simplify Calculation P = 62.4 \cdot 32.2 \cdot \frac{1}{6} 5. Final Calculation P = 334.848 \cdot \frac{1}{6} = 55.808 \, \text{lb/ft}^2

Explanation

1. Understand the Problem<br /> Determine the pressure needed to push water through a trap seal of given height.<br /><br />2. Use Hydrostatic Pressure Formula<br /> **Pressure (P) = \rho \cdot g \cdot h**, where $\rho$ is the density of water, $g$ is acceleration due to gravity, and $h$ is the height of water column.<br /><br />3. Calculate for Two Inches<br /> Convert inches to feet: 2 inches = $\frac{2}{12}$ feet. Use $\rho = 62.4 \, \text{lb/ft}^3$, $g = 32.2 \, \text{ft/s}^2$.<br /> $P = 62.4 \cdot 32.2 \cdot \frac{2}{12}$<br /><br />4. Simplify Calculation<br /> $P = 62.4 \cdot 32.2 \cdot \frac{1}{6}$<br /><br />5. Final Calculation<br /> $P = 334.848 \cdot \frac{1}{6} = 55.808 \, \text{lb/ft}^2$
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