<h1>Problem</h1>
Water flows through 3 inch diameter pipe at a velocity of 8 ft/s. Find the volumetric flow rate in gpm. Round answer to the nearest digit.
<h1>Solution</h1>
We can calculate the volumetric flow rate using the equation <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>Q</mi><mo>=</mo><mi>A</mi><mi>V</mi></mrow><annotation encoding="application/x-tex">Q = AV</annotation></semantics></math>Q=AV, where <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>Q</mi></mrow><annotation encoding="application/x-tex">Q</annotation></semantics></math>Q is the volumetric flow rate, <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>A</mi></mrow><annotation encoding="application/x-tex">A</annotation></semantics></math>A is the cross-sectional area of the pipe, and <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>V</mi></mrow><annotation encoding="application/x-tex">V</annotation></semantics></math>V is the velocity of the fluid.
<div class="math math-display"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>Q</mi><mo>=</mo><mi>A</mi><mi>V</mi><mo>=</mo><mo stretchy="false">(</mo><mi>π</mi><msup><mi>R</mi><mn>2</mn></msup><mo stretchy="false">)</mo><mi>V</mi><mo>=</mo><mo stretchy="false">(</mo><mi>π</mi><mo stretchy="false">(</mo><mfrac><mrow><mn>0.25</mn><mtext> ft</mtext></mrow><mn>2</mn></mfrac><msup><mo stretchy="false">)</mo><mn>2</mn></msup><mo stretchy="false">)</mo><mo stretchy="false">(</mo><mn>8</mn><mtext> ft/s</mtext><mo stretchy="false">)</mo><mo>≃</mo><mn>0.3927</mn><mtext> cfs</mtext></mrow><annotation encoding="application/x-tex">Q = AV = (\pi R^2)V = (\pi (\frac{0.25 \textrm{ ft}}{2})^2)(8 \text{ ft/s}) \simeq 0.3927 \text{ cfs}</annotation></semantics></math>Q=AV=(πR2)V=(π(20.25 ft​)2)(8 ft/s)≃0.3927 cfs</div>
Converting to gpm:
<div class="math math-display"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mn>0.3927</mn><mfrac><menclose notation="updiagonalstrike"><mrow><mi>f</mi><msup><mi>t</mi><mn>3</mn></msup></mrow></menclose><menclose notation="updiagonalstrike"><mi>s</mi></menclose></mfrac><mo stretchy="false">(</mo><mfrac><mrow><mn>60</mn><menclose notation="updiagonalstrike"><mi>s</mi></menclose></mrow><mrow><mn>1</mn><mtext> min</mtext></mrow></mfrac><mo stretchy="false">)</mo><mo stretchy="false">(</mo><mfrac><mrow><mn>7.48052</mn><mtext> gal</mtext></mrow><menclose notation="updiagonalstrike"><mrow><mi>f</mi><msup><mi>t</mi><mn>3</mn></msup></mrow></menclose></mfrac><mo stretchy="false">)</mo><mo>≃</mo><mn>176</mn><mtext> gpm</mtext></mrow><annotation encoding="application/x-tex">0.3927 \frac{\cancel{ft^3}}{\cancel{s}} (\frac{60 \cancel{s}}{1 \text{ min}}) (\frac{7.48052 \text{ gal}}{\cancel{ft^3}}) \simeq 176 \text{ gpm}</annotation></semantics></math>0.3927s<svg width="100%" height="0.83056em"><line x1="0" y1="100%" x2="100%" y2="0" stroke-width="0.046em"></line></svg>​ft3<svg width="100%" height="1.008548em"><line x1="0" y1="100%" x2="100%" y2="0" stroke-width="0.046em"></line></svg>​​(1 min60s<svg width="100%" height="0.83056em"><line x1="0" y1="100%" x2="100%" y2="0" stroke-width="0.046em"></line></svg>​​)(ft3<svg width="100%" height="0.9345479999999999em"><line x1="0" y1="100%" x2="100%" y2="0" stroke-width="0.046em"></line></svg>​7.48052 gal​)≃176 gpm</div>

Volumetric Flow Rate of Pipe