Thermodynamics: An Engineering Approach 8th Edition

Published by McGraw-Hill Education
ISBN 10: 0-07339-817-9
ISBN 13: 978-0-07339-817-4

Chapter 1 - Introduction and Basic Concepts - Problems - Page 42: 1-17

Answer

$V = \frac{\pi \times D^{2}}{4} \times V \times t $

Work Step by Step

Assuming the pool was empty: Since the hose's cross-section is a circle, its area would be: $ A(m^{2}) = \frac{\pi \times D^{2}}{4} $ The hose's flowrate would be: $ v(\frac{m^{3}}{s}) = A(m^{2}) \times V(\frac{m}{s}) $ $ v(\frac{m^{3}}{s}) = \frac{\pi \times D^{2}}{4} \times V $ With a filling time t(s), the volume of the pool would be: $ V(m^{3}) = v(\frac{m^{3}}{s}) \times t(s) $ $ V(m^{3}) = \frac{\pi \times D^{2}}{4} \times V \times t $ P.S. If the pool has an initial volume V_{0} (m^{3}), its total volume would be given by: $ V = \frac{\pi \times D^{2}}{4} \times V \times t + V_{0} $
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