Problemas de mecanica de fluidos

Mechanics of Fluids
Solutions Manual

Mechanics of Fluids
Eighth edition

Solutions manual
Bernard Massey
Reader Emeritus in Mechanical Engineering University College, London

Revised by

John Ward-Smith
Formerly Senior Lecturer in Mechanical Engineering Brunel University

Seventh edition published by Stanley Thornes (Publishers) Ltd in 1998 Eighth edition published 2006 byTaylor & Francis 2 Park Square, Milton Park, Abingdon, Oxon OX14 4RN Simultaneously published in the USA and Canada by Taylor & Francis 270 Madison Ave, New York, NY 10016 Taylor & Francis is an imprint of the Taylor & Francis Group
This edition published in the Taylor & Francis e-Library, 2005. “To purchase your own copy of this or any of Taylor & Francis or Routledge’s collection of thousands ofeBooks please go to www.eBookstore.tandf.co.uk.”

© 2006 Bernard Massey and John Ward-Smith The right of B. S. Massey and J. Ward-Smith to be identified as authors of this work has been asserted by them in accordance with the Copyright Designs and Patents Act 1988. All rights reserved. No part of this book may be reprinted or reproduced or utilised in any form or by any electronic, mechanical, orother means, now known or hereafter invented, including photocopying and recording, or in any information storage or retrieval system, without permission in writing from the publishers. The publisher makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any efforts or omissionsthat may be made. British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging in Publication Data A catalog record for this book has been requested
ISBN 0-203-01231-3 Master e-book ISBN

ISBN 0–415–36204–0 (Print Edition)

Chapter 1

1.1

Since pV = mRT, ∴ V1 =

V1 T1 p2 = V1 T2 p 1

288.15 1.1π (20 m)3 = 56.2 m3 6 233.15 101.3

1.2

=

1.4 × 105 N · m−2 p = = 1.51 kg · m−3 RT 287 J · kg−1 · K−1 × 323.15 K ∂p p − p0 Assume K constant. Then ln( / 0 ) = ∂ K p − p0 81.7 × 106 = 1025 kg · m−3 exp = 0 exp K 2.34 × 109 = 1061 kg · m−3

1.3

K= ∴

1.4

= R=

2 × 10−5 N · s · m−2 µ = 1.333 kg · m−3 = ν 15 × 10−6 m2 · s−1 1.013 × 105 N · m−2 p = 259.2 J · kg−1 · K−1 = T 1.333 kg ·m−3 × 293.15 K 8310 = 32.06 259.2

∴M= 1.5

µ = ν = 400 × 10−6 m2 · s−1 × 850 kg · m−3 = 0.34 Pa · s Velocity gradient = 0.12 m · s−1 = 1200 s−1 0.1 × 10−3 m

Area = π 0.2 × 1.2 m2 = 0.754 m2 Force = 0.754 m2 × 0.34 Pa · s × 1200 s−1 = 307.6 N

2

Solutions manual 1.6 Total force on plate = Area × µ = (0.25 m)2 × 0.7 Pa · s = 1.439 N 1.7 For annulus, radius r, width δr Force = Area × µ× Velocity ωr = 2π rδrµ Clearance c 3 µω ∴ Torque = Force × r = 2π r δr c
R 0

∂u ∂y

+
side A

∂u ∂y

side B

0.15 m · s−1 0.15 m · s−1 + 0.019 m 0.006 m

Total torque = = 1.8 p= h=

2π r3

µω π R4 µω dr = c 2c

π(0.1 m)4 0.14 Pa · s × 2π × 7 rad · s−1 = 7.44 N · m 2 × 0.00013 m

2 × 0.073 N · m−1 2γ = 36.5 Pa = 0.004 m d 4 × 0.073 N · m−1 × 1 4γ cos θ = gd 1000 kg · m−3 ×9.81 N · kg−1 × 0.005 m

1.9

= 0.00595 m = 5.95 mm 1.10 h= 4 × 0.377 N · m−1 × cos 140◦ (13.56 − 1)1000 kg · m−3 × 9.81 N · kg−1 × 0.006 m

= −1.563 mm 1.11 Re = u= 1.12 4Q 4 × 0.0025 m3 · s−1 × 900 kg · m−3 ud = = = 1508 µ π dµ π 0.05 m × 0.038 N · s · m−2 2000 × 0.038 N · s · m−2 2000µ = 1.689 m · s−1 = d 0.05 m × 900 kg · m−3 4 × 0.01 m3 · s−1 4Q = 430 ∴ Laminar = π dµ π 0.08 m × 370 ×10−6 m2 · s−1

Re =

Chapter 2

2.1 2.2

h=

200 × 103 N · m−2 p = = 12.82 m g 1590 kg · m−3 × 9.81 N · kg−1

Pressure depends only on depth below free surface. (a) p = gh = (820 kg · m−3 × 9.81 N · kg−1 )(3 − 0.15) m = 22 930 N · m−2 = 22.93 kPa (b) (c) p = 820 × 9.81 N · m−3 × (3 + 2) m = 40.2 kPa p = 820 × 9.81 N · m−3 × {3 + 2 − (1.2 sin 30◦ + 0.6)} m = 820 × 9.81 × 3.8 N · m−2...