Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world

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2 Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: Pearson Education Limited 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6 10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners. ISBN 10: ISBN 13: British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America

3 Chapter 5 6. Select the reservoir and associated plumbing, including piping, valving, filters and strainers, and other miscellaneous components such as accumulators. 7. Consider factors such as noise levels, horsepower loss, need for a heat exchanger due to generated heat, pump wear, and scheduled maintenance service to provide a desired life of the total system. 8. Calculate the overall cost of the system. Normally the sequence of operation is repeated several times with different sizes and types of components. After the procedure is repeated for several alternative systems, the best overall system is selected for the given application.this process is called optimization. It means determining the ultimate selection of a combination of system components to produce the most efficient overall system at minimum cost commensurate with the requirements of a particular application PUMP PERFORMANCE RATINGS IN METRIC UNITS Performance data for hydraulic pumps are measured and specified in metric units as well as English units. Figure 5-39 shows actual performance data curves for a variable displacement, pressure-compensated vane pump operating at 1200 rpm. The curves give values of flow rate (gpm), efficiency, and power (hp and kw) versus output pressure (psi and bars). This particular pump (see Figure 5-39) can operate at speeds between 1000 and 1800 rpm, is rated at 2540 psi (175 bars), and has a nominal displacement volume of 1.22 in 3 (20 cm 3 or 0.02 L). Although the curves give flow rates in gpm, metric flow rates of liters per minute (Lpm) are frequently specified. EXAMPLE 5-9 A pump has a displacement volume of 100 cm 3. It delivers m 3 /s at 1000 rpm and 70 bars. If the prime mover input torque is 120 N m, a. What is the overall efficiency of the pump? b. What is the theoretical torque required to operate the pump? Solution a. Using Eq. (5-2M), where the volumetric displacement is we have V D 100 cm 3 >rev a 1 m 100 cm b m 3 >rev Q T V D N m 3 >rev2 a rev>sb m3 >s Next, solve for the volumetric efficiency: h y Q A % Q T

4 Hydraulic Pumps Figure English/metric performance curves for variable displacement pressurecompensated vane pump at 1200 rpm. (Courtesy of Vickers, Inc., Troy, Michigan.) Then solve for the mechanical efficiency: h m pq T T A N N>m m 3 >s N # m2a1000 2p 60 rad>sb h m 11,690 N # m>s 12,570 N # m>s % Note that the product T A N gives power in units of N m/s (W) where torque (T A ) has units of N m and shaft speed has units of rad/s. Finally, we solve for the overall efficiency: h o h y h m % 192

5 Chapter 5 b. T T T A h m N # m Thus, due to mechanical losses within the pump, 120 N m of torque are required to drive the pump instead of 112 N m. EXAMPLE 5-10 The pump in Example 5-9 is driven by an electric motor having an overall efficiency of 85%. The hydraulic system operates 12 hours per day for 250 days per year. The cost of electricity is $0.11 per kilowatt hour. Determine a. The yearly cost of electricity to operate the hydraulic system b. The amount of the yearly cost of electricity that is due to the inefficiencies of the electric motor and pump Solution a. First, we calculate the mechanical input power the electric motor delivers to the pump. Per Eq. (3-37), we have Pump input power 1kW2 T A 1N # m2 N 1rpm kw Next, we calculate the electrical input power the electric motor receive: Electric motor input power Since the electric motor output power equals the pump input power we have Electric motor input power Finally, we determine the yearly cost of electricity: yearly cost power rate time per year unit cost of electricity 14.8 kw 12 hr days $ $4884>yr day year kw hr b. The total kw loss equals the kw loss due to the electric motor plus the kw loss due to the pump. Thus, we have Total kw loss kw Electric motor output power Electric motor overall efficiency 12.6 kw kw 193

6 Hydraulic Pumps Yearly cost due to inefficiencies 4.3 $4884>yr $1419>yr 14.8 Since 4.3/ , we conclude that 29% of the total cost of electricity is due to the inefficiencies of the electric motor and pump. This also means that only 71% of the electrical power entering the electric motor is transferred into hydraulic power at the pump outlet port KEY EQUATIONS Gear pump volumetric displacement: V D p 4 1D2 o D 2 i 2L (5-1) Theoretical flow rate of a pump (flow rate a no-leak pump would deliver): English or metric units: Q T V D N (5-2) English units: Q T 1ft 3 >min2 V D 1ft 3 >rev2 N 1rev>min2 (5-2) Special English Q T 1gpm2 V D 1in 3 >rev2 N 1rpm2 (5-2) units: 231 Q T 1m 3 >min2 V D 1m 3 >rev2 N 1rev>min2 (5-2M) Pump volumetric efficiency: h y Q A Q T (5-3) Vane pump volumetric V D p (5-4) displacement: 2 1D C D R 2eL Piston pump volumetric V D Y AD tan 1u2 (5-5) displacement: 194

7 Chapter 5 Piston pump theoretical flow rate Special English Q T 1gpm2 D 1in2 A 1in2 2 N 1rpm2 Y tan 1u2 (5-6) 231 units: Q T a m3 min b D 1m2 A 1m2 2 N a rev b Y tan 1u2 min (5-6M) Pump mechanical efficiency Special English units: h m p 1psi2 Q T 1gpm2>1714 T A 1in # lb2 N 1rpm2>63,000 (5-8) h m p 1Pa2 Q T 1m 3 >s2 T A 1N # m2 N 1rad>s2 (5-8M) English or h m T T (5-9) T metric units: A Theoretical torque of a pump (input torque a frictionless pump would require from the prime mover) V English units: T T 1in # D 1in 3 2 p 1psi2 lb2 (5-10) 2p V T T 1N # D 1m 3 2 p 1Pa2 m2 2p Actual torque a pump receives from the prime mover (5-10M) Special English units: T A 1in # lb2 actual horsepower delivered to pump 63,000 N 1rpm2 (5-11) T A 1N # m2 actual power delivered to pump 1W2 N 1rad>s2 (5-11M) 195