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Content material: - Contents, Pages v-viiiChapter 1 - Parameters, Pages 1-8Chapter 2 - Pump Calculations, Pages 9-18Chapter three - Required facts for Specifying Pumps, Pages 19-20Chapter four - Pump varieties, Pages 21-41Chapter five - standards, Pages 42-44Chapter 6 - Pump Curves, Pages 45-54Chapter 7 - results of Viscosity on Pump functionality, Pages 55-60Chapter eight - Vibration, Pages 61-82Chapter nine - web optimistic Suction Head (NPSH), Pages 66-73Chapter 10 - Pump Shaft Sealing, Pages 74-82Chapter eleven - Pump Bearings, Pages 83-91Chapter 12 - Metallurgy, Pages 92-99Chapter thirteen - Pump Drivers, Pages 100-113Chapter 14 - Gears, Pages 114-120Chapter 15 - Couplings, Pages 121-127Chapter sixteen - Pump Controls, Pages 128-136Chapter 17 - Instrumentation, Pages 137-139Chapter 18 - Documentation, Pages 140-145Chapter 19 - Inspection and trying out, Pages 142-145Chapter 20 - set up and Operation, Pages 146-150Chapter 21 - Troubleshooting, Pages 151-153Appendix 1 - pattern Pump necessities, Pages 154-159Appendix 2 - Centrifugal Pump information Sheet, web page 160Appendix three - inner Combustion Engine information Sheet, web page 161Appendix four - electrical Motor information Sheet, web page 162Appendix five - Centrifugal Pump package deal, web page 163Appendix 6 - greatest plausible Suction Lifts at quite a few Altitudes, web page 164Appendix 7 - advised checklist of owners, Pages 165-174Appendix eight - API-610 Mechanical Seal type Code, web page 175References, web page 176Index, Pages 177-182
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Additional info for An Introduction to Algol 68 through Problems
Hence we store it in a string called 'path'. " then i minusab I; path plusab "N" elif path= " " then stuck := true C entrance is a dead end C else int p =upb path; C now backtrack and prune path C char lastmove = path[p]; if lastmove = "E" then j rninusab I elif lastmove = "S" then i rninusab I elif lastmove = "W' then j plusab l elif lastmove = "N" then i plusab 1 fi; path:= path [I :p- I] fi end' od; if stuck then print("N 0 PATH POSSIBLE") else print (("MAZE SOLVED BY FOLLOWING THE PATH" ,newline,path)) fi There is a problem if the path has length greater than the width of the lineprinter page, since the print routine abandons the output of a string when the end of the line is reached.
1 Perform this algorithm for the date 29 Feb 2000. Apart from the input and output, the algorithm comprises two logically distinct calculations, and their isolation as separate steps makes it easier to follow and check. A self-contained piece of program that takes some data like 'y', or even no data, and defines some operation possibly producing a result, like 'advance days(y)', can be made into a procedure. There are standard procedures like 'sin' or 'sqrt' that can be used in any program by writing 'sin(x)' or 'sqrt(x)' where 'x' is real.
3 Given the declaration 'struct (string nationality, int age, struct(int day,[I :3) char month, int year)birthdate) john', write assignment statements to make John British and born on Christmas day 1940. We now use the procedural approach in writing a program for the calendar problem, including the new language features. The input is a CALENDARS: PROCEDURES 41 date, like 3 MAY 1999 . 'read(date)' will read this row of denotations into their respective fields. The output is to be of the form "3 MAY 1999 IS A MONDAY".