File:Factory and industrial management (1891) (14577783039).jpg

Identifier: factoryindustria15newy (find matches)
Title: Factory and industrial management
Year: 1891 (1890s)
Authors:
Subjects: Engineering Factory management Industrial efficiency
Publisher: New York (etc.) McGraw-Hill (etc.)
Contributing Library: Engineering - University of Toronto
Digitizing Sponsor: University of Toronto

View Book Page: Book Viewer
About This Book: Catalog Entry
View All Images: All Images From Book
Click here to view book online to see this illustration in context in a browseable online version of this book.

Text Appearing Before Image:
emselves (Diagram D) from the water powers to thepoint where they strike the line of the railroad. Under the aboveconditions, if the interest on the additional investment for thesetransmission lines from the cheap source of power to the railwayline in Case i exceeds the difference between Case 3 and Case i incost of power delivered to the sub-station supply feeders, then itis unprofitableJ^to utilize the cheap source of power. Our calculations, therefore, can be confined to the cost of the twotransmission lines in Case i, each loo miles long, from the twosources of water power to the line of railway, and the difference inthe cost of i)roducing power, at the sub-station suj)ply feeders. Ifthe interest on the former exceeds the difference in the latter, then itis unprofitable to use the cheap source of power. In making the calculations, every care has been taken to be fair tothe distant source of power, and wherever then is any room for doubt,the (heap source has been given the benefit.
Text Appearing After Image:
D. RAILWAY BETWEEN NEW YORK AND PHILADELPHIA. DISFANT WATER POWER. 979 980 ELECTRIC TRACTION. We have figured the pounds of copper necessary as follows: Circular mills _ Current X Distance in feet X 20 (Diam. of wires squared) Volts lost in line. ^ , Watts Horse PowerX746 Current = = --^. V Voltage of transmission. Pounds of Copper =:.oi6Xcir. millsXniiles. Therefore : 2 ,, ^ , H. P. X Miles Lbs. of CU.= 1,262,000 r :; r (Voltage of Trans. )X (loss m volts)For 100 miles transmission at 20,000 volts, and a loss of voltage inthe line of 20 per cent, or 4,000 volts, the formula becomes: Lbs. of copper= 158XH.P. This is correct for direct current. For alternating, the inductiveloss would increase the amount of copper required to obtain the sametotal percentage loss, or, if the same amount of copper is used, wouldincrease the total percentage loss. At the same time there would bea saving due to the three-phase wiring system, which would partlycounterbalance the loss due to induction. The a

Note About Images