The Theory of Electric Cables and Networks

📖 This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1908 edition. Excerpt: ... =400x7.5 =3 kws., and P min =200 x 5 = 1 kw. When the resistance z of the earth connexion of the middle main is 2 ohms, we have, by (26), Vt=x,'(F+x) F2= 100 6=16.7 volts approx., and therefore, F/= 216.7 and F3'=--183.3 volts. We easily find by (a), (&'), (c), and (eT), that Cm. = (200--16.7)300/400 x 10 = 13.75 amperes, C'min= (200--16.7)100/200 x 10 = 9.17 amperes, P'mux.-21/'10--(1/6)100'2/10 =3.83 kws., and P'mi„. = 200 x 100/10--(1/6) 1002/10 = 1.83 kws. In this case, the current in the earth connexion is 16.7/2, that is 8.35 amperes. If finally we suppose that the middle main is dead earthed so that x is zero, we have by (a"), (b"), (c"), and (d"), C"mm.= (200+100), 2 x 10= 15 amperes, C"m.= 100/10 = 10 amperes, '. = 200-/10 =4 kws., and, P"mtn =200x100/10 =2 kws. We could have predicted at once that the maximum leakage current would in any case have been less than V/P, that is, 20 amperes and that the maximum leakage power could not have been greater than V'-/F, that is, 4 kilowatts. The more complicated formulae, however, us valuable additional information. final example, we shall take the values obtained by ents made in 1900 on a large supply network in this case 190, Fa=85, F3=--20, and =2.5. the limits between which the maximum current to any of the mains must lie, and also the limits for the leakage power. By formulae (a) and (o), we have, ?„,„.=( 105--85)(105+85)/210 x2.5=7.24 amperes, and CmtB =85 X20/105 x2.5 =6.48 amperes. Whatever may have been the actual values of the fault resistances of the mains, the value of the leakage current to the negative main cannot have been less than 6.48 amperes or greater than 7.24 amperes. Similarly, by (c) and (d), we...