82 Scientific Proceedinc/s^ Royal Dublin Society. 



(4) Calculation of electriccd quantities. 



Eefer all vector quantities to Ci as standard. 



Fo = 4r X 66,000 = 38,000 volts at cos^ = -85, = 32,300 + i x 20,000. 



P. = J X 8,250 = 2,750 lew. 



^ 2,750,000 



^"' = 38,000 X -85 = ^° ^'^P'- 



F", = Fa cosh ml + wCa sinh ml. 



V.coshml = (32,300 + i x 20,000) (-999 + i x '000483) = 32,260 + i x 19,996. 



Cawsinhm/ = 85 x (8-49 + i x 16-1) = 722 + ^ x 1,370. 



.-. V, = 32,982 + i X 21,366 = 39,300 volts to neutral at .sending end. 



V 

 C, = C, cosh ml + — ^ sinh ml. 

 n 



a, cosh ml = 85 X (-999 + i x -000483) = 84-9 + i x -0411. 



Y 



-^ sinh ml = (32,300 + i x 20,000) (- -0625 + i x 113-9) x lO"" = - 2-28 + i x 3-679. 



.-. 6\ = 82-62 + i X 372 = 82-8 amps, per phase at sending end. 



Pj = 32,982 x 82-62 + 21,366 x 3-72 watts = 2,725,000 + 79,500 = 2804 kw. 

 2,750 



/ 33,020- + 21.370- ..,. 

 ^' V32,300' + 20,000- " ' 



It is thus seen that the electrical performance, calculated for maximum load, 

 is very satisfactory. 



(5) Determination of coromi loss. 



Take y = -83 for 19/13 cable. 

 9-96 X 30 

 ^ - 273 + 20 = ^ ^^• 

 Vc = 123,400 X -83 x 1-02 x -23 x 2-67 = 64,300. 



As the working voltage to neutral (39,300) is less than this figure, there will be 

 no corona loss. 



14. Seeing that the example in the last paragraph represents a comparatively 

 short line, and the voltage is not extremely high, it is likely the simpler formula; 

 of paragraph 12, in which /if and S are omitted, would lead to close results. 

 This may be verified. 



Fr = (38,000 x -85 + 85 k -.34 x 25)- + (38,000 x •5-<!7 + 85 x -648 x 25f 



= (32,300 + 722)- + (20,000 + 1,375)- = 1-546 x 10^ ; 

 .-. Fi = 39,300. 



