Journal of electricity, power, and gas . 2 is the corresponding clock diagram. Itwill be noted that the vectors, E; and Ps are reversed 14 JOURNAL OF ELECTRICITY, POWER AND GAS [Vol. XXXII—No. as they were so taken in the delta load mesh ofFig. 1. Assuming a balanced load, E., E= and Es;and I., I1* and l\ are equal, respectively, hence thevectors, h and 1= make angles of 30° with It and IVLet the angle of lag be e. With a balanced load thiswill be the same for all three phases; consequently,the phase difference of E> and I. is 30° — e; and that ofE* and 1= is 30° + e. Hence the power measur
Journal of electricity, power, and gas . 2 is the corresponding clock diagram. Itwill be noted that the vectors, E; and Ps are reversed 14 JOURNAL OF ELECTRICITY, POWER AND GAS [Vol. XXXII—No. as they were so taken in the delta load mesh ofFig. 1. Assuming a balanced load, E., E= and Es;and I., I1* and l\ are equal, respectively, hence thevectors, h and 1= make angles of 30° with It and IVLet the angle of lag be e. With a balanced load thiswill be the same for all three phases; consequently,the phase difference of E> and I. is 30° — e; and that ofE* and 1= is 30° + e. Hence the power measured byW. is \y, = cos (30°and by W=, 3) or cos (e_30°)....(l) Since the circuit is balanced, the Es and Is can-and we may apply the formulae, W«- W= 2 sin 6 sin 30° ?? tan © tan 30° W. + W= 2 cos e cos 30° or \/3 tan e; from which, P. F. = cos arc tan y 3. ()/(W. + W») .. (3) Numerical values may be obtained with greaterfacilitiy, however, by starting again with equations(1) and (2). Using the formulae,. W. = EJ= cos (e + 30°) (2) These equations are plotted, assuming unity val-ues of current and voltage in Fig. 3 with cos e orpower factor as abscissa. It will be noted that thebehavior of W* now becomes clear. It decreases in valuewith the power factor becoming zero at 50per e is 60°, and thereafter negative for largervalues of e. With a non-inductive load or unity P. F.,the meter readings are equal and their sum indicatesthe delivered power; at 50 per cent P. F., one is zeroand the other shows the power; while at zero P. readings are equal and opposite and the entirepower is reactive. The power factor and corresponding phase anglemay be obtained from equations (1) and (2) as fol-lows: By composition and division,W. — Wi cos (e — 30°) — EJ» cos (e + 30°) cos (A -j- B) = cos A. cos B -\- sin A. sin B we have,cos (A - B ) — cos (A + B) = 2 sin A sin B and cos (A — B) -f cos (A + B) = 2 cos A cos BWhence, W= cos e
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