114 ELEMENTS OF ELECTRICAL ENGINEERING. 



ally with reference to the two wattmeters in order that each wattmeter reading may 

 be considered as positive when it represents delivered power. Let e', e" and e'" 

 represent the instantaneous values of the voltages, let i f , i ff and z 7// represent the 

 instantaneous values of the currents in the receiving circuits, and let a and b repre- 

 sent the instantaneous values of the currents in mains I and 2 as shown. Then, 

 taking proper account of signs we have : 



b = i" i"' 



The reading W of the upper wattmeter is equal to the average value of the product 

 of the current a in its current coil and the voltage e' across its voltage coil. 

 That is : 



W = average ae / 

 and similarly 



W" = average be" 



Substituting the above values of a and b in the expression for VV -{- W fr we 

 have : 



W -f W" = average e'i' -f average e"i" + average (e f e"} i'" 



but e f e " is equal to e'" so that : 



W -f W" = average e'i' + average e"i" -f average e"'i"' 



Discussion of arrangement shown in Fig. jojfor the case of harmonic voltages 

 and currents. The sum of the two wattmeter readings, having due regard to the 

 possible necessity of considering the reading of one of the wattmeters as essentially 

 negative, is equal to the total power delivered, whether the voltages and currents 

 be harmonic or not, and whether the system be balanced or not ; but a clear un- 

 derstanding of the necessity for considering one of the readings as negative in certain 

 cases may be most easily attained for the case of a balanced system with harmonic elec- 

 tromotive forces and currents. In order to discuss this case it is necessary to deter- 

 mine the correct clock diagram for representing the facts with due reference to the 

 choice of signs indicated in Fig. 105. To determine this clock diagram consider first 

 that if the positive directions of e> ', e // and e' f ' (and of course of ? y , i" and i'" 

 also) are chosen symmetrically with respect to the receiving circuits (in a given direc- 

 tion around the A-mesh), then the clock diagram is symmetrical like Fig. 99 (the 

 letters a, b and c in Fig. 99 correspond to i',i" and i'" in Fig. 105), and that to 

 reverse the chosen positive direction of e' and z 7 , as shown in Fig. 105, is to reverse 

 the lines in the clock diagram which represent e' and i'. Therefore the clock dia- 

 gram which correctly represents the facts in Fig. 105 is that which is shown in Fig. 106. 

 By carefully considering the signs as represented by the arrows in Fig. 105 we see 

 that a = i / -}-i /// , and b=i" z y// , as shown by the heavy dotted lines in Fig. 

 106. 



Having thus constructed the correct clock diagram we find that the phase difference 

 between a and e' is 30 -f- and that the phase difference between b and e" is 

 30 6, so that : 



W=z.*t*. &(& + 9) 



W" = be"<x* (30 6) 



