qr code vb.net library Fig. 257 in Visual Studio .NET

Maker Code 128 Code Set C in Visual Studio .NET Fig. 257

" " " " Let IAB , IBC , and ICA denote the three phase currents. Also let VAB be the reference vector, and let  be the phase angle between the phase voltages and phase currents, as shown in Fig. 258.

Now, in regard to Fig. 257, the power PER PHASE is equal to Pp Vp Ip cos  hence, TOTAL POWER PT 3Vp Ip cos  441

as in eq. (439), where Vp and Ip are the magnitudes of the phase voltages and phase " currents and cos  is the power factor of each of the three equal impedances Z . Inspection of Fig. 257 shows, however, that the line voltage is EQUAL to the phase voltage in a -connected load. Thus eq. (441) can be written as PT 3VL Ip cos  442

where VL is the magnitude of line voltage. In the equation, however, we d like also to have the current expressed in terms of line current; this can be done as follows. Consider (for example) junction point A in Fig. 257; by Kirchho s current law, the current equation at A is equal to " " " IA ICA IAB 0 thus, " " " IA IAB ICA 443 " Now, for simplicity, let s consider IAB as the reference vector (this will have no e ect on " " the relative magnitudes of the phase and line currents). Then, since ICA lags IAB by 2408, and since the phase currents all have equal magnitudes, eq. (443) becomes (angles in degrees) " IA Ip =0 Ip = 240 Ip =0 Ip =120 cos 0 j sin 0 cos 120 j sin 120 Ip thus, " IA 1:5 j0:8660 Ip or since, from inspection of Fig. 257, jIA j jIL j

444

jIA j jIB j jIC j IL Thus, comparison of eqs. (440) and (444) shows that PT is calculated the same way for either a Y-connected or a -connected load. In the above, be reminded that VL and IL are the rms values of line voltage and line current, and thus PT is the total average power produced in a balanced three-phase load. In section 10.7 we mentioned that the total INSTANTANEOUS POWER in a balanced three-phase system is constant. This interesting and important fact can be proved as follows. In Fig. 257, let va , vb , and vc denote the instantaneous values of the three sinusoidal line voltages; then, letting V 0 denote the three equal peak voltages, the instantaneous values of the three voltage waves are va V 0 sin !t vb V 0 sin !t 1208 * vc V 0 sin !t 2408 V 0 sin !t 1208 Then, since the above voltages work into identical loads, the corresponding instantaneous currents would be, letting I 0 denote the three equal peak currents, ia I 0 sin !t  ib I 0 sin !t  1208 ic I 0 sin !t  1208 where  is the phase angle between the voltage and current waves. Then the total INSTANTANEOUS POWER p is equal to p va ia vb ib vc ic which, upon making the above substitutions, becomes p V 0 I 0 sin !t sin !t  sin !t 1208 sin !t  1208 sin !t 1208 sin !t  1208 Note that the above result seemingly says that p is a function of time t, that is, that p varies from instant-to-instant, thus contradicting the statement we made that p remains

* Since it s always understood that !t 2ft radians, it s basically incorrect to write sin !t 1208); instead, since 1208 120 =180 2=3 rad., we should really write sin !t 2=3 . However, the particular operation above is such that there is no harm in writing 1208 instead of 2=3 radians if we wish to do so.