《电路》（英文版）12-5 Power measurement in three-phase

8 12-5 Power measurement in three-phase system B B W)朱 Three watt-meters method P-P+PB+Pc

§12-5 Power measurement in three-phase system W  W     A B C W  Three watt-meters method P=PA+PB+PC  W  W W     A B C

B w rJo pots D.idt b P=P+PB+Po ( UAria4 +UB ib8 +UGis)dt cW The readings of three watt-meters 丿x=VAN+ BN +yNx vc=van+y nx (UANiaA +UBNibB +Uvic)dt +To(UN(ia4 +ibB +ic)dt AN aA tUBNi bB +UnIo)dt The sum of the average powers taking by the three-phase loads. ∑P watt-meter ∑P load

W   W   W   • a b c C A B ZA ZB ZC N x i i i dt T Bx bB Cx cC T Ax aA ( ) 1 0 =  + +  i dt T p dt T P T Ax aA T A  A  = = 0 0 1 1  P = PA + PB + PC The readings of three watt-meters V Ax V AN V Nx V Bx V BN V Nx VCx VCN V Nx • • • • • • • • • = + = + = + i i i dt T i i i dt T P bB cC T BN bB CN cC Nx aA T AN aA ( ( ) 1 ( ) 1 0 0 = + + + + +       i i i dt T BN bB CN cC T AN aA ( ) 1 0 =  + +  The sum of the average powers taking by the three-phase loads. Pwatt−meter = Pload

B w Let us now consider A the effect of placing point x directly on the one of b*w three lines (x=>B) C cW VBx=0→PB=0 P=P+p DABiaAdt UcBCC =OOAN-UBN iadt +C(Uc -UBN)icdt TJo T TJo DANladt+Dawicdt-DBN (a+ioc )dt L(UANiA +uBvibB+Cicc)dt bB (The sum of the average powers taking by the three-phase loads

Let us now consider the effect of placing point x directly on the one of three lines.(x=>B) W   W   W   • a b c C A B ZA ZB ZC N x = 0 → = 0 • V Bx PB i dt T i dt T P P P T aA CB cC T A C  AB  = + = + 0 0 1 1   i dt T i dt T T BN aA CN BN cC T  AN  = − + − 0 0 ( ) 1 ( ) 1     i dt i dt i i dt T CN cC BN aA cC T AN aA ( ) 1 0 = + − +     i i i dt T BN bB CN cC T AN aA ( ) 1 0 =  + +  bB − i (The sum of the average powers taking by the three-phase loads.)

B矿ZA=2B=ZC12|∠9 w b PA=VARIO4 COS(V AB laA)=v cos( 30+9) bn Pc=VcBIc cos( cb icc)=vl cos(30-9) cos(30°+9) or tan 9=3C-IA PF=cos) tan-IV3 cos30°-9) Pc+PA

If ZA = ZB = ZC =| Z | cos( ) cos(30 )  = = + • • P V I V AB I a A VI A AB a A ^ cos( ) cos(30 )  = = − • • P V I V CB I cC VI C CB cC ^ V bn • V cn • V bc • V ca • V ab • V an  • 30  I aA • I cC •  cos(30 ) cos(30 )    −+ = CA PP C A C A P P P P or +− tan  = 3   +− = − C A C A P P P P PF cos tan 3 1 W   W   W   • abc C A B ZA ZB ZC N x

Example: Pp=2.5kW, PF=0.866(lagging), 2=380V. Find ) Wy Solution: P 2500 L =4.3864 √3V,0.866√3×380×0.866 D =cos0.866=30° Let v==∠0°=220∠0°V ∴Vab=380∠30°VIa4=4.386∠-30°4 cb=380∠90°VIc=4386∠90°A cC N=P=380×4.386c0s60°=831W W=P=380×4386c0s0°=1667W D=P+Pe=2500 bc

Example: PD=2.5kW,PF=0.866(lagging),VL=380V. Find . Wa Wc Solution: A V P I L D L 4.386 3 380 0.866 2500 3 0.866 =   = =   cos 0.866 30 1 = = − Let V an V   0 220 0 3 380 =  =  • V bn • V cn • V bc • V ca • V ab • V an  • 30  I aA • I cC •  V ab V I aA A    = 38030 = 4.386 − 30 • • V cb V I cC A    = 38090 = 4.38690 • • Wa = Pa = 3804.386cos 60 = 831W Wc = Pc = 3804.386cos 0  = 1667W PD = Pa + Pc = 2500W D     • a b c Wc Wa