25. 3상 시스템에서의 전력측정

25. 3상 시스템에서의 전력측정

전력계는 반드시 전압, 전류 둘 다 있어야 하며, 전원, 부하 둘 중 하나에 대해 측정할 수 있어야 한다.

전류코일(Curren coil): 단락회로 \((\mathbf{Z}=0\Omega)\)

전위코일(Potential coil): 개방회로 \((\mathbf{Z}=\infty\Omega)\)

왼쪽 회로에 KVL을 적용하면

\(\mathbf{V}_{1}=10\mathbf{I}_{1}+j5(\mathbf{I}_{1}-\mathbf{I})\), \(\mathbf{V}_{2}=-10\mathbf{I}+j5(\mathbf{I}_{1}-\mathbf{I})\)이고 \(\mathbf{I}=11.18\angle153.4^{\circ}\text{A}\).

오른쪽 전원 \(\mathbf{V}_{2}\)에서 (+)로 흡수되는 전력은 \(P=|\mathbf{V}_{2}||\mathbf{I}|\cos(\text{Arg}\mathbf{V}_{2}-\text{Arg}\mathbf{I})\)이고 여기서 \(\mathbf{V}=V\angle\theta\)일 때 \(\text{Arg}\mathbf{V}=\theta\)이다. 그러면 \(P=100\times11.18\times\cos(0^{\circ}-153.4^{\circ})=-1000\text{W}\).

3상 시스템에서의 전력계

평균전력

전력계 A: \(\displaystyle P_{A}=\frac{1}{T}\int_{0}^{T}{v_{Ax}i_{aA}dt}\)

전력계 B: \(\displaystyle P_{B}=\frac{1}{T}\int_{0}^{T}{v_{Bx}i_{bB}dt}\)

전력계 C: \(\displaystyle P_{C}=\frac{1}{T}\int_{0}^{T}{v_{Cx}i_{cC}dt}\)

부하에 소비되는 전체 평균전력: \(\displaystyle P=P_{A}+P_{B}+P_{C}=\frac{1}{T}\int_{0}^{T}{(v_{Ax}i_{aA}+v_{Bx}i_{bB}+v_{Cx}i_{cC})dt}\)

이때 \(v_{Ax}=v_{AN}+v_{Nx}\), \(v_{Bx}=v_{BN}+v_{Nx}\), \(v_{Cx}=v_{CN}+v_{Nx}\)이므로 \(\displaystyle P=\frac{1}{T}\int_{0}^{T}{(v_{AN}i_{aA}+v_{BN}i_{bB}+v_{CN}i_{cC})dt}+\frac{1}{T}\int_{0}^{T}{v_{Nx}(i_{aA}+i_{bB}+i_{cC})dt}\)이고 KCL로부터 \(i_{aA}+i_{bB}+i_{cC}=0\)이므로 \(\displaystyle P=\frac{1}{T}\int_{0}^{T}{(v_{AN}i_{aA}+v_{BN}i_{bB}+v_{CN}i_{cC})dt}\).

3상 불평형 Y결선

\(\mathbf{V}_{ab}=100\angle0^{\circ}\text{V}\), \(\mathbf{V}_{bc}=100\angle-120^{\circ}\text{V}\), \(\mathbf{V}_{ca}=100\angle-240^{\circ}\text{V}\)

\(\mathbf{V}_{an}=57.7\angle-30^{\circ}\text{V}\), \(\mathbf{V}_{bn}=57.7\angle-150^{\circ}\text{V}\), \(\mathbf{V}_{cn}=57.7\angle-270^{\circ}\text{V}\)

\(\mathbf{Z}_{A}=-j10\Omega\), \(\mathbf{Z}_{B}=j10\Omega\), \(\mathbf{Z}_{C}=10\Omega\)

\(\displaystyle\mathbf{I}_{aA}=\frac{\mathbf{V}_{an}-\mathbf{V}_{nN}}{\mathbf{Z}_{A}}\), \(\displaystyle\mathbf{I}_{bB}=\frac{\mathbf{V}_{bn}-\mathbf{V}_{nN}}{\mathbf{Z}_{B}}\), \(\displaystyle\mathbf{I}_{cC}=\frac{\mathbf{V}_{cn}-\mathbf{V}_{nN}}{\mathbf{Z}_{C}}\).

이때 KCL로부터 \(\mathbf{I}_{aA}+\mathbf{I}_{bB}+\mathbf{I}_{cC}=0\)이므로 \(\displaystyle\mathbf{V}_{nN}=\left(\frac{\mathbf{V}_{an}}{\mathbf{Z}_{A}}+\frac{\mathbf{V}_{bn}}{\mathbf{Z}_{B}}+\frac{\mathbf{V}_{cn}}{\mathbf{Z}_{C}}\right)\left(\frac{1}{\mathbf{Z}_{A}}+\frac{1}{\mathbf{Z}_{B}}+\frac{1}{\mathbf{Z}_{C}}\right)^{-1}=157.7\angle90^{\circ}\text{V}\)이고 \(\mathbf{I}_{aA}=19.32\angle15^{\circ}\text{A}\), \(\mathbf{I}_{bB}=19.32\angle165^{\circ}\text{A}\), \(\mathbf{I}_{cC}=10\angle-90^{\circ}\text{A}\), \(P_{A}=57.7\times19.32\times\cos(-30^{\circ}-15^{\circ})=788.7\text{W}\), \(P_{B}=57.7\times19.32\times\cos(-150^{\circ}-165^{\circ})=788.7\text{W}\), \(P_{C}=57.7\times10\times\cos(-270^{\circ}-90^{\circ})=-577.4\text{W}\).

 

이전력계법

\(P_{1}\)(전력계1)\(=|\mathbf{V}_{AB}||\mathbf{I}_{aA}|\cos(\text{Arg}\mathbf{V}_{AB}-\text{Arg}\mathbf{I}_{aA})=V_{L}I_{L}\cos(30^{\circ}+\theta)\) \(P_{2}\)(전력계2)\(=|\mathbf{V}_{CB}||\mathbf{I}_{cC}|\cos(\text{Arg}\mathbf{V}_{CB}-\text{Arg}\mathbf{I}_{cC})=V_{L}I_{L}\cos(30^{\circ}-\theta)\) \(\displaystyle\frac{P_{1}}{P_{2}}=\frac{\cos(30^{\circ}+\theta)}{\cos(30^{\circ}-\theta)}\), \(\displaystyle\tan\theta=\sqrt{3}\frac{P_{2}-P_{1}}{P_{2}+P_{1}}\)

\(\tan\theta\)크기가 증가: 유도성 임피던스

\(\tan\theta\)크기가 감소: 용량성 임피던스

왼쪽 회로는 (+)상순이고 \(\mathbf{V}_{ab}=230\angle0^{\circ}\text{V}\), \(\mathbf{V}_{bc}=230\angle-120^{\circ}\text{V}\), \(\mathbf{V}_{ca}=230\angle120^{\circ}\text{V}\). \(\mathbf{V}_{ac}=-\mathbf{V}_{ca}=230\angle-60^{\circ}\text{V}\), \(\displaystyle\mathbf{V}_{an}=\frac{\mathbf{V}_{ab}}{\sqrt{3}\angle30^{\circ}}\), \(\displaystyle\mathbf{I}_{aA}=\frac{\mathbf{V}_{an}}{4+j15}=8.554\angle-105.1^{\circ}\text{A}\).

\(P_{1}=|\mathbf{V}_{ac}||\mathbf{I}_{aA}|\cos(\text{Arg}\mathbf{V}_{ac}-\text{Arg}\mathbf{I}_{aA})=230\times8\times554\times\cos(-60^{\circ}+105.1^{\circ})=1389\text{W}\), \(\displaystyle\mathbf{V}_{bB}=\frac{\mathbf{V}_{bn}}{4+j15}=8.554\angle-134.9^{\circ}\text{V}\), \(P_{2}=|\mathbf{V}_{bc}||\mathbf{I}_{bB}|\cos(\text{Arg}\mathbf{V}_{bc}-\text{Arg}\mathbf{I}_{bB})=-512.5\text{W}\).

부하에 흡수되는 평균전력은 \(P=P_{1}+P_{2}=876.5\text{W}\).

참고자료:

Engineering Circuit Analysis 8th edition, Hayt, Kemmerly, Durbin, McGraw-Hill