Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
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electrical_engineering_2:polyphase_networks [2023/03/22 20:32] mexleadmin |
electrical_engineering_2:polyphase_networks [2023/09/19 23:52] (aktuell) mexleadmin |
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| Zeile 1: | Zeile 1: | ||
| - | ====== 7. Polyphase Networks and Power in AC Circuits ====== | + | ====== 7 Polyphase Networks and Power in AC Circuits ====== |
| emphasizing the importance of power considerations | emphasizing the importance of power considerations | ||
| Zeile 23: | Zeile 23: | ||
| Thus, the induced voltage $u(t)$ is given by: | Thus, the induced voltage $u(t)$ is given by: | ||
| \begin{align*} | \begin{align*} | ||
| - | u(t) & | + | u(t) & |
| - | & | + | & |
| - | & | + | & |
| - | & | + | & |
| - | & | + | & |
| - | & | + | & |
| \end{align*} | \end{align*} | ||
| Zeile 34: | Zeile 34: | ||
| Out of the last formula we derived the following instantaneous voltage $u(t)$ | Out of the last formula we derived the following instantaneous voltage $u(t)$ | ||
| \begin{align*} | \begin{align*} | ||
| - | u(t) & | + | u(t) &= \hat{U} |
| - | & | + | & |
| - | & | + | |
| \end{align*} | \end{align*} | ||
| Zeile 145: | Zeile 144: | ||
| - Ohmic load: The instantaneous voltage is in phase with the instantaneous current. The instantaneous power is always non-negative. The average power is $P=U^2/R = {{1}\over{2}} \hat{U}^2/ | - Ohmic load: The instantaneous voltage is in phase with the instantaneous current. The instantaneous power is always non-negative. The average power is $P=U^2/R = {{1}\over{2}} \hat{U}^2/ | ||
| - | - Inductive load: The voltage is ahead of the current. The phase angle is $+90°$ (which also reflects the $+j$ in the inductive impedance $+j\omega L$). The instantaneous is half positive, half negative; the average power is zero (in the simulation not completely visible). | + | - Inductive load: The voltage is ahead of the current. The phase angle is $+90°$ (which also reflects the $+{\rm j}$ in the inductive impedance $+{\rm j}\omega L$). The instantaneous is half positive, half negative; the average power is zero (in the simulation not completely visible). |
| - | - Capacitive load: The voltage is lagging the current. The phase angle is $-90°$ (which also reflects the $-j$ in the capacitive impedance ${{1}\over{j\omega C}}$). The instantaneous is again half positive, half negative; the average power is zero (in the simulation not completely visible). | + | - Capacitive load: The voltage is lagging the current. The phase angle is $-90°$ (which also reflects the $-{\rm j}$ in the capacitive impedance ${{1}\over{{\rm j}\omega C}}$). The instantaneous is again half positive, half negative; the average power is zero (in the simulation not completely visible). |
| < | < | ||
| Zeile 208: | Zeile 207: | ||
| \begin{align*} | \begin{align*} | ||
| - | \underline{S} &= S \cdot e^{j\varphi} \\ | + | \underline{S} &= S |
| - | &= U \cdot I \cdot e^{j\varphi} | + | &= U \cdot I \cdot {\rm e}^{{\rm j}\varphi} |
| \end{align*} | \end{align*} | ||
| Zeile 215: | Zeile 214: | ||
| \begin{align*} | \begin{align*} | ||
| - | \underline{S} & | + | \underline{S} & |
| - | &= \underbrace{U \cdot e^{j\varphi_U}}_{\underline{U}} \cdot \underbrace{I \cdot e^{-j\varphi_I}}_{\underline{I}^*} | + | &= \underbrace{U \cdot {\rm e}^{{\rm j}\varphi_U}}_{\underline{U}} \cdot \underbrace{I \cdot {\rm e}^{-{\rm j}\varphi_I}}_{\underline{I}^*} |
| \end{align*} | \end{align*} | ||
| Zeile 223: | Zeile 222: | ||
| <callout icon=" | <callout icon=" | ||
| - | * $\underline{S} = UI \cdot e^{j\varphi}$ | + | * $\underline{S} = UI \cdot {\rm e}^{{\rm j}\varphi}$ |
| - | * $\underline{S} = UI \cdot (\cos\varphi + j \sin\varphi)$ | + | * $\underline{S} = UI \cdot (\cos\varphi + {\rm j} \sin\varphi)$ |
| - | * $\underline{S} = P + jQ$ | + | * $\underline{S} = P + {\rm j}Q$ |
| * $\underline{S} = \underline{U} \cdot \underline{I}^*$ | * $\underline{S} = \underline{U} \cdot \underline{I}^*$ | ||
| Zeile 327: | Zeile 326: | ||
| Thus, the voltage phasors $\underline{U}_1 ... \underline{U}_m$ form a symmetrical star. \\ | Thus, the voltage phasors $\underline{U}_1 ... \underline{U}_m$ form a symmetrical star. \\ | ||
| Example: A 3-phase system is symmetrical for $\varphi = 360°/3 = 120°$ between the voltages of the windings: | Example: A 3-phase system is symmetrical for $\varphi = 360°/3 = 120°$ between the voltages of the windings: | ||
| - | $\underline{U}_1 = \sqrt{2} \cdot U \cdot e ^{j(\omega t + 0°)}$, | + | $\underline{U}_1 = \sqrt{2} \cdot U \cdot {\rm e} ^{{\rm j}(\omega t + 0°)}$, |
| - | $\underline{U}_2 = \sqrt{2} \cdot U \cdot e ^{j(\omega t - 120°)}$, | + | $\underline{U}_2 = \sqrt{2} \cdot U \cdot {\rm e} ^{{\rm j}(\omega t - 120°)}$, |
| - | $\underline{U}_3 = \sqrt{2} \cdot U \cdot e ^{j(\omega t - 240°)}$ \\ | + | $\underline{U}_3 = \sqrt{2} \cdot U \cdot {\rm e} ^{{\rm j}(\omega t - 240°)}$ \\ |
| < | < | ||
| </ | </ | ||
| Zeile 362: | Zeile 361: | ||
| ==== 7.2.2 Three-Phase System ==== | ==== 7.2.2 Three-Phase System ==== | ||
| + | |||
| + | See also: [[https:// | ||
| The most commonly used polyphase system is the three-phase system. The three-phase system has advantages over a DC system or single-phase AC system: | The most commonly used polyphase system is the three-phase system. The three-phase system has advantages over a DC system or single-phase AC system: | ||
| Zeile 402: | Zeile 403: | ||
| * **String voltages/ | * **String voltages/ | ||
| - | The string voltages/ | + | The string voltages/ |
| - | These voltages are shown in the previous images as $u_\rm U$, $u_\rm V$, $u_\rm W$. | + | These voltages are shown in the previous images as $u_\rm U$, $u_\rm V$, and $u_\rm W$. |
| </ | </ | ||
| * **Phase voltages/ | * **Phase voltages/ | ||
| Zeile 479: | Zeile 480: | ||
| \underline{I}_2 = {{\underline{U}_{2 \rm N}}\over{\underline{Z}_2^\phantom{O}}} \quad , \quad | \underline{I}_2 = {{\underline{U}_{2 \rm N}}\over{\underline{Z}_2^\phantom{O}}} \quad , \quad | ||
| \underline{I}_3 = {{\underline{U}_{3 \rm N}}\over{\underline{Z}_3^\phantom{O}}} \\ \end{align*}</ | \underline{I}_3 = {{\underline{U}_{3 \rm N}}\over{\underline{Z}_3^\phantom{O}}} \\ \end{align*}</ | ||
| - | - The **true power** $P_x$ for each string is given by the apparent power $S_x$ of the string times the indivitual | + | - The **true power** $P_x$ for each string is given by the apparent power $S_x$ of the string times the individual |
| \begin{align*} P_x &= S_x \cdot \cos \varphi_x = U_{\rm S} \cdot I_x \cdot \cos \varphi_x \end{align*} \\ | \begin{align*} P_x &= S_x \cdot \cos \varphi_x = U_{\rm S} \cdot I_x \cdot \cos \varphi_x \end{align*} \\ | ||
| Therefore, the resulting true power for the full load is: \\ | Therefore, the resulting true power for the full load is: \\ | ||
| Zeile 505: | Zeile 506: | ||
| \begin{align*} | \begin{align*} | ||
| \underline{I}_1 &= {{\underline{U}_{\rm 1N}}\over{\underline{Z}_1}} | \underline{I}_1 &= {{\underline{U}_{\rm 1N}}\over{\underline{Z}_1}} | ||
| - | &= &{{231 ~\rm V}\over{10 ~\Omega + j \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 1 {~\rm mH}}} | + | &= &{{231 ~\rm V}\over{10 ~\Omega + {\rm j} \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 1 {~\rm mH}}} |
| - | &= &+23.08 {~\rm A} &- j \cdot 0.72 {~\rm A} &= &23.09 ~{~\rm A} \quad &\angle -1.8° \\ | + | &= &+23.08 {~\rm A} & |
| \underline{I}_2 &= {{\underline{U}_{\rm 2N}}\over{\underline{Z}_2}} | \underline{I}_2 &= {{\underline{U}_{\rm 2N}}\over{\underline{Z}_2}} | ||
| - | &= &{{231 {~\rm V} \cdot \left( -{{1}\over{2}}-j{{1}\over{2}}\sqrt{3}\right)}\over{5 ~ \Omega + {{1}\over{j \cdot 2\pi\cdot 50{~\rm Hz} \cdot 100 {~\rm µF}}}}} | + | &= &{{231 {~\rm V} \cdot \left( -{{1}\over{2}}-{\rm j}{{1}\over{2}}\sqrt{3}\right)}\over{5 ~ \Omega + {{1}\over{{\rm j} \cdot 2\pi\cdot 50{~\rm Hz} \cdot 100 {~\rm µF}}}}} |
| - | &= &+ 5.58 {~\rm A} &- j \cdot 4.50 {~\rm A} &= & 7.17 {~\rm A} \quad &\angle -38.9° \\ | + | &= &+ 5.58 {~\rm A} & |
| \underline{I}_3 &= {{\underline{U}_{\rm 3N}}\over{\underline{Z}_3}} | \underline{I}_3 &= {{\underline{U}_{\rm 3N}}\over{\underline{Z}_3}} | ||
| - | &= & | + | &= & |
| - | &= & | + | &= & |
| \underline{I}_{\rm N} | \underline{I}_{\rm N} | ||
| - | &= \underline{I}_1 + \underline{I}_2 + \underline{I}_3 & & & = &+22.88 {~\rm A} &+ j \cdot 4.77 {~\rm A} &= &23.37 {~\rm A} \quad &\angle +11.8° | + | &= \underline{I}_1 + \underline{I}_2 + \underline{I}_3 & & & = &+22.88 {~\rm A} & |
| \end{align*} </ | \end{align*} </ | ||
| - The true power is calculated by: < | - The true power is calculated by: < | ||
| Zeile 574: | Zeile 575: | ||
| * With the switch $S$, the star potential can short-circuited to the neutral potential; so set $\underline{U}_{\rm SN}=0$. This enables a comparison with the previous four-wire three-phase system. | * With the switch $S$, the star potential can short-circuited to the neutral potential; so set $\underline{U}_{\rm SN}=0$. This enables a comparison with the previous four-wire three-phase system. | ||
| - | Also here the " | + | Also here, the " |
| - | - **Voltages**: | + | - **Voltages**: |
| \begin{align*} | \begin{align*} | ||
| \underline{U}_{\rm 1S} &= \underline{U}_{\rm 1N} - \underline{U}_{\rm SN} \\ | \underline{U}_{\rm 1S} &= \underline{U}_{\rm 1N} - \underline{U}_{\rm SN} \\ | ||
| Zeile 603: | Zeile 604: | ||
| - Since the three-wire system has no current out of the network star point, the **apparent power** $\underline{S}_x$ for each string is given by the string voltage and the current through the string $\underline{S}_x = \underline{U}_{x \rm S} \cdot \underline{I}_x^*$. This leads to an overall apparent power $\underline{S}$ of < | - Since the three-wire system has no current out of the network star point, the **apparent power** $\underline{S}_x$ for each string is given by the string voltage and the current through the string $\underline{S}_x = \underline{U}_{x \rm S} \cdot \underline{I}_x^*$. This leads to an overall apparent power $\underline{S}$ of < | ||
| \begin{align*} | \begin{align*} | ||
| - | \underline{S} &= P + j\cdot Q = \sum_x \underline{S}_x = \sum_x | + | \underline{S} &= P + {\rm j}\cdot Q = \sum_x \underline{S}_x = \sum_x |
| In order to simplify the calculation, | In order to simplify the calculation, | ||
| \begin{align*} | \begin{align*} | ||
| Zeile 621: | Zeile 622: | ||
| \end{align*} | \end{align*} | ||
| - The abolute **reactive power** $Q$ can be calulated by the apparent power: < | - The abolute **reactive power** $Q$ can be calulated by the apparent power: < | ||
| - | \begin{align*} j\cdot Q &= \underline{S} - P \end{align*} | + | \begin{align*} |
| the **collective reactive power** $Q_\Sigma$ is given by the collective apparent power: | the **collective reactive power** $Q_\Sigma$ is given by the collective apparent power: | ||
| \begin{align*} Q &= \sqrt{S_\Sigma^2 - P^2} \end{align*} | \begin{align*} Q &= \sqrt{S_\Sigma^2 - P^2} \end{align*} | ||
| Zeile 635: | Zeile 636: | ||
| \end{align*} \\ | \end{align*} \\ | ||
| Once investigating the numerator $\sum_x \big( {{1}\over{\underline{Z}_x^\phantom{O}}} \cdot \underline{U}_{x \rm N} \big)$, once can see, that it just equals the sum of the phase currents of the four-wire system. So, the numerator equals the (in the three-wire system: fictive) current on the neutral line. \\ | Once investigating the numerator $\sum_x \big( {{1}\over{\underline{Z}_x^\phantom{O}}} \cdot \underline{U}_{x \rm N} \big)$, once can see, that it just equals the sum of the phase currents of the four-wire system. So, the numerator equals the (in the three-wire system: fictive) current on the neutral line. \\ | ||
| - | The numerator is therefore: $22.88 {~\rm A} + j \cdot 4.77 {~\rm A}$ (see calculation for the four-wire system). \\ | + | The numerator is therefore: $22.88 {~\rm A} + {\rm j} \cdot 4.77 {~\rm A}$ (see calculation for the four-wire system). \\ |
| The denominator is: \\ | The denominator is: \\ | ||
| \begin{align*} | \begin{align*} | ||
| - | \sum_x | + | \sum_x |
| - | + {{1}\over{5~\Omega + {{1}\over{j \cdot 2\pi\cdot 50{~\rm Hz} \cdot 100 {~\rm µF} }} }} | + | + {{1}\over{5~\Omega + {{1}\over{{\rm j} \cdot 2\pi\cdot 50{~\rm Hz} \cdot 100 {~\rm µF} }} }} |
| + {{1}\over{20~\Omega }} \\ \\ | + {{1}\over{20~\Omega }} \\ \\ | ||
| - | &= 0.1547 | + | &= 0.1547 |
| The star-voltage $\underline{U}_{\rm SN}$ of the load is: | The star-voltage $\underline{U}_{\rm SN}$ of the load is: | ||
| \begin{align*} | \begin{align*} | ||
| - | \underline{U}_{\rm SN} &= {{22.88 {~\rm A} + j \cdot 4.77 {~\rm A}}\over{0.1547 ~1/\Omega + j \cdot 0.0275 ~1/\Omega}} \\ \\ | + | \underline{U}_{\rm SN} &= {{22.88 {~\rm A} + {\rm j} \cdot 4.77 {~\rm A}}\over{0.1547 ~1/\Omega + {\rm j} \cdot 0.0275 ~1/\Omega}} \\ \\ |
| - | & | + | & |
| Given this star-voltage $\underline{U}_{\rm SN}$ of the load, the phase currents are: \\ | Given this star-voltage $\underline{U}_{\rm SN}$ of the load, the phase currents are: \\ | ||
| \begin{align*} | \begin{align*} | ||
| \underline{I}_1 &= {{\underline{U}_{\rm 1N} - \underline{U}_{\rm SN}}\over{\underline{Z}_1^\phantom{O}}} | \underline{I}_1 &= {{\underline{U}_{\rm 1N} - \underline{U}_{\rm SN}}\over{\underline{Z}_1^\phantom{O}}} | ||
| - | & = & {{231{~\rm V} - 148.7{~\rm V} - j \cdot 4.41 {~\rm V}}\over{10~\Omega + j \cdot 2\pi\cdot 50{~\rm Hz} \cdot 1{~\rm mH} }} | + | & = & {{231{~\rm V} - 148.7{~\rm V} - {\rm j} \cdot 4.41 {~\rm V}}\over{10~\Omega + {\rm j} \cdot 2\pi\cdot 50{~\rm Hz} \cdot 1{~\rm mH} }} |
| - | & = & +8.21{~\rm A} - j \cdot 0.70{~\rm A} & | + | & = & +8.21{~\rm A} |
| \underline{I}_2 &= {{\underline{U}_{\rm 2N} - \underline{U}_{\rm SN}}\over{\underline{Z}_2^\phantom{O}}} | \underline{I}_2 &= {{\underline{U}_{\rm 2N} - \underline{U}_{\rm SN}}\over{\underline{Z}_2^\phantom{O}}} | ||
| - | & = & {{231{~\rm V} \cdot \left( -{{1}\over{2}}-j{{1}\over{2}}\sqrt{3}\right) - 148.7{~\rm V} - j \cdot 4.41 {~\rm V}}\over{5~\Omega + {{1}\over{j \cdot 2\pi\cdot 50{~\rm Hz} \cdot 100{~\rm µF} }}}} | + | & = & {{231{~\rm V} \cdot \left( -{{1}\over{2}}-{\rm j}{{1}\over{2}}\sqrt{3}\right) - 148.7{~\rm V} - {\rm j} \cdot 4.41 {~\rm V}}\over{5~\Omega + {{1}\over{{\rm j} \cdot 2\pi\cdot 50{~\rm Hz} \cdot 100{~\rm µF} }}}} |
| - | & = & +5.00{~\rm A} + j \cdot 9.08{~\rm A} & =& 10.36 {~\rm A} \quad \angle -61.2° \\ | + | & = & +5.00{~\rm A} + {\rm j} \cdot 9.08{~\rm A} & =& 10.36 {~\rm A} \quad \angle -61.2° \\ |
| \underline{I}_3 &= {{\underline{U}_{\rm 3N} - \underline{U}_{\rm SN}}\over{\underline{Z}_3^\phantom{O}}} | \underline{I}_3 &= {{\underline{U}_{\rm 3N} - \underline{U}_{\rm SN}}\over{\underline{Z}_3^\phantom{O}}} | ||
| - | & = & {{231{~\rm V} \cdot \left( -{{1}\over{2}}+j{{1}\over{2}}\sqrt{3}\right) - 148.7{~\rm V} - j \cdot 4.41 {~\rm V}}\over{20~\Omega }} | + | & = & {{231{~\rm V} \cdot \left( -{{1}\over{2}}+{\rm j}{{1}\over{2}}\sqrt{3}\right) - 148.7{~\rm V} - {\rm j} \cdot 4.41 {~\rm V}}\over{20~\Omega }} |
| - | &= & -13.21{~\rm A} + j \cdot 9.78{~\rm A} & =& 16.44{~\rm A} \quad \angle +143.5° \end{align*} \\ </ | + | &= & -13.21{~\rm A} + {\rm j} \cdot 9.78{~\rm A} & =& 16.44{~\rm A} \quad \angle +143.5° \end{align*} \\ </ |
| - The true power is calculated by: \\ \begin{align*} P = 231{~\rm V} \cdot \big( 8.24 {~\rm A} \cdot \cos (0° - (-4.9°))+ 10.36{~\rm A} \cdot \cos (-120° - (-61.2°)) + 16.44 {~\rm A} \cdot \cos (-240° - (+143.5°)\big) = 6.62 {~\rm kW} \end{align*} | - The true power is calculated by: \\ \begin{align*} P = 231{~\rm V} \cdot \big( 8.24 {~\rm A} \cdot \cos (0° - (-4.9°))+ 10.36{~\rm A} \cdot \cos (-120° - (-61.2°)) + 16.44 {~\rm A} \cdot \cos (-240° - (+143.5°)\big) = 6.62 {~\rm kW} \end{align*} | ||
| - The apparent power $\underline{S}$ is: < | - The apparent power $\underline{S}$ is: < | ||
| \begin{align*} | \begin{align*} | ||
| \underline{S} &= \underline{U}_{13} \cdot \underline{I}_1^* + \underline{U}_{23} \cdot \underline{I}_2^* | \underline{S} &= \underline{U}_{13} \cdot \underline{I}_1^* + \underline{U}_{23} \cdot \underline{I}_2^* | ||
| - | | + | &=& 400{~\rm V} \cdot (- {\rm e}^{-{\rm j} \cdot 7/6 \pi} \cdot (8.21{~\rm A} + {\rm j}\cdot 0.70 {~\rm A}) + {\rm e}^{-{\rm j} \cdot 3/6 \pi} \cdot ( 5.00{~\rm A} -{\rm j} \cdot 9.08{~\rm A}) ) |
| - | &= 6.62 {~\rm kW} - j \cdot 3.40 {~\rm kVAr} \\ | + | |
| - | &= \underline{U}_{12} \cdot \underline{I}_1^* + \underline{U}_{32} \cdot \underline{I}_3^* | + | |
| - | &=& 400{~\rm V} \cdot (e^{j \cdot 1/6 \pi} \cdot (8.21{~\rm A} + j \cdot 0.70{~\rm A}) - e^{-j \cdot 3/6 \pi} \cdot (-13.21{~\rm A} -j \cdot 9.78{~\rm A})) | + | |
| - | &= 6.62 {~\rm kW} - j \cdot 3.40 {~\rm kVAr} \\ | + | &= 6.62 {~\rm kW} - {\rm j} \cdot 3.40 {~\rm kVAr} \\ |
| &= \underline{U}_{21} \cdot \underline{I}_2^* + \underline{U}_{31} \cdot \underline{I}_3^* | &= \underline{U}_{21} \cdot \underline{I}_2^* + \underline{U}_{31} \cdot \underline{I}_3^* | ||
| - | | + | &=& 400{~\rm V} \cdot (- {\rm e}^{{\rm j} \cdot 1/6 \pi} \cdot (5.00{~\rm A} - |
| - | &= 6.62 {~\rm kW} - j \cdot 3.40 {~\rm kVAr} \\ | + | &= 6.62 {~\rm kW} - {\rm j} \cdot 3.40 {~\rm kVAr} \\ |
| & = 7.44 {~\rm kVA} \quad \angle -27.2° | & = 7.44 {~\rm kVA} \quad \angle -27.2° | ||
| \end{align*} | \end{align*} | ||
| Zeile 677: | Zeile 678: | ||
| = 8.45 {~\rm kVA} \end{align*} | = 8.45 {~\rm kVA} \end{align*} | ||
| - The reactive power is: < | - The reactive power is: < | ||
| - | \begin{align*} Q &= -j \cdot (\underline{S} - P) = -3.40{~\rm kVAr} \\ | + | \begin{align*} Q &= -{\rm j} \cdot (\underline{S} - P) = -3.40{~\rm kVAr} \\ |
| \end{align*} \\ | \end{align*} \\ | ||
| The collective reactive power is: \\ | The collective reactive power is: \\ | ||
| Zeile 722: | Zeile 723: | ||
| - **Voltages**: | - **Voltages**: | ||
| \begin{align*} | \begin{align*} | ||
| - | \underline{U}_{12} &=& U_{\rm L} \cdot e^{ j\cdot {{1}\over{6}}} \\ | + | \underline{U}_{12} &=& U_{\rm L} \cdot {\rm e}^{ |
| - | \underline{U}_{23} &=& U_{\rm L} \cdot e^{- j\cdot {{3}\over{6}}} \\ | + | \underline{U}_{23} &=& U_{\rm L} \cdot {\rm e}^{- {\rm j}\cdot {{3}\over{6}}} \\ |
| - | \underline{U}_{31} &=& U_{\rm L} \cdot e^{- j\cdot {{7}\over{6}}} | + | \underline{U}_{31} &=& U_{\rm L} \cdot {\rm e}^{- {\rm j}\cdot {{7}\over{6}}} |
| \end{align*}</ | \end{align*}</ | ||
| - **Currents**: | - **Currents**: | ||
| Zeile 747: | Zeile 748: | ||
| \begin{align*} | \begin{align*} | ||
| \boxed{ | \boxed{ | ||
| - | \underline{S} = P + j \cdot Q | + | \underline{S} = P + {\rm j} \cdot Q |
| = U_{\rm L}^2 \cdot \left( {{1}\over{\underline{Z}_{12}^* }} | = U_{\rm L}^2 \cdot \left( {{1}\over{\underline{Z}_{12}^* }} | ||
| + {{1}\over{\underline{Z}_{23}^* }} | + {{1}\over{\underline{Z}_{23}^* }} | ||
| Zeile 753: | Zeile 754: | ||
| \end{align*} \\ | \end{align*} \\ | ||
| The **collective apparent power** $S_\Sigma$ here is the same as for the three-wire or four-wire connection. \\ | The **collective apparent power** $S_\Sigma$ here is the same as for the three-wire or four-wire connection. \\ | ||
| - | In the Delta connection the phase currents $I_x$ have to be calculated since the formula only applies | + | In the Delta connection the phase currents $I_x$ have to be calculated since the formula only applies |
| \begin{align*} | \begin{align*} | ||
| S_\Sigma &= \sqrt{ {{1}\over{3}} (U_{12}^2 + U_{23}^2 + U_{31}^2) } \cdot \sqrt{\sum_x I_x^2} &= U_{\rm L} \cdot \sqrt{\sum_x I_x^2} | S_\Sigma &= \sqrt{ {{1}\over{3}} (U_{12}^2 + U_{23}^2 + U_{31}^2) } \cdot \sqrt{\sum_x I_x^2} &= U_{\rm L} \cdot \sqrt{\sum_x I_x^2} | ||
| \end{align*} | \end{align*} | ||
| - The abolute **reactive power** $Q$ can be calulated by the apparent power: < | - The abolute **reactive power** $Q$ can be calulated by the apparent power: < | ||
| - | \begin{align*} j\cdot Q &= \underline{S} - P \end{align*} | + | \begin{align*} |
| the **collective reactive power** $Q_\Sigma$ is given by the collective apparent power: | the **collective reactive power** $Q_\Sigma$ is given by the collective apparent power: | ||
| \begin{align*} Q &= \sqrt{S_\Sigma^2 - P^2} \end{align*} | \begin{align*} Q &= \sqrt{S_\Sigma^2 - P^2} \end{align*} | ||
| Zeile 773: | Zeile 774: | ||
| The phasors of the string voltages of the network are given as \\ | The phasors of the string voltages of the network are given as \\ | ||
| {{drawio> | {{drawio> | ||
| - | - Based on the string voltages of the network and the given impedances the string currents $\underline{I}_{12}$, | + | - Based on the string voltages of the network and the given impedances the string currents $\underline{I}_{12}$, |
| \begin{align*} | \begin{align*} | ||
| - | \underline{I}_{12} &=& {{ 400 {~\rm V} \cdot\left(+{{1}\over{2}}\sqrt{3}+{{1}\over{2}} \cdot j \right)}\over{ 10~\Omega + j \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 1 {~\rm mH} }} | + | \underline{I}_{12} &=& {{ 400 {~\rm V} \cdot\left(+{{1}\over{2}}\sqrt{3}+{{1}\over{2}} \cdot {\rm j} \right)}\over{ 10~\Omega + {\rm j} \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 1 {~\rm mH} }} |
| - | & | + | & |
| - | \underline{I}_{23} &=& {{400 \cdot j}\over{ 5~\Omega + {{1}\over{j \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 100 {~\rm µF}}} }} | + | \underline{I}_{23} &=& {{400 \cdot {\rm j}}\over{ 5~\Omega + {{1}\over{{\rm j} \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 100 {~\rm µF}}} }} |
| - | & | + | & |
| - | \underline{I}_{31} &=& {{400 {~\rm V} \cdot\left(-{{1}\over{2}}\sqrt{3}+{{1}\over{2}} \cdot j \right)}\over{ 20 ~\Omega}} | + | \underline{I}_{31} &=& {{400 {~\rm V} \cdot\left(-{{1}\over{2}}\sqrt{3}+{{1}\over{2}} \cdot {\rm j} \right)}\over{ 20 ~\Omega}} |
| - | & | + | & |
| By these voltages the phase currents $\underline{I}_x$ can be calculated: \\ | By these voltages the phase currents $\underline{I}_x$ can be calculated: \\ | ||
| \begin{align*} | \begin{align*} | ||
| - | \underline{I}_{1} &=& ( 35.24 {~\rm A} + j \cdot 18.90 {~\rm A}) - (-17.33 {~\rm A} + j \cdot 10.00 {~\rm A}) & | + | \underline{I}_{1} &=& ( 35.24 {~\rm A} + {\rm j} \cdot 18.90 {~\rm A}) - (-17.33 {~\rm A} + {\rm j} \cdot 10.00 {~\rm A}) & |
| &=& 53.32 {~\rm A} \quad &\angle 9.6° \\ | &=& 53.32 {~\rm A} \quad &\angle 9.6° \\ | ||
| - | \underline{I}_{2} &=& ( 12.27 {~\rm A} - j \cdot 1.93 {~\rm A}) - ( 35.24 {~\rm A} + j \cdot 18.90 {~\rm A}) &=& -22.98 {~\rm A} - j \cdot 20.83 {~\rm A} | + | \underline{I}_{2} &=& ( 12.27 {~\rm A} - {\rm j} \cdot 1.93 {~\rm A}) - ( 35.24 {~\rm A} + {\rm j} \cdot 18.90 {~\rm A}) &=& -22.98 {~\rm A} - {\rm j} \cdot 20.83 {~\rm A} |
| &=& -31.01 {~\rm A} \quad &\angle -137.8° \\ | &=& -31.01 {~\rm A} \quad &\angle -137.8° \\ | ||
| - | \underline{I}_{3} &=& (-17.33 {~\rm A} + j \cdot 10.00 {~\rm A}) - ( 12.27 {~\rm A} - j \cdot 1.93 {~\rm A}) &=& -29.59 {~\rm A} + j \cdot 11.93 {~\rm A} | + | \underline{I}_{3} &=& (-17.33 {~\rm A} + {\rm j} \cdot 10.00 {~\rm A}) - ( 12.27 {~\rm A} - {\rm j} \cdot 1.93 {~\rm A}) &=& -29.59 {~\rm A} + {\rm j} \cdot 11.93 {~\rm A} |
| &=& 31.90 {~\rm A} \quad &\angle 158.0° | &=& 31.90 {~\rm A} \quad &\angle 158.0° | ||
| \end{align*} \\</ | \end{align*} \\</ | ||
| Zeile 799: | Zeile 800: | ||
| \begin{align*} | \begin{align*} | ||
| \underline{S} &= \underline{U}_{13} \cdot \underline{I}_1^* + \underline{U}_{23} \cdot \underline{I}_2^* | \underline{S} &= \underline{U}_{13} \cdot \underline{I}_1^* + \underline{U}_{23} \cdot \underline{I}_2^* | ||
| - | &=& 400 {~\rm V} \cdot (- e^{-j \cdot 7/6 \pi} \cdot (52.57 | + | &=& 400 {~\rm V} \cdot (- {\rm e}^{-{\rm j} \cdot 7/6 \pi} \cdot ( 52.57 {~\rm A} - {\rm j} \cdot 8.90 {~\rm A}) + |
| - | &= 24.77 {~\rm kW} - j \cdot 4.41 {~\rm kVAr} \\ | + | {\rm e}^{-{\rm j} \cdot 3/6 \pi} \cdot (-22.98 {~\rm A} + {\rm j} \cdot 20.83 {~\rm A})) |
| + | &= 24.77 {~\rm kW} - {\rm j} \cdot 4.41 {~\rm kVAr} \\ | ||
| &= \underline{U}_{12} \cdot \underline{I}_1^* + \underline{U}_{32} \cdot \underline{I}_3^* | &= \underline{U}_{12} \cdot \underline{I}_1^* + \underline{U}_{32} \cdot \underline{I}_3^* | ||
| - | &=& 400 {~\rm V} \cdot (e^{j \cdot 1/6 \pi} \cdot (52.57 | + | &=& 400 {~\rm V} \cdot ( |
| - | &= 24.77 {~\rm kW} - j \cdot 4.41 {~\rm kVAr} \\ | + | {\rm e}^{-{\rm j} \cdot 3/6 \pi} \cdot (-29.59 {~\rm A} - {\rm j} \cdot 11.93 {~\rm A})) |
| + | &= 24.77 {~\rm kW} - {\rm j} \cdot 4.41 {~\rm kVAr} \\ | ||
| &= \underline{U}_{21} \cdot \underline{I}_2^* + \underline{U}_{31} \cdot \underline{I}_3^* | &= \underline{U}_{21} \cdot \underline{I}_2^* + \underline{U}_{31} \cdot \underline{I}_3^* | ||
| - | &=& 400 {~\rm V} \cdot (- e^{j \cdot 1/6 \pi} \cdot (-22.98 {~\rm A} + j \cdot 20.83 {~\rm A}) + e^{- j \cdot 7/6 \pi} \cdot (-29.59 {~\rm A} - j \cdot 11.93 {~\rm A})) | + | &=& 400 {~\rm V} \cdot (- {\rm e}^{ {\rm j} \cdot 1/6 \pi} \cdot (-22.98 {~\rm A} + {\rm j} \cdot 20.83 {~\rm A}) + |
| - | &= 24.77 {~\rm kW} - j \cdot 4.41 {~\rm kVAr} \\ | + | {\rm e}^{-{\rm j} \cdot 7/6 \pi} \cdot (-29.59 {~\rm A} - {\rm j} \cdot 11.93 {~\rm A})) |
| + | &= 24.77 {~\rm kW} - {\rm j} \cdot 4.41 {~\rm kVAr} \\ | ||
| & = 25.16 {~\rm kVA} \quad \angle -10.09°\end{align*} | & = 25.16 {~\rm kVA} \quad \angle -10.09°\end{align*} | ||
| The collective apparent power is: \\ | The collective apparent power is: \\ | ||
| Zeile 865: | Zeile 869: | ||
| \end{align*} | \end{align*} | ||
| - | For the **series circuit**, the impedances add up like: $R_s + j\cdot X_{Ls} = \underline{Z} $, and $R_s = |\underline{Z}| \cos\varphi$ such as $X_{Ls} = |\underline{Z}| \sin\varphi$. | + | For the **series circuit**, the impedances add up like: $R_s + {\rm j}\cdot X_{Ls} = \underline{Z} $, and $R_s = |\underline{Z}| \cos\varphi$ such as $X_{Ls} = |\underline{Z}| \sin\varphi$. |
| Therefore: | Therefore: | ||
| \begin{align*} | \begin{align*} | ||
| Zeile 874: | Zeile 878: | ||
| - | For the **parallel circuit**, the impedances add up like ${{1}\over{R_p}} + {{1}\over{j\cdot X_{Lp}}}= {{1}\over{\underline{Z}}} $. \\ | + | For the **parallel circuit**, the impedances add up like ${{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}}= {{1}\over{\underline{Z}}} $. \\ |
| The easiest thing is here to use the formulas of $R_s$ and $X_{Ls}$ from before: | The easiest thing is here to use the formulas of $R_s$ and $X_{Ls}$ from before: | ||
| \begin{align*} | \begin{align*} | ||
| - | {{1}\over{R_p}} + {{1}\over{j\cdot X_{Lp}}} &=& {{1}\over{R_s + j\cdot X_{Ls}}} \\ | + | {{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}} &=& {{1}\over{R_s + {\rm j}\cdot X_{Ls}}} \\ |
| - | {{1}\over{R_p}} - j {{1}\over{X_{Lp}}} | + | {{1}\over{R_p}} - {\rm j} {{1}\over{X_{Lp}}} |
| - | &=& {{Z \cdot \cos \varphi - j\cdot Z \cdot \sin \varphi }\over{Z^2}}\\ | + | &=& {{Z \cdot \cos \varphi - {\rm j}\cdot Z \cdot \sin \varphi }\over{Z^2}} \\ |
| - | &=& {{\cos \varphi - j \cdot \sin \varphi }\over{Z}}\\ | + | &=& { {\cos \varphi - {\rm j} \cdot \sin \varphi } |
| \end{align*} | \end{align*} | ||
| Zeile 977: | Zeile 981: | ||
| \begin{align*} | \begin{align*} | ||
| - | \underline{I} &= I_R + j \cdot I_L \\ | + | \underline{I} &= I_R |
| - | &= I \cdot \cos\varphi - j \cdot I \cdot \sin\varphi | + | &= I \cdot \cos\varphi - {\rm j} \cdot I \cdot \sin\varphi |
| \end{align*} | \end{align*} | ||
| Zeile 1090: | Zeile 1094: | ||
| Q &= \Re (U) \cdot \Im (I) \\ | Q &= \Re (U) \cdot \Im (I) \\ | ||
| &= U \cdot {{U}\over{X}} \\ | &= U \cdot {{U}\over{X}} \\ | ||
| - | &= {{U^2}\over{X}} \\ | + | & |
| \end{align*} | \end{align*} | ||
| Zeile 1148: | Zeile 1152: | ||
| \begin{align*} | \begin{align*} | ||
| \underline{S}_{\rm net} &=& \underline{S}_1 | \underline{S}_{\rm net} &=& \underline{S}_1 | ||
| - | &=& P_1 + j \cdot Q_1 | + | &=& P_1 + {\rm j} \cdot Q_1 |
| - | &=& P_1 + P_2 & | + | &=& P_1 + P_2 & |
| - | &=& 2.7 {~\rm kW} + 3.8 {~\rm kW} &+& j \cdot (1.4 {~\rm kVAr} + 3.2 {~\rm kVAr}) \\ | + | &=& 2.7 {~\rm kW} + 3.8 {~\rm kW} & |
| - | &=& 6.5 {~\rm kW} & | + | &=& 6.5 {~\rm kW} & |
| - | &=& P_{\rm net} & | + | &=& P_{\rm net} & |
| \end{align*} \\ | \end{align*} \\ | ||
| As a complex value in Euler representation: | As a complex value in Euler representation: | ||
| \begin{align*} | \begin{align*} | ||
| - | \underline{S}_{\rm net} &=& \sqrt{P_{\rm net}^2 | + | \underline{S}_{\rm net} &=& \sqrt{P_{\rm net}^2 |
| - | \sqrt{(6.5 {~\rm kW})^2+ | + | \sqrt{(6.5 {~\rm kW})^2+ |
| - | 8.0 {~\rm kVA} & | + | 8.0 {~\rm kVA} & |
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| \begin{align*} | \begin{align*} | ||
| - | \underline{S}_{\rm net} &=& 6.5 {~\rm kW}+ j \cdot 4.6 {~\rm kVAr} \\ | + | \underline{S}_{\rm net} &=& 6.5 {~\rm kW} + {\rm j} \cdot 4.6 {~\rm kVAr} \\ |
| - | &=& 8.0 {~\rm kVA} \cdot e^{j \cdot 35°} \\ | + | &=& 8.0 {~\rm kVA} \cdot {\rm e}^{{\rm j} \cdot 35°} \\ |
| \end{align*} | \end{align*} | ||
| </ | </ | ||