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| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
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electrical_engineering_2:polyphase_networks [2023/03/19 12:44] mexleadmin |
electrical_engineering_2:polyphase_networks [2023/09/19 23:52] 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 5: | Zeile 5: | ||
| * three-phase four-wire systems | * three-phase four-wire systems | ||
| - | === 7.0 Recap of complex two-terminal networks === | + | ===== 7.0 Recap of complex two-terminal networks |
| In the last semester, AC current, AC voltage, and their effects have been considered on a circuit that had simply included an AC voltage source. \\ These circuits can be now understood as. | In the last semester, AC current, AC voltage, and their effects have been considered on a circuit that had simply included an AC voltage source. \\ These circuits can be now understood as. | ||
| 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 320: | Zeile 319: | ||
| Various general technical terms in the polyphase system (in German: Mehrphasensystem) will now be briefly discussed. | Various general technical terms in the polyphase system (in German: Mehrphasensystem) will now be briefly discussed. | ||
| - | - A **$m$-phase system** | + | - A **$m$-phase system** |
| - | - An $m$-phase system is **symmetrical** | + | The voltages are generated by a homogenous magnetic field containing $m$ rotating windings, which are arranged with a fixed offset to each other (see <imgref imageNo04> |
| + | < | ||
| + | </ | ||
| + | - An $m$-phase system is **symmetrical** | ||
| + | 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: | ||
| + | $\underline{U}_1 = \sqrt{2} \cdot U \cdot {\rm e} ^{{\rm j}(\omega t + 0°)}$, | ||
| + | $\underline{U}_2 = \sqrt{2} \cdot U \cdot {\rm e} ^{{\rm j}(\omega t - 120°)}$, | ||
| + | $\underline{U}_3 = \sqrt{2} \cdot U \cdot {\rm e} ^{{\rm j}(\omega t - 240°)}$ \\ | ||
| + | < | ||
| + | </ | ||
| - The windings can be concatenated (=linked) in different ways. The most important ways of **concatenation** are: | - The windings can be concatenated (=linked) in different ways. The most important ways of **concatenation** are: | ||
| - All windings are independently connected to a load. This phase system is called **non-interlinked** | - All windings are independently connected to a load. This phase system is called **non-interlinked** | ||
| - | - All windings are connected to each other, then the phase system is called **interlinked**. \\ \\ <WRAP outdent> | + | - All windings are connected to each other, then the phase system is called **interlinked**. |
| + | With interlinking, | ||
| + | The two simulations in <imgref pic20> show a non-interlinked and an interlinked circuit with generator and load in star shape.</ | ||
| + | < | ||
| - | </ | + | </ |
| - | + | ||
| - | - The instantaneous power $p_i(t)$ of a winding $i$ is variable in time. For the instantaneous power $p(t)$ of the $m$-phase system one has to consider all single instantaneous powers of the windings. When this instantaneous power $p(t)$ does not change with time, the polyphase system is called **balanced**. \\ If a balanced load is used, then polyphase systems are balanced with $m\geq3$. \\ \\ For $m\geq3$ and symmetrical load, the following is obtained for the instantaneous power: \\ \\ $\quad \quad p = m \cdot U \cdot I \cdot \cos\varphi = P$ \\ < | + | |
| + | - The instantaneous power $p_i(t)$ of a winding $i$ is variable in time. For the instantaneous power $p(t)$ of the $m$-phase system one has to consider all single instantaneous powers of the windings. When this instantaneous power $p(t)$ does not change with time, the polyphase system is called **balanced**. < | ||
| + | If a balanced load is used, then polyphase systems are balanced with $m\geq3$. \\ \\ | ||
| + | For $m\geq3$ and symmetrical load, the following is obtained for the instantaneous power: \\ \\ | ||
| + | $\quad \quad p = m \cdot U \cdot I \cdot \cos\varphi = P$ \\ | ||
| + | < | ||
| + | </ | ||
| The following simulation shows the power in the different phases of a symmetrical and balanced system. The instantaneous power of each phase is a non-negative sinusoidal function shifted by $0°$, $120°$, and $240°$. | The following simulation shows the power in the different phases of a symmetrical and balanced system. The instantaneous power of each phase is a non-negative sinusoidal function shifted by $0°$, $120°$, and $240°$. | ||
| Zeile 345: | 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 365: | Zeile 383: | ||
| === Three-phase generator === | === Three-phase generator === | ||
| - | * The windings of a three-phase generator are called $\rm U$, $\rm V$, $\rm W$; the winding connections are correspondingly called: $\rm U1$, $\rm U2$, $\rm V1$, $\rm V2$, $\rm W1$, $\rm W2$ (see <imgref imageNo10> | + | * The windings of a three-phase generator are called $\rm U$, $\rm V$, $\rm W$; the winding connections are correspondingly called: $\rm U1$, $\rm U2$, $\rm V1$, $\rm V2$, $\rm W1$, $\rm W2$ (see <imgref imageNo10> |
| - | * The typical **winding connections** | + | < |
| - | * The **phase voltages** | + | * The typical **winding connections** |
| + | < | ||
| + | * The **phase voltages** | ||
| + | \begin{align*} | ||
| + | \color{RoyalBlue }{u_{\rm U}} & \color{RoyalBlue }{= \sqrt{2} U \cdot \cos(\omega t + \alpha - 0 | ||
| + | \color{Green | ||
| + | \color{DarkOrchid}{u_{\rm W}} & \color{DarkOrchid}{= \sqrt{2} U \cdot \cos(\omega t + \alpha - {{4}\over{3}}\pi)} \\ | ||
| + | \color{RoyalBlue }{u_{\rm U}} + \color{Green}{u_{\rm V}} + \color{DarkOrchid}{u_{\rm W}} & = 0 \end{align*}</ | ||
| * The **direction of rotation** | * The **direction of rotation** | ||
| * The three-phase generator with clockwise direction (CW, mathematically negative orientation) shows the phase sequence: $u_{\rm U}$, $u_{\rm V}$, $u_{\rm W}$, Therefore, $u_{\rm V}$ is $120°$ lagging to $u_{\rm U}$. \\ This is the common setup for generators. | * The three-phase generator with clockwise direction (CW, mathematically negative orientation) shows the phase sequence: $u_{\rm U}$, $u_{\rm V}$, $u_{\rm W}$, Therefore, $u_{\rm V}$ is $120°$ lagging to $u_{\rm U}$. \\ This is the common setup for generators. | ||
| Zeile 377: | Zeile 402: | ||
| The lines connected to the generator / load terminals $\rm U1$, $\rm V1$, $\rm W1$ are often called $\rm L1$, $\rm L2$, $\rm L3$ ($\rm L$ for **L**ine or **L**ive = active) outside of the generator or load. \\ It is important to distinguish between the different types of voltages and currents, which depend on the point of view (either onto a three-phase generator/ | The lines connected to the generator / load terminals $\rm U1$, $\rm V1$, $\rm W1$ are often called $\rm L1$, $\rm L2$, $\rm L3$ ($\rm L$ for **L**ine or **L**ive = active) outside of the generator or load. \\ It is important to distinguish between the different types of voltages and currents, which depend on the point of view (either onto a three-phase generator/ | ||
| - | * **String voltages/ | + | * **String voltages/ |
| - | * **Phase voltages/ | + | The string voltages/ |
| + | These voltages are shown in the previous images as $u_\rm U$, $u_\rm V$, and $u_\rm W$. | ||
| + | </ | ||
| + | * **Phase voltages/ | ||
| + | The phase voltages are measured differentially between the lines. The phase voltages are therefore given as $U_{12}$, $U_{23}$, $U_{31}$. \\ | ||
| + | The phase currents are given as the currents through a single line: $I_1$, $I_2$, $I_3$. \\ | ||
| + | The potential of the star point is called **neutral** $\rm N$ </ | ||
| * **Star-voltages** $U_\rm Y$ (alternatively: | * **Star-voltages** $U_\rm Y$ (alternatively: | ||
| < | < | ||
| Zeile 443: | Zeile 474: | ||
| - **Voltages**: | - **Voltages**: | ||
| - | - **Currents**: | + | - **Currents**: |
| - | - The **true power** $P_x$ for each string is given by the apparent power $S_x$ of the string times the indivitual | + | The phase currents are given by the phase impedances and the star-voltages: |
| - | - For the **apparent power** one could think of $S_x$ for each string is given by the string voltage and the current through the string $S_x = U_{\rm S} \cdot I_x$. However, this misses out on the apparent power of the neutral line! \\ Even when considering all four lines a simple addition of all the apparent powers per phase would be problematic: | + | \begin{align*} |
| + | \underline{I}_1 = {{\underline{U}_{1 \rm N}}\over{\underline{Z}_1^\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*}</ | ||
| + | - 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*} \\ | ||
| + | Therefore, the resulting true power for the full load is: \\ | ||
| + | \begin{align*} P = U_{\rm S} \cdot ( I_1 \cdot \cos \varphi_1 + I_2 \cdot \cos \varphi_2 + I_3 \cdot \cos \varphi_3) \end{align*} \\ | ||
| + | The angle $\varphi$ here is given by $\varphi = \varphi_u - \varphi_i$, and hence: \\ | ||
| + | \begin{align*} P = U_{\rm S} \cdot \left( I_1 \cdot \cos (\varphi_{u, | ||
| + | - For the **apparent power** one could think of $S_x$ for each string is given by the string voltage and the current through the string $S_x = U_{\rm S} \cdot I_x$. However, this misses out on the apparent power of the neutral line! < | ||
| + | Even when considering all four lines a simple addition of all the apparent powers per phase would be problematic: | ||
| + | By DIN 40110 the **collective apparent power ** $S_\Sigma$ can be assumed as \\ | ||
| + | \begin{align*} | ||
| + | S_\Sigma &= \sqrt{\sum_x U_{x \rm N}^2+ \underbrace{U_{\rm N}^2}_{=0}} &\cdot & \sqrt{\sum_x I_{x}^2+ I_{\rm N}^2} \\ | ||
| + | &= \sqrt{3} \cdot U_{\rm S} & \cdot & \sqrt{I_1^2 + I_2^2 + I_2^3 + I_{\rm N}^2} \\ \end{align*}</ | ||
| - Given the collective apparent power the **collective reactive power** $Q_\Sigma$ ist given by \\ \begin{align*} | - Given the collective apparent power the **collective reactive power** $Q_\Sigma$ ist given by \\ \begin{align*} | ||
| </ | </ | ||
| Zeile 451: | Zeile 497: | ||
| <panel type=" | <panel type=" | ||
| In the example, this leads to: | In the example, this leads to: | ||
| - | - The star-voltages and the phase voltages are given as \begin{align*} U_{\rm S}=& 231 ~\rm V = U_{\rm 1N} = U_{\rm 2N} = U_{\rm 3N} \\ U_{\rm L}=\sqrt{3} \cdot 231 ~\rm V = & 400 ~\rm V = U_{12} = U_{23} = U_{31} \end{align*} \\ The phasors of the star-voltages are given as: \\ {{drawio> | + | - The star-voltages and the phase voltages are given as < |
| - | - Based on the star-voltages and the given impedances the phase currents are: \\ \begin{align*} \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}}} &= &+23.08 {~\rm A} &- j \cdot 0.72 {~\rm A} &= &23.09 ~{~\rm A} \quad &\angle -1.8° \\ \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}}}}} &= &+ 5.58 {~\rm A} &- j \cdot 4.50 {~\rm A} &= & 7.17 {~\rm A} \quad &\angle -38.9° \\ \underline{I}_3 &= {{\underline{U}_{\rm 3N}}\over{\underline{Z}_3}} &= & | + | \begin{align*} |
| - | - The true power is calculated by: \\ \begin{align*} P = 231 {~\rm V} \cdot \big( 23.09 {~\rm A} \cdot \cos (0° - (-1.8°))+ 7.17 {~\rm A} \cdot \cos (-120° - (-38.9°)) + 11.55 {~\rm A} \cdot \cos (-240° - (-240°)\big) = 8.26 {~\rm kW} \end{align*} | + | U_{\rm S}=& 231 ~\rm V = U_{\rm 1N} = U_{\rm 2N} = U_{\rm 3N} \\ |
| - | - The collective apparent power is: \\ \begin{align*} S_\Sigma & | + | U_{\rm L}=\sqrt{3} \cdot 231 ~\rm V = & 400 ~\rm V = U_{12} = U_{23} = U_{31} |
| - | - The collective reactive power is: \\ \begin{align*} Q_\Sigma & | + | \end{align*} \\ |
| + | The phasors of the star-voltages are given as: \\ {{drawio> | ||
| + | - Based on the star-voltages and the given impedances the phase currents are: < | ||
| + | \begin{align*} | ||
| + | \underline{I}_1 &= {{\underline{U}_{\rm 1N}}\over{\underline{Z}_1}} | ||
| + | | ||
| + | | ||
| + | \underline{I}_2 &= {{\underline{U}_{\rm 2N}}\over{\underline{Z}_2}} | ||
| + | | ||
| + | | ||
| + | \underline{I}_3 &= {{\underline{U}_{\rm 3N}}\over{\underline{Z}_3}} | ||
| + | | ||
| + | | ||
| + | \underline{I}_{\rm N} | ||
| + | | ||
| + | \end{align*} | ||
| + | - The true power is calculated by: < | ||
| + | \begin{align*} | ||
| + | P = 231 {~\rm V} \cdot \big( 23.09 {~\rm A} \cdot \cos (0° - (-1.8°))+ 7.17 {~\rm A} \cdot \cos (-120° - (-38.9°)) + 11.55 {~\rm A} \cdot \cos (-240° - (-240°)\big) | ||
| + | | ||
| + | \end{align*} | ||
| + | - The collective apparent power is: < | ||
| + | \begin{align*} | ||
| + | S_\Sigma & | ||
| + | | ||
| + | \end{align*}</ | ||
| + | - The collective reactive power is: < | ||
| + | \begin{align*} | ||
| + | Q_\Sigma & | ||
| + | | ||
| + | \end{align*}</ | ||
| < | < | ||
| Zeile 465: | Zeile 541: | ||
| In the case of a symmetric load, the situation and the formulas get much simpler: | In the case of a symmetric load, the situation and the formulas get much simpler: | ||
| - The **phase-voltages** $U_\rm L$ and star-voltages $U_{\rm Y} = U_{\rm S}$ are equal to the asymmetric load: $U_{\rm L} = \sqrt{3}\cdot U_{\rm S}$. | - The **phase-voltages** $U_\rm L$ and star-voltages $U_{\rm Y} = U_{\rm S}$ are equal to the asymmetric load: $U_{\rm L} = \sqrt{3}\cdot U_{\rm S}$. | ||
| - | - For equal impedances the absolute value of all **phase currents** $I_x$ are the same: $|\underline{I}_x|= |\underline{I}_{\rm S}| = \left|{{\underline{U}_{\rm S}}\over{\underline{Z}_{\rm S}^\phantom{O}}} \right|$. Since the phase currents have the same absolute value and have the same $\varphi$, they will add up to zero. Therefore there is no current on the neutral line: $I_{\rm N} =0$ | + | - For equal impedances the absolute value of all **phase currents** $I_x$ are the same: $|\underline{I}_x|= |\underline{I}_{\rm S}| = \left|{{\underline{U}_{\rm S}}\over{\underline{Z}_{\rm S}^\phantom{O}}} \right|$. |
| - | - The **true power** is three times the true power of a single phase: $P = 3 \cdot U_{\rm S} I_{\rm S} \cdot \cos \varphi$. Based on the line voltages $U_{\rm L}$, the formula is $P = \sqrt{3} \cdot U_{\rm L} I_{\rm S} \cdot \cos \varphi$ | + | - The **true power** is three times the true power of a single phase: $P = 3 \cdot U_{\rm S} I_{\rm S} \cdot \cos \varphi$. |
| - | - The **(collective) apparent power** - given the formula above - is: $S_\Sigma = \sqrt{3}\cdot U_{\rm S} \cdot \sqrt{3\cdot I_{\rm S}^2} = 3 \cdot U_{\rm S} I_{\rm S}$. This corresponds to three times the apparent power of a single phase. | + | - The **(collective) apparent power** - given the formula above - is: $S_\Sigma = \sqrt{3}\cdot U_{\rm S} \cdot \sqrt{3\cdot I_{\rm S}^2} = 3 \cdot U_{\rm S} I_{\rm S}$. \\ This corresponds to three times the apparent power of a single phase. |
| - The **reactive power** leads to: $Q_\Sigma = \sqrt{S_\Sigma^2 - P^2} = 3 \cdot U_{\rm S} I_{\rm S} \cdot \sin (\varphi)$. | - The **reactive power** leads to: $Q_\Sigma = \sqrt{S_\Sigma^2 - P^2} = 3 \cdot U_{\rm S} I_{\rm S} \cdot \sin (\varphi)$. | ||
| </ | </ | ||
| Zeile 499: | 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**: |
| - | - **Currents**: | + | \begin{align*} |
| - | - Also here, the **true power** $P_x$ for each string is given by: \\ \begin{align*} P_x &= S_x \cdot \cos \varphi_x = U_{\rm S} \cdot I_x \cdot \cos \varphi_x \end{align*} \\ Also here, the resulting true power for the full load is (with $U_{\rm S}$ as the RMS value of the network star-voltage): | + | \underline{U}_{\rm 1S} &= \underline{U}_{\rm 1N} - \underline{U}_{\rm SN} \\ |
| - | - 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*} \underline{S} &= P + j\cdot Q = \sum_x \underline{S}_x = \sum_x | + | \underline{U}_{\rm 2S} &= \underline{U}_{\rm 2N} - \underline{U}_{\rm SN} \\ |
| - | - The abolute **reactive power** $Q$ can be calulated by the apparent power: | + | \underline{U}_{\rm 3S} &= \underline{U}_{\rm 3N} - \underline{U}_{\rm SN} \\ |
| + | \end{align*}</ | ||
| + | - **Currents**: | ||
| + | \begin{align*} | ||
| + | \underline{I}_1 = {{\underline{U}_{\rm 1S}}\over{\underline{Z}_1^\phantom{O}}} \quad , \quad \underline{I}_2 | ||
| + | | ||
| + | | ||
| + | \end{align*} \\ | ||
| + | To get $\underline{U}_{\rm SN}$, one has to combine the individual formulas for $\underline{I}_x$, | ||
| + | \begin{align*} | ||
| + | \underline{U}_{\rm SN} = {{\sum_x \left( \Large{{{1}\over{\underline{Z}_x^\phantom{O}}}} | ||
| + | \cdot \normalsize{\underline{U}_{x \rm N}} \right) }\over{\sum_x \left( \Large{{{1}\over{\underline{Z}_x^\phantom{O}}}} \right) }} | ||
| + | \end{align*}</ | ||
| + | - Also here, the **true power** $P_x$ for each string is given by: < | ||
| + | \begin{align*} | ||
| + | P_x &= S_x \cdot \cos \varphi_x = U_{\rm S} \cdot I_x \cdot \cos \varphi_x | ||
| + | \end{align*} \\ | ||
| + | Also here, the resulting true power for the full load is (with $U_{\rm S}$ as the RMS value of the network star-voltage): | ||
| + | \begin{align*} | ||
| + | P &= U_{\rm S} \cdot ( I_1 \cdot \cos \varphi_1 | ||
| + | | ||
| + | \end{align*}</ | ||
| + | - 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*} | ||
| + | \underline{S} &= P + {\rm j}\cdot Q = \sum_x \underline{S}_x = \sum_x | ||
| + | In order to simplify the calculation, | ||
| + | \begin{align*} | ||
| + | \underline{S} &= \sum_x | ||
| + | | ||
| + | \end{align*} | ||
| + | Given that $\sum_x \underline{I}_x =0$, it is also true, that $\sum_x \underline{I}_x^* =0$ and so $\underline{I}_3^* = -\underline{I}_1^* - \underline{I}_2^*$. \\ | ||
| + | By this, one can further simplify the calculation for the apparent power down to \\ | ||
| + | \begin{align*} | ||
| + | \underline{S} &= \underline{U}_{13} \cdot \underline{I}_1^* + \underline{U}_{23} \cdot \underline{I}_2^* \\ | ||
| + | | ||
| + | | ||
| + | \end{align*} \\ | ||
| + | For the phase voltages it applies that: $\underline{U}_{12} = - \underline{U}_{21}$, | ||
| + | \begin{align*} | ||
| + | S_\Sigma &= \sqrt{ {{1}\over{2}} \cdot (U_{12}^2 + U_{23}^2 + U_{31}^2) } \cdot \sqrt{\sum_x I_x^2} &= \sqrt{3} U_{\rm S} \cdot \sqrt{\sum_x I_x^2} | ||
| + | \end{align*} | ||
| + | - The abolute **reactive power** $Q$ can be calulated by the apparent power: | ||
| + | \begin{align*} | ||
| + | the **collective reactive power** $Q_\Sigma$ is given by the collective apparent power: | ||
| + | \begin{align*} Q &= \sqrt{S_\Sigma^2 - P^2} \end{align*} | ||
| </ | </ | ||
| Zeile 510: | Zeile 630: | ||
| In the example, this leads to: | In the example, this leads to: | ||
| - The phase voltages are given as \begin{align*} U_{\rm L}=\sqrt{3} \cdot 231 {~\rm V} = & 400 {~\rm V} = U_{12} = U_{23} = U_{31} \end{align*} \\ The phasors of the star-voltages of the network are again given as \\ {{drawio> | - The phase voltages are given as \begin{align*} U_{\rm L}=\sqrt{3} \cdot 231 {~\rm V} = & 400 {~\rm V} = U_{12} = U_{23} = U_{31} \end{align*} \\ The phasors of the star-voltages of the network are again given as \\ {{drawio> | ||
| - | - Based on the star-voltages of the network and the given impedances the star-voltage $\underline{U}_{\rm SN}$ of the load can be calculated with: \\ \begin{align*} \underline{U}_{\rm SN} = {{\sum_x \left( \Large{{{1}\over{\underline{Z}_x^\phantom{O}}}} \cdot \normalsize{\underline{U}_{x \rm N}} \right) }\over{\sum_x \left( \Large{{{1}\over{\underline{Z}_x^\phantom{O}}}} \right) }} \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. \\ The numerator is therefore: $22.88 {~\rm A} + j \cdot 4.77 {~\rm A}$ (see calculation for the four-wire system). \\ The denominator is: \\ \begin{align*} \sum_x | + | - Based on the star-voltages of the network and the given impedances the star-voltage $\underline{U}_{\rm SN}$ of the load can be calculated with: < |
| + | \begin{align*} | ||
| + | \underline{U}_{\rm SN} = {{\sum_x \left( \Large{{{1}\over{\underline{Z}_x^\phantom{O}}}} \cdot \normalsize{\underline{U}_{x \rm N}} \right) } | ||
| + | \over{\sum_x \left( \Large{{{1}\over{\underline{Z}_x^\phantom{O}}}} \right) }} | ||
| + | \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. \\ | ||
| + | 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: \\ | ||
| + | \begin{align*} | ||
| + | \sum_x | ||
| + | | ||
| + | | ||
| + | | ||
| + | The star-voltage $\underline{U}_{\rm SN}$ of the load is: | ||
| + | \begin{align*} | ||
| + | \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}} \\ \\ | ||
| + | &= 148.7 | ||
| + | Given this star-voltage $\underline{U}_{\rm SN}$ of the load, the phase currents are: \\ | ||
| + | \begin{align*} | ||
| + | \underline{I}_1 &= {{\underline{U}_{\rm 1N} - \underline{U}_{\rm SN}}\over{\underline{Z}_1^\phantom{O}}} | ||
| + | | ||
| + | | ||
| + | \underline{I}_2 &= {{\underline{U}_{\rm 2N} - \underline{U}_{\rm SN}}\over{\underline{Z}_2^\phantom{O}}} | ||
| + | | ||
| + | | ||
| + | \underline{I}_3 &= {{\underline{U}_{\rm 3N} - \underline{U}_{\rm SN}}\over{\underline{Z}_3^\phantom{O}}} | ||
| + | | ||
| + | | ||
| - 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: \\ \begin{align*} \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 (8.21 {~\rm A} + j \cdot 0.70 {~\rm A} ) + e^{- j \cdot 3/6 \pi} \cdot (5.00{~\rm A} - 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} \\ &= \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 (5.00{~\rm A} - j \cdot 9.08{~\rm A}) + e^{- j \cdot 7/6 \pi} \cdot (-13.21{~\rm A} - j \cdot 9.78{~\rm A})) &= 6.62 {~\rm kW} - j \cdot 3.40 {~\rm kVAr} \\ & = 7.44 {~\rm kVA} \quad \angle -27.2°\end{align*} | + | - The apparent power $\underline{S}$ is: < |
| - | - The reactive power is: \begin{align*} Q &= -j \cdot (\underline{S} - P) = -3.40{~\rm kVAr} \\ \end{align*} \\ The collective reactive power is: \\ \begin{align*} Q_\Sigma & | + | \begin{align*} |
| + | \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} - {\rm j} \cdot 3.40 {~\rm kVAr} \\ | ||
| + | &= \underline{U}_{12} \cdot \underline{I}_1^* + \underline{U}_{32} \cdot \underline{I}_3^* | ||
| + | &=& 400{~\rm V} \cdot ( | ||
| + | | ||
| + | | ||
| + | &=& 400{~\rm V} \cdot (- {\rm e}^{{\rm j} \cdot 1/6 \pi} \cdot (5.00{~\rm A} - | ||
| + | | ||
| + | | ||
| + | \end{align*} | ||
| + | The collective apparent power is: \\ | ||
| + | \begin{align*} | ||
| + | S_\Sigma &= \sqrt{3} U_{\rm S} | ||
| + | &= \sqrt{3} \cdot 231{~\rm V} \cdot \sqrt{(8.24{~\rm A})^2+(10.36{~\rm A})^2+(16.44A)^2} | ||
| + | | ||
| + | - The reactive power is: < | ||
| + | \begin{align*} Q &= -{\rm j} \cdot (\underline{S} - P) = -3.40{~\rm kVAr} \\ | ||
| + | \end{align*} \\ | ||
| + | The collective reactive power is: \\ | ||
| + | \begin{align*} Q_\Sigma & | ||
| + | \end{align*} | ||
| < | < | ||
| Zeile 552: | Zeile 721: | ||
| Again here the " | Again here the " | ||
| - | - **Voltages**: | + | - **Voltages**: |
| - | - **Currents**: | + | \begin{align*} |
| + | \underline{U}_{12} &=& U_{\rm L} \cdot {\rm e}^{ | ||
| + | \underline{U}_{23} &=& U_{\rm L} \cdot {\rm e}^{- {\rm j}\cdot {{3}\over{6}}} \\ | ||
| + | \underline{U}_{31} &=& U_{\rm L} \cdot {\rm e}^{- {\rm j}\cdot {{7}\over{6}}} | ||
| + | \end{align*}</ | ||
| + | - **Currents**: | ||
| + | \begin{align*} | ||
| + | \underline{I}_{1} = \underline{I}_{12} - \underline{I}_{31} \\ | ||
| + | \underline{I}_{2} = \underline{I}_{23} - \underline{I}_{12} \\ | ||
| + | \underline{I}_{3} = \underline{I}_{31} - \underline{I}_{23} \\ | ||
| + | \end{align*} \\ | ||
| + | The string currents can be calculated by the string voltages and the impedances: \\ | ||
| + | \begin{align*} | ||
| + | \underline{I}_{12} = {{\underline{U}_{12}}\over{\underline{Z}_{12}^\phantom{O}}} \quad , \quad | ||
| + | \underline{I}_{23} = {{\underline{U}_{23}}\over{\underline{Z}_{23}^\phantom{O}}} \quad , \quad | ||
| + | \underline{I}_{31} = {{\underline{U}_{31}}\over{\underline{Z}_{31}^\phantom{O}}} \end{align*} | ||
| - Also here, the **true power** can be calculated by adding up the true power of each phase. The faster way (as shown before) is to add up the (complex) apparent power. | - Also here, the **true power** can be calculated by adding up the true power of each phase. The faster way (as shown before) is to add up the (complex) apparent power. | ||
| - | - The **apparent power** $\underline{S}_x$ here is again the sum the (complex) apparent power for each string: | + | - The **apparent power** $\underline{S}_x$ here is again the sum the (complex) apparent power for each string: |
| - | - The abolute **reactive power** $Q$ can be calulated by the apparent power: | + | \begin{align*} |
| + | \underline{S} &= \underline{U}_{12} \cdot \underline{I}_{12}^* | ||
| + | + \underline{U}_{23} \cdot \underline{I}_{23}^* | ||
| + | + \underline{U}_{31} \cdot \underline{I}_{31}^* | ||
| + | \end{align*} | ||
| + | Since $\underline{U}_{12}$, | ||
| + | \begin{align*} | ||
| + | \boxed{ | ||
| + | \underline{S} = P + {\rm j} \cdot Q | ||
| + | | ||
| + | | ||
| + | | ||
| + | \end{align*} \\ | ||
| + | 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 | ||
| + | \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} | ||
| + | \end{align*} | ||
| + | - The abolute **reactive power** $Q$ can be calulated by the apparent power: | ||
| + | \begin{align*} | ||
| + | the **collective reactive power** $Q_\Sigma$ is given by the collective apparent power: | ||
| + | \begin{align*} Q &= \sqrt{S_\Sigma^2 - P^2} \end{align*} | ||
| </ | </ | ||
| <panel type=" | <panel type=" | ||
| In the example, this leads to: | In the example, this leads to: | ||
| - | - The phase voltages are given as \begin{align*} U_{\rm L}=\sqrt{3} \cdot 231 ~\rm V = & 400 ~\rm V = U_{12} = U_{23} = U_{31} \end{align*} \\ The phasors of the string voltages of the network are given as \\ {{drawio> | + | - The phase voltages are given as < |
| - | - Based on the string voltages of the network and the given impedances the string currents $\underline{I}_{12}$, | + | \begin{align*} |
| - | - The true power is calculated by: \\ \begin{align*} P = 231 {~\rm V} \cdot \big( 53.32 {~\rm A} \cdot \cos (0° - (9.6°))- 31.01 {~\rm A} \cdot \cos (-120° - (-137.8°)) + 31.90 {~\rm A} \cdot \cos (-240° - (+158.0°)\big) = 24.77 {~\rm kW} \end{align*} | + | U_{\rm L}=\sqrt{3} \cdot 231 ~\rm V |
| - | - The apparent power $\underline{S}$ is: \\ \begin{align*} \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 |
| - | - The reactive power is: \begin{align*} Q &= |\underline{S} - P| = -4.41 {~\rm kVAr} \\ \end{align*} \\ The collective reactive power is: \\ \begin{align*} Q_\Sigma & | + | = U_{12} = U_{23} = U_{31} |
| + | \end{align*} \\ | ||
| + | The phasors of the string voltages of the network are given as \\ | ||
| + | {{drawio> | ||
| + | - Based on the string voltages of the network and the given impedances the string currents $\underline{I}_{12}$, | ||
| + | \begin{align*} | ||
| + | \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} }} | ||
| + | &=& 35.24 {~\rm A} + {\rm j} \cdot 18.90 {~\rm A} &=& 40 {~\rm A} \quad &\angle 28.2° | ||
| + | \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}}} }} | ||
| + | &=& 12.27 {~\rm A} - {\rm j} \cdot 1.93 {~\rm A} &=& 12.42 {~\rm A} \quad &\angle -8.9° \\ | ||
| + | \underline{I}_{31} &=& {{400 {~\rm V} \cdot\left(-{{1}\over{2}}\sqrt{3}+{{1}\over{2}} \cdot {\rm j} \right)}\over{ 20 ~\Omega}} | ||
| + | &=& -17.33 {~\rm A} + {\rm j} \cdot 10.00 {~\rm A} &=& 20.01 {~\rm A} \quad &\angle 150° \end{align*} \\ | ||
| + | By these voltages the phase currents $\underline{I}_x$ can be calculated: \\ | ||
| + | \begin{align*} | ||
| + | \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}) & | ||
| + | | ||
| + | \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} | ||
| + | | ||
| + | \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} | ||
| + | | ||
| + | \end{align*} \\</ | ||
| + | - The true power is calculated by: < | ||
| + | \begin{align*} | ||
| + | P = 231 {~\rm V} \cdot \big( 53.32 {~\rm A} \cdot \cos ( | ||
| + | | ||
| + | | ||
| + | | ||
| + | - The apparent power $\underline{S}$ is: < | ||
| + | \begin{align*} | ||
| + | \underline{S} &= \underline{U}_{13} \cdot \underline{I}_1^* + \underline{U}_{23} \cdot \underline{I}_2^* | ||
| + | | ||
| + | {\rm e}^{-{\rm j} \cdot 3/6 \pi} \cdot (-22.98 {~\rm A} + {\rm j} \cdot 20.83 {~\rm A})) | ||
| + | | ||
| + | | ||
| + | | ||
| + | {\rm e}^{-{\rm j} \cdot 3/6 \pi} \cdot (-29.59 {~\rm A} - {\rm j} \cdot 11.93 {~\rm A})) | ||
| + | | ||
| + | | ||
| + | | ||
| + | {\rm e}^{-{\rm j} \cdot 7/6 \pi} \cdot (-29.59 {~\rm A} - {\rm j} \cdot 11.93 {~\rm A})) | ||
| + | | ||
| + | | ||
| + | The collective apparent power is: \\ | ||
| + | \begin{align*} | ||
| + | S_\Sigma &= U_{\rm L} | ||
| + | &= \sqrt{3} \cdot 231 {~\rm V} \cdot \sqrt{(53.32 {~\rm A})^2+(31.01 {~\rm A})^2+(31.90 {~\rm A})^2} | ||
| + | | ||
| + | - The reactive power is: < | ||
| + | \begin{align*} Q &= |\underline{S} - P| = -4.41 {~\rm kVAr} \\ | ||
| + | \end{align*} \\ | ||
| + | The collective reactive power is: \\ | ||
| + | \begin{align*} Q_\Sigma & | ||
| + | \end{align*}</ | ||
| < | < | ||
| Zeile 612: | 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 621: | 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 724: | 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 837: | 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 895: | 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*} | ||
| </ | </ | ||