Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
| electrical_engineering_2:polyphase_networks [2023/03/22 20:32] – mexleadmin | electrical_engineering_2:polyphase_networks [2024/11/17 10:30] (aktuell) – [7.2.2 Three-Phase System] mexleadmin | ||
|---|---|---|---|
| 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 201: | Zeile 200: | ||
| Similarly, the currents and voltages can be separated into active, reactive, and apparent values. </ | Similarly, the currents and voltages can be separated into active, reactive, and apparent values. </ | ||
| - | Based on the given formulas the three types of power are connected with each other. Since the apparent power is given by $S=U\cdot I$, the active power $P = U\cdot I \cdot \sin \varphi = S \cdot \sin \varphi $ and the reactive power $Q = S \cdot \cos \varphi $, the relationship can be shown in a triangle (see <imgref imageNo02> | + | Based on the given formulas the three types of power are connected with each other. Since the apparent power is given by $S=U\cdot I$, the active power $P = U\cdot I \cdot \cos \varphi = S \cdot \cos \varphi $ and the reactive power $Q = S \cdot \sin \varphi $, the relationship can be shown in a triangle (see <imgref imageNo02> |
| < | < | ||
| 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: | ||
| * Simple three-phase machines can be used for generation. | * Simple three-phase machines can be used for generation. | ||
| - | * Rotary field machines (e.g. synchronous | + | * Rotary field machines (e.g., synchronous or induction motors) can also be simply connected to a load, converting electrical energy into mechanical energy. |
| * When a symmetrical load can be assumed, the energy flow is constant in time. | * When a symmetrical load can be assumed, the energy flow is constant in time. | ||
| * For energy transport, the voltage can be up-transformed and thus the AC current, as well as the associated power loss (= waste heat), can be reduced. | * For energy transport, the voltage can be up-transformed and thus the AC current, as well as the associated power loss (= waste heat), can be reduced. | ||
| Zeile 382: | 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 $\rm U$, $\rm V$, $\rm W$. \\ Often they are also colour-coded as red, yellow and blue - and consecutively sometimes also called $\rm R$, $\rm Y$, $\rm B$. |
| + | * 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 typical **winding connections** |
| < | < | ||
| * The **phase voltages** | * The **phase voltages** | ||
| Zeile 402: | Zeile 404: | ||
| * **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 409: | Zeile 411: | ||
| The phase currents are given as the currents through a single line: $I_1$, $I_2$, $I_3$. \\ | 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** $\rm N$ </ | The potential of the star point is called **neutral** $\rm N$ </ | ||
| - | * **Star-voltages** $U_\rm Y$ (alternatively: | + | * **Star |
| < | < | ||
| Zeile 432: | Zeile 434: | ||
| The phase voltage is therefore ${{1}\over{\sqrt{3}}} \cdot 400~\rm V \approx 230~\rm V$. The following two simulations show these voltages. | The phase voltage is therefore ${{1}\over{\sqrt{3}}} \cdot 400~\rm V \approx 230~\rm V$. The following two simulations show these voltages. | ||
| - | < | + | < |
| - | <WRAP > | + | <WRAP > |
| ==== 7.2.3 Load and Power in Three-Phase Systems ==== | ==== 7.2.3 Load and Power in Three-Phase Systems ==== | ||
| Zeile 465: | Zeile 467: | ||
| </ | </ | ||
| - | < | + | < |
| <callout title=" | <callout title=" | ||
| Zeile 479: | Zeile 481: | ||
| \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 507: | ||
| \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 539: | 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 related by: $U_{\rm L} = \sqrt{3}\cdot U_{\rm S}$ (equal to the asymmetric load). |
| - 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: $|\underline{I}_x|= |\underline{I}_{\rm S}| = \left|{{\underline{U}_{\rm S}}\over{\underline{Z}_{\rm S}^\phantom{O}}} \right|$. \\ 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: $I_{\rm N} =0$ | ||
| - 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: $P = 3 \cdot U_{\rm S} I_{\rm S} \cdot \cos \varphi$. \\ 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 **(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 564: | Zeile 566: | ||
| </ | </ | ||
| - | < | + | < |
| <callout title=" | <callout title=" | ||
| Zeile 574: | Zeile 576: | ||
| * 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 605: | ||
| - 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 623: | ||
| \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 637: | ||
| \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 679: | ||
| = 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 709: | Zeile 711: | ||
| </ | </ | ||
| - | < | + | < |
| <callout title=" | <callout title=" | ||
| Zeile 722: | Zeile 724: | ||
| - **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 749: | ||
| \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 755: | ||
| \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 775: | ||
| 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 801: | ||
| \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 845: | Zeile 850: | ||
| <panel type=" | <panel type=" | ||
| - | A passive component is fed by a sinusoidal AC voltage with the RMS value $U=230~\rm V$ and $f=50.0~\rm Hz$. The RMS current on this component is $I=5.00~\rm A$ with a phase angle of $\varphi=60°$. | + | A passive component is fed by a sinusoidal AC voltage with the RMS value $U=230~\rm V$ and $f=50.0~\rm Hz$. The RMS current on this component is $I=5.00~\rm A$ with a phase angle of $\varphi=+60°$. |
| - | - Draw the equivalent circuits based on a series and on a parallel circuit. | + | 1. Draw the equivalent circuits based on a series and a parallel circuit. |
| - | - Calculate the equivalent components for both circuits. | + | |
| - | - Calculate the real power, the reactive power, and the apparent power based on the equivalent components for both circuits from 2. . | + | |
| - | - Check the solutions from 3. via direct calculation based on the input in the task above. | + | |
| - | + | ||
| - | <button size=" | + | |
| + | # | ||
| {{drawio> | {{drawio> | ||
| + | # | ||
| - | </ | + | 2. Calculate the equivalent components for both circuits. \\ |
| - | <button size=" | + | # |
| The apparent impedance is: | The apparent impedance is: | ||
| Zeile 865: | Zeile 867: | ||
| \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 873: | Zeile 875: | ||
| \end{align*} | \end{align*} | ||
| + | \\ \\ | ||
| + | For the **parallel circuit**, the impedances add up like ${{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}}= {{1}\over{\underline{Z}}} $ with $\underline{Z} = {{U}\over{I}}\cdot e^{j\cdot \varphi}$. \\ | ||
| - | For the **parallel circuit**, the impedances add up like ${{1}\over{R_p}} + {{1}\over{j\cdot X_{Lp}}}= {{1}\over{\underline{Z}}} $. \\ | + | There are multiple ways to solve this problem. Two ways shall be shown here: |
| - | The easiest thing is here to use the formulas of $R_s$ and $X_{Ls}$ from before: | + | |
| + | === with the Euler representation === | ||
| + | Given the formula $\underline{Z} = {{U}\over{I}}\cdot e^{j\cdot \varphi}$ the following can be derived: | ||
| \begin{align*} | \begin{align*} | ||
| - | {{1}\over{R_p}} + {{1}\over{j\cdot X_{Lp}}} &=& {{1}\over{R_s + j\cdot X_{Ls}}} \\ | + | {{1}\over{\underline{Z}^{\phantom{A}}}} &= {{I}\over{U}}\cdot e^{-j\cdot \varphi} \\ |
| - | {{1}\over{R_p}} - j {{1}\over{X_{Lp}}} & | + | & |
| - | &=& {{Z \cdot \cos \varphi - j\cdot Z \cdot \sin \varphi }\over{Z^2}}\\ | + | &= {{1}\over{Z}}\cdot \left( |
| - | & | + | |
| \end{align*} | \end{align*} | ||
| + | |||
| + | Therefore, the following can be concluded: | ||
| + | \begin{align*} | ||
| + | {{1}\over{Z}}\cdot \cos(\varphi) | ||
| + | - {\rm j}\cdot \sin(\varphi) | ||
| + | \end{align*} | ||
| + | |||
| + | === with the calculated values of the series circuit === | ||
| + | Another way is to use the formulas of $R_s$ and $X_{Ls}$ from before. | ||
| + | |||
| + | \begin{align*} | ||
| + | {{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}} &=& {{1}\over{R_s + {\rm j}\cdot X_{Ls}}} \\ | ||
| + | {{1}\over{R_p}} - {\rm j} {{1}\over{X_{Lp}}} | ||
| + | &=& {{Z \cdot \cos \varphi - {\rm j}\cdot Z \cdot \sin \varphi }\over{Z^2}} \\ | ||
| + | &=& { {\cos \varphi - {\rm j} \cdot \sin \varphi } | ||
| + | \end{align*} | ||
| + | |||
| + | Therefore | ||
| Now, the real and imaginary part is analyzed individually. First the real part: | Now, the real and imaginary part is analyzed individually. First the real part: | ||
| Zeile 888: | Zeile 910: | ||
| \begin{align*} | \begin{align*} | ||
| {{1}\over{R_p}} | {{1}\over{R_p}} | ||
| - | \rightarrow R_p & | + | \rightarrow R_p & |
| \end{align*} | \end{align*} | ||
| \begin{align*} | \begin{align*} | ||
| {{1}\over{X_{Lp}}} | {{1}\over{X_{Lp}}} | ||
| - | \rightarrow X_{Lp} | + | \rightarrow X_{Lp} |
| - | \rightarrow L_p & | + | \rightarrow L_p & |
| \end{align*} | \end{align*} | ||
| - | </ | + | # |
| + | # | ||
| + | For the series circuit: | ||
| + | \begin{align*} | ||
| + | R_s &= {23 ~\Omega} \\ | ||
| + | L_s &= {127 ~\rm mH} \\ | ||
| + | \end{align*} | ||
| - | <button size=" | + | For the parallel circuit: |
| + | \begin{align*} | ||
| + | R_p &= {92 ~\Omega} \\ | ||
| + | L_p &= {169 ~\rm mH} \\ | ||
| + | \end{align*} | ||
| + | # | ||
| + | 3. Calculate the real, reactive, and apparent power based on the equivalent components for both circuits from 2. . \\ | ||
| + | |||
| + | # | ||
| + | The general formula for the apparent power is $\underline{S} = U \cdot I \cdot e^{\rm j\varphi}$. \\ By this, the following can be derived: | ||
| + | \begin{align*} | ||
| + | \underline{S} &= U \cdot I \cdot e^{\rm j\varphi} \\ | ||
| + | &= Z \cdot I^2 \cdot e^{\rm j\varphi} | ||
| + | &= {{U^2}\over{Z}} \cdot e^{\rm j\varphi} &&= {{U^2}\over{\underline{Z}^{*\phantom{I}}}} \\ | ||
| + | \end{align*} | ||
| + | |||
| + | These formulas are handy for both types of circuits to separate the apparent power into real part (real power) and complex part (apparent power): | ||
| + | - for **series circuit**: $\underline{S} =\underline{Z} \cdot I^2 $ with $\underline{Z} = R + {\rm j} X_L$ | ||
| + | - for **parallel circuit**: $\underline{S} ={{U^2}\over{\underline{Z}^{*\phantom{I}}}} $ with ${{1} \over {\underline{Z}^{\phantom{I}}} } = {{1}\over{R}} + {{1}\over{{\rm j} X_L}} \rightarrow {{1} \over {\underline{Z}^{*\phantom{I}}} } = {{1}\over{R}} + {{\rm j}\over{ X_L}} $ | ||
| + | \\ | ||
| + | Therefore: | ||
| ^ ^ series circuit ^ parallel circuit ^ | ^ ^ series circuit ^ parallel circuit ^ | ||
| | active | | active | ||
| - | | reactive power | \begin{align*} Q_s & | + | | reactive power | \begin{align*} Q_s & |
| - | | apparent power | \begin{align*} S_s &= \sqrt{P_s^2 - Q_s^2} \\ &= I^2 \cdot \sqrt{R_s^2 + Z_{Ls}^2} \\ &= 1150 {~\rm VA} \end{align*} | \begin{align*} S_p &= \sqrt{P_s^2 - Q_s^2} \\ &= U^2 \cdot \sqrt{{{1}\over{R_p^2}} + {{1}\over{Z_{Lp}^2}}} \\ &= 1150 {~\rm VA} | + | | apparent power | \begin{align*} S_s &= \sqrt{P_s^2 - Q_s^2} \\ &= I^2 \cdot \sqrt{R_s^2 + X_{Ls}^2} \\ &= 1150 {~\rm VA} \end{align*} | \begin{align*} S_p &= \sqrt{P_s^2 - Q_s^2} \\ &= U^2 \cdot \sqrt{{{1}\over{R_p^2}} + {{1}\over{X_{Lp}^2}}} \\ &= 1150 {~\rm VA} |
| - | </ | + | # |
| + | 4. Check the solutions from 3. via direct calculation based on the input in the task above. \\ | ||
| - | <button size=" | + | <button size=" |
| active power: | active power: | ||
| Zeile 928: | Zeile 977: | ||
| apparent power: | apparent power: | ||
| \begin{align*} | \begin{align*} | ||
| - | Q &= U \cdot I \\ | + | S &= U \cdot I \\ |
| &= 230{~\rm V} \cdot 5{~\rm A} \\ | &= 230{~\rm V} \cdot 5{~\rm A} \\ | ||
| &= 1150 {~\rm VA} | &= 1150 {~\rm VA} | ||
| Zeile 940: | Zeile 989: | ||
| A magnetic coil shows at a frequency of $f=50.0 {~\rm Hz}$ the voltage of $U=115{~\rm V}$ and the current $I=2.60{~\rm A}$ with a power factor of $\cos \varphi = 0.30$ | A magnetic coil shows at a frequency of $f=50.0 {~\rm Hz}$ the voltage of $U=115{~\rm V}$ and the current $I=2.60{~\rm A}$ with a power factor of $\cos \varphi = 0.30$ | ||
| - | - Calculate the real power, the reactive power, and the apparent power . | + | - Calculate the real power, the reactive power, and the apparent power. |
| - Draw the equivalent parallel circuit. Calculate the active and reactive part of the current. | - Draw the equivalent parallel circuit. Calculate the active and reactive part of the current. | ||
| - Draw the equivalent series circuit. Calculate the ohmic and inductive impedance and the value of the inductivity. | - Draw the equivalent series circuit. Calculate the ohmic and inductive impedance and the value of the inductivity. | ||
| Zeile 977: | Zeile 1026: | ||
| \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 1024: | Zeile 1073: | ||
| <button size=" | <button size=" | ||
| - | The active power is $P = 1.80 kW$. \\ | + | The active power is $P = 1.80 ~\rm kW$. \\ \\ |
| - | The apparent power is $S = U \cdot I = 220V \cdot 20A = 4.40 kVA$. \\ | + | The apparent power is $S = U \cdot I = 220 ~\rm V \cdot 20 ~\rm A = 4.40 ~\rm kVA$. \\ \\ |
| - | The reactive power is $Q = \sqrt{S^2 - P^2} = \sqrt{(4.40 kVA)^2 - (1.80 kW)^2} = 4.01 kVar$ \\ | + | The reactive power is $Q = \sqrt{S^2 - P^2} = \sqrt{(4.40 |
| - | The power factor is $\cos \varphi = {{P}\over{S}} = {{1.80 kW}\over{4.40 kVA}} = 0.41$. | + | The power factor is $\cos \varphi = {{P}\over{S}} = {{1.80 |
| </ | </ | ||
| Zeile 1090: | Zeile 1139: | ||
| 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 1197: | ||
| \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*} | ||
| </ | </ | ||
| Zeile 1207: | Zeile 1256: | ||
| </ | </ | ||
| + | |||
| + | # | ||
| + | |||
| + | A three-phase motor is connected to an artificial three-phase system and can be configured in wye or delta configuration. | ||
| + | * The voltage measured on a single coil shall always be $230 ~\rm V$. | ||
| + | * The current measured on a single coil shall always be $10 ~\rm A$. | ||
| + | * The phase shift for every string is $25°$ | ||
| + | |||
| + | - The motor shall be in wye configuration. \\ Write down the string voltage, phase voltage, string current, phase current, and active power | ||
| + | - The motor shall be in delta configuration. \\ Write down the string voltage, phase voltage, string current, phase current, and active power | ||
| + | - Compare the results | ||
| + | # | ||
| + | |||
| + | # | ||
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| + | A three-phase heater with given resistors is connected to the $230~\rm V$/$400~\rm V$ three-phase system. The heater shows purely ohmic behavior and can be configured in wye or delta configuration. \\ | ||
| + | |||
| + | - The heater is configured in a delta configuration and provides a constant heating power of $6 ~\rm kW$. | ||
| + | - Calculate the resistor value of a single string in the heater | ||
| + | - Calculate the RMS values of the string currents and phase currents. | ||
| + | - The heater with the same resistors as in 1. is now configured in a wye configuration. | ||
| + | - Calculate the RMS values of the string currents and phase currents. | ||
| + | - Compare the heating power in delta configuration (1.) and wye configuration (2.) | ||
| + | # | ||
| + | |||
| + | # | ||
| + | |||
| + | A three-phase motor is connected to a three-phase system with a phase voltage of $400 ~\rm V$. The phase current is $16 ~\rm A$ and the power factor $0.9$. \\ | ||
| + | Calculate the active power, reactive power, and apparent power. | ||
| + | |||
| + | # | ||
| + | |||
| + | |||
| + | # | ||
| + | |||
| + | A symmetrical and balanced three-phase motor of a production line shall be configured in a star configuration and provide a power of $17~\rm kW$ with a power factor of $0.75$. The voltage on a single string is measured to be $135 ~\rm V$. \\ | ||
| + | Calculate the string current. | ||
| + | |||
| + | # | ||
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