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 [2023/03/17 15:23] – 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 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 59: | Zeile 58: | ||
| The simplest component to look at for the instantaneous power is the resistor. For this, we start with the basic definition of the instantaneous voltage $u_R(t)$ (which was given in the last semester) as | The simplest component to look at for the instantaneous power is the resistor. For this, we start with the basic definition of the instantaneous voltage $u_R(t)$ (which was given in the last semester) as | ||
| - | \begin{align*} \color{blue}{u_R(t)} &= \sqrt{2}U sin(\omega t + \varphi_u) \end{align*} | + | \begin{align*} \color{blue}{u_R(t)} &= \sqrt{2}U |
| With the defining formula for the resistor, we get: | With the defining formula for the resistor, we get: | ||
| Zeile 76: | Zeile 75: | ||
| \end{align*} | \end{align*} | ||
| - | For the last step the {{https:// | + | For the last step the {{https:// |
| This result is interesting in the following ways: | This result is interesting in the following ways: | ||
| Zeile 89: | Zeile 88: | ||
| We again start with the basic definition of the instantaneous voltage | We again start with the basic definition of the instantaneous voltage | ||
| - | \begin{align*} \color{blue}{u_L(t)} &= \sqrt{2}U sin(\omega t + \varphi_u) \end{align*} | + | \begin{align*} \color{blue}{u_{\rm L}(t)} &= \sqrt{2}U sin(\omega t + \varphi_u) \end{align*} |
| - | With the defining formula for the inductivity, | + | With the defining formula for inductivity, |
| \begin{align*} | \begin{align*} | ||
| - | | + | |
| - | \rightarrow \color{red}{i_L(t)} &= {{1}\over{L}} \int \color{blue}{u_L(t)} {\rm d}t \\ | + | \rightarrow \color{red}{i_{\rm L}(t)} &= {{1}\over{L}} \int \color{blue}{u_{\rm L}(t)} {\rm d}t \\ |
| &= - \sqrt{2} {{U}\over{\omega L}} \cos(\omega t + \varphi_u) | &= - \sqrt{2} {{U}\over{\omega L}} \cos(\omega t + \varphi_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 174: | Zeile 173: | ||
| \begin{align*} | \begin{align*} | ||
| - | p(t) = UI\big( cos\varphi \left( 1- \cos(2(\omega t + \varphi_u)\right) - \sin\varphi \cdot sin(2(\omega t + \varphi_u) \big) \\ | + | p(t) = UI\big( |
| \end{align*} | \end{align*} | ||
| This result is twofold: | This result is twofold: | ||
| - | - The part $ \cos\varphi \; \cdot \; \left( 1- \cos(2(\omega t + \varphi_u)\right)$ results into a non-zero average - explicitly this part is $1$ in average. On average the first part of the formula results in $UI cos\varphi$. | + | - The part $ \cos\varphi \; \cdot \; \left( 1- \cos(2(\omega t + \varphi_u)\right)$ results into a non-zero average - explicitly this part is $1$ in average. On average the first part of the formula results in $UI \cos\varphi$. |
| - | - The part $- \sin\varphi \; \cdot \; \sin(2(\omega t + \varphi_u))$ is zero on average, so the second part of the formula results in zero. The amplitude of the second part is $UI sin\varphi$ | + | - The part $- \sin\varphi \; \cdot \; \sin(2(\omega t + \varphi_u))$ is zero on average, so the second part of the formula results in zero. The amplitude of the second part is $UI \sin\varphi$ |
| <callout icon=" | <callout icon=" | ||
| Zeile 193: | Zeile 192: | ||
| * The reactive power describes the " | * The reactive power describes the " | ||
| * The reactive power is completely regained by the electric circuit. | * The reactive power is completely regained by the electric circuit. | ||
| - | * In order to distinguish the values, the unit of the reactive power is $\rm VAr$ (or $\rm Var$) for **__V__** | + | * To distinguish the values, the unit of the reactive power is $\rm VAr$ (or $\rm Var$) for **__V__** |
| * An **apparent power** | * An **apparent power** | ||
| * The apparent power is the simple multiplication of the RMS values from the current and the voltage. | * The apparent power is the simple multiplication of the RMS values from the current and the voltage. | ||
| 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 232: | Zeile 231: | ||
| The following simulation shows three ohmic-inductive loads, all with an impedance of $|Z| = 1 ~\rm k\Omega$, however with different phase angles $\varphi$. | The following simulation shows three ohmic-inductive loads, all with an impedance of $|Z| = 1 ~\rm k\Omega$, however with different phase angles $\varphi$. | ||
| The diagram on top of each circuit shows the instantaneous **<fc # | The diagram on top of each circuit shows the instantaneous **<fc # | ||
| - | Similar to the last simulation, a pure ohmic resistance would consume an average power of $P=U^2/R = {{1}\over{2}} \hat{U}^2/ | + | Similar to the last simulation, a pure ohmic resistance would consume an average power of $P=U^2/R = {{1}\over{2}} \hat{U}^2/ |
| The three diagrams shall be discussed shortly. | The three diagrams shall be discussed shortly. | ||
| - Phase angle $\varphi = 10°$: Nearly all of the impedance is given by the resistance and therefore the real part of the impedance. The instantaneous voltage is nearly in phase with the current. The instantaneous power is almost always larger than zero. The average power with $17.47~\rm mW$ is about the same as for an ohmic impedance. | - Phase angle $\varphi = 10°$: Nearly all of the impedance is given by the resistance and therefore the real part of the impedance. The instantaneous voltage is nearly in phase with the current. The instantaneous power is almost always larger than zero. The average power with $17.47~\rm mW$ is about the same as for an ohmic impedance. | ||
| - | - Phase angle $\varphi = 60°$: It is clearly visible, that instantaneous voltage and current are out of phase. The instantaneous power is often lower than zero. The ohmic resistor has $500 ~\rm \Omega = {{1}\over{2}}|Z|$, | + | - Phase angle $\varphi = 60°$: It is clearly visible, that instantaneous voltage and current are out of phase. The instantaneous power is often lower than zero. The ohmic resistor has $500 ~\rm \Omega = {{1}\over{2}}|Z|$, |
| - Phase angle $\varphi = 84.28°$: The phase angle is calculated in such a way, that the resistance is only 10% of the amplitude of the impedance $|Z|$. In this case, the load is nearly pure inductive. The instantaneous power is consequently almost half of the time lower than zero. The average power here is also only $10%$ of the power for a pure ohmic impedance. | - Phase angle $\varphi = 84.28°$: The phase angle is calculated in such a way, that the resistance is only 10% of the amplitude of the impedance $|Z|$. In this case, the load is nearly pure inductive. The instantaneous power is consequently almost half of the time lower than zero. The average power here is also only $10%$ of the power for a pure ohmic impedance. | ||
| Zeile 259: | Zeile 258: | ||
| )) | )) | ||
| - | Also the last simulation shows the relation between the phase angle (here: $\alpha$) and instantaneous values, like power, voltage and current. | + | Also, the last simulation shows the relation between the phase angle (here: $\alpha$) and instantaneous values, like power, voltage, and current. |
| < | < | ||
| Zeile 273: | Zeile 272: | ||
| <callout icon=" | <callout icon=" | ||
| - | \begin{align*} pf &= cos \varphi \\ &= {{P}\over{|\underline{S}|}} \end{align*} | + | \begin{align*} pf & |
| The power factor shows how much real power one gets out of the needed apparent power. </ | The power factor shows how much real power one gets out of the needed apparent power. </ | ||
| Zeile 281: | Zeile 280: | ||
| < | < | ||
| - | The usable output power is $P_L = U_L \cdot I \cdot \cos \varphi$. Based on this, the current $\underline{I}$ is: | + | The usable output power is $P_L = U_{\rm L} \cdot I \cdot \cos \varphi$. Based on this, the current $\underline{I}$ is: |
| - | \begin{align*} I = {{P_L}\over{U_L \cdot \cos \varphi}} \end{align*} | + | \begin{align*} I = {{P_L}\over{U_{\rm L} \cdot \cos \varphi}} \end{align*} |
| The power loss of the wire $P_{wire}$ is therefore: | The power loss of the wire $P_{wire}$ is therefore: | ||
| Zeile 289: | Zeile 288: | ||
| \begin{align*} | \begin{align*} | ||
| P_{\rm wire} &= R_{\rm wire} \cdot I^2 \\ | P_{\rm wire} &= R_{\rm wire} \cdot I^2 \\ | ||
| - | & | + | & |
| \end{align*} | \end{align*} | ||
| - | This means: As smaller, the power factor $cos \varphi$, as more power losses $P_{\rm wire}$ will be generated. More power losses $P_{\rm wire}$ lead to more heat up to or even beyond the maximum temperature. | + | This means: As smaller, the power factor $\cos \varphi$, as more power losses $P_{\rm wire}$ will be generated. More power losses $P_{\rm wire}$ lead to more heat up to or even beyond the maximum temperature. |
| Alternatively, | Alternatively, | ||
| Zeile 308: | Zeile 307: | ||
| ===== 7.2 Polyphase Networks ===== | ===== 7.2 Polyphase Networks ===== | ||
| - | In order to transfer power over long distances alternating current and explicitly rotary current are used. Rotary current is the common name for a three-phase current. The first three-phase high voltage power transfer worldwide started in the August of 1891 for the " | + | To transfer power over long distances alternating current and explicitly rotary current are used. Rotary current is the common name for a three-phase current. The first three-phase high voltage power transfer worldwide started in the August of 1891 for the " |
| < | < | ||
| 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: | ||
| * 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. | ||
| - | In order to understand the three-phase system, we have to investigate the different voltages and currents in this system. | + | To understand the three-phase system, we have to investigate the different voltages and currents in this system. |
| For this the three-phase system will be separated into three parts: | For this the three-phase system will be separated into three parts: | ||
| 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 $\rm U$, $\rm V$, $\rm W$. \\ Often they are also colour-coded as red, yellow and blue - and consecutively sometimes also called $\rm R$, $\rm Y$, $\rm B$. |
| - | * The typical **winding connections** | + | * The winding connections are correspondingly called: $\rm U1$, $\rm U2$, $\rm V1$, $\rm V2$, $\rm W1$, $\rm W2$ (see <imgref imageNo10> |
| - | * 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 403: | ||
| 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/ |
| - | * **Star-voltages** $U_\rm Y$ (alternatively: | + | 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 | ||
| < | < | ||
| Zeile 400: | Zeile 432: | ||
| The phase voltages are $\sqrt{3}$ larger than the star-voltage. In Europe, the low-voltage network of electric power distribution is defined by the RMS value of a star-voltage of $400~\rm V$. \\ | The phase voltages are $\sqrt{3}$ larger than the star-voltage. In Europe, the low-voltage network of electric power distribution is defined by the RMS value of a star-voltage of $400~\rm V$. \\ | ||
| - | The phase voltage is therefore ${{1}\over{\sqrt{3}}} \cdot 400~\rm V \approx 230~\m V$. 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 410: | Zeile 442: | ||
| < | < | ||
| - | In order to understand the load in three-phase systems, the power at different types of loads will be investigated: | + | To understand the load in three-phase systems, the power at different types of loads will be investigated: |
| * Load in Wye connection with the three-phase four-wire system | * Load in Wye connection with the three-phase four-wire system | ||
| Zeile 426: | Zeile 458: | ||
| For the four-wire system, the four pins $\rm L1$, $\rm L2$, $\rm L3$, and the neutral line $\rm N$ are used for power transfer. | For the four-wire system, the four pins $\rm L1$, $\rm L2$, $\rm L3$, and the neutral line $\rm N$ are used for power transfer. | ||
| - | This is for example applied for loads in a star configuration, | + | This is for example applied for loads in a star configuration, |
| <panel type=" | <panel type=" | ||
| Zeile 435: | Zeile 467: | ||
| </ | </ | ||
| - | < | + | < |
| <callout title=" | <callout title=" | ||
| Zeile 443: | Zeile 475: | ||
| - **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 $S_x = U_S \cdot I_x$. However, this misses out 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*} | ||
| </ | </ | ||
| <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*} | + | - 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 464: | 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: $|\underline{I}_x|= |\underline{I}_{\rm S}| = \left|{{\underline{U}_{\rm S}}\over{\underline{Z}_{\rm S}^\phantom{O}}} \right|$. |
| - | - 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 $P = \sqrt{3} \cdot U_{\rm L} I_S \cdot \cos \varphi$ | + | - 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 481: | Zeile 558: | ||
| <panel type=" | <panel type=" | ||
| - | The example in the following simulation shows a $50Hz$ / $231V$ three-phase __three-wire system__ with an unbalanced load in Wye connection, with the given impedances. | + | The example in the following simulation shows a $50 ~\rm Hz$ / $231 ~\rm V$ three-phase __three-wire system__ with an unbalanced load in the Wye connection, with the given impedances. |
| * Calculate the phase currents $I_1$, $I_2$, $I_3$, the neutral current $I_\rm N$ | * Calculate the phase currents $I_1$, $I_2$, $I_3$, the neutral current $I_\rm N$ | ||
| Zeile 489: | Zeile 566: | ||
| </ | </ | ||
| - | < | + | < |
| <callout title=" | <callout title=" | ||
| Zeile 499: | 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**: |
| - | - **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} | ||
| + | \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*} | + | - The phase voltages are given as \begin{align*} |
| - | - 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: < |
| - | - 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*} | + | \begin{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*} | + | \underline{U}_{\rm SN} = {{\sum_x \left( \Large{{{1}\over{\underline{Z}_x^\phantom{O}}}} \cdot \normalsize{\underline{U}_{x \rm N}} \right) } |
| - | - 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 & | + | \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 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 (- {\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 528: | Zeile 698: | ||
| The delta connection uses also a three-wire system like the Wye connection without the neutral line. | The delta connection uses also a three-wire system like the Wye connection without the neutral line. | ||
| Internally, the load is now connected in a triangular shape. Therefore the individual string currents differ from the individual phase currents. | Internally, the load is now connected in a triangular shape. Therefore the individual string currents differ from the individual phase currents. | ||
| - | This setup is for the example given for motors when higher torque is needed. In this connection, each string sees $U_L = \sqrt{3}\cdot U_{\rm S}$ and can therefore produce more current and more power. | + | This setup is for the example given for motors when higher torque is needed. In this connection, each string sees $U_{\rm L} = \sqrt{3}\cdot U_{\rm S}$ and can therefore produce more current and more power. |
| In the case of a delta connection, also only the potentials $\rm L1$, $\rm L2$, and $\rm L3$ are provided and used for power transfer. | In the case of a delta connection, also only the potentials $\rm L1$, $\rm L2$, and $\rm L3$ are provided and used for power transfer. | ||
| Zeile 541: | Zeile 711: | ||
| </ | </ | ||
| - | < | + | < |
| <callout title=" | <callout title=" | ||
| Zeile 552: | Zeile 722: | ||
| Again here the " | Again here the " | ||
| - | - **Voltages**: | + | - **Voltages**: |
| - | - **Currents**: | + | \begin{align*} |
| + | \underline{U}_{12} & | ||
| + | \underline{U}_{23} & | ||
| + | \underline{U}_{31} & | ||
| + | \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} & | ||
| + | \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*} | + | - 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 = 231V \cdot \big( 53.32 A \cdot cos (0° - (9.6°))- 31.01A \cdot cos (-120° - (-137.8°)) + 31.90 A \cdot cos (-240° - (+158.0°)\big) = 24.77 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 |
| - | - The reactive power is: \begin{align*} Q &= |\underline{S} - P| = -4.41kVAr | + | = 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 | ||
| + | &=& 35.24 {~\rm A} + {\rm j} \cdot 18.90 {~\rm A} & | ||
| + | \underline{I}_{23} &=& {{400 \cdot {\rm j}}\over{ 5~\Omega + {{1}\over{{\rm j} \cdot 2\pi\cdot | ||
| + | &=& 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 | ||
| + | | ||
| + | \end{align*} \\</ | ||
| + | - The true power is calculated by: < | ||
| + | \begin{align*} | ||
| + | P = 231 {~\rm V} \cdot \big( 53.32 | ||
| + | | ||
| + | | ||
| + | | ||
| + | - 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 & | ||
| + | &= \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 573: | Zeile 831: | ||
| <callout title=" | <callout title=" | ||
| \\ | \\ | ||
| - | In case of a symmetric load the situation and the formulas get much simplier: | + | In the case of a symmetric load, the situation and the formulas get much simpler: |
| - | - The **phase voltages** $U_L$ and string voltages $U_S$ are equal to the asymmetric load: $U_L = U_S$. | + | - The **phase voltages** $U_{\rm L}$ and string voltages $U_{\rm S}$ are equal to the asymmetric load: $U_{\rm L} = U_{\rm S}$. |
| - | - For equal impedances the absolute value of all **phase currents** $I_L = I_x$ are the same: $|\underline{I}_x|= {3}\cdot|\underline{I}_S| = \sqrt{3}\cdot\left|{{\underline{U}_S}\over{\underline{Z}_S^\phantom{O}}}\right|$. Since the phase currents the same absolute value and have the same $\varphi$, they will add up to zero. Therefore there is no current on neutral line: $I_N =0$ | + | - For equal impedances the absolute value of all **phase currents** $I_{\rm L} = I_x$ are the same: $|\underline{I}_x|= {3}\cdot|\underline{I}_{\rm S}| = \sqrt{3}\cdot\left|{{\underline{U}_{\rm S}}\over{\underline{Z}_{\rm S}^\phantom{O}}}\right|$. Since the phase currents |
| - | - | ||
| - | - The **true power** is three times the true power of a single phase: $P = 3 \cdot U_S I_S \cdot cos \varphi$. Based on the line voltages $U_L$, the formula is $P = \sqrt{3} \cdot U_L I_S \cdot cos \varphi$ | + | - 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 | + | - The **(collective) apparent power** - given the formula above - is: $S_\Sigma = \sqrt{3}\cdot |
| - | - The **reactive power** leads to: $Q_\Sigma = \sqrt{S_\Sigma^2 - P^2} = 3 \cdot U_S I_S \cdot sin (\varphi)$. | + | - 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 592: | Zeile 850: | ||
| <panel type=" | <panel type=" | ||
| - | A passive component is fed by a sinosidal | + | A passive component is fed by a sinusoidal |
| - | + | ||
| - | - Draw the equivalent circuits based on a series and on 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=" | + | 1. Draw the equivalent circuits based on a series and a parallel circuit. \\ |
| + | # | ||
| {{drawio> | {{drawio> | ||
| + | # | ||
| - | </ | + | 2. Calculate the equivalent components for both circuits. \\ |
| - | <button size=" | + | # |
| The apparent impedance is: | The apparent impedance is: | ||
| \begin{align*} | \begin{align*} | ||
| - | Z = |\underline{Z}| & | + | Z = |\underline{Z}| & |
| \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}| |
| Therefore: | Therefore: | ||
| \begin{align*} | \begin{align*} | ||
| - | R_s & | + | R_s & |
| - | X_{Ls} & | + | X_{Ls} & |
| - | \rightarrow L_s &=& {{X_{Ls}}\over{2\pi f}} &=& {{{{U}\over{I}} \cdot sin \varphi}\over{2\pi f}} &=& \boldsymbol{127mH} | + | \rightarrow L_s &=& {{X_{Ls}}\over{2\pi f}} &=& {{{{U}\over{I}} \cdot \sin \varphi}\over{2\pi f}} &=& \boldsymbol{127~\rm mH} |
| \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 | + | &= {{1}\over{Z}}\cdot \left( \cos(\varphi) - {\rm j}\cdot \sin(\varphi) \right) &&= {{1}\over{R_p}} - {{\rm j}\over{X_{Lp}}} \\ |
| - | & | + | |
| \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: | ||
| \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 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 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} \end{align*} | |
| - | </ | + | # |
| + | 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: | ||
| \begin{align*} | \begin{align*} | ||
| - | P &= U \cdot I \cdot cos \varphi \\ | + | P &= U \cdot I \cdot \cos \varphi \\ |
| - | & | + | & |
| - | &= 575 W | + | &= 575 {~\rm W} |
| \end{align*} | \end{align*} | ||
| reactive power: | reactive power: | ||
| \begin{align*} | \begin{align*} | ||
| - | Q &= U \cdot I \cdot sin \varphi \\ | + | Q &= U \cdot I \cdot \sin \varphi \\ |
| - | & | + | & |
| - | &= 996 Var | + | &= 996 {~\rm Var} |
| \end{align*} | \end{align*} | ||
| apparent power: | apparent power: | ||
| \begin{align*} | \begin{align*} | ||
| - | Q &= U \cdot I \\ | + | S &= U \cdot I \\ |
| - | & | + | & |
| - | &= 1150 VA | + | &= 1150 {~\rm VA} |
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| Zeile 685: | Zeile 987: | ||
| <panel type=" | <panel type=" | ||
| - | A magnetic coil shows at a frequency of $f=50.0 Hz$ the voltage of $U=115V$ and the current $I=2.60A$ with a power factor of $cos \varphi = 0.30$ | + | A magnetic coil shows at a frequency of $f=50.0 |
| - | - 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 696: | Zeile 998: | ||
| The real power is | The real power is | ||
| \begin{align*} | \begin{align*} | ||
| - | P &= U \cdot I \cdot cos \varphi \\ | + | P &= U \cdot I \cdot \cos \varphi \\ |
| - | & | + | & |
| - | &= 89.7 W | + | &= 89.7 {~\rm W} |
| \end{align*} | \end{align*} | ||
| The reactive power is | The reactive power is | ||
| \begin{align*} | \begin{align*} | ||
| - | Q &= U \cdot I \cdot sin \varphi \\ | + | Q &= U \cdot I \cdot \sin \varphi \\ |
| - | & | + | & |
| - | &= 285 Var | + | &= 285 {~\rm Var} |
| \end{align*} | \end{align*} | ||
| Zeile 711: | Zeile 1013: | ||
| \begin{align*} | \begin{align*} | ||
| S &= U \cdot I \\ | S &= U \cdot I \\ | ||
| - | & | + | & |
| - | &= 299 VA | + | &= 299 {~\rm VA} |
| \end{align*} | \end{align*} | ||
| Zeile 724: | 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*} | ||
| The active and reactive part of the current is therefore: | The active and reactive part of the current is therefore: | ||
| \begin{align*} | \begin{align*} | ||
| - | I_R & | + | I_R & |
| - | I_L &= - 2.60A \cdot \sqrt{1 - 0.30^2} &= 2.48 A | + | I_L &= - 2.60{~\rm A} \cdot \sqrt{1 - 0.30^2} &= 2.48 {~\rm A} |
| \end{align*} | \end{align*} | ||
| Zeile 738: | Zeile 1040: | ||
| <button size=" | <button size=" | ||
| - | Important: The cosine function is ambiguous! Based on $cos \varphi = 0.30$ it is unclear, whether $\varphi$ is positive or negative. \\ | + | Important: The cosine function is ambiguous! Based on $\cos \varphi = 0.30$ it is unclear, whether $\varphi$ is positive or negative. \\ |
| - | Therefore, only based on the power factor it is unclear whether the circuit is ohmic-inductive or ohmic-capacitive! However, this is explicitely | + | Therefore, only based on the power factor it is unclear whether the circuit is ohmic-inductive or ohmic-capacitive! However, this is explicitly |
| {{drawio> | {{drawio> | ||
| \begin{align*} | \begin{align*} | ||
| - | Z_s &= {{U}\over{I}} | + | Z_s &= {{U}\over{I}} |
| - | R_s &= {{U}\over{I}} \cdot cos \varphi &=& {{115V}\over{2.60A}} \cdot 0.30 &=& 13.3 \Omega \\ | + | R_s &= {{U}\over{I}} \cdot \cos \varphi &=& {{115{~\rm V}}\over{2.60{~\rm A}}} \cdot 0.30 &=& 13.3 ~\Omega \\ |
| - | X_{Ls} &= {{U}\over{I}} \cdot sin \varphi &=& {{115V}\over{2.60A}} \cdot \sqrt{1 - 0.30^2} &=& 42.2 \Omega \\ \\ | + | X_{Ls} &= {{U}\over{I}} \cdot \sin \varphi &=& {{115{~\rm V}}\over{2.60{~\rm A}}} \cdot \sqrt{1 - 0.30^2} &=& 42.2 ~\Omega \\ \\ |
| - | L_s & | + | L_s & |
| \end{align*} | \end{align*} | ||
| Zeile 758: | Zeile 1060: | ||
| <panel type=" | <panel type=" | ||
| - | A consumer is connected to a $220V$ / $50Hz$ network. A current of $20.0A$ and a power of $1800W$ is measured. | + | A consumer is connected to a $220~\rm V$ / $50 ~\rm Hz$ network. A current of $20.0~\rm A$ and a power of $1800 ~\rm W$ is measured. |
| - | - What is the value of the active power, the reactive power and the power factor? | + | - What is the value of the active power, the reactive power, and the power factor? |
| - Assume that the consumer is a parallel circuit. | - Assume that the consumer is a parallel circuit. | ||
| - Calculate the resistance and reactance. | - Calculate the resistance and reactance. | ||
| - | - Calculate the necessary inductance / capacitance. | + | - Calculate the necessary inductance/ |
| - Assume that the consumer is a series circuit. | - Assume that the consumer is a series circuit. | ||
| - Calculate the resistance and reactance. | - Calculate the resistance and reactance. | ||
| - | - Calculate the necessary inductance / capacitance. | + | - Calculate the necessary inductance/ |
| <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 780: | Zeile 1082: | ||
| <button size=" | <button size=" | ||
| - | Important: The cosine function is ambiguous! Based on $cos \varphi = 0.30$ it is unclear, whether $\varphi$ is positive or negative. \\ | + | Important: The cosine function is ambiguous! Based on $\cos \varphi = 0.30$ it is unclear, whether $\varphi$ is positive or negative. \\ |
| Therefore, only based on the power factor it is unclear whether the circuit is ohmic-inductive or ohmic-capacitive! | Therefore, only based on the power factor it is unclear whether the circuit is ohmic-inductive or ohmic-capacitive! | ||
| - | The consumer is a parallel circuit of the resistance $R_p$ and the reactance $X_p$ on the voltage $U$. Both values can be calculated based on the real and reactive power: | + | The consumer is a parallel circuit of the resistance $R_\rm p$ and the reactance $X_\rm p$ on the voltage $U$. Both values can be calculated based on the real and reactive power: |
| \begin{align*} | \begin{align*} | ||
| - | P &= {{U^2}\over{R_p}} \rightarrow & R_p &= {{U^2}\over{P}} &= 26.9 \Omega \\ | + | P &= {{U^2}\over{R_\rm p}} \rightarrow & R_p &= {{U^2}\over{P}} &= 26.9 ~\Omega \\ |
| - | Q &= {{U^2}\over{X_p}} \rightarrow & X_p &= {{U^2}\over{Q}} &= 12.1 \Omega \\ | + | Q &= {{U^2}\over{X_\rm p}} \rightarrow & X_p &= {{U^2}\over{Q}} &= 12.1 ~\Omega \\ |
| \end{align*} | \end{align*} | ||
| - | The respective values for a inductance / capacitance are: | + | The respective values for inductance/ |
| \begin{align*} | \begin{align*} | ||
| - | L &= {{X_p}\over{2\pi \cdot f}} &= 38.4 mH \\ | + | L &= {{X_p}\over{2\pi \cdot f}} |
| - | C &= {{1}\over{2\pi \cdot f \cdot X_p}} &= 263 \mu F \\ | + | C &= {{1}\over{2\pi \cdot f \cdot X_p}} &= 263 ~\rm µF \\ |
| \end{align*} | \end{align*} | ||
| Zeile 799: | Zeile 1101: | ||
| <button size=" | <button size=" | ||
| - | The consumer is a series circuit of the resistance $R_s$ and the reactance $X_s$ with the current $I$. Both values can be calculated based on the real and reactive power: | + | The consumer is a series circuit of the resistance $R_\rm s$ and the reactance $X_\rm s$ with the current $I$. Both values can be calculated based on the real and reactive power: |
| \begin{align*} | \begin{align*} | ||
| - | P &= I^2 \cdot R_s | + | P &= I^2 \cdot R_{\rm s} |
| - | Q &= I^2 \cdot X_s | + | Q &= I^2 \cdot X_{\rm s} |
| \end{align*} | \end{align*} | ||
| - | The respective values for a inductance / capacitance are: | + | The respective values for inductance/ |
| \begin{align*} | \begin{align*} | ||
| - | L &= {{X_s}\over{2\pi \cdot f}} &= 31.9 mH \\ | + | L &= {{X_{\rm s}}\over{2\pi \cdot f}} |
| - | C &= {{1}\over{2\pi \cdot f \cdot X_s}} &= 318 \mu F \\ | + | C &= {{1}\over{2\pi \cdot f \cdot X_{\rm s}}} &= 318 ~\rm µF \\ |
| \end{align*} | \end{align*} | ||
| Zeile 817: | Zeile 1119: | ||
| <panel type=" | <panel type=" | ||
| - | An uncompensated ohmic-inductive series circuit shows at $U=230V$, $f=50Hz$ the current $I_{RL}=7A$, $P_{RL}=1.3kW$ | + | An uncompensated ohmic-inductive series circuit shows at $U=230~\rm V$, $f=50 ~\rm Hz$ the current $I_{RL}=7 ~\rm A$, $P_{RL}=1.3 ~\rm kW$ |
| - | The power factor shall be compensated to $cos\varphi = 1$ via a parallel compensation. | + | The power factor shall be compensated to $\cos\varphi = 1$ via a parallel compensation. |
| - | - Calculate the apparent power, the reactive power, the phase angle and the power factor before the compensation. | + | - Calculate the apparent power, the reactive power, the phase angle, and the power factor before the compensation. |
| - | - Calculate the capacity $C$ which have to be connected in parallel | + | - Calculate the capacity $C$ which has to be connected in parallel to get $\cos\varphi=1$. |
| - | <button size=" | + | <button size=" |
| + | \begin{align*} | ||
| + | S | ||
| + | Q | ||
| + | \varphi & | ||
| + | = \arccos\left({{P}\over{S}}\right) | ||
| + | \end{align*} | ||
| - | The inductor $L$ creates the reactive power $Q = Q_L$. In order to compensate a equivalent reactive power $|Q_C| = |Q_L|$ has to be given by a capacitor. The reactive power is given by \begin{align*} Q &= \Re (U) \cdot \Im (I) \\ &= U \cdot {{U}\over{X}} \\ &= {{U^2}\over{X}} \\ \end{align*} | + | The inductor $L$ creates the reactive power $Q = Q_L$. To compensate |
| + | The reactive power is given by: | ||
| + | \begin{align*} | ||
| + | Q &= \Re (U) \cdot \Im (I) \\ | ||
| + | | ||
| + | | ||
| + | \end{align*} | ||
| - | The capacity can therefore be calculated by \begin{align*} X_C &= {{U^2}\over{Q_L}} = {{1}\over{\omega C}} \quad \rightarrow \quad C = {{1}\over{\omega U^2}} \end{align*} | + | The capacity can therefore be calculated by |
| + | \begin{align*} | ||
| + | X_C &= {{U^2}\over{Q_L}} = {{1}\over{\omega C}} \quad \rightarrow \quad C = {{1}\over{\omega U^2}} | ||
| + | \end{align*} | ||
| </ | </ | ||
| - | <button size=" | + | <button size=" |
| + | \begin{align*} | ||
| + | S &= 1.62 {~\rm kVA} \\ | ||
| + | Q &= 0.95 {~\rm kVAr} \\ | ||
| + | \varphi &= +36° \\ \\ | ||
| + | C &= 57.2 ~\rm µF | ||
| + | \end{align*} | ||
| </ | </ | ||
| Zeile 840: | Zeile 1163: | ||
| <panel type=" | <panel type=" | ||
| - | A three-phase power net with a phase voltage of $400V$ has two symmatrical | + | A three-phase power net with a phase voltage of $400 ~\rm V$ has two symmetrical |
| < | < | ||
| - | * Load 1: $P_1 = 2.7 kW$, $cos \varphi_1 = 0.89$ | + | * Load 1: $P_1 = 2.7 ~\rm kW$, $\cos \varphi_1 = 0.89$ |
| - | * Load 1: $P_2 = 3.8 kW$, $cos \varphi_1 = 0.76$ | + | * Load 1: $P_2 = 3.8 ~\rm kW$, $\cos \varphi_1 = 0.76$ |
| 1. Calculate the reactive power $Q_1$ and $Q_2$. \\ | 1. Calculate the reactive power $Q_1$ and $Q_2$. \\ | ||
| <button size=" | <button size=" | ||
| - | First calculate the apparent power: | + | First, calculate the apparent power: |
| \begin{align*} | \begin{align*} | ||
| - | P_1 &=& S_1 \cdot cos \varphi_1 \quad \rightarrow \quad S_1 &=& {{1}\over{cos \varphi_1}} \cdot P_1 &=& {{1}\over{0.89}} \cdot 2.7 kW &=& 3.0 kVA\\ | + | P_1 &=& S_1 \cdot \cos \varphi_1 \quad \rightarrow \quad S_1 &=& {{1}\over{\cos \varphi_1}} \cdot P_1 &=& {{1}\over{0.89}} \cdot 2.7 {~\rm kW} &=& 3.0 {~\rm kVA}\\ |
| - | P_2 &=& S_2 \cdot cos \varphi_2 \quad \rightarrow \quad S_2 &=& {{1}\over{cos \varphi_2}} \cdot P_1 &=& {{1}\over{0.76}} \cdot 3.8 kW &=& 5.0 kVA\\ | + | P_2 &=& S_2 \cdot \cos \varphi_2 \quad \rightarrow \quad S_2 &=& {{1}\over{\cos \varphi_2}} \cdot P_1 &=& {{1}\over{0.76}} \cdot 3.8 {~\rm kW} &=& 5.0 {~\rm kVA}\\ |
| \end{align*} \\ | \end{align*} \\ | ||
| - | For calculating the reactive power, $sin \varphi_{1, | + | For calculating the reactive power, $\sin \varphi_{1, |
| - | * One way would be to calculate $\varphi_{1, | + | * One way would be to calculate $\varphi_{1, |
| - | * Another way is to use the formula $1^2 = sin^2\varphi_{1, | + | * Another way is to use the formula $1^2 = \sin^2\varphi_{1, |
| \begin{align*} | \begin{align*} | ||
| - | Q_1 &=& S_1 \cdot \sqrt{1 - cos^2 \varphi_1} &=& 3.0 kVA \cdot \sqrt{1 - 0.89^2} &=& 1.4 kVAr\\ | + | Q_1 &=& S_1 \cdot \sqrt{1 - \cos^2 \varphi_1} &=& 3.0 {~\rm kVA} \cdot \sqrt{1 - 0.89^2} &=& 1.4 {~\rm kVAr}\\ |
| - | Q_2 &=& S_2 \cdot \sqrt{1 - cos^2 \varphi_2} &=& 5.0 kVA \cdot \sqrt{1 - 0.76^2} &=& 3.2 kVAr\\ | + | Q_2 &=& S_2 \cdot \sqrt{1 - \cos^2 \varphi_2} &=& 5.0 {~\rm kVA} \cdot \sqrt{1 - 0.76^2} &=& 3.2 {~\rm kVAr}\\ |
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| \begin{align*} | \begin{align*} | ||
| - | Q_1 &=& 1.4 kVAr\\ | + | Q_1 &=& 1.4 {~\rm kVAr}\\ |
| - | Q_2 &=& 3.2 kVAr\\ | + | Q_2 &=& 3.2 {~\rm kVAr}\\ |
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| Zeile 873: | Zeile 1196: | ||
| \begin{align*} | \begin{align*} | ||
| - | \underline{S}_{net} &=& \underline{S}_1 | + | \underline{S}_{\rm net} &=& \underline{S}_1 |
| - | &=& P_1 + j \cdot Q_1 &+& P_2 + j \cdot Q_2 \\ | + | &=& P_1 + {\rm j} \cdot Q_1 |
| - | &=& P_1 + P_2 &+& j \cdot (Q_1 + Q_2) \\ | + | &=& P_1 + P_2 |
| - | &=& 2.7 kW + 3.8 kW &+& j \cdot (1.4 kVAr + 3.2 kVAr) \\ | + | &=& 2.7 {~\rm kW} + 3.8 {~\rm kW} & |
| - | &=& 6.5 kW &+& j \cdot 4.6 kVAr \\ | + | &=& 6.5 {~\rm kW} & |
| - | &=& P_{net} &+& j \cdot Q_{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}_{net} &=& \sqrt{P_{net}^2+ | + | \underline{S}_{\rm net} &=& \sqrt{P_{\rm net}^2 |
| - | \sqrt{(6.5 kW)^2+ | + | \sqrt{(6.5 |
| - | 8.0 kVA & | + | |
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| \begin{align*} | \begin{align*} | ||
| - | \underline{S}_{net} &=& 6.5 kW+ j \cdot 4.6 kVAr \\ | + | \underline{S}_{\rm net} &=& 6.5 {~\rm kW} + {\rm j} \cdot 4.6 {~\rm kVAr} \\ |
| - | &=& 8.0 kVA \cdot e^{j \cdot 35°} \\ | + | &=& 8.0 {~\rm kVA} \cdot {\rm e}^{{\rm j} \cdot 35°} \\ |
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| - | 3. Calculate the RMS value of the a single-phase current $I$ which the power net has to provide. \\ | + | 3. Calculate the RMS value of the single-phase current $I$ which the power net has to provide. \\ |
| <button size=" | <button size=" | ||
| - | The apparent power $S_{net}$ is given by $S_{net} = 3 \cdot U \cdot I$, with $U$ as the star-voltage $U_Y = {{1}\over{ \sqrt{3} }} \cdot U_L$. \\ | + | The apparent power $S_{\rm net}$ is given by $S_{\rm net} = 3 \cdot U \cdot I$, with $U$ as the star-voltage $U_Y = {{1}\over{ \sqrt{3} }} \cdot U_{\rm L}$. \\ |
| - | Therefore, the single-phase current $I$ can be calculated from $S_{net}$: | + | Therefore, the single-phase current $I$ can be calculated from $S_{\rm net}$: |
| \begin{align*} | \begin{align*} | ||
| - | S_{net} &=& 3 \cdot U \cdot I \\ | + | S_{\rm net} &=& 3 \cdot U \cdot I \\ |
| - | &=& \sqrt{3} \cdot U_L \cdot I \\ | + | &=& \sqrt{3} \cdot U_{\rm L} \cdot I \\ |
| \end{align*} \\ | \end{align*} \\ | ||
| The current is: | The current is: | ||
| \begin{align*} | \begin{align*} | ||
| - | I &=& {{ S_{net} }\over{ \sqrt{3} \cdot U_L }} \\ | + | I &=& {{ S_{\rm net} }\over{ \sqrt{3} \cdot U_{\rm L} }} \\ |
| - | &=& {{8.0 kVA }\over{ \sqrt{3} \cdot 400V }} \\ | + | &=& {{8.0 {~\rm kVA}}\over{ \sqrt{3} \cdot 400{~\rm V} }} \\ |
| - | & | + | & |
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| \begin{align*} | \begin{align*} | ||
| - | I & | + | I & |
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| - | 4. What is the overall power factor $cos\varphi_{net}$ of the power net? \\ | + | 4. What is the overall power factor $\cos\varphi_{\rm net}$ of the power net? \\ |
| <button size=" | <button size=" | ||
| - | The overall power factor can be calculated from the apparent power and its angle $\varphi_{net}$ (see the Euler representation in task 2). \\ | + | The overall power factor can be calculated from the apparent power and its angle $\varphi_{\rm net}$ (see the Euler representation in task 2). \\ |
| \begin{align*} | \begin{align*} | ||
| - | cos\varphi_{net} = cos (35°) = 0.82 | + | \cos\varphi_{\rm net} = \cos (35°) = 0.82 |
| \end{align*} | \end{align*} | ||
| Zeile 927: | Zeile 1250: | ||
| \begin{align*} | \begin{align*} | ||
| - | cos\varphi_{net} = 0.82 | + | \cos\varphi_{\rm net} = 0.82 |
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| Zeile 933: | 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 | ||
| + | # | ||
| + | |||
| + | # | ||
| + | |||
| + | 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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