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electrical_engineering_2:polyphase_networks [2022/06/19 17:07]
tfischer |
electrical_engineering_2:polyphase_networks [2023/09/19 23:52] (aktuell)
mexleadmin |
| ====== 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 |
| |
| * 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 a 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. |
| |
| * the sinusoidal alternating voltage is produced by the rotation of a coil in a homogeneous magnetic field, and | * the sinusoidal alternating voltage is produced by the rotation of a coil in a homogeneous magnetic field, and |
| <WRAP> <imgcaption imageNo01 | voltage generation in generator></imgcaption> {{drawio>voltagegenerationgenerator.svg}} </WRAP> | <WRAP> <imgcaption imageNo01 | voltage generation in generator></imgcaption> {{drawio>voltagegenerationgenerator.svg}} </WRAP> |
| |
| For the rotation angle $\varphi$ holds: \begin{align*} \varphi(t)=\omega t + \varphi_0 \quad \text{with} \quad \varphi_0=\varphi(t=0) \\ \end{align*} | For the rotation angle $\varphi$ holds: |
| | \begin{align*} \varphi(t)=\omega t + \varphi_0 \quad \text{with} \quad \varphi_0=\varphi(t=0) \\ \end{align*} |
| |
| Thus, the induced voltage $u(t)$ is given by: \begin{align*} u(t) &= \frac{d\Psi}{dt} \\ &= N\cdot\frac{d\Phi}{dt} \\ &= NBA\cdot\frac{d cos \varphi(t)}{dt} \\ &= \hat{\Psi}\cdot\frac{d cos (\omega t + \varphi_0)}{dt} \\ &= -\omega \hat{\Psi}\cdot sin (\omega t + \varphi_0) \\ &= -\hat{U}\cdot sin (\omega t + \varphi_0) \\ \end{align*} | Thus, the induced voltage $u(t)$ is given by: |
| | \begin{align*} |
| | u(t) &= -\frac{{\rm d} \Psi} {{\rm d}t} \\ |
| | &= -N \cdot \frac{{\rm d} \Phi} {{\rm d}t} \\ |
| | &= -NBA\cdot \frac{{\rm d} \cos \varphi(t)} {{\rm d}t} \\ |
| | &= -\hat{\Psi}\cdot\frac{{\rm {\rm d}} \cos (\omega t + \varphi_0)}{{\rm d}t} \\ |
| | &= \omega \hat{\Psi} \cdot \sin (\omega t + \varphi_0) \\ |
| | &= \hat{U} \cdot \sin (\omega t + \varphi_0) \\ |
| | \end{align*} |
| |
| Such single-phase systems are therefore alternating current systems, which use one outgoing line and one return line each for the current conduction. \\ Out of the last formula we derived the following instantaneous voltage $u(t)$ \begin{align*} u(t) &= -\hat{U}\cdot sin (\omega t + \varphi_0) \\ &= \hat{U}\cdot sin (\omega t + \varphi'_0) \\ &= \sqrt{2} U\cdot sin (\omega t + \varphi'_0) \\ \end{align*} | Such single-phase systems are therefore alternating current systems, which use one outgoing line and one return line each for the current conduction. \\ |
| | Out of the last formula we derived the following instantaneous voltage $u(t)$ |
| | \begin{align*} |
| | u(t) &= \hat{U} \cdot \sin (\omega t + \varphi_0) \\ |
| | &= \sqrt{2} U\cdot \sin (\omega t + \varphi_0) \\ |
| | \end{align*} |
| |
| ===== 7.1 Power in AC ===== | ===== 7.1 Power in AC ===== |
| By the end of this section, you will be able to: | By the end of this section, you will be able to: |
| |
| - Know the formula of the instantaneous power of the resistor, inductor and capacitor and be able to determine its values. | - Know the formula of the instantaneous power of the resistor, inductor, and capacitor and be able to determine its values. |
| |
| </callout> | </callout> |
| |
| In the first chapter we mainly focussed on the power given by $P = U \cdot I$. This is however only valid for DC circuits. For AC circuits we have to consider the instantaneous power $p(t)$. The instantaneuos power is given by: | In the first chapter, we mainly focussed on the power given by $P = U \cdot I$. This is however only valid for DC circuits. For AC circuits we have to consider the instantaneous power $p(t)$. The instantaneous power is given by: |
| |
| \begin{align*} p(t) &= \color{blue}{u(t)} \cdot \color{red}{i(t)} \end{align*} | \begin{align*} p(t) &= \color{blue}{u(t)} \cdot \color{red}{i(t)} \end{align*} |
| ==== Ideal Ohmic resistance R ==== | ==== Ideal Ohmic resistance R ==== |
| |
| The simplest component to look at for the instantaneuos 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 \sin(\omega t + \varphi_u) \end{align*} |
| |
| With the defining formula for the resistor we get: | With the defining formula for the resistor, we get: |
| |
| \begin{align*} \color{blue}{u_R(t)} &= R \cdot \color{red}{i(t)} \\ \rightarrow \color{red}{i(t)} &= {{\color{blue}{u_R(t)}}\over{R}} \\ &= \sqrt{2} {{U}\over{R}} sin(\omega t + \varphi_u) \end{align*} | \begin{align*} |
| | \color{blue}{u_R(t)} &= R \cdot \color{red}{i(t)} \\ |
| | \rightarrow \color{red}{i(t)} &= {{\color{blue}{u_R(t)}}\over{R}} \\ |
| | &= \sqrt{2}{ {U} \over{R}} \sin(\omega t + \varphi_u) |
| | \end{align*} |
| |
| This leads to an instantaneous power $p_R(t)$ of | This leads to an instantaneous power $p_R(t)$ of |
| | \begin{align*} |
| | p_R(t) &= \color{blue}{u_R(t)} \cdot \color{red}{i_R(t)} \\ |
| | &= 2\cdot {{U^2}\over{R}} \sin^2(\omega t + \varphi_u) \\ |
| | &= {{U^2}\over{R}}\left(1- \cos\left(2\cdot (\omega t + \varphi_u)\right)\right) \\ |
| | \end{align*} |
| |
| \begin{align*} p_R(t) &= \color{blue}{u_R(t)} \cdot \color{red}{i_R(t)} \\ &= 2\cdot {{U^2}\over{R}} sin^2(\omega t + \varphi_u) \\ &= {{U^2}\over{R}}\left(1- cos\left(2\cdot (\omega t + \varphi_u)\right)\right) \\ \end{align*} | For the last step the {{https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle_formulae|Double-angle formula}} "$\cos(2x) = 1 - 2 sin^2(x)$" was used. |
| | |
| For the last step the {{https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle_formulae|Double-angle formula}} "$cos(2x) = 1 - 2 sin^2(x)$" was used. | |
| |
| This result is interesting in the following ways: | This result is interesting in the following ways: |
| |
| - The part $1- cos (2\cdot (\omega t + \varphi_u) )$ is always non-negative and a shifted sinusidal function between $0...2$. The average value of this part is $1$. | - The part $1- \cos (2\cdot (\omega t + \varphi_u) )$ is always non-negative and a shifted sinusoidal function between $0...2$. The average value of this part is $1$. |
| - The average value of $p_R(t)$ is then: $P_R = {{U^2}\over{R}}$ | - The average value of $p_R(t)$ is then: $P_R = {{U^2}\over{R}}$ |
| - The use of the $\sqrt{2}$ in the definition $\color{blue}{u_R(t)} = \sqrt{2} U sin(\omega t + \varphi_u)$ leads to the average power as $P_R = {{U^2}\over{R}}$. This formula for the power is exactly like the formula for the power in pure DC situations. | - The use of the $\sqrt{2}$ in the definition $\color{blue}{u_R(t)} = \sqrt{2} U \sin(\omega t + \varphi_u)$ leads to the average power as $P_R = {{U^2}\over{R}}$. This formula for the power is exactly like the formula for the power in pure DC situations. |
| |
| ==== Ideal Inductivity L ==== | ==== Ideal Inductivity L ==== |
| |
| The similar approach is done for the inductivity. We again start with the basic definition of the instantaneous voltage | A similar approach is done for the ideal inductivity. |
| | 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 we get: \begin{align*} \color{blue}{u_L(t)} &= L\cdot {{d\color{red}{i_L(t)}}\over{dt}} \\ \rightarrow \color{red}{i_L(t)} &= {{1}\over{L}} \int \color{blue}{u_L(t)} dt \\ &= - \sqrt{2} {{U}\over{\omega L}} cos(\omega t + \varphi_u) \end{align*} | With the defining formula for inductivity, we get: |
| | \begin{align*} |
| | \color{blue}{u_{\rm L}(t)} &= L\cdot {{{\rm d}\color{red} {i_{\rm L}(t)}}\over{{\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) |
| | \end{align*} |
| |
| This leads to an instantaneous power $p_L(t)$ of | This leads to an instantaneous power $p_L(t)$ of |
| |
| \begin{align*} p_L(t) &= \color{blue}{u(t)} \cdot \color{red}{i(t)} \\ &= - 2\cdot {{U^2}\over{\omega L}} \cdot sin(\omega t + \varphi_u) cos(\omega t + \varphi_u) \\ &= - {{U^2}\over{\omega L}} \cdot sin( 2\cdot (\omega t + \varphi_u)) \\ \end{align*} | \begin{align*} |
| | p_L(t) &= \color{blue}{u(t)} \cdot \color{red}{i(t)} \\ |
| | &= - 2\cdot {{U^2}\over{\omega L}} \cdot \sin( \omega t + \varphi_u) \cos(\omega t + \varphi_u) \\ |
| | &= - {{U^2}\over{\omega L}} \cdot \sin( 2\cdot (\omega t + \varphi_u)) \\ |
| | \end{align*} |
| |
| Again a trigonometric identity ({{https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle_formulae|Double-angle formula}} "$sin(2x) = 2 sin(x)cos(x)$") was used. | Again a trigonometric identity ({{https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle_formulae|Double-angle formula}} "$\sin(2x) = 2 \sin(x)\cos(x)$") was used. |
| |
| Also this result is interesting: | Also, this result is interesting: |
| |
| - The part $sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$. | - The part $\sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$. |
| - Therefore, the average value of $p_L(t)=0$ | - Therefore, the average value of $p_L(t)=0$ |
| |
| Also here, we start with the basic definition of the instantaneous voltage | Also here, we start with the basic definition of the instantaneous voltage |
| |
| \begin{align*} \color{blue}{u_C(t)} &= \sqrt{2}U sin(\omega t + \varphi_u) \end{align*} | \begin{align*} \color{blue}{u_C(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \end{align*} |
| |
| With the defining formula for the capacity we get: \begin{align*} \color{red}{i_C(t)} &= C {{d\color{blue}{u_C(t)}}\over{dt}} \\ &= \sqrt{2} U \omega C cos(\omega t + \varphi_u) \end{align*} | With the defining formula for the capacity, we get: |
| | \begin{align*} |
| | \color{red}{i_C(t)} &= C {{{\rm d}\color{blue}{u_C(t)}}\over{{\rm d}t}} \\ |
| | &= \sqrt{2} U \omega C \cos(\omega t + \varphi_u) |
| | \end{align*} |
| |
| This leads to an instantaneous power $p_C(t)$ of | This leads to an instantaneous power $p_C(t)$ of |
| |
| \begin{align*} p_C(t) &= \color{blue}{u_C(t)} \cdot \color{red}{i_C(t)} \\ &= 2\cdot U^2 \omega C \cdot sin(\omega t + \varphi_u) cos(\omega t + \varphi_u) \\ &= + U^2 \omega C \cdot sin( 2\cdot (\omega t + \varphi_u)) \\ \end{align*} | \begin{align*} |
| | p_C(t) &= \color{blue}{u_C(t)} \cdot \color{red}{i_C(t)} \\ |
| | &= 2\cdot U^2 \omega C \cdot \sin( \omega t + \varphi_u) \cos(\omega t + \varphi_u) \\ |
| | &= + U^2 \omega C \cdot \sin( 2\cdot (\omega t + \varphi_u)) \\ |
| | \end{align*} |
| |
| Again this result leads to: | Again this result leads to: |
| |
| - The part $sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$. | - The part $\sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$. |
| - Therefore, also the average value of $p_C(t)=0$ | - Therefore, also the average value of $p_C(t)=0$ |
| |
| * Instantaneuos values of power at $R$, $L$, $C$ | * Instantaneous values of power at $R$, $L$, $C$ |
| * Active, reactive, apparent, and complex power | * Active, reactive, apparent, and complex power |
| |
| This effect can also be seen in the following simulation: The simulation shows three loads, all with an impedance of $|Z| = 1 k\Omega$. The diagram on top of each circuit shows the instantaneous **<fc #00bb00>voltage</fc>**, **<fc #bbbb00>current</fc>** and **<fc #bbbbbb>power</fc>**. | This effect can also be seen in the following simulation: The simulation shows three loads, all with an impedance of $|Z| = 1 ~\rm k\Omega$. |
| | The diagram on top of each circuit shows the instantaneous **<fc #00bb00>voltage</fc>**, **<fc #bbbb00>current</fc>** and **<fc #bbbbbb>power</fc>**. |
| |
| - Ohmic load: The instantaneous voltage is in phase with the instantaneous current. The instantaneous power is always non negative. The average power is $P=U^2/R = {{1}\over{2}} \hat{U}^2/R= {{1}\over{2}}(6V)^2/1k\Omega = 18mW$ | - Ohmic load: The instantaneous voltage is in phase with the instantaneous current. The instantaneous power is always non-negative. The average power is $P=U^2/R = {{1}\over{2}} \hat{U}^2/R= {{1}\over{2}}(6V)^2/1 ~\rm k\Omega = 18 ~\rm mW$ |
| - 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). |
| |
| <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3EZjAZhWALADgwVhZAGyFYBMhKIuEuIGluApgLTIBQATiK2KSKRkjdkfMAHZCUcPDYZSQ5qVyTxk1oSGqphDGwDuwjf0HDexoZH11s5ugE5JAi7PkgUuUROtZwXoTqsbJzc+YMsDFE9JGy1w8AwMWzA7UJNLOSEYr0xE2JAAgHMQ20j+ZSlLABtwFN8VWrCoWDBcOxbuGDhcBIwxMWwxVDbCZChZSEpS0vkfHKksAH0wBcgFgA8sMEhxBdwV2CIlBdYF0lPjloXFlH24G9ZdQUm+achZhPmlhbF1gEMCMCEW4IYHbC5fTpgLCDQh2OHUN52MRYRaQi6PCbcJSSUrMBzxRKaCHrTbbH57TqHPYnM5nVh7ZY3Sn3TDjSiKcq4-FzInLH5rX5bQG7fbLE7LTrI3gU5oXFBstwoHF8aiiD6ZYkbIXk4GEI4086sH4YBZMg4ssQKtDKqhmHl0L78wXbIEysXLZYysHqFyTFBiEB2WgYPWB2jyX2KgP4kN8fERuR+sPgXA+IP8dKkJN2XKpwOJBNZrHdZOsUh2ZOF9lKRI54Tl-MZtgANyxYlSClI7bqUggW3g+SkrGguDEkDEDkihAn2AckhguDYRUU3eCK9C5WcBnXPY5Kj8Vj3th3YSXiscJmtZXnYwiSp77iit7cHhK99P28Bmi8PCfcWST5zLEgRZNE3IHgAxvY+5gReQiiCObRMMwPikNAWB2JAHjyAkIxvGg3CEGw8gBjuWhkbUiQAKorMR47FFopQAXQIA0ZYJHng+97MdRtFrMWPh4sI1CESxULgNAAA6ADOAD2AAWAC2ACWEEyZUsm-AAJmw-HoNwswjNwwbgKhUnScpAB2WkAK4QQALspzaMOpmk6Xp+koXQY7GWJPgoOZEG-AADr8EHKY5zmudpbCyeA8UBhsAKME08CIAc2xSJQMACO4eD+vQ-RYHqoaQvMIABjl9DdPg-QFcV5B8GVmQVU0uU1QV9UlU1zRjCgqGUAAYhAlCYXAKYDqwICDRwjAAI42YwlkQQAnmwQA noborder}} </WRAP> | <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3EZjAZhWALADgwVhZAGyFYBMhKIuEuIGluApgLTIBQATiK2KSKRkjdkfMAHZCUcPDYZSQ5qVyTxk1oSGqphDGwDuwjf0HDexoZH11s5ugE5JAi7PkgUuUROtZwXoTqsbJzc+YMsDFE9JGy1w8AwMWzA7UJNLOSEYr0xE2JAAgHMQ20j+ZSlLABtwFN8VWrCoWDBcOxbuGDhcBIwxMWwxVDbCZChZSEpS0vkfHKksAH0wBcgFgA8sMEhxBdwV2CIlBdYF0lPjloXFlH24G9ZdQUm+achZhPmlhbF1gEMCMCEW4IYHbC5fTpgLCDQh2OHUN52MRYRaQi6PCbcJSSUrMBzxRKaCHrTbbH57TqHPYnM5nVh7ZY3Sn3TDjSiKcq4-FzInLH5rX5bQG7fbLE7LTrI3gU5oXFBstwoHF8aiiD6ZYkbIXk4GEI4086sH4YBZMg4ssQKtDKqhmHl0L78wXbIEysXLZYysHqFyTFBiEB2WgYPWB2jyX2KgP4kN8fERuR+sPgXA+IP8dKkJN2XKpwOJBNZrHdZOsUh2ZOF9lKRI54Tl-MZtgANyxYlSClI7bqUggW3g+SkrGguDEkDEDkihAn2AckhguDYRUU3eCK9C5WcBnXPY5Kj8Vj3th3YSXiscJmtZXnYwiSp77iit7cHhK99P28Bmi8PCfcWST5zLEgRZNE3IHgAxvY+5gReQiiCObRMMwPikNAWB2JAHjyAkIxvGg3CEGw8gBjuWhkbUiQAKorMR47FFopQAXQIA0ZYJHng+97MdRtFrMWPh4sI1CESxULgNAAA6ADOAD2AAWAC2ACWEEyZUsm-AAJmw-HoNwswjNwwbgKhUnScpAB2WkAK4QQALspzaMOpmk6Xp+koXQY7GWJPgoOZEG-AADr8EHKY5zmudpbCyeA8UBhsAKME08CIAc2xSJQMACO4eD+vQ-RYHqoaQvMIABjl9DdPg-QFcV5B8GVmQVU0uU1QV9UlU1zRjCgqGUAAYhAlCYXAKYDqwICDRwjAAI42YwlkQQAnmwQA noborder}} </WRAP> |
| ==== arbitrary two-terminal Component ==== | ==== arbitrary two-terminal Component ==== |
| |
| For an arbitrary component we do not have any defining formula. But, the $u(t)$ and $i(t)$ can generally be defined as: | For an arbitrary component, we do not have any defining formula. But, the $u(t)$ and $i(t)$ can generally be defined as: |
| |
| \begin{align*} \color{blue}{u(t)} &= \sqrt{2}U sin(\omega t + \varphi_u) \\ \color{red }{i(t)} &= \sqrt{2}I sin(\omega t + \varphi_i) \\ \end{align*} | \begin{align*} |
| | \color{blue}{u(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \\ |
| | \color{red }{i(t)} &= \sqrt{2}I \sin(\omega t + \varphi_i) \\ |
| | \end{align*} |
| |
| This leads to an instantaneous power $p(t)$ of | This leads to an instantaneous power $p(t)$ of |
| |
| \begin{align*} p(t) &= \color{blue}{u(t)} \cdot \color{red}{i(t)} \\ &= 2\cdot UI sin(\omega t + \varphi_u) sin(\omega t + \varphi_i) \\ \end{align*} | \begin{align*} |
| | p(t) &= \color{blue}{u(t)} \cdot \color{red}{i(t)} \\ |
| | &= 2\cdot UI \sin(\omega t + \varphi_u) \sin(\omega t + \varphi_i) \\ |
| | \end{align*} |
| |
| The formula can be further simplified with the help of the following equations | The formula can be further simplified with the help of the following equations |
| |
| * $\varphi = \varphi_u - \varphi_i \quad \rightarrow \varphi_i = \varphi_u - \varphi$ | * $\varphi = \varphi_u - \varphi_i \quad \rightarrow \varphi_i = \varphi_u - \varphi$ |
| * $sin(\Box - \varphi) = sin(\Box) cos \varphi - cos(\Box) sin \varphi $ | * $ \sin(\Box - \varphi) = \sin(\Box) \cos \varphi - \cos(\Box) \sin \varphi $ |
| * $2sin\Box sin\Box = 1 - cos(2\Box)$ | * $2\sin\Box \sin\Box = 1 - \cos(2\Box)$ |
| * $2sin\Box cos\Box = sin(2\Box)$ | * $2\sin\Box \cos\Box = \sin(2\Box)$ |
| |
| \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) \\ \end{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) \\ |
| | \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 result into $UI cos\varphi$. | - 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 in average, so the second part of the formula result into 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="fa fa-exclamation" color="red" title="Notice:"> | <callout icon="fa fa-exclamation" color="red" title="Notice:"> |
| A distinction is now made between: | A distinction is now made between: |
| |
| * An **active power** (alternatively real or true power, in German: Wirkleistung): $P = UI cos \varphi$ | * An **active power** (alternatively real or true power, in German: //Wirkleistung//): $P = UI \cos \varphi$ |
| * The active power represents a pulsed energy drain out of the electrical system (commonly by an ohmic resistor). | * The active power represents a pulsed energy drain out of the electrical system (commonly by an ohmic resistor). |
| * The active power transform the electric energy permanently into thermal or mechanical energy | * The active power transforms the electric energy permanently into thermal or mechanical energy |
| * Therefore, the unit of the active power is $Watt$. | * Therefore, the unit of the active power is $\rm Watt$. |
| * A **reactive power** (in German: Blindleistung): $Q = UI sin \varphi$ | * A **reactive power** (in German: //Blindleistung//): $Q = UI \sin \varphi$ |
| * The reactive power describes the "sloshing back and forth" of the energy into the electric and/or magnetic fields. | * The reactive power describes the "sloshing back and forth" of the energy into the electric and/or magnetic fields. |
| * The reactive power is completely regained by the electic circuit. | * The reactive power is completely regained by the electric circuit. |
| * In order to distinguish the values, the unit of the reactive power is $V\!\! Ar$ (or $Var$) for **__V__** olt**__a__** mpere-**__r__** eactive. | * To distinguish the values, the unit of the reactive power is $\rm VAr$ (or $\rm Var$) for **__V__** olt**__a__** mpere-**__r__** eactive. |
| * An **apparent power** (in German Scheinleistung): $S = UI $ | * An **apparent power** (in German //Scheinleistung//): $S = UI $ |
| * 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. |
| * The apparent power shows only what seem to be a value of power, but can deviate from usable power, when indoctors aor capacitors are used in the circuit. | * The apparent power shows only what seems to be a value of power, but can deviate from usable power when inductors or capacitors are used in the circuit. |
| * The unit of the apparent power is $V\!\! A$ for **__V__** olt**__a__** mpere | * The unit of the apparent power is $\rm VA$ for **__V__**olt**__a__**mpere |
| |
| Similarly, the currents and voltages can be separated into active, reactive and apparent values. </callout> | Similarly, the currents and voltages can be separated into active, reactive, and apparent values. </callout> |
| |
| Based on the given formulas the three types of power are connected with each other. Since the apparent power are 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 \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>). |
| |
| <WRAP> <imgcaption imageNo02 | Power Triangle of active, reactive and apparent power></imgcaption> {{drawio>powertriangle.svg}} </WRAP> | <WRAP> <imgcaption imageNo02 | Power Triangle of active, reactive and apparent power></imgcaption> {{drawio>powertriangle.svg}} </WRAP> |
| Generally, the apparent power can also be interpreted as a complex value: | Generally, the apparent power can also be interpreted as a complex value: |
| |
| \begin{align*} \underline{S} &= S \cdot e^{j\varphi} \\ &= U \cdot I \cdot e^{j\varphi} \end{align*} | \begin{align*} |
| | \underline{S} &= S \cdot {\rm e}^{{\rm j}\varphi} \\ |
| | &= U \cdot I \cdot {\rm e}^{{\rm j}\varphi} |
| | \end{align*} |
| |
| Based on the definition of the phase angle $\varphi = \varphi_U - \varphi_I$, this can be divided into: | Based on the definition of the phase angle $\varphi = \varphi_U - \varphi_I$, this can be divided into: |
| |
| \begin{align*} \underline{S} &= U \cdot I \cdot e^{j(\varphi_U - \varphi_I)} \\ &= \underbrace{U \cdot e^{j\varphi_U}}_{\underline{U}} \cdot \underbrace{I \cdot e^{-j\varphi_I}}_{\underline{I}^*} \end{align*} | \begin{align*} |
| | \underline{S} &= U \cdot I \cdot {\rm e}^{ {\rm j}(\varphi_U - \varphi_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*} |
| |
| where $\underline{I}^*$ is the complex conjungated value of $\underline{I}$. | where $\underline{I}^*$ is the complex conjungated value of $\underline{I}$. |
| <callout icon="fa fa-exclamation" color="red" title="Notice:"> The apparent power $\underline{S} $ is given by: | <callout icon="fa fa-exclamation" color="red" title="Notice:"> The apparent power $\underline{S} $ is given by: |
| |
| * $\underline{S} = UI \cdot e^{j\varphi}$ | * $\underline{S} = UI \cdot {\rm e}^{{\rm j}\varphi}$ |
| * $\underline{S} = UI \cdot (cos\varphi + j sin\varphi)$ | * $\underline{S} = UI \cdot (\cos\varphi + {\rm j} \sin\varphi)$ |
| * $\underline{S} = P + jQ$ | * $\underline{S} = P + {\rm j}Q$ |
| * $\underline{S} = \underline{U} \cdot \underline{I}^*$ | * $\underline{S} = \underline{U} \cdot \underline{I}^*$ |
| |
| </callout> | </callout> |
| |
| The following simulation shows three ohmic-inductive loads, all with an impedance of $|Z| = 1 k\Omega$, however with different phase angles $\varphi$. The diagram on top of each circuit shows the instantaneous **<fc #00bb00>voltage</fc>**, **<fc #bbbb00>current</fc>** and **<fc #bbbbbb>power</fc>**. Similar to the last simulation, a pure ohmic resistance would consume an average power of $P=U^2/R = {{1}\over{2}} \hat{U}^2/R= {{1}\over{2}}(6V)^2/1k\Omega = 18mW$. The three diagrams shall be duscussed shortly | 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 #00bb00>voltage</fc>**, **<fc #bbbb00>current</fc>** and **<fc #bbbbbb>power</fc>**. |
| | Similar to the last simulation, a pure ohmic resistance would consume an average power of $P=U^2/R = {{1}\over{2}} \hat{U}^2/R= {{1}\over{2}}(6~\rm V)^2/1~\rm k\Omega = 18~\rm mW$. |
| | The three diagrams shall be discussed shortly. |
| |
| - Phase angle $\varphi = 10°$: Nearly all of the impedance id 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 $17.47mW$ is about the same like 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 $500\Omega = {{1}\over{2}}|Z|$, but does not show half of the voltage! This is due to the fact that the addition has to respect the complex behavior of the values. The complex part is $90°$ perpendicular to the real part - so they generate a right-angled triangle. The average power with $9mW$ is exactly the half of the power for an ohmic impedance, since only the resistance provides a way for consuming power permanently. | - 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 $500 ~\rm \Omega = {{1}\over{2}}|Z|$, but does not show half of the voltage! This is because the addition has to respect the complex behavior of the values. The complex part is $90°$ perpendicular to the real part - so they generate a right-angled triangle. The average power with $9~\rm mW$ is exactly half of the power for an ohmic impedance since only the resistance provides a way for consuming power permanently. |
| - 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 load is nearly pure inductive. The instantaneous power is consequently almost half of the time lower than zero. The average power here also only $10%$ of the power for an 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. |
| |
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| |
| The next simulation enables to play around with the phase angle of an impedance. The circuit on the left side is a bit harder to understand, but consist of a resistive (real) impedance and an complex impedance, which are driven by an AC voltage source. All of these components are parameterizable in such a way that the phase angle can be manipulated by the slider on the right side. \\ In the middle part reflects the time course of: | The next simulation enables us to play around with the phase angle of an impedance. |
| | The circuit on the left side is a bit harder to understand but consists of a resistive (real) impedance and a complex impedance, which are driven by an AC voltage source. |
| | All of these components are parameterizable in such a way that the phase angle can be manipulated by the slider on the right side. \\ In the middle part reflects the time course of: |
| |
| * The instantaneous power $p$ of the **<fc #bb0000>real part (active power)</fc>**, the **<fc #bb6600>imaginary part (reactive power)</fc>** and **<fc #bbbbbb>overall power</fc>**. | * The instantaneous power $p$ of the **<fc #bb0000>real part (active power)</fc>**, the **<fc #bb6600>imaginary part (reactive power)</fc>** and **<fc #bbbbbb>overall power</fc>**. |
| * The instantaneous **<fc #00bb00>voltage</fc>** and **<fc #bbbb00>current</fc>**. | * The instantaneous **<fc #00bb00>voltage</fc>** and **<fc #bbbb00>current</fc>**. |
| |
| On the right-hand side the impedance Phasor is shown (lower diagram). The upper diagram depicts the $u$-$i$-diagram, which would be a perfect line for an pure ohmic resistance (since $u_R = R \cdot i_R$) and a circle for a pure complex impedance (since the phase angle of $\pm 90°$ between $u_{L,C}$ and $i_{L,C}$). The simulation is in this part not completely perfect: The pure line and circle are sometimes not reachable. | On the right-hand side, the impedance Phasor is shown (lower diagram). The upper diagram depicts the $u$-$i$-diagram, which would be a perfect line for a pure ohmic resistance (since $u_R = R \cdot i_R$) and a circle for a pure complex impedance (since the phase angle of $\pm 90°$ between $u_{L, C}$ and $i_{L, C}$). The simulation is in this part not completely perfect: The pure line and circle are sometimes not reachable. |
| |
| The folllowing questions can be solved with this simulation: | The following questions can be solved with this simulation: |
| |
| - How does the amplitude of the active and reactive instantaneous power change, when the phase angle is changed between $-90°...+90°$? | - How does the amplitude of the active and reactive instantaneous power change, when the phase angle is changed between $-90°...+90°$? |
| - What is the phase shift between the active and reactive instantaneous power? | - What is the phase shift between the active and reactive instantaneous power? |
| |
| 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noborder}} </WRAP>((For further development: [[https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCDsB0AcCMAmArAUwLSRABmveYY8ALLMcmNgGxWyJUSJbIjERrr4BQAHiOogCcgkPFgQBVeKNhVWoxCAAKACwCGAZwD2AJ16jIsEEMaJixwXPNIQAYTUAHNQGMAlgBcAnvrER4Vc0QwOX8sa0UASQA7ABMAV2d3VwA3D29EbCwBWCMwDP487HB88wiAWwBvAC0AXy5-IoFkOUI5bNzgnGMi+Gw+-oGuACV+RBzwZEV0QonFIqLiea756GRh0chFMEmNrbIuxRQBkBYl3DWMrJQWzoEzcE7SsoB9B3VtPUvR5uN7u8D7uYhqhXu9dOt2sZFgV8mYzqx4Ss1g1vnJEH8xkZ0eYivlegMCVwvpCMkZprDsEYgahqnUvtdirjxjNzLYiZlRFQittFPgtjtzAAzHQAR3ZWBRMyQuRKIHKtK4AHNjD8eSqWvt5kSbJKKTKlooABQAIwAVGoADoaAB8FpFVuwAEorQB+B3QAnYeAALit8AA9IaBKaHK5TWa1I7xerRIJDuN4HH5AAZaMo7GiGwZoWitNc2O8nVJp4K5UMjPl6Far7+VhU+vyYEKmtozZ14xtku1bW8-OJw6UgsHEAaVAAG1QiUNagANNgZ2pA4gQ2HjY6o8TYEUt-w+SAd+Y3q5o+gD2S96R5FUTzumFM93erzfudhzJwbJRzIeVMfEDZTwsZJ5O2uIgGoprOFoGgmv6YjYCuUbKrWO61nuWp8Jg0hBEYWHENI4RymUVpKGoOjuFahoaK4USoIhnLgK+9GtMsPb8Lebbkh2czGGBppUVEMFwQhXAAO67LM4lwjg+iYH4LAEDI8g2MCxGkeR0GQfxtFcEAA|CalcImpedances]] [[https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCDsB0AcCMAmArAUwLSRABmveYY8ALLMcmNgGxWyJUSJbIjERrr4BQA7iMpCo5+gkFWI5erWNhDpxIofInYppWQvXhEkxNixbZbHUZAAlLogLTZ6MDuTEJdk+BAAKAMboAhgEoAendcbGx4AB0ABwAjPzUZOXsbOQVVPgElBQzEk3iNCWzUy31k5VL8HQko4oMEzQSkWFY5LiA|VarResistor]] [[https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCDsB0AcCMAmArAUwLSRABmveYY8ALLMcmNgGxWyJUSJbIjERrr4BQi2WktEOirEogzsVGiAOgAcefKMkQh4AThWRlIFFVZCuAdxCwRINXtOiz2IydLnLZizjtlYqxB6uePt41oq8FTYSkEargHaZoGOkSZmMdGitogEYXEIei6hEAAUACYAxgUALugFAIZlAJQAVABGCvzaSB5UiKJt+gDCbmbdPuoqqYpUkCMgHSn6ADJcAOZTE+Aq4yoOoanp6zhQvHuhKpX9ouguPsIpdpBgeudOZzk3d0Jgmq8P8QIe6O9Qn3+-gSZ3+PhsPBIYl+Zlg2FCVz2oiqZUap1UEViw1cvH4glCAgRwX0ADk+kA|varLC]] [[https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCDsB0AcCMAmArAUwLSRABmveYY8ALLMcmNgGxWyJUSJbIjERrr4BQi2WkvELCiD0iYqxAAdAA48+I7CHGKhSienlYqA5RKqk9kgGYAnAI5aQOxEOuHhEgGoB9ALYBDAB48k13cK0wsLw4CAeAFQAzgCWAHYAFAkARhEALrIAxgCUEYgRMjHZVjasSgaO6iAyABYxLk5cQA|SineGenerator]])) | <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-xbCBiTDclIGVLBn0YPZIwHb9PkCSNuBjVBxyjbNFFwrMoxjIjc1yFBHFKcoAHsYSNHBT2ZLVySwcZRCQFI-SYggWKHG0OK1RIeIDZkWESbAOIk5BGLEg5-kE8lFIkFAb3MDxrFyMwrFsBwuggFJlGvTkDJAAAzGF4VPGQn30SCrjomE7PkCAaQZWyIAgCVek8y5rhkDA7JEP9HIcej6ESCADQ8xARAgWkbS6ERXEsDTaJ3ABldhfGscIsAAUQ8AxHCwRwwEcDBHDQOwYVqcwvA8IqSocCq6oapriuq2r6sa5rHFIWrUvSrKcp8PLCq6hxIAGxx4G6mEADVr0WzL+ocKqHAIRwKEcJBauW1b1s27aNv2hx9qWla1qm7BKs2+aLoO67jrKjbNs2-pnAcPZvs+L4Ab2779uqkGnqBiGLqGSGytKyAvm+mbIbBn5ke+xpgcx5HofBn6wf+gG+hh77ywAIwAKnMZxvoAPmpux4Xpi6a2phwAH4maLL4MAALiZjAV3LDRyaMLxyYp8wa1x2GHAqDxoGsHxy3MAAaJAVfMQWsBFsXSarKWwZl8xyesWiHDJldvyQHWDe+o3yYcLxLAtq2baNT4nXebRXmSWzrw0rTjB0gp9PO36zsJvpkCBhwMae+Pzp+aOfrhhHppjxOhmT8648TjOs7D86Ccj5O3rJynaecRm9pZ9ngfnAHeb+wXhdF8XKarUvHDlhWldV9XNfLbW271zuw9rE2zZdihrdFsfSonx3ndJy2Z7dwV-i5R4t+ocULKso9-k7R4kAeTQLlpI9vV6VyTy5eQrhuNTFvMOroEdhwfAisNwF-uwjDsgMWAtkf71BgnZZa14botTehVDaC8NrR3KiTawsBJaCyLFVamRg9bu0YBqUQHIKDPD4CaIKL834fy-s5c6cDyyoPQeWTBzgcHNkFNfL0p8nzyHkJfdhnZnjX3vvveEsQnhHDQIkCgpIFAJG5M+a8XgYTWD8L4RwY1wiOAAG6v23KTeWjhagSmvOiFw9DaKWE-lYHwjgACyu5LC+ACEEWi5lHDaXyLYZs0gCCJjOMobAHsNAYAhNwkAAAVLwBggjXlon4GEFRHCuNMSUWi7BoAwhVo4RatEPA+F3EEOsjhrzxLhJYHwrBn6v2MV0IKPCQD-xAV0YBcw-byLsj5SSEDrxQPWnAmqKdWoLyur0qaM146DQcPNGEPSjq3R2rNKZCyhmUK8J-ei61JlfWQY4csphrCmB8LACwhyqzi2JrWSGNVNzmEOec3GxcAZ4OPPFEQBCZ62QoTU2huz9mHOObc6s4su4OHMMnAZJyfDAqLpHJorAdye09JSY+rx+hHlwBMT4rl3SiBNPISyoiwEcj4ByDFogcXzEYNxLYOw1KZSdtAAA4siDJ5hqHJV-okRpQUgrNDsjykRv8goiEfuGRa9ARAiEWuGAw5hpDyBQiAQ60DupDPTvAn6MhvoVUXk7cs5cbHYOsGc4eXhmyYEYGSklKQrV6iCvSywTKWX7nZZtOBxsl76opoaowxryamubESiYzwrUfMpW8bYwxjRUteHwU4GxNQuhtBQRiMbXnCl0Ki5NdoqA6HbAm4CaLOC5t4D0AtSLZLFufHGng5aXR2iEJqHojbwBzDtAW14Bai0ZtaSW7tLb+CfhPIOytUBh3tiUE2tF-pTy6D4MO4M5pan5ijEmSQ5gExJhTHhDMUY5zFj4qQRQo4kCkXDMacMSYliqC4sepYZ6ASXpPTeo9275xnqMGkjJjgAKcEtBWydpa0WZG5FCXlvA+JjmYqIEANEKisCAA noborder}} </WRAP>((For further development: [[https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCDsB0AcCMAmArAUwLSRABmveYY8ALLMcmNgGxWyJUSJbIjERrr4BQAHiOogCcgkPFgQBVeKNhVWoxCAAKACwCGAZwD2AJ16jIsEEMaJixwXPNIQAYTUAHNQGMAlgBcAnvrER4Vc0QwOX8sa0UASQA7ABMAV2d3VwA3D29EbCwBWCMwDP487HB88wiAWwBvAC0AXy5-IoFkOUI5bNzgnGMi+Gw+-oGuACV+RBzwZEV0QonFIqLiea756GRh0chFMEmNrbIuxRQBkBYl3DWMrJQWzoEzcE7SsoB9B3VtPUvR5uN7u8D7uYhqhXu9dOt2sZFgV8mYzqx4Ss1g1vnJEH8xkZ0eYivlegMCVwvpCMkZprDsEYgahqnUvtdirjxjNzLYiZlRFQittFPgtjtzAAzHQAR3ZWBRMyQuRKIHKtK4AHNjD8eSqWvt5kSbJKKTKlooABQAIwAVGoADoaAB8FpFVuwAEorQB+B3QAnYeAALit8AA9IaBKaHK5TWa1I7xerRIJDuN4HH5AAZaMo7GiGwZoWitNc2O8nVJp4K5UMjPl6Far7+VhU+vyYEKmtozZ14xtku1bW8-OJw6UgsHEAaVAAG1QiUNagANNgZ2pA4gQ2HjY6o8TYEUt-w+SAd+Y3q5o+gD2S96R5FUTzumFM93erzfudhzJwbJRzIeVMfEDZTwsZJ5O2uIgGoprOFoGgmv6YjYCuUbKrWO61nuWp8Jg0hBEYWHENI4RymUVpKGoOjuFahoaK4USoIhnLgK+9GtMsPb8Lebbkh2czGGBppUVEMFwQhXAAO67LM4lwjg+iYH4LAEDI8g2MCxGkeR0GQfxtFcEAA|CalcImpedances]] [[https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCDsB0AcCMAmArAUwLSRABmveYY8ALLMcmNgGxWyJUSJbIjERrr4BQA7iMpCo5+gkFWI5erWNhDpxIofInYppWQvXhEkxNixbZbHUZAAlLogLTZ6MDuTEJdk+BAAKAMboAhgEoAendcbGx4AB0ABwAjPzUZOXsbOQVVPgElBQzEk3iNCWzUy31k5VL8HQko4oMEzQSkWFY5LiA|VarResistor]] [[https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCDsB0AcCMAmArAUwLSRABmveYY8ALLMcmNgGxWyJUSJbIjERrr4BQi2WktEOirEogzsVGiAOgAcefKMkQh4AThWRlIFFVZCuAdxCwRINXtOiz2IydLnLZizjtlYqxB6uePt41oq8FTYSkEargHaZoGOkSZmMdGitogEYXEIei6hEAAUACYAxgUALugFAIZlAJQAVABGCvzaSB5UiKJt+gDCbmbdPuoqqYpUkCMgHSn6ADJcAOZTE+Aq4yoOoanp6zhQvHuhKpX9ouguPsIpdpBgeudOZzk3d0Jgmq8P8QIe6O9Qn3+-gSZ3+PhsPBIYl+Zlg2FCVz2oiqZUap1UEViw1cvH4glCAgRwX0ADk+kA|varLC]] [[https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCDsB0AcCMAmArAUwLSRABmveYY8ALLMcmNgGxWyJUSJbIjERrr4BQi2WkvELCiD0iYqxAAdAA48+I7CHGKhSienlYqA5RKqk9kgGYAnAI5aQOxEOuHhEgGoB9ALYBDAB48k13cK0wsLw4CAeAFQAzgCWAHYAFAkARhEALrIAxgCUEYgRMjHZVjasSgaO6iAyABYxLk5cQA|SineGenerator]] |
| | [[http://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3EZgEwGYwDY0YCxpwOyQ4CcBIyJIArDZDQKYC0YYAUMmDiDgBz0peNFOGRD63ACYBDSQBcAVACM2Adx78QuDQLFQ1NAhnAZ61IxWrHIBvgNM8+J+jfU4nmCU5LXb34+bGPvrqgSDBYYIhOlrcYdquhsZM2mFMqMjRqKjJ2nYgKdyJ+emZJQkcYKgxpTTuBRlQINUAFADGTNIAlAD0LTAIADoADkpdbADmPNqovNw4eXBNNpzVs9xgJGWNm5n0mdKTzXMUyNzrFDgu+pyZTJrIBHeNj3sUINIKAM4AlgB2LSUPTA-AUwx+4ymmGaxBMzRyyzYAA8GgRqhghEwxLQCBANgcALbDAA2PzkAFdJAxBl8ANoAeQAFgSALrIrQkKiPHELCgEIT4kDDRnSL7UukABrZACUCmd6NosWdYsseAJ4KqYNQ2LK0cYxEICNRMgbVVcCgNSJzrTbObjrMtoNqUUxcRRZuBLnjRCAAJIEmkABWkACc5DSWr8-gxxi70BBWPQlbRWNVBdLxcGwxG2gB7KMxo7QzTQ1hvFYie4STGNJz7D4KPNfQHA0Hg8bISDkLHmxWvFXcABaHC7ID1FF4hthpu4wp+I+7mlQsJYImX8yFjPnncX+yeBTLfLKIGHO4Kmicq7KApPC-P9E0V5AF8325EnnAWwokCEuyamTFYkGDaOQWmkAAaSBwOkPpkDBH5lC6DtRx7B8k37F8ADMQwARx1OVzUfZBlU0a5zXVDUXCddl0moWgTixXh6PIQVsIYHCGD+NoAE8aVpAAJAAvNkz2hfsnHEv0CQAb0HABfO9oScZSbwzWSFKmZArFObgtP1ciblHD8-0PP9uH9dTFIcYiNhEGyeBAbC8LPPTPxNSc3IcgAZO8Lj-VBjU87gfLPfIJEaCQQDU+S2DwfZtMKeFciKWIDFcxLO0xCp1Fc-Yf0RAA3b9xGKgoEk9MB4DMJoWCdIgCB8DIMAavgfGsaj1GyB0uuOIpbBmE4wguGwrmqfkNmMSr8sPcKAH0wFmghZqRaRKswWbqFmpBZpYeatugMArC0-a4CWpg8FmoNc1UBgQyDWLIDGk4oimoR5SaHBUD2yBlt4Nals2sBoAwHaFuQWbwZ+z7Zq+gZIEhy6vgep7uBYL9XrNL7cC237KuoBgNv2hbdvBoHNqhr7YdgSB1p+gBVHpfRpSQfmkCYQ2kAkDCGzskl66IFnOE4euGu9BeaXmcGhDJIoAYVsTQ8iXPZKkyD8ZdKjX6FaJQFGkGkAD5BhwmlIC6GkAH5TepqqwAALhpMA+ixeDlD1yFLGMAKTW04bDPIdXeaiDXzJkmLRIcb3wDXQLuCcu9XI100Q5AeXVmmXSPONN6PO1kAWkkNp5CYGR5C6ZRuYsVzxdy-rdO0mvc7r5pY4GvqI+1569AubhZdm+kADVucj3meZcYf9izqum9CaehDCWuz0Yt78tKPLVIYSyPyxbT7OXnT-woqqqvw-f3rXy5rgkK-HW1WUsX3c-+37esUGoDVaCv6il9cvf7Ps7gGZZrClFLmEMd4d5ewREqc4CJQ7AJFF8MBp9+xRwfpkP2astIf01NRe+ydAoXyjmRG+VFtTb0TtA5OCI8rgGPsfCBydeZEMllJSyo0xwnFchjcW9A0TfVxpAKk80xA-W2iTCG78cYEBwDjbBONAyIOQXgMaRpnxqxppkXgKsVFjjUbwaoa0tHVBGmgPRAdno0w2CcUxqiLGGKseAAxAtHypAsIlRIYQzAWAqCiEEhjtDYBMCxH0vo-iSHJCBH4BUyTcXZGAfkFASDVH-sEQUstpDDGkG0WJ8THoQACpNH85xqBUEFGEiJUSYlyDiS6SgVB-FygwAmXgxhBSKNAeAj0vMABiiACgkCQOEJMEAOlijYEAA|full simu]] |
| | )) |
| |
| 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. |
| |
| <panel> <imgcaption BildNr00 | Simulation of instantaneous power as a function of phase> </imgcaption> \\ <collapse id="geogebra" collapsed="true"><well> {{url>https://www.geogebra.org/material/iframe/id/yfwfhtn7/width/1000/height/800/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/true/at/preferhtml5 750,600 noborder}} \\ Change the phase angle with the slider under $\alpha$ </well></collapse> | <panel> <imgcaption BildNr00 | Simulation of instantaneous power as a function of phase> </imgcaption> \\ <collapse id="geogebra" collapsed="true"><well> {{url>https://www.geogebra.org/material/iframe/id/yfwfhtn7/width/1000/height/800/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/true/at/preferhtml5 750,600 noborder}} \\ Change the phase angle with the slider under $\alpha$ </well></collapse> |
| === Power Factor Correction === | === Power Factor Correction === |
| |
| Cables and components have to conduct the sum of active and reactive current, but only the active current is used outside of the circuit. Therefore, a common goal is to minimize the reactive part. The technical way to represent this the **{{https://en.wikipedia.org/wiki/power factor|power factor}} ** $pf$ is used. | Cables and components have to conduct the sum of active and reactive currents, but only the active current is used outside of the circuit. Therefore, a common goal is to minimize the reactive part. The technical way to represent this is the **{{https://en.wikipedia.org/wiki/power factor|power factor}} ** $pf$ is used. |
| |
| <callout icon="fa fa-exclamation" color="red" title="Notice:"> The power factor is given by: | <callout icon="fa fa-exclamation" color="red" title="Notice:"> The power factor is given by: |
| |
| \begin{align*} pf &= cos \varphi \\ &= {{P}\over{|\underline{S}|}} \end{align*} | \begin{align*} pf &= \cos \varphi \\ &= {{P}\over{|\underline{S}|}} \end{align*} |
| |
| The power factor shows how much real power one gets out of the needed apparent power. </callout> | The power factor shows how much real power one gets out of the needed apparent power. </callout> |
| |
| How does the power factor show the problematic effects? For this one can investigate situation of a ohmic-inductive load $\underline{Z}_L$ which is connected to a voltage $\underline{U}_0$ source with a wire $R_{wire}$. This circuit is shown in <imgref imageNo03>. | How does the power factor show the problematic effects? For this one can investigate the situation of an ohmic-inductive load $\underline{Z}_L$ which is connected to a voltage $\underline{U}_0$ source with a wire $R_{wire}$. This circuit is shown in <imgref imageNo03>. |
| |
| <WRAP> <imgcaption imageNo03 | Power Factor of a Power Line></imgcaption> {{drawio>PowerFactorPowerLine.svg}} </WRAP> | <WRAP> <imgcaption imageNo03 | Power Factor of a Power Line></imgcaption> {{drawio>PowerFactorPowerLine.svg}} </WRAP> |
| |
| 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: |
| |
| \begin{align*} P_{wire} &= R_{wire} \cdot I^2 \\ &= R_{wire} \cdot {{P_L^2}\over{U_L^2 \cdot cos^2 \varphi}} \end{align*} | \begin{align*} |
| | P_{\rm wire} &= R_{\rm wire} \cdot I^2 \\ |
| | &= R_{\rm wire} \cdot {{P_L^2}\over{U_{\rm L}^2 \cdot \cos^2 \varphi}} |
| | \end{align*} |
| |
| This means: As smaller the power factor $cos \varphi$, as more power losses $P_{wire}$ will be generated. More power losses $P_{wire}$ leads to more heat up to or even beyond the maximum temperature. In order to compensate this, the cross section of the wire have to be increased, which means more copper. | 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. To compensate for this, the cross-section of the wire has to be increased, which means more copper. |
| |
| Alternatively, a bad power factor the can be compensated with a counteracting complex impedance. This compensating impedance has to provide enough power with the opposite sign to cancel out the unwanted reactive power. The following simulation shows an uncompensated circuit and a circuit with power factor correction. In the ladder the voltage on the load resistor is the same, but the current provided by the power supply is smaller. | Alternatively, a bad power factor can be compensated with a counteracting complex impedance. This compensating impedance has to provide enough power with the opposite sign to cancel out the unwanted reactive power. The following simulation shows an uncompensated circuit and a circuit with power factor correction. In the ladder, the voltage on the load resistor is the same, but the current provided by the power supply is smaller. |
| |
| <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3Z5OfATFaYwGZtgCwAc+ArNpAGwWGoXYjkgkj70kCmAtFgFAA2ITgE4K4fPkFExEyBjioA7KM4w4JQktpDIQ-KlL4F6SDwBOkwiH2zOUgjJBDCJHgHdBIq-htTrUN16yfuQSfiYA5uBg6H5YMSSisiYAblEx3mnSUODgSCCJgmDQJAqQCiLYtOVEIonFZgyQoRkhWbJOLgKx4o0S9tmqkIrKg+qaFNq6+iSGxjypnKiQln6Ly205iPD52dzFpeV0VbqEtRgukWsrGVdWCQMBt-3Cov0m7i+BgkvXSTwAxl8wEJ0rJ+jFoEouEImPtCBR1EIwGRsAk8HJIGAGn4cRlgcZwFxogFcWDol93plwXieiZvPRbrZLNxXgUwAB9DmQdlLIgUTnc1TREjs7g8zns7A8enfdZSFmZWSECXcgAezkxIpFgwoqBFYtQ4u5yuw7MGhrpkHocQsvXABX5XJ5TXhArNsGFoo5hv5UplNqkrRtshFTvVJE17vgXI9qGVBqj8HwQhTqbT6f53JFpvNZp4qu+5M4JBhi2YxdEfUsAFUAHb-AD2AFsAA7sWsAZwAhgAXdgAEwAOh2AMIAS1M-wArmOe-ncsySwwi-cqyBx5OZz3h65ZwALYcABQAYsORw3TKZ2P8e2OG7WeNhLLJjxAkh7wB-uCAAJK1-tTjeXb1uwPBAA noborder}} </WRAP> | <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3Z5OfATFaYwGZtgCwAc+ArNpAGwWGoXYjkgkj70kCmAtFgFAA2ITgE4K4fPkFExEyBjioA7KM4w4JQktpDIQ-KlL4F6SDwBOkwiH2zOUgjJBDCJHgHdBIq-htTrUN16yfuQSfiYA5uBg6H5YMSSisiYAblEx3mnSUODgSCCJgmDQJAqQCiLYtOVEIonFZgyQoRkhWbJOLgKx4o0S9tmqkIrKg+qaFNq6+iSGxjypnKiQln6Ly205iPD52dzFpeV0VbqEtRgukWsrGVdWCQMBt-3Cov0m7i+BgkvXSTwAxl8wEJ0rJ+jFoEouEImPtCBR1EIwGRsAk8HJIGAGn4cRlgcZwFxogFcWDol93plwXieiZvPRbrZLNxXgUwAB9DmQdlLIgUTnc1TREjs7g8zns7A8enfdZSFmZWSECXcgAezkxIpFgwoqBFYtQ4u5yuw7MGhrpkHocQsvXABX5XJ5TXhArNsGFoo5hv5UplNqkrRtshFTvVJE17vgXI9qGVBqj8HwQhTqbT6f53JFpvNZp4qu+5M4JBhi2YxdEfUsAFUAHb-AD2AFsAA7sWsAZwAhgAXdgAEwAOh2AMIAS1M-wArmOe-ncsySwwi-cqyBx5OZz3h65ZwALYcABQAYsORw3TKZ2P8e2OG7WeNhLLJjxAkh7wB-uCAAJK1-tTjeXb1uwPBAA noborder}} </WRAP> |
| |
| Another explanation of power factor can be seen here: | Another explanation of the power factor can be seen here: |
| |
| {{youtube>Tv_7XWf96gg}} | {{youtube>Tv_7XWf96gg}} |
| ===== 7.2 Polyphase Networks ===== | ===== 7.2 Polyphase Networks ===== |
| |
| In order to transfer power over long distances alternating current ans explicitely rotary current is used. Rotary current is the common name for three-phase current. The first three-phase high Voltage power transfer worldwide started in the August of 1891 for the "{{https://en.wikipedia.org/wiki/International Electrotechnical Exhibition|International Electrotechnical Exhibition}} ". The power plant in Lauffen (see <imgref imageNo07>) - about 10km away from the uinversity Heilbronn - was therefore the first modern three-phase generator and started the three-phase transimssion networks, which are the power backbone throughout the world. | 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 "{{https://en.wikipedia.org/wiki/International Electrotechnical Exhibition|International Electrotechnical Exhibition}} ". The power plant in Lauffen (see <imgref imageNo07>) - about $10 ~\rm km$ away from the university Heilbronn - was therefore the first modern three-phase generator and started the three-phase transmission networks, which are the power backbone throughout the world. |
| |
| <WRAP> <imgcaption imageNo07 | Worldwide first three phase High Voltage Power Transfer></imgcaption> {{drawio>Lauffen}} </WRAP> | <WRAP> <imgcaption imageNo07 | Worldwide first three-phase High Voltage Power Transfer></imgcaption> {{drawio>Lauffen}} </WRAP> |
| |
| In the following, the way to polyphase networks and explicitely three-phase systems will be described. Be aware, that the term "phase" is used in two different meanings: In the first usage, the phase shift $\varphi$ between voltage and current on a single component is commonly called "phase". The second usage of "phase" is for a single circuit of a "multi circuit" setup, called polyphase network. The ladder terminology will be explained in the following in more detail. | In the following, the way to polyphase networks and explicitly three-phase systems will be described. Be aware, that the term "phase" is used in two different meanings: In the first usage, the phase shift $\varphi$ between voltage and current on a single component is commonly called "phase". The second usage of "phase" is for a single circuit of a "multi-circuit" setup, called a polyphase network. The ladder terminology will be explained in the following in more detail. |
| |
| ==== 7.2.1 Technical terms of the Polyphase Networks ==== | ==== 7.2.1 Technical terms of the Polyphase Networks ==== |
| 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** describes a circuit in which $m$ sinusoidal voltages transport the power. The general term of these systems is polyphase systems. \\ 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>). The induced voltages exhibit the same frequency $f$. \\ <WRAP><imgcaption imageNo04 | Visible Representations of a m-phase System></imgcaption>{{drawio>technicalTermispolySys1}}</WRAP> | - A **$m$-phase system** describes a circuit in which $m$ sinusoidal voltages transport the power. The general term for these systems is polyphase systems. <WRAP> |
| - An $m$-phase system is **symmetrical** when the voltages of the individual windings exhibit the same amplitude and are offset at the same angle to each other ($\varphi = 2\pi/m$). \\ Thus, the voltage phasors $\underline{U}_1 ... \underline{U}_m$ form a symmetrical star. \\ Example: A 3-phase system is symmetrical for $\varphi = 360°/3 = 120°$ beween the voltages of the windings: $\underline{U}_1 = \sqrt{2} \cdot U \cdot e ^{j(\omega t + 0°)}$, $\underline{U}_2 = \sqrt{2} \cdot U \cdot e ^{j(\omega t - 120°)}$, $\underline{U}_3 = \sqrt{2} \cdot U \cdot e ^{j(\omega t - 240°)}$ \\ <WRAP><imgcaption imageNo05 | Visible Representations of the a symmetric and asymmetric System></imgcaption>{{drawio>technicalTermispolySys2}}</WRAP> | 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>). The induced voltages exhibit the same frequency $f$. \\ |
| - The windings can be concatenated (=linked) in different ways. The most importend ways of **concatenation** are: | <WRAP><imgcaption imageNo04 | Visible Representations of a m-phase System></imgcaption>{{drawio>technicalTermispolySys1.svg}}</WRAP> |
| - All windings are independently connected to a load. This phase system is called **non-interlinked** (in German: nicht verkettet). | </WRAP> |
| - All windings are connected to each other, then the phase system is called **interlinked**. \\ \\ <WRAP outdent>With interlinking, fewer wires are needed. Star or ring circuits can be used for daisy chaining. \\ The two simulations in <imgref pic20> show a non-interlinked and an interlinked circuit with generator and load in star shape.</WRAP> <WRAP><imgcaption pic20|Comparison of non-interlinked and interlinked circuits></imgcaption> \\ <collapse id="openAni1" collapsed="true"><well> To show the simulations: click on ''Edit'' >> ''Center Circuit''<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgA3EbFcGmkExp5uvKhErUqKWMTrEEYSCmzFsYBfRgJOIPNPzhCVHlXEUqSS9BUJieLAjsF7xKNB1d9fAWGOiocAspcFgaDA0MJ1ZiQjsMQndPALQRIRFTQIlLQJpoMDwHYmVIcLxCVUgrD11Y8ASQFBpxBvNJK0DtXT96xJYUETBWoPbA7HyaMDsiBGwEPDI1JLYAJ25CDN5+Qf4s+FXaOtSKBszxfbXtxsNsDdEzZDgDmjqwXYRTsUfIZ55exsK-3OTzWcUSQ0Sil4EL2TwAHuACIIwCgtm9kRpaI0BAA7AD2OKYAEscQAXBgrAA2JIA1gwACYAHQAzgBhIkrADGAFciaS2AB3QTCe7CjJ4AQ-IXpbgS9biyWC+WygS3TZmJV-Y5-ZoaoU6uC0EW6qBKq4mq7HKVGExoG20P7W7yTRIvV09a0fV1-L2ip0GETOj1mo6GN20YNCvxUOi8X2x03SkV-GWRsVGZVptUZ7OOzXQ3aOxMBGMivVGnYCK7l7xvATeFiJf3I111RvF8N1w59NPRgJ9ph5qP+Qdxhqj4u+ie+ocZmE9bPW-qDccojIIMRK5fcDcA9eboUDfd7oFKhd3BOLs1-bMJmHW14NNtlpVNFqJN+CQiKoVQ-5-mVrTBf5gKYb9i1A8DPzAn8KBRdNnxEH4EWwDBeCYBwQEIO0MKrbEQAAGTxABDJlmQACgAZVJYiVhZVkCRxBhOVJIkCQASjYBEhCQDDyBQD5BDlAQmhAABxBgmJWYjSTxOiKOo2j6MY5jWI4tggA noborder}} | - An $m$-phase system is **symmetrical** when the voltages of the individual windings exhibit the same amplitude and are offset at the same angle to each other ($\varphi = 2\pi/m$). <WRAP> |
| | 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: |
| | $\underline{U}_1 = \sqrt{2} \cdot U \cdot {\rm e} ^{{\rm j}(\omega t + 0°)}$, |
| | $\underline{U}_2 = \sqrt{2} \cdot U \cdot {\rm e} ^{{\rm j}(\omega t - 120°)}$, |
| | $\underline{U}_3 = \sqrt{2} \cdot U \cdot {\rm e} ^{{\rm j}(\omega t - 240°)}$ \\ |
| | <WRAP><imgcaption imageNo05 | Visible Representations of the a symmetric and asymmetric System></imgcaption>{{drawio>technicalTermispolySys2.svg}}</WRAP> |
| | </WRAP> |
| | - 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** (in German: //nicht verkettet//). |
| | - All windings are connected to each other, then the phase system is called **interlinked**. <WRAP>\\ \\ <WRAP outdent> |
| | With interlinking, fewer wires are needed. Star or ring circuits can be used for daisy chaining. \\ |
| | The two simulations in <imgref pic20> show a non-interlinked and an interlinked circuit with generator and load in star shape.</WRAP> <WRAP><imgcaption pic20|Comparison of non-interlinked and interlinked circuits></imgcaption> \\ |
| | <collapse id="openAni1" collapsed="true"><well> To show the simulations: click on ''Edit'' >> ''Center Circuit''<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgA3EbFcGmkExp5uvKhErUqKWMTrEEYSCmzFsYBfRgJOIPNPzhCVHlXEUqSS9BUJieLAjsF7xKNB1d9fAWGOiocAspcFgaDA0MJ1ZiQjsMQndPALQRIRFTQIlLQJpoMDwHYmVIcLxCVUgrD11Y8ASQFBpxBvNJK0DtXT96xJYUETBWoPbA7HyaMDsiBGwEPDI1JLYAJ25CDN5+Qf4s+FXaOtSKBszxfbXtxsNsDdEzZDgDmjqwXYRTsUfIZ55exsK-3OTzWcUSQ0Sil4EL2TwAHuACIIwCgtm9kRpaI0BAA7AD2OKYAEscQAXBgrAA2JIA1gwACYAHQAzgBhIkrADGAFciaS2AB3QTCe7CjJ4AQ-IXpbgS9biyWC+WygS3TZmJV-Y5-ZoaoU6uC0EW6qBKq4mq7HKVGExoG20P7W7yTRIvV09a0fV1-L2ip0GETOj1mo6GN20YNCvxUOi8X2x03SkV-GWRsVGZVptUZ7OOzXQ3aOxMBGMivVGnYCK7l7xvATeFiJf3I111RvF8N1w59NPRgJ9ph5qP+Qdxhqj4u+ie+ocZmE9bPW-qDccojIIMRK5fcDcA9eboUDfd7oFKhd3BOLs1-bMJmHW14NNtlpVNFqJN+CQiKoVQ-5-mVrTBf5gKYb9i1A8DPzAn8KBRdNnxEH4EWwDBeCYBwQEIO0MKrbEQAAGTxABDJlmQACgAZVJYiVhZVkCRxBhOVJIkCQASjYBEhCQDDyBQD5BDlAQmhAABxBgmJWYjSTxOiKOo2j6MY5jWI4tggA noborder}} |
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| </WRAP> <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgA3EYw8DXsL0I0o4ClSRUUsYnWIIwkFNmLYw8+jASduA-n14o8eURErUJ0ZQmJ4sCGwVvEo0bVx4Gvw0+IvhYGgx1DAdWHht9LR1PMH1BECMTKjMJURpoMGMMYiVIILxCFUhJNxi9QxoUqLFzSVFojwqQJhoTON4Uv3rsTJowGyIEbAQ8MlVXbQAnWjgQHzbqzuQ4NhnFrwRCKWNTeDXaPCXZneSVyAONjopt+ZEU-fWjr2xsZofVmcIUEWurmsUqwA7okql4flQ2lA2CCIYd5j94RcQd8RFDUV5kTcqD4tlRsChoSC8SACS9CVjrmTrj5KUJfgJOjDwIyTiyiWyElUcfdmdcki98hyqUKSbhecTbuLsYldljubKTAqCVR5XMVeyNXTSWh2QKsa12vpDaS3hyTa9eJbTUyQdb7WbrhcZupCShEW8TO77udmf8rYQvYisbF4sQ3cG2AAPFqjFpumi-WigkAAcQYADsGFMAIYAFwA9lMADoAZwAFABlPM5kulgDCBYzWYAxnmAJZNgCU0dJNnj80g5AgIkRABkCzmACZlqs1uuN5sMNudjM9mPKKhMLa0MAibf0UcidsZvPZgA2J4A1gwZw321MWwBXdt5thAA noborder}} </WRAP> </well></collapse> <collapse id="openAni1" collapsed="false"> <button type="warning" collapse="openAni1">To view the simulations: click here!</button> </collapse></WRAP> | </WRAP> <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgA3EYw8DXsL0I0o4ClSRUUsYnWIIwkFNmLYw8+jASduA-n14o8eURErUJ0ZQmJ4sCGwVvEo0bVx4Gvw0+IvhYGgx1DAdWHht9LR1PMH1BECMTKjMJURpoMGMMYiVIILxCFUhJNxi9QxoUqLFzSVFojwqQJhoTON4Uv3rsTJowGyIEbAQ8MlVXbQAnWjgQHzbqzuQ4NhnFrwRCKWNTeDXaPCXZneSVyAONjopt+ZEU-fWjr2xsZofVmcIUEWurmsUqwA7okql4flQ2lA2CCIYd5j94RcQd8RFDUV5kTcqD4tlRsChoSC8SACS9CVjrmTrj5KUJfgJOjDwIyTiyiWyElUcfdmdcki98hyqUKSbhecTbuLsYldljubKTAqCVR5XMVeyNXTSWh2QKsa12vpDaS3hyTa9eJbTUyQdb7WbrhcZupCShEW8TO77udmf8rYQvYisbF4sQ3cG2AAPFqjFpumi-WigkAAcQYADsGFMAIYAFwA9lMADoAZwAFABlPM5kulgDCBYzWYAxnmAJZNgCU0dJNnj80g5AgIkRABkCzmACZlqs1uuN5sMNudjM9mPKKhMLa0MAibf0UcidsZvPZgA2J4A1gwZw321MWwBXdt5thAA noborder}} </WRAP> </well></collapse> <collapse id="openAni1" collapsed="false"> <button type="warning" collapse="openAni1">To view the simulations: click here!</button> </collapse></WRAP></WRAP> |
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| - 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: \\ \\ $\quad \quad p = m \cdot U \cdot I \cdot cos\varphi = P$ \\ <WRAP><imgcaption imageNo06 | Visible Representations of a balanced System></imgcaption>{{drawio>technicalTermispolySys3}}</WRAP> | - 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**. <WRAP> |
| | 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: \\ \\ |
| | $\quad \quad p = m \cdot U \cdot I \cdot \cos\varphi = P$ \\ |
| | <WRAP><imgcaption imageNo06 | Visible Representations of a balanced System></imgcaption>{{drawio>technicalTermispolySys3.svg}}</WRAP> |
| | </WRAP> |
| |
| | 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 sinusidal function shifted by $0°$, $120°$ and $240°$. | |
| <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgzCAMB0l5AWAnC1b0DYrTAJgIz64DskArAmPmWGRtSApCLnIxGQKYC0hAUACUo4XCG4IAHCOH5GzVs0XYyg4Uizip6mXJZt5sJJSRl8kXGCRUTEGCqHN8GZppBOlspnqUJo+CRLESOaIxBIYluTKfADujFLuLAFuzlB8AOZJCcRYuMn4SKKKfABO4Igg2rSi2o7wfAA2csxVZDVYirCQssUIrFmVWGAVtSmx5QhuRM1TReNMjtMLg2lxy-g5M4mQ48OTG1jrmzt98vmpeykeGI1bF21XwjBwPWlll4nVj3Vw82ySMzwqwm8RmAJO-TyUiBlyBjhuTWWMIecOwLyepRByNEqLMvziUNBhMIcwJ+WmxOOfDApBkqQBiVkPWptJahy0WCZwIB2kJtRZSiBAKBXJ2a2hokJcL4AHs3KJZMRGIRmHhOq8IBJmEqlVq3DqQFq+EA noborder}} </WRAP> | <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgzCAMB0l5AWAnC1b0DYrTAJgIz64DskArAmPmWGRtSApCLnIxGQKYC0hAUACUo4XCG4IAHCOH5GzVs0XYyg4Uizip6mXJZt5sJJSRl8kXGCRUTEGCqHN8GZppBOlspnqUJo+CRLESOaIxBIYluTKfADujFLuLAFuzlB8AOZJCcRYuMn4SKKKfABO4Igg2rSi2o7wfAA2csxVZDVYirCQssUIrFmVWGAVtSmx5QhuRM1TReNMjtMLg2lxy-g5M4mQ48OTG1jrmzt98vmpeykeGI1bF21XwjBwPWlll4nVj3Vw82ySMzwqwm8RmAJO-TyUiBlyBjhuTWWMIecOwLyepRByNEqLMvziUNBhMIcwJ+WmxOOfDApBkqQBiVkPWptJahy0WCZwIB2kJtRZSiBAKBXJ2a2hokJcL4AHs3KJZMRGIRmHhOq8IBJmEqlVq3DqQFq+EA noborder}} </WRAP> |
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| Balanced polyphase networks have the lowest wiring costs. | Balanced polyphase networks have the lowest wiring costs. |
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| Polyphase networks are also possible in ring interlinking (also called ring connection, see next simulation). | The structure of a polyphase network can also represent a ring interlinking (also called ring connection, see next simulation). |
| |
| The often used (and also here used) term of three-phase current or alternating current have not be taken literally: If no load is connected, there is no three-phase AC network, but a three-phase **voltage** network. | The often used (and also here used) term of three-phase current or alternating current has not to be taken literally: If no load is connected, there is no three-phase AC network, but a three-phase **voltage** network. |
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| <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgA3cDQkQm8L35RwFKkiopYxOsQRhIKbMWxg59GAk4hsKECjx5uvA0aoRK1cdCUJieLAjsF7xKNC1cUNcz2N8BczErcFgaDDUMJ1ZiQjs-TW0mGiMwPzATQxELcREaaDBDDGJFSHC8QmVICQ9tNMyjb19eIMsJEUSuDICQZNSE0TaRbAKaMDsiBGwEPDIVdy0AJ1o4HoRCKmFzeDZln0ks9YOzZDhdnWxeepXj7J3lwm9-I57ts73V4WxL-zfIc8eAmuL1MdzOAHd9D5-E1aGY2JDYSk+E9kf9IYC4SigQkERQNmsCbooHiXsTrsT0f5yX4tnjusJut0qUzePtBCTIezuuy6ZDrqCKWVOdThWThVTxQIQVkqezQezdFQ5aslRy1SzeGruqCqX1-Prvi08YafkaLsbIebrT9rv8AB69Ga9PQoSwQARPADiDAAdgxFgBDAAuAHtFgAdADOAAoAEoAS19AHNowBhUO+-0AY2DCczAEo2I7pm4mHpCGA9B6oSAADKhwMAE2j8aTqajGazDFz+d9ReWOqyxN17ngcHYQA noborder}} </WRAP> | <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgA3cDQkQm8L35RwFKkiopYxOsQRhIKbMWxg59GAk4hsKECjx5uvA0aoRK1cdCUJieLAjsF7xKNC1cUNcz2N8BczErcFgaDDUMJ1ZiQjs-TW0mGiMwPzATQxELcREaaDBDDGJFSHC8QmVICQ9tNMyjb19eIMsJEUSuDICQZNSE0TaRbAKaMDsiBGwEPDIVdy0AJ1o4HoRCKmFzeDZln0ks9YOzZDhdnWxeepXj7J3lwm9-I57ts73V4WxL-zfIc8eAmuL1MdzOAHd9D5-E1aGY2JDYSk+E9kf9IYC4SigQkERQNmsCbooHiXsTrsT0f5yX4tnjusJut0qUzePtBCTIezuuy6ZDrqCKWVOdThWThVTxQIQVkqezQezdFQ5aslRy1SzeGruqCqX1-Prvi08YafkaLsbIebrT9rv8AB69Ga9PQoSwQARPADiDAAdgxFgBDAAuAHtFgAdADOAAoAEoAS19AHNowBhUO+-0AY2DCczAEo2I7pm4mHpCGA9B6oSAADKhwMAE2j8aTqajGazDFz+d9ReWOqyxN17ngcHYQA noborder}} </WRAP> |
| ==== 7.2.2 Three-Phase System ==== | ==== 7.2.2 Three-Phase System ==== |
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| 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: | See also: [[https://de.mathworks.com/videos/series/what-is-3-phase-power.html|MATHWORKS Onramp Video: What is 3-phase power?]] |
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| | 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: |
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| * Simple three-phase machines can be used for generation. | * Simple three-phase machines can be used for generation. |
| * Rotary field machines (e.g. synchronous motors or induction motors) can also be simply connected to as load, converting the electrical energy into mechanical energy. | * Rotary field machines (e.g. synchronous motors or induction motors) can also be simply connected to a load, converting the 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. |
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| 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. |
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| For this the three-phase system will be separated into three parts: | For this the three-phase system will be separated into three parts: |
| - Loads | - Loads |
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| <WRAP> <imgcaption imageNo09 | Three-Phase System></imgcaption> {{drawio>ThreePhaseSystem}} </WRAP> | <WRAP> <imgcaption imageNo09 | Three-Phase System></imgcaption> {{drawio>ThreePhaseSystem.svg}} </WRAP> |
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| === Three-phase generator === | === Three-phase generator === |
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| * The windings of a three-phase generator are called $U$, $V$, $W$; the winding connections are correspondingly called: $U1$, $U2$, $V1$, $V2$, $W1$, $W2$ (see <imgref imageNo10>). \\ <WRAP> <imgcaption imageNo10 | Motor Terminal></imgcaption> {{drawio>Motorterminal}} </WRAP> | * 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>). \\ <WRAP> |
| * The typical **winding connections** in a three-phase generator are called **Delta connection** (for ring connection) and **Wye connection** (for star connection). This winding connections can simply be changed by reconnecting the motor terminal. In <imgref imageNo11> the two types of winding connections are shown. Often for Wye connection the star configuration is shown and for Delta connection the ring configuration. For Wye-connection is is also possible to have the star point on an separate terminal. \\ <WRAP> <imgcaption imageNo11 | Motor Terminal Setup for the two Connections></imgcaption> {{drawio>MotorterminalConnections}} </WRAP> | <imgcaption imageNo10 | Motor Terminal></imgcaption> {{drawio>Motorterminal.svg}} </WRAP> |
| * The **phase voltages** are given by: \\ \begin{align*} \color{RoyalBlue}{u_U} &\color{RoyalBlue}{= \sqrt{2} U \cdot cos(\omega t + \alpha - 0)} \\ \color{Green}{u_V} & \color{Green}{= \sqrt{2} U \cdot cos(\omega t + \alpha - {{2}\over{3}}\pi)} \\ \color{DarkOrchid}{u_W} & \color{DarkOrchid}{= \sqrt{2} U \cdot cos(\omega t + \alpha - {{4}\over{3}}\pi)} \\ \color{RoyalBlue}{u_U} + \color{Green}{u_V} + \color{DarkOrchid}{u_W} & = 0 \end{align*} | * The typical **winding connections** in a three-phase generator are called **Delta connection** (for ring connection) and **Wye connection** (for star connection). This winding connection can simply be changed by reconnecting the motor terminal. In <imgref imageNo11> the two types of winding connections are shown. For the Wye connection, is often the star configuration shown, and for the Delta connection the ring configuration. For the Wye connection, it is also possible to have the star point on a separate terminal. <WRAP> |
| | <imgcaption imageNo11 | Motor Terminal Setup for the two Connections></imgcaption> {{drawio>MotorterminalConnections.svg}} </WRAP> |
| | * The **phase voltages** are given by: <WRAP> |
| | \begin{align*} |
| | \color{RoyalBlue }{u_{\rm U}} & \color{RoyalBlue }{= \sqrt{2} U \cdot \cos(\omega t + \alpha - 0 )} \\ |
| | \color{Green }{u_{\rm V}} & \color{Green }{= \sqrt{2} U \cdot \cos(\omega t + \alpha - {{2}\over{3}}\pi)} \\ |
| | \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*}</WRAP> |
| * The **direction of rotation** is given by the arrangement of the windings: | * The **direction of rotation** is given by the arrangement of the windings: |
| * The three-phase generator with clockwise direction (CW, mathematically negative orientation) shows the phase sequence: $u_U$, $u_V$, $u_W$, Therefore, $u_V$ is $120°$ lagging to $u_U$. \\ 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. |
| * The three-phase generator with counter-clockwise direction (CCW, mathematically positive orientation) shows the phase sequence: $u_U$, $u_W$, $u_V$, Therefore, $u_V$ is $120°$ ahead of $u_U$. | * The three-phase generator with counter-clockwise direction (CCW, mathematically positive orientation) shows the phase sequence: $u_{\rm U}$, $u_{\rm W}$, $u_{\rm V}$, Therefore, $u_{\rm V}$ is $120°$ ahead of $u_{\rm U}$. |
| * The direction can be changed simply by switching two of the three phases (it does not need to be $u_V$ with $u_W$!). \\ <WRAP> <imgcaption imageNo12 | Direction of Rotation></imgcaption> {{drawio>DirectionOfRotation}} </WRAP> | * The direction can be changed simply by switching two of the three phases (it does not need to be $u_{\rm V}$ with $u_{\rm W}$!). \\ <WRAP> <imgcaption imageNo12 | Direction of Rotation></imgcaption> {{drawio>DirectionOfRotation.svg}} </WRAP> |
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| === Line Conductors === | === Line Conductors === |
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| The lines connected to the generator / load terminals $U1$, $V1$, $W1$ are often called $L1$, $L2$, $L3$ ($L$ 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 depent on the point of view (either onto a three-phase generator/load or the external conductors). | 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/load or the external conductors). |
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| * **String voltages/currents** $U_S$ (alternatively: winding voltages/currents, in German: Strangströme/Strangspannungen): \\ The string voltages/currents are the values measured on the windings - independent on the winding connection. \\ These voltages are shown in the previous images as $u_U$, $u_V$, $u_W$. | * **String voltages/currents** $U_\rm S$, $I_\rm S$ (alternatively: winding voltages/currents, in German: //Strangspannungen/Strangströme//): <WRAP> |
| * **Phase voltages/currents** $U_L$ (alternatively: phase-to-phase voltages/currents, line-to-line voltages/currents, external conductor voltages/currents, in German: Außenleiterströme/Außenleiterspannungen): \\ 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** $N$ | The string voltages/currents are the values measured on the windings - independent of the winding connection. \\ |
| * **Star voltages** (alternatively: phase-to-neutral voltages, line-to-neutral voltages, in German: Sternspannungen): the voltages of the lines can be also measured or used refering to the neutral potential. | These voltages are shown in the previous images as $u_\rm U$, $u_\rm V$, and $u_\rm W$. |
| <WRAP> <imgcaption imageNo13 | Example of an Three-Phase System></imgcaption> {{drawio>ExampleThreePhaseSystem}} </WRAP> | </WRAP> |
| | * **Phase voltages/currents** $U_\rm L$, $I_\rm L$ (alternatively: phase-to-phase voltages/currents, line-to-line voltages/currents, external conductor voltages/currents, in German: //Außenleiterspannungen/Außenleiterströme//): <WRAP> |
| | 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** $\rm N$ </WRAP> |
| | * **Star-voltages** $U_\rm Y$ (alternatively: phase-to-neutral voltages, line-to-neutral voltages, in German: //Sternspannungen//): the voltages of the lines can be also measured or used referring to the neutral potential. |
| | <WRAP> <imgcaption imageNo13 | Example of an Three-Phase System></imgcaption> {{drawio>ExampleThreePhaseSystem.svg}} </WRAP> |
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| The setup with $L1$, $L2$, $L3$ and $N$ is called **three-phase four-wire system**. When only a Delta connection without neutral is connected it is called a **three-phase three-wire system**. The star and phase voltages are given by | The setup with $\rm L1$, $\rm L2$, $\rm L3$ and $\rm N$ is called **three-phase four-wire system**. When only a Delta connection without neutral is connected it is called a **three-phase three-wire system**. The star and phase voltages are given by |
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| {{drawio>StarPhaseVoltageFormula}} | {{drawio>StarPhaseVoltageFormula.svg}} |
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| A phasor diagram can be constructed based on the given voltages. Be aware, that commonly the phasor is shown as vector (i.e. as an arrow starting from the origin or zero). In contrast to this, the voltage in a circuit is shown as a arrow pointing towards the zero potential. | A phasor diagram can be constructed based on the given voltages. Be aware, that commonly the phasor is shown as a vector (i.e. as an arrow starting from the origin or zero). In contrast to this, the voltage in a circuit is shown as an arrow pointing towards the zero potential. |
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| <WRAP> <imgcaption imageNo14 | Phasor-Diagram with Phase and Star Voltages></imgcaption> {{drawio>PhasorDiagramPhaseStar}} </WRAP> | <WRAP> <imgcaption imageNo14 | Phasor-Diagram with Phase- and Star-Voltages></imgcaption> {{drawio>PhasorDiagramPhaseStar.svg}} </WRAP> |
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| For the yellow triangle in <imgref imageNo14> applies: | For the yellow triangle in <imgref imageNo14> applies: |
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| \begin{align*} {{1}\over{2}} U_{31} &= U_{1N} \cdot cos 30° \\ &= U_{1N} \cdot {{\sqrt{3}}\over{2}} \\ \\ U_{31} &= \sqrt{3} \cdot U_{1N} \\ U_{L} &= \sqrt{3} \cdot U_{S} \\ \end{align*} | \begin{align*} |
| | {{1}\over{2}} U_{31} &= U_{\rm 1N} \cdot \cos 30° \\ |
| | &= U_{\rm 1N} \cdot {{\sqrt{3}}\over{2}} \\ \\ |
| | U_{31} &= \sqrt{3} \cdot U_{\rm 1N} |
| | \end{align*} |
| | \begin{align*} \boxed{U_{L} = \sqrt{3} \cdot U_\rm Y } \end{align*} |
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| 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 $400V$. The phase voltage is therefore ${{1}\over{\sqrt{3}}} \cdot 400V \approx 230V$. The following two simulations show these voltages. | 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 ${{1}\over{\sqrt{3}}} \cdot 400~\rm V \approx 230~\rm V$. The following two simulations show these voltages. |
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| ==== 7.2.3 Load and Power in Three-Phase Systems ==== | ==== 7.2.3 Load and Power in Three-Phase Systems ==== |
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| <WRAP> <imgcaption imageNo14 | CEE Connector></imgcaption> {{drawio>CEEConnector}} </WRAP> | <WRAP> <imgcaption imageNo14 | CEE Connector></imgcaption>{{drawio>CEEConnector.svg}} </WRAP> |
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| In order to understand the load in three-phase systems, the power at different types of loads are investigated: | To understand the load in three-phase systems, the power at different types of loads will be investigated: |
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| * Load in Wye connection with three-phase four-wire system | * Load in Wye connection with the three-phase four-wire system |
| * Load in Wye connection with three-phase three-wire system | * Load in Wye connection with the three-phase three-wire system |
| * Load in Delta connection | * Load in Delta connection |
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| \\ | \\ |
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| The four-wire wiring is one of the most common way to connect three-phase systems. The {{wp>IEC 60309}} connector (in German commonly knwon as CEE Stecker) is often used for this connection. The connector has an additional pin $PE$, besides the explained phases $L1$, $L2$, $L3$ and the neutral line $N$. **PE** stand for protective earth and is part of the {{wp>Earthing_system#TT_network|Earthing System}}. In short: PE provides a reference potential, which is everywhere available, due to the fact, that the power plant provide it to the soil ground. | The four-wire wiring is one of the most common ways to connect three-phase systems. The {{wp>IEC 60309}} connector (in German commonly known as //CEE Stecker//) is often used for this connection. The connector has an additional pin $\rm PE$, besides the explained phases $\rm L1$, $\rm L2$, $\rm L3$, and the neutral line $\rm N$. **PE** stand for protective earth and is part of the {{wp>Earthing_system#TT_network|Earthing System}}. In short: $\rm PE$ provides a reference potential, which is everywhere available, due to the fact, that the power plant provides it to the soil ground. |
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| For the four-wire system, the four pins $L1$, $L2$, $L3$ and the neutral line $N$ 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 star configuration, e.g. three-phase motors in Wye connection or three one phase loads, where each load is connected ot a single phase. | This is for example applied for loads in a star configuration, e.g. three-phase motors in Wye connection or three single-phase loads, where each load is connected to a single phase. |
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| <panel type="info" title="Example"> | <panel type="info" title="Example"> |
| The example in the following simulation shows a $50Hz$ / $231V$ three-phase __four-wire system__ with 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 __four-wire system__ with an unbalanced load in the Wye connection, with the given impedances. |
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| * Calculate the phase currents $I_1$, $I_2$, $I_3$, the neutral current $I_N$ | * Calculate the phase currents $I_1$, $I_2$, $I_3$, the neutral current $I_\rm N$ |
| * Calculate the true power, apparent power and reactive power. | * Calculate the true power, apparent power, and reactive power. |
| </panel> | </panel> |
| |
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| |
| <callout title="Voltages - Currents - True Power - Aparrent and Reactive Power"> | <callout title="Voltages - Currents - True Power - Apparent and Reactive Power"> |
| \\ | \\ |
| |
| The following "path": calculate voltages $\rightarrow$ calculate currents $\rightarrow$ calculate true power $\rightarrow$ calculate aparrent and reactive power is the best way to get to all wanted values. | The following "path": calculate voltages $\rightarrow$ calculate currents $\rightarrow$ calculate true power $\rightarrow$ calculate apparent and reactive power is the best way to get to all wanted values. |
| |
| - **Voltages**: It is obvious, that the phase voltages $U_L$ and star voltages $U_S$ are applied by the three-phase network independently of the load. | - **Voltages**: It is obvious, that the phase voltages ($\underline{U}_{12}$, $\underline{U}_{23}$, $\underline{U}_{31}$) and star-voltages ($\underline{U}_{1 \rm N}$, $\underline{U}_{2 \rm N}$, $\underline{U}_{3 \rm N}$) are applied by the three-phase network independently of the load. |
| - **Currents**: For the phase currents it applies that: $\underline{I}_1 + \underline{I}_2 + \underline{I}_3 = \underline{I}_N $ (be aware, that the voltage given in the simulation is only the RMS value without the phase shift). \\ The phase currents are given by the phase impedances and the star voltages: \\ \begin{align*} \underline{I}_1 = {{\underline{U}_{1N}}\over{\underline{Z}_1^\phantom{O}}} \quad , \quad \underline{I}_2 = {{\underline{U}_{2N}}\over{\underline{Z}_2^\phantom{O}}} \quad , \quad \underline{I}_3 = {{\underline{U}_{3N}}\over{\underline{Z}_3^\phantom{O}}} \\ \end{align*} | - **Currents**: For the phase currents it applies that: $\underline{I}_1 + \underline{I}_2 + \underline{I}_3 = \underline{I}_\rm N $ (be aware, that the voltage given in the simulation is only the RMS value without the phase shift). <WRAP> |
| - The **true power** $P_x$ for each string is given by the apparent power $S_x$ of the string times the indivitual phase angle $\varphi_x$ of the string: \\ \begin{align*} P_x &= S_x \cdot cos \varphi_x = U_S \cdot I_x \cdot cos \varphi_x \end{align*} \\ Therefore, the resulting true power for the full load is: \\ \begin{align*} P = U_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_S \cdot \left( I_1 \cdot cos (\varphi_{u,1} - \varphi_{i,1})+ I_2 \cdot cos (\varphi_{u,2} - \varphi_{i,2}) + I_3 \cdot cos (\varphi_{u,3} - \varphi_{i,3})\right) \end{align*} | 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: The apparent power can be either positive or negative. There is the possibility to chancel each other out in the calculation, even when there is an unbalanced impedance given. It is better to use a definition, which can consider all of phase apparent powers. \\ By DIN 40110 the **collective apparent power ** $S_\Sigma$ can be assumed as \\ \begin{align*} S_\Sigma &= \sqrt{\sum_x U_{x N}^2+ \underbrace{U_N^2}_{=0}} &\cdot & \sqrt{\sum_x I_{x}^2+ I_N^2} \\ &=\sqrt{3} \cdot U_S & \cdot & \sqrt{I_1^2 + I_2^2 + I_2^3 + I_N^2} \\ \end{align*} | \begin{align*} |
| - Given the collective apparent power the **collective reactive power** $Q_\Sigma$ ist given by \\ \begin{align*} Q_\Sigma = \sqrt{S_\Sigma^2-P^2} \end{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*}</WRAP> |
| | - The **true power** $P_x$ for each string is given by the apparent power $S_x$ of the string times the individual phase angle $\varphi_x$ of the string: <WRAP> |
| | \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,1} - \varphi_{i,1})+ I_2 \cdot \cos (\varphi_{u,2} - \varphi_{i,2}) + I_3 \cdot \cos (\varphi_{u,3} - \varphi_{i,3})\right) \end{align*} </WRAP> |
| | - 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! <WRAP> |
| | Even when considering all four lines a simple addition of all the apparent powers per phase would be problematic: The apparent power can be either positive or negative. There is the possibility to cancel each other out in the calculation, even when there is an unbalanced impedance given. It is better to use a definition, which can consider all of the phase apparent powers. \\ |
| | 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*}</WRAP> |
| | - Given the collective apparent power the **collective reactive power** $Q_\Sigma$ ist given by \\ \begin{align*} Q_\Sigma = \sqrt{S_\Sigma^2-P^2} \end{align*} |
| </callout> | </callout> |
| |
| <panel type="info" title="Example"> | <panel type="info" title="Example"> |
| In the example this leads to: | In the example, this leads to: |
| - The star voltages and the phase voltages are given as \begin{align*} U_S=& 231V = U_{1N} = U_{2N} = U_{3N} \\ U_L=\sqrt{3} \cdot 231V = & 400V = U_{12} = U_{23} = U_{31} \end{align*} \\ The phasors of the star voltages are given as: \\ {{drawio>FourWireStarPhaseVoltageFormula}} | - The star-voltages and the phase voltages are given as <WRAP> |
| - Based on the star voltages and the given impedances the phase currents are: \\ \begin{align*} \underline{I}_1 &= {{\underline{U}_{1N}}\over{\underline{Z}_1}} &= &{{231V}\over{10\Omega + j \cdot 2\pi\cdot 50Hz \cdot 1mH}} &= &+23.08 A &- j \cdot 0.72 A &= &23.08 A \quad &\angle -1.8° \\ \underline{I}_2 &= {{\underline{U}_{2N}}\over{\underline{Z}_2}} &= &{{231V \cdot \left( -{{1}\over{2}}-j{{1}\over{2}}\sqrt{3}\right)}\over{5\Omega + {{1}\over{j \cdot 2\pi\cdot 50Hz \cdot 100\mu F}}}} &= &+ 5.58 A &- j \cdot 4.50 A &= & 7.17 A \quad &\angle -38.9° \\ \underline{I}_3 &= {{\underline{U}_{3N}}\over{\underline{Z}_3}} &= &{{231V\cdot \left( -{{1}\over{2}}+j{{1}\over{2}}\sqrt{3}\right)}\over{20 \Omega}} &= &-5.78A &+ j \cdot 10.00A &= &11.55 A \quad &\angle -240.0° \\ \\ \underline{I}_N & = \underline{I}_1 + \underline{I}_2 + \underline{I}_3 & & & = &+22.88 A &+ j \cdot 4.77 A &= &23.37 A \quad &\angle +11.8° \end{align*} | \begin{align*} |
| - The true power is calculated by: \\ \begin{align*} P = 231V \cdot \big( 23.08 A \cdot cos (0° - (-1.8°))+ 7.17A \cdot cos (-120° - (-38.9°)) + 11.55 A \cdot cos (-240° - (-240°)\big) = 8.26 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 &=\sqrt{3} \cdot 231V & \cdot & \sqrt{(11.54A)^2 + (7.17A)^2 + (11.54A)^3 + (12.53A)^2} = 14.23 kVA \\ \end{align*} | 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 &=\sqrt{(8.72 kVA)^2 - (5.59 kW)^2} = 11.58kVar \\ \end{align*} | \end{align*} \\ |
| | The phasors of the star-voltages are given as: \\ {{drawio>FourWireStarPhaseVoltageFormula.svg}}</WRAP> |
| | - Based on the star-voltages and the given impedances the phase currents are: <WRAP> |
| | \begin{align*} |
| | \underline{I}_1 &= {{\underline{U}_{\rm 1N}}\over{\underline{Z}_1}} |
| | &= &{{231 ~\rm V}\over{10 ~\Omega + {\rm j} \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 1 {~\rm mH}}} |
| | &= &+23.08 {~\rm A} &- {\rm 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}}-{\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} &- {\rm 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}} |
| | &= &{{231{~\rm V}\cdot \left( -{{1}\over{2}}+{\rm j}{{1}\over{2}}\sqrt{3}\right)}\over{20 ~\Omega}} |
| | &= &-5.78{~\rm A} &+ {\rm j} \cdot 10.00{~\rm A} &= &11.55 {~\rm A} \quad &\angle -240.0° \\ \\ |
| | \underline{I}_{\rm N} |
| | &= \underline{I}_1 + \underline{I}_2 + \underline{I}_3 & & & = &+22.88 {~\rm A} &+ {\rm j} \cdot 4.77 {~\rm A} &= &23.37 {~\rm A} \quad &\angle +11.8° |
| | \end{align*} </WRAP> |
| | - The true power is calculated by: <WRAP> |
| | \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*} </WRAP> |
| | - The collective apparent power is: <WRAP> |
| | \begin{align*} |
| | S_\Sigma &=\sqrt{3} \cdot 231 {~\rm V} & \cdot & \sqrt{(23.09 {~\rm A})^2 + (7.17 {~\rm A})^2 + (11.55 {~\rm A})^3 + (23.37{~\rm A})^2} |
| | = 14.23 {~\rm kVA}\\ |
| | \end{align*}</WRAP> |
| | - The collective reactive power is: <WRAP> |
| | \begin{align*} |
| | Q_\Sigma &=\sqrt{(14.23 {~\rm kVA})^2 - (8.26 {~\rm kW})^2} |
| | = 11.58 {~\rm kVar} \\ |
| | \end{align*}</WRAP> |
| |
| <WRAP> <imgcaption imageNo15 | Load in Wye connection (Four-Wire System) ></imgcaption> {{drawio>PhasorWyeFourWire}} </WRAP> | <WRAP> <imgcaption imageNo15 | Load in Wye connection (Four-Wire System) ></imgcaption> {{drawio>PhasorWyeFourWire.svg}} </WRAP> |
| </panel> | </panel> |
| |
| \\ | \\ |
| |
| 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 star voltages $U_S$ are equal to the asymmetric load: $U_L = \sqrt{3}\cdot U_S$. | - The **phase-voltages** $U_\rm L$ and star-voltages $U_{\rm Y} = U_{\rm S}$ are equal to the asymmetric load: $U_{\rm L} = \sqrt{3}\cdot U_{\rm S}$. |
| - For equal impedances the absolute value of all **phase currents** $I_x$ are the same: $|\underline{I}_x|= |\underline{I}_S| = {{\underline{U}_S}\over{\underline{Z}_S^\phantom{O}}}$. 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** $I_x$ 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: $P = 3 \cdot U_S I_S \cdot cos \varphi$. Based on the line voltages $U_L$, the formula is $P = \sqrt{3} \cdot U_S I_S \cdot cos \varphi$ | - 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: $S_\Sigma = \sqrt{3}\cdot U_S \cdot \sqrt{3\cdot I_S^2} = 3 \cdot U_S I_S$. This corresponds to three times the apaarent 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: $Q_\Sigma = \sqrt{S_\Sigma^2 - P^2} = 3 \cdot U_S I_S \cdot sin (\varphi)$. | - The **reactive power** leads to: $Q_\Sigma = \sqrt{S_\Sigma^2 - P^2} = 3 \cdot U_{\rm S} I_{\rm S} \cdot \sin (\varphi)$. |
| </callout> | </callout> |
| |
| === Load in Wye connection with (Three-Wire System) === | === Load in Wye connection with (Three-Wire System) === |
| \\ | \\ |
| The three-wire system is used the four pins $L1$, $L2$, $L3$ and the neutral line $N$ are used for power transfer. | The three-wire system is used the four pins $\rm L1$, $\rm L2$, $\rm L3$, and the neutral line $\rm N$ is used for power transfer. |
| Sometimes the load do not provide a neutral connection, even when it is in Wye connection - the star point can for example only be provided for measurement with a thin cable. | Sometimes the load does not provide a neutral connection, even when it is in a Wye connection - the star point can for example only be provided for measurement with a thin cable. |
| Additionally, the neutral connection of a load in Wye connection can break an lead to a three wire system. | Additionally, the neutral connection of a load in a Wye connection can break and lead to a three-wire system. |
| |
| In the case of three wire system, only the potentials $L1$, $L2$ and $L3$ are provided and used for power transfer. | In the case of a three-wire system, only the potentials $\rm L1$, $\rm L2$, and $\rm L3$ are provided and used for power transfer. |
| |
| <panel type="info" title="Example"> | <panel type="info" title="Example"> |
| The example in the following simulation shows a $50Hz$ / $231V$ three-phase __three-wire system__ with 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 $I_N$ | * Calculate the phase currents $I_1$, $I_2$, $I_3$, the neutral current $I_\rm N$ |
| * Calculate the true power, apparent power and reactive power. | * Calculate the true power, apparent power, and reactive power. |
| |
| The following simulation has the same impedances in the load, but the load does not provide a neutral connection. | The following simulation has the same impedances in the load, but the load does not provide a neutral connection. |
| <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgzCAMB0l5AWAnC1b0DYrTAdiWBhrgBxIIYlgCMuuWATNSAnCAKyQcCmAtNdQBQADxC8M1JCAFTeyBNIYMW0kiAAKACwCGAZ24AdXQDUA9gBsALtoDm3XSLEll1JSFxdX7FdTUAZU20AEyMAZWsAJyMzK1t7QRsxDAQ1MAQFcRTwDC5cwQB3MXYsNIzWcHSoArFcZVKiksrIQQAHMXKaZTkueuZmZrbeYpAGMC7hhiQsPqrCoaRlUa7CLiWqwdrVDM2GLJmB9qmRki45I4ZINX3q+cWTw8Z75rn2Ba2H94PbkazvhnvroNXos2H8xtIIc0Lrgklk1pk1Bc1Ao-AxBNDYYj7giRsMUWB0ZAYTjfBlklccio-EIMTj6nSLioAHI3ZCMS4fUmzD7-U5gdh3PJzckVDL8upNQQAJUx73ERy5ng4PQYGGglAUuWw7GlsvhGHO4KVnHAqvVJE1YgYsHIYCQ7GokFGBEk7AgMB1MpxvKS5yeEJNYzVGqg7WgvhI+CdiFIGDtkG8HpuIq530Vybh4LBymeeuxwOOQpqi1+m3quaGjCOldF3N4my5AhcFrrE3BF1WRtZq3uPp9zQi0lcP08w8V0gHIzYaw7xzUXB1AGMpz3EWwxlqYPBIOhd+h2mAwNBIydHUhEETcOwFB3BIO+737hvQxdBOYhyXR3d59g4H1qDgSCkKweDsCeYGatUTbvM+6aFLOazPmsuYPiqPZFtB9SzuW1TYZUPo4fBT6VM+OF4HylLwmMiLgjMQhzNQhAjFmbY5oI5FiCscrUe8dGsgqLa8FxcFiIx0yCQ2LbNBxQkCoWnH-PJfEMUxPrdPJ0keB+ppfrWfEyTx9LinpEL0dpsHDv887sVp2GMnhCj6bZT72cRjmmY4cgUIofJMIoECOWoWh6IYugAMIAK4RBE3AAHaWA4oiOjCrgQGAJDeNQlA+P4gQhLoACWsVGAA6gAnqFYWmLFsXcIulgFdVRgGAYsVFQAKpoMV8MF+hGJ13W8CVBUxWEZW6JY3AALYtbF+QFZYmhhJYERFTYRiRdFcUJZ5rB9Bg3hCQq2WBSAACSnl2pluAQEJJBXPQPiOZdBAjG4d2MAw3iOTeL1SEwYrkCMtA5edL0wqqh12osJ2qGDohHcwDA3ZxLbI1gP3gODprEtD4BgFwp0XQjdrMIeUMkGT6VPSALII9Q3hIaaAXSAoYChJddSMVaPRZaDACql2MHJvC7Li31w4L9P-a8VojBgyiYwwHMI-ZjqccDLPUI5KucRDkC3V9U4Y5L3bgMO6n9jcBazmmUnW28uxqNmVQI2B3gXBk7Aoxip1Sw0iICEUbrAwomOuJ5xSIt7wcQKqhOs8xkcZbipze3HId+7tuA3pM7Sy2smM0LtSBI+9CD4G9Eu+CAgu6KJj30phTQQgAZto5j6J52upO9YmmtXagc3MHSVM3kEMcO9JlpKcyIZU-CN7PVrgp0K+Cq7Vp5xchsgxcJs1-7ouWer-D2prNPUHTVpB19siSDeCs0ww1+i4pX1dEwwuD7XnlKGz3Nj5cwPmoI+GJnxnxVFrNmLI5iGQXqqNmy8hLv2xDOJ4NweKpkQZ8TB78sxO2YmxOe7ZsT9ytnPTMXRyFdjnskOU-cRKizhNWQhhErT0MIaLehOFBg4OwZwvYkJWhWioaIxECphF8NXhbee7l+g3BwfSNhyCVEZDkZvUWCo+41y8KDJk3BLD5FMBEAA1stbQUQTAWGsHYBwGw6iCVoCUFs1w5hlgkhKCe69wCCQ0c0VWSBUhEFEWlA0oNaoRRWh3TylJCCZSkIQZEcNwiWKMOoUwRVLA3Gcb49Rq97bS0Zo9ReJRK6Y1CK-JgdRwmi3XJsLOphpAs1WIgSMIBki5HDKGCAw4uBSC4L1UKMRbHcEEE0iWrSUhYE6dgGYEAsBcB-KkqxIy4gOCaQgSZLAdwwlmQ6aAiwQDIkZiwd0ZzwAsB-FkCADpQx3IgEMjaUUYrxQcEAA noborder}} </WRAP> | <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgzCAMB0l5AWAnC1b0DYrTAdiWBhrgBxIIYlgCMuuWATNSAnCAKyQcCmAtNdQBQADxC8M1JCAFTeyBNIYMW0kiAAKACwCGAZ24AdXQDUA9gBsALtoDm3XSLEll1JSFxdX7FdTUAZU20AEyMAZWsAJyMzK1t7QRsxDAQ1MAQFcRTwDC5cwQB3MXYsNIzWcHSoArFcZVKiksrIQQAHMXKaZTkueuZmZrbeYpAGMC7hhiQsPqrCoaRlUa7CLiWqwdrVDM2GLJmB9qmRki45I4ZINX3q+cWTw8Z75rn2Ba2H94PbkazvhnvroNXos2H8xtIIc0Lrgklk1pk1Bc1Ao-AxBNDYYj7giRsMUWB0ZAYTjfBlklccio-EIMTj6nSLioAHI3ZCMS4fUmzD7-U5gdh3PJzckVDL8upNQQAJUx73ERy5ng4PQYGGglAUuWw7GlsvhGHO4KVnHAqvVJE1YgYsHIYCQ7GokFGBEk7AgMB1MpxvKS5yeEJNYzVGqg7WgvhI+CdiFIGDtkG8HpuIq530Vybh4LBymeeuxwOOQpqi1+m3quaGjCOldF3N4my5AhcFrrE3BF1WRtZq3uPp9zQi0lcP08w8V0gHIzYaw7xzUXB1AGMpz3EWwxlqYPBIOhd+h2mAwNBIydHUhEETcOwFB3BIO+737hvQxdBOYhyXR3d59g4H1qDgSCkKweDsCeYGatUTbvM+6aFLOazPmsuYPiqPZFtB9SzuW1TYZUPo4fBT6VM+OF4HylLwmMiLgjMQhzNQhAjFmbY5oI5FiCscrUe8dGsgqLa8FxcFiIx0yCQ2LbNBxQkCoWnH-PJfEMUxPrdPJ0keB+ppfrWfEyTx9LinpEL0dpsHDv887sVp2GMnhCj6bZT72cRjmmY4cgUIofJMIoECOWoWh6IYugAMIAK4RBE3AAHaWA4oiOjCrgQGAJDeNQlA+P4gQhLoACWsVGAA6gAnqFYWmLFsXcIulgFdVRgGAYsVFQAKpoMV8MF+hGJ13W8CVBUxWEZW6JY3AALYtbF+QFZYmhhJYERFTYRiRdFcUJZ5rB9Bg3hCQq2WBSAACSnl2pluAQEJJBXPQPiOZdBAjG4d2MAw3iOTeL1SEwYrkCMtA5edL0wqqh12osJ2qGDohHcwDA3ZxLbI1gP3gODprEtD4BgFwp0XQjdrMIeUMkGT6VPSALII9Q3hIaaAXSAoYChJddSMVaPRZaDACql2MHJvC7Li31w4L9P-a8VojBgyiYwwHMI-ZjqccDLPUI5KucRDkC3V9U4Y5L3bgMO6n9jcBazmmUnW28uxqNmVQI2B3gXBk7Aoxip1Sw0iICEUbrAwomOuJ5xSIt7wcQKqhOs8xkcZbipze3HId+7tuA3pM7Sy2smM0LtSBI+9CD4G9Eu+CAgu6KJj30phTQQgAZto5j6J52upO9YmmtXagc3MHSVM3kEMcO9JlpKcyIZU-CN7PVrgp0K+Cq7Vp5xchsgxcJs1-7ouWer-D2prNPUHTVpB19siSDeCs0ww1+i4pX1dEwwuD7XnlKGz3Nj5cwPmoI+GJnxnxVFrNmLI5iGQXqqNmy8hLv2xDOJ4NweKpkQZ8TB78sxO2YmxOe7ZsT9ytnPTMXRyFdjnskOU-cRKizhNWQhhErT0MIaLehOFBg4OwZwvYkJWhWioaIxECphF8NXhbee7l+g3BwfSNhyCVEZDkZvUWCo+41y8KDJk3BLD5FMBEAA1stbQUQTAWGsHYBwGw6iCVoCUFs1w5hlgkhKCe69wCCQ0c0VWSBUhEFEWlA0oNaoRRWh3TylJCCZSkIQZEcNwiWKMOoUwRVLA3Gcb49Rq97bS0Zo9ReJRK6Y1CK-JgdRwmi3XJsLOphpAs1WIgSMIBki5HDKGCAw4uBSC4L1UKMRbHcEEE0iWrSUhYE6dgGYEAsBcB-KkqxIy4gOCaQgSZLAdwwlmQ6aAiwQDIkZiwd0ZzwAsB-FkCADpQx3IgEMjaUUYrxQcEAA noborder}} </WRAP> |
| |
| <callout title="Voltages - Currents - True Power - Aparrent and Reactive Power"> | <callout title="Voltages - Currents - True Power - Apparent and Reactive Power"> |
| \\ | \\ |
| |
| The simulation differs in the following: | The simulation differs in the following: |
| * The network star voltages $\underline{U}_{1N}$, $\underline{U}_{2N}$, $\underline{U}_{3N}$ related to the neutral potential $\underline{U}_N$, and the load star voltages $\underline{U}_{1S}$, $\underline{U}_{2S}$, $\underline{U}_{3S}$ related to the star potential of the load are separated. | * The network star-voltages $\underline{U}_{\rm 1N}$, $\underline{U}_{\rm 2N}$, $\underline{U}_{\rm 3N}$ related to the neutral potential $\underline{U}_\rm N$, and the load string voltages (= load star-voltages) $\underline{U}_{\rm 1S}$, $\underline{U}_{\rm 2S}$, $\underline{U}_{\rm 3S}$ related to the star potential of the load are separated. |
| * The star point voltage $\underline{U}_{KN}$ is shown. | * The star point voltage $\underline{U}_{\rm SN}$ from the potential of node $\rm S$tar point to $\rm N$eutral is shown. |
| * With the switch $S$, the star potential can short-circuited to the neutral potential; so set $\underline{U}_{KN}=0$. 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 "path": calculate voltages $\rightarrow$ calculate currents $\rightarrow$ calculate true power $\rightarrow$ calculate aparrent and reactive power is the best way to get to all wanted values. | Also here, the "path": calculate voltages $\rightarrow$ calculate currents $\rightarrow$ calculate true power $\rightarrow$ calculate apparent and reactive power is the best way to get to all wanted values. |
| - **Voltages**: Here, only the the phase voltages $U_L$ are applied by the three-phase net, independently of the load. The star voltages of the load $\underline{U}_{xS}$ are not given by the network anymore, since the neutral potential is not provided. The network star voltages and the load star voltages can be connected in the following way: The calculation of the star voltage $\underline{U}_{SN}$ is explained after investigating the currents. \\ \begin{align*} \underline{U}_{1S} &= \underline{U}_{1N} - \underline{U}_{SN} \\ \underline{U}_{2S} &= \underline{U}_{2N} - \underline{U}_{SN} \\ \underline{U}_{3S} &= \underline{U}_{3N} - \underline{U}_{SN} \\ \end{align*} | - **Voltages**: Here, only the phase voltages ($\underline{U}_{12}$, $\underline{U}_{23}$, $\underline{U}_{31}$) are applied by the three-phase net, independently of the load. The star-voltages of the load $\underline{U}_{x \rm S}$ are not given by the network anymore since the neutral potential is not provided. The network star-voltages and the load star-voltages can be connected in the following way: The calculation of the star-voltage $\underline{U}_{\rm SN}$ is explained after investigating the currents. <WRAP> |
| - **Currents**: For the phase currents it applies that: $\underline{I}_1 + \underline{I}_2 + \underline{I}_3 = 0$ (again, voltages given in the simulation are only the RMS value without the phase shift). \\ The phase currents are given by the phase impedances and the star voltages: \\ \begin{align*} \underline{I}_1 = {{\underline{U}_{1S}}\over{\underline{Z}_1^\phantom{O}}} \quad , \quad \underline{I}_2 = {{\underline{U}_{2S}}\over{\underline{Z}_2^\phantom{O}}} \quad , \quad \underline{I}_3 = {{\underline{U}_{3S}}\over{\underline{Z}_3^\phantom{O}}} \end{align*} \\ In order to get $\underline{U}_{SN}$, one has to combine the individual formulas for $\underline{I}_x$, $\underline{U}_{xS}$ and that the $\sum_x \underline{I}_x =0$. This leads to \\ \begin{align*} \underline{U}_{SN} = {{\sum_x \left( \Large{{{1}\over{\underline{Z}_x^\phantom{O}}}} \cdot \normalsize{\underline{U}_{xN}} \right) }\over{\sum_x \left( \Large{{{1}\over{\underline{Z}_x^\phantom{O}}}} \right) }} \end{align*} | \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_S \cdot I_x \cdot cos \varphi_x \end{align*} \\ Also here, the resulting true power for the full load is (with $U_S$ as the RMS value of the network star voltage): \\ \begin{align*} P &= U_S \cdot ( I_1 \cdot cos \varphi_1 + I_2 \cdot cos \varphi_2 + I_3 \cdot cos \varphi_3) \\ &= U_S \cdot \left( I_1 \cdot cos (\varphi_{u,1} - \varphi_{i,1})+ I_2 \cdot cos (\varphi_{u,2} - \varphi_{i,2}) + I_3 \cdot cos (\varphi_{u,3} - \varphi_{i,3})\right) \end{align*} | \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 $\underline{S}_x = \underline{U}_{xS} \cdot \underline{I}_x^*$. 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 \left( \underline{U}_{xS} \cdot \underline{I}_x^* \right) \end{align*} \\ In order to simplify the calculation, it would be better to have a formula based on the network star voltages: \\ \begin{align*} \underline{S} &= \sum_x \left( \underline{U}_{xN} \cdot \underline{I}_x^* \right) + \underline{U}_{xN} \cdot \underbrace{\underline{I}_N}_{=0} \\ &= \sum_x \left( \underline{U}_{xN} \cdot \underline{I}_x^* \right) \\ \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^* \\ &= \underline{U}_{12} \cdot \underline{I}_1^* + \underline{U}_{32} \cdot \underline{I}_3^* \\ &= \underline{U}_{21} \cdot \underline{I}_2^* + \underline{U}_{31} \cdot \underline{I}_3^* \end{align*} \\ For the phase voltages it applies that: $\underline{U}_{12} = - \underline{U}_{21}$, $\underline{U}_{23} = - \underline{U}_{32}$, $\underline{U}_{31} = - \underline{U}_{13}$. For the **collective apparent power** $S_\Sigma$ the formula of differs in the definition (in order to consider the reactive part more for an unbalanced generator). Since we consider here a given balanced network the definition lead to a similar result as based on the four-wire connection: \\ \begin{align*} S_\Sigma &= \sqrt{ {{1}\over{2}} \cdot (U_{12}^2 + U_{23}^2 + U_{31}^2) } \cdot \sqrt{\sum_x I_x^2} &= \sqrt{3} U_S \cdot \sqrt{\sum_x I_x^2} \end{align*} | \underline{U}_{\rm 2S} &= \underline{U}_{\rm 2N} - \underline{U}_{\rm SN} \\ |
| - The abolute **reactive power** $Q$ can be calulated by the apparent power: \\ \begin{align*} j\cdot Q &= \underline{S} - P \end{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*} | \underline{U}_{\rm 3S} &= \underline{U}_{\rm 3N} - \underline{U}_{\rm SN} \\ |
| | \end{align*}</WRAP> |
| | - **Currents**: For the phase currents it applies that: $\underline{I}_1 + \underline{I}_2 + \underline{I}_3 = 0$ (again, voltages given in the simulation are only the RMS value without the phase shift). \\ The phase currents are given by the phase impedances and the star-voltages: <WRAP> |
| | \begin{align*} |
| | \underline{I}_1 = {{\underline{U}_{\rm 1S}}\over{\underline{Z}_1^\phantom{O}}} \quad , \quad \underline{I}_2 |
| | = {{\underline{U}_{\rm 2S}}\over{\underline{Z}_2^\phantom{O}}} \quad , \quad \underline{I}_3 |
| | = {{\underline{U}_{\rm 3S}}\over{\underline{Z}_3^\phantom{O}}} |
| | \end{align*} \\ |
| | To get $\underline{U}_{\rm SN}$, one has to combine the individual formulas for $\underline{I}_x$, $\underline{U}_{x \rm S}$ and that the $\sum_x \underline{I}_x =0$. This leads to \\ |
| | \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*}</WRAP> |
| | - Also here, the **true power** $P_x$ for each string is given by: <WRAP> |
| | \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 + I_2 \cdot \cos \varphi_2 + I_3 \cdot \cos \varphi_3) \\ |
| | &= U_{\rm S} \cdot \left( I_1 \cdot \cos (\varphi_{u,1} - \varphi_{i,1}) + I_2 \cdot \cos (\varphi_{u,2} - \varphi_{i,2}) + I_3 \cdot \cos (\varphi_{u,3} - \varphi_{i,3})\right) |
| | \end{align*}</WRAP> |
| | - 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 <WRAP> |
| | \begin{align*} |
| | \underline{S} &= P + {\rm j}\cdot Q = \sum_x \underline{S}_x = \sum_x \left( \underline{U}_{x \rm S} \cdot \underline{I}_x^* \right) \end{align*} \\ |
| | In order to simplify the calculation, it would be better to have a formula based on the network star-voltages: \\ |
| | \begin{align*} |
| | \underline{S} &= \sum_x \left( \underline{U}_{x \rm N} \cdot \underline{I}_x^* \right) + \underline{U}_{x \rm N} \cdot \underbrace{\underline{I}_{\rm N}}_{=0} \\ |
| | &= \sum_x \left( \underline{U}_{x \rm N} \cdot \underline{I}_x^* \right) \\ |
| | \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^* \\ |
| | &= \underline{U}_{12} \cdot \underline{I}_1^* + \underline{U}_{32} \cdot \underline{I}_3^* \\ |
| | &= \underline{U}_{21} \cdot \underline{I}_2^* + \underline{U}_{31} \cdot \underline{I}_3^* |
| | \end{align*} \\ |
| | For the phase voltages it applies that: $\underline{U}_{12} = - \underline{U}_{21}$, $\underline{U}_{23} = - \underline{U}_{32}$, $\underline{U}_{31} = - \underline{U}_{13}$. For the **collective apparent power,** $S_\Sigma$ the formula differs in the definition (to consider the reactive part more for an unbalanced generator). Since we consider here a given balanced network the definition leads to a similar result as based on the four-wire connection: \\ |
| | \begin{align*} |
| | S_\Sigma &= \sqrt{ {{1}\over{2}} \cdot (U_{12}^2 + U_{23}^2 + U_{31}^2) } \cdot \sqrt{\sum_x I_x^2} &= \sqrt{3} U_{\rm S} \cdot \sqrt{\sum_x I_x^2} |
| | \end{align*} </WRAP> |
| | - The abolute **reactive power** $Q$ can be calulated by the apparent power: <WRAP> |
| | \begin{align*} {\rm j}\cdot Q &= \underline{S} - P \end{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*} </WRAP> |
| </callout> | </callout> |
| |
| <panel type="info" title="Example"> | <panel type="info" title="Example"> |
| In the example this leads to: | In the example, this leads to: |
| - The phase voltages are given as \begin{align*} U_L=\sqrt{3} \cdot 231V = & 400V = U_{12} = U_{23} = U_{31} \end{align*} \\ The phasors of the star voltages of the network are again given as: \\ {{drawio>FourWireStarPhaseVoltageFormula}} | - The phase voltages are given as \begin{align*} U_{\rm L}=\sqrt{3} \cdot 231 {~\rm V} = & 400 {~\rm V} = U_{12} = U_{23} = U_{31} \end{align*} \\ The phasors of the star-voltages of the network are again given as \\ {{drawio>FourWireStarPhaseVoltageFormula.svg}} |
| - Based on the star voltages of the network and the given impedances the star voltage $\underline{U}_{SN}$ of the load can be calculated with : \\ \begin{align*} \underline{U}_{SN} = {{\sum_x \left( \Large{{{1}\over{\underline{Z}_x^\phantom{O}}}} \cdot \normalsize{\underline{U}_{xN}} \right) }\over{\sum_x \left( \Large{{{1}\over{\underline{Z}_x^\phantom{O}}}} \right) }} \end{align*} \\ Once investigating the numerator $\sum_x \big( {{1}\over{\underline{Z}_x^\phantom{O}}} \cdot \underline{U}_{xN} \big)$, 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: $22.88A + j \cdot 4.77A$ (see calculation for the four-wire system). \\ The denominator is: \\ \begin{align*} \sum_x {{1}\over{\underline{Z}_x^\phantom{O}}} &= {{1}\over{10\Omega + j \cdot 2\pi\cdot 50Hz \cdot 1mH }} + {{1}\over{5\Omega + {{1}\over{j \cdot 2\pi\cdot 50Hz \cdot 100\mu F }} }}+ {{1}\over{20\Omega }} \\ \\ &= 0.1547 \cdot 1/\Omega + j \cdot 0.02752 \cdot 1/\Omega \end{align*} \\ The star voltage $\underline{U}_{SN}$ of the load is: \begin{align*} \underline{U}_{SN} &= {{22.88A + j \cdot 4.77A}\over{0.1547 \cdot 1/\Omega + j \cdot 0.0275 \cdot 1/\Omega}} \\ \\ &= 148.7V + j \cdot 4.41 V \end{align*} \\ Given this star voltage $\underline{U}_{SN}$ of the load, the phase currents are: \\ \begin{align*} \underline{I}_1 &= {{\underline{U}_{1N} - \underline{U}_{SN}}\over{\underline{Z}_1^\phantom{O}}} & = & {{231V - 148.7V - j \cdot 4.41 V}\over{10\Omega + j \cdot 2\pi\cdot 50Hz \cdot 1mH }} & = & +8.21A - j \cdot 0.70A &=& 8.24 A \quad \angle -4.9° \\ \underline{I}_2 &= {{\underline{U}_{2N} - \underline{U}_{SN}}\over{\underline{Z}_2^\phantom{O}}} & = & {{231V \cdot \left( -{{1}\over{2}}-j{{1}\over{2}}\sqrt{3}\right) - 148.7V - j \cdot 4.41 V}\over{5\Omega + {{1}\over{j \cdot 2\pi\cdot 50Hz \cdot 100\mu F }}}} & = & +5.00A + j \cdot 9.08A & =& 10.36 A \quad \angle -61.2° \\ \underline{I}_3 &= {{\underline{U}_{3N} - \underline{U}_{SN}}\over{\underline{Z}_3^\phantom{O}}} & = & {{231V \cdot \left( -{{1}\over{2}}+j{{1}\over{2}}\sqrt{3}\right) - 148.7V - j \cdot 4.41 V}\over{20\Omega }} &= & -13.21A + j \cdot 9.78A & =& 16.44A \quad \angle +143.5° \end{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: <WRAP> |
| - The true power is calculated by: \\ \begin{align*} P = 231V \cdot \big( 8.24 A \cdot cos (0° - (-4.9°))+ 10.36A \cdot cos (-120° - (-61.2°)) + 16.44 A \cdot cos (-240° - (+143.5°)\big) = 6.62 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^* &=& 400V \cdot (- e^{-j \cdot 7/6 \pi} \cdot (8.21 A + j \cdot 0.70 A ) + e^{- j \cdot 3/6 \pi} \cdot (5.00A - j \cdot 9.08A) ) &= 6.62 kW - j \cdot 3.40 kVAr \\ &= \underline{U}_{12} \cdot \underline{I}_1^* + \underline{U}_{32} \cdot \underline{I}_3^* &=& 400V \cdot (e^{j \cdot 1/6 \pi} \cdot (8.21A + j \cdot 0.70A) - e^{-j \cdot 3/6 \pi} \cdot (-13.21A -j \cdot 9.78A)) &= 6.62 kW - j \cdot 3.40 kVAr \\ &= \underline{U}_{21} \cdot \underline{I}_2^* + \underline{U}_{31} \cdot \underline{I}_3^* &=& 400V \cdot (- e^{j \cdot 1/6 \pi} \cdot (5.00A - j \cdot 9.08A) + e^{- j \cdot 7/6 \pi} \cdot (-13.21A - j \cdot 9.78A)) &= 6.62 kW - j \cdot 3.40 kVAr \\ & = 7.44 kVA \quad \angle -27.2°\end{align*} \\ The collective apparent power is: \\ \begin{align*} S_\Sigma &= \sqrt{3} U_S \cdot \sqrt{\sum_x I_x^2} \\ &= \sqrt{3} \cdot 231V \cdot \sqrt{(8.24A)^2+(10.36A)^2+(16.44A)^2} = 8.45 kVA \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 &= \underline{S} - P = -3.40kVAr \\ \end{align*} \\ The collective reactive power is: \\ \begin{align*} Q_\Sigma &=\sqrt{(8.44 kVA)^2 - (6.62 kW)^2} = 5.24kVAr \\ \end{align*} | \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 {{1}\over{\underline{Z}_x^\phantom{O}}} &= {{1}\over{10~\Omega + {\rm j} \cdot 2\pi\cdot 50{~\rm Hz} \cdot 1 {~\rm mH} }} |
| | + {{1}\over{5~\Omega + {{1}\over{{\rm j} \cdot 2\pi\cdot 50{~\rm Hz} \cdot 100 {~\rm µF} }} }} |
| | + {{1}\over{20~\Omega }} \\ \\ |
| | &= 0.1547 ~1/\Omega + {\rm j} \cdot 0.02752 ~1/\Omega \end{align*} \\ |
| | 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 {~\rm V} + {\rm j} \cdot 4.41 {~\rm V} \end{align*} \\ |
| | 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}}} |
| | & = & {{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} - {\rm j} \cdot 0.70{~\rm A} &=& 8.24 {~\rm A} \quad \angle -4.9° \\ |
| | \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}}-{\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} + {\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}}} |
| | & = & {{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} + {\rm j} \cdot 9.78{~\rm A} & =& 16.44{~\rm A} \quad \angle +143.5° \end{align*} \\ </WRAP> |
| | - 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: <WRAP> |
| | \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 ( {\rm e}^{{\rm j} \cdot 1/6 \pi} \cdot (8.21{~\rm A} + {\rm j}\cdot 0.70{~\rm A}) - {\rm e}^{-{\rm j} \cdot 3/6 \pi} \cdot (-13.21{~\rm A} -{\rm j} \cdot 9.78{~\rm A})) |
| | &= 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^* |
| | &=& 400{~\rm V} \cdot (- {\rm e}^{{\rm j} \cdot 1/6 \pi} \cdot (5.00{~\rm A} - {\rm j}\cdot 9.08{~\rm A}) + {\rm e}^{- {\rm j} \cdot 7/6 \pi} \cdot (-13.21{~\rm A} - {\rm j} \cdot 9.78{~\rm A})) |
| | &= 6.62 {~\rm kW} - {\rm j} \cdot 3.40 {~\rm kVAr} \\ |
| | & = 7.44 {~\rm kVA} \quad \angle -27.2° |
| | \end{align*} \\ |
| | The collective apparent power is: \\ |
| | \begin{align*} |
| | S_\Sigma &= \sqrt{3} U_{\rm S} \cdot \sqrt{\sum_x I_x^2} \\ |
| | &= \sqrt{3} \cdot 231{~\rm V} \cdot \sqrt{(8.24{~\rm A})^2+(10.36{~\rm A})^2+(16.44A)^2} |
| | = 8.45 {~\rm kVA} \end{align*} </WRAP> |
| | - The reactive power is: <WRAP> |
| | \begin{align*} Q &= -{\rm j} \cdot (\underline{S} - P) = -3.40{~\rm kVAr} \\ |
| | \end{align*} \\ |
| | The collective reactive power is: \\ |
| | \begin{align*} Q_\Sigma &=\sqrt{(8.44 {~\rm kVA})^2 - (6.62 {~\rm kW})^2} = 5.24{~\rm kVAr} \\ |
| | \end{align*} </WRAP> |
| |
| <WRAP> <imgcaption imageNo16 | Load in Wye connection (Three-Wire System) ></imgcaption> {{drawio>PhasorWyeThreeWire}} </WRAP> | <WRAP> <imgcaption imageNo16 | Load in Wye connection (Three-Wire System) ></imgcaption> {{drawio>PhasorWyeThreeWire.svg}} </WRAP> |
| </panel> | </panel> |
| |
| <callout title="For Symmetric Load"> | <callout title="For Symmetric Load"> |
| \\ | \\ |
| The case of a symmetric load in a three-wire system equals the symmetric load in a four-wire system, since the symmetric four-wire system also does not show a current on the neutral line. | The case of a symmetric load in a three-wire system equals the symmetric load in a four-wire system since the symmetric four-wire system also does not show a current on the neutral line. |
| </callout> | </callout> |
| |
| \\ | \\ |
| 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 triangluar 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 example given for motors, when higher torque is needed. In this connection each string sees $U_L = \sqrt{3}\cdot U_S$ 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 delct connection, also only the potentials $L1$, $L2$ and $L3$ 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. |
| |
| <panel type="info" title="Example"> | <panel type="info" title="Example"> |
| The example in the following simulation shows a $50Hz$ / $231V$ three-phase __three-wire system__ with unbalanced load in delta 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 delta connection, with the given impedances. |
| |
| * Calculate the phase currents $I_1$, $I_2$, $I_3$, the neutral current $I_N$ | * Calculate the phase currents $I_1$, $I_2$, $I_3$, the neutral current $I_\rm N$ |
| * Calculate the true power, apparent power and reactive power. | * Calculate the true power, apparent power, and reactive power. |
| |
| The following simulation has the same impedances in the load, but the load does not provide a neutral connection. | The following simulation has the same impedances in the load, but the load does not provide a neutral connection. |
| <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgzCAMB0l5AWAnC1b0DYrTAdiWBhrgBxIIYlgCMuuWATNSAnCAKyQcCmAtNdQBQADxC8M1JCAFTeyBNIYMW0kiAAKACwCGAZ24AdXQDUA9gBsALtoDm3XSLEMkWakrHUyi9is8gAytYATkZmVrb2gjZiGAhqYAgK4nHgGFzpggDuTikJSQxgynlQWU6x4Ik58ZWQggAOToXgbo1FlczMtQ28DCkFyj19LtIjtdm87EjK-WKEXDNdZQqe+eW9ah0l3ciMJFxywwyQG6OlE1MgDHtiO5fXY2KTyis3wy+L59Mpn3dcmx9PS5sH4zf6CI64GJ9JrJNRHNQKAAyDHBkEhsN+ULh7CwSLAqPRsQ2JCSROkaRUiKEEKxFVJuSOKgAcmdbvDXq4SSVxmzrrwwOxpvczmTivzBXSSgAlWkvcRvLl-DhcQoYaCUBTpbDsQQyjEzeWMJpKzjgBhqjVQJywchgJDsaiQAoESTsCAwHV6slXfYYQ73Eam1XqxU3aCeEj4J2IUgYO2QHwekUpOWA97JuEwwELDOYkHC8bmhTrJaSh6DOHDCtls5F1RJa7pwvQgYCabG2tNH05TG1IJia69faN0OOwT9zPKOEkNRcHUAYxA07U8UZ6Vg8HQW-QTiYOAQOITaATuAgPTg46X064M7NWqOgnMA-mrGfd1n2DgzHYCGgUwYpDUHGCQCqQ3Jvi81DtIqpSTuScEPLe3ZAde4GDKkN4YeBUHLFyKH1uBt5zNIlTEbUeD7MRBqFHBmxCOMNGXDCJY5hRswUnKjGQSM9EcgR-IcTBDFXPxdbpmx4pCpRInITxZyMd2PRdsKEnXGAhBvsRdGCBJjFigKbTLHJwnVKsCjFOWakabwg63uRuCYepWDxBxclsURGlEa5HSOHIFCKNJHQFL4ahaHohi6AAwgArkEQTcAAdpYDiiDZPhuBAbjPAgkK4SAiKmNoAAmRgAJYJUYAAi3DhEYkWmAlCXcPOlilQ1RgGAYCWZKVliaEYgRBOVNh1bF8VJSlNyQB0GA+PybyUCFIAAJK+Xa6WnrMt60LiJHSGtBCXO4-IkIwDA+MsxYHVITBJHaxa0Etq2pXakLmnNdrTIteXPbMSDMABZ5gFyAG7VB4AHZChTop94BgFwP0Hcw6kfSQyPA74CgsqlNKFPskgqsFl3gNju7mdQZ5MEUQFLQAqr5SjYgMTCMOwF2qCA9M4wIlyTB45CXBgyjEwwpM9C0TD4-alwU5j0hi0oN2QJTj1HGDaj04WTQ0MzykZOM2bAmmQmPBcJYgh2qXsCQPhHEk7CbTSeVc48i1MMwExujLCjE24vk4ti6JsxA5oI3tBT+zbvP7A7Ide87vk5cWzg3HzMzEzQif-UdAw5Td5102cakUvwLRkWcbZYaX7bKOWvRV-X3blpXMwIbWxbXEhBYkbhl3mRStSpbNcNqBMKoIMwvsoqluDo76WDqRAiOiNQPiL0CcPqytjhMBUyML5UIv4tkq6ORpiHxJ5p8lKIKBw8siBw7lHO-Y6a8IM8jIJOz4OZ6Y0hLytHbSMIBYjrk6OARQVopBcDCvoUIFhrB2EEP-dm8xECnVAa+GAmwIBYEwlwQI2gQgmEQREBw-8DwqARpAJAkIwFAl-HCJcLA14sHdOwyBKQEbxA4BAh0kC4ERRinFRKyVBBAA noborder}} </WRAP> | <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgzCAMB0l5AWAnC1b0DYrTAdiWBhrgBxIIYlgCMuuWATNSAnCAKyQcCmAtNdQBQADxC8M1JCAFTeyBNIYMW0kiAAKACwCGAZ24AdXQDUA9gBsALtoDm3XSLEMkWakrHUyi9is8gAytYATkZmVrb2gjZiGAhqYAgK4nHgGFzpggDuTikJSQxgynlQWU6x4Ik58ZWQggAOToXgbo1FlczMtQ28DCkFyj19LtIjtdm87EjK-WKEXDNdZQqe+eW9ah0l3ciMJFxywwyQG6OlE1MgDHtiO5fXY2KTyis3wy+L59Mpn3dcmx9PS5sH4zf6CI64GJ9JrJNRHNQKAAyDHBkEhsN+ULh7CwSLAqPRsQ2JCSROkaRUiKEEKxFVJuSOKgAcmdbvDXq4SSVxmzrrwwOxpvczmTivzBXSSgAlWkvcRvLl-DhcQoYaCUBTpbDsQQyjEzeWMJpKzjgBhqjVQJywchgJDsaiQAoESTsCAwHV6slXfYYQ73Eam1XqxU3aCeEj4J2IUgYO2QHwekUpOWA97JuEwwELDOYkHC8bmhTrJaSh6DOHDCtls5F1RJa7pwvQgYCabG2tNH05TG1IJia69faN0OOwT9zPKOEkNRcHUAYxA07U8UZ6Vg8HQW-QTiYOAQOITaATuAgPTg46X064M7NWqOgnMA-mrGfd1n2DgzHYCGgUwYpDUHGCQCqQ3Jvi81DtIqpSTuScEPLe3ZAde4GDKkN4YeBUHLFyKH1uBt5zNIlTEbUeD7MRBqFHBmxCOMNGXDCJY5hRswUnKjGQSM9EcgR-IcTBDFXPxdbpmx4pCpRInITxZyMd2PRdsKEnXGAhBvsRdGCBJjFigKbTLHJwnVKsCjFOWakabwg63uRuCYepWDxBxclsURGlEa5HSOHIFCKNJHQFL4ahaHohi6AAwgArkEQTcAAdpYDiiDZPhuBAbjPAgkK4SAiKmNoAAmRgAJYJUYAAi3DhEYkWmAlCXcPOlilQ1RgGAYCWZKVliaEYgRBOVNh1bF8VJSlNyQB0GA+PybyUCFIAAJK+Xa6WnrMt60LiJHSGtBCXO4-IkIwDA+MsxYHVITBJHaxa0Etq2pXakLmnNdrTIteXPbMSDMABZ5gFyAG7VB4AHZChTop94BgFwP0Hcw6kfSQyPA74CgsqlNKFPskgqsFl3gNju7mdQZ5MEUQFLQAqr5SjYgMTCMOwF2qCA9M4wIlyTB45CXBgyjEwwpM9C0TD4-alwU5j0hi0oN2QJTj1HGDaj04WTQ0MzykZOM2bAmmQmPBcJYgh2qXsCQPhHEk7CbTSeVc48i1MMwExujLCjE24vk4ti6JsxA5oI3tBT+zbvP7A7Ide87vk5cWzg3HzMzEzQif-UdAw5Td5102cakUvwLRkWcbZYaX7bKOWvRV-X3blpXMwIbWxbXEhBYkbhl3mRStSpbNcNqBMKoIMwvsoqluDo76WDqRAiOiNQPiL0CcPqytjhMBUyML5UIv4tkq6ORpiHxJ5p8lKIKBw8siBw7lHO-Y6a8IM8jIJOz4OZ6Y0hLytHbSMIBYjrk6OARQVopBcDCvoUIFhrB2EEP-dm8xECnVAa+GAmwIBYEwlwQI2gQgmEQREBw-8DwqARpAJAkIwFAl-HCJcLA14sHdOwyBKQEbxA4BAh0kC4ERRinFRKyVBBAA noborder}} </WRAP> |
| |
| <callout title="Voltages - Currents - True Power - Aparrent and Reactive Power"> | <callout title="Voltages - Currents - True Power - Apparent and Reactive Power"> |
| \\ | \\ |
| |
| * The strings of the load are now connected to two phases and not to a star point anymore. | * The strings of the load are now connected to two phases and not to a star point anymore. |
| * The network phase voltages $\underline{U}_{12}$, $\underline{U}_{23}$, $\underline{U}_{31}$ equals to the string voltages of the load. | * The network phase voltages $\underline{U}_{12}$, $\underline{U}_{23}$, $\underline{U}_{31}$ equals to the string voltages of the load. |
| * The star point that the voltages toward is is here not relevant, and cannot be connected in any way to the strings. | * The star point that the voltages are toward is in this case not relevant, and cannot be connected in any way to the strings. |
| |
| Again here the "path": calculate voltages $\rightarrow$ calculate currents $\rightarrow$ calculate true power $\rightarrow$ calculate aparrent and reactive power is the best way to get to all wanted values. | Again here the "path": calculate voltages $\rightarrow$ calculate currents $\rightarrow$ calculate true power $\rightarrow$ calculate apparent and reactive power is the best way to get to all wanted values. |
| - **Voltages**: Here, the string voltages of the load are applied by the three phase net: \\ \begin{align*} \underline{U}_{12} &=& U_L \cdot e^{j\cdot {{1}\over{6}}} \\ \underline{U }_{23} &=& U_L \cdot e^{- j\cdot {{3}\over{6}}} \\ \underline{U }_{31} &=& U_L \cdot e^{- j\cdot {{7}\over{6}}} \end{align*} | - **Voltages**: Here, the string voltages of the load are applied by the three-phase net: <WRAP> |
| - **Currents**: For the phase currents one can focus on the nodes between the phase lines and the strings. An incomming single phase current onto a note divides into two string 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*} | \begin{align*} |
| - Also here, the **true power** can be calculated adding up the true power of each phase. The faster way (as shown before) is to add up the (complex) apparent power. | \underline{U}_{12} &=& U_{\rm L} \cdot {\rm e}^{ {\rm j}\cdot {{1}\over{6}}} \\ |
| - The **apparent power** $\underline{S}_x$ here is again the sum the (complex) apparent power for each string: \\ \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 $\underline{U}_{12}$, $\underline{U}_{23}$, and $\underline{U}_{31}$ are given by the three-phase network, a further simplification lead to: \\ \begin{align*} \boxed{\underline{S} = P + j \cdot Q = U_L^2 \cdot \left( {{1}\over{\underline{Z}_{12}^* }} + {{1}\over{\underline{Z}_{23}^* }} + {{1}\over{\underline{Z}_{31}^* }}\right) } \end{align*} \\ The **collective apparent power** $S_\Sigma$ here is the same as for the three-wire or four-wire connection. \\ In the Delta configuration the phase currents $I_x$ have to be calculated, since the formula only applies for them: \\ \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_L \cdot \sqrt{\sum_x I_x^2} \end{align*} | \underline{U}_{23} &=& U_{\rm L} \cdot {\rm e}^{- {\rm j}\cdot {{3}\over{6}}} \\ |
| - The abolute **reactive power** $Q$ can be calulated by the apparent power: \\ \begin{align*} j\cdot Q &= \underline{S} - P \end{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*} | \underline{U}_{31} &=& U_{\rm L} \cdot {\rm e}^{- {\rm j}\cdot {{7}\over{6}}} |
| | \end{align*}</WRAP> |
| | - **Currents**: For the phase currents one can focus on the nodes between the phase lines and the strings. An incomming single-phase current onto a note divides into two string currents: <WRAP> |
| | \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*} </WRAP> |
| | - 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: <WRAP> |
| | \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 $\underline{U}_{12}$, $\underline{U}_{23}$, and $\underline{U}_{31}$ are given by the three-phase network, a further simplification lead to: \\ |
| | \begin{align*} |
| | \boxed{ |
| | \underline{S} = P + {\rm j} \cdot Q |
| | = U_{\rm L}^2 \cdot \left( {{1}\over{\underline{Z}_{12}^* }} |
| | + {{1}\over{\underline{Z}_{23}^* }} |
| | + {{1}\over{\underline{Z}_{31}^* }}\right) } |
| | \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 to them: \\ |
| | \begin{align*} |
| | S_\Sigma &= \sqrt{ {{1}\over{3}} (U_{12}^2 + U_{23}^2 + U_{31}^2) } \cdot \sqrt{\sum_x I_x^2} &= U_{\rm L} \cdot \sqrt{\sum_x I_x^2} |
| | \end{align*} </WRAP> |
| | - The abolute **reactive power** $Q$ can be calulated by the apparent power: <WRAP> |
| | \begin{align*} {\rm j}\cdot Q &= \underline{S} - P \end{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*} </WRAP> |
| </callout> | </callout> |
| |
| <panel type="info" title="Example"> | <panel type="info" title="Example"> |
| In the example this leads to: | In the example, this leads to: |
| - The phase voltages are given as \begin{align*} U_L=\sqrt{3} \cdot 231V = & 400V = U_{12} = U_{23} = U_{31} \end{align*} \\ The phasors of the star voltages of the network are again given as: \\ {{drawio>FourWireStarPhaseVoltageFormula}} | - The phase voltages are given as <WRAP> |
| - Based on the string voltages of the network and the given impedances the string currents $\underline{I}_{12}$, $\underline{I}_{23}$, $\underline{I}_{31}$ of the load can be calculated : \\ \begin{align*} \underline{I}_{12} &=& {{ 400V \cdot\left(+{{1}\over{2}}\sqrt{3}+{{1}\over{2}} \cdot j \right)}\over{ 10\Omega + j \cdot 2\pi\cdot 50Hz \cdot 1mH }} &=& 35.24A + j \cdot 18.90A &=& 40A \quad &\angle 28.2° \\ \underline{I}_{23} &=& {{400 \cdot j}\over{ 5\Omega + {{1}\over{j \cdot 2\pi\cdot 50Hz \cdot 100\mu F }} }} &=& 12.27A + j \cdot -1.93A &=& 12.42A \quad &\angle -8.9° \\ \underline{I}_{31} &=& {{400V \cdot\left(-{{1}\over{2}}\sqrt{3}+{{1}\over{2}} \cdot j \right)}\over{ 20 \Omega}} &=& -17.33A + j \cdot 10.00A &=& 20.01A \quad &\angle 150° \end{align*} \\ By these voltages the phase currents $\underline{I}_x$ can be calculated: \\ \begin{align*} \underline{I}_{1} &=& (35.24A + j \cdot 18.90A) - (-17.33A + j \cdot 10.00A) &=& 52.57A + j \cdot 8.90A &=& 53.32A \quad &\angle 9.6° \\ \underline{I}_{2} &=& (12.27A + j \cdot -1.93A) - (35.24A + j \cdot 18.90A ) &=& -22.98A - j \cdot 20.83A &=& -31.01A \quad &\angle -137.8° \\ \underline{I}_{3} &=& (-17.33A + j \cdot 10.00A) - (12.27A + j \cdot -1.93A) &=& -29.59 + j \cdot 11.93 &=& 31.90A \quad &\angle 158.0° \end{align*} \\ | \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^* &=& 400V \cdot (- e^{-j \cdot 7/6 \pi} \cdot (52.57 A - j \cdot 8.90 A ) + e^{- j \cdot 3/6 \pi} \cdot (-22.98A + j \cdot 20.83A) ) &= 24.77 kW - j \cdot 4.41 kVAr \\ &= \underline{U}_{12} \cdot \underline{I}_1^* + \underline{U}_{32} \cdot \underline{I}_3^* &=& 400V \cdot (e^{j \cdot 1/6 \pi} \cdot (52.57 A - j \cdot 8.90 A) - e^{-j \cdot 3/6 \pi} \cdot (-29.59A - j \cdot 11.93A)) &= 24.77 kW - j \cdot 4.41 kVAr \\ &= \underline{U}_{21} \cdot \underline{I}_2^* + \underline{U}_{31} \cdot \underline{I}_3^* &=& 400V \cdot (- e^{j \cdot 1/6 \pi} \cdot (-22.98A + j \cdot 20.83A) + e^{- j \cdot 7/6 \pi} \cdot (-29.59A - j \cdot 11.93A)) &= 24.77 kW - j \cdot 4.41 kVAr \\ & = 25.16 kVA \quad \angle -10.09°\end{align*} \\ The collective apparent power is: \\ \begin{align*} S_\Sigma &= \sqrt{3} U_S \cdot \sqrt{\sum_x I_x^2} \\ &= \sqrt{3} \cdot 231V \cdot \sqrt{(53.32A)^2+(31.01A)^2+(31.90A)^2} = 27.78 kVA \end{align*} | = & 400 ~\rm V |
| - The reactive power is: \begin{align*} Q &= \underline{S} - P = -4.41kVAr \\ \end{align*} \\ The collective reactive power is: \\ \begin{align*} Q_\Sigma &=\sqrt{(27.78 kVA)^2 - (24.77 kW)^2} = 12.59kVAr \\ \end{align*} | = U_{12} = U_{23} = U_{31} |
| | \end{align*} \\ |
| | The phasors of the string voltages of the network are given as \\ |
| | {{drawio>ThreeWireStringPhaseVoltageFormula.svg}}</WRAP> |
| | - Based on the string voltages of the network and the given impedances the string currents $\underline{I}_{12}$, $\underline{I}_{23}$, $\underline{I}_{31}$ of the load can be calculated: <WRAP> |
| | \begin{align*} |
| | \underline{I}_{12} &=& {{ 400 {~\rm V} \cdot\left(+{{1}\over{2}}\sqrt{3}+{{1}\over{2}} \cdot {\rm j} \right)}\over{ 10~\Omega + {\rm j} \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 1 {~\rm mH} }} |
| | &=& 35.24 {~\rm A} + {\rm j} \cdot 18.90 {~\rm A} &=& 40 {~\rm A} \quad &\angle 28.2° \\ |
| | \underline{I}_{23} &=& {{400 \cdot {\rm j}}\over{ 5~\Omega + {{1}\over{{\rm j} \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 100 {~\rm µF}}} }} |
| | &=& 12.27 {~\rm A} - {\rm j} \cdot 1.93 {~\rm A} &=& 12.42 {~\rm A} \quad &\angle -8.9° \\ |
| | \underline{I}_{31} &=& {{400 {~\rm V} \cdot\left(-{{1}\over{2}}\sqrt{3}+{{1}\over{2}} \cdot {\rm j} \right)}\over{ 20 ~\Omega}} |
| | &=& -17.33 {~\rm A} + {\rm j} \cdot 10.00 {~\rm A} &=& 20.01 {~\rm A} \quad &\angle 150° \end{align*} \\ |
| | By these voltages the phase currents $\underline{I}_x$ can be calculated: \\ |
| | \begin{align*} |
| | \underline{I}_{1} &=& ( 35.24 {~\rm A} + {\rm j} \cdot 18.90 {~\rm A}) - (-17.33 {~\rm A} + {\rm j} \cdot 10.00 {~\rm A}) &=& 52.57 {~\rm A} + {\rm j} \cdot 8.90 {~\rm A} |
| | &=& 53.32 {~\rm A} \quad &\angle 9.6° \\ |
| | \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° \\ |
| | \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° |
| | \end{align*} \\</WRAP> |
| | - The true power is calculated by: <WRAP> |
| | \begin{align*} |
| | P = 231 {~\rm V} \cdot \big( 53.32 {~\rm A} \cdot \cos ( 0° - (9.6°) ) |
| | - 31.01 {~\rm A} \cdot \cos (-120° - (-137.8°) ) |
| | + 31.90 {~\rm A} \cdot \cos (-240° - (+158.0°) \big) |
| | = 24.77 {~\rm kW} \end{align*}</WRAP> |
| | - The apparent power $\underline{S}$ is: <WRAP> |
| | \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 ( 52.57 {~\rm A} - {\rm j} \cdot 8.90 {~\rm A}) + |
| | {\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^* |
| | &=& 400 {~\rm V} \cdot ( {\rm e}^{ {\rm j} \cdot 1/6 \pi} \cdot ( 52.57 {~\rm A} - {\rm j} \cdot 8.90 {~\rm A}) - |
| | {\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^* |
| | &=& 400 {~\rm V} \cdot (- {\rm e}^{ {\rm j} \cdot 1/6 \pi} \cdot (-22.98 {~\rm A} + {\rm j} \cdot 20.83 {~\rm A}) + |
| | {\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*} \\ |
| | The collective apparent power is: \\ |
| | \begin{align*} |
| | S_\Sigma &= U_{\rm L} \cdot \sqrt{\sum_x I_x^2} \\ |
| | &= \sqrt{3} \cdot 231 {~\rm V} \cdot \sqrt{(53.32 {~\rm A})^2+(31.01 {~\rm A})^2+(31.90 {~\rm A})^2} |
| | = 27.78 {~\rm kVA} \end{align*} </WRAP> |
| | - The reactive power is: <WRAP> |
| | \begin{align*} Q &= |\underline{S} - P| = -4.41 {~\rm kVAr} \\ |
| | \end{align*} \\ |
| | The collective reactive power is: \\ |
| | \begin{align*} Q_\Sigma &=\sqrt{(27.78 {~\rm kVA})^2 - (24.77 {~\rm kW})^2} = 12.59 {~\rm kVAr} \\ |
| | \end{align*}</WRAP> |
| |
| <WRAP> <imgcaption imageNo17 | Load in Delta connection ></imgcaption> {{drawio>PhasorDelta}} </WRAP> | <WRAP> <imgcaption imageNo17 | Load in Delta connection ></imgcaption>{{drawio>PhasorDelta}.svg}} </WRAP> |
| </panel> | </panel> |
| |
| <callout title="For Symmetric Load"> | <WRAP hide> |
| | <callout title="For Symmetric Load in Delta connection "> |
| \\ | \\ |
| The case of a symmetric load in a three-wire system equals the symmetric load in a four-wire system, since the symmetric four-wire system also does not show a current on the neutral line. | In the case of a symmetric load, the situation and the formulas get much simpler: |
| | - 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** $I_{\rm L} = I_x$ 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 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: $I_N =0$ |
| | - |
| | - 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 **(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: $Q_\Sigma = \sqrt{S_\Sigma^2 - P^2} = 3 \cdot U_{\rm S} I_{\rm S} \cdot \sin (\varphi)$. |
| </callout> | </callout> |
| | </WRAP> |
| | |
| | === Overview of the Power in different Connections === |
| | |
| | <WRAP> <imgcaption imageNo20 | Overview of the Power in different Connections></imgcaption> {{drawio>OverviewPower.svg}} </WRAP> |
| | |
| |
| ===== Excercises ===== | ===== Excercises ===== |
| <panel type="info" title="Exercise 7.1.1 Power and Power Factor I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> | <panel type="info" title="Exercise 7.1.1 Power and Power Factor I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> |
| |
| A passive component is fed by a sinosidal AC voltage with the RMS value $U=230V$ and $f=50Hz$. The RMS current on this component is $I=5A$ 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. | - Draw the equivalent circuits based on a series and on a parallel circuit. |
| - Calculate the equivalent components for both circuits. | - 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.. | - 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. | - Check the solutions from 3. via direct calculation based on the input in the task above. |
| | |
| | <button size="xs" type="link" collapse="Loesung_7_1_1_1_Rechnung">{{icon>eye}} Result for 1.</button><collapse id="Loesung_7_1_1_1_Rechnung" collapsed="true"> |
| | |
| | {{drawio>electrical_engineering_2:Sol711EquivCirc.svg}} |
| | |
| | </collapse> |
| | |
| | <button size="xs" type="link" collapse="Loesung_7_1_1_2_Rechnung">{{icon>eye}} Result for 2.</button><collapse id="Loesung_7_1_1_2_Rechnung" collapsed="true"> |
| | |
| | The apparent impedance is: |
| | \begin{align*} |
| | Z = |\underline{Z}| &={{U}\over{I}}= {{230~\rm V}\over{5~\rm A}} = 46 ~\Omega \\ |
| | \end{align*} |
| | |
| | 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: |
| | \begin{align*} |
| | R_s &=&{{U}\over{I}} \cdot \cos \varphi &=& {{230~\rm V}\over{5.00~\rm A}} \cdot \cos 60° &=& \boldsymbol{23.0 ~\Omega} & \\ |
| | X_{Ls} &=&{{U}\over{I}} \cdot \sin \varphi &=& {{230~\rm V}\over{5.00~\rm A}} \cdot \sin 60° &=& 39.8 ~\Omega &= \omega \cdot L_s \\ |
| | \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*} |
| | |
| | |
| | For the **parallel circuit**, the impedances add up like ${{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}}= {{1}\over{\underline{Z}}} $. \\ |
| | The easiest thing is here to use the formulas of $R_s$ and $X_{Ls}$ from before: |
| | |
| | \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}}} &=& {{R_s - {\rm j}\cdot X_{Ls}}\over{R_s^2 + X_{Ls}^2}} \\ |
| | &=& {{Z \cdot \cos \varphi - {\rm j}\cdot Z \cdot \sin \varphi }\over{Z^2}} \\ |
| | &=& { {\cos \varphi - {\rm j} \cdot \sin \varphi } \over{Z}} \\ |
| | \end{align*} |
| | |
| | Now, the real and imaginary part is analyzed individually. First the real part: |
| | |
| | \begin{align*} |
| | {{1}\over{R_p}} &=& {{\cos \varphi}\over{Z}} \\ |
| | \rightarrow R_p &=& {{Z}\over{\cos \varphi}} &=& {{46 ~\Omega}\over{\cos 60°}} = \boldsymbol{92 ~\Omega} |
| | \end{align*} |
| | |
| | \begin{align*} |
| | {{1}\over{X_{Lp}}} &= {{\sin \varphi}\over{Z}} & \\ |
| | \rightarrow X_{Lp} &= {{Z}\over{\sin \varphi}} = {{46 ~\Omega}\over{\sin 60°}} = 53.1 ~\Omega \\ |
| | \rightarrow L_p &= {{46 ~\Omega}\over{2\pi \cdot 50~\rm Hz \cdot \sin 60°}} &= \boldsymbol{169 ~\rm mH} |
| | \end{align*} |
| | |
| | </collapse> |
| | |
| | |
| | <button size="xs" type="link" collapse="Loesung_7_1_1_3_Endergebnis">{{icon>eye}} Result for 3. </button><collapse id="Loesung_7_1_1_3_Endergebnis" collapsed="true"> |
| | |
| | ^ ^ series circuit ^ parallel circuit ^ |
| | | active power | \begin{align*} P_s &= R_s \cdot I^2 \\ &= 23.0 ~\Omega \cdot (5{~\rm A})^2 \\ &= 575 {~\rm W} \end{align*} | \begin{align*} P_p &= {{U_p^2}\over{R_p}} \\ &= {{(230{~\rm V})^2}\over{92~\Omega}} = 575 {~\rm W} \end{align*} | |
| | | reactive power | \begin{align*} Q_s &= Z_{Ls} \cdot I^2 \\ &= 39.8 ~\Omega \cdot (5{~\rm A})^2 \\ &= 996 {~\rm Var} \end{align*} | \begin{align*} Q_p &= {{U_p^2}\over{Z_{Lp}}} \\ &= {{(230{~\rm V})^2}\over{53.1 ~\Omega}} = 996 {~\rm Var} \end{align*} | |
| | | 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} \end{align*} | |
| | |
| | </collapse> |
| | |
| | |
| | <button size="xs" type="link" collapse="Loesung_7_1_1_4_Endergebnis">{{icon>eye}} Result for 4. </button><collapse id="Loesung_7_1_1_4_Endergebnis" collapsed="true"> |
| | |
| | active power: |
| | \begin{align*} |
| | P &= U \cdot I \cdot \cos \varphi \\ |
| | &= 220{~\rm V} \cdot 5{~\rm A} \cos 60° \\ |
| | &= 575 {~\rm W} |
| | \end{align*} |
| | |
| | reactive power: |
| | \begin{align*} |
| | Q &= U \cdot I \cdot \sin \varphi \\ |
| | &= 230{~\rm V} \cdot 5{~\rm A} \sin 60° \\ |
| | &= 996 {~\rm Var} |
| | \end{align*} |
| | |
| | apparent power: |
| | \begin{align*} |
| | Q &= U \cdot I \\ |
| | &= 230{~\rm V} \cdot 5{~\rm A} \\ |
| | &= 1150 {~\rm VA} |
| | \end{align*} |
| | </collapse> |
| |
| </WRAP></WRAP></panel> | </WRAP></WRAP></panel> |
| <panel type="info" title="Exercise 7.1.2 Power and Power Factor II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> | <panel type="info" title="Exercise 7.1.2 Power and Power Factor II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> |
| |
| A magnetic coil shows at a frequency of $f=50Hz$ the voltage of $U=115V$ and the current $I=2.6A$ 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 inductivity. | - Draw the equivalent series circuit. Calculate the ohmic and inductive impedance and the value of the inductivity. |
| | |
| | |
| | <button size="xs" type="link" collapse="Loesung_7_1_2_1_Rechnung">{{icon>eye}} Result for 1.</button><collapse id="Loesung_7_1_2_1_Rechnung" collapsed="true"> |
| | |
| | The real power is |
| | \begin{align*} |
| | P &= U \cdot I \cdot \cos \varphi \\ |
| | &= 115{~\rm V} \cdot 2.6{~\rm A} \cdot 0.3 \\ |
| | &= 89.7 {~\rm W} |
| | \end{align*} |
| | |
| | The reactive power is |
| | \begin{align*} |
| | Q &= U \cdot I \cdot \sin \varphi \\ |
| | &= 115{~\rm V} \cdot 2.6{~\rm V} \cdot \sqrt{1 - 0.3^2} \\ |
| | &= 285 {~\rm Var} |
| | \end{align*} |
| | |
| | The apparent power is |
| | \begin{align*} |
| | S &= U \cdot I \\ |
| | &= 115{~\rm V} \cdot 2.6{~\rm A} \\ |
| | &= 299 {~\rm VA} |
| | \end{align*} |
| | |
| | </collapse> |
| | |
| | <button size="xs" type="link" collapse="Loesung_7_1_2_2_Rechnung">{{icon>eye}} Result for 2.</button><collapse id="Loesung_7_1_2_2_Rechnung" collapsed="true"> |
| | |
| | {{drawio>electrical_engineering_2:Sol7122EquivCirc.svg}} |
| | |
| | The complex current $\underline{I}$ is given as: |
| | |
| | \begin{align*} |
| | \underline{I} &= I_R + {\rm j} \cdot I_L \\ |
| | &= I \cdot \cos\varphi - {\rm j} \cdot I \cdot \sin\varphi |
| | \end{align*} |
| | |
| | The active and reactive part of the current is therefore: |
| | \begin{align*} |
| | I_R &= 2.60{~\rm A} \cdot 0.30 &= 0.78 {~\rm A} \\ |
| | I_L &= - 2.60{~\rm A} \cdot \sqrt{1 - 0.30^2} &= 2.48 {~\rm A} |
| | \end{align*} |
| | |
| | </collapse> |
| | |
| | <button size="xs" type="link" collapse="Loesung_7_1_2_3_Rechnung">{{icon>eye}} Result for 3.</button><collapse id="Loesung_7_1_2_3_Rechnung" collapsed="true"> |
| | |
| | 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 explicitly given in the problem definition. |
| | |
| | {{drawio>electrical_engineering_2:Sol7123EquivCirc.svg}} |
| | |
| | \begin{align*} |
| | Z_s &= {{U}\over{I}} &=& {{115{~\rm V}}\over{2.60{~\rm A}}} &=& 44.2 ~\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 &=& {{115{~\rm V}}\over{2.60{~\rm A}}} \cdot \sqrt{1 - 0.30^2} &=& 42.2 ~\Omega \\ \\ |
| | |
| | L_s &= 134~\rm mH |
| | |
| | \end{align*} |
| | |
| | </collapse> |
| |
| </WRAP></WRAP></panel> | </WRAP></WRAP></panel> |
| <panel type="info" title="Exercise 7.1.3 Power and Power Factor III"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> | <panel type="info" title="Exercise 7.1.3 Power and Power Factor III"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> |
| |
| A consumer is connected to a $220V$ network. A current of $20A$ 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/capacitance. |
| - 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/capacitance. |
| | |
| | |
| | <button size="xs" type="link" collapse="Loesung_7_1_3_1_Rechnung">{{icon>eye}} Result for 1.</button><collapse id="Loesung_7_1_3_1_Rechnung" collapsed="true"> |
| | |
| | The active power is $P = 1.80 kW$. \\ |
| | The apparent power is $S = U \cdot I = 220V \cdot 20A = 4.40 kVA$. \\ |
| | The reactive power is $Q = \sqrt{S^2 - P^2} = \sqrt{(4.40 kVA)^2 - (1.80 kW)^2} = 4.01 kVar$ \\ |
| | The power factor is $\cos \varphi = {{P}\over{S}} = {{1.80 kW}\over{4.40 kVA}} = 0.41$. |
| | |
| | </collapse> |
| | |
| | <button size="xs" type="link" collapse="Loesung_7_1_3_2_Rechnung">{{icon>eye}} Result for 2.</button><collapse id="Loesung_7_1_3_2_Rechnung" collapsed="true"> |
| | |
| | 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! |
| | |
| | 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*} |
| | P &= {{U^2}\over{R_\rm p}} \rightarrow & R_p &= {{U^2}\over{P}} &= 26.9 ~\Omega \\ |
| | Q &= {{U^2}\over{X_\rm p}} \rightarrow & X_p &= {{U^2}\over{Q}} &= 12.1 ~\Omega \\ |
| | \end{align*} |
| | |
| | The respective values for inductance/capacitance are: |
| | \begin{align*} |
| | L &= {{X_p}\over{2\pi \cdot f}} &= 38.4 {~\rm mH} \\ |
| | C &= {{1}\over{2\pi \cdot f \cdot X_p}} &= 263 ~\rm µF \\ |
| | \end{align*} |
| | |
| | </collapse> |
| | |
| | <button size="xs" type="link" collapse="Loesung_7_1_3_3_Rechnung">{{icon>eye}} Result for 3.</button><collapse id="Loesung_7_1_3_3_Rechnung" collapsed="true"> |
| | |
| | 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*} |
| | P &= I^2 \cdot R_{\rm s} \rightarrow & R_{\rm s} &= {{P}\over{I^2}} &= 4.50 ~\Omega \\ |
| | Q &= I^2 \cdot X_{\rm s} \rightarrow & X_{\rm s} &= {{Q}\over{I^2}} &= 10.0 ~\Omega \\ |
| | \end{align*} |
| | |
| | The respective values for inductance/capacitance are: |
| | \begin{align*} |
| | L &= {{X_{\rm s}}\over{2\pi \cdot f}} &= 31.9 {~\rm mH} \\ |
| | C &= {{1}\over{2\pi \cdot f \cdot X_{\rm s}}} &= 318 ~\rm µF \\ |
| | \end{align*} |
| | |
| | </collapse> |
| |
| </WRAP></WRAP></panel> | </WRAP></WRAP></panel> |
| <panel type="info" title="Exercise 7.1.4 Power and Power Factor IV"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> | <panel type="info" title="Exercise 7.1.4 Power and Power Factor IV"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> |
| |
| 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 in order to get $cos\varphi=1$. | - Calculate the capacity $C$ which has to be connected in parallel to get $\cos\varphi=1$. |
| |
| <button size="xs" type="link" collapse="Loesung_7_1_4_1_Rechnung">{{icon>eye}} Solution</button><collapse id="Loesung_7_1_4_1_Rechnung" collapsed="true"> \begin{align*} S &= U \cdot I_{RL} \\ Q &= \sqrt{S^2 - P_{RL}^2} \\ \varphi &= atan\left({{Q}\over{P}}\right) = arccos\left({{P}\over{S}}\right) \end{align*} | <button size="xs" type="link" collapse="Loesung_7_1_4_1_Rechnung">{{icon>eye}} Solution</button><collapse id="Loesung_7_1_4_1_Rechnung" collapsed="true"> |
| | \begin{align*} |
| | S &= U \cdot I_{RL} \\ |
| | Q &= \sqrt{S^2 - P_{RL}^2} \\ |
| | \varphi &= \arctan\left({{Q}\over{P}}\right) |
| | = \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 for 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 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*} |
| |
| </collapse> | </collapse> |
| |
| <button size="xs" type="link" collapse="Loesung_7_1_4_1_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_7_1_4_1_Endergebnis" collapsed="true"> \begin{align*} S &= 1.62 kVA \\ Q &= 0.95 kVAr \\ \varphi &= +36° \\ \\ C &= 57.2 \mu F \end{align*} | <button size="xs" type="link" collapse="Loesung_7_1_4_1_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_7_1_4_1_Endergebnis" collapsed="true"> |
| | \begin{align*} |
| | S &= 1.62 {~\rm kVA} \\ |
| | Q &= 0.95 {~\rm kVAr} \\ |
| | \varphi &= +36° \\ \\ |
| | C &= 57.2 ~\rm µF |
| | \end{align*} |
| |
| </collapse> | </collapse> |
| | |
| | </WRAP></WRAP></panel> |
| | |
| | <panel type="info" title="Exercise 7.2.1 Three-Phase Load"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> |
| | |
| | A three-phase power net with a phase voltage of $400 ~\rm V$ has two symmetrical ohmic-inductive loads connected to it. The true power and the power factor are given as follows: |
| | |
| | <WRAP> <imgcaption imageNoEx721 | Three-Phase Net with Two Loads></imgcaption> {{drawio>SetupThreePhaseNetworkEx1.svg}} </WRAP> |
| | |
| | * Load 1: $P_1 = 2.7 ~\rm kW$, $\cos \varphi_1 = 0.89$ |
| | * Load 1: $P_2 = 3.8 ~\rm kW$, $\cos \varphi_1 = 0.76$ |
| | |
| | 1. Calculate the reactive power $Q_1$ and $Q_2$. \\ |
| | <button size="xs" type="link" collapse="Loesung_7_2_1_1_Rechnung">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_1_1_Rechnung" collapsed="true"> |
| | First, calculate the apparent power: |
| | \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 {~\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 {~\rm kW} &=& 5.0 {~\rm kVA}\\ |
| | \end{align*} \\ |
| | |
| | For calculating the reactive power, $\sin \varphi_{1,2}$ are needed. There are different ways to get this: |
| | * One way would be to calculate $\varphi_{1,2}$ first from $\arccos \left(\cos \varphi_{1,2} \right)$ |
| | * Another way is to use the formula $1^2 = \sin^2\varphi_{1,2} + \cos^2\varphi_{1,2}$. |
| | \begin{align*} |
| | 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 {~\rm kVA} \cdot \sqrt{1 - 0.76^2} &=& 3.2 {~\rm kVAr}\\ |
| | \end{align*} |
| | </collapse> <button size="xs" type="link" collapse="Loesung_7_2_1_1_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_7_2_1_1_Endergebnis" collapsed="true"> |
| | \begin{align*} |
| | Q_1 &=& 1.4 {~\rm kVAr}\\ |
| | Q_2 &=& 3.2 {~\rm kVAr}\\ |
| | \end{align*} |
| | </collapse>\\ |
| | |
| | 2. Which (complex) apparent power does the net have to provide for both loads combined? \\ |
| | <button size="xs" type="link" collapse="Loesung_7_2_1_2_Rechnung">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_1_2_Rechnung" collapsed="true"> |
| | |
| | \begin{align*} |
| | \underline{S}_{\rm net} &=& \underline{S}_1 &+& \underline{S}_2 \\ |
| | &=& P_1 + {\rm j} \cdot Q_1 &+& P_2 + {\rm j} \cdot Q_2 \\ |
| | &=& P_1 + P_2 &+& {\rm j} \cdot (Q_1 + Q_2) \\ |
| | &=& 2.7 {~\rm kW} + 3.8 {~\rm kW} &+& {\rm j} \cdot (1.4 {~\rm kVAr} + 3.2 {~\rm kVAr}) \\ |
| | &=& 6.5 {~\rm kW} &+& {\rm j} \cdot 4.6 {~\rm kVAr} \\ |
| | &=& P_{\rm net} &+& {\rm j} \cdot Q_{\rm net} \\ |
| | \end{align*} \\ |
| | |
| | As a complex value in Euler representation: |
| | \begin{align*} |
| | \underline{S}_{\rm net} &=& \sqrt{P_{\rm net}^2 + Q_{\rm net}^2 } &\cdot& {\rm e}^{{\rm j} \cdot \arctan ({{Q_{\rm net}}\over{P_{\rm net}}})} \\ |
| | \sqrt{(6.5 {~\rm kW})^2+ (4.6 {~\rm kVAr})^2} &\cdot& {\rm e}^{{\rm j} \cdot \arctan ({{4.6 }\over{6.5 }})} \\ |
| | 8.0 {~\rm kVA} &\cdot& {\rm e}^{{\rm j} \cdot 35°} \\ |
| | \end{align*} |
| | </collapse><button size="xs" type="link" collapse="Loesung_7_2_1_2_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_7_2_1_2_Endergebnis" collapsed="true"> |
| | \begin{align*} |
| | \underline{S}_{\rm net} &=& 6.5 {~\rm kW} + {\rm j} \cdot 4.6 {~\rm kVAr} \\ |
| | &=& 8.0 {~\rm kVA} \cdot {\rm e}^{{\rm j} \cdot 35°} \\ |
| | \end{align*} |
| | </collapse>\\ |
| | |
| | 3. Calculate the RMS value of the single-phase current $I$ which the power net has to provide. \\ |
| | <button size="xs" type="link" collapse="Loesung_7_2_1_3_Rechnung">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_1_3_Rechnung" collapsed="true"> |
| | 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 $S_{\rm net}$: |
| | \begin{align*} |
| | S_{\rm net} &=& 3 \cdot U \cdot I \\ |
| | &=& \sqrt{3} \cdot U_{\rm L} \cdot I \\ |
| | \end{align*} \\ |
| | |
| | The current is: |
| | \begin{align*} |
| | I &=& {{ S_{\rm net} }\over{ \sqrt{3} \cdot U_{\rm L} }} \\ |
| | &=& {{8.0 {~\rm kVA}}\over{ \sqrt{3} \cdot 400{~\rm V} }} \\ |
| | &=& 12{~\rm A} \\ |
| | \end{align*} |
| | </collapse><button size="xs" type="link" collapse="Loesung_7_2_1_3_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_7_2_1_3_Endergebnis" collapsed="true"> |
| | |
| | \begin{align*} |
| | I &=& 12 ~\rm A \\ |
| | \end{align*} |
| | </collapse>\\ |
| | |
| | 4. What is the overall power factor $\cos\varphi_{\rm net}$ of the power net? \\ |
| | <button size="xs" type="link" collapse="Loesung_7_2_1_4_Rechnung">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_1_4_Rechnung" collapsed="true"> |
| | 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*} |
| | \cos\varphi_{\rm net} = \cos (35°) = 0.82 |
| | \end{align*} |
| | |
| | </collapse><button size="xs" type="link" collapse="Loesung_7_2_1_4_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_7_2_1_4_Endergebnis" collapsed="true"> |
| | |
| | \begin{align*} |
| | \cos\varphi_{\rm net} = 0.82 |
| | \end{align*} |
| | </collapse> |
| | |
| |
| </WRAP></WRAP></panel> | </WRAP></WRAP></panel> |