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| - | ===== Exercises ===== | ||
| - | # | ||
| - | # | ||
| - | |||
| - | A transformer has $N_1=1200$ turns and $N_2=300$ turns. | ||
| - | The primary RMS voltage is $U_1=230~{\rm V}$. | ||
| - | The secondary side supplies a load current $I_2=2.0~{\rm A}$. | ||
| - | |||
| - | 1. Calculate the turns ratio $n$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The turns ratio of an ideal transformer is defined as: | ||
| - | \begin{align*} | ||
| - | n=\frac{N_1}{N_2} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the given values: | ||
| - | \begin{align*} | ||
| - | n | ||
| - | &= | ||
| - | \frac{1200}{300} | ||
| - | \\ | ||
| - | &= | ||
| - | 4 | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | n=4 | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 2. Calculate the ideal secondary voltage $U_2$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | For an ideal transformer, | ||
| - | \begin{align*} | ||
| - | n=\frac{U_1}{U_2} | ||
| - | \end{align*} | ||
| - | |||
| - | Therefore: | ||
| - | \begin{align*} | ||
| - | U_2 | ||
| - | &= | ||
| - | \frac{U_1}{n} | ||
| - | \\ | ||
| - | &= | ||
| - | \frac{230~{\rm V}}{4} | ||
| - | \\ | ||
| - | &= | ||
| - | 57.5~{\rm V} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | U_2=57.5~{\rm V} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 3. Calculate the magnitude of the ideal primary current $I_1$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | For the ideal transformer, | ||
| - | \begin{align*} | ||
| - | I_1=\frac{I_2}{n} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | I_1 | ||
| - | &= | ||
| - | \frac{2.0~{\rm A}}{4} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.50~{\rm A} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | I_1=0.50~{\rm A} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 4. State whether this is a step-up or step-down transformer. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | Compare the primary and secondary voltages: | ||
| - | \begin{align*} | ||
| - | U_1 &= 230~{\rm V} | ||
| - | \\ | ||
| - | U_2 &= 57.5~{\rm V} | ||
| - | \end{align*} | ||
| - | |||
| - | Since | ||
| - | \begin{align*} | ||
| - | U_2<U_1 | ||
| - | \end{align*} | ||
| - | |||
| - | the transformer reduces the voltage. | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | The transformer is a step-down transformer. | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | # | ||
| - | |||
| - | |||
| - | # | ||
| - | # | ||
| - | |||
| - | Two coils are wound on the same ideal magnetic core. | ||
| - | The main magnetic reluctance is | ||
| - | |||
| - | \begin{align*} | ||
| - | R_{\rm mH}=2.0\cdot 10^6~\frac{1}{\rm H}. | ||
| - | \end{align*} | ||
| - | |||
| - | The number of turns is $N_1=500$ and $N_2=100$. | ||
| - | |||
| - | 1. Calculate $L_{1{\rm H}}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The main-flux inductance of coil 1 is: | ||
| - | \begin{align*} | ||
| - | L_{1{\rm H}} | ||
| - | = | ||
| - | \frac{N_1^2}{R_{\rm mH}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | L_{1{\rm H}} | ||
| - | &= | ||
| - | \frac{500^2}{2.0\cdot 10^6~1/{\rm H}} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.125~{\rm H} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | L_{1{\rm H}}=0.125~{\rm H} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 2. Calculate $L_{2{\rm H}}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The main-flux inductance of coil 2 is: | ||
| - | \begin{align*} | ||
| - | L_{2{\rm H}} | ||
| - | = | ||
| - | \frac{N_2^2}{R_{\rm mH}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | L_{2{\rm H}} | ||
| - | &= | ||
| - | \frac{100^2}{2.0\cdot 10^6~1/{\rm H}} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.0050~{\rm H} | ||
| - | \\ | ||
| - | &= | ||
| - | 5.0~{\rm mH} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | L_{2{\rm H}} | ||
| - | = | ||
| - | 0.0050~{\rm H} | ||
| - | = | ||
| - | 5.0~{\rm mH} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 3. Calculate $M$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The mutual inductance is: | ||
| - | \begin{align*} | ||
| - | M | ||
| - | = | ||
| - | \frac{N_1N_2}{R_{\rm mH}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | M | ||
| - | &= | ||
| - | \frac{500\cdot 100}{2.0\cdot 10^6~1/{\rm H}} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.025~{\rm H} | ||
| - | \\ | ||
| - | &= | ||
| - | 25~{\rm mH} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | M | ||
| - | = | ||
| - | 0.025~{\rm H} | ||
| - | = | ||
| - | 25~{\rm mH} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 4. Check whether the units are correct. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The reluctance is given in: | ||
| - | \begin{align*} | ||
| - | [R_{\rm mH}]=\frac{1}{\rm H} | ||
| - | \end{align*} | ||
| - | |||
| - | The number of turns is dimensionless. Therefore: | ||
| - | \begin{align*} | ||
| - | \left[ | ||
| - | \frac{N^2}{R_{\rm mH}} | ||
| - | \right] | ||
| - | = | ||
| - | \frac{1}{1/ | ||
| - | = | ||
| - | {\rm H} | ||
| - | \end{align*} | ||
| - | |||
| - | The same argument applies to the mutual inductance: | ||
| - | \begin{align*} | ||
| - | \left[ | ||
| - | \frac{N_1N_2}{R_{\rm mH}} | ||
| - | \right] | ||
| - | = | ||
| - | {\rm H} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | The unit is correct because | ||
| - | \begin{align*} | ||
| - | \frac{1}{1/ | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | # | ||
| - | |||
| - | |||
| - | # | ||
| - | # | ||
| - | |||
| - | Two coils are wound on the same magnetic core. The shared main magnetic path has the reluctance | ||
| - | |||
| - | \begin{align*} | ||
| - | R_{\rm mH}=1.6\cdot 10^6~\frac{1}{\rm H}. | ||
| - | \end{align*} | ||
| - | |||
| - | The numbers of turns are | ||
| - | |||
| - | \begin{align*} | ||
| - | N_1=400, | ||
| - | \qquad | ||
| - | N_2=100. | ||
| - | \end{align*} | ||
| - | |||
| - | The leakage inductances are | ||
| - | |||
| - | \begin{align*} | ||
| - | L_{1\sigma}=4.0~{\rm mH}, | ||
| - | \qquad | ||
| - | L_{2\sigma}=0.30~{\rm mH}. | ||
| - | \end{align*} | ||
| - | |||
| - | Calculate: | ||
| - | |||
| - | * $L_{1{\rm H}}$ | ||
| - | * $L_{2{\rm H}}$ | ||
| - | * $M$ | ||
| - | * the total self-inductances $L_1$ and $L_2$ | ||
| - | * the coupling coefficient $k=\frac{M}{\sqrt{L_1L_2}}$ | ||
| - | |||
| - | 1. Calculate $L_{1{\rm H}}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The main-flux self-inductance of coil 1 is: | ||
| - | \begin{align*} | ||
| - | L_{1{\rm H}} | ||
| - | = | ||
| - | \frac{N_1^2}{R_{\rm mH}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | L_{1{\rm H}} | ||
| - | &= | ||
| - | \frac{400^2}{1.6\cdot 10^6~1/{\rm H}} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.100~{\rm H} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | L_{1{\rm H}}=0.100~{\rm H}=100~{\rm mH} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 2. Calculate $L_{2{\rm H}}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The main-flux self-inductance of coil 2 is: | ||
| - | \begin{align*} | ||
| - | L_{2{\rm H}} | ||
| - | = | ||
| - | \frac{N_2^2}{R_{\rm mH}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | L_{2{\rm H}} | ||
| - | &= | ||
| - | \frac{100^2}{1.6\cdot 10^6~1/{\rm H}} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.00625~{\rm H} | ||
| - | \\ | ||
| - | &= | ||
| - | 6.25~{\rm mH} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | L_{2{\rm H}} | ||
| - | = | ||
| - | 0.00625~{\rm H} | ||
| - | = | ||
| - | 6.25~{\rm mH} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 3. Calculate $M$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The mutual inductance is: | ||
| - | \begin{align*} | ||
| - | M | ||
| - | = | ||
| - | \frac{N_1N_2}{R_{\rm mH}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | M | ||
| - | &= | ||
| - | \frac{400\cdot 100}{1.6\cdot 10^6~1/{\rm H}} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.025~{\rm H} | ||
| - | \\ | ||
| - | &= | ||
| - | 25~{\rm mH} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | M | ||
| - | = | ||
| - | 0.025~{\rm H} | ||
| - | = | ||
| - | 25~{\rm mH} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 4. Calculate the total self-inductances $L_1$ and $L_2$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The total self-inductance is the sum of main-flux inductance and leakage inductance. | ||
| - | |||
| - | For coil 1: | ||
| - | \begin{align*} | ||
| - | L_1 | ||
| - | &= | ||
| - | L_{1{\rm H}}+L_{1\sigma} | ||
| - | \\ | ||
| - | &= | ||
| - | 100~{\rm mH}+4.0~{\rm mH} | ||
| - | \\ | ||
| - | &= | ||
| - | 104~{\rm mH} | ||
| - | \end{align*} | ||
| - | |||
| - | For coil 2: | ||
| - | \begin{align*} | ||
| - | L_2 | ||
| - | &= | ||
| - | L_{2{\rm H}}+L_{2\sigma} | ||
| - | \\ | ||
| - | &= | ||
| - | 6.25~{\rm mH}+0.30~{\rm mH} | ||
| - | \\ | ||
| - | &= | ||
| - | 6.55~{\rm mH} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | L_1 &= 104~{\rm mH} | ||
| - | \\ | ||
| - | L_2 &= 6.55~{\rm mH} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 5. Calculate the coupling coefficient $k=\frac{M}{\sqrt{L_1L_2}}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The coupling coefficient is: | ||
| - | \begin{align*} | ||
| - | k | ||
| - | = | ||
| - | \frac{M}{\sqrt{L_1L_2}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values in henry: | ||
| - | \begin{align*} | ||
| - | k | ||
| - | &= | ||
| - | \frac{0.025~{\rm H}}{\sqrt{0.104~{\rm H}\cdot 0.00655~{\rm H}}} | ||
| - | \\ | ||
| - | & | ||
| - | 0.96 | ||
| - | \end{align*} | ||
| - | |||
| - | The coupling is strong, but not ideal, because leakage inductances are present. | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | k\approx 0.96 | ||
| - | \end{align*} | ||
| - | |||
| - | The coupling is strong, but not ideal. | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | # | ||
| - | |||
| - | |||
| - | # | ||
| - | # | ||
| - | |||
| - | A transformer has $n=5$. | ||
| - | The secondary winding resistance is $R_2=0.20~\Omega$ and the secondary leakage reactance is $X_{2\sigma}=0.35~\Omega$. | ||
| - | |||
| - | Calculate the values $R'_2$ and $X' | ||
| - | |||
| - | 1. Calculate $R' | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | When a resistance is referred from the secondary side to the primary side, it is multiplied by $n^2$: | ||
| - | \begin{align*} | ||
| - | R'_2 | ||
| - | = | ||
| - | n^2R_2 | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | R'_2 | ||
| - | &= | ||
| - | 5^2\cdot 0.20~\Omega | ||
| - | \\ | ||
| - | &= | ||
| - | 25\cdot 0.20~\Omega | ||
| - | \\ | ||
| - | &= | ||
| - | 5.0~\Omega | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | R' | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 2. Calculate $X' | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The secondary leakage reactance is also referred to the primary side by multiplying with $n^2$: | ||
| - | \begin{align*} | ||
| - | X' | ||
| - | = | ||
| - | n^2X_{2\sigma} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | X' | ||
| - | &= | ||
| - | 5^2\cdot 0.35~\Omega | ||
| - | \\ | ||
| - | &= | ||
| - | 25\cdot 0.35~\Omega | ||
| - | \\ | ||
| - | &= | ||
| - | 8.75~\Omega | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | X' | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 3. Check the unit. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The turns ratio $n$ is dimensionless: | ||
| - | \begin{align*} | ||
| - | [n]=1 | ||
| - | \end{align*} | ||
| - | |||
| - | Therefore, multiplying by $n^2$ does not change the unit: | ||
| - | \begin{align*} | ||
| - | [R'_2] &= \Omega | ||
| - | \\ | ||
| - | [X' | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | The unit remains $\Omega$, because $n$ is dimensionless. | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | # | ||
| - | |||
| - | |||
| - | # | ||
| - | # | ||
| - | |||
| - | A transformer has a rated primary current $I_{1{\rm N}}=10~{\rm A}$ and a short-circuit voltage $u_{\rm k}=5~\%$. | ||
| - | |||
| - | 1. Calculate the continuous short-circuit current $I_{1{\rm k}}$ when rated primary voltage is applied. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The short-circuit current can be estimated from the rated current and the relative short-circuit voltage: | ||
| - | \begin{align*} | ||
| - | I_{1{\rm k}} | ||
| - | = | ||
| - | I_{1{\rm N}}\cdot \frac{100~\%}{u_{\rm k}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | I_{1{\rm k}} | ||
| - | &= | ||
| - | 10~{\rm A}\cdot \frac{100~\%}{5~\%} | ||
| - | \\ | ||
| - | &= | ||
| - | 200~{\rm A} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | I_{1{\rm k}}=200~{\rm A} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 2. Estimate the initial peak short-circuit current $i_{\rm p}$ using $i_{\rm p}\approx 2.54 I_{1{\rm k}}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The initial peak current is estimated by: | ||
| - | \begin{align*} | ||
| - | i_{\rm p} | ||
| - | \approx | ||
| - | 2.54\cdot I_{1{\rm k}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the continuous short-circuit current: | ||
| - | \begin{align*} | ||
| - | i_{\rm p} | ||
| - | & | ||
| - | 2.54\cdot 200~{\rm A} | ||
| - | \\ | ||
| - | &= | ||
| - | 508~{\rm A} | ||
| - | \end{align*} | ||
| - | |||
| - | The short-circuit current is much larger than the rated current. Protection devices must be selected accordingly. | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | i_{\rm p}\approx 508~{\rm A} | ||
| - | \end{align*} | ||
| - | |||
| - | Protection devices must be selected accordingly. | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | # | ||
| - | |||
| - | |||
| - | # | ||
| - | # | ||
| - | |||
| - | A single-phase transformer supplies an actuator driver. | ||
| - | Rated data and equivalent circuit data are: | ||
| - | |||
| - | \begin{align*} | ||
| - | U_{1{\rm N}}& | ||
| - | & | ||
| - | U_{2{\rm N}}& | ||
| - | & | ||
| - | I_{2{\rm N}}& | ||
| - | \\ | ||
| - | R_1& | ||
| - | & | ||
| - | X_{1\sigma}& | ||
| - | \\ | ||
| - | R_2& | ||
| - | & | ||
| - | X_{2\sigma}& | ||
| - | \end{align*} | ||
| - | |||
| - | Assume $n=\frac{U_{1{\rm N}}}{U_{2{\rm N}}}$. The magnetizing branch is neglected for the loaded operating point. | ||
| - | |||
| - | 1. Calculate $n$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The turns ratio is estimated from the rated voltages: | ||
| - | \begin{align*} | ||
| - | n | ||
| - | = | ||
| - | \frac{U_{1{\rm N}}}{U_{2{\rm N}}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | n | ||
| - | &= | ||
| - | \frac{230~{\rm V}}{23~{\rm V}} | ||
| - | \\ | ||
| - | &= | ||
| - | 10 | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | n=10 | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 2. Refer $R_2$ and $X_{2\sigma}$ to the primary side. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | Secondary quantities are referred to the primary side by multiplying them with $n^2$: | ||
| - | \begin{align*} | ||
| - | R'_2 &= n^2R_2 | ||
| - | \\ | ||
| - | X' | ||
| - | \end{align*} | ||
| - | |||
| - | With $n=10$: | ||
| - | \begin{align*} | ||
| - | R'_2 | ||
| - | &= | ||
| - | 10^2\cdot 0.012~\Omega | ||
| - | \\ | ||
| - | &= | ||
| - | 1.2~\Omega | ||
| - | \\[4pt] | ||
| - | X' | ||
| - | &= | ||
| - | 10^2\cdot 0.018~\Omega | ||
| - | \\ | ||
| - | &= | ||
| - | 1.8~\Omega | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | R'_2 &= 1.2~\Omega | ||
| - | \\ | ||
| - | X' | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 3. Calculate $R_{\rm k}$ and $X_{\rm k}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The short-circuit equivalent values are the sums of the primary quantities and the referred secondary quantities: | ||
| - | \begin{align*} | ||
| - | R_{\rm k} | ||
| - | &= | ||
| - | R_1+R' | ||
| - | \\ | ||
| - | X_{\rm k} | ||
| - | &= | ||
| - | X_{1\sigma}+X' | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | R_{\rm k} | ||
| - | &= | ||
| - | 1.2~\Omega+1.2~\Omega | ||
| - | \\ | ||
| - | &= | ||
| - | 2.4~\Omega | ||
| - | \\[4pt] | ||
| - | X_{\rm k} | ||
| - | &= | ||
| - | 1.8~\Omega+1.8~\Omega | ||
| - | \\ | ||
| - | &= | ||
| - | 3.6~\Omega | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | R_{\rm k} &= 2.4~\Omega | ||
| - | \\ | ||
| - | X_{\rm k} &= 3.6~\Omega | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 4. Calculate the primary rated current magnitude $I_{1{\rm N}}$ using the ideal current ratio. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | For an ideal transformer, | ||
| - | \begin{align*} | ||
| - | I_{1{\rm N}} | ||
| - | = | ||
| - | \frac{I_{2{\rm N}}}{n} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | I_{1{\rm N}} | ||
| - | &= | ||
| - | \frac{5.0~{\rm A}}{10} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.50~{\rm A} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | I_{1{\rm N}}=0.50~{\rm A} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 5. Estimate the magnitude of the internal voltage drop $U_{\rm k}\approx |\underline{Z}_{\rm k}|I_{1{\rm N}}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | First calculate the magnitude of the short-circuit impedance: | ||
| - | \begin{align*} | ||
| - | |\underline{Z}_{\rm k}| | ||
| - | = | ||
| - | \sqrt{R_{\rm k}^2+X_{\rm k}^2} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | |\underline{Z}_{\rm k}| | ||
| - | &= | ||
| - | \sqrt{(2.4~\Omega)^2+(3.6~\Omega)^2} | ||
| - | \\ | ||
| - | &= | ||
| - | 4.33~\Omega | ||
| - | \end{align*} | ||
| - | |||
| - | Now calculate the internal voltage drop: | ||
| - | \begin{align*} | ||
| - | U_{\rm k} | ||
| - | & | ||
| - | |\underline{Z}_{\rm k}|I_{1{\rm N}} | ||
| - | \\ | ||
| - | &= | ||
| - | 4.33~\Omega\cdot 0.50~{\rm A} | ||
| - | \\ | ||
| - | &= | ||
| - | 2.17~{\rm V} | ||
| - | \end{align*} | ||
| - | |||
| - | This is a primary-side voltage drop. On the secondary side: | ||
| - | \begin{align*} | ||
| - | \frac{2.17~{\rm V}}{10} | ||
| - | = | ||
| - | 0.217~{\rm V} | ||
| - | \end{align*} | ||
| - | |||
| - | For a $23~{\rm V}$ actuator supply this is small but not zero. | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | |\underline{Z}_{\rm k}| &= 4.33~\Omega | ||
| - | \\ | ||
| - | U_{\rm k} & | ||
| - | \end{align*} | ||
| - | |||
| - | Secondary-side equivalent: | ||
| - | \begin{align*} | ||
| - | U_{\rm k,2}\approx 0.217~{\rm V} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | # | ||
| - | |||
| - | |||
| - | # | ||
| - | # | ||
| - | |||
| - | A transformer has the rated primary data | ||
| - | |||
| - | \begin{align*} | ||
| - | U_{1{\rm N}}=230~{\rm V}, | ||
| - | \qquad | ||
| - | I_{1{\rm N}}=2.0~{\rm A}. | ||
| - | \end{align*} | ||
| - | |||
| - | The short-circuit equivalent impedance referred to the primary side is | ||
| - | |||
| - | \begin{align*} | ||
| - | R_{\rm k}=1.5~\Omega, | ||
| - | \qquad | ||
| - | X_{\rm k}=4.0~\Omega. | ||
| - | \end{align*} | ||
| - | |||
| - | Calculate: | ||
| - | |||
| - | * $|\underline{Z}_{\rm k}|$ | ||
| - | * the rated short-circuit voltage $U_{1{\rm k}}$ | ||
| - | * the relative short-circuit voltage $u_{\rm k}$ | ||
| - | * the prospective continuous short-circuit current $I_{1{\rm k}}$ for rated primary voltage | ||
| - | * the approximate first peak current $i_{\rm peak}\approx 2.54 I_{1{\rm k}}$ | ||
| - | |||
| - | 1. Calculate $|\underline{Z}_{\rm k}|$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The short-circuit impedance magnitude is: | ||
| - | \begin{align*} | ||
| - | |\underline{Z}_{\rm k}| | ||
| - | = | ||
| - | \sqrt{R_{\rm k}^2+X_{\rm k}^2} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | |\underline{Z}_{\rm k}| | ||
| - | &= | ||
| - | \sqrt{(1.5~\Omega)^2+(4.0~\Omega)^2} | ||
| - | \\ | ||
| - | &= | ||
| - | 4.27~\Omega | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | |\underline{Z}_{\rm k}|=4.27~\Omega | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 2. Calculate the rated short-circuit voltage $U_{1{\rm k}}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The primary voltage required to drive rated current through the short-circuited transformer is: | ||
| - | \begin{align*} | ||
| - | U_{1{\rm k}} | ||
| - | = | ||
| - | |\underline{Z}_{\rm k}|I_{1{\rm N}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | U_{1{\rm k}} | ||
| - | &= | ||
| - | 4.27~\Omega\cdot 2.0~{\rm A} | ||
| - | \\ | ||
| - | &= | ||
| - | 8.54~{\rm V} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | U_{1{\rm k}}=8.54~{\rm V} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 3. Calculate the relative short-circuit voltage $u_{\rm k}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The relative short-circuit voltage is: | ||
| - | \begin{align*} | ||
| - | u_{\rm k} | ||
| - | = | ||
| - | \frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\% | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | u_{\rm k} | ||
| - | &= | ||
| - | \frac{8.54~{\rm V}}{230~{\rm V}}\cdot 100~\% | ||
| - | \\ | ||
| - | &= | ||
| - | 3.71~\% | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | u_{\rm k}=3.71~\% | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 4. Calculate the prospective continuous short-circuit current $I_{1{\rm k}}$ for rated primary voltage. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The prospective continuous short-circuit current is: | ||
| - | \begin{align*} | ||
| - | I_{1{\rm k}} | ||
| - | = | ||
| - | I_{1{\rm N}}\cdot \frac{100~\%}{u_{\rm k}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | I_{1{\rm k}} | ||
| - | &= | ||
| - | 2.0~{\rm A}\cdot \frac{100}{3.71} | ||
| - | \\ | ||
| - | &= | ||
| - | 53.9~{\rm A} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | I_{1{\rm k}}=53.9~{\rm A} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 5. Calculate the approximate first peak current $i_{\rm peak}\approx 2.54 I_{1{\rm k}}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The approximate first peak current is: | ||
| - | \begin{align*} | ||
| - | i_{\rm peak} | ||
| - | \approx | ||
| - | 2.54 I_{1{\rm k}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the short-circuit current: | ||
| - | \begin{align*} | ||
| - | i_{\rm peak} | ||
| - | & | ||
| - | 2.54\cdot 53.9~{\rm A} | ||
| - | \\ | ||
| - | &= | ||
| - | 137~{\rm A} | ||
| - | \end{align*} | ||
| - | |||
| - | Even though the rated current is only $2.0~{\rm A}$, a short-circuit fault could lead to a much larger current until protection reacts. | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | i_{\rm peak}\approx 137~{\rm A} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | # | ||
| - | |||
| - | |||
| - | # | ||
| - | # | ||
| - | |||
| - | A transformer has the turns ratio | ||
| - | |||
| - | \begin{align*} | ||
| - | n=10. | ||
| - | \end{align*} | ||
| - | |||
| - | The short-circuit equivalent parameters referred to the primary side are | ||
| - | |||
| - | \begin{align*} | ||
| - | R_{\rm k}=2.4~\Omega, | ||
| - | \qquad | ||
| - | X_{\rm k}=3.6~\Omega. | ||
| - | \end{align*} | ||
| - | |||
| - | The secondary load current is | ||
| - | |||
| - | \begin{align*} | ||
| - | I_2=4.0~{\rm A}. | ||
| - | \end{align*} | ||
| - | |||
| - | The load has the power factor | ||
| - | |||
| - | \begin{align*} | ||
| - | \cos\varphi=0.8 | ||
| - | \end{align*} | ||
| - | |||
| - | and is inductive. | ||
| - | |||
| - | Estimate the voltage drop on the secondary side using | ||
| - | |||
| - | \begin{align*} | ||
| - | \Delta U_1 | ||
| - | \approx | ||
| - | I_1\left(R_{\rm k}\cos\varphi+X_{\rm k}\sin\varphi\right) | ||
| - | \end{align*} | ||
| - | |||
| - | and | ||
| - | |||
| - | \begin{align*} | ||
| - | \Delta U_2\approx \frac{\Delta U_1}{n}. | ||
| - | \end{align*} | ||
| - | |||
| - | 1. Calculate the primary current magnitude $I_1$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The primary current magnitude is approximately: | ||
| - | \begin{align*} | ||
| - | I_1 | ||
| - | = | ||
| - | \frac{I_2}{n} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | I_1 | ||
| - | &= | ||
| - | \frac{4.0~{\rm A}}{10} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.40~{\rm A} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | I_1=0.40~{\rm A} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 2. Determine $\sin\varphi$ for the inductive load. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | For an inductive load with $\cos\varphi=0.8$: | ||
| - | \begin{align*} | ||
| - | \sin\varphi | ||
| - | = | ||
| - | \sqrt{1-\cos^2\varphi} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the value: | ||
| - | \begin{align*} | ||
| - | \sin\varphi | ||
| - | &= | ||
| - | \sqrt{1-0.8^2} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.6 | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | \sin\varphi=0.6 | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 3. Estimate the primary-side voltage drop $\Delta U_1$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | Use the Kapp approximation: | ||
| - | \begin{align*} | ||
| - | \Delta U_1 | ||
| - | & | ||
| - | I_1\left(R_{\rm k}\cos\varphi+X_{\rm k}\sin\varphi\right) | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | \Delta U_1 | ||
| - | & | ||
| - | 0.40~{\rm A} | ||
| - | \left( | ||
| - | 2.4~\Omega\cdot 0.8 | ||
| - | + | ||
| - | 3.6~\Omega\cdot 0.6 | ||
| - | \right) | ||
| - | \\ | ||
| - | &= | ||
| - | 0.40~{\rm A} | ||
| - | \left( | ||
| - | 1.92~\Omega+2.16~\Omega | ||
| - | \right) | ||
| - | \\ | ||
| - | &= | ||
| - | 1.63~{\rm V} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | \Delta U_1\approx 1.63~{\rm V} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 4. Estimate the secondary-side voltage drop $\Delta U_2$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The secondary-side voltage drop is: | ||
| - | \begin{align*} | ||
| - | \Delta U_2 | ||
| - | \approx | ||
| - | \frac{\Delta U_1}{n} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | \Delta U_2 | ||
| - | & | ||
| - | \frac{1.63~{\rm V}}{10} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.163~{\rm V} | ||
| - | \end{align*} | ||
| - | |||
| - | The secondary voltage decreases by approximately $0.16~{\rm V}$ for this operating point. | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | \Delta U_2\approx 0.163~{\rm V} | ||
| - | \end{align*} | ||
| - | |||
| - | The secondary voltage decreases by approximately $0.16~{\rm V}$. | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | # | ||
| - | |||
| - | |||
| - | # | ||
| - | # | ||
| - | |||
| - | A transformer has the rated primary voltage | ||
| - | |||
| - | \begin{align*} | ||
| - | U_{1{\rm N}}=230~{\rm V}. | ||
| - | \end{align*} | ||
| - | |||
| - | Its rated primary current is | ||
| - | |||
| - | \begin{align*} | ||
| - | I_{1{\rm N}}=3.0~{\rm A}. | ||
| - | \end{align*} | ||
| - | |||
| - | At rated voltage and no-load operation, the magnetizing current is approximately | ||
| - | |||
| - | \begin{align*} | ||
| - | I_{\rm m, | ||
| - | \end{align*} | ||
| - | |||
| - | The short-circuit voltage is | ||
| - | |||
| - | \begin{align*} | ||
| - | u_{\rm k}=6.0~\%. | ||
| - | \end{align*} | ||
| - | |||
| - | Assume that the magnetizing current is approximately proportional to the applied voltage. | ||
| - | |||
| - | 1. Calculate the short-circuit test voltage $U_{1{\rm k}}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The short-circuit test voltage is: | ||
| - | \begin{align*} | ||
| - | U_{1{\rm k}} | ||
| - | = | ||
| - | \frac{u_{\rm k}}{100~\%}\cdot U_{1{\rm N}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | U_{1{\rm k}} | ||
| - | &= | ||
| - | 0.06\cdot 230~{\rm V} | ||
| - | \\ | ||
| - | &= | ||
| - | 13.8~{\rm V} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | U_{1{\rm k}}=13.8~{\rm V} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 2. Estimate the magnetizing current $I_{\rm m,k}$ during the short-circuit test. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The magnetizing current is assumed to be proportional to the voltage: | ||
| - | \begin{align*} | ||
| - | I_{\rm m,k} | ||
| - | = | ||
| - | I_{\rm m,N}\cdot \frac{U_{1{\rm k}}}{U_{1{\rm N}}} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | I_{\rm m,k} | ||
| - | &= | ||
| - | 0.12~{\rm A}\cdot \frac{13.8~{\rm V}}{230~{\rm V}} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.0072~{\rm A} | ||
| - | \\ | ||
| - | &= | ||
| - | 7.2~{\rm mA} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | I_{\rm m,k} | ||
| - | = | ||
| - | 0.0072~{\rm A} | ||
| - | = | ||
| - | 7.2~{\rm mA} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 3. Compare $I_{\rm m,k}$ with $I_{1{\rm N}}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | Compare the short-circuit magnetizing current with the rated current: | ||
| - | \begin{align*} | ||
| - | \frac{I_{\rm m, | ||
| - | = | ||
| - | \frac{0.0072~{\rm A}}{3.0~{\rm A}} | ||
| - | \end{align*} | ||
| - | |||
| - | Calculate the ratio: | ||
| - | \begin{align*} | ||
| - | \frac{I_{\rm m, | ||
| - | &= | ||
| - | 0.0024 | ||
| - | \\ | ||
| - | &= | ||
| - | 0.24~\% | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | \frac{I_{\rm m, | ||
| - | = | ||
| - | 0.24~\% | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 4. Explain why the magnetizing branch can be neglected. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | During the short-circuit test, the applied voltage is much smaller than the rated voltage: | ||
| - | \begin{align*} | ||
| - | U_{1{\rm k}} \ll U_{1{\rm N}} | ||
| - | \end{align*} | ||
| - | |||
| - | Since the magnetizing current is assumed to be approximately proportional to the applied voltage, the magnetizing current is also very small: | ||
| - | \begin{align*} | ||
| - | I_{\rm m, | ||
| - | \end{align*} | ||
| - | |||
| - | Compared with the rated current: | ||
| - | \begin{align*} | ||
| - | I_{1{\rm N}}=3.0~{\rm A} | ||
| - | \end{align*} | ||
| - | |||
| - | the magnetizing current is only $0.24~\%$. | ||
| - | |||
| - | So during the short-circuit test, almost all current flows through the short-circuit path consisting of $R_{\rm k}$ and $X_{\rm k}$. | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | The magnetizing branch $R_{\rm Fe}\parallel jX_{1{\rm H}}$ can usually be neglected in the short-circuit test. | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | # | ||
| - | |||
| - | |||
| - | # | ||
| - | # | ||
| - | |||
| - | A transformer has the turns ratio $n=10$. The short-circuit equivalent parameters referred to the primary side are $R_{\rm k}=2.0~\Omega$, | ||
| - | |||
| - | The secondary load current is $I_2=5.0~{\rm A}$. The load is ohmic-inductive with $\cos\varphi=0.8$. | ||
| - | |||
| - | The no-load current on the primary side is approximately $I_{10}=0.03~{\rm A}$. | ||
| - | \\ \\ | ||
| - | 1. Calculate the load-related primary current $I' | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The load-related primary current is: | ||
| - | \begin{align*} | ||
| - | I'_2 | ||
| - | = | ||
| - | \frac{I_2}{n} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | I'_2 | ||
| - | &= | ||
| - | \frac{5.0~{\rm A}}{10} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.50~{\rm A} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | I' | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 2. Estimate the primary-side voltage drop. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The voltage drop is estimated with: | ||
| - | \begin{align*} | ||
| - | \Delta U_1 | ||
| - | \approx | ||
| - | I'_2 | ||
| - | \left( | ||
| - | R_{\rm k}\cos\varphi | ||
| - | + | ||
| - | X_{\rm k}\sin\varphi | ||
| - | \right) | ||
| - | \end{align*} | ||
| - | |||
| - | For $\cos\varphi=0.8$: | ||
| - | \begin{align*} | ||
| - | \sin\varphi | ||
| - | &= | ||
| - | \sqrt{1-\cos^2\varphi} | ||
| - | \\ | ||
| - | &= | ||
| - | \sqrt{1-0.8^2} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.6 | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | \Delta U_1 | ||
| - | & | ||
| - | 0.50~{\rm A} | ||
| - | \left( | ||
| - | 2.0~\Omega\cdot 0.8 | ||
| - | + | ||
| - | 4.0~\Omega\cdot 0.6 | ||
| - | \right) | ||
| - | \\ | ||
| - | &= | ||
| - | 0.50~{\rm A} | ||
| - | \left( | ||
| - | 1.6~\Omega+2.4~\Omega | ||
| - | \right) | ||
| - | \\ | ||
| - | &= | ||
| - | 2.0~{\rm V} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | \Delta U_1\approx 2.0~{\rm V} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 3. Calculate the secondary-side voltage drop $\Delta U_2\approx \frac{\Delta U_1}{n}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The secondary-side voltage drop is: | ||
| - | \begin{align*} | ||
| - | \Delta U_2 | ||
| - | \approx | ||
| - | \frac{\Delta U_1}{n} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | \Delta U_2 | ||
| - | & | ||
| - | \frac{2.0~{\rm V}}{10} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.20~{\rm V} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | \Delta U_2\approx 0.20~{\rm V} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 4. Estimate an upper bound for the neglected voltage drop caused by $I_{10}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | First calculate the magnitude of the short-circuit impedance: | ||
| - | \begin{align*} | ||
| - | |\underline{Z}_{\rm k}| | ||
| - | = | ||
| - | \sqrt{R_{\rm k}^2+X_{\rm k}^2} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | |\underline{Z}_{\rm k}| | ||
| - | &= | ||
| - | \sqrt{(2.0~\Omega)^2+(4.0~\Omega)^2} | ||
| - | \\ | ||
| - | &= | ||
| - | 4.47~\Omega | ||
| - | \end{align*} | ||
| - | |||
| - | The upper bound for the voltage drop caused by the no-load current is: | ||
| - | \begin{align*} | ||
| - | \Delta U_{1,10} | ||
| - | &\leq | ||
| - | |\underline{Z}_{\rm k}|I_{10} | ||
| - | \\ | ||
| - | &= | ||
| - | 4.47~\Omega\cdot 0.03~{\rm A} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.134~{\rm V} | ||
| - | \end{align*} | ||
| - | |||
| - | On the secondary side: | ||
| - | \begin{align*} | ||
| - | \Delta U_{2,10} | ||
| - | &\leq | ||
| - | \frac{0.134~{\rm V}}{10} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.0134~{\rm V} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | |\underline{Z}_{\rm k}| &= 4.47~\Omega | ||
| - | \\ | ||
| - | \Delta U_{1,10} &\leq 0.134~{\rm V} | ||
| - | \\ | ||
| - | \Delta U_{2,10} &\leq 0.0134~{\rm V} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 5. Decide whether the short-circuit equivalent circuit is sufficient for this load estimate. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The load-related secondary-side voltage drop is: | ||
| - | \begin{align*} | ||
| - | \Delta U_2\approx 0.20~{\rm V} | ||
| - | \end{align*} | ||
| - | |||
| - | The estimated neglected secondary-side voltage drop caused by the no-load current is at most: | ||
| - | \begin{align*} | ||
| - | \Delta U_{2, | ||
| - | \end{align*} | ||
| - | |||
| - | This is small compared with the load-related drop of $0.20~{\rm V}$. | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | For this engineering estimate, the short-circuit equivalent circuit is sufficient. | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | # | ||
| - | |||
| - | |||
| - | # | ||
| - | # | ||
| - | |||
| - | A transformer has the rated primary voltage of $U_1=230~{\rm V}$ and and the turns ratio $n=10$. A resistive load draws $I_2=4.0~{\rm A}$. \\ | ||
| - | For the real transformer, | ||
| - | The iron loss is approximately $P_{\rm Fe}=1.5~{\rm W}$. \\ | ||
| - | Assume a resistive load with $\cos\varphi=1$. | ||
| - | \\ \\ | ||
| - | 1. Calculate the ideal secondary voltage $U_{2,\rm ideal}$. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | For the ideal transformer: | ||
| - | \begin{align*} | ||
| - | U_{2,\rm ideal} | ||
| - | = | ||
| - | \frac{U_1}{n} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | U_{2,\rm ideal} | ||
| - | &= | ||
| - | \frac{230~{\rm V}}{10} | ||
| - | \\ | ||
| - | &= | ||
| - | 23.0~{\rm V} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | U_{2,\rm ideal}=23.0~{\rm V} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 2. Calculate the load-related primary current $I' | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The load-related primary current is: | ||
| - | \begin{align*} | ||
| - | I'_2 | ||
| - | = | ||
| - | \frac{I_2}{n} | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | I'_2 | ||
| - | &= | ||
| - | \frac{4.0~{\rm A}}{10} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.40~{\rm A} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | I' | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 3. Estimate the real secondary voltage. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | For a resistive load, the approximate primary-side voltage drop is: | ||
| - | \begin{align*} | ||
| - | \Delta U_1 | ||
| - | \approx | ||
| - | I' | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | \Delta U_1 | ||
| - | & | ||
| - | 0.40~{\rm A}\cdot 2.4~\Omega | ||
| - | \\ | ||
| - | &= | ||
| - | 0.96~{\rm V} | ||
| - | \end{align*} | ||
| - | |||
| - | The corresponding secondary-side voltage drop is: | ||
| - | \begin{align*} | ||
| - | \Delta U_2 | ||
| - | & | ||
| - | \frac{\Delta U_1}{n} | ||
| - | \\ | ||
| - | &= | ||
| - | \frac{0.96~{\rm V}}{10} | ||
| - | \\ | ||
| - | &= | ||
| - | 0.096~{\rm V} | ||
| - | \end{align*} | ||
| - | |||
| - | Thus the real secondary voltage is approximately: | ||
| - | \begin{align*} | ||
| - | U_{2,\rm real} | ||
| - | & | ||
| - | 23.0~{\rm V}-0.096~{\rm V} | ||
| - | \\ | ||
| - | &= | ||
| - | 22.90~{\rm V} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | \Delta U_1 & | ||
| - | \\ | ||
| - | \Delta U_2 & | ||
| - | \\ | ||
| - | U_{2,\rm real} & | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 4. Estimate the copper losses. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | The copper losses are: | ||
| - | \begin{align*} | ||
| - | P_{\rm Cu} | ||
| - | \approx | ||
| - | R_{\rm k}(I' | ||
| - | \end{align*} | ||
| - | |||
| - | Insert the values: | ||
| - | \begin{align*} | ||
| - | P_{\rm Cu} | ||
| - | & | ||
| - | 2.4~\Omega\cdot (0.40~{\rm A})^2 | ||
| - | \\ | ||
| - | &= | ||
| - | 0.384~{\rm W} | ||
| - | \end{align*} | ||
| - | |||
| - | The real transformer also has iron losses: | ||
| - | \begin{align*} | ||
| - | P_{\rm Fe}=1.5~{\rm W} | ||
| - | \end{align*} | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | \begin{align*} | ||
| - | P_{\rm Cu}\approx 0.384~{\rm W} | ||
| - | \end{align*} | ||
| - | |||
| - | Additionally: | ||
| - | \begin{align*} | ||
| - | P_{\rm Fe}=1.5~{\rm W} | ||
| - | \end{align*} | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | 5. Compare ideal and real transformer behavior. | ||
| - | <WRAP group> | ||
| - | <WRAP half column rightalign> | ||
| - | # | ||
| - | <WRAP leftalign> | ||
| - | For the ideal transformer: | ||
| - | \begin{align*} | ||
| - | U_{2,\rm ideal}=23.0~{\rm V} | ||
| - | \end{align*} | ||
| - | |||
| - | For the real transformer: | ||
| - | \begin{align*} | ||
| - | U_{2,\rm real}\approx 22.90~{\rm V} | ||
| - | \end{align*} | ||
| - | |||
| - | The real transformer has copper losses and iron losses: | ||
| - | \begin{align*} | ||
| - | P_{\rm Cu}& | ||
| - | \\ | ||
| - | P_{\rm Fe}& | ||
| - | \end{align*} | ||
| - | |||
| - | So the main differences are: | ||
| - | |||
| - | * the ideal transformer has exactly $U_2=23.0~{\rm V}$, the real transformer has a slightly lower voltage, | ||
| - | * the ideal transformer has no losses, the real transformer has copper and iron losses, | ||
| - | * the ideal transformer has no leakage voltage drop, the real transformer has a load-dependent voltage drop. | ||
| - | |||
| - | For this operating point the transformer is close to ideal, but not exactly ideal. | ||
| - | </ | ||
| - | # | ||
| - | </ | ||
| - | <WRAP half column> | ||
| - | # | ||
| - | The real transformer differs from the ideal transformer by: | ||
| - | |||
| - | * a slightly lower secondary voltage, | ||
| - | * copper and iron losses, | ||
| - | * a load-dependent voltage drop. | ||
| - | |||
| - | For this operating point it is close to ideal, but not exactly ideal. | ||
| - | # | ||
| - | </ | ||
| - | </ | ||
| - | |||
| - | # | ||