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``` > Why are the drawings bad? I draw them with a mouse ### Etc #### FIRST-PASS CHECKS - Double check all are in the correct phase! Multiplications and divisions by $\sqrt{3}$ or $3$ where necessary must be checked! Try annotating everything that does not have an associated phase. - Check conjugate in current. $\bar{S}=\bar{V}\bar{I}^*$ - Check transformer parameters are referred to the proper side #### Y-$\Delta$ transformation (Balanced case) $Z_\Delta=3Z_Y$ ### Types of power factors (From `ENSC2003`) Where $\bar{S}=|\bar{S}|\angle\varphi$: $$ \varphi = \arctan\left(\frac{Q}{P}\right) = \theta_v-\theta_i$$ | | Lagging | Leading | Unity | | ----------- | -------------- | ------------- | ------------ | | Voltage | Current behind | Current ahead | In phase | | Load type | Inductive | Capacitive | Resistive | | $Q$ | $Q>0$ | $Q<0$ | $Q=0$ | | $\varphi$ | $\varphi>0°$ | $\varphi<0°$ | $\varphi=0°$ | | PF [Load] | $[0,1)$ | $[0,1)$ | $1$ | | PF [Source] | $[0,-1)$ | $[0,-1)$ | $-1$ | ## Power types in induction motor | Type | Description | Equivalent terms | | ------------------ | ---------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------- | | Input power | Power into machine. $V_T=V_{3\phi}$, $I_L=I_{3\phi}$ | $P_\text{in}$, $\sqrt{3}V_TI_L\cos(\theta)$ | | Output power | Mechanical output power of the machine, excludes losses | $P_\text{out}$, $P_\text{load}$ | | Converted power | Total electrical power converted to mechanical power, includes useful power and mechanical losses inside machine | $P_\text{conv}$, $P_\text{converted}$, $P_\text{mech}$, $P_\text{developed}$, $\tau_\text{mech}\times\omega_m$ | | Airgap power | Power transmitted over airgap. | $P_\text{AG}$, $\tau_\text{mech}\times\omega_s$ | | Mechanical loss | Power lost to friction and windage | $P_\text{mechanical loss}$, $P_\text{F\\\&W}$, $P_\text{friction and windage}$ | | Core loss | Power lost in machine magnetic material due to hysteresis loss and eddy currents | $P_\text{core}$ | | Rotor copper loss | Due to resistance of rotor windings | $P_r$, $P_\text{RCL}$ | | Stator copper loss | Due to resistance of stator windings | $P_s$, $P_\text{SCL}$ | | Miscellaneous loss | Add 1% to losses to account for other unmeasured losses | $P_\text{misc}$, $P_\text{stray}$ |
$$ \begin{align} P_\text{in}&=P_\text{SCL}+P_\text{RCL}+P_\text{core}+P_\text{F\\\&W}+P_\text{misc}+P_\text{out}\\ P_\text{AG}&=P_\text{RCL}+P_\text{F\\\&W}+P_\text{misc}+P_\text{out}\\ P_\text{mech}&=P_\text{F\\\&W}+P_\text{misc}+P_\text{out} \end{align} $$ Note - assume loss is 0 if not mentioned! | Type | Description | Symbols | | ------------------------- | ------------------------------------------- | --------------------------------------- | | Load torque, Shaft torque | Torque experienced by load after all losses | $\tau_\text{load}$, $\tau_\text{shaft}$ | ## $3\phi$ induction motor ### Etc. - Slip speed $N_\text{slip}=N_{s\text{ (sync)}}-N_r=sN_{s\text{ (sync)}}$ - "1/4 of rated load" != "1/4 times full load" - Means 1/4 of full load slip as it is in the linear region. Accounts for the minimum load. - Rated power stated in machine specification refers to the output power $P_\text{out}$, and excludes all losses. - Speed regulation using machine speed: $$\text{SR}=\frac{N_{r,\text{NL}}-N_{r,\text{FL}}}{N_{r,\text{FL}}}$$ ### Diagram
### Equivalent model #### Assumptions - $x_m\approx X_m$ - $R_c\ggg X_m\Rightarrow r_c\lll x_m$ - $x_m=\frac{{R_c}^2}{{R_c}^2+{X_m}^2}X_m\approx\frac{\cancel{{R_c}^2}}{\cancel{{R_c}^2}}X_m=X_m$ - $r_c\approx {X_m}^2/R_c$ - $R_c\ggg X_m\Rightarrow r_c\lll x_m$ - $r_c=\frac{{X_m}^2}{{R_c}^2+{X_m}^2}R_c\approx\frac{{X_m}^2}{{R_c}^2}R_c=\frac{{X_m}^2}{R_c}$ ### Diagram
### DC test #### $\Delta$ machine $$R_s=\frac{3}{2}\cdot\frac{V_{\text{DC},3\phi}}{I_{\text{DC},3\phi}}$$
#### Y machine $$R_s=\frac{1}{2}\cdot\frac{V_{\text{DC},3\phi}}{I_{\text{DC},3\phi}}$$
### No-load test #### Assumptions - $P_\text{out}=0$ - No output power as no load. - $R_r/s=\infty$ and $I_r=0$ - Infinite rotor resistance, ignore rotor path. #### Diagram Using assumptions, remove rotor part of circuit and only consider stator and magnetizing path.
### Blocked rotor test #### Assumptions - Ignore magnetizing path, $I_m=0$ - $I_r\ggg I_m$ as $R_r/s\ggg Z_m$ - $R_r/s=R_r$, $s=1$ - Slip is $1$ as rotor is blocked. - $x_s=x_r'$ - Same number of turns in stator and rotor - and $x_r=f_0/f_\text{BL} \times x_r'$ - Note: $x_r'$ is the inductance at $f_\text{BL}$, the blocked rotor test frequency which is less than the nominal frequency $f_0$ #### Diagram Ignore magnetizing path
--- ## Single-phase induction motor ### Diagram
### Blocked-rotor #### Diagram
### No-load #### Diagram
## Synchronous machine ### Etc $$E_A=V_{1\phi}-I_A(R_A+j X_s)$$ $$I_A=\text{CONJUGATE}\left(\frac{|S_{3\phi}|\angle\arccos(x)}{3V_{1\phi}}\right), \begin{cases}x=+\text{PF} && \text{lagging} \\ x=-\text{PF} && \text{leading}\end{cases}$$ ### Voltage regulation $$\text{VR}=\frac{|V_\text{NL}|-|V_\text{FL}|}{|V_\text{FL}|}=\frac{|E_A|-|V_{1\phi,\text{rated}}|}{|V_{1\phi,\text{rated}}|}$$ - $V_\text{FL}$ is the full-load voltage which is the full-load/maximum rated voltage at the output terminal. - Calculate $E_A$ at full load by calculating the current as shown above. - $V_\text{NL}$ is the no-load voltage, which in the no-load case will be $E_A$. | No-load | Full-load | | ---------------------------------------------------------------- | ---------------------------------------------------------------- | |
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| | Power factor | Voltage regulation | | ------------ | ------------------ | | Lagging | Positive | | Unity | Near 0 | | Leading | Negative | ### Open and short circuit test #### **Note** - double-check if the axis refers to per-phase or line voltage/current. | Open-circuit test | Short-circuit test | | ----------------------------------------------------------------- | ---------------------------------------------------------------- | |
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| ### Power flow $P_\text{out}$ is the rated power $$P_\text{out}=S_\text{rated}\times \text{PF}$$ $$ \begin{align} P_\text{in}&=P_\text{copper}+P_\text{core}+P_\text{F\\\&W}+P_\text{misc}+P_\text{out}\\ P_\text{mech}&=P_\text{F\\\&W}+P_\text{misc}+P_\text{out} \end{align} $$ --- ### Magnetic circuit analogy | Magnetic circuit | name | | Electrical circuit | name | | ----------------------------------- | ---------------------------- | --- | ------------------------ | ----------------------- | | $$\mathcal F$$ | Magnetomotive force [A-turn] | | $$\mathcal E$$ | Electromotive force [V] | | $$\mathcal R$$ | Reluctance [1/H] | | $$R$$ | Resistance [$\Omega$] | | $$\Phi$$ | Magnetic flux [Wb] | | $$I$$ | Current [A] | | $$\mathcal P=\frac{1}{\mathcal R}$$ | Permeance [H] | | $$G=\frac{1}{R}$$ | Conductivity [$\mho$] | | $$\mathcal F=\Phi\mathcal R$$ | Hopkinson's law | | $$V=IR$$ | Ohm's law | | $$\mathcal R=\frac{l}{\mu A}$$ | | | $$R=\frac{l}{\sigma A}$$ | --- ### Transformers $$Z_P=Z_S\left(\frac{N_P}{N_S}\right)^2=Z_S n^2$$ #### Maximum power. If load is resistive ($jX_\text{load}=0$) then for maximum power transfer: $$R_\text{load}=|{R_\text{src}}^2+j{X_\text{src}}^2|$$ #### Parameter identification
| Open-circuit test | Short-circuit test | | ---------------------------------------------------------------- | ---------------------------------------------------------------- | |
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| #### Voltage regulation $$\text{VR}=\frac{|V_\text{NL,P}|-|V_\text{rated,P}|}{|V_\text{rated,P}|}=\frac{|V_\text{in}|-|V_\text{rated,P}|}{|V_\text{rated,P}|}$$ Ignore shunt resistance. Refer from primary side. Use KVL to determine $V_\text{in}$. Voltage regulation is typically small. $$|V_\text{in}|=|V_\text{rated,P}+I_\text{L,P}\cdot\bar Z|$$
### DC machine | Separately excited machine | Shunt excited | Series excited | | ---------------------------------------------------- | ----------------------------------------------------------------- | --------------------------------------------------------------------------------- | |
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| | | Similar torque-speed characteristic to separately-excited machine | High torque per ampere. Used in high-torque applications | | Requires two independent voltage sources | | Do not run unloaded - infinite speed at 0 torque as $\omega\propto 1/\sqrt{\tau}$ | | Motor control using $R_f$ | Motor control using $R_F$ | Motor control using $V_T$. | #### Starting DC motors $R_A$ might need to be adjusted so it is high initially in large DC motors, as the starting current is high since there is no back-emf created by $E_A$. #### Magnetizating curve When a question specifies the field current or $R_\text{adj}$, refer to magnetization curve. Magnetizating curve is valid at a specific speed $n_{m1}$, and the curve is used to find $E_{A1}$. Using the load condition to find the armature current $I_A=\tau_\text{ind}/(K\varPhi)$, $V_A$ can be used to find a second induced EMF $E_{A2}$. Using $E_{A2}$ find the speed $n_{m2}$ by scaling $n_{m1}$ by $E_{A2}/E_{A1}$. #### Idk $$P_\text{mech}=E_AI_A$$ No-load separately excited machine. Assuming no mechanical losses. $$E_A=V_A\text{ (No load)}$$ $$I_A=0\text{ (No load)}$$ Armature reaction causes increase in speed and causes instability as the core saturates near the poles. Can be reduced with compensating winding which is in series with the armature coil. $$K\Phi\omega=E_A$$ $$K\Phi I_A=\tau$$ For shunt motor $$K\Phi=\frac{V_T-R_AI_A}{\omega}$$ $$\tau=K\Phi I_A=\frac{V_T-R_AI_A}{\omega}I_A$$ Assume no saturation, speed locked(?): This doesn't seem right. We are meant to use the machine constant and the proportionality of current to magnetic flux. $$\frac{E_{A2}}{E_{A1}}=\frac{I_{f2}}{I_{f1}}$$