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Why are the drawings bad?

I draw them with a mouse

Etc

FIRST-PASS CHECKS

Y-Δ\Delta transformation (Balanced case)

ZΔ=3ZYZ_\Delta=3Z_Y

Types of power factors (From ENSC2003)

Where Sˉ=Sˉφ\bar{S}=|\bar{S}|\angle\varphi:

φ=arctan(QP)=θvθi \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
QQ Q>0Q>0 Q<0Q<0 Q=0Q=0
φ\varphi φ>0°\varphi>0° φ<0°\varphi<0° φ=0°\varphi=0°
PF [Load] [0,1)[0,1) [0,1)[0,1) 11
PF [Source] [0,1)[0,-1) [0,1)[0,-1) 1-1

Power types in induction motor

Type Description Equivalent terms
Input power Power into machine. VT=V3ϕV_T=V_{3\phi}, IL=I3ϕI_L=I_{3\phi} PinP_\text{in}, 3VTILcos(θ)\sqrt{3}V_TI_L\cos(\theta)
Output power Mechanical output power of the machine, excludes losses PoutP_\text{out}, PloadP_\text{load}
Converted power Total electrical power converted to mechanical power, includes useful power and mechanical losses inside machine PconvP_\text{conv}, PconvertedP_\text{converted}, PmechP_\text{mech}, PdevelopedP_\text{developed}, τmech×ωm\tau_\text{mech}\times\omega_m
Airgap power Power transmitted over airgap. PAGP_\text{AG}, τmech×ωs\tau_\text{mech}\times\omega_s
Mechanical loss Power lost to friction and windage Pmechanical lossP_\text{mechanical loss}, PF&WP_\text{F\\\&W}, Pfriction and windageP_\text{friction and windage}
Core loss Power lost in machine magnetic material due to hysteresis loss and eddy currents PcoreP_\text{core}
Rotor copper loss Due to resistance of rotor windings PrP_r, PRCLP_\text{RCL}
Stator copper loss Due to resistance of stator windings PsP_s, PSCLP_\text{SCL}
Miscellaneous loss Add 1% to losses to account for other unmeasured losses PmiscP_\text{misc}, PstrayP_\text{stray}

Pin=PSCL+PRCL+Pcore+PF&W+Pmisc+PoutPAG=PRCL+PF&W+Pmisc+PoutPmech=PF&W+Pmisc+Pout \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 τload\tau_\text{load}, τshaft\tau_\text{shaft}

3ϕ3\phi induction motor

Etc.

Diagram

Equivalent model

Assumptions

Diagram

DC test

Δ\Delta machine

Rs=32VDC,3ϕIDC,3ϕR_s=\frac{3}{2}\cdot\frac{V_{\text{DC},3\phi}}{I_{\text{DC},3\phi}}

Y machine

Rs=12VDC,3ϕIDC,3ϕR_s=\frac{1}{2}\cdot\frac{V_{\text{DC},3\phi}}{I_{\text{DC},3\phi}}

No-load test

Assumptions

Diagram

Using assumptions, remove rotor part of circuit and only consider stator and magnetizing path.

Blocked rotor test

Assumptions

Diagram

Ignore magnetizing path


Single-phase induction motor

Diagram

Blocked-rotor

Diagram

No-load

Diagram

Synchronous machine

Etc

EA=V1ϕIA(RA+jXs)E_A=V_{1\phi}-I_A(R_A+j X_s)

IA=CONJUGATE(S3ϕarccos(x)3V1ϕ),{x=+PFlaggingx=PFleadingI_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

VR=VNLVFLVFL=EAV1ϕ,ratedV1ϕ,rated\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}}|}

No-load Full-load
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

Power flow

PoutP_\text{out} is the rated power

Pout=Srated×PFP_\text{out}=S_\text{rated}\times \text{PF}

Pin=Pcopper+Pcore+PF&W+Pmisc+PoutPmech=PF&W+Pmisc+Pout \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
F\mathcal F
Magnetomotive force [A-turn]
E\mathcal E
Electromotive force [V]
R\mathcal R
Reluctance [1/H]
RR
Resistance [Ω\Omega]
Φ\Phi
Magnetic flux [Wb]
II
Current [A]
P=1R\mathcal P=\frac{1}{\mathcal R}
Permeance [H]
G=1RG=\frac{1}{R}
Conductivity [\mho]
F=ΦR\mathcal F=\Phi\mathcal R
Hopkinson's law
V=IRV=IR
Ohm's law
R=lμA\mathcal R=\frac{l}{\mu A}
R=lσAR=\frac{l}{\sigma A}

Transformers

ZP=ZS(NPNS)2=ZSn2Z_P=Z_S\left(\frac{N_P}{N_S}\right)^2=Z_S n^2

Maximum power.

If load is resistive (jXload=0jX_\text{load}=0) then for maximum power transfer:

Rload=Rsrc2+jXsrc2R_\text{load}=|{R_\text{src}}^2+j{X_\text{src}}^2|

Parameter identification

Open-circuit test Short-circuit test

Voltage regulation

VR=VNL,PVrated,PVrated,P=VinVrated,PVrated,P\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 VinV_\text{in}.

Voltage regulation is typically small.

Vin=Vrated,P+IL,PZˉ|V_\text{in}|=|V_\text{rated,P}+I_\text{L,P}\cdot\bar Z|

DC machine

Separately excited machine Shunt excited Series excited
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 ω1/τ\omega\propto 1/\sqrt{\tau}
Motor control using RfR_f Motor control using RFR_F Motor control using VTV_T.

Starting DC motors

RAR_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 EAE_A.

Magnetizating curve

When a question specifies the field current or RadjR_\text{adj}, refer to magnetization curve. Magnetizating curve is valid at a specific speed nm1n_{m1}, and the curve is used to find EA1E_{A1}. Using the load condition to find the armature current IA=τind/(KΦ)I_A=\tau_\text{ind}/(K\varPhi), VAV_A can be used to find a second induced EMF EA2E_{A2}. Using EA2E_{A2} find the speed nm2n_{m2} by scaling nm1n_{m1} by EA2/EA1E_{A2}/E_{A1}.

Idk

Pmech=EAIAP_\text{mech}=E_AI_A

No-load separately excited machine. Assuming no mechanical losses.

EA=VA (No load)E_A=V_A\text{ (No load)}

IA=0 (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.

KΦω=EAK\Phi\omega=E_A

KΦIA=τK\Phi I_A=\tau

For shunt motor

KΦ=VTRAIAωK\Phi=\frac{V_T-R_AI_A}{\omega}

τ=KΦIA=VTRAIAωIA\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.

EA2EA1=If2If1\frac{E_{A2}}{E_{A1}}=\frac{I_{f2}}{I_{f1}}