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Why three-phase electric power is awesome

The Unofficial ELEC3105 Final Exam Formula Sheet Edit

Magnetic Circuits Edit

Ampere's Law $ \oint H \cdot dl = NI $

$ \phi = \phi_{max} \sin(\omega t) $

$ v = N \frac{d \phi}{d t} $

$ H = \frac{1}{\mu}B $

$ \Re = \frac{l}{\mu_0 S} $

$ \phi = \frac{N I}{\Re} $

Magnetising current $ I_m = (\phi_c \Re_c + \phi_g \Re_c) \frac{1}{N} = (\frac{\phi_c l_c}{\mu_c S_c} + \frac{\phi_g l_g}{\mu_g S_g}) \frac{1}{N} $

Magnetising inductance $ L_m = \frac{N \phi_c}{I_m} = \frac{\lambda}{i_m} $

$ \mu_0 = 4 \pi \times 10^{-7} $

AC Power Computation Edit

Single Phase Edit

Real power $ P = V I \cos \phi $

Reactive power $ Q = V I \sin \phi $ (in VAR)

Complex power $ = S = V I^* = P + jQ $

Apparent power $ = |S| = \sqrt{P^2 + Q^2} = V I $

Three Phase Edit

$ P = 3 V_p I_p \cos \phi = \sqrt{3} V_l I_l \cos \phi $

where $ V_l $ and $ I_l $ are line-to-line voltage and current for delta-connected loads, and $ V_p $ and $ I_p $ are phase voltage and current for star-connected loads.

Transformers Edit

$ E = 4.44 N f \phi_{max} = 4.44 N f B_{max} S_c $

Referring to primary side:

$ a = \frac{N_1}{N_2} $

$ V'_2 = a V_2 $

$ I'_2 = \frac{1}{a} I_2 $

$ R'_2 = a^2 R_2 $

$ L'_2 = a^2 L_2 $

Efficiency $ \eta = \frac{P_{out}}{P_{in}} = \frac{P_{out}}{P_{out} + P_{loss}} $

% Voltage regulation = $ \frac{V_{1} - V_{2rated}}{V_{2rated}} \times 100 $

Electromechanical Energy Conversion Edit

In a rotary system:

$ Torque = \frac{1}{2} i^{2} \frac{dL}{d\theta} $


where:

$ i $ = current

$ L $ = inductance

$ \theta $ = angle of displacement

DC Machines Edit

$ E = k_{E}' \phi \omega = K_{E} \omega $ where $ K_{E} $ = (area of coil) x B x (number of coils)


$ T = k_{T}' \phi i_a = K_T i_a $ where $ K_{T} $ = (area of coil) x B x (number of coils)


The mechanical power used by the load is

$ P = T \omega $

and the power delivered to the load is

$ P = E i_a $

The total loop equation for the armature is

$ V = R_a i_a + k_{E}' \phi \omega $

and the equation for the field coil is simply $ V = R_f i_f $.

Induction Machines Edit

$ n_s = \frac{f_s}{p} $

Thevenin equivalent circuit:

$ V_{Th} = \frac{X_m V_1}{\sqrt{R_1^2 + ( X_1 + X_m )^2}} \approx \frac{X_m V_1}{X_1 + X_m} $

$ Z_{Th} = \frac{j X_m (R_1 + jX_1)}{R_1 + j(X_1 + X_m)} $ but that is pretty bad, try $ R_{Th} \approx R_1 $ and $ X_{Th} \approx X_1 $

Slip for maximum produced torque:

$ s_{Tmax} = \frac{R_2'}{\sqrt{R_{Th}^2 + (X_{Th} + X_2')^2}} $

Power Edit

(Input power) - (Stator copper loss) = (Air gap power)

(Air gap power) - (Rotor copper loss) = (Developed mechanical power)

(Dev. mechanical power) - (Windage and friction losses) = (Output Power)

(Windage and friction losses) = $ \frac{3 {I'}_{nl}^{2} {R'}_{2} ( 1 - s_{nl} ) }{s_{nl}} $

Synchronous Machines Edit

As with induction machines, $ n_s = \frac{f_s}{p} $

The synchronous impedance is calculated from the open circuit test voltage and short-circuit test current: $ Z_s = \frac{E_{OCC}}{I_{SCC}} = R_a + j X_s $.

Often, $ R_a \ll X_s $ so $ Z_s \approx X_s $.

The important power equation is:

$ P = \frac{3 V_a E_a}{X_s} \sin \delta $

where $ \delta $ is the load angle.

Power Electronics Edit

For buck converter:

$ V_o = D V_d $

where D is the duty cycle.

For boost converter:

$ V_o = \frac{V_d}{1-D} $