Locked rotor, rotor speed and synchronous speed

From Quora Profile

When AC voltage is applied to stator windings, stator AC currents begin to flow through windings. Stator AC currents lead to changing magnetic fields around stator windings, which means magnetic field strength, magnetic flux and magnetic flux density (B) due to the closed loops (coils). Stator magnetic flux induces voltage on rotor bars ( or windings), then rotor currents begin to flow and rotor currents lead to changing magnetic field around rotor windings after the rotation of rotor. Now we have stator magnetic field and rotor magnetic field.

Rotor changing magnetic field probably causes emf on stator windings, but there will be net magnetic flux and net voltage in induction machine. Rotor windings are current carrying conductors in the stator’s changing magnetic field due to AC voltage and according to Lorentz Law, rotor begins to rotate in mechanical speed. Slip occurs because there will be delay or difference between stator magnetic field speed and rotor mechanical speed.

Regarding rotor mechanical speed and rotor electrical frequency:

n_rm=rotor mechanical speed

f_re=rotor electrical frequency

Related to stator electrical frequency and stator’s magnetic field rotating speed:

f_se=stator electrical frequency = Applied voltage frequency ( 50 Hz, or 60 Hz)

n_synch= stator rotating magnetic field speed

f_se=n_synch*P/120

P=number of poles of machine

slip= (stator rotating magnetic field speed - rotor mechanical speed) / stator rotating magnetic field speed

that is,

s=(n_synch-n_rm)/n_synch

s=1-n_rm/n_synch

and rotor mechanical speed:

n_rm=(1-s)n_synch

s=(n_synch-n_rm)/n_synch if n_rm=0 then s=1 which means rotor is stationary or locked rotor but stator has rotating magnetic field due to the applied AC voltage, which induces emf on rotor windings

Now frequency equations:

Rotor electrical frequency=slip x stator electrical frequency

f_re=s.f_se

If you use the equations for “s” and “f_se”

f_re=((n_synch-n_rm)/n_synch)*(n_synch*P/120)

f_re=(n_synch-n_rm)*P/120

It is the relation between electrical frequency of rotor and mechanical speed of rotor and stator rotating magnetic field speed.

Then again locked rotor conditions, s=1, and n_rm =0, but according to ”f_re=(n_synch-n_rm)*P/120″ there will be “f_re”that is electrical rotor frequency due to the “n_synch” which means rotating magnetic field due to the applied AC voltage to stator terminals. Rotating magnetic field induces voltage on rotor windings. These are maximum rotor voltage and max. rotor electrical frequency when locked rotor conditions.

If n_rm=n_sycnh, then “s=(n_synch-n_rm)/n_synch” and slip will be zero.

f_re=s.f_se, and s=0 then there will be no electrical frequency of rotor.

Voltage equations are;

E_r=rotor voltage and E_rmax = max. rotor voltage ( in case of locked rotor)

E_r=s.E_rmax

and if s=0 then there will be no rotor voltage when rotor mechanical speed and stator rotating magnetic speed is equal.

Instantaneous Power Equation, Active and Reactive Power

 

Nominal T Method of Transmission Lines with Medium Length

The equations and phasor diagram of T method of medium length transmission lines will be obtained. For this purpose the figure shows the one phase of three phase lines.

T Method Equations

The circuit can be separated in three parts based on inputs and outputs to obtain equations for T method.

Basic equation in matrice will be obtained by A, B, C, D parameters.

\[\left( \begin{array}{l}Vs\\{I_S}\end{array} \right) = \left( {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right).\left( \begin{array}{l}{V_R}\\{I_R}\end{array} \right)\]

In section 1 output of the circuit will be V1 and Is1 when inputs are Vs and Is. However, consider all the electrical parameters in complex form a + jb, which means there will be a phase angle and magnitude of the currents and voltages. 

\[\begin{array}{l} - {V_S} + {I_{S1}}\frac{Z}{2} + {V_1} = 0\\{V_S} = {V_1} + \frac{Z}{2}{I_{S1}}\\{I_S} = 0.{V_1} + {I_{S1}}\\\left( \begin{array}{l}{V_S}\\{I_S}\end{array} \right) = \left( {\begin{array}{*{20}{c}}1&{\frac{Z}{2}}\\0&1\end{array}} \right)\left( \begin{array}{l}{V_1}\\{I_{S1}}\end{array} \right)\end{array}\]

In section 2, if we use the inputs as V₁ and I_s1 which previously are the outputs of section 1, then the outputs will be V₂ and I_s2. As it is seen V₂=V₀ but no need to use for the matrice. V₀ will be used for phasor diagram later.

\[\begin{array}{l}{V_1} = {V_2} + 0.{I_{S2}}\\{I_{S1}} = {I_Y} + {I_{S2}}\\{I_{S1}} = Y.{V_2} + {I_{S2}}\\\left( \begin{array}{l}{V_1}\\{I_{S1}}\end{array} \right) = \left( {\begin{array}{*{20}{c}}1&0\\Y&1\end{array}} \right)\left( \begin{array}{l}{V_2}\\{I_{S2}}\end{array} \right)\end{array}\]

Remember the basic formula.

\[\begin{array}{l}V = IZ\\I = V/Z\\Y = 1/Z\\I = VY\end{array}\]

In section 3, V₂ and I_s2 , the outputs of section 2, will be the inputs and finally I_R and V_R will be the outputs in similar way.

\[\begin{array}{l}{V_2} = {V_R} + \frac{Z}{2}{I_R}\\{I_{S2}} = 0.{V_R} + {I_R}\\\left( \begin{array}{l}{V_2}\\{I_{S2}}\end{array} \right) = \left( {\begin{array}{*{20}{c}}1&{\frac{Z}{2}}\\0&1\end{array}} \right)\left( \begin{array}{l}{V_R}\\{I_R}\end{array} \right)\end{array}\]

Consider all matrices again, and replace the values into previous ones.

\[\left( \begin{array}{l}{V_1}\\{I_{S1}}\end{array} \right) = \left( {\begin{array}{*{20}{c}}1&0\\Y&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&{\frac{Z}{2}}\\0&1\end{array}} \right)\left( \begin{array}{l}{V_R}\\{I_R}\end{array} \right)\]

\[\left( \begin{array}{l}{V_S}\\{I_S}\end{array} \right) = \left( {\begin{array}{*{20}{c}}1&{\frac{Z}{2}}\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\Y&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&{\frac{Z}{2}}\\0&1\end{array}} \right)\left( \begin{array}{l}{V_R}\\{I_R}\end{array} \right)\]

\[\left( \begin{array}{l}{V_S}\\{I_S}\end{array} \right) = \left( {\begin{array}{*{20}{c}}{1 + \frac{{ZY}}{2}}&{0 + \frac{Z}{2}}\\{0 + Y}&{0 + 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&{\frac{Z}{2}}\\0&1\end{array}} \right)\left( \begin{array}{l}{V_R}\\{I_R}\end{array} \right)\]

\[\left( \begin{array}{l}{V_S}\\{I_S}\end{array} \right) = \left( {\begin{array}{*{20}{c}}{1 + \frac{{ZY}}{2}}&{\left( {1 + \frac{{ZY}}{2}} \right).\frac{Z}{2} + \frac{Z}{2}}\\Y&{1 + \frac{{ZY}}{2}}\end{array}} \right)\left( \begin{array}{l}{V_R}\\{I_R}\end{array} \right)\]

\[\begin{array}{l}\left( {1 + \frac{{ZY}}{2}} \right).\frac{Z}{2} + \frac{Z}{2} = \frac{Z}{2}\left( {1 + \frac{{ZY}}{2} + 1} \right)\\ = Z\left( {1 + \frac{{ZY}}{4}} \right)\end{array}\]

T method equation :

\[\left( \begin{array}{l}{V_S}\\{I_S}\end{array} \right) = \left( {\begin{array}{*{20}{c}}{1 + \frac{{ZY}}{2}}&{Z\left( {1 + \frac{{ZY}}{4}} \right)}\\Y&{1 + \frac{{ZY}}{2}}\end{array}} \right)\left( \begin{array}{l}{V_R}\\{I_R}\end{array} \right)\]

Phasor Diagram of T Method

Equations to obtain phasors. Impedance will be in complex form with angle and magnitude as well. 

\[\left| {\frac{Z}{2}} \right|\angle \theta \]

\[\left| {{I_R}} \right|\angle - {\phi _R}\]

Inductive current lags reference voltage by negative sign angle. V_R receiving voltage is the reference voltage for phasor diagram.

Based on right side of the circuit, voltage drop can be calculated by V₀.

\[\overline {{V_0}} = \overline {{V_R}} + \overline {\Delta {V_2}} = \left| {{V_0}} \right|\angle {\delta _0}\]

Sending current with receiving current can be written in complex form. Notice that admitance with V₀ which means the angle of the current flowing through admitance I_Y angle is 90 degree.

\[\begin{array}{l}\overline {{I_S}} = \overline {{I_Y}} + \overline {{I_R}} = \overline Y .\overline {{V_0}} + \overline {{I_R}} \\\overline {{I_S}} = \left| Y \right|\angle 90^\circ .\left| {{V_0}} \right|\angle {\delta _0} + \left| {{I_R}} \right|\angle - {\phi _R}\\\overline {{I_S}} = \left| {Y{V_0}} \right|\angle \left( {90^\circ + {\delta _0}} \right) + \left| {{I_R}} \right|\angle - {\phi _R}\end{array}\]

Voltage drop in left side of the circuit:

 \[\overline {\Delta {V_1}} = \overline {{I_S}} \frac{{\overline Z }}{2} = \left| {{I_S}} \right|\angle - {\phi _S}\left| {\frac{Z}{2}} \right|\angle \theta = \left| {{I_S}\frac{Z}{2}} \right|\angle \theta - {\phi _S}\]

Sending voltage with left side voltage drop and V₀ . Consider V_S angle with respect to V_R receiving voltage which is reference voltage for phasor diagram.

\[{\overline V _S} = \overline {{V_0}} + \overline {\Delta {V_1}} = \left| {{V_S}} \right|\angle {\delta _{}}\]

Phasor diagram can be drawn now.