Brushless AC
Lets start with a brushless AC machine first & Field Orientated Control.
So BLAC machine's have their stator windings sinusoidally distributed (higher concentration close to the tooth, lower for the outer turns)
Now to control such machine is relatively complex and you can use a field orientated control ( F.O.C. ).Using Park's transform (Clark + rotation) the 3phase sinus stator current's can be transformed 1st into a rotating 2phase representation
Park Transforms - General
$$
I_{\alpha \beta 0 } = \frac{2}{3}\begin{bmatrix}
1 & \frac{-1}{2} & \frac{-1}{2} \\
0 & \frac{\sqrt{3}}{2} &\frac{-\sqrt{3}}{2} \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2}
\end{bmatrix}
$$
or simply
\$I_\alpha = I_a \$
\$I_\beta = \frac{2I_b + Ia}{\sqrt{3}} \$
And these phasors can further be reduced to two DC quantities via a rotating frame of reference transform
$$
\begin{bmatrix}
I_d\\
I_q
\end{bmatrix} = \begin{bmatrix}
Cos(\Theta ) & Sin(\Theta )\\
-Sin(\Theta ) & Cos(\Theta )
\end{bmatrix} \cdot \begin{bmatrix}
I_\alpha\\
I_\beta
\end{bmatrix}
$$
These two DC terms (with Id usually controlled to 0, unless field weakening is desired) are simple inputs to two classic PI loops to control Id and Iq (the output of which is Vd & Vq).
This Vd & Vq, via inverse Clark&PArk produce the 3phase voltage that will be applied to the stator (insert SVM or SPWM)
Great, it works
Brushless DC & general Permanent Magnet machine equations
Thing is BLDC machines produce higher torque than a sinusoidally wound stator (downto the higher concentration of winding around the teeth to produce the flattened line-line profile). The downside is the torque ripple. This is mostly due to simplifying the control downto which equally limits the effective bandwidth of the controller.
A park-like transform can be applied to overcome some of these shortcomings as the aim is to produce a stator stimulus closer to the airgap profile - tpyically a Quazi squarewave (30-120-30) is superimposed onto a trapezoidal backEMF (60-60-60).
First of all you need the machine equation for a BLDC machine (which is the same for a BLAC machine).
$$
\begin{bmatrix}
V_a\\
V_b\\
V_c
\end{bmatrix} = R_s \begin{bmatrix}
i_a\\
i_b\\
i_c
\end{bmatrix} + L\frac{\mathrm{d} }{\mathrm{d} t}\begin{bmatrix}
i_a\\
i_b\\
i_c
\end{bmatrix} + \begin{bmatrix}
e_a\\
e_b\\
e_c
\end{bmatrix}$$
\$e_a, e_b, e_c\$ are the 3 backEMF's that are generated for a rotating rotor. Via Faraday's law
$$
\varepsilon = - \frac{\mathrm{d} \Phi _B}{\mathrm{d} t}
$$
BackEMF is the rate of change of flux
$$
\frac{\mathrm{d} \Phi }{\mathrm{d} t} = \frac{\mathrm{d} \Phi }{\mathrm{d} \Theta}\frac{\mathrm{d} \Theta }{\mathrm{d} t} = \frac{\mathrm{d} \Phi }{\mathrm{d} \Theta}\omega
$$
$$
e = \omega {\Phi}'
$$
Thus:
$$
\begin{bmatrix}
e_a\\
e_b\\
e_c
\end{bmatrix} = \omega_e \frac{\mathrm{d} }{\mathrm{d} \vartheta_e}\begin{bmatrix}
\Phi_a\\
\Phi_b\\
\Phi_c
\end{bmatrix} = \omega_e \begin{bmatrix}
{\Phi}'_a\\
{\Phi}'_b\\
{\Phi}'_c
\end{bmatrix}$$
As the total torque produced is the summation of the 3 phases producing torque (factoring in the pole-pair count):
Torque equation
\$T_e = P({\Phi}'_a i_a + {\Phi}'_b i_b + {\Phi}'_b i_b)\$
Now, the Park transform can be applied to voltage, currents and flux, thus:
$$
T_e = \frac{3}{2}P({\Phi}'_\alpha i_\alpha + {\Phi}'_\beta i_\beta + {\Phi}'_0 i_0) = \frac{3}{2}P({\Phi}'_d i_d + {\Phi}'_q i_q + {\Phi}'_0 i_0) = \frac{3}{2}P{\Phi}'_q i_q
$$
For a BLAC machine \${\Phi}'_q\$ is a simple sinus profile (and thus a lookup table, CORDIC... can be used) but for a BLDC an appropriate lookup table of flux vs angle is required specific for the machine being used.
This is correct.
\$K_t\$ is approximately equal to \$K_e\$ (approximatly as \$K_e\$ is defined as open-terminal voltage and \$K_t\$ is defined at rated current & the machine may be saturating)
\$K_v\$ is the reciprocal of \$K_e\$
The units of \$K_t\$ are \$\frac{Nm}{A}\$
The units of \$K_e\$ are \$\frac{V}{\omega}\$ (NOTE: from a magnetic point of view this is peak line-line voltage...)
so... 2100\$\frac{rpm}{v}\$ == 219.911 \$\frac{rad/s}{V}\$ = \$K_e\$
Thus taking the reciprical of \$K_e\$
\$K_e = \frac{1}{K_v}\$ = 0.0045473
Best Answer
When talking about BLDC motors (and by BLDC I assume you mean a brushless, synchronous motor with a trapezoidal phase back-emf), there is no standard in the motor world as to whether \$ e\$ refers to line to line back-emf or phase back-emf, or whether it refers to peak to peak or RMS values. Similarly, there is no standard for what \$ K_e\$ means exactly. Often \$ K_e\$ means peak line-to-line back-emf constant. Good datasheets will explicitly tell you but many won't.
Your confusion between the back-emf constant for brushed DC motors and BLDC motors is common. Very little is published about the difference and many, many people just assume that what works for brushed DC motors also works for BLDC motors. I can explain the difference to you but first we need to understand the back-emf constant for brushed DC motors.
Consider your equation \$e = K_e * \omega_m\$. In a brushed DC motor there are a number of coils that produce close to a triangular-shaped back emf. But the commutator effectively acts as a rectifier to create a DC back-emf. So in your equation, \$ e\$ is a DC value.
In a brushed DC motor you can also say that \$T * \omega_m = e * i\$, where \$ i\$ is the DC current. If you substitute your equation into this equation you get \$ T = K_t * i\$, where \$K_t = K_e\$. All this tells us is that the back-emf constant and the torque constant of an ideal DC motor are equal.
When you start looking at brushless motors (generally with 3 phases), you have to take a couple of things into consideration. First, we don't have a commutator that serves to rectify the back-emf to DC. A brushless motor could have a trapezoidal back-emf or a sinusoidal back-emf or something in between. Second, our current isn't DC into a brushless motor. Typically a brushless motor will either be sinusoidal or it will be more of a square-wave that is positive for 120 electrical degrees, off for 60 electrical degrees, negative for 120 electrical degrees, and then off for another 60 electrical degrees.
As it turns out, if you look at a BLDC motor that has a trapezoidal back-emf and is controlled with a square-wave current as described in the previous paragraph, the equations for the back-emf constant and the torque constant are exactly the same as for an ideal brushed DC motor. The only caveat is that \$ K_e\$, \$ K_t\$, and \$ e\$ must be defined as in terms of their peak line-to-line values and \$ i \$ as a peak line current.
These equations do not necessarily hold for other back-emf and current waveforms. For example, you cannot say that \$K_t = K_e\$ if you have a sinusoidal back-emf and sinusoidal currents, assuming we are still talking about peak line-to-line back-emf and peak line current.
You can read more about these relationships in James Meavey's thesis found here or in Hendershot and Miller's book Design of Brushless Permanent-Magnet Machines.