Deriving power formulas for transmission lines

Question

I'm trying to derive formulas for active power (\$P_s\$) and reactive power (\$Q_s\$) at the sending end of a transmission line shown below:

Where: \$E_s\$ and \$E_r\$ are voltage magnitudes

\$X\$ is the unit line reactance

\$\alpha\$ and \$0°\$ are the load angles

I think the equations are these, but I'm unsure of how to take into account the reactance values.

Would line 2 and 3 be in parallel then in series with the rest?

And would the load angle be the sum of those shown in the diagram (in this case \$\alpha+0°\$)?

$$ S_r = P_r + jQ_r = E_r\dot{I^*}$$

$$ = E_r\left[\frac{E_scos\delta+jE_ssin\delta-E_r}{jX}\right]$$

$$ = \frac{E_sE_r}{X}sin\delta + j\frac{E_sE_rcos\delta-E_r^2}{X}$$

$$\boxed{ P_r = P_s = \frac{E_sE_r}{X}sin\delta \\ Q_r = \frac{E_sE_rcos\delta-E_r^2}{X} \\ Q_s = \frac{E_s^2-E_sE_rcos\delta}{X} }$$

Any help with this would be greatly appreciated!

Answer 1

Yes, combine all of the reactances into a single equivalent value between the sources.

The load angle you show in your pasted equations (\$\delta\$) is the angle by which the sending end voltage leads the receiving end voltage. In your one-line drawing, this is \$-\alpha\$.

It seems to me the easiest way to derive the power transfer equation is to let the receiving end bus be the reference at \$0^°\$ and your sending end have the phase angle.

Below I use \$\theta\$ for the phase angle and \$V\$ for receiving end voltage.

You know,

$$ \overline{I}=\frac{Ee^{j\theta}-V}{jX}$$

and you know,

$$ \overline{S}=Ee^{j\theta}\text{ }{\overline{I}}^* $$

So, you just simplify the above equation using Euler's \$e^{j\theta}=cos\theta+jsin\theta\$, collect terms, and you have it.