GATE EE 2024 Set 1 — Question 41

Given a lossless power network with voltage magnitudes at all buses equal to 1 pu, and small voltage phase angle differences. This allows us to use the DC load flow approximation for real power flow.The real power flow from bus $i$ to bus $j$ is given by $P_{ij} = \frac{|V_i||V_j|}{X_{ij}} \sin(\delta_i - \delta_j)$.
Since $|V_i| = |V_j| = 1$ pu and $(\delta_i - \delta_j)$ is small, $\sin(\delta_i - \delta_j) \approx (\delta_i - \delta_j)$.
So, $P_{ij} \approx \frac{\delta_i - \delta_j}{X_{ij}}$.Let $\delta_1, \delta_2, \delta_3$ be the phase angles at buses 1, 2, and 3, respectively.From the figure, the line reactances are:
$X_{12} = \alpha x$ (between bus 1 and bus 2)
$X_{13} = \beta x$ (between bus 1 and bus 3)
$X_{23} = \gamma x$ (between bus 2 and bus 3)Given that the real power flow from bus 1 to bus 2 is 0 pu:
$P_{12} = 0 \implies \frac{\delta_1 - \delta_2}{X_{12}} = 0 \implies \frac{\delta_1 - \delta_2}{\alpha x} = 0$.
Since $\alpha x \neq 0$, this implies $\delta_1 - \delta_2 = 0 \implies \delta_1 = \delta_2$.Now, consider the bus injections $P_1$ and $P_2$. The arrows indicate power flowing out of the buses (loads or net injections).
$P_1$ is the net power injection at bus 1. It flows to bus 2 and bus 3.
$P_1 = P_{12} + P_{13}$. Since $P_{12} = 0$, we have:
$P_1 = P_{13} = \frac{\delta_1 - \delta_3}{X_{13}} = \frac{\delta_1 - \delta_3}{\beta x}$.$P_2$ is the net power injection at bus 2. It flows to bus 1 and bus 3.
$P_2 = P_{21} + P_{23}$. Since $P_{21} = -P_{12} = 0$, we have:
$P_2 = P_{23} = \frac{\delta_2 - \delta_3}{X_{23}} = \frac{\delta_2 - \delta_3}{\gamma x}$.Since we found $\delta_1 = \delta_2$, we can substitute $\delta_1$ for $\delta_2$ in the expression for $P_2$:
$P_2 = \frac{\delta_1 - \delta_3}{\gamma x}$.Given that $P_1 = mP_2$:
$\frac{\delta_1 - \delta_3}{\beta x} = m \left(\frac{\delta_1 - \delta_3}{\gamma x}\right)$.Assuming $\delta_1 \neq \delta_3$ (otherwise $P_1 = P_2 = 0$, which is a trivial case and doesn't lead to a relationship between $\beta$ and $\gamma$), we can cancel the common term $(\delta_1 - \delta_3)$ and $x$ from both sides:
$\frac{1}{\beta} = \frac{m}{\gamma}$
$\gamma = m\beta$.The final answer is $\boxed{\gamma = m\beta}$.