Question

For the balanced 3-phase transmission line shown, consider the following cases:

Case-1: \(|V_1|=1.1\) p.u., \(|V_2|=0.9\) p.u., \(Z=0.75\angle0^{\circ}\) p.u. and \(\theta_{12}=\theta_1-\theta_2=0^{\circ}\)

Case-2: \(|V_1|=1.1\) p.u., \(|V_2|=0.9\) p.u., \(Z=0.75\angle90^{\circ}\) p.u. and \(\theta_{12}=\theta_1-\theta_2=90^{\circ}\)

Which of the following statements is/are correct about real power loss and reactive power loss in the line?

• A. Real power loss in Case-1 is more than that in Case-2
• B. Real power loss in Case-2 is more than that in Case-1
• C. Reactive power loss in Case-1 is more than that in Case-2
• D. Reactive power loss in Case-2 is more than that in Case-1

Hint:

Split the line impedance into its resistive and reactive parts; each part produces only its own kind of I-squared-Z loss.

Answer 1

The Correct Option is A, D

Step 1: Use power balance instead of the direct $I^2Z$ formula.
The power leaving the sending end minus the power arriving at the receiving end equals the power lost in the line:
\[ S_{loss}=S_1-S_2=V_1I^*-V_2I^* \]
where $I=(V_1-V_2)/Z$.

Step 2: Work out Case-1.
Here $Z=0.75$ (real) and $\theta_{12}=0$, so take $V_1=1.1$, $V_2=0.9$, both real.
\[ I=\frac{1.1-0.9}{0.75}=0.267 \]
Since everything is real, $I^*=I=0.267$.
\[ S_1=V_1I^*=1.1\times0.267=0.293 \]
\[ S_2=V_2I^*=0.9\times0.267=0.240 \]
\[ S_{loss,1}=0.293-0.240=0.053\text{ p.u.} \]
This is entirely real, so $P_{loss,1}=0.053$ and $Q_{loss,1}=0$.

Step 3: Work out Case-2.
Here $Z=j0.75$ and $\theta_{12}=90^\circ$, so take $V_2=0.9\angle0^\circ=0.9$ and $V_1=1.1\angle90^\circ=j1.1$.
\[ I=\frac{j1.1-0.9}{j0.75}=\frac{-0.9+j1.1}{j0.75} \]
Multiply the numerator and denominator by $-j$: since $(-0.9)(-j)=0.9j$ and $(j1.1)(-j)=1.1$, the numerator becomes $1.1+j0.9$, and the denominator becomes $0.75$. So
\[ I=\frac{1.1+j0.9}{0.75}=1.467+j1.2 \]
\[ I^*=1.467-j1.2 \]

Step 4: Compute $S_1$ and $S_2$ for Case-2.
\[ S_1=V_1I^*=(j1.1)(1.467-j1.2)=1.32+j1.614 \]
\[ S_2=V_2I^*=0.9(1.467-j1.2)=1.32-j1.08 \]

Step 5: Subtract to get the loss.
\[ S_{loss,2}=S_1-S_2=(1.32+j1.614)-(1.32-j1.08)=j2.694 \]
So $P_{loss,2}=0$ and $Q_{loss,2}=2.694$.

Step 6: Compare the two cases.
Real loss: $0.053$ in Case-1 versus $0$ in Case-2, so Case-1 has the bigger real loss. Reactive loss: $0$ in Case-1 versus $2.694$ in Case-2, so Case-2 has the bigger reactive loss.

Step 7: Conclude.
\[ \boxed{\text{(A) Real loss bigger in Case-1, and (D) Reactive loss bigger in Case-2 are both correct}} \]