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: Write the general formulas needed.
The current flowing in the line is
\[ I=\frac{V_1-V_2}{Z} \]
and the complex power lost in the line impedance is
\[ S_{loss}=|I|^2Z=P_{loss}+jQ_{loss} \]
So the real power loss depends on the resistive part of \(Z\), and the reactive power loss depends on the reactive part of \(Z\), both scaled by \(|I|^2\).

Step 2: Analyze Case-1.
Here \(Z=0.75\angle0^{\circ}\), a purely resistive line: \(R_1=0.75\), \(X_1=0\). Since \(\theta_{12}=0^{\circ}\), take \(\theta_1=\theta_2=0^{\circ}\), since only the difference matters. Then
\[ V_1-V_2=1.1-0.9=0.2 \]
\[ I_1=\frac{0.2}{0.75}=0.267\text{ p.u.} \]
\[ |I_1|^2=0.0711 \]
Since \(Z\) is purely resistive,
\[ P_{loss,1}=|I_1|^2R_1=0.0711\times0.75=0.0533\text{ p.u.} \]
\[ Q_{loss,1}=|I_1|^2X_1=0 \]

Step 3: Analyze Case-2.
Here \(Z=0.75\angle90^{\circ}=j0.75\), a purely reactive line: \(R_2=0\), \(X_2=0.75\). Since \(\theta_{12}=90^{\circ}\), take \(\theta_2=0^{\circ}\) and \(\theta_1=90^{\circ}\). Then
\[ V_1=1.1\angle90^{\circ}=j1.1,\qquad V_2=0.9\angle0^{\circ}=0.9 \]
\[ V_1-V_2=-0.9+j1.1 \]
\[ |V_1-V_2|=\sqrt{0.9^2+1.1^2}=\sqrt{2.02}=1.421 \]
\[ |I_2|=\frac{1.421}{0.75}=1.895\text{ p.u.} \]
\[ |I_2|^2=3.591 \]
Since \(Z\) is purely reactive here,
\[ P_{loss,2}=|I_2|^2R_2=0 \]
\[ Q_{loss,2}=|I_2|^2X_2=3.591\times0.75=2.693\text{ p.u.} \]

Step 4: Compare real power losses.
\[ P_{loss,1}=0.0533\text{ p.u.},\qquad P_{loss,2}=0 \]
Since \(0.0533>0\), the real power loss in Case-1 is more than in Case-2. So statement (A) is correct and statement (B) is incorrect.

Step 5: Compare reactive power losses.
\[ Q_{loss,1}=0,\qquad Q_{loss,2}=2.693\text{ p.u.} \]
Since \(2.693>0\), the reactive power loss in Case-2 is more than in Case-1. So statement (D) is correct and statement (C) is incorrect.

Step 6: Explain the underlying reason.
When the line is purely resistive (Case-1), all of the \(I^2Z\) loss appears as real power. When the line is purely reactive (Case-2), all of that loss appears as reactive power. This is why Case-1 has real loss but no reactive loss, while Case-2 has reactive loss but no real loss.

Step 7: Final conclusion.
\[ \boxed{\text{Correct statements: (A) Real power loss in Case-1 is more, and (D) Reactive power loss in Case-2 is more}} \]