Question 6 of 8: Quarter-wave line: generator EMF variation with load power

Question 6: Quarter-wave line: generator EMF variation with load power (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An ideal 60 Hz source feeds a quarter-wavelength, 1000 $\Omega$ line through a 1:5 step-up transformer; the far end is a purely resistive load held at 100 kV whose absorbed power falls from 5 MW to 4 MW.

Find. The percentage change in generator EMF needed to hold $V_L$ at 100 kV as the load power falls from 5 MW to 4 MW.

Approach. Confirm the line really is a quarter wavelength, then use the quarter-wave inversion in its current form — the input voltage of a $\lambda/4$ line is $Z_0$ times the load current — to convert each load condition into a required primary EMF, and compare the two.

1. Check the electrical length. $$\lambda = \frac{v_p}{f} = \frac{3\times10^{8}}{60} = 5\times10^{6}\ \text{m} = 5000\ \text{km}, \qquad \frac{\lambda}{4} = 1250\ \text{km}$$ which matches the stated line length exactly, so $\beta\ell = \pi/2$.
2. The quarter-wave relations. For $\beta\ell = \pi/2$ the ABCD chain reduces to $A = D = 0$, $B = jZ_0$, $C = j/Z_0$, so $$V_{\text{in}} = jZ_0 I_L, \qquad I_{\text{in}} = j\frac{V_L}{Z_0}$$ Taking magnitudes, $|V_{\text{in}}| = Z_0|I_L|$. This is the current form of the impedance relation the paper supplies, $Z(s)\,Z(s\pm\lambda/4) = Z_0^{2}$ — a quarter-wave line is an inverter, so a constant-voltage load appears at the input as a constant-current source and vice versa.
3. Load conditions. The load is resistive and its terminal voltage is pinned at 100 kV, so $I_L = P_L/V_L$ and $R_L = V_L^2/P_L$: $$P_L = 5\ \text{MW}:\quad I_L = \frac{5\times10^{6}}{10^{5}} = 50.0\ \text{A}, \quad R_L = 2000\ \Omega$$$$P_L = 4\ \text{MW}:\quad I_L = \frac{4\times10^{6}}{10^{5}} = 40.0\ \text{A}, \quad R_L = 2500\ \Omega$$
4. Voltage required at the line input. Applying $|V_{\text{in}}| = Z_0|I_L|$ to each case, $$|V_{\text{in}}|_{5\,\text{MW}} = 1000(50.0) = 50.0\ \text{kV}, \qquad |V_{\text{in}}|_{4\,\text{MW}} = 1000(40.0) = 40.0\ \text{kV}$$ As a check, $Z_{\text{in}} = Z_0^2/R_L$ gives 500 $\Omega$ and 400 $\Omega$, and $\sqrt{P_L Z_{\text{in}}}$ reproduces the same two voltages.
5. Refer back through the transformer. The ideal 1:5 step-up transformer divides the required primary EMF by 5: $$E_{5\,\text{MW}} = \frac{50.0\ \text{kV}}{5} = 10.0\ \text{kV}, \qquad E_{4\,\text{MW}} = \frac{40.0\ \text{kV}}{5} = 8.0\ \text{kV}$$ (The generator has zero internal impedance, so its EMF is its terminal voltage — no regulation drop to allow for.)
6. Percentage variation. Referred to the 5 MW operating point, $$\frac{\Delta E}{E} = \frac{10.0 - 8.0}{10.0} = 0.20 \;\Longrightarrow\; \boxed{20\ \%\ \text{reduction in generator EMF}}$$ Expressed the other way round (raising the load from 4 MW to 5 MW) the same change is a 25 % increase on the 8.0 kV value.

The reason the answer is so clean is worth stating: on a quarter-wave line the input voltage tracks the load current, and at fixed load voltage the current is directly proportional to power. The EMF therefore scales exactly as the load power, so a 5:4 power ratio is a 5:4 voltage ratio, i.e. 20 %. This is the same inversion that makes a quarter-wave line a natural constant-current driver, and it is why $\lambda/4$ sections are used as impedance transformers and as fault-tolerant feeds in practice. At 60 Hz, of course, a genuine 1250 km quarter-wave line is an idealisation — real long-distance AC lines are compensated well short of that length — but the algebra is the standard textbook exercise.