--- title: "Reversible power flow on a DC line" subject: "Power Electronics" chapter: 7 key: "semester_7/power_electronics:reversible-power-flow-dc-line" source: https://www.bctengineer.com/q/semester_7/power_electronics/reversible-power-flow-dc-line site: BCT Engineering --- # Reversible power flow on a DC line Power Electronics · Chapter 7: HVDC power transmission - Asked in: 2073 Shrawan, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/reversible-power-flow-dc-line - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-7 ## Question With the help of suitable circuit diagram, describe the operation of reversible power flow on DC line. ## Short answer In an HVDC line with thyristor converters, power reversal is achieved by reversing voltage polarity, not current direction. Normally α1<90° (rectifier) and α2>90° (inverter); for reversal, α1>90° and α2<90°, making Vd negative while Id direction stays fixed. Current flows only if α1+α2<180°, and for low line resistance α2≈180°–α1. ## Answer In a DC (HVDC) line with thyristor converters at both ends, the direction of current cannot reverse because thyristors conduct only one way. Power flow is reversed by **reversing the polarity of the DC voltage**, which is done by changing the firing angles: the rectifier becomes an inverter and the inverter becomes a rectifier. #### Circuit ```text AC sys 1 AC sys 2 | Conv 1 R, Ld Conv 2 | +--[3-ph full]--+--/\/\--+--[3-ph full]--+ bridge Vd1 (+) Id -> (+) Vd2' bridge | | +---------------+ Vd1 = Vdo cos a1 Vd2' = -Vdo cos a2 ``` Both converters are three-phase fully controlled bridges whose average output is $V_{do}\cos\alpha$: positive for $0 < \alpha < 90°$ (rectifier) and negative for $90° < \alpha < 180°$ (inverter). #### Operation Let both converters have the same no-load voltage $V_{do} = \dfrac{3\sqrt{2}V_L}{\pi}$ and the line resistance be $R$. Converter 1 gives $V_{d1} = V_{do}\cos\alpha_1$. Converter 2 is connected in opposition, so its terminal voltage seen by the line is $V_{d2}' = -V_{do}\cos\alpha_2$. Thyristors allow current $I_d$ in **one direction only**: $$ I_d = \frac{V_{d1} - V_{d2}'}{R} = \frac{V_{do}(\cos\alpha_1 + \cos\alpha_2)}{R} \ge 0 $$ Power sent by converter 1 and received by converter 2: $$ P_1 = V_{do}\cos\alpha_1\,I_d, \qquad P_2 = -V_{do}\cos\alpha_2\,I_d, \qquad P_1 - P_2 = I_d^2R $$ **Case 1: power from system 1 to system 2.** - $\alpha_1 < 90°$: converter 1 is a rectifier ($V_{d1} > 0$, $P_1 > 0$). - $\alpha_2 > 90°$: converter 2 is an inverter ($\cos\alpha_2 < 0$, $P_2 > 0$ delivered to system 2). - Line voltage is positive at the top conductor. **Case 2: power from system 2 to system 1 (reversal).** - Make $\alpha_1 > 90°$: converter 1 becomes an inverter ($V_{d1} < 0$). - Make $\alpha_2 < 90°$: converter 2 becomes a rectifier. - The **current direction stays the same**; only the **polarity of the DC line voltage reverses**, so $P = V_dI_d$ changes sign. **Condition for current flow in both cases:** $$ \cos\alpha_1 + \cos\alpha_2 > 0 \;\Rightarrow\; \cos\alpha_1 > \cos(180° - \alpha_2) \;\Rightarrow\; \alpha_1 + \alpha_2 < 180° $$ For a line of small resistance, $I_dR$ is small, so $$ \alpha_1 + \alpha_2 \approx 180°, \quad \text{i.e. } \alpha_2 \approx 180° - \alpha_1 $$ The magnitude of $I_d$ (and power) is set by how much $\alpha_1 + \alpha_2$ is less than $180°$; the direction of power is set by which converter has $\alpha < 90°$. | Mode | $\alpha_1$ | $\alpha_2$ | Line polarity | Power | |---|---|---|---|---| | Normal | $< 90°$ (rectifier) | $> 90°$ (inverter) | + | 1 → 2 | | Reversed | $> 90°$ (inverter) | $< 90°$ (rectifier) | − | 2 → 1 | #### Practical sequence for reversal 1. Reduce the current order to a minimum value. 2. Increase $\alpha_1$ beyond $90°$ and decrease $\alpha_2$ below $90°$ gradually; the line voltage passes through zero and reverses. 3. Raise the current order again; power now flows from system 2 to system 1. No switching of conductors is needed and reversal takes only a fraction of a second, which is an important advantage of HVDC links. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In an HVDC link with thyristor converters, how is the direction of power flow reversed? A. By reversing the DC current direction using the thyristors B. By reversing the DC voltage polarity through the firing angles C. By physically switching the line conductors D. By swapping the AC supply phase sequence Answer: B. Thyristors conduct one way only, so current cannot reverse. Changing the firing angles reverses the DC voltage polarity and hence the sign of P = Vd*Id. (Key point: How reversal is achieved) 2. A three-phase fully controlled bridge with average output Vdo cos(alpha) operates as an inverter when: A. 0 < alpha < 90° B. alpha = 0° C. 90° < alpha < 180° D. alpha is exactly 90° only Answer: C. For 90° < alpha < 180°, cos(alpha) is negative so the average DC voltage is negative and the converter inverts. (Key point: Converter mode versus firing angle) 3. To reverse power from system 1 -> 2 to system 2 -> 1, the firing angles are changed so that: A. alpha1 > 90° and alpha2 < 90° B. alpha1 < 90° and alpha2 > 90° C. both alpha1 and alpha2 < 90° D. both alpha1 and alpha2 > 90° Answer: A. Converter 1 must become an inverter (alpha1 > 90°) and converter 2 a rectifier (alpha2 < 90°); the current direction stays the same while line voltage reverses. (Key point: Reversed-mode firing angles) 4. For a low-resistance line carrying current, the firing angles of the two converters satisfy: A. alpha1 + alpha2 > 180° B. alpha1 + alpha2 is approximately 180° but slightly less C. alpha1 = alpha2 = 90° D. alpha1 - alpha2 = 180° Answer: B. Id = Vdo(cos a1 + cos a2)/R must be positive, requiring a1 + a2 < 180°; with small R, a1 + a2 is approximately 180°, and the gap sets the current magnitude. 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