Optimal Reactive Power Allocation in Large-Scale Grid-Connected Photovoltaic Systems

Abstract

In this paper, an optimal strategy is proposed for the reactive power allocation in large-scale grid-connected photovoltaic systems. Grid-connected photovoltaic systems with direct current to alternating current inverters are able to supply active power to the utility grid as well as reactive power. The active power, extracted by the direct current to alternating current inverters, is usually controlled to be around the maximum power point of the photovoltaic array attached to it. For large-scale grid-connected photovoltaic systems with multiple direct current to alternating current inverters, due to the limited apparent power transfer capability of each inverter, the reactive power needs to be allocated among the direct current to alternating current inverters in a proper way. The proposed method achieves the maximum reactive power transfer capability of the entire system by applying classic Lagrange multiplier method. The sufficient conditions of the optimal reactive power allocation strategy are provided and mathematically proved. The proposed optimal reactive power allocation strategy is then tested in a case study against a sample large-scale grid-connected photovoltaic system.

Appendix: Proofs of Theorems and Lemmas

Proof of Theorem 3.1

Here, we use Lagrange multiplier method [17] to solve the problem given in (6) and only consider the positive part. Let $Q = \begin{bmatrix}Q_1,\ldots ,Q_m \end{bmatrix}^{\mathrm {T}}$, the Lagrangian function is constructed as follows,

$$\begin{aligned} \begin{aligned} L(Q,\lambda ,\mu ) =&-\sum _{i=1}^m \left( C^{\mathrm {LMT}}_i - \frac{\sqrt{P_i^2 + Q_i^2}}{3|V|} \right) + \lambda \left( \sum _{i=1}^m Q_i - Q_D\right) \\&+ \sum _{i=1}^m \mu _i \bigg (Q_i -\sqrt{9|V|^2\left( C^{\mathrm {LMT}}_i\right) ^2 - P^2_i} \bigg ) \\ \end{aligned} \end{aligned}$$

(15)

where $\lambda$, $\mu _j$, $j=1,\ldots ,m$ are Lagrange multipliers. As for this case we assume the reactive power of the ith inverter $Q_i$ satisfies (5) with strictly inequalities, the inequality constraints are inactive. Hence, the Lagrangian function in (15) becomes

$$\begin{aligned} L(Q,\lambda ) = -\sum _{i=1}^m \left( C^{\mathrm {LMT}}_i - \frac{\sqrt{P_i^2 + Q_i^2}}{3|V|} \right) + \lambda \left( \sum _{i=1}^m Q_i - Q_D\right) \end{aligned}$$

(16)

Let the gradient of the Lagrangian function (16) $\nabla _Q L(Q,\lambda ) = 0$, we have

$$\begin{aligned} \frac{Q_i}{3|V|\sqrt{P_i^2 + Q_i^2}} = -\lambda , i=1,\ldots ,m \end{aligned}$$

(17)

From (17), we know that $Q_i$ and $\lambda$ have opposite signs, and |V| and $P_i$ are both positive, so we obtain

$$\begin{aligned} Q_i = -\frac{3|V|P_i\lambda }{\sqrt{1-9|V|^2\lambda ^2}} \end{aligned}$$

(18)

If we substitute (18) into $\sum _{i=1}^m Q_i = Q_D$, we obtain one equation with $\lambda$ as the only variable,

$$\begin{aligned} -\frac{3|V|P_i\lambda }{\sqrt{1-9|V|^2\lambda ^2}}\sum _{i=1}^m P_i - Q_D = 0 \end{aligned}$$

(19)

By solving (19), we have

$$\begin{aligned} \lambda ^2 = \frac{Q_D^2}{9|V|^2\bigg ( (\sum _{i=1}^m P_i)^2 + Q_D^2 \bigg )} \end{aligned}$$

As $Q_i$ and $\lambda$ have opposite signs, it is obvious that $Q_D$ and $\lambda$ also have opposite signs. Then, $\lambda$ is expressed as follows,

$$\begin{aligned} \lambda = - \frac{Q_D}{3|V|\sqrt{(\sum _{i=1}^m P_i)^2 + Q_D^2}} \end{aligned}$$

(20)

Substituting (20) in (17), we have the reactive power $Q_i^*$, $i=1,\ldots ,m$,

$$\begin{aligned} Q_i^* = \frac{P_i}{\sum _{i=1}^m P_i} Q_D, i=1,\ldots ,m \end{aligned}$$

(21)

To guarantee $Q_i^*$ in (21) is the optimal reactive power for the ith inverter, and we need to the Hessian of the Lagrangian function to be positive definite [17]. The Hessian of the Lagrangian function is

$$\begin{aligned} \nabla _{QQ}L(Q^*,\lambda ^*) = \begin{bmatrix} \frac{P_1^2}{3|V|\left( P_1^2 +Q_1^2\right) ^{3/2}}&0&\cdots&0 \\ 0&\frac{P_2^2}{3|V|\left( P_2^2 +Q_2^2\right) ^{3/2}}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&\frac{P_m^2}{3|V|\left( P_m^2 + Q_m^2\right) ^{3/2}} \end{bmatrix} \end{aligned}$$

where $\lambda ^*$ is the one given in (20). For all $y \ne 0$ such that $\nabla (\sum _{i=1}^m Q_i - Q_D)^{\mathrm {T}} y = 0$, we have

$$\begin{aligned} y^{\mathrm {T}} \nabla _{QQ} L\left( Q^*,\lambda ^*\right) y = \sum _{i=1}^m \frac{P_i^2}{3|V|\left( P_i^2 + Q_i^2\right) ^{3/2}}y_i^2 > 0 \end{aligned}$$

Hence, the Hessian of the Lagrangian function is positive definite. So $Q_i^*$ given by (21) is the optimal reactive power profile. To let the inactive inequalities assumption hold, we need $Q_i^*$ to satisfy the first inequality of (5) (for positive $Q_i$). Then, we have

$$\begin{aligned} \frac{P_i}{\sum _{i=1}^m P_i} Q_D \le \sqrt{9|V|^2\left( C^{\mathrm {LMT}}_i\right) ^2 - P^2_i} ,\ i=1,\ldots ,m \end{aligned}$$

(22)

As (22) needs to hold for all inverters, we obtain

$$\begin{aligned} \begin{aligned} Q_D \le \min _{i=1,\ldots ,m}\left\{ \frac{\sqrt{9|V|^2\left( C^{\mathrm {LMT}}_i\right) ^2 - P^2_i}}{P_i}\sum _{i=1}^m P_i\right\} \\ \end{aligned} \end{aligned}$$

for positive $Q_i$, similarly we can prove the negative part, which proves (8). $\square$

Proof of Lemma 3.1

Suppose that all the inverters are already sorted in the order given by (10) and in such an order the reactive power of the first $r-1$ inverters already hit their upper bounds. Now consider the assumption that the reactive power of the rth inverter does not reach its upper bound, i.e., $Q_r < Q^{\mathrm {max}}_r$, and the reactive power of the $(r+1)$th inverter hits its upper bound, i.e., $Q_{r+1} = Q^{\mathrm {max}}_{r+1}$. As indicated by the assumption, the reactive powers $Q_r$ and $Q_i$, $i = r+2,\ldots ,m$, do not hit their upper bounds, and according to Theorem 3.1 for these $m-r$ reactive powers, we have

$$\begin{aligned} Q_i = \frac{Q_D - \sum _{i=1}^{r-1} Q_i^{\mathrm {max}} - Q_{r+1}^{\mathrm {max}}}{\sum _{i=1}^m P_i - \sum _{i=1}^{r-1} P_i - P_{r+1}} P_i,\ i = r, r+2,\ldots ,m \end{aligned}$$

(23)

For the rth inverter, we substitute (23) into $Q_r < Q^{\mathrm {max}}_r$, then we have

$$\begin{aligned} \frac{Q_D - \sum _{i=1}^{r-1} Q_i^{\mathrm {max}} - Q_{r+1}^{\mathrm {max}}}{\sum _{i=1}^m P_i - \sum _{i=1}^{r-1} P_i - P_{r+1}} P_r < Q^{\mathrm {max}}_r \end{aligned}$$

(24)

The $(r+1)$th inverter’s reactive power $Q_{r+1}$, by the assumption, hits the upper bounds. If we apply Theorem 3.1 and calculate $Q_{r+1}$ by using a manner similar to (23), the reactive power $Q_{r+1}$ will exceed the upper bound $Q^{\mathrm {max}}_{r+1}$. Based on this, we have such inequality

$$\begin{aligned} \frac{Q_D - \sum _{i=1}^{r-1} Q_i^{\mathrm {max}}}{\sum _{i=1}^m P_i - \sum _{i=1}^{r-1} P_i} P_r \ge Q^{\mathrm {max}}_{r+1} \end{aligned}$$

(25)

As $P_i >0$, $i = 1,\ldots ,m$, from (24) and (25) we obtain

$$\begin{aligned} \frac{Q_D - \sum _{i=1}^{r-1} Q_i^{\mathrm {max}} - Q_{r+1}^{\mathrm {max}}}{\sum _{i=1}^m P_i - \sum _{i=1}^{r-1} P_i - P_{r+1}} < \frac{Q^{\mathrm {max}}_r}{P_r} \end{aligned}$$

(26)

and

$$\begin{aligned} \frac{Q_D - \sum _{i=1}^{r-1} Q_i^{\mathrm {max}}}{\sum _{i=1}^m P_i - \sum _{i=1}^{r-1} P_i} \ge \frac{Q^{\mathrm {max}}_{r+1}}{P_{r+1}} \end{aligned}$$

(27)

Subtract the left-hand side of (26) by the left-hand side of (27), we obtain the following inequality

$$\begin{aligned}&\frac{Q_D - \sum _{i=1}^{r-1}Q^{\mathrm {max}}_i - Q^{\mathrm {max}}_{r+1}}{\sum _{i=1}^m P_i - \sum _{i=1}^{r-1}P_i - P_{r+1}} - \frac{Q_D - \sum _{i=1}^{r-1}Q^{\mathrm {max}}_i}{\sum _{i=1}^m P_i - \sum _{i=1}^{r-1}P_i} \nonumber \\&\quad = \frac{(Q_D - \sum _{i=1}^{r-1}Q^{\mathrm {max}}_i)P_{r+1} - (\sum _{i=1}^m P_i - \sum _{i=1}^{r-1}P_i) Q^{\mathrm {max}}_{r+1}}{(\sum _{i=1}^m P_i - \sum _{i=1}^{r-1}P_i - P_{r+1})(\sum _{i=1}^m P_i - \sum _{i=1}^{r-1}P_i)} \ge 0 \end{aligned}$$

(28)

The reason that (28) holds is that from (25) we know

$$\begin{aligned} \bigg (Q_D - \sum _{i=1}^{r-1} Q^{\mathrm {max}}_i\bigg )P_{r+1} \ge \bigg (\sum _{i=1}^m P_i - \sum _{i=1}^{r-1} P_i \bigg ) Q^{\mathrm {max}}_{r+1} \end{aligned}$$

and the denominator of the second line of (28) is obviously positive. Hence,

$$\begin{aligned} \frac{Q_D - \sum _{i=1}^{r-1}Q^{\mathrm {max}}_i - Q^{\mathrm {max}}_{r+1}}{\sum _{i=1}^m P_i - \sum _{i=1}^{r-1}P_i - P_{r+1}} \ge \frac{Q_D - \sum _{i=1}^{r-1}Q^{\mathrm {max}}_i}{\sum _{i=1}^m P_i - \sum _{i=1}^{r-1}P_i} \end{aligned}$$

(29)

From (25), (26), and (29), we have the following inequality,

$$\begin{aligned} \frac{Q^{\mathrm {max}}_r}{P_r} > \frac{Q_D - \sum _{i=1}^{r-1}Q^{\mathrm {max}}_i - Q^{\mathrm {max}}_{r+1}}{\sum _{i=1}^m P_i - \sum _{i=1}^{r-1}P_i - P_{r+1}} \ge \frac{Q_D - \sum _{i=1}^{r-1}Q^{\mathrm {max}}_i}{\sum _{i=1}^m P_i - \sum _{i=1}^{r-1}P_i} \ge \frac{Q^{\mathrm {max}}_{r+1}}{P_{r+1}} \end{aligned}$$

(30)

The inequality in (30) shows $\frac{Q^{\mathrm {max}}_r}{P_r}>\frac{Q^{\mathrm {max}}_{r+1}}{P_{r+1}}$ which contradicts the order in (10). Hence, the assumption that $Q_r < Q^{\mathrm {max}}_r$, while $Q_{r+1} = Q^{\mathrm {max}}_{r+1}$ is invalid. Then, we conclude that when $Q_D > 0$ the first r inverters’ reactive power $Q_i$, $i = 1,\ldots ,r$ in the order given in (10) hit their upper bounds. For the rest $m-r$ inverters, the following inequality holds,

$$\begin{aligned} \frac{P_i}{\sum _{i=r+1}^m P_i} \bigg (Q_D - \sum _{i=1}^r Q^{\mathrm {max}}_j \bigg ) < Q^{\mathrm {max}}_i,\ i = r+1,\ldots ,m \end{aligned}$$

(31)

Thus, r is the minimum number that makes (31) hold. $\square$

Proof of Theorem 3.2

We use Lagrange multiplier method [17] to prove this theorem. The Lagrangian function is the one given in (15). We have two cases. The reactive power demand $Q_D>0$. For this case, all the inverters are in the order given in (10). By Lemma 3.1, we know that those r inverters with reactive power that hits the upper bound are the first r inverters in that order. Hence, the inequality constraints

$$\begin{aligned} g_i(Q_i) = Q_i -\sqrt{9|V|^2\left( C^{\mathrm {LMT}}_i\right) ^2 - P^2_i} \le 0,\,i = 1,\ldots ,r \end{aligned}$$

are active. For $i = 1,\ldots ,r$, we have

$$\begin{aligned} Q_i = Q^{\mathrm {max}}_i,\ i = 1,\ldots ,r \end{aligned}$$

(32)

and by taking the gradient of (15) we have

$$\begin{aligned} \mu _i = -\frac{Q_i}{ 3|V|\sqrt{P^2_i + Q^2_i}} - \lambda ,\ i = 1,\ldots ,r \end{aligned}$$

(33)

For those $m-r$ inverters with inactive inequality constraints, we have

$$\begin{aligned} \frac{Q_i}{3|V|\sqrt{P^2_i + Q^2_i}} + \lambda = 0 ,\ i=r+1,\ldots ,m \end{aligned}$$

(34)

Also, we have the equality constraints which we are

$$\begin{aligned} \sum _{i=1}^r Q_i + \sum _{i=r+1}^m Q_i - Q_D = 0 \end{aligned}$$

(35)

From (34), we obtain,

$$\begin{aligned} Q_i = - \frac{3|V|P_i\lambda }{\sqrt{1 - 9|V|^2\lambda ^2}},\ \mathrm {for\ } i=r+1,\ldots ,m \end{aligned}$$

(36)

Substitute (32) and (36) in (35), and we obtain

$$\begin{aligned} \begin{aligned}&\sum _{i=1}^r Q_i - \frac{3|V|\lambda }{\sqrt{1 - 9|V|^2\lambda ^2}} \sum _{i=r+1}^m P_i - Q_D = 0 \\&\frac{3|V|\lambda }{\sqrt{1 - 9|V|^2\lambda ^2}} = - \frac{Q_D - \sum _{i=1}^r Q_i}{\sum _{i=r+1}^m P_i} \\&9|V|^2\lambda ^2\left( \sum _{i=r+1}^m P_i\right) ^2 = \left( Q_D - \sum _{i=1}^r Q_i\right) ^2 \left( 1-9|V|^2\lambda ^2\right) \\&9|V|^2\lambda ^2\bigg (\left( \sum _{i=r+1}^m P_i\right) ^2 + \left( Q_D - \sum _{i=1}^r Q_i\right) ^2 \bigg ) = \bigg (Q_D - \sum _{i=1}^r Q_i\bigg )^2 \\ \end{aligned} \end{aligned}$$

(37)

From the second line of (37), we know that $\lambda$ has the opposite sign of $Q_D - \sum _{i=1}^r Q_i$. In this case, $\lambda$ is negative. As $Q_i = Q_i^{\mathrm {max}}$, $i =1,\ldots ,r$. Hence,

$$\begin{aligned} \lambda = - \frac{Q_D - \sum _{i=1}^r Q_i^{\mathrm {max}}}{3|V| \sqrt{(\sum _{i=r+1}^m P_i)^2 + (Q_D - \sum _{i=1}^r Q_i^{\mathrm {max}})^2}} \end{aligned}$$

(38)

Substitute (32) and (38) in (36), and we obtain (12).

For this case,

$$\begin{aligned} \mu _j>0,\,\forall j \in A(Q) \end{aligned}$$

(39)

where $A(Q) = \{j\ |\ g_j(Q) = 0\}$ is the index set that the inequality constraints are active. Now we show the reason why (39) holds. We assume some inverters’ reactive powers hit their upper bounds, (8) does not hold. Then, consider the inverters in the order given by (10). For $i = 1,\ldots ,r$, $Q_i = Q_i^{\mathrm {max}}$, then

$$\begin{aligned} Q_i^{\mathrm {max}} < \frac{P_i}{\sum _{i=1}^m P_i} Q_D,\ i=1,\ldots ,r \end{aligned}$$

The reactive power $Q_i$, $i = 1,\ldots ,r$, reaches its upper bound, so the amount of reactive power $\frac{P_i}{\sum _{i=1}^m P_i} Q_D - Q_i^{\mathrm {max}}$, $i=1,\ldots ,r$, will be allocated on other inverters. Hence,

$$\begin{aligned} Q_j \ge \frac{P_i}{\sum _{i=1}^m P_i} Q_D,\ j=r+1,\ldots ,m \end{aligned}$$

(40)

Then, we obtain

$$\begin{aligned} \frac{Q_i^{\mathrm {max}}}{P_i} < \frac{Q_j}{P_j},\ i=1,\ldots ,r,\ j=r+1,\ldots ,m \end{aligned}$$

(41)

From (33) and (38), $\mu ^*_i$ is expressed as

$$\begin{aligned} \begin{aligned} \mu _i^* =&\frac{Q_D - \sum _{i=1}^r Q_i^{\mathrm {max}}}{3|V| \sqrt{\left( \sum _{i=r+1}^m P_i\right) ^2 + \left( Q_D - \sum _{i=1}^r Q_i^{\mathrm {max}}\right) ^2}} \\&- \frac{Q_i^{\mathrm {max}}}{ 3|V|\sqrt{P^2_i + \left( Q^{\mathrm {max}}_i\right) ^2}},\,i = 1,\ldots ,r\\ \end{aligned} \end{aligned}$$

(42)

Since $Q_D - \sum _{i=1}^r Q_i^{\mathrm {max}} > 0$ and $Q^{\mathrm {max}}_i > 0$, we can turn (42) into the following form,

$$\begin{aligned} \begin{aligned} \mu _i^* =&\frac{1}{3|V| \sqrt{\bigg (\frac{\sum _{i=r+1}^m P_i}{Q_D - \sum _{i=1}^r Q_i^{\mathrm {max}}}\bigg )^2 + 1}} -\frac{1}{ 3|V|\sqrt{\bigg (\frac{P_i}{Q_i^{\mathrm {max}}}\bigg )^2 + 1}},\ i = 1,\ldots ,r \\ \end{aligned} \end{aligned}$$

(43)

Now consider the denominators of those two terms in (43). From (12), we know that

$$\begin{aligned} \frac{P_j}{Q_j}= \frac{\sum _{i=r+1}^m P_i}{Q_D - \sum _{i=1}^r Q_i^{\mathrm {max}}},\,j = r+1,\ldots ,m \end{aligned}$$

From (41), we know that

$$\begin{aligned} \frac{P_j}{Q_j} < \frac{P_i}{Q_i^{\mathrm {max}}} \end{aligned}$$

Hence, in (43) the denominator of the first term is smaller than the denominator of the second term. Then, we conclude that $\mu _j > 0$, $\forall j \in A(Q)$. For all $y\ne 0$ such that $\nabla h(Q)^{\mathrm {T}}y = 0$, and $\nabla g_j(Q)^{\mathrm {T}}y = 0$, $\forall j\in A(Q)$, we have

$$\begin{aligned} \begin{aligned}&\nabla _{QQ}L(Q^*,\lambda ^*,\mu ^*) = \\&\begin{bmatrix} \frac{P_1^2}{3|V|\left( Q_1^2 +P_1^2\right) ^{3/2}}&0&\cdots&0 \\ 0&\frac{P_2^2}{3|V|\left( Q_2^2 +P_2^2\right) ^{3/2}}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&\frac{P_m^2}{3|V|\left( Q_m^2 + P_m^2\right) ^{3/2}} \end{bmatrix} \\ \end{aligned} \end{aligned}$$

(44)

and

$$\begin{aligned} y^{\mathrm {T}}\nabla _{QQ} L(Q^*, \lambda ^*,\mu ^*)y = \sum _{i=1}^m\frac{P_i^2}{3|V|\left( Q_i^2 + P_i^2\right) ^{3/2}}y_i^2 > 0 \end{aligned}$$

(45)

The Hessian of the Lagrangian function is positive definite. Hence, the reactive power profile given by (12) is the optimal allocation reactive power profile when (8) is not satisfied. $\square$