n ie 1t x 1t y 1 x 0 y 0 we imagine that xi

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Unformatted text preview: j (with positive meaning in the direction of the line as defined in A, negative meaning power flow in the opposite direction). Node i has voltage phase angle φi , and external power input si . (If a generator is attached to node i we have si > 0; if a load is attached we have si < 0; if the node has neither, si = 0.) Neglecting power losses in the lines, and assuming power is conserved at each node, we have Ap = s. (We must have 1T s = 0, which means that the total power pumped into the network by generators balances the total power pulled out by the loads.) The line power flows are a nonlinear function of the difference of the phase angles at the nodes they connect to: pj = κj sin(φk − φl ), where line j goes from node k to node l. Here κj is a known positive constant (related to the inductance of the line). We can write this in matrix form as p = diag(κ) sin(AT φ), where sin is applied elementwise. The DC power flow equations are Ap = s, p = diag(κ) sin(AT φ). In the power analysis problem, we a...
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