Consider the three bus system shown below-If the second row of the bus admittance matrix is Y21-12-13- 104-04-p-u-Y22-24-23- -75-95- p-u andY23-12-13- 104-04-p-u- - then the Jacobian matrix element- P2-2-Assume V20-1- j0 p-u- and -30-0

According to static load flow equation, ∂ P_2∂δ_2=∂∂δ_2 [|V_2||V_1||V_12|cos(θ_12+δ_1 δ_2)+|V_2|^2|Y_22|cos (θ_22)+|V_2||V_3||Y_23|cos(θ_23+δ_3^0 δ_2)] ∂ P_2∂ ...

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Byju's Answer

Standard XII

Physics

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Consider the ...

Question

Consider the three bus system shown below,

If the second row of the bus admittance matrix is $Y_{21} = 12.13 ∠ {104.04}^{\circ} p . u . Y_{22} = 24.23 ∠ - {75.95}^{\circ}$ p.u and
$Y_{23} = 12.13 ∠ {104.04}^{\circ}$ p.u. , then the Jacobian matrix element
$(\frac{\partial P_{2}}{\partial δ_{2}}) =$

(Assume $V_{2}^{0}$ =1+ j0 p.u. and $δ_{3}^{0}$ =0)
24.47

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Solution

The correct option is A 24.47
According to static load flow equation,

$\frac{\partial P_{2}}{\partial δ_{2}} = \frac{\partial}{\partial δ_{2}} [| V_{2} | | V_{1} | | V_{12} | cos (θ_{12} + δ_{1} - δ_{2}) + | V_{2} |^{2} | Y_{22} | cos (θ_{22}) + | V_{2} | | V_{3} | | Y_{23} | cos (θ_{23} + δ_{3}^{0} - δ_{2})]$

$\frac{\partial P_{2}}{\partial δ_{2}} = | V_{2} | | V_{1} | | V_{12} | sin (θ_{12} + δ_{1} - δ_{2}) + | V_{2} | | V_{3} | | Y_{23} | sin (θ_{23}) + (θ_{23} + δ^{0} - δ_{2})$

$= 2 [1 \times 1.04 \times 12.13 sin ({104.04}^{\circ})] = 24.47$

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Consider the three bus system shown below, If the second row of the bus admittance matrix is Y21=12.13∠104.04∘p.u.Y22=24.23∠−75.95∘ p.u and Y23=12.13∠104.04∘p.u. , then the Jacobian matrix element (∂P2∂δ2)= (Assume V02=1+ j0 p.u. and δ03=0)24.47