Using the results of Application Examples 7.15 to 7.17, evaluate the correction bus vector ΔU¯^0 for the system of Fig. E7.14.1. Update the mismatch vector and comment on the convergence of the harmonic load flow.

Question

Using the results of Application Examples 7.15 to 7.17, (7.15, 7.16, 7.17)evaluate the correction bus vector [latex]\Delta \bar{U}^0[/latex] for the system of Fig. E7.14.1. Update the mismatch vector and comment on the convergence of the harmonic load flow.

Accepted Answer

Step #5 of Harmonic Load Flow. Evaluate [latex]\Delta \bar{U}^0 ext { and update } \Delta \bar{U}^0[/latex].
The correction bus vector [latex]\Delta \bar{U}^0[/latex] is computed as follows:

[latex]\Delta \bar{U}^0=\overline{\left(J^0 ight)}^{-1}\left[\begin{array}{l} 3.43472 E-6 \ -1.33365 E-6 \ -7.62939 E-6 \ -3.81470 E-6 \ 2.99421 E-2 \ 2.97265 E-4 \ 4.217029 E-6 \ -200.0 \ -4.09782 E-8 \ 2.384186 E-7 \ -1.93715 E-7 \ -5.960464 E-8 \ 0.29935 \ -3.03515 E-3 \ -1.78814 E-7 \ 8.86619 E-7 \end{array} ight]=\left[\begin{array}{c} 1.20815 E-6 \ -8.59742 E-7 \ 9.552948 E-6 \ 1.54111 E-6 \ 1.184274 E-5 \ 2.3572 E-6 \ 1.46588 E-3 \ 0.10000228 \ 2.38763 E-2 \ 0.1005407 \ 0.2042002 \ 0.1021379 \ 0.249194 \ 0.1027635 \ -- \ -- \end{array} ight][/latex]

The updated bus vector is

[latex]\begin{gathered} \bar{U}^1=\bar{U}^0-\Delta \bar{U}^0=(-2.68332 E-4,0.99759,-2.80029 E-3,0.99640,-3.4257 E \ -3,0.99594,-1.465844 E-3,-2.28 E-6,-2.38763 E-2,-5.407 E-4, \ -0.2042,-2.138 E-3,-0.24919,-2.7635 E-3,0,0)^t . \end{gathered}[/latex]

Note that all harmonic voltage magnitudes are negative. To prevent negative magnitudes, the corresponding phase angles are shifted by π radians; therefore,

[latex]\begin{gathered} \bar{U}^1=(-2.68332 E-4,0.99759,-2.80029 E-3 \ 0.99640,-3.4257 E-3,0.99594,3.14013,2.28 E-6 \ 3.11772,5.407 E-4,2.93738,2.138 E-3,2.89239,2.7635 E-3,0,0)^t \end{gathered}[/latex]

The new (updated) nonlinear device currents are [latex]G_{r, 4}^{(1)}=0.256629, G_{i, 4}^{(1)}=-0.1011142, G_{r, 4}^{(5)}=0.2965726 ext { and } G_{i, 4}^{(5)}=-2.59613 E-3[/latex]. The new (updated) mismatch vector is

[latex]\begin{gathered} \Delta \bar{W}^1=(5.342 E-6,-3.800 E-7,5.722 E-6, \ 3.815 E-6,-1.403 E-3,4.332 E-4,-2.269 E-3, \ -3.300 E-2,-2.677 E-3,6.475 E-3,-1.021 E-3, \ -2.105 E-3,0.3023,3.015 E-2,-6.264 E-4,-1.473 E-4)^t \end{gathered}[/latex]

This mismatch vector is not small enough and the solution has not converged. However, the iterative procedure will converge in eight iterations. The final solution is (all magnitudes are in percentage of the base values and all angles are in degrees):
Fundamental power flow output solution (Fig. E7.14.1):

	Bus voltage		Bus generation (G) and load (L) powers				Line power³
From Bus	\|v\| (%)	[latex]\delta (^{\circ})[/latex]	[latex]P_g (\%)[/latex]	[latex]Q_G (\%)[/latex]	[latex]P_L (\%)[/latex]	To Bus	[latex]P_{Line} (\%)[/latex]	[latex]Q_{Line} (\%)[/latex]
1	100.0	0.00	35.16	20.90	0.00	0.00	2	13.35	10.81
							4	21.81	10.09
2	99.76	-0.01	0.00	0.00	10.00	10.00	1	-13.33	-10.78
							3	3.32	0.78
3	99.64	-0.16	0.00	0.00	0.00	0.00	2	-3.32	-0.77
							4	3.32	0.77
4	99.58	-0.19	0.00	0.00	25.00	10.00	3	-3.32	-0.77
							1	-21.75	-9.98

Harmonic power flow output solution for harmonic number 5, frequency=300 Hz (Fig. E7.14.1):

	Bus voltage			[latex]Line\ power^a[/latex]		[latex]Line\ current^a[/latex]
From bus	\|v\| (%)	[latex]\delta[/latex] (%)	To bus	[latex]P_{Line}[/latex] ($)	[latex]Q_{line}[/latex] ($)	Magnitude (%)	Angle [latex](^{\circ}[/latex]
1	0.0148	-90.57	2	0.000015	–0.000675	4.563896	-1.88
			4	–0.000015	–0.003707	25.042787	-0.33
			Neutral Shunt	0.000000	0.004382	29.605302	179.43
2	0.2472	-102.44	1	0.002067	0.01109	4.563896	178.12
			3	–0.002067	–0.011090	4.563895	-1.88
3	2.0731	-95.66	2	0.006233	0.094407	4.563895	178.12
			4	–0.006233	–0.094406	4.563865	-1.88
4	2.5315	-96.01	3	0.008315	0.115235	4.563865	178.12
			1	0.06273	0.630848	25.042787	179.67
			Neutral nonlinear	–0.071046	–0.746085	29.605308	–0.57

Total current/power:

		Line current				Line power
From bus	To bus	Fundamental value (%)	rms value (%)	Peak value (%)	[latex]THD_{i}[/latex] (%)	p (%)	Q (%)	D (%)	S (%)
1	2	17.18	17.78	19.26	0.9665	13.35	10.81	4.57	17.78
1	4	24.03	34.71	46.94	0.6923	21.81	10.09	25.04	34.71
2	1	17.18	17.78	19.26	0.9665	-13.32	-10.77	4.59	17.73
2	3	3.42	5.70	7.90	0.6000	3.32	0.77	4.56	5.69
3	2	3.42	5.70	7.90	0.6000	-3.32	-0.68	4.57	5.69
3	4	3.42	5.71	7.90	0.6000	3.32	0.68	4.57	5.69
4	3	3.42	5.71	7.90	0.6000	-3.31	-0.65	4.57	5.69
4	1	24.03	34.71	46.94	0.6923	-21.69	-9.35	25.25	34.57
4	Neutral nonlinear	27.04	40.09	54.76	0.6744	24.93	9.25	29.8	39.94

Bus voltage summary:

Bus number	Fundamental value (%)	rms value (%)	Peak value (%)	[latex]THD_v[/latex] (%)
1	100.00	100.00	100.00	1.0000
2	99.76	99.76	99.71	1.0000
3	99.63	99.65	99.97	0.9998
4	99.58	99.61	100.11	0.9997

Question 7.AE.18: Using the results of Application Examples 7.15 to 7.17, eval......

Related Answered Questions

Using the results of Application Examples 7.15 and 7.16, evaluate the Jacobian matrix for the system of Fig. E7.14.1. ...

Verified Answer:

Verified Answer:

Using the results of Application Example 7.15, evaluate the mismatch vector for the system of Fig. E7.14.1. ...

Verified Answer:

For the system of Fig. E7.8.1, compute the correction voltage vectorΔx^-0, updating the mismatch power vector, and check the convergence of the load flow algorithm (using a convergence tolerance of 0.0001). ...

Verified Answer:

For the system of Fig. E7.8.1, compute the Jacobian matrix. ...

Verified Answer:

Verified Answer:

A simple (few buses) power system (Fig. E7.8.1) is used to illustrate in detail the Newton– Raphson load flow analysis. All impedances are in pu and the base apparent power is Sbase=1 kVA. Compute the fundamental admittance matrix for this system. ...

Verified Answer:

Verified Answer:

Verified Answer:

In order to demonstrate the entire procedure for the computation of the inverse of a matrix the following 3×3 matrix will be used A^-=[1 3 -1 2 -4 2 3 1 1] ...

Verified Answer: