Question 16.23: In Figure 16–27, the source at bus 1 supplies two load buses......
In Figure 16–27, the source at bus 1 supplies two load buses through transmission lines with ZW1 = 45 + j250 Ω/phase and ZW2 = 50 + j330 Ω/phase. Line 1 connects the source bus to a load at bus 2 that draws a complex power of S2 = 1.5 + j0.5 MVA. Line 2 connects bus 2 to a load at bus 3 that draws a complex power of S3 = 2 + j1.5 MVA. Assuming that the line voltage at bus 3 is VL3 = 115 kV(rms), find the bus voltages VL1 and VL2, the line currents IL1 and IL2, and the source power output needed to produce the specified load powers S2 and S3.
For the given values of VL3 and S3, we find the line current in line 2 as
I_{L2} = \frac{|S_{3}|}{\sqrt{3} V_{L3}} = \frac{|(2 + j1.5) × 10^{6}|}{\sqrt{3} × 115 × 10^{3}} = 12.55 A(rms)
The power lost in the line 2 is
S_{W2} = 3I^{2}_{L2}Z_{W2} = 3 × (12.55)^{2}(50 + j330)
= 23.6 + j156 kVA
At bus 2 the input power to line 2 supplies its losses SW2 plus the bus 3 load S3, namely
SW2 + S3 = 2.024 + j1.656 MVA
At bus 2, the line current into line 2 is IL2. Since the input power to line 2 is SW2 + S3, the line voltage at bus 2 is found to be
V_{L2} = \frac{|S_{W2} + S_{3}|}{\sqrt{3} I_{L2}} = \frac{|(2.024 + j1.656) × 10^{6}|}{\sqrt{3} × 12.55} = 120.3 kV(rms)
At bus 2, the output power of line 1 supplies the bus 2 load plus the input power into line 2, namely
S2 + (SW2 + S3) = 3.524 + j2.156 MVA
The line voltage at the output of line 1 is VL2, hence the line current in line 1 is
I_{L1} = \frac{|S_{2} + S_{W2} + S_{3}|}{\sqrt{3} V_{L2}} = \frac{|(3.524 + j2.156) × 10^{6}|}{\sqrt{3} × 120.3 × 10^{3}} = 19.83 A(rms)
Given the line current IL1 the total complex power lost in line 1 is
S_{W1} = 3I^{2}_{L1}Z_{W1} = 3 × (19.83)^{2}(45 + j250)
= 53.1 + j295 kVA
Finally we arrive at bus 1, where the input power to line 1 must equal its losses SW1, plus its output at the bus 2. This power is supplied by the source at bus 1, hence
S1 = SW1 + [S2 + (SW2 + S3)] = 3.577 + j2.451 MVA
At bus 1 the line current is IL1, hence the line voltage is found to be
V_{L1} = \frac{|S_{1}|}{\sqrt{3} I_{L1}} = \frac{|(3.577 + j2.451) × 10^{6}|}{\sqrt{3} × 19.83} = 126.2 kV(rms)
In round numbers, the conditions VL1 = 126.2 kV(rms), IL1 = 19.8 A(rms), VL2 = 120.3 kV(rms), IL2 = 12.5 A(rms), and VL3 = 115 kV(rms) will produce the required load power flow. This set of conditions is not unique. For example, the set VL1 = 153 kV(rms), IL1 = 15.9A(rms), VL2 = 148 kV(rms), IL2 = 10 A(rms), and VL3 = 144 kV(rms) will produce the same load power flow.