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 Z_{W1} = 45 + j250 Ω/phase and Z_{W2} = 50 + j330 Ω/phase. Line 1 connects the source bus to a load at bus 2 that draws a complex power of S_{2} = 1.5 + j0.5 MVA. Line2 connects bus 2 to a load at bus 3 that draws a complex power of S_{3} = 2 + j1.5 MVA. Assuming that the line voltage at bus 3 is V_{L3} = 115 kV(rms), find the bus voltages V_{L1} and V_{L2}, the line currents I_{L1} and I_{L2}, and the source power output needed to produce the specified load powers S_{2} and S_{3}.
For the given values of V_{L3} and S_{3}, 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 S_{W2} plus the bus 3 load S_{3}, namely
S_{W2} + S_{3} = 2.024 + j1.656 MVA
At bus 2 the line current into line 2 is I_{L2}. Since the input power to line 2 is S_{W2} + S_{3}, 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
S_{2} + (S_{W2} + S_{3} ) = 3.524 + j2.156 MVA
The line voltage at the output of line 1 is V_{L2}, 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 I_{L1} 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 S_{W1}, plus its output at the bus 2. This power is supplied by the source at bus 1, hence
S_{1} = S_{W1} + [S_{2} + (S_{W2} + S_{3})] = 3.577 + j2.451 MVA
At bus 1 the line current is I_{L1}, 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}= 126.2k V (rms)
In round numbers, the conditions V_{L1} = 126.2 kV (rms), I_{L1} = 19.8 A (rms), V_{L2} = 120.3 kV (rms), I_{L2} = 12.5 A (rms), and V_{L3} = 115 kV (rms) will produce the required load power flow. This set of conditions is not unique. For example, the set V_{L1} = 153 kV (rms), I_{L1} = 15.9 A (rms), V_{L2} = 148 kV (rms), I_{L2} = 10 A (rms), and V_{L3} = 144 kV (rms) will produce the same load power flow.