Question

Figure 3 shows a simplified schematic of a single phase equivalent circuit diagram of inverter coupled embedded generation, where Vinv is the voltage at the inverter connection point (assume this to be 240VZ0°) Vpcc is the voltage at the point of common coupling Vgnd is the grid voltage X the reactance in the inverter connection (3.6 m) R the resistance in the inverter connection (55 mO) Xgrid is the reactance of the distribution grid (200 m) Rend is the resistance of the distribution grid (300 m) Xg Xgrta Rgntd Inv Inverter Vxs- VRs VxgnaVRgna + + + Vgnd VPcc inv Figure 3 Simplified single phase equivalent circuit diagram of inverter coupled embedded generation, showing the grid impedance (Zgrid=Rgnd+jXgnd) between the grid (modelled as Vgnd) and the point of common coupling (PCC) as well as the impedance between the inverter coupled generation and PCC The inverter used for connection of embedded generation is a single phase 240 VAC inverter, with a power rating of 10 kVA Case 1: The inverter first operates at rated power and the power factor is set to cos 0.8 so that the inverter current is leading the inverter voltage, as shown in the phasor diagram in Fig 4. 11 Viny

Answer 1

Total impedance, z = (R_s + R_grid) + j (x_s + x_grid) = (55 + 300) + j (3.6 + 200) m ohm = 355 + j 203.6 m ohm = 0.4092 < 29.835 degree ohm v_inv = 240 < 0 volt (reference) s_inr = 10 KVA < power rating > case 1: (i) operated at rated power (ii) operated at leading pf of cos phi = 0.8 therefore I_inv/rated = 10 times 10^3/240 A = 41.667 A therefore I_inv = 41.6670 < 36.86 degree A [cos^-1 0.8 = 36.86 leading w.r.t 240 < 0 volts] Now, z = 0.40920 < 29.835 degree V therefore voltage drop, I_inv z = 17.05 < 66.695 degree volts now by KVL in the loop, V_grid < -8 = V_in < 0 - i_inv z = 240 < 0 - 17.05 < 66.695 = (240 - 6.745) - j 15.659 volts = 233.78 < -3.84 degree volts therefore delta = 3.84 degree, N_grid is lagging behind v_inv by angle delta