Let's solve the problem step by step. --- ## **Given Data** - **Single Phase Loads:** 25 kW each (assume three, one per phase) - **Motor M1:** 50 kVA, cos θ = 0.60 lagging (3-phase, 208 V) - **Motor M2:** 160 kVA, cos θ = 0.70 lagging (3-phase, 208 V) - **Secondary Voltages:** 208 V (line-to-line), 120 V (line-to-neutral) - **Primary Voltage:** 2400 V (delta) - **Transformer:** Y-connected secondary --- ## **Step 1: Calculate Currents in the Secondary Windings** ### **A. Single-Phase Loads** Each single-phase load = 25 kW at 120 V (assume unity power factor unless stated): \[ I_{single-phase} = \frac{P}{V} = \frac{25000}{120} = 208.33~\text{A} \quad \text{(per phase)} \] --- ### **B. Motor Loads (3-phase)** #### **Motor M1** - \( S_1 = 50\,\text{kVA} \) - \( V_{LL} = 208\,\text{V} \) - \(\cos \theta = 0.60 \) \[ I_{M1} = \frac{S_1}{\sqrt{3} V_{LL}} = \frac{50000}{\sqrt{3} \times 208} = \frac{50000}{360.8} = 138.5~\text{A} \] #### **Motor M2** - \( S_2 = 160\,\text{kVA} \) - \( V_{LL} = 208\,\text{V} \) - \(\cos \theta = 0.70 \) \[ I_{M2} = \frac{S_2}{\sqrt{3} V_{LL}} = \frac{160000}{\sqrt{3} \times 208} = \frac{160000}{360.8} = 443.5~\text{A} \] --- ### **C. Total Currents in Each Secondary Winding** Each phase winding carries: - The single-phase load (one per phase) - 3-phase load current (divide equally among phases) \[ I_{sec,\,phase} = I_{single-phase} + \left(I_{M1} + I_{M2}\right) \] But for 3-phase loads: The phase current is equal to line current (for Y connection): \[ I_{phase,\,3\text{-}\phi} = I_{line,\,3\text{-}\phi} \] So, **per phase total:** \[ I_{total,\,phase} = I_{single-phase} + I_{M1} + I_{M2} \] \[ I_{total,\,phase} = 208.3 + 138.5 + 443.5 = \boxed{790.3~\text{A}} \] --- ## **Step 2: Line Currents and Power Factor on the 2400 V Side** ### **A. Total Apparent Power on Secondary Side** - Single-phase loads: 3 × 25 kW = 75 kW (assume unity PF, so 75 kVA) - Motor M1: 50 kVA, PF = 0.60 - Motor M2: 160 kVA, PF = 0.70 \[ S_{total} = S_{single-phase} + S_{M1} + S_{M2} = 75 + 50 + 160 = 285\,\text{kVA} \] #### **Total Real Power \(P_{total}\):** \[ P_{M1} = 50 \times 0.60 = 30\,\text{kW} \] \[ P_{M2} = 160 \times 0.70 = 112\,\text{kW} \] \[ P_{total} = 75 + 30 + 112 = 217\,\text{kW} \] #### **Total Reactive Power \(Q_{total}\):** \[ Q_{M1} = 50 \times \sin(\cos^{-1}(0.60)) = 50 \times 0.8 = 40\,\text{kVAR} \] \[ Q_{M2} = 160 \times \sin(\cos^{-1}(0.70)) = 160 \times 0.7141 = 114.25\,\text{kVAR} \] \[ Q_{total} = 0 + 40 + 114.25 = 154.25\,\text{kVAR} \] --- ### **B. Total Apparent Power** \[ S_{total} = \sqrt{P_{total}^2 + Q_{total}^2} \] \[ S_{total} = \sqrt{217^2 + 154.25^2} = \sqrt{47089 + 23793} = \sqrt{70882} = 266.5\,\text{kVA} \] --- ### **C. Total Line Current on Primary Side (2400 V)** \[ I_{primary} = \frac{S_{total}}{\sqrt{3} \times V_{primary}} \] \[ I_{primary} = \frac{266500}{\sqrt{3} \times 2400} = \frac{266500}{4156.9} = 64.1\,\text{A} \] --- ### **D. Power Factor on Primary Side** \[ \text{PF}_{primary} = \frac{P_{total}}{S_{total}} = \frac{217}{266.5} = 0.814 \] --- ## **Final Answers** ### **a. Currents in the secondary windings** - **Per phase (secondary winding):** \(\boxed{790.3~\text{A}}\) ### **b. Line currents and power factor on the 2400 V side** - **Line current:** \(\boxed{64.1~\text{A}}\) - **Power factor:** \(\boxed{0.814~\text{lagging}}\) --- **Let me know if you need phasor diagrams or further breakdowns!**