# Step-by-Step Solution Given: - **Transformer Rating**: 5000 kVA, 3-phase - **Voltage**: 6.6 kV (primary, delta), 33 kV (secondary, star) - **No load loss**: 15 kW - **Full-load loss (Total loss)**: 50 kW - **Impedance drop at full load**: 7% - **Load delivered**: 3200 kW at 0.8 pf (power factor), at 33 kV (secondary side) We are asked to **calculate the primary voltage** (at the 6.6 kV side) under these load conditions. --- ## 1. **Calculate Full-Load Current on Secondary Side** \[ \text{Full-load kVA} = 5000\, \text{kVA} \] \[ I_{FL,2} = \frac{\text{Full-load kVA} \times 1000}{\sqrt{3} \times V_{2}} = \frac{5000 \times 1000}{\sqrt{3} \times 33000} = \frac{5000000}{57154} \approx 87.45\, \text{A} \] --- ## 2. **Calculate Actual Load Current on Secondary Side** \[ \text{Load delivered} = 3200\, \text{kW} \text{ at } 0.8\, \text{pf} \] \[ S_{\text{load}} = \frac{3200\, \text{kW}}{0.8} = 4000\, \text{kVA} \] \[ I_{2} = \frac{4000 \times 1000}{\sqrt{3} \times 33000} = \frac{4000000}{57154} \approx 70.0\, \text{A} \] --- ## 3. **Determine Per-Unit Load Fraction** \[ \text{Load fraction} = \frac{70.0}{87.45} \approx 0.80 \] --- ## 4. **Calculate Load Loss at Actual Load** Load losses (mainly copper losses) vary as the square of the load: \[ \text{Load loss at actual load} = (\text{full load loss} - \text{no-load loss}) \times (\text{Load fraction})^2 \] \[ = (50 - 15)\,\text{kW} \times (0.80)^2 = 35 \times 0.64 = 22.4\, \text{kW} \] Total loss at this load: \[ \text{Total loss} = \text{no-load loss} + \text{load loss at actual load} = 15 + 22.4 = 37.4\, \text{kW} \] --- ## 5. **Calculate Output and Input Power** \[ \text{Output (delivered)} = 3200\, \text{kW} \] \[ \text{Input} = \text{Output} + \text{Losses} = 3200 + 37.4 = 3237.4\, \text{kW} \] --- ## 6. **Impedance Drop at Actual Load** Impedance drop at full load is **7%** (of rated voltage), at 0.8 power factor. Let’s calculate the **impedance voltage drop** at 0.8 per-unit load. Impedance drop (in percent) at a given load: \[ \text{Voltage drop} = \text{Percentage drop at full load} \times \text{Load fraction} \] \[ = 7\% \times 0.8 = 5.6\% \] But since impedance drop is a vector sum of resistive and reactive drops, and power factor affects it, let's be precise. If \( Z_{pu} = 0.07 \) per unit at full load, the voltage drop at less than full load and any power factor is: \[ \Delta V = I_2 \left[ R\cos\phi + X\sin\phi \right] \] But typically, with only percent impedance given, we use: \[ \text{Voltage drop (\%)} = \text{Impedance \%} \times \text{Load fraction} \times (\cos\theta) \] But for more accuracy, let's split impedance into resistance and reactance. Let \( x \) = reactance percent, \( r \) = resistance percent At full load, total losses are: - No-load loss = Iron loss = 15 kW - Full-load loss = 50 kW (includes iron + copper loss) - So, copper loss at full load = 50 - 15 = 35 kW The per-unit resistance drop at full load: \[ \text{Full-load copper loss} = 3 I_{FL,1}^2 R_{phase} \] But with percent impedance (Z%), and knowing the copper loss, we can use: \[ \text{Impedance (Z)}^2 = R_{pu}^2 + X_{pu}^2 \] And: \[ \text{Copper loss at full load} = I_{FL}^2 R \] Let’s find \( R_{pu} \): \[ \text{Per-phase full-load current} = 87.45\, \text{A} \] \[ \text{Per-phase voltage (secondary)} = \frac{33000}{\sqrt{3}} = 19052\, \text{V} \] \[ \text{Base power per phase} = \frac{5000}{3} = 1666.67\, \text{kVA} \] \[ \text{Base impedance (secondary)} = \frac{(19052)^2}{1666667} \approx 0.218\, \Omega \] \[ \text{Copper loss at full load} = 35\, \text{kW} \] \[ \text{Total copper loss} = 3 \times I_{FL}^2 \times R_{phase} \] \[ R_{phase} = \frac{35000}{3 \times (87.45)^2} = \frac{35000}{3 \times 7647.4} = \frac{35000}{22942.2} \approx 1.526\, \Omega \] \[ R_{pu} = \frac{1.526}{0.218} \approx 7.0 \] But this is 7 per unit, which is too high; let's use % base: \[ \text{Base MVA} = 5\, \text{MVA} \] \[ \text{Base voltage} = 33\, \text{kV} \] \[ \text{Base impedance} = \frac{33^2}{5} = 217.8\, \Omega \] But that's for single-phase. For three-phase, \[ Z_{base} = \frac{V_{LL}^2}{S_{base}} \] \[ = \frac{(33,000)^2}{5,000,000} = \frac{1,089,000,000}{5,000,000} = 217.8\, \Omega \] So, \[ R_{pu} = \frac{1.526}{217.8} = 0.007 \text{ (or 0.7\% per unit)} \] So, \[ Z_{pu} = 0.07 \text{ (7\%)} \] \[ R_{pu} = 0.007 \text{ (0.7\%)} \] \[ X_{pu} = \sqrt{(0.07)^2 - (0.007)^2} = \sqrt{0.0049 - 0.000049} = \sqrt{0.004851} \approx 0.0696 \text{ (6.96\%)} \] --- ## 7. **Voltage Drop at Actual Load and 0.8 pf** For any power factor, the per-unit voltage drop: \[ \Delta V_{pu} = (\text{Load fraction}) \times [R_{pu} \cos\phi + X_{pu} \sin\phi] \] Given: - Load fraction: 0.8 - \(\cos\phi = 0.8\) - \(\sin\phi = 0.6\) - \(R_{pu} = 0.007\) - \(X_{pu} = 0.0696\) \[ \Delta V_{pu} = 0.8 [0.007 \times 0.8 + 0.0696 \times 0.6] = 0.8 [0.0056 + 0.0418] = 0.8 \times 0.0474 = 0.0379 \] So, in percent: \[ \Delta V = 0.0379 \times 100 = 3.79\% \] --- ## 8. **Calculate the Primary Voltage Required** The transformer is delivering 33 kV (secondary, line-to-line) at the load terminals, which is after the voltage drop. The **primary voltage must be such that after accounting for this drop, the secondary voltage is 33 kV**. Let \( V_{2} \) = 33,000 V (at load), \( V_{1}' \) = ideal primary voltage required. The total actual secondary voltage at transformer terminals: \[ V_{2, \text{tr}} = V_{2} + \Delta V \] But since voltage drop is a percent of rated voltage, \[ \Delta V = 3.79\% \text{ of } 33,000 = 0.0379 \times 33,000 = 1,250\, \text{V} \] So, the transformer **secondary winding** must supply: \[ V_{2, \text{tr}} = 33,000 + 1,250 = 34,250\, \text{V} \] To maintain 33,000 V at the load, transformer secondary must be at 34,250 V. The transformer is ideal, so the ratio holds: \[ \frac{V_1}{V_2} = \frac{6,600}{33,000} = 0.2 \] So, \[ V_1 = V_2 \times 0.2 \] But since the transformer must supply 34,250 V at the secondary, the primary voltage must be: \[ V_{1, \text{required}} = 34,250 \times 0.2 = 6,850\, \text{V} \] --- ## 9. **Final Answer** ### **The required primary voltage is:** \[ \boxed{6,850\, \text{V}} \] --- ## **Summary Table** | Quantity | Value | |--------------------------|---------------------------| | Load delivered | 3,200 kW @ 0.8 pf | | Secondary voltage (load) | 33,000 V | | Load current (secondary) | 70.0 A | | Load fraction | 0.8 | | Voltage drop (secondary) | 1,250 V | | Secondary voltage (tr) | 34,250 V | | Required primary voltage | 6,850 V | --- ## **Answer** > **The required primary (6.6 kV side) voltage to maintain 33 kV at the secondary load terminals when supplying 3,200 kW at 0.8 pf is** > > \[ > \boxed{6,850\, \text{V}} > \] > > (rounded to the nearest 10 V)