Let's solve the problem step by step: ## **Given Data** - **Generator Voltage, \( V_{\text{LL}} \) = 23 kV** - **Generator MVA, \( S_{\text{base}} \) = 69.5 MVA** - **Sub-transient Reactance, \( X''_1 \) = 0.17 p.u.** - **Negative Sequence Reactance, \( X_2 \) = 0.32 p.u.** - **Zero Sequence Reactance, \( X_0 \) = 0.29 p.u.** - **Neutral grounded, zero fault impedance** - **Type of Fault: Double line-to-ground (b-c-g)** - **Assume generator is unloaded** --- ## **Step 1: Per Unit Value Calculations** Use the per unit (pu) system. The fault current for a double line-to-ground (DLG or LL-G) fault is: \[ I_{\text{fault, pu}} = \sqrt{3} \cdot I_{a, pu} \] But the current in phase *a* (the unfaulted phase) is: \[ I_a = 0 \] The symmetrical component method gives the sequence network connection for a LL-G fault as: \[ I_{b,\text{fault, pu}} = I_{c,\text{fault, pu}} = \frac{3V_{pre\text{-}fault, pu}}{X_1 + X_2 + X_0} \] Assuming the prefault voltage is 1.0 p.u. \[ I_{\text{fault, pu}} = \frac{3 \times 1.0}{X_1 + X_2 + X_0} \] \[ I_{\text{fault, pu}} = \frac{3}{0.17 + 0.32 + 0.29} = \frac{3}{0.78} = 3.846 \text{ p.u.} \] --- ## **Step 2: Convert to Actual (Amps)** The base current is: \[ I_{\text{base}} = \frac{S_{\text{base}}}{\sqrt{3} V_{\text{LL}}} \] Where: - \( S_{\text{base}} = 69.5 \times 10^6 \) VA - \( V_{\text{LL}} = 23 \times 10^3 \) V \[ I_{\text{base}} = \frac{69,500,000}{\sqrt{3} \times 23,000} \] \[ I_{\text{base}} = \frac{69,500,000}{39,797.97} \] \[ I_{\text{base}} \approx 1,746 \text{ A} \] --- ## **Step 3: Fault Current in Amps** \[ I_{\text{fault}} = I_{\text{fault, pu}} \times I_{\text{base}} \] \[ I_{\text{fault}} = 3.846 \times 1,746 = 6,718 \text{ A} \] --- ## **Step 4: Convert to kA** \[ I_{\text{fault}} = \frac{6,718}{1000} = \boxed{6.72 \text{ kA}} \] --- ### **Final Answer** **The rms value of the fault current is \(\boxed{6.72 \text{ kA}}\).**