# Step-by-Step Solution ## Given Data - **Distance (Length of feeder)**, \( L = 500 \) ft - **System type:** 3-wire, 480 V, 3-phase - **Load:** \( = 250) kVA- **Power:** \( pf = .90 \) (lagging) - **Conductor type:** THW copper - **Conduit:** Assume steel - **Tasks:** (a) Feeder and conduit size for ampacity (b) Voltage drop in feeder (c) Feeder and conduit size for max 1% voltage drop --- ## (a) Feeder and Conduit Size (Ampacity) ### 1. Calculate Full-Load Current The three-phase current is: \[ I = \frac{S}{\sqrt{3} \cdot V \cdot pf} \] But \( S \) is *apparent* power, so: \[ I = \frac{S}{\sqrt{3} \cdot V} \] \[ I = \frac{250,000}{\sqrt{3} \times 480} \] \[ I = \frac{250,000}{831.8} \] \[ I \approx 300.5 \text{ A} \] ### 2. Feeder Ampacity Selection Per NEC, feeders should be sized for 125% of continuous load (unless otherwise specified): \[ I_\text{feeder} = 1.25 \times 300.5 = 375.6 \text{ A} \] ### 3. Select THW Copper Conductor Size From NEC Table 310.16 (2023): - **THW Copper @ 75°C:** - 500 kcmil: 380 A - 400 kcmil: 335 A - 350 kcmil: 310 A **Select:** - **500 kcmil THW copper** (ampacity 380 A) ### 4. Conduit Size Per NEC Chapter 9, Table 5 (for THW insulation): - **500 kcmil THW copper:** .707 in² per wire Three wires needed for 3-phase, 3-wire: \[ 3 \times .707 = 2.121 \text{ in}^2 \] NEC Table 4 (conduit fill, 40% max for 3 conductors): - **3" Trade Size RMC (Rigid Metal Conduit):** 2.864 in² (40% fill) **Select:** - **3" RMC steel conduit** --- ### **Summary (a)** - **Feeder:** 3 × 500 kcmil THW copper - **Conduit:** 3" RMC steel --- ## (b) Voltage Drop in the Feeder ### 1. Use Voltage Drop Formula (Three-Phase, Balanced Load) \[ \Delta V = \sqrt{3} \cdot I \cdot (R \cos\theta + X \sin\theta) \cdot L \] Where: - \( I = 300.5 \) A (from above) - \( \cos\theta = .90 \), \( \sin\theta = \sqrt{1 - .9^2} = .4359 \) - \( L = 500 \) ft (one-way length) - \( R \) and \( X \) per 100 ft for 500 kcmil THW copper in steel conduit: - **R (resistance):** .022 ohms/100 ft - **X (reactance):** .016 ohms/100 ft ### 2. Calculate Total Resistance and Reactance For 500 ft: \[ R_{500} = .022 \times \frac{500}{100} = .011 \ \Omega \] \[ X_{500} = .016 \times \frac{500}{100} = .008 \ \Omega \] ### 3. Plug Values into Formula \[ \Delta V = \sqrt{3} \times 300.5 \times [.011 \times .90 + .008 \times .4359] \] First, calculate inside the brackets: \[ .011 \times .90 = .0099 \] \[ .008 \times .4359 = .0035 \] \[ .0099 + .0035 = .0134 \] Now, \[ \Delta V = 1.732 \times 300.5 \times .0134 \] \[ = 520.4 \times .0134 \] \[ = 6.97 \text{ V} \] ### 4. Percentage Voltage Drop \[ \text{Percent drop} = \frac{6.97}{480} \times 100\% \] \[ = 1.45\% \] --- ### **Summary (b)** - **Voltage drop:** \( 6.97 \) V (\( 1.45\% \) of 480 V) --- ## (c) Feeder and Conduit Size for 1% (Max) Voltage Drop ### 1. Maximum Allowable Voltage Drop \[ \Delta V_\text{max} = 1\% \times 480V = 4.8 \text{ V} \] ### 2. Rearranged Formula for Resistance & Reactance \[ \Delta V = \sqrt{3} \cdot I \cdot (R \cos\theta + X \sin\theta) \] \[ R \cos\theta + X \sin\theta = \frac{\Delta V}{\sqrt{3} \cdot I} \] Plug in values: \[ \frac{4.8}{1.732 \times 300.5} = \frac{4.8}{520.4} = .00923 \] So, \[ R \cos\theta + X \sin\theta = .00923 \] Recall: \[ R \cos\theta + X \sin\theta = (R_{500}) \cdot .90 + (X_{500}) \cdot .4359 \] Let’s solve for the maximum allowed \( R_{500} \) (assuming \( X_{500} \) is proportional to \( R_{500} \) based on wire size). But, more practically, we can try the next larger standard size: **700 kcmil**. #### Table values for 700 kcmil THW copper in steel conduit: - \( R = .016 \, \Omega/100\,\text{ft} \) - \( X = .013 \, \Omega/100\,\text{ft} \) For 500 ft: - \( R_{500} = .016 \times .5 = .008 \, \Omega \) - \( X_{500} = .013 \times .5 = .0065 \, \Omega \) \[ R \cos\theta + X \sin\theta = .008 \times .90 + .0065 \times .4359 \] \[ = .0072 + .00283 = .01003 \] \[ \Delta V = 1.732 \times 300.5 \times .01003 = 520.4 \times .01003 = 5.22\,\text{V} \] **Still too high. Try 900 kcmil:** - \( R = .0127 \, \Omega/100\,\text{ft} \) - \( X = .011 \, \Omega/100\,\text{ft} \) For 500 ft: - \( R_{500} = .0127 \times .5 = .00635 \, \Omega \) - \( X_{500} = .011 \times .5 = .0055 \, \Omega \) \[ = .00635 \times .90 + .0055 \times .4359 = .005715 + .002397 = .00811 \] \[ \Delta V = 1.732 \times 300.5 \times .00811 = 520.4 \times .00811 = 4.22\,\text{V} \] \[ \frac{4.22}{480} \times 100\% = .88\% \] Now below 1%. ### 3. Conduit Size for 900 kcmil - **900 kcmil THW copper:** 1.27 in² per wire \( 3 \times 1.27 = 3.81 \) in² **4" RMC** (40% fill = 4.214 in²) is appropriate. --- ### **Summary (c)** - **Feeder:** 3 × 900 kcmil THW copper - **Conduit:** 4" RMC steel --- # **Final Answers** ## (a) Feeder and Conduit Size (Ampacity) - **Feeder:** 3 × 500 kcmil THW copper - **Conduit:** 3" RMC steel ## (b) Voltage Drop in the Feeder - **Voltage drop:** 6.97 V (\( 1.45\% \) of 480 V) ## (c) Feeder and Conduit Size for 1% Voltage Drop - **Feeder:** 3 × 900 kcmil THW copper - **Conduit:** 4" RMC steel --- **References:** - NEC Table 310.16 (ampacities) - NEC Chapter 9, Table 5 (conduit fill) - Standard resistance/reactance tables for copper conductors in steel conduit