VIPSolutionsGive step-by-step solution with explanation and final answer:& cB 3% 22 org . 0. — a Ss 2s SE p> 5 3 a 8%9 3 SEE gE = IE oa 25 SEES i oi Sol Wr 588 SEEE OE Ea it £500 & High $58 32S | cI Saag fa = S =
Question:
Give step-by-step solution with explanation and final answer:
& cB 3% 22 org .
0. —
a Ss 2s SE p> 5 3 a
8%9 3 SEE gE = IE oa
25 SEES i oi Sol Wr
588 SEEE OE Ea
it £500
& High
$58 32S |
cI Saag
fa = S =
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Asked by: Aman
Created at: 2025-09-25 07:32:44
Asked by: Aman
Created at: 2025-09-25 07:32:44
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To solve the problems in the given image, let's break it down step-by-step:
### Given Data
1. **Feeder Load**: 100 kVA at .8 power factor (lagging).
2. **Additional Synchronous Motor**:
- Capacity: 600 kVA
- Absorbs: 60 kW when unloaded
- Requires: 270 kW to drive its pump
- Efficiency: 90%
### Questions
#### 2.1
##### 2.1.1 Total Line Current in the Feeder
1. **Original Load**:
- Apparent Power (\(S_1\)): 100 kVA
- Power Factor (\(pf_1\)): .8 (lagging)
- Real Power (\(P_1\)): \(S_1 \times pf_1 = 100 \times .8 = 800 \text{ kW}\)
2. **Synchronous Motor Load**:
- Real Power (\(P_2\)): 270 kW
- Efficiency: 90%
- Input Power: \(\frac{P_2}{\text{Efficiency}} = \frac{270}{.9} = 300 \text{ kW}\)
3. **Total Real Power**: \(P_{\text{total}} = P_1 + P_2 = 800 + 300 = 110 \text{ kW}\)
4. **Total Apparent Power**:
- Reactive Power (\(Q_1\)) for the original load: \(Q_1 = S_1 \times \sin(\cos^{-1}(pf_1)) = 100 \times \sin(\cos^{-1}(.8)) = 600 \text{ kVAR}\)
- \(S_{\text{total}} = \sqrt{P_{\text{total}}^2 + Q_1^2} = \sqrt{110^2 + 600^2} \approx 125 \text{ kVA}\)
5. **Line Current**:
- Voltage: 660 V
- \(I = \frac{S_{\text{total}}}{\sqrt{3} \times V} = \frac{125 \times 100}{\sqrt{3} \times 660} \approx 109.3 \text{ A}\)
##### 2.1.2 Power Factor of the Feeder
- \(pf_{\text{total}} = \frac{P_{\text{total}}}{S_{\text{total}}} = \frac{110}{125} = .88\)
#### 2.2
##### 2.2.1 Total Line Current in the Feeder (Unloaded Motor)
1. **Unloaded Motor Power**: 60 kW
2. **Total Real Power (Unloaded)**: \(P_{\text{unloaded}} = P_1 + 60 = 800 + 60 = 860 \text{ kW}\)
3. **Total Apparent Power (Unloaded)**:
- \(S_{\text{unloaded}} = \sqrt{P_{\text{unloaded}}^2 + Q_1^2} = \sqrt{860^2 + 600^2} \approx 105 \text{ kVA}\)
4. **Line Current**:
- \(I_{\text{unloaded}} = \frac{S_{\text{unloaded}}}{\sqrt{3} \times V} = \frac{105 \times 100}{\sqrt{3} \times 660} \approx 91.8 \text{ A}\)
##### 2.2.2 Power Factor of the Feeder (Unloaded Motor)
- \(pf_{\text{unloaded}} = \frac{P_{\text{unloaded}}}{S_{\text{unloaded}}} = \frac{860}{105} \approx .819\)
### Final Answers
- **2.1.1** Total Line Current: \(109.3 \text{ A}\)
- **2.1.2** Power Factor: \(.88\)
- **2.2.1** Total Line Current (Unloaded): \(91.8 \text{ A}\)
- **2.2.2** Power Factor (Unloaded): \(.819\)
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