Questions: Uma instalação tem as seguintes cargas: I. 20 kVA, fator de potência 0,8 atrasado; II. 15 kW, fator de potência 0,5 adiantado; III. 50 KVA, 30 kVAr indutivo. Qual é a menor potência que o transformador da subestação dessa instalação deve ter para atender a todas as cargas simultaneamente? A 112,5kVA B 75 kVA C 300 kVA D 30 kVA E 150 kVA

Solution Steps

For each load, we need to calculate the complex power \( S = P + jQ \), where \( P \) is the real power and \( Q \) is the reactive power.

• Apparent power \( S_1 = 20 \, \text{kVA} \)
• Power factor \( \text{pf}_1 = 0.8 \) (lagging)

Real power \( P_1 = S_1 \times \text{pf}_1 = 20 \times 0.8 = 16 \, \text{kW} \)

Reactive power \( Q_1 = \sqrt{S_1^2 - P_1^2} = \sqrt{20^2 - 16^2} = \sqrt{400 - 256} = \sqrt{144} = 12 \, \text{kVAr} \)

• Real power \( P_2 = 15 \, \text{kW} \)
• Power factor \( \text{pf}_2 = 0.5 \) (leading)

Apparent power \( S_2 = \frac{P_2}{\text{pf}_2} = \frac{15}{0.5} = 30 \, \text{kVA} \)

Reactive power \( Q_2 = \sqrt{S_2^2 - P_2^2} = \sqrt{30^2 - 15^2} = \sqrt{900 - 225} = \sqrt{675} = 25.98 \, \text{kVAr} \)

Since the power factor is leading, \( Q_2 = -25.98 \, \text{kVAr} \).

• Apparent power \( S_3 = 50 \, \text{kVA} \)
• Reactive power \( Q_3 = 30 \, \text{kVAr} \) (inductive)

Real power \( P_3 = \sqrt{S_3^2 - Q_3^2} = \sqrt{50^2 - 30^2} = \sqrt{2500 - 900} = \sqrt{1600} = 40 \, \text{kW} \)

Sum the real and reactive powers of all loads to find the total complex power.

Total real power \( P_{\text{total}} = P_1 + P_2 + P_3 = 16 + 15 + 40 = 71 \, \text{kW} \)

Total reactive power \( Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 12 - 25.98 + 30 = 16.02 \, \text{kVAr} \)

The total apparent power \( S_{\text{total}} \) is given by:

\[ S_{\text{total}} = \sqrt{P_{\text{total}}^2 + Q_{\text{total}}^2} = \sqrt{71^2 + 16.02^2} = \sqrt{5041 + 256.6404} = \sqrt{5297.6404} \approx 72.78 \, \text{kVA} \]

Final Answer