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ankoles [38]
3 years ago
8

At the end of a power distribution system, a certain feeder supplies three distribution transformer, each one supplying a group

of customers whose connected loads are as under: Transformer Load Demand factor Diversity factor Transformer No. 1 10kW 0.65 1.5 Transformer No. 2 12kW 0.6 3.5 Transformer No. 3 15kW 0.7 1.5 If the diversity factor among the transformer is 1.3; find the maximum load on the feeder.
Engineering
1 answer:
murzikaleks [220]3 years ago
6 0

Answer:

≈ 18.62 kw

Explanation:

Given data :

Diversity factor among transformer = 1.3

No of transformers       load      demand factor   diversity factor

Transformer 1               10 kw      0.65                    1.5

Transformer 2              12 kw      0.6                       3.5

Transformer 3               15 kw      0.7                      1.5

calculate the maximum load on the feeder

demand factor = max demand /  load

Max demand = demand factor * load

max demands:

Transformer 1 = 0.65 * 10 = 6.5

Transformer 2 = 0.6 * 12 = 7.2

Transformer 3 = 0.7 * 15 = 10.5

Diversity factor = ( summation of max demands ) / maximum load on feeder

  1.3 = ( 6.5 +7.2 + 10.5 ) / max load on feeder

hence : max load on feeder = 24.2 / 1.3 = 18.62 kw

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2 years ago
A company purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate
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1) The probability of at least 1 defective is approximately 45.621%

2) The probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is approximately 16.0212%

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The defective rate of the device = 3%

Therefore, the probability that a selected device will be defective, p = 3/100

The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;

The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97

The probability of 0 defective in 20 = ₂₀C₀(0.03)⁰·(0.97)²⁰ ≈ 0.543794342927

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The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;

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Therefore, the probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is 16.0212%

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