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devlian [24]
3 years ago
14

A particular cloud-to-ground lightning strike lasts 500 µµsec and delivers 30 kA across a potential difference of 100 MV. Assu

me voltage and current are constant over the duration of the strike.
Required:
a. How much charge is delivered by the lightning bolt?
b. How much power does the lightning deliver?
c. How much total energy is delivered by the lightning strike? Give your answer in both J and Wh, using the appropriate Si prefix in each case.
d. If the average cost of electricity to residential consumers in Oregon is $0.098/kWh, what is the residential retail value of the energy delivered by the strike?
e. If it were possible to harvest all of the energy delivered by a lighting strike, how many lightning events with the given characteristics would be required to power an average US home for a year. Assume an average monthly residential energy consumption of 900 kWh.
Engineering
1 answer:
sasho [114]3 years ago
7 0

Answer:

a) 15 C charge was delivered by the lightening bolt

b) the lightning delivered 3.0 × 10¹² W of power

c)

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is 416666.67 Wh

d)

the residential retail value of the energy delivered by the strike is $ 40.83

e)

a total of 26 lightening strikes would be required to power an average US home for a year.

Explanation:

Given that;

the lighting strike lasted for t ( time ) = 500 μsecs = 500×10⁻⁶ s

Current I = 30 kA = 30×10³ A

voltage V = 100 mV = 100×10⁶ v

a)

we know that; I = Q/t

so Q = I × t

we substitute

Q =  30×10³ × 500×10⁻⁶

Q = 15 C

Therefore 15 C charge was delivered by the lightening bolt

b)

Power P = V × I

we substitute

Power P = 100×10⁶ × 30×10³

P = 3.0 × 10¹² W

Therefore, the lightning delivered 3.0 × 10¹² W of power

c)

we know that; Power = Energy / Time

Energy = Power × Time

we substitute

E = 3.0 × 10¹²  × 500×10⁻⁶

E = 1500 × 10⁶ J

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is;

⇒ 1500×10⁶ / 3600 Wh

= 416666.67 Wh

d)

given that;  1 KWh → $ 0.098

energy delivered by the strike = 416666.67 Wh = 416.66667 KWh

so the residential retail value of the energy delivered by the strike will be;

416.66667 KWh × $ 0.098

= $ 40.83

∴ the residential retail value of the energy delivered by the strike is $ 40.83

e)

Given that; average monthly residential energy consumption is 900 kWh.

for a year; energy consumption = 12 × 900 kWh = 10,800 KWh = 10800000 Wh

Now

1 lightening strike ⇒ 416666.67 Wh

x lightening strike ⇒ 10800000 Wh

x = 10800000 / 416666.67

x = 25.9199 ≈ 26

Therefore; a total of 26 lightening strikes would be required to power an average US home for a year.  

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Answer:

  • <u>0.59</u>

Explanation:

The <em>batch</em> is <em>rejected</em> if any of the <em>random samples are found defective</em>, or, what is the same, it will be accepted only if all 15 samples are good.

The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.

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By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.

The joint probability of all the samples are good is the product of each probability:

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The result is: 0.41278 ≈ 0.41

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Explanation:

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