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Keith_Richards [23]
3 years ago
12

Explain the process of energy conversion by describing how energy was converted from the windmill design brief. Discuss the diff

erent forms of energy and what technology was used to convert the energy from one form to another.
Engineering
1 answer:
cupoosta [38]3 years ago
6 0

Answer:

Wind energy is converted to Mechanical energy  which is then converted in to  electrical energy

Explanation:

In a wind mill the following energy conversions take place

a) Wind energy is converted into Mechanical energy (rotation of rotor blades)

b) Mechanical energy is converted into electrical energy (by using electric motor)

This electrical energy is then used for transmission through electric lines.

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Find the derivative of y = sin(ln(5x2 − 2x))
pickupchik [31]

Answer:

y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)

Explanation:

Let y = \sin[\ln(5\cdot x^{2}-2\cdot x)] and we proceed to find the derivative by the following steps:

1) y = \sin[\ln(5\cdot x^{2}-2\cdot x)] Given

2) y = \sin [\ln[x\cdot (5\cdot x - 2)]] Distributive property

3) y = \sin[\ln x + \ln (5\cdot x - 2 )] \ln (a\cdot b) = \ln a + \ln b

4) y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)  \frac{d}{dx} (\sin x) = \cos x/\frac{d}{dx}(\ln x) = \frac{1}{x}/\frac{d}{dx}(c\cdot x^{n}) = n\cdot c\cdot x^{n-1}/Rule of chain/Result

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3 years ago
I. The time till failure of an electronic component has an Exponential distribution and it is known that 10% of components have
drek231 [11]

Answer:

(a) The mean time to fail is 9491.22 hours

The standard deviation time to fail is 9491.22 hours

(b) 0.5905

(c) 3.915 × 10⁻¹²

(d) 2.63 × 10⁻⁵

Explanation:

(a) We put time to fail = t

∴ For an exponential distribution, we have f(t) = \lambda e^{-\lambda t}

Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);

P(t \leq 1000) = \int\limits^{1000}_0 {\lambda e^{-\lambda t}} \, dt = \dfrac{e^{1000\lambda}-1}{e^{1000\lambda}} = 0.1

e^(1000·λ) - 0.1·e^(1000·λ) = 1

0.9·e^(1000·λ) = 1

1000·λ = ㏑(1/0.9)

λ = 1.054 × 10⁻⁴

Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours

The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours

b) Here we have to integrate from 5000 to ∞ as follows;

p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [  -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905

(c) The Poisson distribution is presented as follows;

P(x = 3) = \dfrac{\lambda ^x e^{-x}}{x!}  = \frac{(1.0532 \times 10^{-4})^3 e^{-3} }{3!}  = 3.915\times 10^{-12}

p(x = 3) = 3.915 × 10⁻¹²

d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours

The Cumulative Distribution Function is given as follows;

p( t ≤ 1/4) CDF = 1 - e^{-\lambda \times t} = 1 - e^{-1.054 \times 10 ^{-4} \times 1/4} = 2.63 \times 10 ^{-5}.

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