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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.

You might be interested in
How deep does electrical conduit need to be buried?
GenaCL600 [577]
In general, bury metal conduits at least 6 inches below the soil surface. You may also run them at a depth of 4 inches under a 4-inch concrete slab. Under your driveway, the conduits must be below a depth of 18 inches, and under a public road or alleyway, they must be buried below 24 inches.
7 0
3 years ago
A system consists of a disk rotating on a frictionless axle
kakasveta [241]

The system includes a disk rotating on a frictionless axle and a bit of clay transferring towards it, as proven withinside the determine above.

<h3>What is the angular momentum?</h3>

The angular momentum of the device earlier than and after the clay sticks can be the same.

Conservation of angular momentum the precept of conservation of angular momentum states that the whole angular momentum is usually conserved.

  1. Li = Lf where;
  2. li is the preliminary second of inertia
  3. If is the very last second of inertia
  4. wi is the preliminary angular velocity
  5. wf is the very last angular velocity
  6. Li is the preliminary angular momentum
  7. Lf is the very last angular momentum

Thus, the angular momentum of the device earlier than and after the clay sticks can be the same.

Read more about the frictionless :

brainly.com/question/13539944

#SPJ4

8 0
2 years ago
A 1 250 kg car moving at a velocity of 30 km/hr along EDSA is accelerated by a force of 1 700 N. What will be its velocity after
Talja [164]
<h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>

Explanation:

<h2>Given:</h2>

m = 1,250 kg

v_i = 30 km/hr

F = 1,700 N

t = 10 s

<h2>Required:</h2>

Final velocity

<h2>Equation:</h2><h3>Force</h3>

F = ma

where: F - force

m - mass

a - acceleration

<h3>Acceleration</h3>

a = \frac{v_f \:-\:v_i}{t}

where: a - acceleration

v_i - initial velocity

v_f - final velocity

t - time elapsed

<h2>Solution:</h2><h3>Solve for acceleration using the formula for force</h3>

F = ma

Substitute the value of F and m

(1700 N) = (1250 kg)(a)

a = \frac{1700\:N}{1250\:N}

a = 1.36 m/s²

<h3>Solve for final velocity using the formula for acceleration</h3>
  • Convert 30 km/hr to m/s

= \frac{30\:km}{hr}\:×\:\frac{1000\:m}{1\:m}\:×\:\frac{1\:hr}{3600\:s}

= 8.33 m/s

  • Substitute the value of a, v_i and t

a = \frac{v_f \:-\:v_i}{t}

1.36\: m/s² \:= \:\frac{v_f \:-\:8.33\:m/s}{10\:s}

(10 \:s)1.36\: m/s² \:= \:v_f \:-\:8.33\:m/s

v_f\: =\: (10 \:s)1.36 \:m/s²\: + \:8.33\:m/s

v_f \: =\: 13.6 \:m/s \:+\: 8.33\:m/s

v_f\: =\:  21.93\: m/s

  • Convert to km/hr

= \frac{21.93\:m}{s}\:×\:\frac{1\:km}{1000\:m}\:×\:\frac{3600\:s}{1/:hr}

= 78.95\: km/hr

<h2>Final answer</h2><h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>
6 0
3 years ago
Describe in your own words the three strengthening mechanisms discussed in this chapter (i.e., grain size reduction, solid-solut
Kobotan [32]

Explanation:

Strengthening by grain size reduction

  • It is based on the fact that dislocations will experience hindrances while trying to move from a grain into the next because of abrupt change in orientation of planes.
  • Hindrances can be two types: forcible change of slip direction, and discontinuous slip plane.
  • Smaller the grain size, often a dislocation encounters a hindrance. Yield strength of material will be increased.
  • Yield strength is related to grain size (diameter, d ) as Hall Petch relation:

                \sigma_{y}=\sigma_{i}+k d^{-1 / 2}

Strengthening by Grain size reduction (contd..)

  • Grain size reduction improves not only strength, but also the toughness of many alloys.
  • If d is average grain diameter, S_{v} is grain boundary area per unit volume, N_{L} is mean number of intercepts of grain boundaries per unit length of test line, N_{A} is number of grains per unit area on a polished surface:

                S_{v}=2 N_{L} \quad d=\frac{3}{S_{v}}=\frac{3}{2 N_{L}} \quad d=\sqrt{\frac{6}{\pi V_{A}}}

  • Grain size can also be measured by comparing the grains at a fixed magnification with standard grain size charts.
  • Other method: Use of ASTM grain size number (Z). It is related to grain diameter, (in mm) as follows:

                 D=\frac{1}{100} \sqrt{\frac{645}{2^{6-1}}}

Solid solution strengthening

  • Impure foreign atoms in a single phase material produces lattice strains which can anchor the dislocations.
  • Effectiveness of this strengthening depends on two factors size difference and volume fraction of solute. Solute atoms interact with dislocations in many ways:

                                 - elastic interaction

                                 - modulus interaction

                                 - stacking-fault interaction

                                 - electrical interaction

                                 - short-range order interaction

                                - long-range order interaction

3 0
4 years ago
A coal-fired power plant equipped with a SO2 scrubber is required to achieve an overall SO2 removal efficiency of 85%. The exist
3241004551 [841]

Answer:

bypassed fraction B will be B= 0.105 (10.5%)

Explanation:

doing a mass balance of SO₂ at the exit

total mass outflow of SO₂ = remaining SO₂ from the scrubber outflow + bypass stream of SO₂

F*(1-er) =  Fs*(1-es) + Fb

where

er= required efficiency

es= scrubber efficiency

Fs and Fb = total mass inflow of  SO₂ to the scrubber and to the bypass respectively

F= total mass inflow of  SO₂

and from a mass balance at the inlet

F= Fs+ Fb

therefore the bypassed fraction B=Fb/F is

F*(1-er) =  Fs*(1-es) + Fb

1-er= (1-B)*(1-es) +B

1-er = 1-es - (1-es)*B + B

(es-er) = es*B

B= (es-er)/es = 1- er/es

replacing values

B= 1- er/es=1-0.85/0.95 = 2/19 = 0.105 (10.5%)

6 0
3 years ago
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