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3 years ago

Give me the description of - Feedforward control loops with an example.

Engineering

1 answer:

ValentinkaMS [17]3 years ago

5 0

Answer:

Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop

Explanation:

Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop. Although Feedforward control seems to be a very attractive idea, it imposes a high responsibility on both the system developer and the operator to examine and consider mathematically the effect of disruptions on the process concerned.

example of feedforward is

Shower

which consist of following control points

Hear toilet flush (measurement)

Customize water to compensate

feedback refers to that point when water turns hot before the configuration changes

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Question 2: (a) In your own words, clearly distinguish and differentiate between Ethics in Engineering and Ethics in Computing (

zlopas [31]

Engineering ethics is not without abstraction, but in contrast with computing, it is animated by a robust and active movement concerned with the seamless identification of ethics with practice.

<h3 /><h3>What is engineering?</h3>

This is a branch of science and technology concerned with the design, building, and use of engines, machines, and structures that uses scientific principles.

Comparing ethics in engineering and ethics in computing:

Engineering ethics are a set of rules and guidelines. While computing ethics deals with procedures, values and practices.
In engineering ethics, engineers must adhere to these rules as a moral obligation to their profession While in computing ethics, the ethics govern the process of consuming computer technology.
Following these ethics for the two professions will NOT cause damage, but disobeying them causes damage.

Some practical examples in the computing field:

Avoid using the computer to harm other people such as creating a bomb or destroying other people's work.
Users also should not use a computer for stealing activities like breaking into a bank or company.
Make sure a copy of the software had been paid for by the users before it is used.

Some practical examples in the engineering field:

Integrity for oneself.
Respect for one another.
Pursuit of excellence and accountability.

Hence, Engineering ethics is the field of system of moral principles that apply to the practice of engineering and following them is important to the profession.

Read more about <em>engineering</em> here:

brainly.com/question/17169621

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7 0

2 years ago

Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2

motikmotik

Answer:

The answer is " $112.97 \ \frac{ft}{s}$ "

Explanation:

Air flowing into the $p_1 = 20 \ \frac{lbf}{in^2}$

Flow rate of the mass $m = 230.556 \frac{lbm}{s}$

inlet temperature $T_1 = 700^{\circ} F$

Pipeline $A= 5 \times 4 \ ft$

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

$\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}$

$= \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}$

$V= \frac{mv}{A}$

$= \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}$

8 0

3 years ago

Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?

Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

x_average = ∑ $x_{i}$ / n

The tolerance or error is the current value over the mean value per 100

Δx₁ = x₁ / x_average

tolerance = | 100 -Δx₁ 100 |

bars indicate absolute value

let's look for these values for each case

x_average = (2.1700000+ 2.258571429) / 2

x_average = 2.2142857145

fluctuation for x₁

Δx₁ = 2.17000 / 2.2142857145

Tolerance = 100 - 97.999999991

Tolerance = 2.000000001%

fluctuation x₂

Δx₂ = 2.258571429 / 2.2142857145

Δx2 = 1.02

tolerance = 100 - 102.000000009

tolerance 2.000000001%

x_average = (2.2 + 2.29) / 2

x_average = 2,245

fluctuation x₁

Δx₁ = 2.2 / 2.245

Δx₁ = 0.9799554

tolerance = 100 - 97,999

Tolerance = 2.00446%

fluctuation x₂

Δx₂ = 2.29 / 2.245

Δx₂ = 1.0200445

Tolerance = 2.00445%

x_average = (2.211445 +2.3) / 2

x_average = 2.2557225

Δx₁ = 2.211445 / 2.2557225 = 0.9803710

tolerance = 100 - 98.0371

tolerance = 1.96%

Δx₂ = 2.3 / 2.2557225 = 1.024624

tolerance = 100 -101.962896

tolerance = 1.96%

x_average = (2.20144927 + 2.29130435) / 2

x_average = 2.24637681

Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

tolerance = 100 - 98.000043

tolerance = 2.000002%

Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

tolerance = 2.0000002%

x_average = (2 +2,3) / 2

x_average = 2.15

Δx₁ = 2 / 2.15 = 0.93023

tolerance = 100 -93.023

tolerance = 6.98%

Δx₂ = 2.3 / 2.15 = 1.0698

tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0

3 years ago

Determine whether or not each of the following four transaction execution histories is serializable. If a history is serializabl

ludmilkaskok [199]

Answer:

Option D. w1[x] w2[u] w2[y] w1[y] w3[x] w3[u] w1[z]

Explanation:

The execution in the option D is correct. This is because there is more than one reasonable criterion.

8 0

3 years ago

Question 7.1: Two possible overhead valve combustion chambers are being considered – the first has two valves; the second has fo

AleksandrR [38]

Answer:

1) The adoption of the second design we can see that the total valve perimeter is increased by 60.8%

2) Increase in flow are : 29%

3) Additional benefits in using 4 valves per cylinder:

a)For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

Explanation:

1) Perimeter of the first possible overhead valve combustion chamber with two valves:

P₂ = πd = π × 23 = 72.26mm

Perimeter of the second possible overhead valve combustion chamber with four valves:

P₄ = π2d = π × 18.5 × 2 = 116.24 mm

If second design is adopted, percentage increase = ((P₄ - P₂)/P₂)×100

= ((116.24 - 72.26)/72.26)×100 = 0.6086 ×100 = 60.86%

Therefore, the total valve perimeter is shown to have increased by 60.8%

2) Formula for flow Area (A) = P × L = πkd²

Area of the first possible overhead valve combustion chamber with two valves: A₂ = πkd² = πk(23)² = 1662k mm²

Area of the first possible overhead valve combustion chamber with four valves: A₄ = πkd² = 2πk(18.5)² = 2150k mm²

The percentage increase in flow area: ((A₄ - A₂)/A₄)×100 = ((2150 - 1662)/2150)×100 = 29%

3) The additional benefits of using are:

a) For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

7 0

3 years ago