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4 years ago

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

Engineering

1 answer:

ValentinkaMS [17]4 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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An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 6 kW. De

Paraphin [41]

Answer:

a. 2.08, b. 1110 kJ/min

Explanation:

The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. <u>Assume that the air conditioner operates steadily.</u>

a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is

COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat

COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08

The COP of this air conditioner is 2.08.

b. The rate of heat discharged to the outside air is determined from the energy balance.

Q(H) = Q(L) + W(net in)

Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min

The rate of heat transfer to the outside air is 1110 kJ for every minute.

5 0

3 years ago

What is a problem that technology can help solve that problem?

Maslowich

Seeing what the other side of the world is doing right now

8 0

3 years ago

Read 2 more answers

5. Create a function named second_a that uses a list comprehension. The function will take a single integer parameter n. Find ev

Mekhanik [1.2K]

Answer:

//Program was implemented using C++ Programming Language

// Comments are used for explanatory purpose

#include<iostream>

using namespace std;

unsigned int second_a(unsigned int n)

{

int r,sum=0,temp;

int first;

for(int i= 1; I<=n; i++)

{

first = n;

//Check if first digit is 3

// Remove last digit from number till only one digit is left

while(first >= 10)

{

first = first / 10;

}

if(first == 3) // if first digit is 3

{

//Check if n is palindrome

temp=n; // save the value of n in a temporary Variable

while(n>0)

{

r=n%10; //getting remainder

sum=(sum*10)+r;

n=n/10;

}

if(temp==sum)

cout<<n<<" is a palindrome";

else

cout<<n<<" is not a palindrome";

}

Explanation:

The above code segments is a functional program that checks if a number that starts with digit 3 is Palindromic or not.

The program was coded using C++ programming language.

The main method of the program is omitted.

Comments were used for explanatory purpose.

8 0

3 years ago

What is hardness and how is it generally tested?

drek231 [11]

Answer:

Hardness is understood as the property of materials in general to resist the penetration of an indenter under load, so that the hardness represents the resistance of the material to the plastic deformation located on its surface.

Explanation:

Hardness of a material is understood as the resistance that the material opposes to its permanent surface plastic deformation by scratching or penetration. It is always true that the hardness of a material is inversely proportional to the footprint that remains on its surface when a force is applied.

In this sense, the hardness of a material can also be defined as that property of the surface layer of the material to resist any elastic deformation, plastic or destruction due to the action of local contact forces caused by another body (called indenter or penetrator), harder, of certain shape and dimensions, which does not suffer residual deformations during contact.

That is, hardness is understood as the property of materials in general to resist the penetration of an indenter under load, so that the hardness represents the resistance of the material to the plastic deformation located on its surface.

The following conclusions can be drawn from the previous definition of hardness:

1) hardness, by definition, is a property of the surface layer of the material, and is not a property of the material itself;

2) the methods of hardness by indentation presuppose the presence of contact efforts, and therefore, the hardness can be quantified within a scale;

3) In any case, the indenter or penetrator must not undergo residual deformations during the test of hardness measurement of the body being tested.

To determine the hardness of the materials, durometers with different types of tips and ranges of loads are used on the various materials. Below are the most commonly used tests to determine the hardness of the materials.

Rockwell hardness :

It refers to the Rockwell hardness test, a method with which the hardness or resistance of a material to be penetrated is calculated. It is characterized by being a fast and simple method that can be applied to all types of materials. An optical reader is not required.

Brinell hardness :

Brinell hardness is a scale that is used to determine the hardness of a material through the indentation method, which consists of penetrating with a hardened steel ball tip into the hard material, a load and for a certain time.

This test is not very precise but easy to apply. It is one of the oldest and was proposed in 1900 by Johan August Brinell, a Swedish engineer.

Vickers hardness:

Vickers hardness is a test that is used in all types of solid and thin or soft materials. In this test, a square-shaped pyramid-shaped diamond and a 136° vertex angle are placed on the penetrating equipment.

In this test the hardness measurement is performed by calculating the diagonal penetration lengths.

However, its result is not read directly on the equipment used, therefore, the following formula must be applied to determine the hardness of the material: HV = 1.8544 · F / (dv2).

3 0

3 years ago

A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. I

Liula [17]

Answer:

fire brick / common brick : 1218 °C
common brick / magnesia : 1019 °C
magnesia / steel : 90.06 °C
heat loss: 4644 kJ/m^2/h

Explanation:

The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...

R = d/k

so the thermal resistances of the layers of furnace wall are ...

R₁ = 0.200/4 = 0.05 °C·m²·h/kJ

R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ

So, the total thermal resistance is ...

R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)

The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C

so, the fire brick interface temperature at the common brick is ...

1450 -232 = 1218 °C

For the next layers, the interface temperatures are ...

common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

_____

<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

5 0

3 years ago