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2 years ago

Which is an example of a subsystem and system relationship?

Engineering

1 answer:

Zanzabum2 years ago

6 0

The example of a subsystem and system relationship is option

A) mechanical toggle, Xbox controller

Explanation:

Controls and Controllers are known as the mechanical, electro-mechanical, or electronic devices which are used to input signals to change the conditions or values which are in processes or access to buildings, operators, gated areas, etc.
When you press and hold the given Shift button, these buttons which are chosen becomes an alternate inputs.
Normally, there are three controllers(PID) which are used in any of the process industries inorder to maintain and control the required setpoint.
an example of a subsystem and system relationship is a mechanical toggle, Xbox controller
Controllers improve the steadiness and state of accuracy by decreasing the steady state error. When the steady state accuracy gets better, the stability also improves.
Controllers are also used to help in reducing the unwanted offsets which are produced by the system.

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A rigid 10-L vessel initially contains a mixture of liquid and vapor water at 100 °C, with a quality factor of 0.123. The mixtur

masya89 [10]

Answer:

$Q_{in} = 46.454\,kJ$

Explanation:

The vessel is modelled after the First Law of Thermodynamics. Let suppose the inexistence of mass interaction at boundary between vessel and surroundings, changes in potential and kinectic energy are negligible and vessel is a rigid recipient.

$Q_{in} = U_{2} - U_{1}$

Properties of water at initial and final state are:

State 1 - (Liquid-Vapor Mixture)

$P = 101.42\,kPa$

$T = 100\,^{\textdegree}C$

$\nu = 0.2066\,\frac{m^{3}}{kg}$

$u = 675.761\,\frac{kJ}{kg}$

$x = 0.123$

State 2 - (Liquid-Vapor Mixture)

$P = 476.16\,kPa$

$T = 150\,^{\textdegree}C$

$\nu = 0.2066\,\frac{m^{3}}{kg}$

$u = 1643.545\,\frac{kJ}{kg}$

$x = 0.525$

The mass stored in the vessel is:

$m = \frac{V}{\nu}$

$m = \frac{10\times 10^{-3}\,m^{3}}{0.2066\,\frac{m^{3}}{kg} }$

$m = 0.048\,kg$

The heat transfer require to the process is:

$Q_{in} = m\cdot (u_{2}-u_{1})$

$Q_{in} = (0.048\,kg)\cdot (1643.545\,\frac{kJ}{kg} - 675.761\,\frac{kJ}{kg} )$

$Q_{in} = 46.454\,kJ$

3 0

2 years ago

Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carb

Nezavi [6.7K]

Answer:

$y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02$

Explanation:

Hello,

a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:

- $\xi_1$ and $\xi_2$ : extent of the reactions (2).

- $F_{O_2}^2$ , $F_{CH_4}^2$ , $F_{H_2O}^2$ , $F_{HCHO}^2$ and $F_{CO_2}^2$ : Molar flows at the second stream (5).

On the other hand, we've got the following equations:

- $F_{O_2}^2=50mol/s-\xi_1-2\xi_2$ : oxygen mole balance.

- $F_{CH_4}^2=50mol/s-\xi_1-\xi_2$ : methane mole balance.

- $F_{H_2O}^2=\xi_1+2\xi_2$ : water mole balance.

- $F_{HCHO}^2=\xi_1$ : formaldehyde mole balance.

- $F_{CO_2}^2=\xi_2$ : carbon dioxide mole balance.

Thus, the degrees of freedom are:

$DF=7unknowns-5equations=2$

It means that we need two additional equations or data to solve the problem.

b. Here, the two missing data are given. For the fractional conversion of methane, we define:

$0.900=\frac{\xi_1+\xi_2}{50mol/s}$

And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:

$0.860=\frac{F_{HCHO}^2}{50mol/s}$

In such a way, one realizes that the output formaldehyde's molar flow is:

$F_{HCHO}^2=0.860*50mol/s=43mol/s$

Which is equal to the first reaction extent $\xi_1$ , therefore, one computes the second one from the fractional conversion of methane as:

$\xi_2=0.900*50mol/s-\xi_1\\\xi_2=0.900*50mol/s-43mol/s\\\xi_2=2mol/s$

Now, one computes the rest of the output flows via:

- $F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s$

- $F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s$

- $F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s$

- $F_{HCHO}^2=43mol/s$

- $F_{CO_2}^2=2mol/s$

The total output molar flow is:

$F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s$

Therefore the output stream composition turns out into:

Best regards.

7 0

2 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

Typical noise associated with failed cv joint

igor_vitrenko [27]

Answer:

A worn inner CV joint often makes a clunking noise during starts and stops.

3 0

2 years ago

Why does teachers grade things that are not due yet

pav-90 [236]

I think because if you’ve already turned it in they might as well grade asap instead of waiting

4 0

3 years ago

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