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frosja888 [35]
3 years ago
12

Which of these is least likely a step in replacing a failed compressor?

Engineering
2 answers:
Akimi4 [234]3 years ago
4 0

Answer:

D. Inspecting the muffler

Explanation:

HVAC/refigeration systems are sealed systems that use refrigerant not combustion. A muffler is found on combustion systems.

You would disconnect,  flush, seal, and possibly replace the condenser if the condenser is found to be failing.

IgorLugansk [536]3 years ago
3 0

<u>Answer:</u>

C. Replacing the condenser is<u> least likely</u> a step in replacing a failed compressor.

<u>Explanation:</u>

To answer the above question, we need to understand the work of Air-cooling systems. There are 2 main units, compressor and condenser. Compresser and condenser are connected by the airtight pipe, and the required gas is injected in the system.

Compressor compresses the gas, compresed gas moves to condenser, condenser cools the gas, and the gas is expanded again, and then heat transfer takes place in indoor unit. Again the compressor compresses the gas, and process goes on.

So here, compressor is failed, so we need to inspect the muffler first of all to detect the issue, and definitely we need to remove all the gas from the system which means flushing the system, also hoses and other valves needs to be disconnected. However, replacing the condenser doesn't seems to be required to replace the failed compressor, as both are totally different units.

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The complete stress distribution obtained by superposing the stresses produced by an axial force and a bending moment is correct
Zanzabum

The complete stress distribution obtained by superposing the stresses produced by an axial force and a bending moment is correctly represented by F/A - (My)/(Iz).

<h3>What is the distribution of pressure at some stage in bending?</h3>

Compressive and tensile forces expand withinside the path of the beam axis beneath neath bending loads. These forces set off stresses at the beam. The most compressive pressure is observed on the uppermost fringe of the beam whilst the most tensile pressure is positioned on the decrease fringe of the beam.

The bending pressure is computed for the rail through the equation Sb = Mc/I, wherein Sb is the bending pressure in kilos in keeping with rectangular inch, M is the most bending second in pound-inches, I is the instant of inertia of the rail in (inches)4, and c is the space in inches from the bottom of rail to its impartial axis.

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7 0
2 years ago
Retaining<br>Function of<br>Wall​
goblinko [34]

Answer:

A retaining of a wall is a protective structure, first and foremost.

Explanation:

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7 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

a)

    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%

b)

    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%

c)

   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%

d)

   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%

e)

   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
4 years ago
Reaming is used for which three of the following functions: (a) accurately locate a hole position, (b) create a stepped hole, (c
Svetlanka [38]

Answer:

b

Explanation:

4 0
3 years ago
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