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3 years ago

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

Answer:

C. Replacing the condenser is least likely a step in replacing a failed compressor.

Explanation:

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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Technician a says that diesel engines can produce more power because air in fuel or not mix during the intake stroke. Technician

mariarad [96]

Answer:

Technician be says that diesel engines produce more power because they use excess air to burn feel who is correct

Explanation:

He is correct as many engines are run by diesel. It produces more power as that is how cars produce more power.

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2 years ago

An existing building is suffering from cracks in the exterior walls. The investigating engineer wants to ensure that the foundat

jek_recluse [69]

Answer:

$18 ft^{2}$

Explanation:

Soil bearing pressure= $\frac {Load}{Area}$

Since we're given pressure of 2500 psf and load of 45000 pounds

The area= $\frac {45000}{2500}=18$

Therefore, the smallest area of safe footings should not be less than $18 ft^{2}$

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3 years ago

Carbon resistors often come as a brown cylinder with colored bands. These colored bands can be read to determine the manufacture

alexandr1967 [171]

Answer:

a) 4.7 kΩ, +/- 5%

b) 2.0 MΩ, +/- 20%

Explanation:

a) If the resistor has the following combination of color bands:

1) Yellow = 1st digit = 4

2) Violet = 2nd digit = 7

3) Red = multiplier = 10e2

4) Gold = tolerance = +/- 5%

this means that the resistor has 4700 Ω (or 4.7 kΩ), with 5% tolerance.

b) Repeating the process for the following combination of color bands:

1) Red = 1st digit = 2

2) Black = 2nd digit = 0

3) Green = multiplier = 10e5

4) Nothing = tolerance = +/- 20%

This combination represents to a resistor of 2*10⁶ Ω (or 2.0 MΩ), with +/- 20% tolerance.

7 0

3 years ago

A nozzle receives an ideal gas flow with a velocity of 25 m/s, and the exit at 100 kPa, 300 K velocity is 250 m/s. Determine the

Margaret [11]

Given Information:

Inlet velocity = Vin = 25 m/s

Exit velocity = Vout = 250 m/s

Exit Temperature = Tout = 300K

Exit Pressure = Pout = 100 kPa

Required Information:

Inlet Temperature of argon = ?

Inlet Temperature of helium = ?

Inlet Temperature of nitrogen = ?

Answer:

Inlet Temperature of argon = 360K

Inlet Temperature of helium = 306K

Inlet Temperature of nitrogen = 330K

Explanation:

Recall that the energy equation is given by

$C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2)$

Where Cp is the specific heat constant of the gas.

Re-arranging the equation for inlet temperature

$T_{in} = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p} + T_{out}$

For Argon Gas:

The specific heat constant of argon is given by (from ideal gas properties table)

$C_p = 520 \:\: J/kg.K$

So, the inlet temperature of argon is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520} + 300$

$T_{in} = \frac{1}{2} \times 119 + 300$

$T_{in} = 360K$

For Helium Gas:

The specific heat constant of helium is given by (from ideal gas properties table)

$C_p = 5193 \:\: J/kg.K$

So, the inlet temperature of helium is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193} + 300$

$T_{in} = \frac{1}{2} \times 12 + 300$

$T_{in} = 306K$

For Nitrogen Gas:

The specific heat constant of nitrogen is given by (from ideal gas properties table)

$C_p = 1039 \:\: J/kg.K$

So, the inlet temperature of nitrogen is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039} + 300$

$T_{in} = \frac{1}{2} \times 60 + 300$

$T_{in} = 330K$

Note: Answers are rounded to the nearest whole numbers.

5 0

3 years ago

At what stage of development is an engineering team the most productive?

Alexxx [7]

Answer:

Explanation:

A because you are continuing to keep moving and thinking.

3 0

3 years ago

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