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hram777 [196]
1 year ago
8

When the liquid line is restricted, the supply of refrigerant to the metering device is reduced. What is the effect on suction p

ressure and superheat
Physics
1 answer:
Nimfa-mama [501]1 year ago
8 0

Answer:

The suction pressure decreases and the superheat increases when the liquid line is restricted and the supply of refrigerant to the metering device is reduced.

Explanation:

1. The five components of refrigeration are:

  • Fluid refrigerant
  • Compressor
  • Condenser coil
  • Evaporator coil
  • Expansion device.

       The compressor limits the vapor released by the refrigerant. This            

       causes a rise in pressure (in refrigerant), which then pushes the  

       vapor into the coils on the outside of the refrigerator.

2. Now when the cooler air meets the warm gas present in the coils, it

   gets converted into liquid form.

3. Thus, when the liquid form is at high pressure, the refrigerant then  

   cools down as it flows through the coils placed in the fridge ( in both

   freezing and normal sections).

4. The refrigerant also absorbs the warm air present in the fridge, which  

   causes it to evaporate and flow back through the compressor and the

   cycle repeats in the same form.

Thus, when the liquid line is restricted and the supply of refrigerant to the metering device is reduced it causes a decrease in suction pressure and an increase in superheat.

Learn more about refrigeration here:

<u>brainly.com/question/9046279</u>

#SPJ4

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A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor
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Answer:

Explanation:

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\phi  = \texttt {magnetic flux}

E=\frac{\Delta (BA) }{\Delta  t}

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substitute

\texttt {substitute}\  \frac{1}{2} \Delta \theta R^2 \ \ for \Delta  A

E=B\frac{(\Delta  \theta R^3/2)}{\Delta  t} \\\\=\frac{1}{2} BR^2\omega

The magnetic field of the earth is oriented at 14.42

\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5

we plug in the values in the equation above

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E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega

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Answer:

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