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anyanavicka [17]
2 years ago
6

How often do you need to add freon to a central air conditioning unit.

Engineering
1 answer:
Veseljchak [2.6K]2 years ago
5 0

Answer:

If everything is working properly, your AC should never need a refrigerant. In fact, a central air conditioner should never need refrigerant added unless there's a refrigerant leak. We'll explain how an AC uses refrigerant and what to do if you think your system is low on refrigerant. OR When you check your air conditioner unit, or your outdoor central air unit with the fan, and see an ice building up on the refrigerant line, then you may be low on freon. ... This will cause the surrounding moisture on the refrigerant line to freeze up. This is a sign you need freon.

Explanation:

(this is according to plumbing services website)

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Answer:

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7 0
3 years ago
One of the best ways to increase engine power and control detonation and preignition is to?
Reika [66]

Answer:

use water injection.

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1 year ago
An electrochemical cell is composed of pure nickel and pure iron electrodes immersed in solutions of their divalent ions. If the
xenn [34]

Answer:C  0.12 V

Explanation:

Given

Concentration of Fe^{2+} M_1=0.40 M

Concentration of Ni^{2+} M_2=0.002 M

Standard Potential for Ni and Fe are V_2=-0.25 V  and V_1=-0.44 V

\Delta V=V_2-V_1-\frac{0.0592}{2}\log (\frac{M_1}{M_2})

\Delta V=-0.25-(-0.44)-\frac{0.0592}{2}\log (\frac{0.4}{0.002})

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7 0
3 years ago
Acoke can with inner diameter(di) of 75 mm, and wall thickness (t) of 0.1 mm, has internal pressure (pi) of 150 KPa and is suffe
NemiM [27]

Answer:

All 3 principal stress

1. 56.301mpa

2. 28.07mpa

3. 0mpa

Maximum shear stress = 14.116mpa

Explanation:

di = 75 = 0.075

wall thickness = 0.1 = 0.0001

internal pressure pi = 150 kpa = 150 x 10³

torque t = 100 Nm

finding all values

∂1 = 150x10³x0.075/2x0,0001

= 0.5625 = 56.25mpa

∂2 = 150x10³x75/4x0.1

= 28.12mpa

T = 16x100/(πx75x10³)²

∂1,2 = 1/2[(56.25+28.12) ± √(56.25-28.12)² + 4(1.207)²]

= 1/2[84.37±√791.2969+5.827396]

= 1/2[84.37±28.33]

∂1 = 1/2[84.37+28.33]

= 56.301mpa

∂2 = 1/2[84.37-28.33]

= 28.07mpa

This is a 2 d diagram donut is analyzed in 2 direction.

So ∂3 = 0mpa

∂max = 56.301-28.07/2

= 14.116mpa

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3 years ago
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