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omeli [17]
3 years ago
15

What happens to the amperage draw of a condensing unit on a split AC system if the liquid line is restricted

Engineering
1 answer:
cupoosta [38]3 years ago
6 0

Answer:

The amperage draw of the condensing unit will be low.

Explanation:

A condensing unit is made up of a compressor and condenser, while an evaporating unit is made up of an evaporator coil.

A split AC system is a type of air conditioner system that has a condensing unit which is placed separately from the evaporative coil unit. Then the two units are connected to each other via a copper tube containing refrigerants.

The liquid line connects the condenser to the evaporator, and if this liquid line is restricted, the amp consumed by the condensing unit will be low.

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A bar having a length of 5 in. and cross-sectional area of 0. 7 in.2 is subjected to an axial force of 8000 lb. If the bar stret
andrew11 [14]

The modulus of elasticity is 28.6 X 10³ ksi

<u>Explanation:</u>

Given -

Length, l = 5in

Force, P = 8000lb

Area, A = 0.7in²

δ = 0.002in

Modulus of elasticity, E = ?

We know,

Modulus of elasticity, E = σ / ε

Where,

σ is normal stress

ε is normal strain

Normal stress can be calculated as:

σ = P/A

Where,

P is the force applied

A is the area of cross-section

By plugging in the values, we get

σ = \frac{8000 X 10^-^3}{0.7}

σ = 11.43ksi

To calculate the normal strain we use the formula,

ε = δ / L

By plugging in the values we get,

ε = \frac{0.002}{5}

ε = 0.0004 in/in

Therefore, modulus of elasticity would be:

E = \frac{11.43}{0.004} \\\\E = 28.6 X 10^3 ksi

Thus, modulus of elasticity is 28.6 X 10³ ksi

6 0
3 years ago
A civil engineer is studying a left-turn lane that is long enough to hold seven cars. Let X be the number of cars in the line at
BartSMP [9]

Answer:

a) C= 1/120

b) P(X>=5) = 0.333

Explanation:

The attached file contains the explanation for the answers

7 0
3 years ago
Read 2 more answers
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solong [7]

Answer:

yh and I said aaaaaaaaaaaa

Explanation:

7 0
3 years ago
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The atmosphere within a room is at 70 °F dry-bulb temperature, 50 percent degree of saturation, and 14.696 psia pressure. The in
Gre4nikov [31]

Answer:

Given that the temperature of the window is below the dew point it will condensate.

Explanation:

A psychrometric chart (like the one attached) will give you the information needed. This chart is for 14.696 psia.

On the bottom horizontal axes you have the dry-bulb temperature, in this case 70°F, going up from this point you can reach the 50% relative humidity curve (red point on chart), going horizontally from this point to the 100% relative humidity you get the dew point temperature (the point at which moisture will condensate) (blue point on chart). In this case the dew point is 50°C. Given that the temperature of the window is below the dew point it will condensate.

6 0
3 years ago
Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi
qaws [65]

Answer:

\Delta V = 209.151\,L, \Delta C = 217.517\,USD

Explanation:

The drag force is equal to:

F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A

Where C_{D} is the drag coefficient and A is the frontal area, respectively. The work loss due to drag forces is:

W = F_{D}\cdot \Delta s

The reduction on amount of fuel is associated with the reduction in work loss:

\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s

Where F_{D,1} and F_{D,2} are the original and the reduced frontal areas, respectively.

\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s

The change is work loss in a year is:

\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)

\Delta W = 2.043\times 10^{9}\,J

\Delta W = 2.043\times 10^{6}\,kJ

The change in chemical energy from gasoline is:

\Delta E = \frac{\Delta W}{\eta}

\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}

\Delta E = 6.81\times 10^{6}\,kJ

The changes in gasoline consumption is:

\Delta m = \frac{\Delta E}{L_{c}}

\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }

\Delta m = 154.772\,kg

\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }

\Delta V = 209.151\,L

Lastly, the money saved is:

\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )

\Delta C = 217.517\,USD

4 0
3 years ago
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