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3 years ago

When is the compressor able to compress the refrigerant?

Engineering

1 answer:

Dimas [21]3 years ago

3 0

Answer:

The compressor receives low pressure gas from the evaporator and converts it to high pressure gas. As the gas is compressed, the temperature rises. The hot refrigerant gas then flows to the condenser.

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Calculate the angle of banking on a bend of 100m radius so that vehicles can travel round the bend at 50km/hr without side thrus

saw5 [17]

Answer:

11.125°

Explanation:

Given:

Radius of bend, R = 100 m

Speed around the bend = 50 Km/hr = $\frac{5}{18}\times50$ = 13.89 m/s

Now,

We have the relation

$\tan\theta=\frac{v^2}{gR}$

where,

θ = angle of banking

g is the acceleration due to gravity

on substituting the respective values, we get

$\tan\theta=\frac{13.89^2}{9.81\times100}$

$\tan\theta=0.1966$

θ = 11.125°

3 0

3 years ago

In an apartment the interior air temperature is 20°C and exterior air temperatures is 5°C. The wall has inner and outer surface

ELEN [110]

Answer:

20 W/m², 20 W/m², -20 W/m²

Yes, the wall is under steady-state conditions.

Explanation:

Air temperature in room = 20°C

Air temperature outside = 5°C

Wall inner temperature = 16°C

Wall outer temperature = 6°C

Inner heat transfer coefficient = 5 W/m²K

Outer heat transfer coefficient = 20 W/m²K

Heat flux = Concerned heat transfer coefficient × (Difference of the temperatures of the concerned bodies)

q = hΔT

Heat flux from the interior air to the wall = heat transfer coefficient of interior air × (Temperature difference between interior air and exterior wall)

⇒ Heat flux from the interior air to the wall = 5 (20-6) = 20 W/m²

Heat flux from the wall to the exterior air = heat transfer coefficient of exterior air × (Temperature difference between wall and exterior air)

⇒Heat flux from the wall to the exterior air = 20 (6-5) = 20 W/m²

Heat flux from the wall to the interior air = heat transfer coefficient of interior air × (Temperature difference between wall and interior air)

⇒Heat flux from the wall and interior air = 5 (16-20) = -20 W/m²

Here the magnitude of the heat flux are same so the wall is under steady-state conditions.

7 0

3 years ago

Air is compressed by a 40-kW compressor from P1 to P2. The air temperature is maintained constant at 25Â°C during this process a

AlexFokin [52]

Answer:

the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Explanation:

Given the data in the question;

From the first law of thermodynamics;

dQ = dU + dW ------ let this be equation 1

where dQ is the heat transfer, dU is internal energy and dW is the work done.

from the question, the process is isothermal ( internally reversible process )

Thus, the change in internal energy is 0

dU = 0

given that; Air is compressed by a 40-kW compressor from P1 to P2

since it is compressed, dW = -40 kW

we substitute into equation 1

dQ = 0 + ( -40 kW )

dQ = -40 kW

Now, change in entropy of air is;

ΔS $_{air$ = dQ / T

given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K

so we substitute

ΔS $_{air$ = -40 kW / 298.15 K

ΔS $_{air$ = -0.13416 ≈ -0.1342 kW/K

Therefore, the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

7 0

3 years ago

Solid Isomorphous alloys strength

adell [148]

Lol please give me points

5 0

3 years ago

A horizontal curve on a 4-lane highway (two lanes on each direction with no median) has a superelevation of 6% and a central ang

kolbaska11 [484]

Answer:

Explanation: see attachment below

8 0

3 years ago