How Is Energy Transferred in a Refrigeration Compressor System?

[onemoretomorrow]

Hi guys i really need some help on this question.

Refrigerant 134a enters the compressor of a refrigeration system as saturated vapour at 0.14MPa and leaves as superheated vapour at 0.8MPa and 50 degrees celsius as a rate of 0.04kg/s. The suction area of the compressor is 10cm^2 and the discharge area 5cm^2. Determine the rates od energy transfer by mass into and out of the compressor, neglecting potential energy. Note the order of magnitude of the enthalpy and kinetic energy terms.

I can't seem to be able to get the value for the mass. can someone help? thanks

 

[Astronuc]

Determine the rates od energy transfer by mass into and out of the compressor

One does not need mass, since the problem seems to be asking for energy on a per unit mass basis.

The kinetic energy per unit mass is simply V²/2. Thermodynamic properties are usually given on a per unit mass basis, e.g. specific enthalpy, h (J/kg), or internal energy, u (J/kg).

Mass flow rate is given - 0.04kg/s - and assuming the continuity equation, as steady-state, mass rate (in) = mass rate (out), otherwise there would be an accumulation (if mass flow in exceeds mass flow out) of the system.

 

[piercas]

Sure, I would be happy to help with your question. In order to calculate the rates of energy transfer, we need to use the first law of thermodynamics, which states that energy cannot be created or destroyed, it can only change forms. In this case, we are looking at the energy transfer by mass into and out of the compressor.

To calculate the mass, we can use the ideal gas law equation, PV = mRT, where P is the pressure, V is the volume, m is the mass, R is the gas constant, and T is the temperature. We have the pressure and temperature values, as well as the volume, which is the area of the compressor. We can also assume that the gas is an ideal gas, since it is a refrigerant.

Using the given values, we can calculate the mass as follows:

m = (PV)/(RT) = (0.14MPa * 10cm^2) / (8.314 kJ/mol*K * 323.15K) = 0.00017 kg

Now, to determine the rates of energy transfer, we need to calculate the change in enthalpy and kinetic energy between the inlet and outlet of the compressor. The change in enthalpy can be calculated using the ideal gas equation, h = u + Pv, where h is the enthalpy, u is the internal energy, P is the pressure, and v is the specific volume.

For the inlet conditions, we have:

h1 = u1 + P1v1 = 0 + (0.14MPa * 0.0017m^3/kg) = 238 kJ/kg

For the outlet conditions, we have:

h2 = u2 + P2v2 = 0 + (0.8MPa * 0.00034m^3/kg) = 272 kJ/kg

Therefore, the change in enthalpy is:

Δh = h2 - h1 = 272 kJ/kg - 238 kJ/kg = 34 kJ/kg

Next, we need to calculate the change in kinetic energy. Since the mass flow rate is constant, the kinetic energy change can be calculated using the following equation:

ΔKE = (m * (V2^2 - V1^2)) / 2

Where m is the mass flow rate, V2 is the outlet velocity, and V1 is the inlet