Refrigerant 134a enters an air conditioner compressor at 4 bar, 20°C, and is compressed at steady state to 12 bar, 80°C. The volumetric flow rate of the refrigerant entering is 5.5 m3/min. The work input to the compressor is 82.5 kJ per kg of refrigerant flowing. Neglecting kinetic and potential energy effects, determine the magnitude of the heat transfer rate from the compressor, in kW.

Respuesta :

Answer:

The magnitude of the heat transfer rate from the compressor is 45.40 kW

Explanation:

Initial pressure of refrigerant = 4 bar

Final pressure of refrigerant = 12 bar

From steam table,

Internal energy at 4 bar (U1) = 2554 kJ/kg

Internal energy at 12 bar (U2) = 2588 kJ/kg

Change in internal energy (∆U) = U2 - U1

= 2588 - 2554

= 34 kJ

Work input (W) = 82.5 kJ/kg

Quantity of heat transfer (Q)

= ∆U + W

= 34 + 82.5

= 116.5 kJ/kg

Volumetric flow rate of refrigerant = 5.5 m^3/min

= 5.5/60

= 0.0917 m^3/s

Density of refrigerant = 4.25 kg/m^3

Mass flow rate = density × volumetric flow rate

= 4.25 kg/m^3 × 0.0917 m^3/s

= 0.38973 kg/s

Q = 116.5 kJ/kg × 0.38973 kg/s

= 45.40 kJ/s

= 45.40 kW

The magnitude of the heat transfer rate from the compressor is 45.40 kW

The magnitude of the heat transfer rate from the compressor is 45.40 kW

  • The calculation is as follows:

Given that,

The initial pressure of refrigerant = 4 bar

The final pressure of refrigerant = 12 bar

Now

From the steam table,

Internal energy at 4 bar (U1) = 2554 kJ/kg

Internal energy at 12 bar (U2) = 2588 kJ/kg

So, Change in internal energy (∆U) is

= U2 - U1

= 2588 - 2554

= 34 kJ

And, the Work input (W) = 82.5 kJ/kg

Now

Quantity of heat transfer (Q)

= ∆U + W

= 34 + 82.5

= 116.5 kJ/kg

Now

Volumetric flow rate of refrigerant is

[tex]= 5.5 m^3/min\\\\= 5.5\div 60\\\\= 0.0917 m^3/s[/tex]

And, Density of refrigerant = 4.25 kg/m^3

Now

[tex]Mass\ flow\ rate = density \times volumetric\ flow\ rate\\\\= 4.25 kg/m^3 \times 0.0917 m^3/s[/tex]

= 0.38973 kg/s

And,

[tex]Q = 116.5 kJ/kg \times 0.38973 kg/s[/tex]

= 45.40 kJ/s

= 45.40 kW

Learn more: https://brainly.com/question/15517759?referrer=searchResults