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
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