Answer:
(a) [tex]m_{R-134a}=0.0338kg/s[/tex]
(b) [tex]Q_H=7.03kW[/tex]
(c) [tex]COP=4.39[/tex]
Explanation:
Hello,
(a) In this part, we must know that the energy provided by the water equals the energy gained by the refrigerant-134a, thus:
[tex]m_{R-134a}(h_2-h1)=m_{H_2O}Cp_{H_2O}*\Delta T_{H_2O}[/tex]
Now, water's heat capacity is about 4.18kJ/kg°C and the enthalpies at both the first and second state for the refrigerant-134a are computed as shown below, considering the first state as a vapor-liquid mixture (VLM) at 12°C and the second state as a saturated vapor (ST) at the same conditions:
[tex]h_{1/12^0C/VLM}=68.18kJ/kg+0.15*189.09kJ/kg=96.54kJ/kg\\h_{2,SV}=257.27kJ/kg[/tex]
Next, solving the mass of water one obtains:
[tex]m_{R-134a}=\frac{m_{H_2O}Cp_{H_2O}*\Delta T_{H_2O}}{(h_2-h1)}=\frac{0.065kg/s*4.18kJ/kg^0C*(60^0C-40^0C)}{257.27kJ/kg-96.54kJ/kg} \\m_{R-134a}=0.0338kg/s[/tex]
(b) Now, the energy balance allows us to compute the heat supply:
[tex]Q_L+W_{in}=Q_H\\Q_H=0.0338kg/s*(257.27kJ/kg-96.54kJ/kg)+1.6kW\\Q_H=7.03kW[/tex]
(c) Finally, the COP (coefficient of performance) is computed via:
[tex]COP=\frac{Q_H}{W_{in}}=\frac{7.03kW}{1.6kW}\\COP=4.39[/tex]
Best regards.
