Question

A heat pump with refrigerant-134a as the working fluid is used to keep aspace at 25°C by absorbing heat from geothermal water that enters the evaporator at 60°C at a rate of 0.065 kg/s and leaves at 40°C. Refrigerant enters the evaporator at 12°C with a quality of 15 percent and leaves at the same pressure as saturated vapor.If the compressor consumes 1.6 kW of power, determine (a) the mass flow rate of the refrigerant, (b) the rate of heat supply, (c) the COP.

Answer 1

Answer:12

Explanation:

Answer 2

Answer:

(a) $m_{R-134a}=0.0338kg/s$

(b) $Q_H=7.03kW$

(c) $COP=4.39$

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:

$m_{R-134a}(h_2-h1)=m_{H_2O}Cp_{H_2O}*\Delta T_{H_2O}$

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:

$h_{1/12^0C/VLM}=68.18kJ/kg+0.15*189.09kJ/kg=96.54kJ/kg\\h_{2,SV}=257.27kJ/kg$

Next, solving the mass of water one obtains:

$m_{R-134a}=\frac{m_{H_2O}Cp_{H_2O}*\Delta T_{H_2O}}{(h_2-h1)}=\frac{0.065kg/s*4.18kJ/kg^0C*(60^0C-40^0C)}{257.27kJ/kg-96.54kJ/kg} \\m_{R-134a}=0.0338kg/s$

(b) Now, the energy balance allows us to compute the heat supply:

$Q_L+W_{in}=Q_H\\Q_H=0.0338kg/s*(257.27kJ/kg-96.54kJ/kg)+1.6kW\\Q_H=7.03kW$

(c) Finally, the COP (coefficient of performance) is computed via:

$COP=\frac{Q_H}{W_{in}}=\frac{7.03kW}{1.6kW}\\COP=4.39$

Best regards.