A heat pump with refrigerant-134a as the working fluid is used to keep a space at 25°C by absorbing heat from geothermal water that enters the evaporator at 60°C at a rate of 0.415 kg/s and leaves at 40°C. The refrigerant enters the evaporator at 12°C with a quality of 15 percent and leaves at the same pressure as saturated vapor. The compressor consumes 1.6 kW of power. Use data from the tables.

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, and
(d) the minimum power input to the compressor for the same rate of heat supply.

a) 0.0338 kg/s

b) 7.04 KW

c) 4.4

d) 0.74 kW

Explanation:

The mass flow rate of the refrigerant is calculated from the energy balance of the water and the refrigerant. The enthalpy values are taken from table A-11  

Q_w = Q_r

m_w*c_p*ΔT_w = m_r(h_2-h_1)

m_r = m_w*c_p*ΔT_w/h_2 - (h_liq,12 + qh_evap,12)

      = 0.065*4.18*20K/(257.33 - (68.17+0.15*189.16))kJ/kg

      = 0.0338 kg/s

 The rate of heat supply is calculated from the rejected heat and the work:

Q_h = Q_L + W

       = m_r(h_2-h_1)+W

       = 7.04 KW

The COP is calculated from the heat supply and work:  

COP = Q_h/W

        = 4.4

The minimum power is determined from the maximum COP:  

W_min = Q_h/COP_max

            = Q_h/1/1-T_L/T_H

            = 0.74 kW

answer:

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explanation:

i need points bih

answer:

just reverse the digits.

: )