A heat exchanger is designed to cool 2 m3s of air at 100 kPa

A heat exchanger is designed to cool 2 m^3/s of air at 100 kPa and 227 degree C to 25 degree C by using R-134a at -12 degree C and 200 kPa. The mass flow rate of the refrigerant is 1.28 kg/s. The heat exchanger is fully insulated. The process is steady state. Neglecting any pressure drops in air and refrigerant side: determine the refrigerant outlet temperature the heat transfer rate from the air to refrigerant Air has not constant specific heats.

Solution

From air properties at 100 kPa and 227 deg C we get v1_air = 1.44 m^3/kg, h1_air = 504 kJ/kg

From air properties at 100 kPa and 25 deg C we get v2_air = 0.856 m^3/kg, h2_air = 298 kJ/kg

From R-134a properties at 200 kPa and -12 deg C we get v1_ref = 0.00075 m^3/kg, h1_ref = 35.9 kJ/kg

a)

Air mass flow rate = 2/1.44 = 1.389 kg/s

By energy balance,

Heat added to refrigerant = Heat lost by air

1.28*(h2 - 35.9) = 1.389*(504 - 298)

h2 = 259.42 kJ/kg

From R-134a properties at 200 kPa and h = 259.42 kJ/kg, we get T = 7.47 deg C

b)

Heat transfer rate = 1.389*(504 - 298)

= 286.13 kW