An air conditioning unit as shown with pressure temperature

An air conditioning unit as shown with pressure, temperature, and relative humidity data. Calculate the heat transfer per kilogram of dry air. Neglect changes in kinetic energy.

Solution

Given

Inlet air water vapor

Pressure: 105kpa

Temp: 30 degree C

Humidity : 80%

Assuming 10M3/min air is entering (V₁)

h₁= 83.31 kJ/kg (from psychrometric chart)

w₁=0.02079

v₁=0.856 m3/kg          

Outlet air water vapor

Pressure: 100kpa

Temp: 15 degree C

Humidity : 95%

h₂=40.99 kJ/kg (from psychrometric chart)

w₂=0.01024

Enthalpy of saturated liguid water at 15 degree C

h_w = 58.8 KJ/kg (from psychrometric chart)

        m_a= V1/v1

                                                =10/0.856

                                                =11.68 kg/min

water mass balance

                        m_w=m_a( w₁-w₂)

                        = 11.68 (0.02079-0.01024)

                        =0.1232

Rate of heat from energy equation:

                                    Q= m(h₁-h₂)-m_wh_w

= 11.68*(83.31-40.99)-0.1232*40.99

= 494.29 – 5.04

=489.24 kj/min

therefore heat transfer per kg of dry air = Q/m_w

                                                            = 489.24/0.1232

                                                            =3971 KJ/kg