One kg of air at 35°C DBT and 60% R.H.

Question

Adiabatic mixing : 

One kg of air at 35°C DBT and 60% R.H. is mixed with 2 kg of air at 20°C DBT and 13°C dew point temperature. Calculate the specific humidity of the mixture.

Answer 1

For the air at 35°C DBT and 60% R.H. : 

Corresponding to 35ºC, from steam tables, 

p_vs = 0.0563 bar

Relative humidity, φ = \(\cfrac{p_v}{p_{vs}}\)

p_v = φ p_vs = 0.6 × 0.0563 = 0.0338 bar

= 0.0214 kg/kg of dry air

Corresponding to 0.0338 bar, from steam tables,

For the air at 20°C DBT and 13°C dew point temperature :

p_v is the vapour pressure corresponding to the saturation pressure of steam at 13ºC.

∴ p_v = 0.0150 bar

= 0.00935 kg/kg of dry air

Enthalpy, h = c_pt_db + Wh_vapour 

= 1.005 × 20 + 0.00935 [h_g + 1.88 (t_db – t_dp)] 

= 20.1 + 0.00935 [2538.1 + 1.88 (20 – 13)] 

= 43.95 kJ/kg of dry air

Now enthalpy per kg of moist air

= 58.54 kJ/kg of moist air

Mass of vapour/kg of moist air

= 0.01316 kg/kg of moist air

Specific humidity of mixture

= \(\cfrac{0.01316}{1-0.01316}\) = 0.01333 kg/kg of dry air.