ragga_muffin
Mechanical
- Aug 3, 2019
- 3
Hi,
I am trying to figure out the mass of air I will need to evaporate 310 kg of water.
The air properties are:
T= 80degree C
Relative Humidity=5.7%
specific Humidity = 0.017kg/kg
specific volume= 1.02 m^3/kg
enthalpy= 140.72kj/kg
The water removal rate is 44.28kg/hr
I am not sure how to approach this so any help will be appreciated.
I am trying to figure out the mass of air I will need to evaporate 310 kg of water.
The air properties are:
T= 80degree C
Relative Humidity=5.7%
specific Humidity = 0.017kg/kg
specific volume= 1.02 m^3/kg
enthalpy= 140.72kj/kg
The water removal rate is 44.28kg/hr
I am not sure how to approach this so any help will be appreciated.