 #1
StoningTree
 2
 0
 Homework Statement:
 A room should be cooled by 3 Kelvin using evaporation. The temperature of the water is 15 ° C. Furthermore, the volume of the room is 180m ^ 3, at a temperature of 25 ° C and a humidity of 50%. Hom much water is needed?
 Relevant Equations:

x=0.622*j*ps/(pj*ps)
h = cpa t + x [cpw t + hwe
First i calculated the latent heat of vaporation: h_vap(15°C)= 2465.4 kJ/kg
In the next step i calculated how much water is already in the air:
x=0.622*j*ps/(pj*ps)=0.00988 kg/kg
where:
j=50%
ps=0.03171 bar
p=1.01325 bar
Now that i know that i can calculate the enthalpy of humid air:
h_hum = cpa t + x [cpw t + hwe] = 50.3 kJ/kg
where:
cpa = 1.006 kj/(kg*C)
cpw = 1.86 kj/(kg*C)
hwe = 2500 kJ/kg
t = 25
With 25° air temperature and 50% humidity the density is at around: 1.177 kg/m^3
> h_hum=59.2 kJ/m^3
Now if i want to cool the room from 25°C to 23°C m=3*180*59.2/2465.4 = 12.9 kg water is required?
In the next step i calculated how much water is already in the air:
x=0.622*j*ps/(pj*ps)=0.00988 kg/kg
where:
j=50%
ps=0.03171 bar
p=1.01325 bar
Now that i know that i can calculate the enthalpy of humid air:
h_hum = cpa t + x [cpw t + hwe] = 50.3 kJ/kg
where:
cpa = 1.006 kj/(kg*C)
cpw = 1.86 kj/(kg*C)
hwe = 2500 kJ/kg
t = 25
With 25° air temperature and 50% humidity the density is at around: 1.177 kg/m^3
> h_hum=59.2 kJ/m^3
Now if i want to cool the room from 25°C to 23°C m=3*180*59.2/2465.4 = 12.9 kg water is required?