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- 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/(p-j*ps)

h = cpa t + x [cpw t + hwe

First i calculated the latent heat of vaporation: h_vap(15°C)=

In the next step i calculated how much water is already in the air:

x=0.622*j*ps/(p-j*ps)=

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] =

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=

Now if i want to cool the room from 25°C to 23°C m=3*180*59.2/2465.4 =

**2465.4 kJ/kg**In the next step i calculated how much water is already in the air:

x=0.622*j*ps/(p-j*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?