Question

On a summer day in New Orleans, Louisiana, the pressure is 1 atm: the temperature is 32°C; and the relative humidity is 95 percent. This air is to be conditioned to 24°C and 60 percent relative humidity. Determine the amount of cooling, in kJ, required and water removed, in kg, per 1000 m3 of dry air processed at the entrance to the system.

Answer 1

Answer:

Δw =7.95 kg/1000m^3

q = 62362.3 kg/1000m^3

Explanation:

To solve this problem we first need to use the psycrometric chart to determine the enthalpy h1, specific volume vi and absolute humidity col by using the given temperature T1 = 32°C and the relative humidity Ф1 = 95%.  

h_1 =  106.5 kJ/kg

v_1 = 0.91 m^3/kg

w_1 = 0.02905

We will also need the enthalpy h2 and the absolute humidity w_2 at the exit point. We will again use the pyscrometric chart and the given temperature T_2 = 24°C. From the problem we also know that the exit relative humidity is = 60%.  

h2 = 52.6 kJ/kg

w_2 = 0.01119

We need to express the final results in units per 1000 m^3. To do that we will need the mass m of this volume of air V and to calculate that we will use the given pressure p = 1 atm = 101.3 kPa.  

m = R_a*T_1/V.p

m = 1000*101.3/0.287*305K

m = 1157 kg

Because it is a closed system, the amount of water removed Δw can be calculated as:  

Δw =w_1 - w_2

Δw =0.02905- 0.01119

Δw =0.00687 kg/kg* 1157kg/1000m^3

Δw =7.95 kg/1000m^3

From the energy balance equation we can calculate the specific heat q removed from the air.  

q = h_1 - h_2

q = 106.5 kJ/kg - 52.6 kJ/kg

q = 53.9 kJ/kg * 1157kg/1000m^3

q = 62362.3 kg/1000m^3