Answer:
Δw =7.95 kg/1000m^3
q = 62362.3 kg/1000m^3
Step-by-step 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