Answer:
a)The specific humidity is _0.00881_kg H2O/kg dry air.
b)The relative humidity is _59.8_ %.
c)The dry-bulb temperature is _20.2_ °C.
d)The volume flow rate of the mixture is _39.77_m3/min
Explanation:
First let determine the enthalpy h1 and absolute humidity W1 from the psychometric chart using temperature T1 = 35°C and relative humidity Q1 = 30%
h1 = 62.2kJ/kg
V1 = 0.89m³/kg
W1 = 0.01054
The volume flow rate V'1 = 15m³/min, the mass flow rate m'1 is ,
m'1 = V'1/v1 = 15/0.89 = 16.85kg/min
Also, From temperature T2 = 12°C and relative humidity Q2 = 90% and volume flow rate V'2 = 25m³/min
h2 = 31.9kJ/kg
V2 = 0.82m³/kg
W2 = 0.00785
the mass flow rate m'2is ,
m'2 = V'2/v2 = 25/0.82 = 30.49kg/min
To get the absolute humidity W3 and the enthalpy h3,
m'1/m'2 = (W2-W3)/(W3-W1)
16.85/30.49 = (0.00785-W3)/(W3-0.01054)
0.5526(W3-0.01054) = (0.00785-W3)
0.5526W3 - 0.00582 = 0.00785-W3
0.5526W3 + W3 = 0.00785 + 0.00582
1.5526W3 = 0.01367
W3 = 0.01367/1.5526 = 0.00881
absolute humidity W3 = 0.00881
m'1/m'2 = (h2-h3)/(h3-h1)
16.85/30.49 = (31.9 - h3)/(h3 - 62.2)
0.5526(h3 - 62.2) = (31.9 - h3)
0.5526h3 - 34.37 = 31.9 - h3
0.5526h3 + h3 = 31.9 + 34.37
1.5526h3 = 66.27
h3 = 66.27/1.5526 = 42.68kJ/kg
enthalpy h3 = 42.68kJ/kg
Use W3 and h3 to determine temperature T3, relative humidity Q3 and and specific volume V3 all through psychometric chart
T3 = 20.2°C
Q3 = 59.8%
V3 = 0.84m³/kg
By last of mass conversations, we can determine m3,
m3 = m1 + m2
m3 = 16.85 + 30.49 = 47.34kg/min
Then, the volume flow rate V'3 can be found by,
V'3 = m3*V3
V'3 = 47.34*0.84 = 39.77m³/min.
volume flow rate V'3 = 39.77m³/min.
