Question

Two airstreams are mixed steadily and adiabatically. The first stream enters at 35°C and 30 percent relative humidity at a rate of 15 m3/min, while the second stream enters at 12°C and 90 percent relative humidity at a rate of 25 m3/min. Assuming that the mixing process occurs at a pressure of 90 kPa, determine the specific humidity, the relative humidity, the dry-bulb temperature, and the volume flow rate of the mixture. Use data from the tables.

a)The specific humidity is __kg H2O/kg dry air.
b)The relative humidity is __ %.
c)The dry-bulb temperature is __ °C.
d)The volume flow rate of the mixture is __m3/min.

Answer 1

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

Step-by-step 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.