Question 13.4: Air at 25°C, 40% RH is mixed adiabatically with air at 45ºC,......

Step 1: Draw the psychrometric chart from the chart we can determined as it is adiabatic mixing two stream.
Step 2: By using psychrometric chart, take a look at horizontal axis to find DBT of 25°C. Move a pencil up this line to meet the intersection with the exponential line 40% relative humidity. Mark it as 1.
Step 3: Similarly take a look at horizontal axis to find DBT of 45°C. Move a pencil up this line to meet the intersection with the exponential line 40% relative humidity. Mark it as 2.
Step 4: From the psychrometric chart we can calculate the terms like enthalpy H_1,  H_2 specific humidity, W_1,  W_2
Step 5: Join the line 1–2. By using the formula of adiabatic equation we can calculate enthalpy H_3 and specific humidity W_3.
By this way the point 3 is located and the psychrometric chart is drawn.
Step 6: From the psychrometric chart we got the values like

H_1 = 45  kJ/kg,  H_2 = 109  kJ/kg,
W_1 = 0.007  kg vapour/kg dry air, W_2 = 0.024  kg vapour/kg dry air,
G_1 = 1 and G_2 = 2.

Step 7: The adiabatic equation of mixed two stream are

G_1 + G_2 = G_3
G_1W_1 + G_2W_2 = G_3W_3
G_1H_1 + G_2H_2 = G_3H_3

By simplifying the above equations we get the result as,

\frac{W_2 – W_3}{W_3 – W_1} = \frac{H_2 – H_3}{H_3 – H_1} = \frac{G_1}{G_2}
\frac{W_2 – W_3}{W_3 – W_1} = \frac{G_1}{G_2}
\frac{0.0024 – W_3}{W_3 – 0.007} = \frac{1}{2}
(0.0024 – W_3)2 = W_3 – 0.007
0.0055 = 3W_3
W_3 = 0.0018  kg vapour/kg dry air.
\frac{H_2 – H_3}{H_3 – H_1} = \frac{G_1}{G_2}
\frac{108.7 – H_3}{H_3 – 45.29} = \frac{1}{2}
(108.7 – H_3)2 = H_3 – 45.29
262.61 = 3H_3
H_3 = 87.5  kJ/kg dry air.

Therefore, the final condition of air is:

W_3 = 0.0018  kg vapour/kg dry air
H_3 = 87.5  kJ/kg dry air