Question 13.1: Atmospheric air at 1.0132 bar has a dry bulb temperature (DB......

Step 1: Determine the specific humidity at exit

W_2 = \frac{0.622p_s}{p-p_s}

The saturation pressure at wbt is 28°C is calculated using,

P = 0.61078  \exp \left\lgroup\frac{17.27 T}{T + 237 .3} \right\rgroup
p_s = 0.037798  bar
W_2 = \frac{0.622 \times 0.03779}{1.0132 – 0.037798}
W_2 = 0.0240  kg vapour/kg dry air

Step 2: Determine the specific humidity

W_1 = \frac{c_{pa} (T_2 – T_1) + W_2 \cdot H_{fg_2}}{H_{w_1} – H_{f_2}}

The values of latent heat of steam H_{fg_2}, specific enthalpy of saturated air H_{f_2}, specific enthalpy of saturated steam H_{w_1} are determined from the steam table.
From the temperature 28°C the values of:

H_{fg_2} = 2434.59  kJ/kg
H_{f_2} = 117.38  kJ/kg

From the temperature 34°C:

H_{w_1} = 2562.79  kJ/kg
W_1 = \frac{1.005 (28 – 34) + (0.0240 \times 2434.59)}{2562.79 – 117.38}
W_1 = \frac{-6.03 + 58.4301}{2445.41}
W_1 = 0.0214  kg vapour/kg dry air

Step 3: Partial pressure of water vapour

W_1 = \frac{0.622 p_w}{p – p_w}
0.0214 = \frac{0.622 \times p_w}{1.0132 – p_w}
\frac{0.0214}{0.622} = \frac{p_w}{1.0132 – p_w}
\frac{1.0132 – p_w}{p_w} = 29.065
1.0132 – p_w = 29.065  p_w
p_w = 0.033  bar

Step 4: Relative humidity RH

RH = \frac{p_w}{p_s}

Saturation pressure at 34°C is determine by:

P= 0.61078  \exp \left\lgroup\frac{17.27 T}{T+237.3} \right\rgroup
p_s = 0.0531
RH = \frac{p_w}{p_s}
RH = \frac{0.033}{0.0531}
RH = 0.6214

Step 5: Degree of saturation

\mu = \frac{W}{W_s} = \frac{p_w}{p_s} \frac{p – p_s}{p – p_w}
\mu = \frac{0.033}{0.0531} \frac{1.0132 – 0.0531}{1.0132 – 0.033}
\underline{\mu } = 0.608