Dew point temperature of air at one atmospheric pressure

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Dew point temperature of air at one atmospheric pressure (1.013 bar) is 18°C. The air dry bulb temperature is 30°C. The saturation pressure of water at 18°C and 30°C are 0.02062 bar and 0.04241 bar respectively. The specific heat of air and water vapour respeetively are 1.005 and 1.88 kJ/kgK and the latent heat of vaporization water of 0°C is 2500 kJ/kg. The specific humidity (kg/kg of dry air) and enthalpy (kJ/kg of dry air) of this moist air respectively, are

1. 0.01051, 52.64
2. 0.01291, 63.15
3. 0.01481, 78.60
4. 0.01532, 81.40

Correct Option: B

Given: p = 1.013 bar, (p_υ)_tdp = 0.02062

Specific humidity = 0.622	P_υ
	p - P_υ

=	0.622 × 0.02062	= 0.01291 kg/kg of dry air
	1.013 - 0.02062

h = 1.022 t_d + w(h_fgdp + 2.3t_dp)
= 63.15 kJ/kg of dry air.
Alternative method
p_ε = Saturation pressure = 0.02062 bar
p_a = Partial pressure of air = 0.04241 bar
t_d = Dry bulb temperature = 30º
t_w = Wet bulb temperature = 18º
Specific humidity,

W =	p_υ	× 0.622 = 0.01292 kg/kg of dry air.
	p - p_υ

Total enthalpy,
h = ^cp_a t_d + wh_s + w_c p_s (t_d – t_ω)
= 1.005 × 30 + 0.0129 × 2500 + 0.01292 × 1.88(30 – 18)
= 63.15 kJ/kg of dry air.